bibkey: liu2026tailcompensation authors: Vincent Liu year: 2026 title: “Certified Weil Positivity Beyond the Unit Window: Source-Exact Block-Schur and Tail-Compensation Bounds for the Riemann Zeta Function” doi: null url: https://github.com/luciferyu666/certified-weil-positivity/releases/tag/v1.0-mcom-submission claim: The author-submitted manuscript states full complex Weil-form coercivity at physical half-widths 1 and 17/16. Source-proof parameter applications supply prime, tail and complement inputs at c=9. The conditional c9 retained comparison has producer and separate-session sign executions on the same hash-bound inputs. A positive constant prime reference and pointwise exterior positivity are incompatible at band256 for c>=27; the same cofinal strategy requires growing bands. Independent matrix regeneration and moment containment remain unverified; cofinal positivity remains unproved. strata_touched: [] license: citation-only triage: anchor

Fixed-window positivity and retained tail compensation

The primary source is the author’s release v1.0-mcom-submission, submitted to Mathematics of Computation on 14 September 2026. The inspected source is pinned to commit b6cd2183c1e79c6c27a34267812a7b2d73ed1b59: manuscript TeX and submitted PDF. The downloaded PDF has SHA-256 91126eee6ceb5315a4a40a4d4b2f34d058a71432ae18a8aaf47146f71abf418e, matching the pinned README’s submitted-original hash. This is an author-posted submission; journal acceptance, a DOI and external human reproduction are not asserted by the release.

The theorem statements and analytic interfaces below were inspected in the pinned source. The numerical package was not executed, and the source’s historical verification labels are not independent evidence supplied by this note. The complete package is the separately attached w200-frozen-artifact.zip, rather than GitHub’s automatic source archive. The rights statement grants no blanket open-source license; this note provides citations and mathematical applications, without copying its implementation or certificates.

Source normalization and support

In section 2, equations (1)–(2), the legal domain is ${\mathcal{D}}_a=C_c^{\infty }((-a,a);\mathbb{C})$, with

$$F_f(z)={\int }_{\mathbb{R}}f(u)e^{-izu}du,H_f(x)=\Re {\int }_{\mathbb{R}}f(v+x)\overline{f(v)}dv.$$

Writing $A(t)=\Re \psi (1/4+it/2)-\log \pi$, the complete form is

$$Q(f)=2\Re (F_f(i/2)\overline{F_f(-i/2)})+\frac{1}{2\pi }{\int }_{\mathbb{R}}A(t)|F_f(t){|}^{2}dt-\sum _{n\ge 2}\frac{2\Lambda (n)}{\sqrt{n}}{H}_f(\log n).$$

Both poles, Gamma and every contributing prime power are retained. There is no imposed Mellin-vanishing condition, parity restriction or RH hypothesis. For a smooth test with this support, only $\log n<2a$ contributes. The minus sign in $F_f$ reverses the project’s plus-sign Fourier variable; the even Gamma bracket and the paired poles retain the same quadratic normalization. For even tests, $F_f(i/2)=F_f(-i/2)$ and the pole term is $2|F_f(i/2){|}^2$.

Theorem A, section 3, equation (3), states

$$Q(f)\ge 2^{-151}\Vert f{\Vert }_2^2(f\in {\mathcal{D}}_1).$$

Theorem B, section 4, equation (6), states

$$Q(f)\ge 2^{-49162}\Vert f{\Vert }_2^2(f\in {\mathcal{D}}_{17/16}).$$

The bounded-operator proof covers both parity sectors and the entire orthogonal complement, then applies its conclusion to the original smooth domain. These are the manuscript’s full-form claims, rather than claims about a positive finite compression alone. Their computer-assisted certificates have not been independently reproduced here. The comparison with Zhu’s versioned half-width $0.8$ result concerns support range, not a stronger coercivity constant.

The positive tail remains in the retained matrix

At the source’s fixed half-width $a=17/16$ and Fourier-band cutoff $\Omega =256$, equations (7)–(9) write

$$Q(f)=⟨f,(M+K)f⟩+T_{\mathrm{tail}}(f),T_{\mathrm{tail}}(f)=\frac{1}{2\pi }{\int }_{|t|>256}(A(t)-7/2)|F_f(t){|}^{2}dt.$$

Here $M$ retains all paired prime shifts for $n\in \{2,3,4,5,7,8\}$, and $K$ retains the pole kernel and the central Gamma band. The source proves the quantitative tail input

$$T_{\mathrm{tail}}(f)\ge 2^{-49162}\Vert f{\Vert }_2^2+⟨f,Uf⟩,U=81|h_0⟩⟨h_0|+27|h_1⟩⟨h_1|.$$

If $B_{256}$ is the sinc-kernel band operator and $v_0,v_1$ are the first two normalized Legendre modes, these retained vectors are

$$h_j=(I-B_{256}-2^{-49158}I)v_j.$$

They are tail-filtered vectors. Replacing them by $v_j$ would change the estimate. The compression of $U$ enters the source-error finite sign test in section 6, equation (22). Its two integer matrices cover the two 224-dimensional parity sectors, with the complement and coupling charged separately. This is a reusable positive-tail mechanism, not a new general Schur principle. The numerical constants and certified matrices belong to the specified support and band parameters.

Reflection commutes with $B_{256}$, so $h_0$ is even and $h_1$ is odd. On even tests the second rank-one contribution vanishes and the retained contribution is $81|⟨h_0,f⟩{|}^2$. This is a restriction of the source’s estimate at $17/16$, not a positivity statement at a larger support.

Matching the first new FIB window

In the project’s Fourier convention the interval has length $\ell =\log c$ and physical half-width $a=\ell /2$. For $c_0=c_1=3$ and $c_{r+2}=c_{r+1}{c}_r$, the first new cutoff is $c_2=9$. Consequently

$$a_9=\log 3>17/16,e^{17/8}\approx 8.3729<9,a_9-17/16\approx \mathrm{0.0361123.}$$

Theorem B does not cover this window. At the exact $c=9$ endpoint, the shift $\log 9=2a_9$ has zero overlap, so the same prime-power list $\{2,3,4,5,7,8\}$ applies. The missing step is an estimate for the enlarged interval’s actual operator, retained matrix and infinite complement. An unchanged prime list does not transport the old certificate or its tiny margin. The next FIB cutoff $c_3=27$, with $a_{27}=3\log 3/2$, also lies outside the stated range.

