Spectral synthesis for the Báez–Duarte step space

The step space and the synthesis theorem

For $a \ge0$, set

We use the normalization

The Mellin transform

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

For $f \in S$, write $f_m$ for its value on $C_m$ before taking the Mellin transform. Thus

are, respectively, the norm and an orthogonal basis of $S$. This is the Mellin formulation of the Báez–Duarte system [1, 2].

For each integer $l \ge2$, define

The singularity at $s = 1$ is removable. Indeed, $G_l$ is the Mellin transform of the negative of the step function

Here $\{\cdot\}$ denotes fractional part. On $C_m$, $q_l = m/l - \lfloor m/l\rfloor$, so $G_l \in S$.

Let $\mathbb{Z}^{+}$ be the set of zeros of $\zeta$ in $\Omega$, and let $m(\rho)$ be the multiplicity of $\rho$. Define

Interior derivative evaluations are continuous on $H^2(\Omega)$, so $J$ is closed. The formula for $G_l$ gives $V\subseteq J$.

The theorem identifies the closed subspace determined by the off-line zero jets. It does not assert that $\mathbb{Z}^{+}$ is empty. The finite-zero case and a conditional infinite-zero case are treated in [5]; the proof below does not use either result.

Throughout, implied constants may depend on explicitly fixed compact parameter ranges. We use the standard inner–outer factorization and Smirnov boundary theorem for half-plane Hardy spaces [4]. In particular, a function in the Smirnov class with square-integrable boundary values belongs to $H^2$, and division by a Blaschke product whose zeros are canceled preserves the $H^2$ norm.

Uniform fractional regularity of Mellin steps

Proof. Put $u=e^{-y}$ and extend all functions of $y$ by zero to $y<0$. The function $t\mapsto f(\tfrac{1}{2}+a+it)$ is the Fourier transform of

where

If $M_m=\|h_{a,m}\|_2^2$, then

For $0<\alpha<1/2$, the Fourier characterization of the fractional seminorm is

where

The identity follows from Plancherel and Tonelli; see also [3]. 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 (9) with one variable outside $I$ is therefore bounded by

Indeed, integration in the outside variable produces the factors $(y-L)^{-2\alpha}/(2\alpha)$ and $(R-y)^{-2\alpha}/(2\alpha)$, while $\|h_{a,m}\|_\infty^2\le CM_m/\ell$. For $y,v\in I$, the mean value theorem gives

Consequently the contribution from $I\times I$ is at most $CM_m\ell^{2-2\alpha}\le CM_m\ell^{-2\alpha}$. Multiplying these estimates by $d_\alpha\ll\alpha$ yields

uniformly for $0<\alpha\le\alpha_0$.

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,

Using (8) and (11), the total cost is bounded by

Finite truncations converge in the corresponding fractional norm, so the same estimate holds for every $f\in S$. Since $(1+|t|)^{2\alpha}\le C(1+|t|^{2\alpha})$ uniformly in $\alpha$, this proves (6) 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 $|\frac{3}{2}+a+it|^\alpha$ and $(1+|t|)^\alpha$ are uniformly comparable in the prescribed range. The Smirnov boundary theorem and (6) now imply (7).

Remark 2.2. 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 2.1, inner division gives

In particular, the constant does not depend on the number or separation of the zeros of $B$.

Reflection of the completed zeta function

Write

The function $X$ is even, entire of order one, and satisfies $X(\overline{z})=\overline{X(z)}$. Its zeros lie in $|\operatorname{Re}z|<1/2$ and none are real. We use the usual canonical product and zero-counting formula for $\xi$ [6].

Let $\nu(\beta,\gamma)$ be the multiplicity of $X$ at $\beta+i\gamma$. For $0\le a<1/2$, define

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 $\operatorname{Re}z>a$ 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

The quotient extends to an entire function. It replaces the zeros $\beta+i\gamma$, with $\beta>a$, by $2a-\beta\pm i\gamma$; every other zero stays in place. In particular, $D_a$ has no zeros in $\operatorname{Re}z>a$.

For clarity, the canonical factor of $X$ associated with an off-line quadruple $\{\pm\beta\pm i\gamma\}$, $\beta>0$, is

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.

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

If $\beta>a$, put $r_1=x+\beta$ and $r_2=x+\beta-2a$. Both are positive, and

It follows that $r_1^2+r_2^2\ge2r_1r_2\ge2\beta^2$, whence

The same comparison applies at $-\gamma$. Multiplying the comparisons over finite products and passing to their locally uniform limits proves (3.5). □

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

For $x\in I$, the distances $r$ belong to a fixed compact subinterval of $(0,\infty)$. Thus

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

By conjugation it suffices to consider $t\ge10$. For $\gamma>2t$,

and summing by (20) gives $O(\log t)$. The ranges $\gamma<t/2$ and $3t/2<\gamma\le2t$ also contribute $O(\log t)$, since each kernel is $O(1/t)$ there. On the remaining range $[t/2,3t/2]$, the terms $K_1(t+\gamma)$ satisfy the same bound.

