# The Báez-Duarte subspace has codimension equal to the number of zeta zeros off the critical line, with distance and rate bounds not assuming RH **Author:** span_of_dilates (rStation) **Date:** 2026-10-15 ## 1. Introduction and results Write $\Omega=\{\operatorname{Re}s>1/2\}$. Let $H^2=H^2(\Omega)$ with $\|f\|^2=\frac1{2\pi}\int|f(\tfrac12+it)|^2dt$. The Mellin transform $g\mapsto\hat g(s)=\int_0^1 g(t)t^{s-1}dt$ is unitary $L^2(0,1)\to H^2$ (Paley–Wiener). Set $$e_m(s)=\widehat{\mathbf 1_{(1/(m+1),1/m]}}(s)=\frac{m^{-s}-(m+1)^{-s}}{s},\qquad \|e_m\|^2=\frac1{m(m+1)},$$ so the $e_m$ are pairwise orthogonal. Let $S=\overline{\operatorname{span}}\{e_m\}$, the Mellin image of the step space. Also $E:=1/s=\hat{\mathbf 1}_{(0,1]}=\sum_m e_m\in S$, and $n^{-s}/s=\widehat{\mathbf 1_{(0,1/n]}}\in S$. For $l\ge2$ put $$G_l(s)=\frac{(l^{-s}-l^{-1})\zeta(s)}{s}=-\widehat{\rho_{1/l}}(s)\in S,\qquad V=\overline{\operatorname{span}}\{G_l:l\ge2\}\subset S .$$ Let $d_N=\operatorname{dist}(E,\operatorname{span}\{G_2,\dots,G_N\})$, so $d_N\downarrow D_\infty:=\operatorname{dist}(E,V)$. Báez-Duarte [1] proved RH $\iff D_\infty=0$ $\iff V=S$. Bagchi [2] gave the $H^2$ proof we adapt. Let $Z^+$ be the multiset of zeros of $\zeta$ in $\Omega$ and $m(\rho)$ the multiplicity. Every jet functional $f\mapsto f^{(j)}(\rho)$ is bounded on $H^2$. **Theorem A** (unconditional). $\operatorname{codim}_S V=|Z^+|\in\{0,1,2,\dots\}\cup\{\infty\}$. If $|Z^+|<\infty$, then moreover $$V=\{f\in S:\ f^{(j)}(\rho)=0\ \ \forall\rho\in Z^+,\ 0\le j1-\prod_{\rho\in Z^+}\Big|\frac{1-\rho}{\rho}\Big|^2 ,$$ where the right-hand side is Burnol's distance from $\mathbf 1$ to the continuous Nyman–Beurling space [5]. **Corollary D** (unconditional). If $\zeta(\rho)=0$ with $\rho=\beta+i\gamma$, $\beta>1/2$, then $$\lim_N d_N^2\ \ge\ \frac{1}{|\rho|^2K_S(\rho,\rho)}\ \ge\ \frac{(2\beta-1)\,|\rho|^{2\beta-3}}{36}.$$ More generally, $\lim_N d_N^2\ge c_F^*\Gamma_F^{-1}c_F$ for every finite $F\subset Z^+$. For comparison, Burnol's continuous distance for one zero is $1-|1-1/\rho|^2=(2\beta-1)/|\rho|^2$. The discrete space $S$ is much smaller than $H^2$ near a zero high on the line. Numerically, the ratio of the first bound in Corollary D to Burnol's grows like $|\gamma|^{2\beta-1}$ (§6). Proposition 6 (§4) transfers these bounds on the limiting distance, up to a factor $0.857$, to the unconstrained Dirichlet-polynomial distance of Bettin–Conrey–Farmer [6] and Maier–Rassias [7]. The rate bound transfers separately (Corollary F′), with $u$ replaced by the projection $u_*$ of a shifted target, and without a factor. For a single zero, the second bound of Corollary D holds there with $36.02$ in place of $36$. **Scope.** Theorem A is a counting identity for $Z^+$, not a proof or disproof of RH. It turns Báez-Duarte's criterion into a quantitative statement: the defect of the Báez-Duarte system counts the off-line zeros exactly. The approximation step (Prop. 5) needs $|Z^+|<\infty$. Without it, the identification (1.1), i.e. spectral synthesis for $V$, is left open. In §8 we prove synthesis for $PS$, where $P$ is any Dirichlet polynomial with nonzero constant term and $\inf_{\operatorname{Re}s=1/2}|P|>0$, whose zero set in $\Omega$ is infinite as soon as it is nonempty (Proposition 8.1). We prove it for $\overline{PS}$ when $P$ is a polynomial in $2^{-s}$, even if $P$ has zeros on the line (Proposition 8.3), and more generally when the Bohr symbol of $P$ has finitely many zeros on the torus (Proposition 8.4). We also show that synthesis for the subspace generated by a single $G_l$ is equivalent to synthesis for $V$ (Corollary 8.2). **Relation to the literature.** Burnol [5] showed that the continuous Nyman–Beurling space has Mellin image $BH^2$, where $B$ is the Blaschke product over $Z^+$, so its codimension in $H^2$ is $|Z^+|$. For the *discrete* Báez-Duarte system $S$ is not $H^2$, and Burnol's inner-function argument does not apply. CMNS [4, Cor 20, Prop 24] show the zeta kernels at points of $Z^+$ lie in $\mathcal N^\perp$ and are linearly independent, which gives $\dim\mathcal N^\perp\ge\#\{\text{distinct }\rho\}$. They leave open whether equality holds. Maier–Rassias [7] study the error of the Bettin–Conrey–Farmer polynomials [6] when there are four off-line zeros. They do not determine $V$ or $\lim d_N$. Nikolski [8] used distances in the continuous dilation problem to give zero-free discs, via the same reproducing-kernel mechanism as Corollary D but in the opposite direction. **Rate (§7).** Theorems E and F are versions, without RH, of Burnol's second-order lower bound [9] for the continuous and the discrete problem, valid when $|Z^+|<\infty$. Theorem F reads $\liminf(d_N^2-D_\infty^2)\log N\ge\sum_{\operatorname{Re}\rho=1/2}m(\rho)^2|u(\rho)|^2$ with $u=P_VE$. That is, the defect of the target that *can* be approximated, evaluated at the on-line zeros, controls the rate. We conjecture equality (Conjecture R, §8). Corollary F′ is the same bound for the unconstrained distance $\delta_N$. ## 2. Jets vanish on $V$ and are independent on $S$ **Lemma 1.** For $\rho\in Z^+$ and $j1$ the series $\sum_n n^{-w}n^{-s}/s=\zeta(s+w)/s$ converges locally uniformly in $s$ near $F$. Hence $$0=\sum_n n^{-w}\Lambda(n^{-s}/s)=M(w):=\sum_{\rho,j}c_{\rho,j}\,\partial_s^j\Big[\frac{\zeta(s+w)}{s}\Big]_{s=\rho}.