The existing same-symbol coupling allowance keeps a common signed cross interval, every middle mode and the second-jet remainder. A sufficient consumer still needs the actual retained form to dominate their complete Schur cost. The source’s positive-tail correction suggests retaining an available arithmetic-compatible positive contribution in that form; it does not establish that domination at $c=9$, an induction step or cofinal support positivity. Failure of a particular upper allowance to fit would not refute positivity or RH.

A source-formula application to the actual prime block at $c=9$

The source’s weighted Schur formula in Appendix B.4, “Weighted Schur bound including prime power eight,” can be reused without its old support-dependent constants. At the new physical half-width $a=\log 3$, let $I=(-a,a)$, $\mathcal{H}=L^2(I;\mathbb{C})$ and

$$(S_{d}f)(u)=1_I(u+d)f(u+d),C_{\mathrm{p}}=\sum _{n\in \{2,3,4,5,7,8\}}{w}_n(S_{\log n}+S_{-\log n}),w_n=\frac{\Lambda (n)}{\sqrt{n}}.$$

These are the actual compressed translations on the whole Hilbert space, with $S_d^{\ast }=S_{-d}$. The $n=9$ endpoint translation is zero almost everywhere. Put $M_9=(7/2)I-C_{\mathrm{p}}$. The following application gives a lower bound for this prime comparison block, not for $Q$.

Choose $q(u)=1+(5/8)(u/a{)}^2$. The source formula bounds the absolute quadratic form by the weighted row sum:

$$|⟨f,C_{\mathrm{p}}f⟩|\le {\int }_I{r}_q(u)|f(u){|}^{2}du,r_q(u)=\sum _n{w}_n\frac{1_I(u+\log n)q(u+\log n)+1_I(u-\log n)q(u-\log n)}{q(u)}.\text{\hspace{1em}(A1)}$$

This formula uses complex weighted Young and translation of the adjoint term; its validity does not require the source’s $17/16$ width, finite-dimensional tests or RH. The function $r_q$ is even. New outward support cells and rational bounds are required at $a=\log 3$.

Use the source’s logarithm grid endpoints $l_p^{-}<\log p<l_p^{+}$ and root lower endpoints $k_n/10^6\le \sqrt{n}$:

For $n=2,3,4,5,7,8$, the respective $k_n$ are $1414213,1732050,2000000,2236067,2645751,2828427$. For $n=p^j$, define

$$w_n^{+}=\frac{l_p^{+}}{k_n/10^6},t_n^{-}=\frac{j{l}_p^{-}}{l_3^{+}},t_n^{+}=\frac{j{l}_p^{+}}{l_3^{-}}.$$

Set $t_3^{-}=t_3^{+}=1$ instead: this shift equals the actual half-width exactly. The weight remains $\log 3/\sqrt{3}$. Define

$$x_1=t_4^{-}-1,x_2=1-t_2^{-},x_3=t_5^{-}-1,x_4=t_7^{-}-1,x_5=t_8^{-}-1.$$

They are, respectively, $287681/1098613$, $405466/1098613$, $510824/1098613$, $847297/1098613$, $980828/1098613$, in increasing order inside $(0,1)$. On $x=u/a\in [0,1]$, the following outward lists activate negative shifts early and retain positive shifts late:

The positive $n=3$ shift is present only at the single endpoint $x=0$ under a closed-interval convention, hence contributes nothing to the $L^2$ integral. The rows cover every other support switch; no prime-power weight is deleted.

For each row, with $b=5/8$ and $R=27/10$, sum its listed terms with multiplicity to form

$$W=\sum w_n^{+},B=\sum _{\mathrm{negative}}{w}_n^{+}{t}_n^{-}-\sum _{\mathrm{positive}}{w}_n^{+}{t}_n^{+},D=\sum w_n^{+}(t_n^{+}{)}^2.$$

Then, on that whole cell,

$$(1+b{x}^2)(R-r_q(ax))\ge P(x):=b(R-W)x^2+2bBx+(R-W-bD).\text{\hspace{1em}(A2)}$$

The listed quadratic-minimum bounds were evaluated using exact fractions, at both endpoints and any interior vertex of a convex quadratic. Each minimum strictly exceeds $1/50$. Since $1+b{x}^2\le 13/8$, (A1)–(A2) give

$$|⟨f,C_{\mathrm{p}}f⟩|\le \left(\frac{27}{10}-\frac{4}{325}\right)\Vert f{\Vert }^2,\boxed{\frac{4}{5}I⪯M_9⪯\frac{31}{5}I.}\text{\hspace{1em}(A3)}$$

The bound applies to the complete complex Hilbert space, and therefore its even subspace. This is a paper application of an existing weighted Schur formula with a new exact scalar parameter calculation; it is not a new kernel theorem, reproduction of a fixed-window certificate or claim of priority. The pole and Gamma band are still in $K$, and the positive exterior-frequency contribution remains in $T_{\mathrm{tail}}$. Neither $M_9\succ 0$ nor its invertibility establishes positivity of $M_9+K+U$ or $Q$.

The following independent scalar replay produces the six quadratic minima using only exact fractions. The logarithm and root enclosures are the cited analytic inputs; this program verifies their downstream rational comparison, not the source’s full analytic proof or a complete Weil certificate.

from fractions import Fraction as F

beta, target = F(5, 8), F(27, 10)
log_lo = {2: 693147, 3: 1098612, 5: 1609437, 7: 1945910}
log_hi = {2: 693148, 3: 1098613, 5: 1609438, 7: 1945911}
root_lo = {2: 1414213, 3: 1732050, 4: 2000000,
           5: 2236067, 7: 2645751, 8: 2828427}
powers = {2: (2, 1), 3: (3, 1), 4: (2, 2),
          5: (5, 1), 7: (7, 1), 8: (2, 3)}
weight = {n: F(log_hi[p], root_lo[n])
          for n, (p, j) in powers.items()}
shift_lo = {n: F(1) if n == 3 else F(j * log_lo[p], log_hi[3])
            for n, (p, j) in powers.items()}
shift_hi = {n: F(1) if n == 3 else F(j * log_hi[p], log_lo[3])
            for n, (p, j) in powers.items()}
ends = [F(0), shift_lo[4] - 1, 1 - shift_lo[2],
        shift_lo[5] - 1, shift_lo[7] - 1, shift_lo[8] - 1, F(1)]
assert all(left < right for left, right in zip(ends, ends[1:]))
negative = [[2, 3], [2, 3, 4], [2, 3, 4], [2, 3, 4, 5],
            [2, 3, 4, 5, 7], [2, 3, 4, 5, 7, 8]]
positive = [[2], [2], [], [], [], []]
floors = [F(2, 5), F(13, 100), F(11, 10),
          F(4, 25), F(1, 50), F(4, 25)]

for i, (left, right) in enumerate(zip(ends, ends[1:])):
    neg, pos = negative[i], positive[i]
    W = sum((weight[n] for n in neg + pos), F(0))
    B = (sum((weight[n] * shift_lo[n] for n in neg), F(0))
         - sum((weight[n] * shift_hi[n] for n in pos), F(0)))
    D = sum((weight[n] * shift_hi[n] ** 2 for n in neg + pos), F(0))
    A, E, C = beta * (target - W), 2 * beta * B, target - W - beta * D
    points = [left, right]
    if A > 0 and left < -E / (2 * A) < right:
        points.append(-E / (2 * A))
    minimum = min(A * x * x + E * x + C for x in points)
    assert minimum > floors[i] >= F(1, 50)
    print(i + 1, minimum)

