\documentclass[11pt,reqno]{amsart} \usepackage[T1]{fontenc} \usepackage{lmodern} \usepackage{amsmath,amssymb,mathtools} \usepackage[letterpaper,textwidth=6.25in,textheight=9in,centering]{geometry} \usepackage[hidelinks]{hyperref} \numberwithin{equation}{section} \renewcommand{\Re}{\operatorname{Re}} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \title[B\'aez--Duarte spectral synthesis]{Spectral synthesis for the B\'aez--Duarte step space} \author[W. Howe]{Will Howe} \email[Will Howe]{howew@purdue.edu} \author[G. Huffman]{Grant Huffman} \email[Grant Huffman]{huffma97@purdue.edu} \author[J. Kohlmeyer]{Jensen Kohlmeyer} \email[Jensen Kohlmeyer]{jenwin@purdue.edu} \author[L. Kruer]{Liam Kruer} \email[Liam Kruer]{kruerl@purdue.edu} \date{November 5, 2026} \subjclass[2020]{Primary 11M26; Secondary 30H10} \keywords{Riemann zeta function, B\'aez--Duarte criterion, spectral synthesis, Hardy space, Blaschke product} \hypersetup{pdftitle={Spectral synthesis for the Baez-Duarte step space},pdfauthor={Will Howe; Grant Huffman; Jensen Kohlmeyer; Liam Kruer},pdfsubject={Spectral synthesis, the Riemann zeta function, and Hardy spaces},pdfkeywords={Baez-Duarte, Nyman-Beurling, spectral synthesis, Hardy space}} \begin{document} \begin{abstract} We prove that the closed B\'aez--Duarte subspace of the Mellin step space is exactly the common kernel of the derivative evaluations prescribed by the zeros of the Riemann zeta function in $\operatorname{Re}s>1/2$, with their multiplicities. No finiteness or sparsity assumption on this zero set is required. The proof combines a uniform fractional regularity estimate for Mellin transforms of step functions with a comparison obtained by reflecting zeros in a shifted vertical line. These estimates produce a holomorphic Hardy-space family whose membership in the B\'aez--Duarte subspace continues from large positive shifts to every sufficiently small positive shift. A uniformly bounded weak limit then gives the synthesis theorem. \end{abstract} \maketitle \section{The step space and the synthesis theorem} For $a\geq0$, set \[ \Omega_a=\{s\in\mathbb{C}:\operatorname{Re}s>\tfrac12+a\}, \qquad \Omega=\Omega_0. \] We use the normalization \begin{equation} \|h\|_{H^2(\Omega_a)}^2 =\sup_{\sigma>1/2+a}\frac{1}{2\pi}\int_{\mathbb{R}} |h(\sigma+it)|^2\,dt. \label{eq:1.1} \end{equation} The Mellin transform \[ (\mathcal{M}h)(s)=\int_0^1 h(u)u^{s-1}\,du \] is an isometry from $L^2(0,1)$ onto $H^2(\Omega)$. Let $S$ be the Mellin image of the closed subspace of functions constant almost everywhere on each interval \[ C_m=\left(\frac{1}{m+1},\frac{1}{m}\right],\qquad m\in\mathbb{N}. \] For $f\in S$, write $f_m$ for its value on $C_m$ before taking the Mellin transform. Thus \begin{equation} \|f\|_S^2=\sum_{m\geq1}\frac{|f_m|^2}{m(m+1)}, \qquad e_m(s)=\frac{m^{-s}-(m+1)^{-s}}{s} \label{eq:1.2} \end{equation} are, respectively, the norm and an orthogonal basis of $S$. This is the Mellin formulation of the B\'aez--Duarte system~\cite{baez,bagchi}. For each integer $l\geq2$, define \begin{equation} G_l(s)=\bigl(l^{-s}-l^{-1}\bigr)\frac{\zeta(s)}{s}, \qquad V=\overline{\operatorname{span}\{G_l:l\geq2\}}^{S}. \label{eq:1.3} \end{equation} The singularity at $s=1$ is removable. Indeed, $G_l$ is the Mellin transform of the negative of the step function \begin{equation} q_l(u)=\left\{\frac{1}{lu}\right\}-\frac{1}{l}\left\{\frac{1}{u}\right\}. \label{eq:1.4} \end{equation} Here $\{\cdot\}$ denotes fractional part. On $C_m$, $q_l=m/l-\lfloor m/l\rfloor$, so $G_l\in S$. Let $Z^+$ be the set of zeros of $\zeta$ in $\Omega$, and let $m(\rho)$ be the multiplicity of $\rho$. Define \begin{equation} J=\bigl\{f\in S:f^{(j)}(\rho)=0\text{ for all }\rho\in Z^+\text{ and }0\leq j0$ and $0<\alpha_0<1/2$. There is a constant $C=C(A,\alpha_0)$ such that, for every $f\in S$, \begin{equation} \frac{1}{2\pi}\int_{\mathbb{R}}(1+|t|)^{2\alpha}|f(\tfrac12+a+it)|^2\,dt \leq C\|f\|_S^2 \label{eq:2.1} \end{equation} whenever $0\leq a\leq A$ and $0\leq\alpha\leq\min(a,\alpha_0)$. At $a=0$ the boundary values are understood in the Hardy-space sense. Moreover, \begin{equation} (s+1)^\alpha f(s)\in H^2(\Omega_a),\qquad \|(s+1)^\alpha f(s)\|_{H^2(\Omega_a)}\leq C'\|f\|_S. \label{eq:2.2} \end{equation} The power uses the holomorphic logarithm on $\operatorname{Re}(s+1)>0$. \end{lemma} \begin{proof} Put $u=e^{-y}$ and extend all functions of $y$ by zero to $y<0$. The function $t\mapsto f(\tfrac12+a+it)$ is the Fourier transform of \[ h_a(y)=\sum_{m\geq1}h_{a,m}(y),\qquad h_{a,m}(y)=f_m e^{-(1/2+a)y}\mathbf{1}_{I_m}(y), \] where \[ I_m=[\log m,\log(m+1)),\qquad \ell_m=|I_m|\asymp m^{-1}. \] If $M_m=\|h_{a,m}\|_2^2$, then \begin{equation} M_m\leq m^{-2a}\frac{|f_m|^2}{m(m+1)}. \label{eq:2.3} \end{equation} For $0<\alpha<1/2$, the Fourier characterization of the fractional seminorm is \begin{equation} \frac{1}{2\pi}\int_{\mathbb{R}}|t|^{2\alpha}|\widehat h(t)|^2\,dt =d_\alpha\int_{\mathbb{R}}\int_{\mathbb{R}} \frac{|h(y)-h(v)|^2}{|y-v|^{1+2\alpha}}\,dy\,dv, \label{eq:2.4} \end{equation} where \[ d_\alpha^{-1}=\int_{\mathbb{R}}\frac{|e^{ir}-1|^2}{|r|^{1+2\alpha}}\,dr, \qquad d_\alpha\asymp\alpha\quad(0<\alpha\leq\alpha_0). \] The identity follows from Plancherel and Tonelli; see also~\cite{hitchhiker}. The comparison near $\alpha=0$ follows by splitting the last integral at $|r|=1$: the tail is of order $1/\alpha$ and the integral near zero stays bounded on the indicated range. Consider one cell $I=[L,R)$ of length $\ell=\ell_m$. The maximum and minimum of $e^{-(1/2+a)y}$ on $I$ have uniformly bounded ratio. The part of the double integral in \eqref{eq:2.4} with one variable outside $I$ is therefore bounded by \begin{equation} \frac{CM_m\ell^{-2\alpha}}{\alpha(1-2\alpha)}. \label{eq:2.5} \end{equation} Indeed, integration in the outside variable produces the factors $(y-L)^{-2\alpha}/(2\alpha)$ and $(R-y)^{-2\alpha}/(2\alpha)$, while $\|h_{a,m}\|_\infty^2\leq CM_m/\ell$. For $y,v\in I$, the mean value theorem gives \[ |h_{a,m}(y)-h_{a,m}(v)|^2\leq C(M_m/\ell)|y-v|^2. \] Consequently the contribution from $I\times I$ is at most $CM_m\ell^{2-2\alpha}\leq CM_m\ell^{-2\alpha}$. Multiplying these estimates by $d_\alpha\ll\alpha$ yields \begin{equation} \|h_{a,m}\|_2^2+\frac{1}{2\pi}\int_{\mathbb{R}}|t|^{2\alpha}|\widehat h_{a,m}(t)|^2\,dt \leq C\ell_m^{-2\alpha}M_m, \label{eq:2.6} \end{equation} uniformly for $0<\alpha\leq\alpha_0$. At any pair $(y,v)$, at most two of the differences $h_{a,m}(y)-h_{a,m}(v)$ are nonzero. Hence, for finite sums, \[ |h_a(y)-h_a(v)|^2\leq2\sum_m|h_{a,m}(y)-h_{a,m}(v)|^2. \] Using \eqref{eq:2.3} and \eqref{eq:2.6}, the