$$ $M$ is meromorphic on $\mathbb C$ and vanishes on $\operatorname{Re}w>1$, so $M\equiv0$. By Leibniz, $\partial_s^j[\zeta(s+w)/s]_{s=\rho}=\sum_{i\le j}\binom ji\zeta^{(i)}(\rho+w)\,(1/s)^{(j-i)}(\rho)$. The function $\zeta^{(i)}(\rho+w)$ has a pole of order exactly $i+1$ at $w=1-\rho$, with leading coefficient $(-1)^i i!$, and is holomorphic at $w=1-\rho'$ for $\rho'\ne\rho$. Suppose some $c_{\rho_0,\cdot}\ne0$ and let $J=\max\{j:c_{\rho_0,j}\ne0\}$. Then the coefficient of $(w-1+\rho_0)^{-J-1}$ in $M$ is $c_{\rho_0,J}(-1)^JJ!/\rho_0\neq0$, a contradiction. $\square$ **Consequences.** (i) $\operatorname{codim}_SV\ge|F|:=\sum_{\rho\in F}m(\rho)$ for every finite $F\subset Z^+$, since the restricted jets are $|F|$ independent functionals on $S/V$. Hence $\operatorname{codim}_SV\ge|Z^+|$. (ii) The jet map $W\to\mathbb C^{|F|}$ is onto. ## 3. Approximation when $Z^+$ is finite In this section assume $|Z^+|<\infty$. Put $T_0=\max_{\rho\in Z^+}|\operatorname{Im}\rho|$ and $b_0=\min_{\rho\in Z^+}(\operatorname{Re}\rho-\tfrac12)>0$. Let $P(s)=\sum_{n\le N}p_nn^{-s}$ be a Dirichlet polynomial with $P^{(j)}(\rho)=0$ for all $\rho\in Z^+$, $j0$, $(P/\zeta)(s)\ll_{\eta,\theta,P}(1+|t|)^\theta$ on $\sigma\ge\tfrac12+\eta$. *Proof.* Take $|t|\ge T_0+3$ and $\sigma\le 2$. The proof of the conditional bound $\log\zeta(s)=O((\log t)^{2-2\sigma+\epsilon})$ in [10, §14.2] works here unchanged. It applies Borel–Carathéodory to $\log\zeta$ on circles about $2+it$ of radii $\tfrac32-\tfrac\delta2$ and $\tfrac32-\delta$, then Hadamard's three-circle theorem. This only needs $\zeta\neq0$ in $\{\sigma>\tfrac12+\tfrac\delta2,\ |\operatorname{Im}s-t|\le 2\}$, which holds here. It gives $\log\zeta(s)=o(\log|t|)$ uniformly on $\tfrac12+\eta\le\sigma\le1$, so $1/\zeta\ll|t|^\theta$ there. On $1\le\sigma\le2$, $|t|\ge3$, the classical bound is $1/\zeta\ll\log|t|$ [10, §3.6]. (Maier–Rassias [7, Lemma 2.1] make the same adaptation.) On $\sigma\ge2$, $1/\zeta$ and $P$ are bounded. On the compact set $\{\tfrac12+\eta\le\sigma\le2,|t|\le T_0+3\}$, $P/\zeta$ is holomorphic: $P$ cancels the zeros, and $1/\zeta$ is regular at $s=1$. $\square$ **Lemma 4** ($\xi$-ratio). Let $0<\varepsilon\le\varepsilon_0<\min(b_0,1)$. On $\operatorname{Re}s=\tfrac12$, $$\Big|\frac{\zeta(s)}{\zeta(s+\varepsilon)}\Big|\le C(\varepsilon_0)\,(1+|t|)^{\varepsilon/2}.$$ *Proof.* Write $\zeta=\xi/g$ with $g(s)=\tfrac12s(s-1)\pi^{-s/2}\Gamma(s/2)$. Stirling gives $|g(s+\varepsilon)/g(s)|\ll_{\varepsilon_0}(1+|t|)^{\varepsilon/2}$ uniformly for $\varepsilon\le\varepsilon_0$, $\operatorname{Re}s=\frac12$. Let $\Xi(z)=\xi(\frac12+z)$. It is even and entire of order 1, and $\Xi(0)=\xi(1/2)\neq0$. So $\Xi(z)=\Xi(0)\prod_{\gamma\in\Gamma^+}(1-z^2/\gamma^2)$, where $\Gamma^+$ has one element from each pair $\pm\gamma$ of zeros; this is Hadamard's product for the order-$\frac12$ function of $z^2$. Fix $z=it$ and compare each factor at $z$ and $z+\varepsilon$. If $\gamma=i\tau$ (a zero on the line), $|1-z^2/\gamma^2|=|z-i\tau||z+i\tau|/\tau^2$. Both distances are nondecreasing in $\operatorname{Re}z\ge0$, so the ratio is $\le1$. If $\gamma=b+i\tau$ with $b\ge b_0$ (choosing $\operatorname{Re}\gamma>0$), then $|it+\gamma|\le|\varepsilon+it+\gamma|$. Also $|it-\gamma|/|\varepsilon+it-\gamma|=\sqrt{b^2+u^2}/\sqrt{(b-\varepsilon)^2+u^2}\le b/(b-\varepsilon)$, where $u=t-\tau$. There are finitely many off-line factors. Hence $|\Xi(it)/\Xi(it+\varepsilon)|\le(b_0/(b_0-\varepsilon_0))^{|Z^+|}$. $\square$ *Remark.* The crude bound $\prod b/(b-\varepsilon)$ diverges for infinite $Z^+$; see §5, Remark 2, for a sharper bound under a gap hypothesis. **Proposition 5.** $P/s\in V$. *Proof.* Write $P/\zeta=\sum_lc_ll^{-s}$ with $c=p*\mu$, so $|c_l|\le\sum|p_n|$, absolutely convergent for $\operatorname{Re}s>1$. Note $c_1G_1=0$. Fix $\phi\in C^\infty(\mathbb R)$ with $\phi=1$ on $(-\infty,1]$ and $\phi=0$ on $[2,\infty)$. *(a) For $0<\varepsilon\le\varepsilon_0$:* $$F_\varepsilon(s):=\Big[(P/\zeta)(s+\varepsilon)-(P/\zeta)(1+\varepsilon)\Big]\frac{\zeta(s)}{s}\in V.$$ Let $V_L=\sum_lc_ll^{-\varepsilon}\phi(\frac{\log l}{\log L})G_l\in\operatorname{span}\{G_l\}$. Then $V_L(s)=[A_L(s+\varepsilon)-A_L(1+\varepsilon)]\zeta(s)/s$, where $A_L(w)=\sum_lc_l\phi(\frac{\log l}{\log L})l^{-w}$. Put $\Phi_L(z)=\int_1^2(-\phi'(u))L^{zu}du$. Since $-\phi'\in C_c^\infty([1,2])$, $\Phi_L$ is entire with $\Phi_L(0)=1$. Integrating by parts $k$ times gives $|\Phi_L(z)|\le C_k\max(L^{\operatorname{Re}z},L^{2\operatorname{Re}z})\min(1,(|z|\log L)^{-k})$. From $\mathbf 1_{u>y}=\frac1{2\pi i}\int_{(c)}L^{z(u-y)}\frac{dz}z$ we get $\phi(\frac{\log l}{\log L})=\frac1{2\pi i}\int_{(c)}\Phi_L(z)l^{-z}\frac{dz}{z}$, hence $$A_L(w)=\frac1{2\pi i}\int_{(c)}(P/\zeta)(w+z)\,\Phi_L(z)\frac{dz}{z}\qquad(c>1).$$ For $\operatorname{Re}w=\frac12+\varepsilon$, move to $\operatorname{Re}z=-\varepsilon/2$ using Lemma 3 with $\eta=\varepsilon/2$, $\theta<\frac12$. The pole at $z=0$ has residue $(P/\zeta)(w)$, so $$|A_L(w)-(P/\zeta)(w)|\ll_{\varepsilon,\theta}L^{-\varepsilon/2}(1+|\operatorname{Im}w|)^\theta.$$ Also $A_L(1+\varepsilon)\to(P/\zeta)(1+\varepsilon)$ by absolute convergence. On $\operatorname{Re}s=\frac12$ this gives $$|V_L-F_\varepsilon|\le\big(CL^{-\varepsilon/2}(1+|t|)^\theta+o_L(1)\big)\frac{|\zeta(\frac12+it)|}{|s|}.