bibkey: freitas2006li
authors: Pedro Freitas
year: 2006
title: A Li-type criterion for zero-free half-planes of Riemann's zeta function
doi: 10.1112/S0024610706022599
url: https://arxiv.org/abs/math/0507368v1
claim: Freitas supplies the normalized generalized Li coefficients and their generating-function interface; a uniform finite signed transfer approximates the actual critical Fibonacci matrix from the unconditional tau equal to two data, without transferring positivity.
strata_touched: []
license: citation-only
triage: anchor

Generalized Li data and a signed return to the critical matrix

The primary is Pedro Freitas, arXiv:math/0507368v1, version 18 July 2005, preceding the publication in Journal of the London Mathematical Society 73(2), 399–414 (2006). The locators below refer to this preprint; the DOI supplies publication metadata. The source’s criterion, generating-function interface and differential system are existing results, not new parts of the Fibonacci model. The application below controls approximation error, without a new bound on the critical signed probes, Lean certification or a priority claim.

TL1. Exact normalization and an existing analytic interface

Freitas’s definition (1.3) and Lemma 3.1 use

$${\alpha }_n(\tau )=\frac{1}{\tau }{\Lambda }_n(\tau ),{\Lambda }_n(\tau )=\underset{T\to \infty }{\lim }\sum _{\begin{array}{c}\rho \in Z(\zeta )\\ |\Im \rho |\le T\end{array}}\left[1-{\left(\frac{\rho }{\rho -\tau }\right)}^n\right].$$

Symmetric zero pairing agrees with the real conjugate-paired limit. Functional symmetry gives ${\Lambda }_n(1)={\lambda }_n$, the original classical coefficient. Theorem 1 gives the zero-free half-plane criterion and, unconditionally, ${\alpha }_n(\tau )\ge 0$ for $\tau \ge 2$. Theorem 4.3, equation (4.1), already gives the differential system; in the unnormalized convention it is $\tau {\partial }_{\tau }{\Lambda }_n=n({\Lambda }_{n+1}-{\Lambda }_n)$. These results are reused, not re-proved.

Choose compatible local logarithms and write $L_{\tau }(z)=\log \xi (\tau /(1-z))$. The exact parameter application of Lemma 3.1(ii) is

$$L_{\tau }(z)-L_{\tau }(0)=\sum _{n\ge 1}\frac{{\Lambda }_n(\tau )}{n}{z}^n.\text{\hspace{1em}(TL1)}$$

The conventional constant factor in $\xi$ changes only the constant term. In particular $L_2$ has a holomorphic logarithm on $|z|<1$, since $\Re (2/(1-z))>1$ there. Use the actual nonnegative coefficients ${\Lambda }_m(2)$ and the published bound ${\Lambda }_m(2)=O(m\log (m+1))$ supplied by Palojärvi’s Theorem 2.1.

TL2. A signed boundary re-expansion

For $0<r<1$ put $v_r=2r/(1+r)$ and ${\tau }_r=2/(1+r)$. Near $z=0$,

$$L_2(r(2z-1))=L_{{\tau }_r}(v_{r}z).$$

The left-hand Taylor series converges absolutely when $r(1+2|z|)<1$. Taking the coefficient of $z^n/n$ in TL1 gives the exact identity

$$S_n(r):=\sum _{m\ge n}{2}^n(-1{)}^{m-n}(\genfrac{}{}{0ex}{}{m-1}{n-1})r^m{\Lambda }_m(2)=v_r^n{\Lambda }_n({\tau }_r).\text{\hspace{1em}(TL2)}$$

The polynomial bound on ${\Lambda }_m(2)$ also proves absolute convergence for each fixed $r<1$ and $n$. The limit $S_n(r)\to {\lambda }_n$ for fixed $n$ as $r\uparrow 1$ is a boundary re-expansion, not a positivity transport or a uniform estimate for growing $n$.

TL3. One common finite transfer with uniform error

For each integer $c\ge 1$ prescribe

$$q_c=e^{-c^2},r_c=1-q_c,M_c=\lceil e^{2c^2}\rceil ,$$

and use the same actual data, regularization and integer cutoff for all $1\le n\le c$:

$$A_n(c)=\sum _{m=n}^{M_c}{2}^n(-1{)}^{m-n}(\genfrac{}{}{0ex}{}{m-1}{n-1})r_c^m{\Lambda }_m(2).\text{\hspace{1em}(TL3)}$$

There are constants $C,c_0$ independent of $n,c$ such that

$$\boxed{\underset{1\le n\le c}{\max }|A_n(c)-{\lambda }_n|\le C{e}^{-c^2/2},c\ge c_0.}\text{\hspace{1em}(TL4)}$$