The actual band complement at $a=\log 3$

Appendix B fixes $L=17/16$, $\Omega =256$ and $\Omega L=272$ at its outset. Its band-complement and tail statements therefore cannot simply be instantiated at another width. The following application checks the width dependencies in B.1–B.3 and constructs the new operators; it does not use the old finite sign certificate.

Keep $a=\log 3$ and the full complex space $\mathcal{H}=L^2((-a,a);\mathbb{C})$. Write $c_{\mathrm{band}}=256a$, distinct from the arithmetic cutoff $c=9$. The cited logarithm enclosure gives

$$272<c_{\mathrm{band}}<282,a<11/10.$$

Define the actual band operator and its exterior Fourier energy by

$$(B_{9}f)(u)={\int }_{-a}^a\frac{\sin (256(u-v))}{\pi (u-v)}f(v)dv,{\mathcal{E}}_9(f)=\frac{1}{2\pi }{\int }_{|t|>256}|F_f(t){|}^{2}dt,$$

with kernel value $256/\pi$ on the diagonal. Plancherel gives $0⪯B_9⪯I$ and ${\mathcal{E}}_9(f)=⟨f,(I-B_9)f⟩$ for every $f\in \mathcal{H}$.

In B.1, the Fourier-derivative bound for a unit vector becomes $|F_f^{(j)}(t)|\le a^j\sqrt{2a}<2a^j$. It differentiates $F_f$ in frequency; it requires no derivatives of $f$. The fixed frequency band, the $N=4096$ exterior nodes with spacing $h=1/32$, their distance bound $768$, and the Lagrange basis sum $<2^{24570}$ do not depend on the support width. The changed interpolation remainder is controlled by

$$\frac{3\cdot 768a}{4096}<\frac{99}{160}<\frac{5}{8},\frac{2(768a{)}^{4096}}{4096!}<2(5/8{)}^{4096}<2^{-2047}<1/32.$$

The middle power bound uses $(5/8{)}^2<1/2$. Thus the source’s real-phase interpolation argument still gives, when ${\mathcal{E}}_9(f)<1/2$,

$$\frac{1}{16}<16\sqrt{{\mathcal{E}}_9(f)}{2}^{24570}+\frac{1}{32},\sqrt{{\mathcal{E}}_9(f)}>2^{-24579}.$$

The other energy case is immediate. Consequently this source-proof application supplies the actual new-window bounded-operator input

$$I-B_9⪰\delta I,\delta =2^{-49158},S_9=I-B_9-\delta I,0⪯S_9⪯I.\text{\hspace{1em}(A5)}$$

This conclusion covers the entire complex Hilbert space. The exterior Gamma-weighted form used below is finite on the original smooth legal domain; its bounded comparison (A5) does not assert finiteness of that weighted integral for arbitrary $L^2$ vectors. This is an application of the inspected proof with a new remainder comparison, not a new uncertainty principle or a kernel-verified declaration.

Positive tail compensation at the new width

At this same $a=\log 3$, define the newly normalized modes and filtered vectors

$$v_{0,9}(u)=(2a{)}^{-1/2},v_{1,9}(u)=\sqrt{3/(2a)}u/a,h_{j,9}=S_9{v}_{j,9},e_j=⟨v_{j,9},(I-B_9)v_{j,9}⟩.$$

The exterior-weight estimate in B.2 uses only $|t|\ge 256$ and the background $7/2$, so it supplies $T_{\mathrm{tail},9}(f)\ge C{\mathcal{E}}_9(f)$ with $C=123/1280>1/16$. This is the source’s actual exterior integral, with all poles, contributing prime powers and the central Gamma band retained elsewhere in $Q$.

B.3’s normalized Fourier integrals, evaluated with $c_{\mathrm{band}}=256a$, supply independent lower and upper bounds:

$$\frac{c_{\mathrm{band}}-1}{\pi c_{\mathrm{band}}^2}\le e_0\le \frac{c_{\mathrm{band}}+1}{\pi c_{\mathrm{band}}^2},\frac{3(c_{\mathrm{band}}-2)}{\pi c_{\mathrm{band}}^2}\le e_1\le \frac{3/c_{\mathrm{band}}+6/c_{\mathrm{band}}^2+2/c_{\mathrm{band}}^3}{\pi }.$$

All four envelopes decrease on $[272,282]$. Using the source’s $157/50<\pi <22/7$, the lower envelopes at $282$ and upper envelopes at $272$ give

Both lower bounds exceed $2^{-11}>\delta$. Thus $e_j-\delta >0$ is established independently of the upper estimates. Reflection commutes with $S_9$, making $h_{0,9}$ even, $h_{1,9}$ odd and $⟨v_{0,9},S_9{v}_{1,9}⟩=0$. The source’s complex square completion in the positive form of $S_9$ therefore applies:

$$⟨f,S_{9}f⟩\ge \sum _{j=0}^1\frac{|⟨h_{j,9},f⟩{|}^2}{e_j-\delta }.$$

Since $C/(e_0-\delta )>81$, $C/(e_1-\delta )>27$ and $C\delta >\delta /16$, this gives the actual new-window tail input

$$\boxed{T_{\mathrm{tail},9}(f)\ge {\tau }_9\Vert f{\Vert }^2+⟨f,U_{9}f⟩,{\tau }_9=2^{-49162},U_9=81|h_{0,9}⟩⟨h_{0,9}|+27|h_{1,9}⟩⟨h_{1,9}|,}\text{\hspace{1em}(A6)}$$

for every $f\in C_c^{\infty }((-a,a);\mathbb{C})$. On even tests the odd contribution vanishes. The displayed constants match the old calibration because the new parameter bounds justify them, while $B_9$, the modes and both filtered vectors are new-width objects. This is a paper application of the inspected analytic proof, independently reviewed; it is not a reproduced numerical certificate, a new kernel theorem or positivity of the complete form.