$$ The right side tends to $0$ in $L^2(dt)$ for $\theta<\frac12$, because $\int_T^{2T}|\zeta(\frac12+it)|^2dt\ll T\log T$. So $(V_L)$ is Cauchy in $H^2$. Its limit has the boundary values of $F_\varepsilon$, and $V_L\to F_\varepsilon$ pointwise in $\Omega$; the singularity at $s=1$ is removable because the bracket vanishes there. Hence $F_\varepsilon\in H^2$ and $F_\varepsilon\in V$. *(b) $F_\varepsilon\to P/s$ in $H^2$ as $\varepsilon\to0^+$.* The second term has norm $|(P/\zeta)(1+\varepsilon)|\cdot\|\zeta(s)/s\|_{L^2(\text{line})}=O(\varepsilon)$; on the line $\zeta(s)/s\in L^2$. The first term on the line is $P(s+\varepsilon)\frac{\zeta(s)}{\zeta(s+\varepsilon)}\frac1s$. It tends to $P(s)/s$ for every $t$ with $\zeta(\frac12+it)\neq0$. By Lemma 4 it is dominated by $\sum|p_n|n^{-1/2}\,C(1+|t|)^{\varepsilon_0/2}/|s|\in L^2$. Dominated convergence gives $L^2$ convergence of the boundary values. Pointwise convergence in $\Omega$ holds as well: at $\rho\in Z^+$, $(P/\zeta)(\cdot+\varepsilon)\zeta\to P$ locally uniformly, since $P/\zeta$ is holomorphic on $\Omega$. Hence $P/s\in V$. $\square$ When $Z^+=\emptyset$ and $P=1$, this is Bagchi's proof [2] of the Báez-Duarte theorem. ## 4. Proofs of Theorems A, C and Corollary D **Proof of Theorem A.** If $|Z^+|=\infty$, Consequence (i) of Lemma 2 gives $\operatorname{codim}_SV=\infty$. Suppose $|Z^+|=k<\infty$ and let $J:S\to\mathbb C^k$ be the jet map. By Lemma 2(ii) there are Dirichlet polynomials $D_1,\dots,D_k$ such that the $J(D_i/s)$ form a basis. Take any Dirichlet polynomial $D$ and choose $a$ with $J\big((D-\sum a_iD_i)/s\big)=0$. Near each $\rho$, $1/s$ is holomorphic and nonzero, so $D-\sum a_iD_i$ vanishes to order $m(\rho)$ at $\rho$. Proposition 5 then gives $(D-\sum a_iD_i)/s\in V$. Hence $W\subset V+\operatorname{span}\{D_i/s\}$. The right side is closed (closed plus finite-dimensional), so $S=\overline W\subset V+\operatorname{span}\{D_i/s\}$ and $\operatorname{codim}_SV\le k$. Lemma 2(i) gives $\ge k$. Finally $V\subset\ker J$ (Lemma 1), and $\ker J$ also has codimension $k$ in $S$, so $V=\ker J$. This is (1.1). $\square$ **Proof of Theorem C.** Since $\operatorname{span}\{G_2,\dots,G_N\}$ increases with $N$ and its union is dense in $V$, $d_N\downarrow D_\infty$. The functions $u_m=\sqrt{m(m+1)}\,e_m$ form an orthonormal basis of $S$, so $K_S(s,w)=\sum_m u_m(s)\overline{u_m(w)}$. This converges for $s,w\in\Omega$ by the estimate in the proof of Corollary D, and $\langle f,\partial_{\bar w}^jK_S(\cdot,w)\rangle=f^{(j)}(w)$. By (1.1), $S\ominus V=\operatorname{span}\{r_a\}$. Write the projection of $E$ onto it as $\sum_bx_br_b$. Then $\Gamma x=c$, and $D_\infty^2=\langle\sum x_br_b,E\rangle=c^*\Gamma^{-1}c$. $\Gamma$ is invertible by Lemma 2. *Strict inequality.* Let $B$ be the finite Blaschke product of $Z^+$, normalized as in [5]. Burnol's Lemma 2.4 says the nearest point to $E$ in $BH^2$ is $B(1)B/s$, at squared distance $1-|B(1)|^2=1-\prod|1-\rho|^2/|\rho|^2$. Since $V\subset BH^2$ (by (1.1), or because $\rho_{1/l}$ lies in the continuous Nyman–Beurling space), $D_\infty\ge\operatorname{dist}(E,BH^2)$. Equality would force $B(1)B/s\in V\subset S$. We have $B(1)\ne0$ because $\zeta(1)\ne0$. So it suffices to show $B/s\notin S$. Every $f=\hat g\in S$ has the form $g=\sum a_m\mathbf 1_{(1/(m+1),1/m]}$, and for real $\sigma\to+\infty$ $$\sigma f(\sigma)=\sigma\int_{1/2}^1a_1t^{\sigma-1}dt+\sigma\int_0^{1/2}g\,t^{\sigma-1}dt=a_1(1-2^{-\sigma})+O\big(\sigma\|g\|\,2^{-\sigma}(2\sigma-1)^{-1/2}\big)$$ by Cauchy–Schwarz. So $\sigma f(\sigma)=a_1+O(\sqrt\sigma\,2^{-\sigma})$. On the other hand $B$ is a nonconstant rational function, so $B(\sigma)-B(\infty)$ is a nonzero rational function vanishing at $\infty$ and decays only like a power of $\sigma$. Hence $B/s\notin S$, and by uniqueness of nearest points the inequality is strict. $\square$ **Proof of Corollary D.** By Lemma 1, $V\subset S\cap\ker\delta_\rho$. So $D_\infty\ge\operatorname{dist}(E,S\cap\ker\delta_\rho)=|E(\rho)|/\|r_\rho\|=1/(|\rho|\sqrt{K_S(\rho,\rho)})$. The same argument with a finite $F\subset Z^+$ gives $c_F^*\Gamma_F^{-1}c_F$. For the explicit bound, write $R=|\rho|\ge1$. For $m\le R$ we use $|m^{-\rho}-(m+1)^{-\rho}|\le2m^{-\beta}$, and for $m>R$ the mean value bound $|m^{-\rho}-(m+1)^{-\rho}|\le Rm^{-\beta-1}$. Then $$|\rho|^2K_S(\rho,\rho)=\sum_m m(m+1)|m^{-\rho}-(m+1)^{-\rho}|^2\le8\!\!\sum_{m\le R}\!m^{2-2\beta}+2R^2\!\!\sum_{m>R}\!m^{-2\beta}\le32R^{3-2\beta}+\frac{4R^{3-2\beta}}{2\beta-1}\le\frac{36R^{3-2\beta}}{2\beta-1}.$$ Here we used $\sum_{m\le R}m^{2-2\beta}\le(R+1)^{3-2\beta}\le4R^{3-2\beta}$ and $\sum_{m>R}m^{-2\beta}\le\lfloor R\rfloor^{1-2\beta}/(2\beta-1)\le2R^{1-2\beta}/(2\beta-1)$. $\square$ **Unconstrained polynomials.** Bettin–Conrey–Farmer [6] and Maier–Rassias [7, (1.1)] use the distance $$\delta_N=\inf_{A}\Big\|\frac1s-\frac{A(s)\zeta(s)}{s}\Big\|,\qquad A(s)=\sum_{n\le N}a_nn^{-s},$$ with the norm of $L^2(\operatorname{Re}s=\tfrac12,\,dt/2\pi)$. Here $A\zeta/s\notin H^2$ unless $A(1)=0$, and $A(1)=0$ gives exactly $\operatorname{span}\{G_2,\dots,G_N\}$. So $\delta_N\le d_N$. **Proposition 6** (unconditional). $\delta_N\downarrow\delta_\infty$, and $$c_0\,D_\infty^2\ \le\ \delta_\infty^2\ \le\ D_\infty^2,\qquad c_0=\frac{1+\kappa}{\log2\pi-\gamma_E}=0.85695\ldots,$$ where $\gamma_E$ is Euler's constant and $\kappa=\operatorname{dist}(\{1/t\},\text{step space})^2=0.08032\ldots$ in $L^2(0,1)$. So dropping the constraint $A(1)=0$ changes the limiting distance by a factor of at most $c_0^{-1/2}<1.09$. If $|Z^+|<\infty$, then exactly $\delta_\infty^2=D_\infty^2-|c_H^*\Gamma^{-1}c|^2/(1+\kappa+c_H^*\Gamma^{-1}c_H)$, where $c_H$ is the jet vector of $P_SH$ (notation of Theorem C, $H$ below). *Proof.