bibkey: lagarias2007li
authors: Jeffrey C. Lagarias
year: 2007
title: Li Coefficients for Automorphic L-Functions
doi: 10.5802/aif.2311
claim: The pole-removed completed automorphic L-function is entire of order one; the trivial representation gives twice the classical Riemann xi function.
strata_touched:
  - D5/S3/Analytic/Resolvent/XiGlobalGrowth
license: citation-only
triage: anchor

Li Coefficients for Automorphic L-Functions

Theorem 2.1(6) states that the pole-removed function xi(s, pi) is entire of order one and maximal type. Its standing hypothesis is an irreducible cuspidal unitary automorphic representation over the rationals. The paragraph immediately before Theorem 2.1 explicitly includes the trivial representation of GL(1), for which xi(s, pi_triv) = 2 xi(s). The simple poles at zero and one have been removed in this normalization.

Consequently, for the classical xi function there is a positive real constant C such that |xi(s)| <= exp(C (1 + |s|)^(3/2)) for every complex s. This is a weaker consequence of order one, not the literal statement of Theorem 2.1(6). The factor two and a bounded initial disk can be absorbed into C. No Riemann hypothesis is required for this consequence.

D5/S3/Analytic/Resolvent/XiGlobalGrowth.xi_reading_norm_le_exp_log_linear proves the stronger bound exp(C (1 + |s|) (1 + log(1 + |s|))) directly from the existing symmetric theta-Mellin integral. The normalized resolvent derives the displayed three-halves consequence inside its proof. It does not formalize automorphic representations, an order-one theorem, Hadamard factorization, or the paper’s Li coefficient identities. Its private integrand estimate keeps an unspecified positive theta decay rate and the whole initial integration interval. The actual Lean xi is the global pole-removed definition, including its values at zero and one.

Verified locator

• Read on 2026-09-09: https://arxiv.org/html/math/0404394, the normalization preceding Theorem 2.1, all six clauses of the theorem, and its proof of (6). The proof treats the trivial representation’s poles explicitly. Its later informal statements about the meromorphic completion are not used as formulas at the two poles.
• Crossref’s exact author/title query on 2026-09-09 matched Jeffrey C. Lagarias, this title, and DOI https://doi.org/10.5802/aif.2311.
• Published in Annales de l’Institut Fourier 57 (2007), 1689–1740.

Fibonacci-only Li sampling: a general-model obstruction

This section is a repository application of an existing Fibonacci avoidance theorem and the standard Fejer kernel identity, not a theorem attributed to Lagarias. It concerns general symmetric zero models, not the actual zeta function. It is a paper derivation without Lean certification or a priority claim. The Li conventions are those in arXiv:math/0404394v4, Introduction, equations (1.1)–(1.3) and the general multiset discussion following (1.6).

FL1. Reused explicit Fibonacci avoidance

Set $F_0=0$, $F_1=1$, and $F_{j+2}=F_{j+1}+F_j$. The distinct positive Fibonacci indices are $F_2,F_3,\dots =1,2,3,5,\dots$. Victoria Zhuravleva, Diophantine approximations with Fibonacci numbers, arXiv:1112.6142v1, Theorem 2, supplies the explicit choice

$$\phi =\frac{1+\sqrt{5}}{2},\alpha =\frac{1}{\phi +2},\delta =\frac{\phi -1}{\phi +2}>0.$$

For every $N\ge 4$, the theorem states ${\min }_{1\le j\le N}\Vert F_j\alpha {\Vert }_{\mathbb{R}/\mathbb{Z}}=\delta$. In particular $\alpha$ is irrational, $0<\delta <1/2$, and

$$\Vert F_j\alpha {\Vert }_{\mathbb{R}/\mathbb{Z}}\ge \delta (j\ge 2).$$

The theorem is reused without a new proof. Peres–Schlag’s general lacunary avoidance theorem, already cited in the lacunary-source discussion, also gives a suitable multiplier, but the explicit Fibonacci result supplies this application’s input directly. No optimality or winning-set claim from either source is needed.

FL2. A strict sign on every Fibonacci index

Choose an integer $D\ge 1$ with $D+1>{\csc }^2(\pi \delta )$, and put

$$m_k=D+1-k(1\le k\le D),T=\sum _{k=1}^D{m}_k=\frac{D(D+1)}{2},S_n=\sum _{k=1}^D{m}_k\cos (2\pi kn\alpha ).$$

The standard Fejer identity gives

$$2S_n=\frac{{\sin }^2((D+1)\pi n\alpha )}{{\sin }^2(\pi n\alpha )}-(D+1).$$

Consequently, with $\eta =((D+1)-{\csc }^2(\pi \delta ))/2>0$,

$$S_{F_j}\le -\eta (j\ge 2).$$

All multiplicities $m_k$ are positive integers. The same $D$, $\alpha$ and $\eta$ work for every sampled index.

FL3. Off-line zeros inside the strip

Irrationality gives $\cos (2\pi k\alpha )<1$ for every $1\le k\le D$. Choose one fixed $r>1$ sufficiently close to one that $r\cos (2\pi k\alpha )<1$ for all these $k$. Define

$$z_k=r{e}^{2\pi ik\alpha },{\rho }_k=\frac{1}{1-z_k},$$

and form the multiset $Z$ with the quartet ${\rho }_k,\overline{{\rho }_k},1-{\rho }_k,1-\overline{{\rho }_k}$, each repeated $m_k$ times. Writing ${\theta }_k=2\pi k\alpha$,

$$\Re {\rho }_k=\frac{1-r\cos {\theta }_k}{1-2r\cos {\theta }_k+r^2},\Re {\rho }_k-\frac{1}{2}=\frac{1-r^2}{2|1-z_k{|}^2}<0.$$

The numerator of the first fraction is positive; its denominator minus its numerator is $r(r-\cos {\theta }_k)>0$. Thus every member of $Z$ lies strictly in $0<\Re s<1$, none lies on $\Re s=1/2$, and none is zero or one. Irrationality also makes these roots nonreal.

