A Fabry obstruction to the separated meromorphic Pick construction of the Riemann xi function

1. Statement and relation to the KPS programme

Konstantopoulos, Patie and Sarkar [1] introduce the entire functions

$$ J_{u\phi(u)}(t)=\sum_{n=0}^{\infty} \frac{(-1)^nt^{2n}}{n!W_\phi(n+1)},\qquad W_\phi(1)=1,\quad W_\phi(z+1)=\phi(z)W_\phi(z). $$

Here $\phi$ is a Bernstein function, with the additional hypotheses specified in their respective results. Their Theorem 4.4 gives only real zeros when $\phi$ belongs to the stronger class $B_{P1}$ of meromorphic Pick functions with the following separation property. The zeros are $-\lambda_k$, the corresponding poles are $-\lambda_k-1$, and

$$ 0\leq\lambda_1,\qquad \lambda_{k+1}>\lambda_k+1. \tag{1} $$

The zeros and poles are simple. This pairing is an extra hypothesis; it does not follow merely from being a Pick function. We use the meromorphic class in [1, (4.14)], rather than a class allowing branch cuts. Degenerate finite sequences are treated separately below.

Let

$$ \xi(s)=\tfrac12s(s-1)\pi^{-s/2}\Gamma(s/2)\zeta(s), \qquad F_\xi(t)=\frac{\xi(1/2+it)}{\xi(1/2)}. $$

Theorem 1. There is no $\phi\in B_{P1}$ such that

$$ F_\xi(t)=J_{u\phi(u)}(t). \tag{2} $$

This excludes the separated meromorphic Pick sufficient mechanism for the normalized xi function. It does not exclude the larger class $D_P$ posed in [1, Section 4.1], the root-zero subclass generated by $B_J$, or other mechanisms for obtaining real zeros. In particular, it has no implication that RH is false.

We prove a general obstruction and then verify its hypotheses for the theta kernel. The mechanism is that regular variation of index one forces almost unit density of separated Mellin zeros. Sampling these zeros forces a density-zero set of changes of sign in the Taylor coefficients of the Fourier density. The classical Fabry sign-change theorem then forces a singularity at a point where that density is analytic.

2. General obstruction

Let $p$ be an even probability density on $\mathbb R$. Assume that it is analytic near zero, is positive there, and satisfies

$$ \int_{\mathbb R}e^{A|x|}p(x)\,dx<\infty \quad\text{for every }A>0. $$

Write

$$ p(x)=\sum_{n=0}^{\infty}c_nx^{2n},\qquad g(y)=\sum_{n=0}^{\infty}c_ny^n, $$

and set

$$ M(s)=\int_0^\infty x^{2s}p(x)\,dx,\qquad M(0)=\tfrac12. $$

The initial domain of $M$ is $\operatorname{Re}s>-1/2$. For real $s>1/2$, define

$$ \phi_p(s)=2(2s-1)\frac{M(s-1)}{M(s)}. \tag{3} $$

Theorem 2. Suppose that $g$ has finite positive Taylor radius $R$, that its sum extends holomorphically to a neighborhood of $+R$, and that $\phi_p$ is regularly varying of index one:

$$ \frac{\phi_p(as)}{\phi_p(s)}\longrightarrow a \quad(s\longrightarrow\infty),\qquad a>0. \tag{4} $$

If, in addition,

$$ \phi_p(s)\longrightarrow\infty, \qquad \phi_p(s)/s\longrightarrow0, \tag{5} $$

then the characteristic function of $p$ cannot equal $J_{u\phi(u)}$ for $\phi\in B_{P1}$.

The conditions in (5) remove the finite-pole cases. They hold for the xi density. They are not needed in the infinite paired case itself.