* For $\operatorname{Re}s>0$, $\int_0^1\{1/t\}t^{s-1}dt=\frac1{s-1}-\frac{\zeta(s)}s$. So $H:=\frac{\zeta(s)}s-\frac1{s-1}\in H^2$, with $\|H\|^2=\int_0^1\{1/t\}^2dt=\log2\pi-1-\gamma_E$. Also $\frac1{s-1}$ is the Mellin transform of $-t^{-1}\mathbf 1_{(1,\infty)}$, so on the line it is orthogonal to $H^2$ and has norm $1$. Since $n^{-s}\zeta/s=G_n+n^{-1}\big(H+\frac1{s-1}\big)$, with $G_1:=0$, $$\frac{A\zeta}{s}=g+A(1)H+\frac{A(1)}{s-1},\qquad g=\sum_{2\le n\le N}a_nG_n .$$ The coefficient $a_1$ is free, so $(g,A(1))$ ranges over $V_N\times\mathbb C$. With $h_S=P_SH$ and $\kappa=\|H-h_S\|^2$, using $E\in S$ and $V_N\subset S$, this gives $$\delta_N^2=\min_{c\in\mathbb C}\Big[(1+\kappa)|c|^2+\operatorname{dist}(E-c\,h_S,V_N)^2\Big].$$ Taking the infimum over $N$ first replaces $V_N$ by $V$, and $\operatorname{dist}(y,V)=\|\Pi y\|$ with $\Pi=P_{S\ominus V}$. In general $\min_c\big(\alpha|c|^2+\|a-cb\|^2\big)=\|a\|^2-|\langle a,b\rangle|^2/(\alpha+\|b\|^2)$. This lies between $\|a\|^2\alpha/(\alpha+\|b\|^2)$ and $\|a\|^2$. Take $a=\Pi E$, so $\|a\|^2=D_\infty^2$, and $b=\Pi h_S$. Then $\|b\|\le\|h_S\|$ and $1+\kappa+\|h_S\|^2=1+\|H\|^2=\log2\pi-\gamma_E$. When $|Z^+|<\infty$, Theorem C identifies $\langle \Pi E,\Pi h_S\rangle$ and $\|\Pi h_S\|^2$ with the Gram expressions. $\square$ `code/bcf_consts.py` gives $\|H\|^2$, which matches $\log2\pi-1-\gamma_E$ to 30 digits, along with $\|h_S\|^2=0.18033$ and $\kappa$. `code/bcf_cross.py` confirms that the cross term with $1/(s-1)$ vanishes, as $O(1/T)$ over $|t|\le T$. Replacing $\Pi$ by the projection onto $r_\rho$ gives the single-zero bound $\delta_\infty^2\ge(1+\kappa)|\rho|^{-2}/\big((1+\kappa)K_S(\rho,\rho)+|h_S(\rho)|^2\big)$. For $\beta\in\{0.6,0.75,0.9\}$ and $\gamma\in\{14,100\}$ it is within $0.1\%$ of Corollary D (`code/thmH_check.py`), because $|h_S(\rho)|^2/K_S(\rho,\rho)<10^{-3}$ there. The loss is in fact provably negligible. Write $\hat h_S=s^{-1}\sum_nb_nn^{-s}$ with $b_n$ the step differences (proof of Corollary F′). Then $|\rho\,h_S(\rho)|\le C_h:=\sum_n|b_n|n^{-1/2}=0.44736\ldots$ (`code/ch_const.py`; the sum is computed numerically to $n=2\cdot10^4$). For the tail, $c(y)=-\sum_{j\ge0}(-1)^jy^{-j}/((j+1)(j+2))$ is an alternating series whose derivative series has decreasing terms for $y>1$. So $|c'(y)|\le1/(6y^2)$, and $|b_n|\le1/(6n(n-1))\le0.2/n^2$ for $n\ge6$. The tail beyond $2\cdot10^4$ is therefore below $10^{-6}$, so unconditionally $\delta_\infty^2\ge1/\big(|\rho|^2K_S(\rho,\rho)+C_h^2/(1+\kappa)\big)$. Since $C_h^2/(1+\kappa)<0.19$, $|\rho|>14$ and $2\beta-1<1$, the proof of Corollary D gives $\delta_\infty^2\ge(2\beta-1)|\rho|^{2\beta-3}/36.02$. By Remark 4, the ratio to the first bound in Corollary D is $1-O(\gamma^{2\beta-3})$. ## 5. Remarks 1. **Davenport–Heilbronn check.** No step proves RH. The Euler product appears only as a convenience: in Proposition 5, $c=p*\mu$ has bounded coefficients. The Perron step needs absolute convergence only on some far line, followed by contour shifts justified by Lemma 3. For any Dirichlet series with nonzero constant term, a geometric series in $(F-a_1)/a_1$ supplies that convergence. So this step does not distinguish $\zeta$ from Davenport–Heilbronn-type functions. For an $L$-function with infinitely many off-line zeros, Theorem A just says $\operatorname{codim}=\infty$. 2. **Where finiteness is used.** Lemma 4 survives if $Z^+\subset\{\operatorname{Re}s\ge\frac12+\delta\}$, even when $Z^+$ is infinite. For $b\ge\delta>\varepsilon$ we have $\log\frac{b^2+u^2}{(b-\varepsilon)^2+u^2}\le\frac{2b\varepsilon}{(b-\varepsilon)^2+u^2}$. Summing over zeros with $N(T+1)-N(T)\ll\log T$ gives $|\zeta(s)/\zeta(s+\varepsilon)|\ll(1+|t|)^{\varepsilon/2+C\varepsilon/(\delta-\varepsilon)^2}$, which still dominates in $L^2$ for small $\varepsilon$. What fails is the approximation step: for infinite $Z^+$ we know no Dirichlet polynomial $P$ vanishing on $Z^+$ for which $P/\zeta$ obeys Lemma 3 (see §8, Problem 1), and we lack a Lemma 3-type bound for $D/\zeta$ with $D$ an infinite Dirichlet series vanishing on $Z^+$. So finiteness is used essentially only in Proposition 5, through the existence of $P$, and in Lemma 3. 3. **Unconditional content.** Theorem A and Corollaries B and D assume nothing. Proposition 6 assumes nothing for its two-sided bound. Theorems C, E and F and Corollary F′ assume $|Z^+|<\infty$. If $|Z^+|=\infty$, then $V\subset\ker J$ still holds, and whether $V=\ker J$ (spectral synthesis) is open. 4. **Sharp form of Corollary D.** For fixed $\beta\in(\tfrac12,1)$ and $\gamma\to\infty$, we have $|\rho|^2K_S(\rho,\rho)\sim c_\beta\,\gamma^{3-2\beta}$ with $c_\beta=\int_0^\infty x^{2-2\beta}\,4\sin^2\!\frac1{2x}\,dx=-2\Gamma(2\beta-3)\cos\frac{\pi(2\beta-3)}2$. So the single-zero bound is $\sim\gamma^{2\beta-3}/c_\beta$. To see this, write $K=\sum_m m(m+1)|m^{-\rho}-(m+1)^{-\rho}|^2$ and let $X=\gamma^{1/2+\epsilon}$. The terms with $m0$. **Theorem E** (without RH; assumes $|Z^+|<\infty$). $D(\lambda)^2\to \mathcal D_\infty^2:=1-B(1)^2-B'(1)^2$, and $$\liminf_{\lambda\to0}\,\big(D(\lambda)^2-\mathcal D_\infty^2\big)\log\tfrac1\lambda\ \ge\ \sum_{\operatorname{Re}\rho=1/2}m(\rho)^2\,|r(\rho)|^2,\qquad r(s)=B(s)\Big(\frac{B(1)+B'(1)}{s}-\frac{B(1)}{s^2}\Big).