The same upper envelopes give $\Vert h_{0,9}\Vert <1/29$ and $\Vert h_{1,9}\Vert <3/50$, since $S_9^2⪯S_9$ implies $\Vert h_{j,9}{\Vert }^2\le e_j-\delta$. These norm bounds do not provide finite-column approximation errors for the new vectors.

This exact scalar replay checks only the new parameter comparisons. Its logarithm and $\pi$ enclosures are the cited analytic inputs; it does not reexecute the original finite matrices or prove the analytic interpolation and square-completion suppliers.

from fractions import Fraction as F

log3_lo, log3_hi = F(1098612, 10**6), F(1098613, 10**6)
assert F(17, 16) < log3_lo < log3_hi < F(11, 10)
assert F(272) < 256 * log3_lo < 256 * log3_hi < F(282)
ratio = F(3 * 768, 4096) * F(11, 10)
assert ratio == F(99, 160) < F(5, 8)
assert F(5, 8)**2 < F(1, 2)

band_lo, band_hi = F(272), F(282)
pi_lo, pi_hi, C = F(157, 50), F(22, 7), F(123, 1280)
lower = [(band_hi - 1) / (pi_hi * band_hi**2),
         3 * (band_hi - 2) / (pi_hi * band_hi**2)]
upper = [(band_lo + 1) / (pi_lo * band_lo**2),
         (3 / band_lo + 6 / band_lo**2 + 2 / band_lo**3) / pi_lo]
assert lower == [F(1967, 1749528), F(245, 72897)]
assert upper == [F(6825, 5807744), F(2794825, 789853184)]
assert all(value > F(1, 2**11) for value in lower)
assert 49158 > 11
assert C > F(1, 16)
assert 81 * upper[0] < C and 27 * upper[1] < C
assert upper[0] < F(1, 29**2) and upper[1] < F(9, 50**2)
for j in range(2):
    print(j, lower[j], upper[j])

The actual even projection at $c=9$

The existing Appendix C projection and block-error formulas can be applied at the new width after checking their parameter dependencies. This supplies explicit projection, cross and complementary-block inputs for (A4); the retained sign still requires proof.

Let $P_N$ be the full complex Legendre projection onto degrees below $N=448$. On the even Hilbert space take its restriction $P$ and the orthonormal embedding $E$ with columns

$$E_j(u)=\sqrt{\frac{4j+1}{2a}}{P}_{2j}(u/a),0\le j<224,a=\log 3,$$

where $P_{2j}$ is the standard Legendre polynomial. Thus $P=E{E}^{\ast }$ is the actual even projection. The same ellipse of radius $3/2$ has imaginary semiaxis $5/12$. Appendix C’s Chebyshev and best-approximation estimate, with the new width, gives

$$\underset{|t|\le 256}{\sup }\Vert (I-P_N)e^{itu}{\Vert }^2\le 72a\exp ((640/3)a)(2/3{)}^{896}<\frac{396}{5}(68/25{)}^{235}(2/3{)}^{896}<2^{-178}.\text{\hspace{1em}(A8)}$$

Here $a<11/10$, $(640/3)(11/10)=704/3<235$ and the source’s $e<68/25$ justify the outward comparison; the last step is exact rational arithmetic. Use $r=2^{-89}$, rather than the old-width $2^{-95}$. On the even space the band vectors are $w_t={\Pi }_{\mathrm{even}}{e}^{itu}=\cos (tu)$, so

$$\Vert (I-P)w_t\Vert \le \Vert (I-P_N)e^{itu}\Vert <r,\Vert w_t\Vert \le \sqrt{2a}<3/2.$$

This does not assert a small residual for $e^{itu}$ under the even projection on the full Hilbert space.

The source pole estimate also has explicit width conditions. They remain valid: $a/2<1$, $e^{a/2}<2$ and $\sqrt{2a}<3/2$. Degree-$447$ Taylor approximation of $a_{\pm }(u)=e^{\pm u/2}$ gives $\Vert (I-P_N)a_{\pm }\Vert <3/448!$ and $\Vert a_{\pm }\Vert <3$. The same rank-one difference estimate therefore yields

$$\Vert K_{\mathrm{pole}}-P_N{K}_{\mathrm{pole}}{P}_N\Vert <36/448!\le 36/2^{447}<2^{-440}=:p.$$

Restriction gives this upper allowance on the even space. Its pole operator is $2|\cosh (u/2)⟩⟨\cosh (u/2)|$, hence positive there. The pole allowance $p$ is an upper-error input, not a negative-complement charge.

For the actual $V_9=K_9+U_9$, put $R_9=V_9-P{V}_{9}P$. The source’s support-independent band-weight input is

$$\kappa =\frac{1}{2\pi }{\int }_{-256}^{256}|A(t)-7/2|dt<896.$$

The rank-one band integral and (A8) bound its cross block by $1344r$ and its complementary norm by $896r^2$. Since $v_{0,9}\in \mathrm{ran}P$, the source’s Cauchy–Schwarz and Plancherel estimate gives $\Vert (I-P)h_{0,9}\Vert <10r$. Thus the actual even tail cross block is bounded by $(810/29)r<28r$, and its complementary norm by $8100r^2$. The pole and tail complementary blocks are both positive. Consequently the following source-proof application pays the actual even blocks:

$$\Vert (R_9{)}_{10}\Vert <1372r+p,(R_9{)}_{11}⪰-896r^{2}I,\Vert (R_9{)}_{11}\Vert <8996r^2+p.\text{\hspace{1em}(A9)}$$

The positivity of the even pole is used only in the lower complementary bound. Its upper allowance remains in the cross and norm bounds. These bounds cover the entire complement inside the even Hilbert space; they do not assert an odd-sector result or supply the finite sign in (A4).

For this common embedding, (A4) can use $e=1372r+p$, $n=896r^2$ and $h=8996r^2+p$, with $n<1/2<4/5$. The source’s sharper prime input $m_{\ast }=264/325$, $b_{\ast }=2011/325$ can also be used directly in its general equations (18)–(22). The actual finite $J_9,D_9$, an upper matrix bound for $G_9$, their directed source errors, and the resulting finite sign test remain unpaid. The ordinary compression $E^{\ast }{M}_{9}E$ is an upper bound for $G_9^{-1}$, so it cannot replace a lower bound for that inverse compression. The prime-coupling term $D_9$ does not vanish merely because the retained space is finite-dimensional. A positive floating compression would not discharge these obligations.