Here is the uniform parameter argument. Since $\xi (1)\ne 0$, one fixed zero-free complex neighborhood of $1$ supplies $\eta >0$, $0<R_0<1$ and bounded compatible logarithms for $1\le \tau \le 1+\eta$, $|z|\le R_0$. Cauchy estimates for TL1 and its $\tau$ derivative give

$$|{\Lambda }_n(\tau )|\le B_{0}n{R}_0^{-n},|{\partial }_{\tau }{\Lambda }_n(\tau )|\le B_{1}n{R}_0^{-n}.$$

As ${\tau }_{r_c}-1=1-v_{r_c}=q_c/(2-q_c)\le q_c$, all large $c$ have

$$|S_n(r_c)-{\lambda }_n|\le q_c(B_{1}n+B_0{n}^2)R_0^{-n}\le C_1{c}^2{R}_0^{-c}{e}^{-c^2}\le C_2{e}^{-c^2/2},1\le n\le c.$$

The zero-free neighborhood is local at the nonzero point $1$; no zero-free critical strip or RH premise has been introduced.

For the omitted series, the published growth supplier implies ${\Lambda }_m(2)\le C_3{m}^2$. If $m>M_c$ and $1\le n\le c$, the absolute summand of TL2 is at most $C_3{2}^c{m}^{c+1}{e}^{-q_{c}m}$. Uniformly for $m\ge M_c$ and all sufficiently large $c$,

$$c\log 2+(c+1)\log m\le q_{c}m/2.$$

Indeed, dividing by $m$ and using decreasing $\log m/m$ bounds the left side divided by $m$ by $M_c^{-1}[c\log 2+(c+1)\log M_c]=O(c^3{e}^{-2c^2})=o(q_c)$. Thus the whole omitted series is at most

$$C_3\sum _{m>M_c}{e}^{-q_{c}m/2}\le 4C_3{q}_c^{-1}{e}^{-q_c{M}_c/2}\le 4C_3{e}^{c^2-e^{c^2}/2}\le C_4{e}^{-c^2/2}.$$

The denominator bound uses $1-e^{-q_c/2}\ge q_c/4$. Combining this with the regularization error proves TL4. This is a paper application of the existing analytic and polynomial suppliers. The constants are existential, the cutoff is large, and no efficient computation, numerical execution or cutoff optimization is asserted.

TL4. The same Fibonacci consumer and its unpaid sign

For the original adjacent Fibonacci $a,b,c=a+b$, let $K_j^A$ be the original real matrix with ${\lambda }_a,{\lambda }_b,{\lambda }_c$ replaced by $A_a(c),A_b(c),A_c(c)$. TL4 and the row-sum bound give

$$\boxed{\Vert K_j^A-K_j{\Vert }_{\mathrm{op}}\le 5C{e}^{-c^2/2}}\text{for every sufficiently large }j.\text{\hspace{1em}(TL5)}$$

These are the entire actual Li coefficients of the same $\xi$; this representation drops no arithmetic channel. It can be combined with the existing $o(c)$ transfer from the actual arithmetic middle. It does not bound either of the signed probes in CS29.

For comparison, the same-shaped matrix built from ${\Lambda }_b(2)$ and ${\Lambda }_c(2)$ is unconditionally positive semidefinite. For each $z_{\rho }=\rho /(\rho -2)$, the classical Poisson probability on the unit circle has cosine moments $\Re z_{\rho }^k$. Its real Gram matrix on $(1-e^{ib\theta },1-e^{ic\theta })$ has entries $2(1-\Re z_{\rho }^b)$, $2(1-\Re z_{\rho }^c)$ and $1-\Re z_{\rho }^b-\Re z_{\rho }^c+\Re z_{\rho }^{c-b}$. Finite symmetric zero sums are positive semidefinite; the convergent coefficient limit gives the asserted comparison matrix. This applies the classical Poisson identity, rather than inferring a matrix sign from scalar positivity.

The inversion in TL3 has signed weights. Set $w_{n,m}=2^n(-1{)}^{m-n}(\genfrac{}{}{0ex}{}{m-1}{n-1})r_c^m$ for $m\ge n$, and zero otherwise. The exact original matrix map gives $K_j^A={\sum }_{m=1}^{M_c}{\Lambda }_m(2)W_m$. Already at $m=a<b<c$ its coefficient matrix is

$$W_a=\left(\begin{array}{cc}0& -w_{a,a}\\ -w_{a,a}& 0\end{array}\right),w_{a,a}=2^a{r}_c^a>0.$$

It has eigenvalues $\pm w_{a,a}$. Hence this representation supplies no termwise positive semidefinite comparison. This does not assert that the total $K_j^A$ is nonpositive or that all joint information in the $\tau =2$ data is insufficient. The remaining task is a signed bound on the actual common sequence in TL3, strong enough for the original probes. The approximation bound supplies no such bound; Robin, endpoint half-variance and RH remain unresolved.