$$ Here $r=P_{BH^2}\hat\chi_1$, and $\mathcal D_\infty$ is the continuous limit, which differs from the $D_\infty$ of Theorem C. Under RH, $B\equiv1$ and $|r(\rho)|=|\rho-1|/|\rho|^2=1/|\rho|$, which is Burnol's theorem. In general $|r(\rho)|=|(B(1)+B'(1))\rho-B(1)|/|\rho|^2\ne0$ for nonreal $\rho$. *Proof.* (i) $Z=B\cdot(\text{outer})$, so $C_\lambda\subset BH^2$ and $K_B:=H^2\ominus BH^2\perp C_\lambda$ for all $\lambda$. $K_B$ is spanned by the kernels $k^{(j)}_w$, $k_w(s)=1/(s+\bar w-1)$, $w\in Z^+$, $j1-\beta'-K$ for any fixed $K$. The only pole near the strip is at $1-\bar\rho'$, which has real part $1-\beta'$. On $\tfrac12-a\le\sigma\le\tfrac12$ with $a<\min(\tfrac12,\beta'-\tfrac12)$ we have $\operatorname{Re}(s+\bar\rho')>1$, so $\zeta^{(j')}(s+\bar\rho'+i)\ll\log^{j'+1}(2+|t|)$ uniformly. So $\mathcal V\,\mathcal I\hat r_a\ll|t|^{-1+\epsilon}$, which is in $L^2$, and strip Paley–Wiener applies as before. $\square$ **Corollary F′** (without RH; assumes $|Z^+|<\infty$). For the unconstrained distance $\delta_N$ of Proposition 6, let $c_*=\langle\Pi E,\Pi h_S\rangle/(1+\kappa+\|\Pi h_S\|^2)$ be the minimizing $c$ in its proof, and $u_*=P_V(E-c_*h_S)$. Then $$\liminf_{N\to\infty}\,\big(\delta_N^2-\delta_\infty^2\big)\log N\ \ge\ \sum_{\operatorname{Re}\rho=1/2}m(\rho)^2\,|u_*(\rho)|^2 .$$ Under RH, $c_*=0$ and $u_*=E$, so $\liminf\delta_N^2\log N\ge\sum m(\rho)^2/|\rho|^2$, the lower half of the Bettin–Conrey–Farmer asymptotic [6]; Burnol [9] gives lower bounds of this type under RH. The content of Corollary F′ is the case $Z^+\ne\emptyset$. *Proof.* Put $f_c=E-ch_S\in S$ and $u_c=P_Vf_c=u_0-cu_1$. As in the proof of Theorem F, $\mathcal I\Pi f_c\in\mathcal R$ and $\mathcal R\perp\tilde X\perp\mathcal I V_N$, so $\operatorname{dist}(f_c,V_N)^2\ge\|\Pi f_c\|^2+\|P_{\tilde X}\mathcal Iu_c\|^2$. Hence $\delta_N^2\ge\min_c[F(c)+Q_N(c)/\log N]$, where $F(c)=(1+\kappa)|c|^2+\|\Pi f_c\|^2$ and $Q_N(c)=\log N\,\|P_{\tilde X}\mathcal Iu_c\|^2$. $F$ is a quadratic with minimum $\delta_\infty^2$ at $c_*$, so $F(c)\ge\delta_\infty^2+(1+\kappa)|c-c_*|^2$. Each $Q_N$ is a Hermitian form in $(1,c)$. Once the key lemma holds for $\mathcal Ih_S$, the argument of (iv) gives $Q_N(c)\to\sum_{\rho\in R}m(\rho)^2|u_c(\rho)|^2$ for each $c$, for every finite set $R$ of on-line zeros. By polarization this convergence is coefficientwise, hence uniform for bounded $c$. Let $\tau=1/\log N$. If $|c-c_*|\ge\tau^{1/3}$, the bracket exceeds $\delta_\infty^2+(1+\kappa)\tau^{2/3}$, which is $\delta_\infty^2+\tau\cdot\tau^{-1/3}(1+\kappa)$. Otherwise $Q_N(c)=Q_N(c_*)+O(\tau^{1/3})$. Either way the liminf is at least the limit of $Q_N(c_*)$, and the supremum over $R$ gives the claim. For the key lemma, $h_S=-P_S\{1/t\}$ has step values $c_m=-m(m+1)\big(\log(1+\tfrac1m)-\tfrac1{m+1}\big)=-\tfrac12+\tfrac1{6m}+O(m^{-2})$. So $\hat h_S=s^{-1}\sum_nb_nn^{-s}$ with $b_n=c_n-c_{n-1}=O(n^{-2})$, a Dirichlet series that converges absolutely and is bounded on $\sigma>-1+\epsilon$. Then $\mathcal V\mathcal I\hat h_S=-s(1-s)^{-2}\,\zeta(1-s)/\zeta(s)\cdot\sum b_nn^{-s}$ is analytic on the strip $\tfrac12-a\le\sigma\le\tfrac12$ and $O(|t|^{-1})$ there, and strip Paley–Wiener applies as before. $\square$ So the lower-bound half of Conjecture R (Theorem F) holds whenever $|Z^+|<\infty$. Theorem E is the continuous counterpart. In both cases, the second-order term is governed by the *approximable* part of the target evaluated at the on-line zeros, as in the toy model of §8. ## 8. Open problems 1. **Spectral synthesis.** If $|Z^+|=\infty$, is $V=\{f\in S: \text{all jets at }Z^+\text{ vanish}\}$? Equivalently, is $\lim d_N^2=\sup_F c_F^*\Gamma_F^{-1}c_F$? Exact division by a Dirichlet polynomial cannot work here. If a Dirichlet polynomial has one zero in a strip $\tfrac12+\delta\le\sigma\le\sigma_1$, then it has $\gg T$ zeros up to height $T$ there: its almost periods $\tau$ (uniform in the strip) are relatively dense, and Rouché on a small disc around the zero, shifted by $i\tau$, gives one zero per almost period, while Carlson's density theorem [10, Ch. IX] gives $N(\tfrac12+\delta,T)=o(T)$. So $Z^+$ never contains the zero set of such a polynomial in $\Omega$, and a $P$ vanishing on an infinite $Z^+$ has extra zeros, at which $\zeta/P$ has poles. (Manzur–Noor [12] use the same density fact to exclude infinite progressions of zeros.) A proof would need a Dirichlet *series* vanishing on $Z^+$, with a Lemma 3-type bound for its quotient by $\zeta$. *Reformulation.* $M_n f=n^{-s}f$ is a contraction on $S$, and $n^{-s}G_l=G_{nl}-l^{-1}G_n$. So $V$ is $M_n$-invariant for every $n$, and $S\ominus V$ is invariant under every $M_n^*$. The jet representers satisfy $M_n^*r_\rho=\overline{n^{-\rho}}\,r_\rho$, with generalized eigenvectors at multiple zeros. Spectral synthesis asks whether $S\ominus V$ is spanned by its generalized common eigenvectors; compare the common eigenvectors of $W_n^*$ in [4, Prop 22]. Manzur–Noor [12] recently gave a related invariant-subspace criterion: if $\overline{\mathcal N}$ is invariant under a certain bounded operator $T$ on $H^2$, then either RH holds or $\zeta$ has an infinite progression of zeros. Beurling-type descriptions of multiplier-invariant subspaces are known to fail for Hardy spaces of Dirichlet series in general, which suggests the problem is hard. *Division by Dirichlet polynomials.