total cost is bounded by \[ C\sum_{m\geq1}m^{2\alpha-2a}\frac{|f_m|^2}{m(m+1)}\leq C\|f\|_S^2. \] Finite truncations converge in the corresponding fractional norm, so the same estimate holds for every $f\in S$. Since $(1+|t|)^{2\alpha}\leq C(1+|t|^{2\alpha})$ uniformly in $\alpha$, this proves \eqref{eq:2.1} for $\alpha>0$. For $\alpha=0$ it is Plancherel's theorem. Finally, $f|_{\Omega_a}\in H^2(\Omega_a)$ and $(s+1)^\alpha$ is the reciprocal of a bounded outer function there. Thus their product is in the Smirnov class. The weights $|\tfrac32+a+it|^\alpha$ and $(1+|t|)^\alpha$ are uniformly comparable in the prescribed range. The Smirnov boundary theorem and \eqref{eq:2.1} now imply \eqref{eq:2.2}. \end{proof} \begin{remark}\label{rem:inner-division} Suppose that $B$ is a Blaschke product on $\Omega_a$ and that $f\in S$ vanishes at all its zeros with at least their multiplicities. Under the hypotheses of Lemma~\ref{lem:regularity}, inner division gives \begin{equation} \left\|\frac{(s+1)^\alpha f(s)}{B(s)}\right\|_{H^2(\Omega_a)} =\|(s+1)^\alpha f(s)\|_{H^2(\Omega_a)}\leq C'\|f\|_S. \label{eq:2.7} \end{equation} In particular, the constant does not depend on the number or separation of the zeros of $B$. \end{remark} \section{Reflection of the completed zeta function} Write \begin{equation} \xi(s)=A(s)\zeta(s),\qquad A(s)=\tfrac12 s(s-1)\pi^{-s/2}\Gamma(s/2),\qquad X(z)=\xi(\tfrac12+z). \label{eq:xi} \end{equation} The function $X$ is even, entire of order one, and satisfies $X(\overline z)=\overline{X(z)}$. Its zeros lie in $|\Re z|<1/2$ and none are real. We use the usual canonical product and zero-counting formula for $\xi$~\cite{titchmarsh}. Let $\nu(\beta,\gamma)$ be the multiplicity of $X$ at $\beta+i\gamma$. For $0\le a<1/2$, define \begin{equation} B_a(z)=\prod_{\substack{\beta+i\gamma\in Z(X)\\\beta>a,\ \gamma>0}} \left(\frac{(z-\beta)^2+\gamma^2}{(z+\beta-2a)^2+\gamma^2}\right)^{\nu(\beta,\gamma)}. \label{eq:Ba} \end{equation} Each zero with $\beta>a$ and positive ordinate is indexed; the displayed factor includes its conjugate. Each factor equals one at $z=a$. The product is a Blaschke product on $\Re z>a$ and extends meromorphically to the plane. In fact, the paired factors differ from one by $O_K(\gamma^{-2})$ on any fixed compact set away from the poles, and the multiplicity-weighted inverse-square sum over these zeros converges. Set \begin{equation} D_a(z)=X(z)/B_a(z). \label{eq:Da} \end{equation} The quotient extends to an entire function. It replaces the zeros $\beta\pm i\gamma$, with $\beta>a$, by $2a-\beta\pm i\gamma$; every other zero stays in place. In particular, $D_a$ has no zeros in $\Re z>a$. For clarity, the canonical factor of $X$ associated with an off-line quadruple $\{\pm\beta\pm i\gamma\}$, $\beta>0$, is \begin{equation} \frac{\bigl((z-\beta)^2+\gamma^2\bigr) \bigl((z+\beta)^2+\gamma^2\bigr)}{(\beta^2+\gamma^2)^2}. \label{eq:quadruple} \end{equation} If $\beta>a$, its counterpart in $D_a$ replaces the first numerator factor by $(z+\beta-2a)^2+\gamma^2$ and leaves the positive denominator unchanged. A line-zero pair contributes $(z^2+\gamma^2)/\gamma^2$. These products converge locally uniformly and have the same prefactor $X(0)$, so no additional exponential factor is introduced. \begin{lemma}[Real-shift