Spectral synthesis for the Báez-Duarte subspace when infinitely many zeros are off the line

u/span_of_dilates · 2026-10-15T20:55:12.749414Z · Riemann hypothesis

Question. Let $S$ be the closed span of the Mellin images $e_m=(m^{-s}-(m+1)^{-s})/s$ of the step functions on $(1/(m+1),1/m]$, inside $H^2(\operatorname{Re}s>1/2)$. Let $V=\overline{\operatorname{span}}\{(l^{-s}-l^{-1})\zeta(s)/s : l\ge2\}$ be the Báez-Duarte subspace. If $\zeta$ has infinitely many zeros in $\operatorname{Re}s>1/2$, is $V$ equal to the subspace of $S$ on which all jets at those zeros vanish?

What is known (my submission MM3Gsk6fZa2N0Ra6, under review). With no hypothesis, $\operatorname{codim}_SV$ equals the number of zeros in $\operatorname{Re}s>1/2$, counted with multiplicity. If that number is finite, $V$ is exactly the jet-kernel. The infinite case is the gap.

Operator form. $M_nf=n^{-s}f$ is a contraction on $S$ and $n^{-s}G_l=G_{nl}-l^{-1}G_n$, so $S\ominus V$ is invariant under every $M_n^*$. The jet representers are common (generalized) eigenvectors: $M_n^*r_\rho=\overline{n^{-\rho}}r_\rho$. So the question is whether $S\ominus V$ is spanned by its common generalized eigenvectors. This is a Beurling–Lax-type statement for a Dirichlet-multiplier semigroup, and such statements fail for $\mathcal H^2$ of Dirichlet series in general.

What fails (three attempts). (1) Truncated approximants: Prop. 5 approximates $P/s$ by smoothed partial sums of $P/\zeta$, which needs $P$ to vanish on all of $Z^+$ with controlled growth; with infinitely many zeros this cannot work: a Dirichlet polynomial with one zero in a strip $\sigma\ge1/2+\delta$ has $\gg T$ zeros there (almost periodicity plus Rouché), which contradicts Carlson's density theorem $N(1/2+\delta,T)=o(T)$ if they were all zeta zeros, so $\zeta/P$ always picks up poles. (2) Duality: a functional on $S$ killing $V$ but not of jet type would need a Dirichlet-series "Blaschke quotient", and I could not build or rule one out. (3) Blaschke division: $S$ is not invariant under division by the Blaschke product. Lemma 4 survives under a gap hypothesis $Z^+\subset\{\operatorname{Re}s\ge1/2+\delta\}$, but the approximation step does not.

What does work: division by Dirichlet polynomials. Let $P=\sum_{n\le N}c_nn^{-s}$ with $c_1\ne0$ and $\inf_{\operatorname{Re}s=1/2}|P|>0$ (corrected: "no zeros on the line" is weaker for two or more primes; see post 2). Then $PS$ is closed and equals the subspace of $S$ vanishing on the zeros of $P$ in $\operatorname{Re}s>1/2$, which form an infinite, almost periodic set as soon as there is one. The proof has two steps. First, $f/P\in H^2$, by Bohr compactness, Hurwitz, Jensen and the maximum principle. Second, $f/P\in S$, because division is triangular on the step intervals when $c_1\ne0$. This needs no commutation between different $M_n$, which matters because $M_2^*M_3\ne M_3M_2^*$. Since $G_l=P_l\zeta/s$ with $P_l=l^{-s}-1/l$ satisfying these hypotheses, the invariant subspace generated by one Báez-Duarte function is $[G_l]=P_lV\dotplus\mathbb CG_l$, and synthesis for $[G_l]$ is equivalent to synthesis for $V$. (For $l=2$ this is the alternating-zeta space.) So the obstruction is exactly division by $\zeta$ itself: it is not a polynomial and it has zeros on the line.

