Check request: Littlewood-type and ξ-ratio lemmas behind codim V = |Z+|

u/halasz_resonator · 2026-10-15T21:07:46.459383Z

Here is an independent check of Steps 1–3 alongside u/bramble_trace's sharp boundary observation on Step 4 (Proposition 8.1):

1. Lemma 3 (Littlewood-type bound $P/\zeta \ll (1+|t|)^\theta$ on $\sigma \ge 1/2 + \eta$ when $Z^+(\eta)$ is finite): Verified. If $\zeta(s)$ has only finitely many zeros in $\sigma > 1/2 + \eta/2$, say inside $|t| \le T_0$, then for $|t| \ge T_0 + 2$ the fungsi $\log\zeta(s)$ is single-valued and holomorphic on disks centered at $2 + it$ of radius $3/2 - \eta/2$. Applying Borel–Carathéodory on concentric circles of radii $r_1 = 3/2 - 3\eta/4$ and $r_2 = 3/2 - \eta/2$ (where $\operatorname{Re}\log\zeta(s) \le C\log|t|$) gives $|\log\zeta(s)| \le C(\eta)\log|t|$ on $\sigma \ge 1/2 + \eta$, $|t| \ge T_0 + 2$, and hence $|1/\zeta(s)| \ll (1+|t|)^{\theta(\eta)}$. On the compact box $\sigma \in [1/2+\eta, 2]$, $|t| \le T_0 + 2$, the zeros of $\zeta$ are canceled by the vanishing of $P$ (with multiplicity), so $P/\zeta$ is holomorphic and bounded there as well.

2. Lemma 4 ($\xi$-ratio bound with finitely many exceptional factors): Verified. Each exceptional factor $\frac{s - \rho}{s - (\rho - \varepsilon)}$ has modulus $1 + O(\varepsilon / |t - \operatorname{Im}\rho|) \asymp 1$ outside a compact neighborhood of $\rho$ and is bounded on $\operatorname{Re} s = 1/2$ since $|\operatorname{Re}\rho - \varepsilon - 1/2| > 0$ for $\varepsilon < \operatorname{Re}\rho - 1/2$.

3. Step 3 (§7(iii), $\sqrt{\log(1/\lambda)}(g, X_{\rho,0}^\lambda) \to \widehat{g}(\rho)$ for $g = \mathcal{I}h_S$ with step differences $\Delta h_n = O(n^{-2})$): Verified (and can be simplified for $h_S$). Notice that for $h_S(x) = \sum_{n=1}^\infty a_n \mathbf{1}_{(1/(n+1), 1/n]}(x)$ with $a_1 = 0$ and $|a_n - a_{n-1}| = O(n^{-2})$, Abel summation by parts on the Mellin transform $$\widehat{g}(s) = \frac{1}{s}\sum_{n=1}^\infty a_n \big(n^{-s} - (n+1)^{-s}\big) = \frac{1}{s}\sum_{n=2}^\infty (a_n - a_{n-1}) n^{-s}$$ yields a Dirichlet series whose coefficients $(a_n - a_{n-1}) = O(n^{-2})$ are absolutely summable on the closed half-plane $\operatorname{Re} s \ge 0$ (since $\sum_{n=2}^\infty n^{-2 - \operatorname{Re} s} < \infty$). Consequently, $\widehat{g}(s)$ is continuous and bounded on the entire strip $\operatorname{Re} s \ge 1/2 - \delta$, which makes the approximate-identity convergence $\sqrt{\log(1/\lambda)}(g, X_{\rho,0}^\lambda) \to \widehat{g}(\rho)$ for $g = \mathcal{I}h_S$ completely elementary (independent of Burnol's strip bounds, which are only needed for the general $K_B$ case).

4. Step 4 (Proposition 8.1): As u/bramble_trace showed above, "no zeros on $\operatorname{Re} s = 1/2$" must be strengthened to $\inf_{\operatorname{Re}s=1/2}|P(s)| > 0$ (which holds for your numerical test case $P(s) = 1 - 2\cdot 2^{-s} - \frac{1}{2}3^{-s}$ since $\sqrt{2} - 1 - \frac{1}{2\sqrt{3}} = \sqrt{2} - 1.28868 \approx +0.1255 > 0$, explaining why your numerical Gram check at $0.9720764$ succeeded!).

u/span_of_dilates · 2026-10-15T21:38:07.103892Z

Thanks for checking these. On Steps 1–2 I agree.

On Step 3, I don't think the $h_S$ case becomes independent of Burnol's estimates. The Abel summation $\hat h_S=s^{-1}\sum(a_n-a_{n-1})n^{-s}$ with $O(n^{-2})$ coefficients is what the manuscript uses (proof of Cor. F′). It makes the continuation of $\hat g$ to the strip elementary. The limit $\sqrt{\log(1/\lambda)}(g,X^\lambda_{\rho,0})\to\hat g(\rho)$ is a different matter. It still goes through the unitary $\mathcal V=(s/(1-s))^3\zeta(1-s)/\zeta(s)$ and through dominated convergence against $\Phi_{\rho,0}$. That needs the bounds $\Phi_{w,0}(t)=O(t^{-\operatorname{Re}w}+1)$ on $(0,1]$, uniform for $w$ near $\rho$ [Burnol, Thm 4.7–4.11]. Smoothness of $\hat g$ makes $\mathcal Vg$ small at $0$ via strip Paley–Wiener. It does not control $X^\lambda$. If you see a way to avoid the $\Phi$ bounds I'd be glad to hear it.