The following exact scalar replay checks the new ellipse, width and coefficient comparisons. It uses the source analytic suppliers above and does not prove those suppliers or reconstruct a numerical certificate.

from fractions import Fraction as F

assert F(396, 5) * F(68, 25)**235 * F(2, 3)**896 < F(1, 2**178)
assert F(68, 25)**11 < 2**20  # e^(a/2) < 2 for a < 11/10
assert F(22, 10) < F(9, 4)   # sqrt(2a) < 3/2
assert F(810, 29) < 28
r, p, m = F(1, 2**89), F(1, 2**440), F(264, 325)
e, n, h = 1372*r+p, 896*r*r, 8996*r*r+p
assert 896 + 8100 == 8996
assert 1344 + 28 == 1372
assert n < F(1, 2) < F(4, 5) < m
print('new-width band, even block and pole parameter comparisons passed')

Reusing the support-independent Gamma moments

Appendix D.1, Integrated digamma moments, supplies the scalar moments

$${\vartheta }_q=\frac{256}{\pi }{\int }_0^1{x}^{2q}(A(256x)-7/2)dx,0\le q\le 1023.$$

Their definition contains no support half-width. The release packet w200-pub-2026-09-14/reproduction/release-run/certificates/moments.json has SHA-256 f8cb5c681a22755b980d2e98d781353fe9ce058fe33eb8a7753585e2c52b2f93, matching the pinned frozen-manifest.json. Its 1,024 ordered rows all have hi-lo=3. The checked packet-wide field parameters.outputBits=1024 specifies the $2^{-1024}$ grid for every row; the rows have no separate outputBits field. Thus the packet midpoints are ${\hat{\vartheta }}_q=({\mathrm{lo}}_q+{\mathrm{hi}}_q)/2^{1025}$. The checked parameters also specify the common band $256$ and background $7/2$. Its metadata records the original $L=17/16$, which is absent from the moment formula and is not a width to retain in the new kernel.

The small reviewer-materials archive has SHA-256 e5547b885d3df9113895877032ba861235adb159cd5b816c9d9f6080d22fbf41 and contains that manifest. Member size and SHA-256 were checked after selective retrieval of the moment packet; the full large archive hash was not checked. These are data-identity and format checks. The mathematical premise that every interval contains its moment, and hence $|{\vartheta }_q-{\hat{\vartheta }}_q|<2^{-1023}$, is the author’s Appendix D.1 claim. Its producer and oracle were not reexecuted here. The author’s programs and certificate data remain ignored, read-only local input and are not redistributed in this repository. The identity checks establish neither moment containment nor a certificate result.

A new-width kernel from the same scalar input

Assuming the moment containment just specified, Appendix D.2, A full-operator kernel enclosure, applies with the actual $a=\log 3$. Put $z=512a<563$ and define

$$k_{0,9}(2ay)=\sum _{q=0}^{1023}\frac{2a^{2q}+(-1{)}^q{z}^{2q}{\hat{\vartheta }}_q}{(2q)!}{y}^{2q}.\text{\hspace{1em}(A10)}$$

This constructs a new kernel; it does not rescale the author’s old matrix. On $|y|\le 1$, the moment replacement costs less than $2^{-1023}\cosh 563<2^{-210}$, because $(68/25{)}^{563}<2^{813}$ and $e<68/25$. The band absolute-weight input $\kappa <896$ and $2\cosh a=10/3<4$ bound the exact-moment Taylor remainder by

$$914\frac{563^{2048}}{2048!}<2^{-758}.$$

Indeed $(68/25)563/2048<3/4$, so the corresponding factorial estimate in Appendix D.2 applies. Consequently the full kernel and convolution operators obey the conditional bounds

$$\underset{|x|\le 2a}{\sup }|k(x)-k_{0,9}(x)|<2^{-209},\Vert K_9-K_{0,9}\Vert <2a{2}^{-209}<2^{-207}.\text{\hspace{1em}(A11)}$$

These bounds are independent of retained dimension. The source Binet remainder is already paid inside its moment intervals and is not subtracted again.

For the actual band operator, apply Appendix D.5, The complete tail columns, using its new-width polynomial

$$b_{0,9}(2ay)=\frac{256}{\pi }\sum _{q=0}^{1023}\frac{(-1{)}^q{z}^{2q}}{(2q+1)!}{y}^{2q}.$$

The sinc remainder and the same outward comparison give

$$\Vert B_9-B_{0,9}\Vert \le \frac{z^{2049}}{\pi 2049!}<2^{-768}.\text{\hspace{1em}(A12)}$$

Thus $h_{0,9}^{(0)}=(I-B_{0,9}-\delta I)v_{0,9}$ approximates the actual filtered vector with norm error less than $2^{-768}$. Since $\Vert h_{0,9}\Vert <1/29$, its rank-one update error is at most $81(2\Vert h_{0,9}\Vert 2^{-768}+2^{-1536})<2^{-765}$. Together with (A11), a matrix assembled from $K_{0,9}+81|h_{0,9}^{(0)}⟩⟨h_{0,9}^{(0)}|$ has analytic operator error less than $2^{-206}$, before paying its own directed arithmetic and center rounding. This is conditional source-input reuse and new-width assembly, not a reproduced author certificate, a finite sign test or a kernel-verified result.

The needed scalar comparisons can be replayed without regenerating any moment:

from fractions import Fraction as F

assert 512 * F(1098613, 10**6) < 563
assert F(68, 25)**563 < 2**813
assert F(68, 25) * 563 / 2048 < F(3, 4)
assert F(68, 25) * 563 / 2049 < F(3, 4)
assert F(3, 4)**4 < F(1, 3)
assert 3**512 > 2**768 and 914 < 2**10
assert F(11, 5) * F(1, 2**209) < F(1, 2**207)
assert 81 * (F(2, 29) * F(1, 2**768) + F(1, 2**1536)) < F(1, 2**765)
print('new-width kernel and actual filtered-band allowances passed')

The remaining retained-matrix consumer at $c=9$

The source’s Certification Theorem, section 6, equations (17)–(22), now has a legitimate prime-block input $m=4/5$, $b=31/5$ at this new window. In particular $M_9$ is boundedly invertible and $\Vert M_9^{-1}\Vert \le 5/4$. The same already evaluated bound (A3), before rounding, also permits $m_{\ast }=264/325$, $b_{\ast }=2011/325$ and $\Vert M_9^{-1}\Vert \le 325/264$; these are parameter substitutions, not another prime-block calculation. The conservative parameters below suffice to state the remaining obligation.