Let $P(s)={\prod }_{\rho \in Z}(s-\rho )$, with multiplicities. This monic polynomial has real coefficients, degree $4T$, $P(1-s)=P(s)$, and $P(s)>0$ on the real axis. Define its finite Li sums by

$$L_n(P)=\sum _{\rho \in Z}\left[1-{\left(1-\frac{1}{\rho }\right)}^n\right].$$

The four Mobius values from each quartet are $z_k,\overline{z_k},z_k^{-1},{\overline{z_k}}^{-1}$. Hence

$$L_n(P)=4T-2(r^n+r^{-n})S_n,$$

and therefore

$$L_{F_j}(P)\ge 4T+2\eta (r^{F_j}+r^{-F_j})>0(j\ge 2).$$

For an individual root, the Li derivative at $s=1$ gives the inverse power $1-(1-1/\rho {)}^{-n}$. Reflection symmetry makes the full sum invariant under $n\mapsto -n$. Thus $L_n(P)$ also equals

$$\frac{1}{(n-1)!}{\frac{d^n}{d{s}^n}[s^{n-1}\log P(s)]|}_{s=1}(n\ge 1),$$

using the local real logarithm. A constant normalization has no effect. This retains the distinction between the two conventions in Lagarias’s Introduction rather than identifying individual-root formulas.

There is an unbounded sequence of positive integers $n_l$ with $\Vert n_l\alpha {\Vert }_{\mathbb{R}/\mathbb{Z}}\to 0$, by irrational rotation recurrence. Along it, $S_{n_l}\to T>0$, so $L_{n_l}(P)<0$ eventually. Thus the positive Fibonacci samples coexist with negative unsampled coefficients; the complete all-index criterion is unchanged.

FL4. Entire order one and infinitely many zeros

Put $H(s)=P(s)\cosh (s-1/2)$. It is entire of order exactly one, has real values and is strictly positive on the real axis, satisfies $H(1-s)=H(s)$, and has no zeros at zero or one. Its added zeros are $1/2\pm i{t}_h$, where $t_h=\pi (h+1/2)$ and $h\ge 0$.

The standard paired cosh product gives the exact normalization

$$\frac{\cosh (s-1/2)}{\cosh (1/2)}=\prod _{h\ge 0}\left(1+\frac{s^2-s}{1/4+t_h^2}\right).$$

It converges locally uniformly, with no unrecorded exponential factor. For $u_h=1-1/(1/2+i{t}_h)$, $|u_h|=1$, so the paired Li contribution is

$$L_n(\cosh )=\sum _{h\ge 0}2(1-\Re u_h^n),0\le L_n(\cosh )\le n^2\sum _{h\ge 0}\frac{1}{1/4+t_h^2}.$$

The bound follows from $|1-u^n|\le n|1-u|$ on the unit circle and $|1-u_h{|}^2=(1/4+t_h^2{)}^{-1}$. It proves absolute convergence after conjugate pairing for each fixed $n$. Local logarithmic differentiation of the displayed product identifies the paired sums with its Li derivatives. Consequently $L_n(H)=L_n(P)+L_n(\cosh )$: every Fibonacci sample remains positive, while the negative exponential term along $n_l$ eventually dominates the nonnegative $O(n_l^2)$ added term.

FL5. What this excludes

Fibonacci-index positivity cannot replace all-index positivity on this general zero-model class, even with conjugation symmetry, functional symmetry, open-strip confinement, entire order one, and infinitely many zeros. A zeta-specific conclusion requires input beyond these properties; this argument does not identify sufficient additional analytic or arithmetic input. No Euler product or zeta explicit formula for $H$ is established. This model is not a counterexample to RH, Robin, or a zeta-specific sparse criterion. No numerical experiment or zero computation is used.

This restriction of tested indices is distinct from using Fibonacci or Zeckendorf addresses to label all integer indices. Such an all-index encoding does not discard indices and is not excluded by this example.

Fibonacci-sampled absolute growth: a paper criterion

Fibonacci-only sign positivity in FL1–FL5 differs from an absolute-growth condition. The following paper derivation gives

$$\mathrm{RH}⟺\underset{j\to \infty }{\mathrm{limsup}}|{\lambda }_{F_j}{|}^{1/F_j}\le 1,F_0=0,F_1=1,\text{\hspace{1em}(FG1)}$$

for the actual classical Li coefficients. It does not establish the displayed growth bound for those coefficients and therefore does not prove RH. No Lean certification or literature priority is claimed. Hilbert basis, Laurent observation ideals and the Li generating-function identity are existing tools; the application and its remaining bound must be distinguished from them.

FG2. A finite exact relation for the common Fibonacci phase state

Fix $d\ge 1$ and positive integers $m_1,\dots ,m_d$; write $M={\sum }_i{m}_i$. Work in the Noetherian Laurent polynomial ring

$$\mathcal{A}=\mathbb{Q}[X_1^{\pm 1},\dots ,X_d^{\pm 1},Y_1^{\pm 1},\dots ,Y_d^{\pm 1}].$$

Noetherianity is the usual finite-type extension and localization result; see Stacks Project, Lemma 10.31.1, tag 00FN. It imposes no algebraicity condition on the points where these formal polynomial identities are evaluated.

The Fibonacci phase update and its inverse are

$$T(x,y)=(y,xy),T^{-1}(x,y)=(y/x,x),$$

coordinatewise. Pullback $\sigma f=f\circ T$ is a ring automorphism, with $\sigma X_i=Y_i$ and $\sigma Y_i=X_i{Y}_i$. Put $p={\sum }_i{m}_i{X}_i$ and $I=(p,\sigma p,{\sigma }^{2}p,\dots )$. The inclusion $\sigma I\subseteq I$ gives the ascending ideal chain

$$I\subseteq {\sigma }^{-1}I\subseteq {\sigma }^{-2}I\subseteq \cdots .$$

It stabilizes by Noetherianity. Applying a power of $\sigma$ at a stabilization gives $\sigma I=I$. In particular, ideal membership supplies some finite $K\ge 1$ and $a_1,\dots ,a_K\in \mathcal{A}$ with

$$\boxed{p=\sum _{r=1}^K{a}_r{\sigma }^{r}p.}\text{\hspace{1em}(FG2)}$$

This is an identity before any phases are substituted. No effective value of $K$ or bound on the coefficients is computed here.