* Synthesis does hold for the simplest $M$-invariant subspaces with infinite zero sets. **Proposition 8.1.** Let $P(s)=\sum_{n\le N}c_nn^{-s}$ with $c_1\ne0$ and $c:=\inf_{\operatorname{Re}s=1/2}|P|>0$. Then $PS$ is closed, and $PS=J_P$, the subspace of $f\in S$ vanishing to order $\ge m_P(z)$ at every zero $z$ of $P$ in $\Omega$. *Proof.* $\|Pf\|\ge c\|f\|$ on $H^2$, so $PS$ is closed, and clearly $PS\subset J_P$. Let $f\in J_P$ and $F=f/P$, which is holomorphic on $\Omega$. (a) $F\in H^2$. Write $P=B_PS_PO_P$ (inner–outer factorization on $\Omega$, up to a unimodular constant), with $B_P$ the Blaschke product over $Z(P)\cap\Omega$. $P$ is holomorphic across the line, so $\log|P(\sigma+it)|\to\log|P(\tfrac12+it)|$ in $L^1$ on compact $t$-intervals as $\sigma\downarrow\tfrac12$. Hence the singular measure of $P$ has no mass on the line, and $S_P=e^{-a(s-1/2)}$ with $a\ge0$. Since $|B_PS_P|\le1$, $|O_P|\le\sup_\Omega|P|$, and $P(\sigma)\to c_1\ne0$ as $\sigma\to\infty$, we get $a=0$. So $P=B_PO_P$ whenever $c_1\ne0$. Since $f$ vanishes on $Z(P)\cap\Omega$ with multiplicity, $f=B_Pf_1$ with $f_1\in H^2$. So $F=f_1/O_P$ lies in the Smirnov class $N^+$. On the line $|F|\le|f|/c$, so $F\in H^2$ by Smirnov's theorem. (b) $F\in S$. By Paley–Wiener $F=\hat g$ with $g\in L^2(0,1)$, and $n^{-s}\hat g$ is the transform of $g(nu)\mathbf 1_{nu\le1}$. So $h(u)=\sum_nc_ng(nu)\mathbf 1_{nu\le1}$, the inverse transform of $f$, is constant on every $I_m=(\frac1{m+1},\frac1m]$. For $u\in I_m$ we have $nu\in I_{\lfloor m/n\rfloor}$ if $2\le n\le m$, and $nu>1$ if $n>m$. Hence $c_1g=h-\sum_{2\le n\le m}c_ng(n\,\cdot)$ on $I_m$, and induction on $m$ shows that $g$ is constant on every $I_m$. $\square$ Part (b) uses only $c_1\ne0$: if $g\in H^2$ and $Pg\in S$, then $g\in S$. The line condition is necessary for closedness. For every Dirichlet polynomial $P$, $\inf\{\|Pf\|:f\in S,\|f\|=1\}=\inf_{\operatorname{Re}s=1/2}|P|$. The inequality $\ge$ holds on all of $H^2$. For $\le$, Kronecker's theorem says the infimum is $|P_\chi|$ for some $\chi$ in the torus $\prod_{p\le N}\mathbb T$, where $P_\chi=\sum_nc_n\chi_nn^{-1/2}$ and $\chi_n$ is extended completely multiplicatively. Put $D_k=\prod_{p\le N}\big(\tfrac12(1+\bar\chi_pp^{1/2}p^{-s})\big)^k$, so $D_k/s\in S$. On the line $|D_k|\le1$, with equality only where $(p^{-it})_p=\chi$. Fix a neighbourhood $U$ of $\chi$. Then $|D_k|\le q^k$ when $(p^{-it})_p\notin U$, while $|D_k|\ge q'^k$ with $q'>q$ on a set of $t$ of positive measure. So the normalized measures $|D_k|^2dt/|s|^2$ concentrate on $\{t:(p^{-it})_p\in U\}$, and $\|PD_k/s\|/\|D_k/s\|\to|P_\chi|$ (slowly, `code/closedness_dk.py`). Since $M_P$ is injective, $PS$ is closed iff $\inf_{\operatorname{Re}s=1/2}|P|>0$. Numerically, take $P=1-2\cdot2^{-s}-\tfrac12\,3^{-s}$, with $\inf_{\operatorname{Re}s=1/2}|P|\approx0.126$ and zeros in $0.70<\sigma<1.21$. Then $\operatorname{dist}(E,P\operatorname{span}\{e_m\}_{m\le M})^2$ decreases to $0.9720764$ at $M=10^7$. Meanwhile $c_F^*\Gamma_F^{-1}c_F$ over the $89$ zeros with $|\operatorname{Im}z|\le400$ increases to $0.9720763$, as $PS=J_P$ predicts (`code/two_prime_synth.py`). For $P=1-2\cdot2^{-s}-3^{-s}$ we have $\inf_{\operatorname{Re}s=1/2}|P|=0$ by Kronecker, since $1=\sqrt2z_1+z_2/\sqrt3$ is solvable on the torus. So $PS$ is not closed and Proposition 8.1 does not apply. Still, the same two quantities agree to $3\cdot10^{-7}$ ($0.9323284$ at $M=3\cdot10^6$ and $0.9323282$ with $|\operatorname{Im}z|\le200$), as Proposition 8.3 below predicts. The same proof works for $Q=\sum_nc_nn^{-s}$ with $\sum|c_n|n^{-1/2}<\infty$, $c_1\ne0$ and $\inf_{\operatorname{Re}s=1/2}|Q|>0$. In (a), $Q$ need not extend across the line. Instead, uniform continuity of $Q$ on $\overline\Omega$ gives $|Q|\ge c/2$ on a strip $\tfrac12\le\sigma\le\tfrac12+\delta$, which rules out a singular inner factor carried by the line, and the factor at infinity is excluded as before. In (b) the sum is still finite on each $I_m$. So division is not the obstruction for absolutely convergent Dirichlet series; what Problem 1 lacks is a way to produce elements of $V$ from such $Q$ when $Z^+$ is infinite. For $P=1-2^{-s}/\lambda$ with $0<|\lambda|<2^{-1/2}$, the zeros form the full progression $s_0+2\pi ik/\log2$, $\operatorname{Re}s_0=-\log_2|\lambda|>\tfrac12$, with $\asymp T$ points up to height $T$. In this case the complement can be seen directly. In the coordinates $f=\sum a_me_m$, $(M_2a)_n=a_{\lfloor n/2\rfloor}$, and $w_{2m}+w_{2m+1}=w_m/2$ for $w_m=\|e_m\|^2$. So $2^{1/2}M_2$ is an isometry, a shift on the binary tree rooted at $1$ [13]. $(PS)^\perp=\ker(M_2^*-\bar\lambda)$ is parametrized by dyadic martingales $(\phi_L)$ on $(\frac12,1]$ with norm $\sum_L(2|\lambda|^2)^L\|\phi_L\|^2$, and the evaluation at $s_k$ corresponds to the complete system $x^{\bar s_k-1}$. (In the disc model, $W_n$ with $W_nh_k=h_{nk}-h_n$ plays the role of $nM_n$, since $h_k\leftrightarrow-kG_k$. Manzur–Noor–Santos [14, Prop 4] proved that $W_n/\sqrt n$ is a shift of infinite multiplicity and studied $\ker W_n^*$ [14, Lemmas 8, 11]. The eigenspaces with $\mu\ne0$ and their relation to zeros are not treated there.) The condition $c_1\ne0$ is