On the actual interval $(-\log 3,\log 3)$, the compact self-adjoint operator $K_9$ has the source’s kernel

$$k(u-v)=2\cosh ((u-v)/2)+\frac{1}{2\pi }{\int }_{-256}^{256}(A(t)-7/2)e^{it(u-v)}dt.$$

Together with the prime-block decomposition and the now supplied tail input (A6), this gives

$$Q(f)\ge 2^{-49162}\Vert f{\Vert }^2+⟨f,(M_9+K_9+U_9)f⟩(f\in C_c^{\infty }((-\log 3,\log 3);\mathbb{C})).\text{\hspace{1em}(A7)}$$

The sign of the bounded term in (A7) is not established. Use the same actual interval, orthonormal retained embedding $E$ and projection $P=E{E}^{\ast }$ throughout. Reflection invariance permits restricting all operators, norms and complements to the even Hilbert space for the existing even-test RH route. In that space $U_9=81|h_{0,9}⟩⟨h_{0,9}|$; the even pole contribution is also nonnegative, but the central Gamma band remains payable. Put $V_9=K_9+U_9$, $J_9=E^{\ast }{V}_{9}E$, $R_9=V_9-E{J}_9{E}^{\ast }$ and

$$G_9=E^{\ast }{M}_9^{-1}E,D_9=E^{\ast }{M}_9^{2}E-(E^{\ast }{M}_{9}E{)}^2.$$

Suppose the actual new-window blocks satisfy $\Vert (R_9{)}_{10}\Vert \le e$, $(R_9{)}_{11}⪰-nI$, $\Vert (R_9{)}_{11}\Vert \le h$, with $0\le n<4/5$. For $\theta ,\chi >0$, direct parameter substitution in the existing source theorem makes the following a sufficient target:

$$G_9^{-1}+J_9⪰\left(e\theta +\frac{(1+\chi )e^2}{4/5-n}\right)I+\frac{25}{16}\left(e/\theta +n+\frac{(1+{\chi }^{-1})h^2}{4/5-n}\right)D_9.\text{\hspace{1em}(A4)}$$

This condition would imply $M_9+V_9⪰0$, and (A7) would then give $Q(f)\ge 2^{-49162}\Vert f{\Vert }^2$ on the legal even tests in this window when all objects are restricted to the even space. It is an application of the published block criterion, not an established inequality (A4). The actual entries of $K_9$ and the filtered-vector columns, their directed source errors, the inverse-compression bound and the finite sign test remain payable. The tail input is supplied by (A5)–(A6), and the explicit even embedding has its cross and complementary-block allowances in (A8)–(A9). No retained matrix or approximation error from the $17/16$ certificate has been transported to $\log 3$; the matching tail numbers have their separate parameter proof above. Using another retained basis requires identifying the same form and transporting all these objects together. Even a completed $c=9$ sign test would still leave the subsequent cofinal support layers required for RH.

A common 256-mode consumer

For an actual assembly using 256 even Legendre columns, take $E_j(u)=\sqrt{(4j+1)/(2a)}{P}_{2j}(u/a)$, $0\le j<256$, and redefine $P,J,R,A,B,G,D$ together with this embedding. This is distinct from the 224-mode instance above. The kernel bounds (A10)–(A12) are dimension-independent and remain applicable; its projection allowances must be calculated for the new space.

The Bernstein ellipse of parameter $2$ has imaginary semiaxis $3/4$. Degree-$511$ Chebyshev truncation of $\cos (tax)$ is even, so its degree is at most $510$ and it belongs to this retained space. The coefficient tail gives

$$r:=\underset{|t|\le 256}{\sup }\Vert (I-P)\cos (tu)\Vert \le \sqrt{2a}{e}^{192a}{2}^{-510}<2^{-200}.$$

Here $a<11/10$ implies $192a<211.2<212$, hence $e^{192a}<(68/25{)}^{212}$. Together with $\sqrt{2a}<3/2$ and $(3/2)(68/25{)}^{212}<2^{310}$, this proves the last outward comparison. Taylor truncation of $\cosh (u/2)$ through degree $510$ gives $p_{\perp }:=\Vert (I-P)\cosh (u/2)\Vert <3/512!<2^{-509}$. For $c(u)=\cosh (u/2)$, its actual norm satisfies $\Vert c{\Vert }^2=a+\sinh a=\log 3+4/3<73/30<4$. Thus the even pole cross block is at most $2\Vert Pc\Vert p_{\perp }<4p_{\perp }$; its complementary block is positive with norm $2p_{\perp }^2$. Using $\Vert h_{0,9}\Vert <1$, the same band and tail estimates as (A9) give

$$\Vert R_{10}\Vert <2154r+4p_{\perp },R_{11}⪰-896r^{2}I,\Vert R_{11}\Vert <8996r^2+2p_{\perp }^2.$$

Thus $e=2^{-188}$, $n=2^{-390}$ and $h=2^{-386}$ are valid conservative allowances for this common embedding. With $m=264/325$ and $\theta =\chi =1$, the source’s coefficients satisfy $\alpha <2^{-187}$ and $\beta <2^{-186}$. This is another source-proof parameter application, with no finite sign inferred from the smaller projection error.

The unnumbered inverse-residual identity in section 4’s proof of Theorem B supplies the inverse-compression upper bound without knowing the target sign. Equation (11) there lists old-window error constants; those constants are not used here. For any trial matrix $X$, choose $\mu =4/5<m$ and put

$${\mathcal{W}}_{\mu }(A,B;X)=X+X^{\ast }-X^{\ast }AX+{\mu }^{-1}(I-AX-X^{\ast }A+X^{\ast }BX).$$

The source residual identity gives ${\mathcal{W}}_{\mu }(A,B;X)⪰G$. This is a parameterized obligation: choose a trial $X$ and a supplied bound $s\ge \Vert X\Vert$, and Hermitian centers satisfying $\Vert A-A_0\Vert \le {\epsilon }_A$ and $\Vert B-B_0\Vert \le {\epsilon }_B$. No numerical $s$, ${\epsilon }_A$ or ${\epsilon }_B$ from the old window is transferred. Then

$$W={\mathcal{W}}_{\mu }(A_0,B_0;X)+\left[(s^2+2s/\mu ){\epsilon }_A+(s^2/\mu ){\epsilon }_B\right]I⪰G\succ 0.$$