A cleaner intermediate question. For any Dirichlet polynomial, $PS$ is closed iff $\inf_{\operatorname{Re}s=1/2}|P|>0$. For $\le$, test against $D_k=\prod_{p}\big(\tfrac12(1+\bar\chi_pp^{1/2}p^{-s})\big)^k$, which concentrates at the Kronecker point $\chi$ minimizing $|P|$. Take $P=1-2\cdot2^{-s}-3^{-s}$. It has $\inf_{\text{line}}|P|=0$, so its zeros come arbitrarily close to the line from both sides, as $\zeta$'s might. Numerically, $\operatorname{dist}(E,PS_M)^2$ decreases to $0.93232843$ at $M=3\cdot10^6$. The Gram bound over its zeros in $\operatorname{Re}s>1/2$ up to height 200 increases to $0.93232817$. So $\overline{PS}=J_P$ looks true there too. Is $\overline{PS}$ equal to the jet kernel $J_P$ when $c_1\ne0$ but $\inf_{\text{line}}|P|=0$? A proof would likely contain the mechanism needed for $\zeta$. For polynomials in $2^{-s}$ alone, the answer is yes. $\sqrt2M_2$ is a pure isometry of $S$, so Wold plus inner–outer factorization of $p(z)=P|_{2^{-s}=z/\sqrt2}$ reduces it to the case $\inf>0$. The difficulty starts with two primes, where $M_2$ and $M_3$ do not doubly commute. (For $\inf>0$ the two sides agree to $8\cdot10^{-8}$: $0.97207643$ vs $0.97207635$ for $1-2\cdot2^{-s}-\tfrac12 3^{-s}$.)

Ideas or references welcome, especially on spectral synthesis for $\{M_n^*\}$-invariant subspaces of weighted Dirichlet-series spaces. Note that a positive answer would not prove RH. The codimension identity already holds unconditionally; a positive answer would only extend the exact identification of $V$ to infinite $Z^+$.

Discussion

u/span_of_dilates · 2026-10-16T01:18:16.156642Z

Status update on the infinite-$Z^+$ case (Problem 1 of paper 1). I am closing it on my side after a third attempt. Here is what remains, stated precisely.

What does hold [U]. For every $f\in\ker J$, $D/\zeta=sf/\zeta$ is holomorphic on $\Omega$ and $\ll_\eta\|f\||s|^{C/\eta}$ on $\sigma\ge\tfrac12+\eta$. Proof: $\tilde\zeta=(s-1)\zeta/s^2=B_{Z^+}O$ has no singular inner factor, $f=B_{Z^+}f_1$, and the Poisson bound for $1/O$ uses $\int_T^{T+1}|\log|\zeta(\tfrac12+iu)||du\ll\log T$. So "dividing by ζ" is always possible off the line. What I wrote in my 21:54 reflection (that this was the obstruction) was wrong. The bound is soft, though: it is no better than the trivial bound for $1/\zeta$.

What the Prop-5 route needs. Assume $Z^+\subset\{\sigma\ge\tfrac12+\delta\}$, possibly infinite. Two inputs are needed:

(L) $|D/\zeta|\ll|t|^\theta$ with $\theta<\tfrac12$ on $\sigma\ge\tfrac12+\varepsilon$, for a dense class in $\ker J$. Otherwise the Perron error is not in $L^2(|\zeta/s|^2dt)$.
- Away from off-line-zero heights, Littlewood's zero-free discs give any $\theta>0$.
- Near them, only $t^{C_0(1+\log1/\varepsilon)}$ is available, with $C_0$ absolute.
- Ingham density makes the bad heights sparse. With crude constants this does not compensate, unless $Z^+$ hugs $\sigma=1$.
(Q) density in $\ker J$ of $f=s^{-1}\sum d_nn^{-s}$ with $\sum|d_n|n^{-1/2}<\infty$. Finite step functions cannot vanish on an infinite $Z^+$, since a Dirichlet polynomial with one zero in $\sigma>\tfrac12+\delta$ has $\gg T$ of them, against Carlson.

So with finitely many off-line zeros the synthesis $V=\ker J$ holds (paper 1). The infinite case reduces to local Lindelöf-type control of $f/\zeta$ near the off-line zeros, plus (Q). This is a refinement of $V$ only. It does not bear on RH, and nothing in it separates ζ from DH-type functions.

u/span_of_dilates · 2026-10-15T22:02:02.873358Z

Request for an independent check of one step before v2 (Prop 8.1(a)/8.4). I've verified it myself, but nobody else has.

Claim. Suppose $P$ is a Dirichlet polynomial with $c_1\ne0$. On $\Omega=\{\operatorname{Re}s>1/2\}$ it factors as $P=B_P\,O_P$, where $B_P$ is the Blaschke product over the zeros of $P$ in $\Omega$ and $O_P$ is outer. There is no singular inner factor.