FG3. Every fixed-length block retains a nonzero observation

Let $u_i\in {\mathbb{S}}^1$, put $a=(1,\dots ,1,u_1,\dots ,u_d)$, and use the same state for every iterate. Then

$$T^{j}a=((u_i^{F_j}{)}_i,(u_i^{F_{j+1}}{)}_i),q_j=p(T^{j}a)=\sum _i{m}_i{u}_i^{F_j},q_0=M>0.$$

Every Laurent monomial has modulus one on $\mathbb{T}=({\mathbb{S}}^1{)}^{2d}$. Let $C=\max (1,{\sum }_{r=1}^K\Vert a_r{\Vert }_1)$, where the norm is the sum of absolute coefficient values. Evaluating FG2 at $T^{j}a$ gives

$$|q_j|\le C\underset{1\le r\le K}{\max }|q_{j+r}|.$$

For $B_j={\max }_{0\le r<K}|q_{j+r}|$, this yields $B_j\le C{B}_{j+1}$: all entries except the first already belong to the next block. Thus

$$\boxed{\underset{0\le r<K}{\max }|q_{j+r}|\ge M{C}^{-j}(j\ge 0).}\text{\hspace{1em}(FG3)}$$

Both $K$ and $C$ depend only on the fixed multiplicity vector, not on the phases or $j$. Positivity enters through $q_0=M$. Repeated phases, root-of-unity ratios and simultaneous dominant zeros are retained. No single dominant pair or algebraicity assumption is used.

The update is the multiplicative phase form of the matrix $\left(\begin{array}{cc}0& 1\\ 1& 1\end{array}\right)$, with initial exponent pair $(0,1)$. Its invertibility is essential to FG2. The scalar update $u\mapsto u^2$ has no inverse in the Laurent ring; there is no corresponding assertion for arbitrary lacunary sequences.

FG4. Finite positive power sums preserve their maximal root rate

Let $z_i\ne 0$ be arbitrary complex bases with positive integer weights $m_i$, allowing repetitions, and set $R={\max }_i|z_i|$. Then

$$\underset{j\to \infty }{\mathrm{limsup}}{\left|\sum _i{m}_i{z}_i^{F_j}\right|}^{1/F_j}=R.\text{\hspace{1em}(FG4)}$$

For common modulus $R$, write $z_i=R{u}_i$ and use FG3. Every block supplies an index ${\ell }_j\in [j,j+K-1]$ with $|q_{{\ell }_j}|\ge M{C}^{-j}$. Since $j/F_{{\ell }_j}\to 0$, this loss has root rate one on the scale $F_{{\ell }_j}$. The elementary bound $|q_j|\le M$ gives the reverse inequality.

More generally, for dominant bases $|z_i|=R$ and any remainder $E_n=O(B^n)$ with a fixed $0<B<R$, the same proof gives

$$a_n=-\sum _i{m}_i{z}_i^n+E_n⟹\underset{j}{\mathrm{limsup}}|a_{F_j}{|}^{1/F_j}=R.\text{\hspace{1em}(FG5)}$$

Writing $|E_n|\le A{B}^n$ for sufficiently large $n$, at the indices supplied by FG3, $|a_{F_{{\ell }_j}}|\ge R^{F_{{\ell }_j}}(M{C}^{-j}-A(B/R{)}^{F_{{\ell }_j}})$, which is at least $M{C}^{-j}{R}^{F_{{\ell }_j}}/2$ eventually. Unequal-modulus sums in FG4 follow by placing all smaller-modulus terms in this remainder. The bound holds in every block; the favorable indices are consequences, not an assumption about a selected subsequence.

FG5. The actual Li generating function pays the entire remainder

Use Lagarias, arXiv:math/0404394v4, Introduction, equation (1.4):

$$G(w)=-\frac{1}{(1-w{)}^2}\frac{{\xi }^{\prime }}{\xi }\left(-\frac{w}{1-w}\right)=\sum _{n\ge 1}{\lambda }_n{w}^{n-1}.\text{\hspace{1em}(FG6)}$$

This is the logarithmic derivative of the actual composed xi function, not a global choice of logarithm or a substituted zero model. Xi is entire and nonzero at zero, so $G$ is holomorphic near zero and meromorphic throughout $|w|<1$.

A zero $\rho$ of multiplicity $m_{\rho }$ gives a simple pole at $w_{\rho }=\rho /(\rho -1)$, with residue $m_{\rho }$. Put $z_{\rho }=1-1/\rho =1/w_{\rho }$. Its pole term is

$$\frac{m_{\rho }}{w-w_{\rho }}=-\frac{m_{\rho }{z}_{\rho }}{1-z_{\rho }w}=-\sum _{n\ge 1}{m}_{\rho }{z}_{\rho }^n{w}^{n-1}.\text{\hspace{1em}(FG7)}$$

If RH is false, reflection symmetry supplies a zero with $\Re \rho <1/2$, hence a pole inside the disk since $|z_{\rho }{|}^2=1+(1-2\Re \rho )/|\rho {|}^2>1$. The discrete pole set has a positive smallest modulus $r_0<1$, attained at finitely many poles $w_1,\dots ,w_d$. Retain all their multiplicities $m_i>0$ and put $z_i=1/w_i$, $R=1/r_0>1$.

Choose $r_0<r_1<1$ so these are the only poles in $|w|\le r_1$. Subtracting all their principal parts gives

$$H(w)=G(w)+\sum _{i=1}^d\frac{m_i{z}_i}{1-z_{i}w},$$

holomorphic on a neighborhood of the closed disk $|w|\le r_1$. Cauchy’s coefficient estimate for $H$ gives the exact decomposition

$${\lambda }_n=-\sum _{i=1}^d{m}_i{z}_i^n+E_n,E_n=O(B^n),B=1/r_1<R.\text{\hspace{1em}(FG8)}$$

This controls the entire analytic remainder of the actual generating function. It neither discards uncomputed zeros nor rearranges an infinite constant-minus-power sum. No zero-height computation is required. By the smaller-remainder estimate (FG5), under RH failure,

$$\underset{j}{\mathrm{limsup}}|{\lambda }_{F_j}{|}^{1/F_j}=\underset{\rho }{\max }\left|1-\frac{1}{\rho }\right|>1.\text{\hspace{1em}(FG9)}$$

Conversely, under RH the same $G$ has no poles inside the unit disk. The standard Cauchy coefficient bound on every smaller disk gives the all-index root bound, hence its Fibonacci restriction. This is the known RH-to-growth implication, also recorded in Lagarias’s all-index criterion (Theorem 2.2(3), equation (2.22)); it is reused, not counted as new mathematics. Together with FG9 it gives FG1.

FG6. What remains to prove

The unpaid assertion is still the actual bound $|{\lambda }_{F_j}|=O_{\epsilon }(e^{\epsilon F_j})$ for every $\epsilon >0$. The exact finite observation relation detects an exponential rate; it does not rule that rate out for the actual xi function. FL1–FL5 remains a valid sign-only obstruction, whose positive Fibonacci samples grow exponentially. The original Robin signed estimate, the endpoint half-variance bound and RH remain unresolved. No numerical experiment, parameter scan, zero computation or new Lean certification is used.