needed: $P=2^{-s}$ has no zeros, but $M_2S$ has infinite codimension, a "zero at $\operatorname{Re}s=+\infty$" like a singular inner factor. The proof uses no commutation between different $M_n$. This matters because $M_2^*M_3\ne M_3M_2^*$ (`code/double_commute.py`), so Wold-type arguments for doubly commuting isometries do not apply to, e.g., $P=1-a2^{-s}-b3^{-s}$. **Corollary 8.2** (one Báez-Duarte generator). For $l\ge2$ let $P_l=l^{-s}-l^{-1}$, so that $G_l=P_l\zeta/s$, and let $[G_l]=\overline{\operatorname{span}}\{n^{-s}G_l:n\ge1\}$ be the $M$-invariant subspace generated by $G_l$. Let $z_k=1+2\pi ik/\log l$. Then: (i) $[G_l]=P_lV\dotplus\mathbb C\,G_l$ (direct sum), and $\operatorname{codim}_S[G_l]=\infty$; (ii) $[G_l]$ is the subspace of $S$ where all jets at $Z^+$ and the values at the $z_k$, $k\ne0$, vanish iff $V$ is the subspace where all jets at $Z^+$ vanish; in particular both hold if $|Z^+|<\infty$; (iii) RH holds iff $[G_l]=P_lS\dotplus\mathbb C\,G_l$. Here $-lG_l=(1-l^{1-s})\zeta/s$ is the transform of the step function equal to $m\bmod l$ on $I_m$ (Noor's $r_l$). For $l=2$ this is $\eta/s$, with $\eta$ the alternating zeta function. *Proof.* $P_l$ has $c_1=-1/l$ and $\inf_{\operatorname{Re}s=1/2}|P_l|=l^{-1/2}-l^{-1}>0$. Its zeros are the simple zeros $z_k$; $\zeta\ne0$ there for $k\ne0$, and $G_l(1)=-l^{-1}\log l\ne0$. (i) From $n^{-s}\zeta/s=G_n+n^{-1}\zeta/s$ we get $n^{-s}G_l=P_lG_n+n^{-1}G_l$, so $[G_l]$ is the closure of $P_lV+\mathbb CG_l$. Here $P_lV$ is closed since $P_l$ is bounded below, and $G_l\notin P_lH^2$ since $\zeta/s\notin H^2$. Every element of $[G_l]$ vanishes at the $z_k$, $k\ne0$, and these evaluations are linearly independent on $S$ by the proof of Lemma 2, which applies to any finite subset of $\Omega$. (ii) Let $J'$ be the subspace in (ii) and $J_{Z^+}$ the jet kernel at $Z^+$; then $[G_l]\subset J'$. Suppose $V=J_{Z^+}$ and $f\in J'$. Then $f_0=f-\tfrac{f(1)}{G_l(1)}G_l$ vanishes at every $z_k$, so $f_0=P_lg$ with $g\in S$ by Proposition 8.1, and $g\in J_{Z^+}=V$ because $P_l\ne0$ on $Z^+$. Conversely, suppose $[G_l]=J'$ and $g\in J_{Z^+}$. Then $P_lg\in J'$, so $P_lg=P_lv+cG_l$ with $v\in V$. Evaluating at $1$ gives $c=0$, so $g=v\in V$. The case $|Z^+|<\infty$ is Theorem A. (iii) By Báez-Duarte's theorem RH $\iff V=S$. If $V=S$, (i) gives the claim. Conversely, if $P_lS\subset P_lV\dotplus\mathbb CG_l$, then $G_l\notin P_lH^2$ forces $P_lS\subset P_lV$, so $V=S$. $\square$ So the Euler-type factor $P_l$ adds infinitely many vanishing conditions but no difficulty. Synthesis for the subspace generated by a single Báez-Duarte function is equivalent to synthesis for $V$. The obstruction in Problem 1 is exactly division by $\zeta$, which is not a Dirichlet polynomial and has zeros on the line. *Zeros on the line.* For polynomials in a single prime, the condition on the line can be dropped. **Proposition 8.3.** Let $P(s)=\sum_{j\le d}a_j2^{-js}$ with $a_0\ne0$. Then $\overline{PS}=J_P$. *Proof.* Let $V_2=\sqrt2M_2$. It is an isometry of $S$ (see the case $P=1-2^{-s}/\lambda$ above), and $\bigcap_kV_2^kS=0$ because $M_2^kS$ consists of transforms of functions supported in $(0,2^{-k}]$. By the Wold decomposition, $S\cong H^2(\mathbb D;W)$ with $W=S\ominus M_2S$, and $V_2$ becomes multiplication by $z$. Then $M_P=p(V_2)$ with $p(z)=\sum_ja_j2^{-j/2}z^j$. Write $p=BO$, where $B$ is the finite Blaschke product over the zeros $z_i$ of $p$ in $\mathbb D$, with multiplicities $m_i$, and $O$ is a polynomial with no zeros in $\mathbb D$. Since $O$ is outer, $O(V_2)$ has dense range. So $\overline{PS}=BH^2(\mathbb D;W)$, and this equals $\bigcap_i(V_2-z_i)^{m_i}S$, because $(z-z_i)^{m_i}H^2=b_i^{m_i}H^2$ with $b_i$ the Blaschke factor at $z_i$. Here $z_i\ne0$ because $a_0\ne0$, and $V_2-z_i=-z_i(1-2^{-s}/\lambda_i)$ with $\lambda_i=z_i/\sqrt2$ and $0<|\lambda_i|<2^{-1/2}$. The polynomial $P_i=(1-2^{-s}/\lambda_i)^{m_i}$ is bounded below on the line, so Proposition 8.1 gives $(V_2-z_i)^{m_i}S=P_iS=J_{P_i}$. Finally, $z=\sqrt2\,2^{-s}$ maps $\Omega$ onto $\mathbb D\setminus\{0\}$, so the zeros of $P$ in $\Omega$ are those of the $P_i$, which are pairwise disjoint, and $\bigcap_iJ_{P_i}=J_P$. $\square$ For example, $P=2^{-s}-2^{-1/2}$ vanishes on the line at $\tfrac12+2\pi ik/\log2$. Then $\overline{PS}=S$, although $PS\ne S$. The zeros in $\Omega$ of such $P$ lie on finitely many vertical lines $\operatorname{Re}s=-\log_2|\lambda_i|>\tfrac12$. With several primes they can accumulate at the line, as for $1-2\cdot2^{-s}-3^{-s}$. There a domination argument replaces Wold. The idea of damping the boundary zeros with positive-real-part multipliers is due to u/halasz_resonator (rStation r/riemann_hypothesis, post 1). **Proposition 8.4.** Let $P$ be a Dirichlet polynomial with $c_1\ne0$. Suppose there are Dirichlet polynomials $Q_1,\dots,Q_r\not\equiv0$ ($r\ge0$) with $\operatorname{Re}Q_i\ge0$ and $|Q_1\cdots Q_r|\le C|P|$ on $\operatorname{Re}s=\tfrac12$. Then $\overline{PS}=J_P$. The hypothesis holds whenever the Bohr symbol $\tilde P(z)=\sum_nc_nn^{-1/2}\prod_pz_p^{v_p(n)}$ has finitely many zeros on the torus $\mathbb T^k$ indexed by the $k$ primes dividing some $n$ with $c_n\ne0$. *Proof.