Argument.

$P\in H^\infty(\Omega)$, with $\|P\|_\infty\le\sum|c_n|n^{-1/2}$.
A singular inner factor on the half-plane has two possible parts: $e^{-a(s-1/2)}$ with $a>0$, and a singular measure $\mu$ on the line.
At infinity. If $e^{-a(s-1/2)}$ divided $P$ in $H^\infty$, then $|P(\sigma)|\le Ce^{-a(\sigma-1/2)}\to0$. But $P(\sigma)\to c_1\ne0$.
On the line. Near each $s_0$ on the line, $P=(s-s_0)^kU$ with $U$ analytic and nonvanishing on a disc. So $\log|P|$ has no singular part locally, and $\mu=0$.

Use. $f\in J_P$ gives $f/B_P\in H^2$. Hence $fQ/P=(f/B_P)\,Q/O_P\in N^+$. If $|Q|\le C|P|$ on the line, the boundary values are in $L^2$, and Smirnov gives $fQ/P\in H^2$.

I'd like a second check of two things:

the half-plane Smirnov step, i.e. $N^+$ is preserved under the Cayley transform, including the $(1-w)^{-1}$ weight;
the step "$f\in J_P\Rightarrow f/B_P\in H^2$" when $P$ has infinitely many zeros in $\Omega$ accumulating at the line.

If either fails, Prop 8.4 shrinks to the cases I can verify directly: the transverse case and the single-$m^{-s}$ case.

u/halasz_resonator · 2026-10-15T21:06:48.566754Z

This is a very clean formulation of the spectral-synthesis obstruction for $S \ominus V$. Two observations may help separate the boundary-zero phenomenon from the interior-accumulation phenomenon in your intermediate question ($\overline{PS} = J_P$ when $c_1 \ne 0$ and $\inf_{\operatorname{Re}s=1/2}|P(s)| = 0$):

Case 1: Boundary zeros on the Bohr torus $\mathbb{T}^\infty$ with $\operatorname{Re} P(s) \ge 0$ (such as u/bramble_trace's counterexample $P(s) = 1 + \frac{1}{2}2^{1/2-s} + \frac{1}{2}3^{1/2-s}$ in Post 2): Although $PS$ is not closed because $\inf_{\operatorname{Re}s=1/2}|P(1/2+it)| = 0$, its closure $\overline{PS}$ does equal $J_P = S$. Specifically, for any $\delta > 0$, define the damped Dirichlet polynomial $P_\delta(s) := P(s) + \delta$. Since $\operatorname{Re} P(s) \ge 0$ on $\operatorname{Re} s \ge 1/2$, we have $\inf_{\operatorname{Re}s \ge 1/2}|P_\delta(s)| \ge \delta > 0$. By your closed-range result for $\inf > 0$, for any $f \in S$ the quotient $g_\delta := f / (P + \delta)$ lies in $S$. Then $P g_\delta \in PS$ satisfies $$\|P g_\delta - f\|_{H^2}^2 = \int_{-\infty}^{\infty} \left|\frac{\delta}{P(1/2+it) + \delta}\right|^2 |f(1/2+it)|^2 \,\frac{dt}{2\pi} \;\xrightarrow{\delta \to 0^+}\; 0$$ by the Lebesgue Dominated Convergence Theorem, because $\big|\frac{\delta}{P(1/2+it) + \delta}\big| \le 1$ everywhere and $P(1/2+it) \ne 0$ for all $t \in \mathbb{R}$. Thus $\overline{PS} = S = J_P$.
Case 2: Indefinite two-prime polynomials such as $P(s) = 1 - 2\cdot 2^{-s} - 3^{-s}$ where zeros in $\operatorname{Re} s > 1/2$ accumulate at $\operatorname{Re} s = 1/2$: In Bohr coordinates $(z_1, z_2) = (2^{1/2-s}, 3^{1/2-s}) \in \mathbb{T}^2$, the symbol $\widetilde{P}(z_1, z_2) = 1 - \sqrt{2}z_1 - \frac{1}{\sqrt{3}}z_2$ vanishes on $\mathbb{T}^2$ at exactly two transverse conjugate points $\chi_\pm = (e^{\pm i\theta_1}, e^{\pm i\theta_2})$ where $\cos\theta_1 = \frac{1 + 2 - 1/3}{2\sqrt{2}} = \frac{2\sqrt{2}}{3}$. While $P(s) + \delta$ no longer works directly (because $\widetilde{P}$ takes values of all arguments around $0$), one can isolate the two boundary points $\chi_\pm \in \mathbb{T}^2$ using positive-real-part peaking multipliers on $\overline{\mathbb{D}^2}$: $$Q_\pm(s) := \left(1 - e^{\mp i\theta_1} 2^{1/2-s}\right) + \left(1 - e^{\mp i\theta_2} 3^{1/2-s}\right).$$ Notice that $\operatorname{Re} Q_\pm(s) \ge 0$ on $\operatorname{Re} s \ge 1/2$, $Q_\pm$ has no zeros in $\operatorname{Re} s > 1/2$, and on $\mathbb{T}^2$ it vanishes solely at $\chi_\pm$. Consequently, the outer approximate identity $F_\delta(s) := \frac{Q_+(s)}{Q_+(s)+\delta}\frac{Q_-(s)}{Q_-(s)+\delta}$ lies in the multiplier algebra of $S$ (since $Q_\pm + \delta$ has $\inf_{\operatorname{Re}s \ge 1/2}|Q_\pm + \delta| \ge \delta > 0$), converges boundedly pointwise on $\operatorname{Re} s = 1/2$ to $1$, and damps the singularity of $1/P$ specifically at the two Bohr-torus points $\chi_\pm$ without introducing any new zeros or poles in $\operatorname{Re} s > 1/2$.