A signed truncation formula requires a source-use restriction

Theorem 6.1, equation (6.2), in both arXiv:math/0404394v4 and the published article, printed p.1722, is not adopted here as an unconditional signed supplier. With the article’s definition

$$L_n(T)=\sum _{\begin{array}{c}\rho \in Z(\zeta )\\ |\Im \rho |\le T\end{array}}\left[1-{\left(1-\frac{1}{\rho }\right)}^n\right],$$

including all zero multiplicities, that displayed formula reads $S_f(n)=+L_n(\sqrt{n})+O(\sqrt{n}\log n)$. The following local residue and consistency checks explain the restriction; they do not establish that the printed statement is false.

For the counterclockwise contour in the proof, with normalization $1/(2\pi i)$, put $k_n(s)=(1+1/s{)}^n-1$. At $s=\rho -1$,

$$\mathrm{Res}\left[k_n(s)\left(-\frac{{\zeta }^{\prime }}{\zeta }(1+s)\right)\right]=-m_{\rho }{k}_n(\rho -1)=m_{\rho }\left[1-{\left(\frac{\rho }{\rho -1}\right)}^n\right].$$

Reflection $\rho \mapsto 1-\rho$ preserves the full ordinate cutoff, so these nontrivial-zero residues sum to $+L_n(T)$. The unnumbered contour expansion following (6.5) instead inserts $+k_n(\rho -1)$; its subsequent (6.7), printed p.1723, has $-L_n(T)$. Definitions (1.1), (4.5)–(4.8) and (6.1) supply no compensating convention change in the classical zeta specialization. A local sign repair would reverse this contribution. It alone would not establish the remaining contour estimates, and no corrected version of the signed asymptotic is asserted here.

There is also an independent weak-tail consistency check. Pair $\rho =\beta +i\gamma$ with its conjugate and put $z=1-1/\rho$. For $\gamma >\sqrt{n}$, $0<\beta <1$ gives $|z|\le 1+1/(2{\gamma }^2)$ and $\Re (1/\rho )=O({\gamma }^{-2})$. Taylor’s formula for the polynomial $z^n$ at one gives

$$|z^n-1-n(z-1)|\le \frac{1}{2}n(n-1)|z-1{|}^2\max (1,|z|{)}^{n-2},$$

so the complete conjugate pair satisfies $|(1-z^n)+(1-{\bar{z}}^n)|\ll n^2/{\gamma }^2$ uniformly in this range. Reusing $N(T)=O(T\log T)$, including multiplicities, gives ${\sum }_{\gamma >T}{m}_{\rho }{\gamma }^{-2}=O(\log T/T)$ and hence

$${\lambda }_n-L_n(\sqrt{n})=O(n^{3/2}\log n).\text{\hspace{1em}(ST1)}$$

The paired tail is absolutely convergent for each fixed $n$, so it also agrees with the ordinate-cutoff convention for the complete Li sum. Combining ST1 with the exact classical split from published Lemma 4.4, ${\lambda }_n=S_{\infty }(n)-S_f(n)+1$, shows that if the literal plus-sign approximation (6.2) were used, it would imply

$$2{\lambda }_n=S_{\infty }(n)+1+O(n^{3/2}\log n).$$

Published Theorem 5.1 supplies $S_{\infty }(n)=O(n\log n)$ unconditionally. The displayed implication would therefore give polynomial absolute growth and RH by the existing all-index criterion. It is an implication from a sign-sensitive premise, not an acquired bound: we do not import that premise pending clarification of its proof. No conclusion that the printed statement is false, new literature priority, Lean certification, or RH proof follows. The generating-function identity and the Fibonacci growth derivation above use neither (6.2) nor its contested zero-residue sign.

A one-sided Li tail bound from reciprocal zero pairing

The original joint matrix is the $(G_b,G_c)$ specialization of published Theorem 3.1, equation (3.3), printed p.1705, where $G_n(s)=1-(1-1/s{)}^n$. Its entries are ${\lambda }_n+{\lambda }_{-m}-{\lambda }_{n-m}$; zeta reflection gives ${\lambda }_{-m}={\lambda }_m$. This existing joint identity is used directly.

This is a paper application of the classical Li zero sum, the functional equation and the unit-circle Gram identity already used in the generalized Li interface. The explicit fourth-moment allowance is reused from VH5–VH6. No corrected version of the sign-sensitive contour formula (6.2) is assumed. The application estimates the actual tail’s possible negative contribution; it supplies no sign for the complete Li matrix, Robin or RH. It has no new Lean certification or literature-priority claim.

Let $a,b$ be positive integers and $c=a+b$. For an integer $H\ge 5$ put

$${\lambda }_n^{\mathrm{head}}(H)=\sum _{|\Im \rho |\le H}{m}_{\rho }\left[1-{\left(1-\frac{1}{\rho }\right)}^n\right],{\lambda }_n^{\mathrm{tail}}(H)={\lambda }_n-{\lambda }_n^{\mathrm{head}}(H).$$

The sum is over distinct nontrivial zeros, with their full analytic multiplicities. Zeros at height exactly $H$ stay in the head. Use the original linear matrix map

$$\mathcal{M}(x_a,x_b,x_c)=\left(\begin{array}{cc}2x_b& x_b+x_c-x_a\\ x_b+x_c-x_a& 2x_c\end{array}\right),K_{a,b}^{\mathrm{tail}}(H)=\mathcal{M}({\lambda }_a^{\mathrm{tail}},{\lambda }_b^{\mathrm{tail}},{\lambda }_c^{\mathrm{tail}}).$$

Write $S_4^{+}(H)={\sum }_{\Im \rho >H}{m}_{\rho }/(\Im \rho {)}^4$. The existing allowance, for $L_H=\lfloor {\log }_2(H+4)\rfloor +1$, is

$$\begin{array}{rl}{\mathcal{R}}_4(H)& =\frac{256}{H}\left[\left(\frac{2}{H^3}+\frac{1}{2H^2}\right)L_H+\frac{2/H^2+1/H}{H+4}\right],\\ S_4^{+}(H)& \le \frac{1}{2}{\mathcal{R}}_4(H).\end{array}\text{\hspace{1em}(RT1)}$$

The factor $1/2$ uses conjugation: VH6 already counts both ordinate signs. Its supplier is finite_inverse_fourth_tail_le, applied to the actual ordinate and multiplicity with $A=128$, $T=H$, $L=L_H$, and zeroData_large_window_count, then to the supremum of finite subsets. No spectral-ball containment or newly verified zero height is required.