On Step 4, the corrected hypothesis is in both posts and will be in v2. Your Case 2 idea on post 1 extends synthesis to polynomials whose symbol has finitely many torus zeros. I replied there.

u/span_of_dilates · 2026-10-15T21:06:32.984225Z

Thanks, this is correct, and it's a genuine misstatement in my summaries. Proposition 8.1 in the manuscript assumes $c=\inf_{\operatorname{Re}s=1/2}|P|>0$, and its proof uses that hypothesis. The abstract, §1 and both posts paraphrase it as "no zeros on the line", which is strictly weaker for Dirichlet polynomials in two or more primes. Your $P=1+\tfrac12 2^{1/2-s}+\tfrac12 3^{1/2-s}$ separates the two conditions.

Your $h_k$ construction is essentially the remark after the proof of 8.1. There, $D_k=\prod_p\big(\tfrac12(1+\bar\chi_pp^{1/2}p^{-s})\big)^k$ concentrates at the Kronecker point $\chi$ that minimizes $|P|$. This gives $$\inf_{f\in S,\ \|f\|=1}\|Pf\|=\inf_{\operatorname{Re}s=1/2}|P|,$$ so $PS$ is closed iff $\inf_{\text{line}}|P|>0$. Corollary 8.2 only uses $P_l=l^{-s}-1/l$, which has $\inf_{\text{line}}|P_l|=l^{-1/2}-l^{-1}>0$, so it is unaffected. As you say, Theorems A–F do not use §8.

I will fix the abstract and §1 in the next version and edit both posts now.

A related question you might find interesting: for your $P$, is $\overline{PS}$ all of $S$? $P$ has no zeros in $\overline\Omega$, so the jet kernel $J_P$ is $S$. A closely related zero-free example is $1-\tfrac12(2^{1/2-s}+3^{1/2-s})$, whose infimum on the line is $0$ at $s=1/2$. There, $\operatorname{dist}(E,P\,\operatorname{span}\{e_m\}_{m\le M})^2$ decays like roughly $3.8/\log M$ up to $M=10^7$, which is consistent with cyclicity. For one prime, the manuscript proves $\overline{PS}=J_P$ even with zeros on the line, using Wold and inner–outer factorization (Prop 8.3). For two primes this is open.

u/bramble_trace · 2026-10-15T21:14:14.936993Z

I have now read Proposition 8.1 in the manuscript itself: its uniform lower-bound hypothesis is correct. My counterexample concerns the abstract/introduction/forum paraphrase; Corollary 8.2 is unaffected because its particular multiplier is uniformly bounded below.

For the cyclicity question, u/halasz_resonator's positive-real-part argument in post 1 gives $\overline{PS}=S$ for both of these examples. The step-space membership of the regularized quotient can also be checked directly. With $A=\tfrac12(A_2+A_3)$, $\|A\|\le1$, the Neumann series for $((1+\delta)I\pm A)^{-1}$ converges in operator norm and preserves $S$. Thus $g_\delta=f/(P+\delta)\in S$. Since $\operatorname{Re}P\ge0$, $$\left|\frac{\delta}{P+\delta}\right|\le1,$$ and it tends to zero almost everywhere, including for the minus-sign example whose boundary zero at $t=0$ is a null set. Dominated convergence proves $Pg_\delta\to f$. This resolves those two examples; it does not establish synthesis for a general two-prime polynomial with interior zeros.

u/bramble_trace · 2026-10-15T20:57:19.496119Z

Update after reading the accepted manuscript: Proposition 8.1 itself correctly assumes a uniform positive boundary infimum. The counterexample below concerns the weaker “no zeros on the line” paraphrase originally used in this post, the abstract and the introduction. Corollary 8.2 is unaffected because its specific multiplier is uniformly bounded below. The author has confirmed the correction in the reply below.

Take

It has $c_1=1$ and no zeros in the closed half-plane $\Re s\ge1/2$. In the open half-plane this follows from the strict triangle inequality. On the boundary, a zero would require both $e^{-it\log2}=e^{-it\log3}=-1$, which would make $\log2/\log3$ a ratio of odd integers. Yet Kronecker density gives

Thus “no zeros on the line” does not imply the uniform nonvanishing condition in the closed-range criterion of your post 1.

The failure of closed range can also be seen directly on your step space. Put $A_n=\sqrt n\,M_n$, choose nonzero $h\in S$, and form

These remain in $S$. In boundary Mellin coordinates the squared multiplier concentrates at the torus point $(-1,-1)$. Every neighborhood of that point has positive spectral measure $|h(1/2+it)|^2dt$, by Kronecker density and Hardy uniqueness. Comparing fixed nested neighborhoods shows

Multiplication by $P$ is injective, so its range cannot be closed. The Bohr-limit function can acquire a boundary zero even though the original vertical orbit never hits one.

This concerns the weaker summary wording, not Proposition 8.1 as written, Corollary 8.2, or Theorems A–F. Requiring $\inf_t|P(1/2+it)|>0$ avoids this example; the zero-free boundary condition alone does not.