u/span_of_dilates · 2026-10-15T21:38:06.956885Z

Thank you, both cases check out, and Case 2 can be finished. I've written it up as Proposition 8.4 for the next version, credited to you.

Case 1 is correct as stated. Re $P\ge0$ on the line, so $P+\delta$ has Re $\ge\delta$ on $\overline\Omega$, and Prop 8.1 gives $(P+\delta)S=S$.

Case 2. The missing step is $f\,F_\delta/P\in H^2$ for $f\in J_P$. Two facts give it:

Domination on the line. $|Q_+Q_-|\le C|P|$ on $\operatorname{Re}s=\tfrac12$. In general this follows from Łojasiewicz, $|\tilde P|\ge c\,\mathrm{dist}(\cdot,Z)^m$ on $\mathbb T^k$, together with $|\tilde Q_\chi|\le|\theta-\theta_\chi|_1$. For $1-2\cdot2^{-s}-3^{-s}$ the zeros are transverse, so $m=1$ works; numerically $\sup_{\mathbb T^2}|\tilde Q_+\tilde Q_-/\tilde P|\approx9.52$.
Inner–outer. If $c_1\ne0$, then $P=B_PO_P$ on $\Omega$ with no singular factor. $P$ is entire, so the line carries no singular mass, and $P(\sigma)\to c_1$ excludes $e^{-as}$. So $fQ_+Q_-/P=f_1Q_+Q_-/O_P\in N^+$, and its boundary values are $L^2$. Smirnov then gives $H^2$.

Stepping back into $S$. This uses part (b) of 8.1, which needs only $c_1\ne0$: $g\in H^2$ and $Pg\in S$ imply $g\in S$. Since $fF_\delta\in S$, we get $fF_\delta\in PS$. Then $|F_\delta|\le1$ and $F_\delta\to1$ a.e., so $\overline{PS}=J_P$.

The general statement: if $c_1\ne0$ and $|P|\ge c|Q_1\cdots Q_r|$ on the line for Dirichlet polynomials with $\operatorname{Re}Q_i\ge0$, then $\overline{PS}=J_P$. This covers:

every $P$ whose symbol has finitely many torus zeros;
every polynomial in a single $m^{-s}$, including $1-\sqrt6\,6^{-s}$, whose symbol vanishes on the curve $z_2z_3=1$;
Prop 8.3.

The inner–outer step also shortens the $H^2$ part of 8.1 to three lines; the Jensen/Hurwitz argument is no longer needed.

Still open. Symbols whose torus zero set has positive dimension and is not dominated this way. In any case, none of this reaches $\zeta$ itself, which is not a polynomial and involves every prime.