Keep each reflected pair’s common phase

For an upper zero $\rho =\beta +i\gamma$, $\gamma >H$, set

$$z_{\rho }=1-\frac{1}{\rho }=e^{d_{\rho }}{w}_{\rho },|w_{\rho }|=1,d_{\rho }=\log |z_{\rho }|.$$

The strip $0<\beta <1$ gives

$$|d_{\rho }|=\frac{1}{2}\left|\log \frac{{\gamma }^2+(1-\beta {)}^2}{{\gamma }^2+{\beta }^2}\right|\le \frac{1}{2{\gamma }^2}.$$

Indeed the logarithm is $1/{\gamma }^2$-Lipschitz on the two positive arguments and their distance is $|1-2\beta |\le 1$. The functional equation and conjugation preserve multiplicities and the upper tail under $\rho \mapsto 1-\overline{\rho }$. They give exactly

$$z_{1-\overline{\rho }}=\frac{1}{\overline{z_{\rho }}},d_{1-\overline{\rho }}=-d_{\rho },w_{1-\overline{\rho }}=w_{\rho }.$$

Define the comparison coefficients on these same upper zeros by

$$C_n(H)=2\sum _{\Im \rho >H}{m}_{\rho }(1-\Re w_{\rho }^n).$$

They converge for each fixed $n$: $|w_{\rho }-1|\le 2/\gamma$, $2(1-\Re w_{\rho }^n)=|1-w_{\rho }^n{|}^2\le 4n^2/{\gamma }^2$, and the classical reciprocal-square zero tail is summable. On every finite interval $H<\gamma \le Y$, average over the reflected permutation before taking $Y\to \infty$. The exact difference is

$${\Delta }_n(H):={\lambda }_n^{\mathrm{tail}}(H)-C_n(H)=-2\sum _{\Im \rho >H}{m}_{\rho }(\cosh (n{d}_{\rho })-1)\Re w_{\rho }^n.$$

Thus the first radial variation cancels on the actual reflected zero set. From $\cosh x-1\le x^2{e}^{|x|}/2$, for $1\le n\le c$,

$$|{\Delta }_n(H)|\le \frac{n^2}{4}{e}^{c/(2H^2)}{S}_4^{+}(H)\le \frac{n^2}{8}{e}^{c/(2H^2)}{\mathcal{R}}_4(H).\text{\hspace{1em}(RT2)}$$

The error series is absolutely convergent. This argument reorders no unpaired infinite zero sum and discards no off-line partner.

The original mixed entry survives the comparison

For each upper zero the matrix $\mathcal{M}(2(1-\Re w^a),2(1-\Re w^b),2(1-\Re w^c))$ is twice the real Gram matrix on $(1-w^b,1-w^c)$, since $c-b=a$. It is positive semidefinite. The convergent sum therefore gives $\mathcal{M}(C_a,C_b,C_c)⪰0$. For the error matrix $E=\mathcal{M}({\Delta }_a,{\Delta }_b,{\Delta }_c)$, RT2 and the maximum absolute row sum give

$$\Vert E{\Vert }_{\mathrm{op}}\le \frac{a^2+b^2+3c^2}{8}{e}^{c/(2H^2)}{\mathcal{R}}_4(H)\le \frac{c^2}{2}{e}^{c/(2H^2)}{\mathcal{R}}_4(H)=:{\eta }_c(H).$$

Consequently the complete actual high-zero matrix satisfies

$$\boxed{K_{a,b}^{\mathrm{tail}}(H)⪰-{\eta }_c(H)I_2.}\text{\hspace{1em}(RT3)}$$

The two scalar directions needed by CS29 also follow directly, with

$$\begin{array}{rl}{\lambda }_c^{\mathrm{tail}}(H)& \ge -\frac{c^2}{8}{e}^{c/(2H^2)}{\mathcal{R}}_4(H),\\ 2{\lambda }_a^{\mathrm{tail}}(H)+2{\lambda }_b^{\mathrm{tail}}(H)-{\lambda }_c^{\mathrm{tail}}(H)& \ge -\frac{2a^2+2b^2+c^2}{8}{e}^{c/(2H^2)}{\mathcal{R}}_4(H).\end{array}\text{\hspace{1em}(RT4)}$$

For the second inequality the comparison direction is nonnegative because it equals $(2,-1)\mathcal{M}(C_a,C_b,C_c)(2,-1{)}^{\mathsf{T}}/2$. The same zeros, phases and cutoff occur in both directions.

A sublinear negative allowance, with the unknown head retained

For $c\ge 25$ take the single symbolic choice $H_c=\lceil \sqrt{c}\rceil$. Since ${\mathcal{R}}_4(H)=O(\log H/H^3)$ and $c/H_c^2\le 1$,

$${\eta }_c(H_c)=O(\sqrt{c}\log c)=o(c),K_{a,b}⪰K_{a,b}^{\mathrm{head}}(H_c)-{\eta }_c(H_c)I_2.\text{\hspace{1em}(RT5)}$$

No zero computation, height scan or change to an arithmetic cutoff is used. At adjacent Fibonacci indices, CS28 already gives the original actual arithmetic middle, with the complete $h^{\ast }=H+U-B$, all prime powers and its original endpoints. If its stated allowance is ${ϵ}_{\mathrm{mid}}(c)=o(c)$, then

$$\boxed{K_j^{\mathrm{mid}}⪰K_j^{\mathrm{head}}(H_c)-[{\eta }_c(H_c)+{ϵ}_{\mathrm{mid}}(c)]I_2.}\text{\hspace{1em}(RT6)}$$

RT3 is a lower bound, not an operator-norm bound on the tail itself. The head in RT5–RT6 contains every actual zero through the growing height $H_c$, including unverified off-line zeros and their multiplicities; no positivity or polynomial lower bound for that head has been acquired. With a fixed verified height instead, the allowance retains its factor $e^{c/(2H^2)}$ and supplies no cofinal polynomial budget. The original head probes, CS29, Robin and RH remain unresolved.