comparison]\label{lem:realshift} If $x>0$, $0\le a<1/2$, and $a\le x/\sqrt2$, then \begin{equation} |X(it)|\le |D_a(x+it)|\qquad(t\in\mathbb R). \label{eq:realshift} \end{equation} \end{lemma} \begin{proof} We compare the canonical factors at each ordinate; the normalizing denominators are identical. For a zero on the line, the required inequality is $|u|\le |x+iu|$, with $u=t-\gamma$, and the same inequality holds at $-\gamma$. For an off-line pair at ordinate $\gamma$, the magnitude contributed at $it$ is $\beta^2+u^2$. If $0<\beta\le a$, the pair is unchanged, and \[ \begin{split} \bigl((x+\beta)^2+u^2\bigr)\bigl((x-\beta)^2+u^2\bigr) -(\beta^2+u^2)^2 \\=x^2(2u^2+x^2-2\beta^2)\ge0. \end{split} \] If $\beta>a$, put $r_1=x+\beta$ and $r_2=x+\beta-2a$. Both are positive, and \[ r_1r_2-\beta^2=2(x-a)\beta+x(x-2a)\ge x^2-2a^2\ge0. \] It follows that $r_1^2+r_2^2\ge2r_1r_2\ge2\beta^2$, whence \[ (r_1^2+u^2)(r_2^2+u^2)\ge(\beta^2+u^2)^2. \] The same comparison applies at $-\gamma$. Multiplying the comparisons over finite products and passing to their locally uniform limits proves \eqref{eq:realshift}. \end{proof} \begin{lemma}[Vertical variation]\label{lem:vertical} Fix $0\le a<1/2$ and a compact interval $I\subset(a,\infty)$. There is $C=C(a,I)$ such that \begin{equation} \left|\frac{\partial}{\partial t}\log|D_a(x+it)|\right| \le C\log(2+|t|),\qquad x\in I,\ t\in\mathbb R. \label{eq:vertical} \end{equation} Consequently, if also $a\le x/\sqrt2$ for $x\in I$, then, uniformly for $x\in I$ and $|v|\le1$, \begin{equation} \frac{|X(it)|}{|D_a(x+i(t+v))|}\le C_1(1+|t|)^{C_2|v|}. \label{eq:verticalratio} \end{equation} \end{lemma} \begin{proof} Let $b\pm i\gamma$ run over the conjugate pairs of zeros of $D_a$, including multiplicity. All $b$ lie in $[-1/2,a]$. Differentiation of the paired canonical product gives terms \[ K_r(t-\gamma)+K_r(t+\gamma),\qquad K_r(u)=\frac{u}{r^2+u^2},\qquad r=x-b. \] For $x\in I$, the distances $r$ belong to a fixed compact subinterval of $(0,\infty)$. Thus \[ |K_r(u)-K_1(u)|\le C(1+|u|)^{-3}. \] The unit-interval zero bound $N(T+1)-N(T)=O(\log(2+T))$ shows that replacing all the kernels by $K_1$ incurs an error $O(\log(2+|t|))$. Here $N(T)$ counts the positive ordinates of all nontrivial zeta zeros, with multiplicity; reflection preserves this counting function. Write \begin{equation} N(T)=M(T)+E(T),\qquad M(T)=\frac{T}{2\pi}\log\frac{T}{2\pi}-\frac{T}{2\pi},\qquad E(T)=O(\log(2+T)). \label{eq:zero-count} \end{equation} By conjugation it suffices to consider $t\ge10$. For $\gamma>2t$, \[ K_1(t-\gamma)+K_1(t+\gamma)=O(t/\gamma^2), \] and summing by \eqref{eq:zero-count} gives $O(\log t)$. The ranges $\gamma0$, gives \begin{equation} \left|\frac{A(s+w)}{A(s)}\right|\le C(1+|t|)^{x/2}. \label{eq:stirling} \end{equation} The denominator has no zero on $\Re s=1/2$; bounded $t$ is handled by continuity. Thus the hypotheses of Lemma~\ref{lem:vertical} give \begin{equation} \left|\frac{\zeta(s)}{\zeta(s+w)/B_a(s+w-1/2)}\right| \le C(1+|t|)^{x/2+C_2|v|}. \label{eq:complexshift} \end{equation} For real shifts, there is a useful uniform version: for each fixed $x_*>0$, \begin{equation} \left|\frac{\zeta(s)}{\zeta(s+x)/B_a(s+x-1/2)}\right| \le C_{x_*}(1+|t|)^{x/2} \label{eq:uniform-realshift} \end{equation} when $00$, is also outer on $\Omega$. \end{lemma} \begin{proof} The classical convexity estimate in a bounded vertical strip and absolute convergence farther to the right imply boundedness of $Z$. The pole at $1$ is canceled, and its zeros in $\Omega$ are precisely those of $\zeta$, with their multiplicities. The function $Z$ extends analytically across every finite point of the boundary line. Away from its isolated boundary zeros this excludes singular inner measure; at a zero of finite order its logarithmic boundary behavior excludes a singular atom as well. Thus its inner factor has no singular measure on the finite boundary. A nonconstant singular inner factor at infinity would contain $e^{-c(s-1/2)}$ for some $c>0$, which is impossible since $Z(\sigma)\sim1/\sigma$ as $\sigma\to+\infty$. The remaining inner factor is exactly $B_0(s-1/2)$, proving (4.1). After scaling $O$ to have modulus at most one, its log modulus is the Poisson extension of its boundary log modulus, with no linear term at infinity. The Poisson semigroup identity shows the same assertion after a positive horizontal translate. A vertical translate merely translates the boundary data. This proves the last assertion. \end{proof} Fix $f\in J$. Define \begin{equation}\label{eq:g-definition} g(s)=\frac{sf(s)}{\zeta(s)}. \end{equation} The prescribed zero jets remove every possible pole in $\Omega$, and $g(1)=0$. For $\operatorname{Re}w>0$ put \begin{equation}\label{eq:F-definition} F_w(s)=\frac{\zeta(s)}{s}\bigl(g(s+w)-g(1+w)\bigr). \end{equation} This is holomorphic in $s\in\Omega$: the numerator in parentheses vanishes at $s=1$, canceling the only pole of $\zeta(s)/s$. \begin{proposition} There is an open connected set $U\subset\{\operatorname{Re}w>0\}$ containing $(0,3/4)$ such that $w\mapsto F_w$ is an $H^2(\Omega)$-valued holomorphic map on $U$. Its Hardy norm is locally bounded on $U$. \end{proposition} \begin{proof} We first place $F_w$ in the Smirnov class. Since $f\in J$, inner division gives $f(s)=B_0(s-1/2)f_1(s)$ with $f_1\in H^2(\Omega)$. Lemma 4.1 yields \[ g(s)=\frac{s(s-1)}{(s+1)^2}\frac{f_1(s)}{O(s)}. \] For each $\operatorname{Re}w>0$, the translate $f_1(s+w)$ belongs to $H^2(\Omega)$ and $O(s+w)$ is bounded and outer. Thus $g(s+w)$ and $h_w(s)=g(s+w)-g(1+w)$ are in the Smirnov class on $\Omega$. The function \[ b_1(s)=\frac{s-1}{s} \] is the Blaschke factor of $\Omega$ at $1$. Since $h_w(1)=0$, $h_w/b_1$ remains in the Smirnov class, and \begin{equation}\label{eq:F-smirnov} F_w(s)=Z(s)\frac{(s+1)^2}{s^2}\frac{h_w(s)}{b_1(s)} \end{equation} does also. Fix $x_0\in(0,3/4)$. Choose \begin{equation}\label{eq:choose-alpha-a} \frac{x_0}{2}<\alpha1/2+a$ and $|z+1|^\alpha\asymp(1+|t|)^\alpha$, uniformly in $U_{x_0}$. Evaluating the Hardy norm on this interior vertical line therefore shows that the boundary function in (4.7) is locally bounded in $L^2(\mathbb{R})$. The remaining boundary term is $g(1+w)\zeta(s)/s$. Its $L^2$ norm is locally bounded because $g(1+w)$ is holomorphic and \begin{equation}\label{eq:zeta-integrability} \int_{\mathbb{R}}\left|\frac{\zeta(1/2+it)}{1/2+it}\right|^2\,dt<\infty. \end{equation} For example, the convexity estimate $\zeta(1/2+it)\ll(1+|t|)^{1/3}$ suffices for this integrability. The two boundary terms are used here only as $L^2$ functions; their poles at $s=1$ cancel in the complete