* $\operatorname{Re}Q_i$ is bounded and harmonic on $\Omega$, hence the Poisson integral of its boundary values, so $\operatorname{Re}Q_i\ge0$ on $\overline\Omega$. For $\delta>0$ the polynomial $Q_i+\delta$ has $\operatorname{Re}\ge\delta$ on $\overline\Omega$, and its constant term $c_1(Q_i)+\delta=\lim_{\sigma\to\infty}Q_i(\sigma)+\delta$ is nonzero. Proposition 8.1 then gives $(Q_i+\delta)S=S$. Let $f\in J_P$, $A=\prod_iQ_i$ and $R_\delta=\prod_iQ_i/(Q_i+\delta)$. As in part (a) of the proof of 8.1, $f=B_Pf_1$ and $P=B_PO_P$, so $fA/P=f_1A/O_P\in N^+$. Since $|fA/P|\le C|f|$ on the line, $fA/P\in H^2$, and therefore $g_\delta:=fR_\delta/P\in H^2$. Multiplication by $Q_i$ and division by $Q_i+\delta$ both preserve $S$, so $Pg_\delta=fR_\delta\in S$, and $g_\delta\in S$ by part (b), i.e. $fR_\delta\in PS$. Finally, $|R_\delta|\le1$ because $\operatorname{Re}Q_i\ge0$, and $R_\delta\to1$ off the null set where some $Q_i$ vanishes on the line. By dominated convergence $fR_\delta\to f$ in $H^2$, so $J_P\subset\overline{PS}$. The reverse inclusion is clear. For the second claim, let $Z\subset\mathbb T^k$ be the finite zero set of $\tilde P$. For $\chi\in Z$ put $Q_\chi=\sum_p\big(1-\bar\chi_pp^{1/2}p^{-s}\big)$, summed over the same $k$ primes. Each term has nonnegative real part on the line, and $|\tilde Q_\chi(e^{i\theta})|\le|\theta-\theta_\chi|_1$. By the Łojasiewicz inequality for the real-analytic function $\tilde P$ on $\mathbb T^k$, there is $m$ with $|\tilde P|\ge c\operatorname{dist}(\cdot,Z)^m$. So $A=\prod_{\chi\in Z}Q_\chi^m$ satisfies $|\tilde A|\le C|\tilde P|$ on $\mathbb T^k$, hence $|A|\le C|P|$ on the line. $\square$ Proposition 8.4 contains Proposition 8.3. For $P=p(m^{-s})$ with any integer $m\ge2$, the factors $1-\bar z_j\sqrt m\,m^{-s}$ over the zeros $z_j$ of $p(m^{-1/2}z)$ on $\mathbb T$, repeated with multiplicity, dominate $P$ when $p(0)\ne0$. This covers $1-\sqrt6\,6^{-s}$, whose symbol vanishes on the curve $z_2z_3=1$. It also covers $1-2\cdot2^{-s}-3^{-s}$, whose symbol $1-\sqrt2z_2-z_3/\sqrt3$ vanishes transversally at two conjugate points of $\mathbb T^2$. There, numerically, $\sup_{\mathbb T^2}|\tilde Q_+\tilde Q_-/\tilde P|\approx9.52$ (`code/torus_domination.py`). It covers the zero-free example $1+\tfrac12 2^{1/2-s}+\tfrac12 3^{1/2-s}$ of u/bramble_trace, which has $\inf_{\text{line}}|P|=0$, $J_P=S$ and $Q=P$. The case left open is a symbol whose torus zero set has positive dimension but is not dominated by products of positive-real-part factors. The argument does not reach $\zeta$, whose Bohr symbol involves all primes and is not a polynomial. For comparison, in the Hardy space of Dirichlet series with square-summable coefficients (the $\ell^2$ dilation problem), a Dirichlet polynomial is cyclic iff its Bohr lift has no zeros in the open polydisc. This is due to Neuwirth–Ginsberg–Newman [17]; see also Hedenmalm–Lindqvist–Seip [18] and Nikolski [19]. Those results concern zero-free polynomials. Propositions 8.1–8.4 instead identify $\overline{PS}\subset S$ as a jet kernel when $P$ has zeros in $\Omega$. For the analytic-function-space framework of [3, 4] and the zero-free half-plane criteria derived from it, see Ghosh–Kremnizer–Noor–Santos [15] and Manzur–Noor–Quintero [16]. 2. **Rate.** Under FOZ (finitely many off-line zeros, $|Z^+|<\infty$), $V_N\subset V$ gives $d_N^2=D_\infty^2+\operatorname{dist}(u,V_N)^2$. Theorem F gives the lower bound. We conjecture that it is sharp: *Conjecture R* [FOZ]. With $u=P_VE$, $$(d_N^2-D_\infty^2)\log N\ \longrightarrow\ \sum_{\operatorname{Re}\rho=1/2}m(\rho)^2\,|u(\rho)|^2 .$$ Under RH with simple zeros this is the Bettin–Conrey–Farmer asymptotic, which they prove only under an extra hypothesis on $\sum1/|\zeta'(\rho)|^2$ [6]. Their result is for $\delta_N$. It transfers to $d_N$ whenever the near-optimal polynomials $A_N$ satisfy $A_N(1)=o((\log N)^{-1/2})$. Replacing $a_1$ by $a_1-A_N(1)$ enforces $A_N(1)=0$ and changes the distance by at most $|A_N(1)|\,\|\zeta/s\|_{L^2(\text{line})}$. For instance, the weights $\mu(n)(1-\log n/\log N)$ give $A_N(1)=(\log N)^{-1}\sum_{n\le N}\mu(n)n^{-1}\log(N/n)\sim1/\log N$ (`code/bcf_AN1.py`). Toy-model evidence: in Burnol's disc setting [9, Thm 1.4] with $Q(z)=(1-z/\omega)(1-z/a)$, $|\omega|=1$, $|a|<1$, we computed $E(N)=\min_{\deg A\le N}\|1-QA\|^2$ numerically, in double precision (`code/toy_rate2.py`). For nine choices of $(\omega,a)$ and $N\le1600$, $N(E(N)-E_\infty)\to|a|^2=|(P_{bH^2}1)(\omega)|^2$ to three digits. That is, the projection onto the *approximable* part, evaluated at the boundary zero, governs the rate. This is the analogue of $|u(\rho)|^2$. The toy statement can also be derived. Let $u=P_{bH^2}1=\overline{b(0)}\,b$. Then $1-u\perp bH^2\supset Q\mathcal P_N$, so $E(N)-E_\infty=\operatorname{dist}(u,Q\mathcal P_N)^2$. Division by $b$ is isometric, and $Q/b=-(1-z/\omega)(1-\bar az)/a$. Hence $E(N)-E_\infty=\min_A\|\overline{b(0)}-(1-z/\omega)(1-\bar az)A\|^2$. This is Burnol's problem [9, Thm 1.4 and Note 2.2] with one root on the circle and one outside the closed disc, and its limit is $|b(0)|^2=|a|^2$. The toy proof uses the fact that $\mathcal P_N$ is stable under the division by $b$ step, since everything is a polynomial in $z$. The same mechanism works inside $S$ when $P$ is a polynomial in $2^{-s}$. Then $\sqrt2M_2$ is an isometry with wandering vector $e_1$, $E=\sum_j2^{-j/2}(\sqrt2M_2)^je_1$, and for $M=2^d-1$ we have $\operatorname{dist}(E,P\operatorname{span}\{e_m\}_{m\le M})^2=\tfrac12\min_{\deg A