The actual high-zero Fibonacci block carries a positive budget

Keep RT1–RT6’s actual zeta zero sums, their original two-coordinate matrix $\mathcal{M}$, all multiplicities and the inclusive head convention. Let $a\le b$ be adjacent positive Fibonacci numbers and $c=a+b\ge 1024$. Use the single cutoff $H=c$: the head contains $|\Im \rho |\le c$ and the tail contains $|\Im \rho |>c$. Put $$S_4^{+}(c)=\sum _{\Im \rho >c}\frac{m_{\rho }}{(\Im \rho {)}^4}.$$ This is the same convergent positive zero tail used in RT1. The added estimate is an actual positive lower bound for this complete tail, rather than another estimate for the unitary-comparison error: $$\boxed{K_{a,b}^{\mathrm{tail}}(c)⪰\frac{c^4}{2^{19}}{S}_4^{+}(c)I_2\succ 0.}\text{\hspace{1em}(PTL1)}$$ It uses the existing reflected-pair identity and unit-circle Gram matrix. The classical zero-counting asymptotic below is reused, not reproved. This is a paper application without new Lean certification or a literature-priority claim; it supplies no sign for the unknown head.

Retain a uniformly positive Gram determinant

For the SAME upper zero $\rho =\beta +i\gamma$, $\gamma >c$, write $1-1/\rho =e^{d_{\rho }+i{\theta }_{\rho }}$ as in RT2. The open critical strip gives $${\theta }_{\rho }=\arctan \frac{\beta }{\gamma }+\arctan \frac{1-\beta }{\gamma },\frac{1}{2\gamma }\le {\theta }_{\rho }\le \frac{1}{\gamma },|d_{\rho }|\le \frac{1}{2{\gamma }^2}.\text{\hspace{1em}(PTL2)}$$ The angle bounds use $x/(1+x^2)\le \arctan x\le x$ for $x\ge 0$. In particular $0<c{\theta }_{\rho }<1$.

Set $A=a{\theta }_{\rho }/2$, $B=b{\theta }_{\rho }/2$ and $C=c{\theta }_{\rho }/2$, so $C-B=A$. The unit-circle comparison matrix for the coefficient triple $2(1-\cos (n{\theta }_{\rho }))$ is exactly $$G_{\rho }=8\left(\begin{array}{cc}{\sin }^{2}B& \sin B\sin C\cos A\\ \sin B\sin C\cos A& {\sin }^{2}C\end{array}\right).$$ All three angles lie in $(0,1/2)$. Thus $$\det G_{\rho }=64{\sin }^{2}A{\sin }^{2}B{\sin }^{2}C>0,\mathrm{tr}{G}_{\rho }=8({\sin }^{2}B+{\sin }^{2}C).$$ Using $\sin x\ge x/2$ on $[0,1/2]$, $\sin x\le x$, and ${\lambda }_{\min }\ge \det /\mathrm{tr}$ for a positive $2$-by-$2$ matrix gives $${\lambda }_{\min }(G_{\rho })\ge \frac{a^2{b}^2{c}^2{\theta }_{\rho }^4}{128(b^2+c^2)}\ge \frac{a^2{b}^2{c}^2}{2048(b^2+c^2){\gamma }^4}\ge \frac{c^4}{147456{\gamma }^4}\ge \frac{c^4}{2^{18}{\gamma }^4}.\text{\hspace{1em}(PTL3)}$$ Here adjacency gives $a\ge c/3$, $b\ge c/2$ and $b^2+c^2\le 2c^2$. No phase independence or verification that this zero lies on the critical line is assumed.

Pay the radial error on the same reflected pair

Reflection of the upper zero, $\rho \mapsto 1-\bar{\rho }$, preserves the ordinate and multiplicity, preserves ${\theta }_{\rho }$ and changes $d_{\rho }$ to $-d_{\rho }$. RT2 therefore represents the complete tail by summing the matrices from the reflected-average coefficient triple $$T_n(\rho )=2(1-\cosh (n{d}_{\rho })\cos (n{\theta }_{\rho })),n=a,b,c.$$ This is an equality AFTER reflection averaging over the actual upper-zero set; it does not replace each unaveraged zero response by an independently chosen phase.

For $1\le n\le c$, the existing radial-error estimate gives $$|T_n(\rho )-2(1-\cos (n{\theta }_{\rho }))|\le n^2{d}_{\rho }^2{e}^{c|d_{\rho }|}\le \frac{n^2}{4{\gamma }^4}{e}^{c/(2{\gamma }^2)}.$$ Write $E_{\rho }=\mathcal{M}(T_a,T_b,T_c)-G_{\rho }$. The original matrix map and its maximum absolute row sum give $$\Vert E_{\rho }{\Vert }_{\mathrm{op}}\le \frac{a^2+b^2+3c^2}{4{\gamma }^4}{e}^{c/(2{\gamma }^2)}\le \frac{2c^2}{{\gamma }^4}\le \frac{c^4}{2^{19}{\gamma }^4}.$$ The last inequality uses $c\ge 1024$; the exponential is at most two because $\gamma >c$. Combining this with PTL3 yields $$\mathcal{M}(T_a(\rho ),T_b(\rho ),T_c(\rho ))⪰\frac{c^4}{2^{19}{\gamma }^4}{I}_2.\text{\hspace{1em}(PTL4)}$$ Sum first over $c<\gamma \le Y$, where the reflected permutation is exact, and then let $Y\to \infty$ using RT2–RT3’s convergence. All upper-zero multiplicities remain in this sum. This proves PTL1. Strict positivity uses the classical infinitude of zeta zeros and $S_4^{+}(c)>0$.

Use the positive tail on the original arithmetic middle

Let $N_{+}(T)$ count the actual upper zeros with $0<\Im \rho \le T$, with analytic multiplicity. The same tail and its contained annulus give $$\frac{c^4}{2^{19}}{S}_4^{+}(c)\ge \frac{N_{+}(2c)-N_{+}(c)}{2^{23}}.$$ Reuse Lagarias, arXiv:math/0404394v4, Theorem 2.1(4), equations (2.29)–(2.30), specialized to the trivial representation ($N=1$, conductor $Q=1$). This is the unconditional Riemann–von Mangoldt formula in the upper-zero convention: $$N_{+}(T)=\frac{T}{2\pi }\log \frac{T}{2\pi }-\frac{T}{2\pi }+O(\log T).$$ It follows that $$\frac{N_{+}(2c)-N_{+}(c)}{2^{23}}=\left(\frac{1}{2^{24}\pi }+o(1)\right)c\log c.\text{\hspace{1em}(PTL5)}$$ The source’s strict upper endpoint can be changed to this inclusive convention: its formula bounds a jump’s multiplicity by $O(\log T)$, so the same error term remains valid. Zeros at $c$ remain in the head; zeros at $2c$ stay in this annulus. No ordinate scan, zero certification or RH premise is used.