The only remaining sum is

The contribution of $dE$ is $O(\log t)$ by integration by parts, because $K_1'$ is integrable on $\mathbb{R}$ and the endpoint terms are bounded by $O(\log t/t)$. For the main term, the constant $M'(t)$ integrates to zero by oddness on this symmetric interval, and

Its remaining integral is therefore $O(1)$. The estimates on a fixed bounded range of $t$ follow from local convergence and the positive lower bound on $r$. This proves (18).

Integrating (18) between $t$ and $t+v$ and applying Lemma 3.1 proves (19). □

We record the form of these comparisons that will be used below. For $s=1/2+it$ and $w=x+iv$, (13) gives

Stirling’s formula, uniformly for $w$ in a fixed compact subset of $\operatorname{Re}w>0$, gives

The denominator has no zero on $\operatorname{Re}s=1/2$; bounded $t$ is handled by continuity. Thus the hypotheses of Lemma 3.2 give

For real shifts, there is a useful uniform version: for each fixed $x_*>0$,

when $0<x\le x_*$, $0\le a<1/2$, and $a\le x/\sqrt{2}$. Indeed, the comparison in Lemma 3.1 has constant one, while Stirling’s formula is uniform for $0\le x\le x_*$. All ratios are understood by removing their removable singularities.

A holomorphic family in the Hardy space

We first justify the boundary-to-Hardy step needed for the family.

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. □

Fix $f \in J$. Define

The prescribed zero jets remove every possible pole in $\Omega$, and $g(1) = 0$. For $\operatorname{Re} w > 0$ put

This is holomorphic in $s \in\Omega$: the numerator in parentheses vanishes at $s = 1$, canceling the only pole of $\zeta(s)/s$.

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

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

is the Blaschke factor of $\Omega$ at $1$. Since $h_w(1) = 0$, $h_w/b_1$ remains in the Smirnov class, and

does also.

Fix $x_0 \in(0,3/4)$. Choose

Choose a small compact interval $I$ about $x_0$ with $a<x/\sqrt{2}$ for $x\in I$. By making a complex neighborhood $U_{x_0}$ smaller if necessary, Lemma 3.2 gives, for $w=x+iv\in U_{x_0}$,

On $\operatorname{Re}s=1/2$, factor the first term of (27) as

The rational factor is uniformly bounded. By (23) and (30), the middle factor is bounded by $C(1+|t|)^\alpha$. Choose a fixed $\alpha_0\in(\alpha,1/2)$. The zeros of $B_a(s-1/2)$ are among the zeros canceled by $f$, so Lemma 2.1 and (12) give

For $z=s+w$ we have $\operatorname{Re}z=1/2+x>1/2+a$ 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 (31) 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

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 (27). Combining the boundary bound with (28), 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. □

Membership, continuation, and the zero-shift limit

Proof. First suppose $f$ has finite step support. Then

is a Dirichlet polynomial. On $\operatorname{Re}s>1$,

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

To justify norm convergence, use (1.4). On $(0,1/\ell]$ one has $|q_\ell|\le1+1/\ell$, while on $[1/\ell,1]$,

Hence

The series in (5.1) is therefore absolutely convergent in $H^2(\Omega)$, since $\sum_\ell\ell^{-x-1/2}<\infty$. 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

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. □

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

We next show that

Set

Then $\alpha<a<x/\sqrt{2}$. On the boundary $s=1/2+it$, (4.7) and the uniform real-shift estimate (3.12) imply

Lemma 2.1 and inner division, with fixed upper bounds for $a$ and $\alpha$, make the right-hand side uniformly bounded in $L^2(\mathbb{R})$. Explicitly,

and the vertical line $z=s+x$ lies inside $\Omega_{2x/3}$. The comparison of $|z+1|^{x/2}$ with $(1+|t|)^{x/2}$ is uniform for $0<x<1/4$.

For the subtraction term, the simple pole of $\zeta$ at $1$ and boundedness of evaluation near $1$ give

Its boundary norm therefore tends to zero by (32). Since each $F_x$ already belongs to $H^2(\Omega)$, the boundary estimates prove (35).

Finally, $g$ is holomorphic on $\Omega$ and $g(1)=0$. Thus, at every $s\in\Omega\setminus\{1\}$,

including the canceled zeta zeros by holomorphic continuation. At $s=1$ the removable value is $F_x(1)=g'(1+x)$, which tends to $g'(1)=f(1)$. Hence $F_x(s)\to f(s)$ at every point of $\Omega$. The uniform Hardy bound and density of the linear span of the reproducing kernels imply

By (34) and weak closedness of the closed linear subspace $V$, we obtain $f\in V$. Together with $V\subseteq J$, this proves the theorem. □

Acknowledgment of AI assistance

The authors acknowledge assistance from OpenAI’s AI tools in refining the argument, checking intermediate steps, and preparing the mathematical exposition.

Email address, Will Howe: [email protected]

Email address, Grant Huffman: [email protected]

Email address, Jensen Kohlmeyer: [email protected]

Email address, Liam Kruer: [email protected]