Consequently $W^{-1}⪯G^{-1}$. In particular, the direction is suitable for a sufficient lower comparison. Let $J_0$ be a Hermitian center of the compression of the polynomial kernel plus its actual polynomial-filtered rank-one update. If ${\rho }_J$ bounds the directed finite assembly and center-rounding error of that complete update, define ${\epsilon }_J:=2^{-206}+{\rho }_J$. The first term pays both analytic replacements in (A11)–(A12), including the rank-one filtered-vector error; ${\rho }_J$ must pay all remaining arithmetic, column-evaluation and rank-one rounding errors. Then $\Vert J-J_0\Vert <{\epsilon }_J$. If instead a separately rounded filtered column is used to construct $J_0$, its induced rank-one error must also enter ${\rho }_J$. With $D_0=B_0-A_0^2$, $b=2011/325$ and ${\epsilon }_D={\epsilon }_B+(2b+{\epsilon }_A){\epsilon }_A$, the parameterized finite target is

$$W^{-1}+J_0-{\epsilon }_{J}I-2^{-187}I-2^{-186}(D_0+{\epsilon }_{D}I)⪰0.\text{\hspace{1em}(A13)}$$

All three centers and their supplied errors must belong to the same actual 256-mode embedding. In particular, $B_0$ must approximate $E^{\ast }{M}_9^{2}E$, rather than $A_0^2$; clipping occurs before composing the shifts. The combined analytic allowance $2^{-206}$ enters ${\epsilon }_J$ only under the stated author-moment premise. Source positivity of the original window does not settle (A13). This note supplies its criterion and error interfaces, with no kernel theorem or originality claim.

The actual c9 reproduction package publishes project-authored programs, hash-bound new-width matrix inputs and producer and separate-session conditional runs of (A13). The separate native Codex CLI session used the same program and matrix inputs, reusing the producer’s environment after directly checking the pinned versions. Its full wrapper exited zero at 1536 bits; inspection of its new result confirmed all 256 serialized pivot lower endpoints strictly positive, no directed negative witness, and the exact Frobenius bound 32 by recomputing the squared norm from all 65,536 trial integers. The compact execution record binds that fresh result by its hash. The recorded minimum pivot lower bound is not an eigenvalue bound. Moment containment remains an author premise; the retained-target source contract was reviewed separately, while independent implementation and matrix regeneration remain unverified. The package reuses the existing inputs rather than replaying the author’s old certificate. This is independent execution evidence for the same conditional sufficient comparison, with no new mathematics or kernel proof; the cofinal support positivity required for RH remains unproved.

from fractions import Fraction as F

assert F(3, 2) * F(68, 25)**212 < 2**310
assert 192 * F(11, 10) < 212
assert F(11, 10) + F(4, 3) == F(73, 30) < 4
r, p = F(1, 2**200), F(1, 2**509)
e, n, h = F(1, 2**188), F(1, 2**390), F(1, 2**386)
assert 2154*r + 4*p < e
assert 896*r*r < n and 8996*r*r + 2*p*p < h
m = F(264, 325)
assert e + 2*e*e/(m-n) < F(1, 2**187)
assert (e+n)/m**2 + 2*h*h/(m**2*(m-n)) < F(1, 2**186)
print('common256-mode projection and consumer allowances passed')

The next FIB layer needs a different prime reference

The source’s inverse-residual comparison requires a positive reference operator. Its constant-background splitting is therefore subject to a separate scale condition. At $c>1$, put $a_c=\frac{1}{2}\log c$ and use the actual even space $L_{\mathrm{even}}^2((-a_c,a_c);\mathbb{C})$. With the same compressed shifts as above, define

$$C_{\mathrm{p},c}=\sum _{n<c}\frac{\Lambda (n)}{\sqrt{n}}(S_{\log n}+S_{-\log n}),M_{c,\beta }=\beta I-C_{\mathrm{p},c}.$$

Only prime powers contribute. The normalized constant vector $u_c=(\log c{)}^{-1/2}{1}_{(-a_c,a_c)}$ gives the exact overlap identity

$$\rho (c):=⟨u_c,C_{\mathrm{p},c}{u}_c⟩=2\sum _{n<c}\frac{\Lambda (n)}{\sqrt{n}}\left(1-\frac{\log n}{\log c}\right).\text{\hspace{1em}(A14)}$$

The endpoint shift is zero almost everywhere. Each positive-part summand increases with $c$, so $\rho$ is nondecreasing. In particular, $M_{c,\beta }⪰mI$ with $m>0$ requires $\beta \ge \rho (c)+m$. The constant vector belongs to the actual even Hilbert space; density also transports a strictly negative reference quadratic to even smooth compactly supported tests. This tests the reference operator, not the complete Weil form.

At the next FIB cutoff $c=27$, retaining just $n\in \{2,3,4,5,7,8,9,11\}$ already gives $\rho (27)>4$. The other prime-power terms are nonnegative and need no recomputation for this lower comparison. Use the preceding source logarithm enclosures for $2,3,5,7$, the elementary $239/100<\log 11<12/5$, and root upper bounds as in the scalar replay below. For $n=p^j$, each retained summand is bounded below by

$$\frac{2l_p^{-}}{r_n^{+}}\left(1-\frac{j{l}_p^{+}}{3l_3^{-}}\right)>0.\text{\hspace{1em}(A15)}$$

This is an application of the existing compressed-translation formula and scalar test, rather than another prime matrix or certificate. The following exact replay verifies the new downstream comparison and the elementary enclosure for $\log 11$; the other logarithm enclosures remain the previously cited analytic inputs. For $0<x<N+2$, the first omitted exponential-series term is $x^{N+1}/(N+1)!$ and every subsequent ratio is at most $x/(N+2)<1$, which justifies the geometric-tail upper bound used with $x=239/100$ and $N=12$.

from fractions import Fraction as F
from math import factorial

lo = {2: F(693147, 10**6), 3: F(1098612, 10**6),
      5: F(1609437, 10**6), 7: F(1945910, 10**6), 11: F(239, 100)}
hi = {2: F(693148, 10**6), 3: F(1098613, 10**6),
      5: F(1609438, 10**6), 7: F(1945911, 10**6), 11: F(12, 5)}
roots = {2: F(1414214, 10**6), 3: F(1732051, 10**6), 4: F(2),
         5: F(2236068, 10**6), 7: F(2645752, 10**6),
         8: F(2828428, 10**6), 9: F(3), 11: F(10, 3)}
powers = {2: (2, 1), 3: (3, 1), 4: (2, 2), 5: (5, 1),
          7: (7, 1), 8: (2, 3), 9: (3, 2), 11: (11, 1)}

def exp_partial(x, N):
    return sum((x**k / factorial(k) for k in range(N + 1)), F(0))

N = 12
x = lo[11]
exp_upper = (exp_partial(x, N)
             + x**(N + 1) / factorial(N + 1) / (1 - x / (N + 2)))
assert exp_upper < 11 < exp_partial(hi[11], N)
assert all(r * r >= n for n, r in roots.items())
terms = [2 * lo[p] / roots[n] * (1 - j * hi[p] / (3 * lo[3]))
         for n, (p, j) in powers.items()]
assert all(term > 0 for term in terms)
assert sum(terms, F(0)) > 4
print('the next FIB layer requires beta > 4 for a positive prime reference')

Consequently every $c\ge 27$ requires $\beta >4$ for a strictly positive constant-background reference, including all subsequent cutoffs of the specified FIB sequence. This rules out reusing the $7/2$ prime background; it does not assert negativity of $Q$.