For the SAME $K_j^{\mathrm{mid}}$ and original arithmetic endpoints in CS28, retain its actual common-matrix approximation allowance ${ϵ}_{\mathrm{mid}}(c)=o(c)$. PTL1 gives $$\boxed{K_j^{\mathrm{mid}}⪰K_j^{\mathrm{head}}(c)+\left[\frac{c^4}{2^{19}}{S}_4^{+}(c)-{ϵ}_{\mathrm{mid}}(c)\right]I_2.}\text{\hspace{1em}(PTL6)}$$ Consequently every fixed $\kappa <1/(2^{24}\pi )$ gives $K_j^{\mathrm{mid}}⪰K_j^{\mathrm{head}}(c)+\kappa c\log c{I}_2$ for all sufficiently large adjacent Fibonacci triples.

The head at this cutoff contains all actual zeros through height $c$, including any unverified off-line zeros and their multiplicities. Its lower bound remains unknown. PTL6 would pay a negative head of size smaller than the displayed positive budget, but does not establish such a head bound. Compared with RT5–RT6’s head at $\lceil \sqrt{c}\rceil$, this is a different decomposition of the SAME full matrix; no order comparison of the two heads is asserted. No positivity of the complete Li matrix, full retained Schur minimum, all-integer Robin, endpoint half-variance target or RH follows.

Reuse the joint zero sum to reduce the unpaid Li head

Keep the actual Fibonacci matrix, all zero multiplicities and the inclusive-head convention of RT1–RT6 and PTL1–PTL6. Put $$L=\log c,{\Omega }_c=(L/\log L{)}^{1/3},h_c=\lceil \sqrt{\frac{c}{2L}}\rceil ,R=57.54,\nu =\frac{1}{20R}.$$ Work for sufficiently large adjacent positive Fibonacci $a\le b$, $a+b=c$, so $h_c\ge 3\cdot 10^{12}$ and $h_c<c$. No numerical starting value of $c$ is certified.

For every integer cutoff $h_c\le H\le c$, the complete actual tail has $$\boxed{K_{a,b}^{\mathrm{tail}}(H)⪰[{\mu }_c-{\mathcal{E}}_c]I_2,{\mu }_c=\frac{c^4}{2^{19}}{S}_4^{+}(c),{\mathcal{E}}_c\ll c{e}^{-\nu {\Omega }_c}=o(c).}\text{\hspace{1em}(DHR1)}$$ Thus every fixed $0<\kappa <1/(2^{24}\pi )$ gives $K_{a,b}^{\mathrm{tail}}(H)⪰\kappa c\log c{I}_2$ eventually, uniformly over these cutoffs. The smaller exact head still grows without bound. This is a paper application of an owned zero-sum bound, without new Lean certification or a literature-priority claim.

Use the existing weighted sum at an auxiliary parameter

The joint density and zero-free application already bounds the intermediate positive quantity $$B_2(A;T)=\sum _{\gamma >T}\frac{m_{\rho }(A^{\beta -1/2}+A^{1/2-\beta })}{{\gamma }^2}.$$ For $\ell =\log A$, $\Omega (A)=(\ell /\log \ell {)}^{1/3}$ and $T_A=A^{1/4}{e}^{-\Omega (A)/(28R)}$, its bound is $$B_2(A;T_A)\ll \ell A^{-1/8}{e}^{\Omega (A)/(28R)}+{\ell }^3{e}^{-\Omega (A)/(14R)}+{\ell }^3{A}^{-1/14}.\text{\hspace{1em}(DHR2)}$$ This intermediate estimate uses the positive real parameter $A$ and the actual upper-zero multiset; no source-selection property of $A$ enters its proof. Here $A_c=c^2$ is an auxiliary parameter.

The uniform explicit Ingham input and Ford zero-free support in that application are reused as stated there. Their analytic proofs and finite verified-height input remain external premises; global RH is not a premise. No density-layer proof or zero computation is repeated for the Li consumer.

Dominate the Li radial charge pointwise

For the SAME upper zero put ${\delta }_{\rho }=|\beta -1/2|$. RT2’s logarithmic calculation gives $|d_{\rho }|\le {\delta }_{\rho }/{\gamma }^2$. After its multiplicity-preserving reflection average, the index $n\le c$ error is at most $n^2{d}_{\rho }^2{e}^{c|d_{\rho }|}$. The original mixed entry and its row sum therefore give the nonnegative allowance $${\mathcal{E}}_c:=4c^2\sum _{\gamma >h_c}\frac{m_{\rho }{\delta }_{\rho }^2}{{\gamma }^4}{e}^{c{\delta }_{\rho }/{\gamma }^2}.\text{\hspace{1em}(DHR3)}$$ It bounds the unit-phase comparison error on every sub-band of this tail. Critical-line zeros have zero charge, and the fourth-moment supplier in RT1 makes the sum finite for fixed $c$.

For $\gamma >h_c$, one has $c/{\gamma }^2\le c/h_c^2\le 2L$, $4{\delta }_{\rho }^2\le 1$, and $$e^{c{\delta }_{\rho }/{\gamma }^2}\le (c^2{)}^{{\delta }_{\rho }}\le (c^2{)}^{\beta -1/2}+(c^2{)}^{1/2-\beta }.$$ Consequently, on the SAME zeros and heights, $$\boxed{{\mathcal{E}}_c\le \frac{c^2}{h_c^2}{B}_2(c^2;h_c)\le 2cL{B}_2(c^2;T_{c^2}).}\text{\hspace{1em}(DHR4)}$$ The last inequality uses positivity of the summands and the eventual comparison $$\frac{h_c}{T_{c^2}}\ge \frac{e^{\Omega (c^2)/(28R)}}{\sqrt{2L}}⟶\infty .$$ This is an upper-bound comparison for one common positive zero sum. All reflected partners and analytic multiplicities remain.

Substituting $A=c^2$, $\ell =2L$ in DHR2 yields $$\begin{array}{rl}{\mathcal{E}}_c\ll & c^{3/4}{L}^2{e}^{\Omega (c^2)/(28R)}+c{L}^4{e}^{-\Omega (c^2)/(14R)}+c^{6/7}{L}^4\\ \ll & c{e}^{-{\Omega }_c/(20R)}=o(c).\end{array}\text{\hspace{1em}(DHR5)}$$ Here $\Omega (c^2)/{\Omega }_c\to 2^{1/3}$, ${\Omega }_c/\log L\to \infty$ and ${\Omega }_c=o(L)$. The first and third terms have a fixed power saving in $c$; the middle term absorbs the displayed logarithms. No effective starting clock or new density theorem is claimed.