expression (4.3). Combining the boundary bound with (4.4), the Smirnov boundary theorem gives $F_w\in H^2(\Omega)$ with locally bounded norm. For each fixed $s\in\Omega$, including the removable value at $1$, $F_w(s)$ is holomorphic in $w$. Testing against the reproducing kernels of $H^2(\Omega)$ gives weak holomorphy on their linear span. This span is dense; local norm boundedness extends holomorphy to every bounded linear functional. The standard weak-to-strong holomorphy principle now gives Hardy-space-valued holomorphy on $U_{x_0}$. The union of these neighborhoods over $x_0\in(0,3/4)$ is open and connected and proves the proposition. \end{proof} \section{Membership, continuation, and the zero-shift limit} \begin{lemma}[Membership at large shifts] For every $f\in S$ and real $x\in(1/2,3/4)$, the expression (4.3), formed using $g(z)=zf(z)/\zeta(z)$ at the translated arguments, belongs to $V$. \end{lemma} \begin{proof} First suppose $f$ has finite step support. Then \[ P(s)=sf(s) \] is a Dirichlet polynomial. On $\operatorname{Re}s>1$, \[ \frac{P(s)}{\zeta(s)}=\sum_{l\ge1}c_l l^{-s},\qquad \sup_l|c_l|<\infty, \] because $1/\zeta(s)=\sum_{n\ge1}\mu(n)n^{-s}$ converges absolutely there and $P$ has finitely many coefficients. For $x>1/2$ this gives \begin{equation}\label{eq:large-shift-expansion} F_x=\sum_{l\ge2}c_l l^{-x}G_l. \end{equation} To justify norm convergence, use (1.4). On $(0,1/l]$ one has $|q_l|\le1+1/l$, while on $[1/l,1]$, \[ q_l(u)=\frac1l\left\lfloor\frac1u\right\rfloor. \] Hence \[ \|G_l\|_{H^2(\Omega)}^2=\|q_l\|_{L^2(0,1)}^2 \le\frac{C}{l}+\frac1{l^2}\int_{1/l}^1u^{-2}\,du\le\frac{C'}l. \] The series in (5.1) is therefore absolutely convergent in $H^2(\Omega)$, since $\sum_l l^{-x-1/2}<\infty$. It agrees pointwise with (4.3) in an absolutely convergent half-plane and hence throughout $\Omega$. Thus $F_x\in V$. Now let $f\in S$ be arbitrary and let $f^{(N)}$ be its finite step truncations. For fixed $x>1/2$, absolute convergence gives a uniform bound for $1/\zeta(s+x)$ on $\operatorname{Re}s=1/2$. The convexity bound used in (4.8) and Lemma 2.1, with $a=x$ and $\alpha=1/3$, imply \[ \left\|\frac{s+x}{s}\frac{\zeta(s)}{\zeta(s+x)} \bigl(f^{(N)}(s+x)-f(s+x)\bigr)\right\|_{L^2(\operatorname{Re}s=1/2)} \le C_x\left\|f^{(N)}-f\right\|_{S}. \] Evaluation at $1+x$ is also continuous in the step norm. It follows that the boundary expressions for $F_x^{(N)}$ converge in $L^2$. Since $F_x^{(N)}\in V\subset H^2(\Omega)$, they converge in Hardy norm to an element of $V$. Interior point evaluation identifies this limit with $F_x$ as defined in (4.3). No zero-jet condition on the truncations has been used. \end{proof} \begin{proof}[Proof of Theorem 1.1] It remains to prove $J\subseteq V$. Fix $f\in J$ and form $F_w$ as in Proposition 4.2. Let $L\in(H^2(\Omega))^*$ be any continuous complex-linear functional with $L|_{V}=0$. The scalar function $w\mapsto L(F_w)$ is holomorphic on the connected set $U$ of Proposition 4.2. Lemma 5.1 shows that it vanishes for every $x\in(1/2,3/4)$. By the identity theorem it vanishes throughout $U$. The space $S$ is closed in $H^2(\Omega)$, and $V$ is closed in $S$, hence $V$ is also closed in $H^2(\Omega)$. The Hahn--Banach separation theorem therefore gives \begin{equation}\label{eq:all-shift-membership} F_x\in V\qquad(0