Raising the background conflicts with the fixed-band tail

For a variable background $\beta$ at band cutoff $\Omega$, the same exact splitting has exterior weight $A(t)-\beta$. Retaining the source mechanism’s pointwise condition $A(t)-\beta \ge 0$ for $|t|>\Omega$ requires

$$\beta \le A(\Omega ).$$

Continuity gives this necessary boundary inequality even when the frequency endpoint itself is excluded. This is a condition on that particular tail mechanism; no assertion is made that every lower bound for a signed exterior integral requires a pointwise nonnegative multiplier.

At $\Omega =256$, the elementary upper bound $A(256)<4$ suffices. The primary formula DLMF 5.9.13, for $\Re z>0$, is

$$\psi (z)=\log z-{\int }_0^{\infty }h(t)e^{-zt}dt,h(t)=\frac{1}{1-e^{-t}}-\frac{1}{t}.$$

The inequalities $1-e^{-t}<t<e^t-1$ give $0<h(t)<1$, hence $|\psi (w)-\log w|\le 1/\Re w$. Apply the digamma recurrence, DLMF 5.5.2, to $z=1/4+128i$ and $w=z+64$. Every subtracted reciprocal has positive real part, so

$$\Re \psi (z)=\Re \psi (w)-\sum _{k=0}^{63}\Re \frac{1}{z+k}\le \Re \psi (w).$$

Now $|w|<144$, $\Re w>64$ and $\pi >3$ give

$$A(256)<\log 48+\frac{1}{64}<4.\text{\hspace{1em}(A16)}$$

For the last comparison, $e^4>{\sum }_{k=0}^9{4}^k/k!>54$ and $e^{1/64}<64/63$ imply $e^{4-1/64}>54\cdot 63/64>48$. Its rational replay is:

from fractions import Fraction as F
from math import factorial

assert F(257, 4)**2 + 128**2 < 144**2
assert sum((F(4)**k / factorial(k) for k in range(10)), F(0)) > 54
assert F(54) * F(63, 64) > 48
print('the band256 pointwise tail mechanism requires beta < 4')

Thus, for every $c\ge 27$, no real $\beta$ can satisfy both $M_{c,\beta }⪰mI$ for some $m>0$ and the pointwise exterior-weight condition at band $256$. Merely raising the constant background cannot continue this fixed-band comparison through the next FIB layer. This is a paper-level obstruction to the stated splitting and sufficient criterion; it refutes neither full Weil positivity nor RH.

Necessary band growth for the same cofinal strategy

The classical unconditional prime number theorem, in the form $\Psi (t)={\sum }_{n\le t}\Lambda (n)\sim t$, and partial summation give

$$S(t):=\sum _{n\le t}\frac{\Lambda (n)}{\sqrt{n}}=\frac{\Psi (t)}{\sqrt{t}}+\frac{1}{2}{\int }_1^t\frac{\Psi (u)}{u^{3/2}}du\sim 2\sqrt{t}.$$

This is the existing half-weighted Mangoldt sum discussed in Chirre–Helfgott’s source application; no new prime-number theorem or stronger error term is used. Writing $\log (c/n)={\int }_n^{c}dt/t$ in (A14) and exchanging the finite positive sum with the integral gives

$$\rho (c)=\frac{2}{\log c}{\int }_1^c\frac{S(t)}{t}dt\sim \frac{8\sqrt{c}}{\log c}.\text{\hspace{1em}(A17)}$$

The positive-integral representation matters: a coarse error in two separately estimated weighted sums need not survive their leading cancellation, whereas $S(t)\sim 2\sqrt{t}$ directly yields (A17).

Suppose the same strategy uses $M_{c,{\beta }_c}⪰m_{c}I$ with $m_c>0$ and $A(t)-{\beta }_c\ge 0$ for all $|t|>{\Omega }_c$, along a specified cofinal family of cutoffs. These are conditions on the actual common splitting. They require

$$A({\Omega }_c)\ge {\beta }_c\ge \rho (c)+m_c.$$

In particular ${\Omega }_c\to \infty$. The digamma asymptotic DLMF 5.11.2, applied in a fixed sector containing $1/4+it/2$, gives $A(t)=\log (t/(2\pi ))+o(1)$ as $t\to +\infty$. Therefore the strategy necessarily satisfies

$$\underset{c\to \infty }{\mathrm{liminf}}\frac{\log {\Omega }_c\log c}{\sqrt{c}}\ge 8.\text{\hspace{1em}(A18)}$$

The limit is taken through that asserted family; parameters at intervening cutoffs are not required.

For the specified FIB sequence, write $c_r=3^{k_r}$ with $k_0=k_1=1$ and $k_{r+2}=k_{r+1}+k_r$. Then (A18) requires

$$\log {\Omega }_r\ge (8-o(1))\frac{3^{k_r/2}}{k_r\log 3}.$$

Thus even a band growing polynomially in $c$ is insufficient for these two simultaneous conditions. This is a necessary growth bound for a constant-background reference with a pointwise nonnegative exterior multiplier. It is not a lower bound for the cost of proving RH, and a band satisfying it still supplies no retained sign or cofinal positivity.

The existing joint pole–prime localization identity provides another representation: it combines the continuous prime main term with the pole and Gamma terms before estimating the signed Chebyshev remainder. Its common-test remainder still lacks the required favorable bound. Alternatives to the present obstruction must supply such a joint estimate, control a signed exterior band, or justify another reference operator without a globally positive $\beta I-C_{\mathrm{p},c}$. None is supplied by raising $\beta$ alone, reusing the scalar moment packet at a different band, or changing the FIB coordinates. These conclusions are source applications and interface analysis, not originality claims or a proof of RH.

The reusable spectral construction, ground-state theorem and negative example, effective prolate concentration bounds, semilocal trace remainder, archimedean signed Sonin comparison and semilocal dual pairing are catalogued with their source hypotheses and parameter maps. They should be reused in searching for an actual common-form estimate. Their different operator, normalization and sign contracts do not supply the retained comparison or its cofinal extension automatically.