Keep the positive budget with the smaller original head

For each $h_c\le H\le c$, use the exact SAME-source decomposition $$K_{a,b}^{\mathrm{tail}}(H)=K_{a,b}^{\mathrm{tail}}(c)+G_{H,c}+E_{H,c},G_{H,c}⪰0,\Vert E_{H,c}{\Vert }_{\mathrm{op}}\le {\mathcal{E}}_c.\text{\hspace{1em}(DHR6)}$$ Here $G_{H,c}$ is RT3’s unit-phase Gram sum on $H<\gamma \le c$. Reflection preserves this finite band exactly. PTL1 pays the complete tail at $c$, and DHR5 pays the whole finite-band error. This proves DHR1 without subtracting separate lower bounds for two tails or assuming an order on their heads.

At $H=h_c$, retain CS28’s original arithmetic-middle allowance ${ϵ}_{\mathrm{mid}}(c)=o(c)$. The original matrix obeys $$\boxed{K_j^{\mathrm{mid}}⪰K_j^{\mathrm{head}}(h_c)+[{\mu }_c-{\mathcal{E}}_c-{ϵ}_{\mathrm{mid}}(c)]I_2.}\text{\hspace{1em}(DHR7)}$$ The same eventual $\kappa c\log c$ budget is therefore available with $h_c\sim \sqrt{c/(2\log c)}$ and unchanged arithmetic endpoints.

No lower bound for $K_j^{\mathrm{head}}(h_c)$ has been acquired. It contains every actual zero through $h_c$, including any unverified off-line zeros; finite verification does not verify this unbounded head. CS29’s original weak signed probes, full Li/Schur positivity, the same-source Robin strict upper, strict Robin for every $n>5040$, and RH remain unresolved.

The complete descended tail has the classical Li main term

Keep DHR1–DHR7’s actual zeros, multiplicities, inclusive head and $h_c=\lceil \sqrt{c/(2\log c)}\rceil$. The radial allowance remains ${\mathcal{E}}_c\ll c{e}^{-\nu {\Omega }_c}=o(c)$ with the same external premises and no effective starting $c$. Let $H_n$ be the already-used archimedean term from CS4: $$H_n=\frac{n}{2}(\log n+q)+O(1),q={\gamma }_E-1-\log (2\pi ).$$ Here ${\gamma }_E$ is Euler’s constant, not a zero ordinate.

The complete tail satisfies the two-sided uniform application $$\boxed{\underset{1\le n\le c}{\max }|{\lambda }_n^{\mathrm{tail}}(h_c)-H_n|=o(c).}\text{\hspace{1em}(DLM1)}$$ This uses an existing phase-only asymptotic and the owned actual radial estimate. It does not use the sign-sensitive contour formula discussed above or identify the auxiliary phase-only sequence with the original Li coefficients.

Reuse the ordinate-sequence theorem

Use Juan Arias de Reyna, Asymptotics of Keiper-Li coefficients, DOI 10.7169/facm/1317045228, Theorem 1, Section 3, pp.4–5 of the author version dated 31 January 2011. The theorem accepts any nondecreasing positive sequence, including repetitions, with counting law $$N(t)=\frac{t}{2\pi }\log \frac{t}{2\pi }-\frac{t}{2\pi }+O(\log t).$$ List the ordinates of ALL actual upper zeros, repeating each according to its analytic multiplicity. PTL5’s existing unconditional count supplies this hypothesis. Put $${\vartheta }_{\gamma }=2\arctan \frac{1}{2\gamma },C_n^{\circ }=2\sum _{\gamma >0}{m}_{\rho }[1-\cos (n{\vartheta }_{\gamma })].$$ The source’s $J_n=(4/n){\sum }_j{\sin }^2(n\arctan (1/(2{\gamma }_j)))$ is exactly $C_n^{\circ }/n$. Thus its general theorem supplies one fixed sequence $(y_n)\in {\ell }^2$ with $$C_n^{\circ }=\frac{n}{2}(\log n+q)+n{y}_n.\text{\hspace{1em}(DLM2)}$$ We use this general ordinate-sequence theorem, not the paper’s RH-conditional identification with actual Keiper coefficients in Corollary 2. The theorem and its proof are reused as an external paper-level premise; neither its counting-integral proof nor its Mellin calculation is repeated here.

Connect the actual phase and the smaller cutoff

For the SAME actual upper zero put ${\delta }_{\rho }=|\beta -1/2|$ and $w_{\rho }=e^{i{\theta }_{\rho }}$ as in RT2. For $\gamma >h_c\ge 5$, $${\theta }_{\rho }=\arctan \frac{\gamma }{{\gamma }^2-\beta (1-\beta )},0\le {\vartheta }_{\gamma }-{\theta }_{\rho }\le \frac{{\delta }_{\rho }^2}{{\gamma }^3}.\text{\hspace{1em}(DLM3)}$$ Indeed ${\vartheta }_{\gamma }$ is the same arctangent at $v=1/4$, whereas ${\theta }_{\rho }$ uses $v=\beta (1-\beta )=1/4-{\delta }_{\rho }^2$. On $0\le v\le 1/4$, $$\frac{d}{dv}\arctan \frac{\gamma }{{\gamma }^2-v}=\frac{\gamma }{({\gamma }^2-v{)}^2+{\gamma }^2}\le {\gamma }^{-3}.$$ The actual count also gives ${\sum }_{\gamma >h}{m}_{\rho }{\gamma }^{-3}=O(\log h/h^2)$. Comparing the two tail phases therefore costs $O(c\log h_c/h_c^2)$ uniformly for $n\le c$. The deleted part of $C_n^{\circ }$ is at most $4N_{+}(h_c)$. DHR3 bounds the actual radial error by ${\mathcal{E}}_c/4$ per coefficient. Together with the owned $H_n$ asymptotic, these give $$\underset{1\le n\le c}{\max }|{\lambda }_n^{\mathrm{tail}}(h_c)-H_n-n{y}_n|\ll {\mathcal{E}}_c+h_c\log h_c+c\log h_c/h_c^2+1=o(c).\text{\hspace{1em}(DLM4)}$$ All phase comparisons retain the actual ordinates and multiplicities; only the auxiliary sequence has its phases projected to the circle. Every zero at the cutoff stays in the actual head.