2.1. Identification of the entire Mellin factor

Suppose, for contradiction, that such a representation exists. The product law in [1, Lemma 5.8], specialized to the root-zero case, identifies its probability distribution as

$$ X\ \stackrel{d}{=}\ \sqrt{I_\phi}\,A, $$

where $A$ has the arcsine density $(\pi\sqrt{4-x^2})^{-1}$ on $(-2,2)$, independently of the subordinator exponential functional $I_\phi$. The standard Mellin formula is

$$ \mathbb E I_\phi^s=\frac{\Gamma(s+1)}{W_\phi(s+1)}. $$

These facts, including the distinguished zero-free Bernstein-gamma function $W_\phi$, are developed in [1, Section 5.2] and [2]. For $\operatorname{Re}s>0$ they give

$$ \frac{M(s)}{M(0)} =\mathbb E|X|^{2s} =\frac{4^s\Gamma(s+1/2)}{\sqrt\pi W_\phi(s+1)}. \tag{6} $$

Thus

$$ E(s):=\frac{\sqrt\pi\,4^{-s}M(s)}{M(0)\Gamma(s+1/2)} =\frac1{W_\phi(s+1)}. \tag{7} $$

This identification uses equality of probability laws, not uniqueness for a recurrence or interpolation at the positive integers. In particular, it introduces no unproved exclusion of periodic factors.

Taking the ratio of (6) at $s$ and $s-1$ proves $\phi(s)=\phi_p(s)$ for $s>1/2$. Hence (4) and (5) apply to the proposed Bernstein function itself.

Taylor subtraction at zero continues $M$ meromorphically to the plane. Its only possible poles are simple poles at $-n-1/2$, with residues $c_n/2$. For example, split the integral at a positive number inside the analytic radius of $p$, subtract a finite Taylor polynomial on the lower interval, and increase its degree. The upper interval is entire in $s$ by the moment hypotheses. The zeros of $1/\Gamma(s+1/2)$ cancel all these poles, so $E$ is entire. Evaluating (7) at the canceled poles gives the exact identity

$$ c_n=\frac{M(0)(-1)^n}{\sqrt\pi\,4^nn!} E(-n-1/2). \tag{8} $$

2.2. The Pick intervals have density one

A meromorphic Pick function with finitely many poles is rational, apart from its allowed nonnegative linear term. This follows directly from its Herglotz representation: its representing measure is supported on the finitely many poles, and the remaining entire Pick term is affine. Along the positive axis it is consequently bounded, or has a positive limiting ratio to $s$. Both possibilities contradict (5). There must therefore be an infinite paired sequence as in (1).

In the upper half-plane the principal logarithm $L=\log\phi$ has bounded harmonic imaginary part between zero and $\pi$. Its boundary imaginary part equals $\pi$ exactly on the intervals where $\phi$ is negative. With

$$ U=\bigcup_{k\geq1}(\lambda_k,\lambda_k+1), $$

these are the reflected intervals $-U$. The Poisson representation of a bounded harmonic function has its bounded boundary function as density; there is no linear harmonic term or singular boundary measure. Differentiating its holomorphic representation gives, for $s>0$,

$$ \frac{\phi'(s)}{\phi(s)} =\int_0^\infty\frac{\mathbf1_U(t)}{(s+t)^2}\,dt. \tag{9} $$

This also follows from the logarithmic Pick representation in the theory of complete Bernstein functions [3]. The bounded-harmonic argument specifies why no additional term can occur here.

Bernstein concavity and (4) imply

$$ \frac{\phi(2s)-\phi(s)}{\phi(s)} \leq\frac{s\phi'(s)}{\phi(s)} \leq\frac{2(\phi(s)-\phi(s/2))}{\phi(s)}, $$

so the middle expression tends to one. Since $\int_0^\infty s(s+t)^{-2}\,dt=1$, equation (9) yields

$$ \int_0^\infty\frac{s\mathbf1_{U^c}(t)}{(s+t)^2}\,dt=o(1). $$

The weight on $[0,s]$ is at least $1/(4s)$. Therefore

$$ |[0,s]\setminus U|=o(s),\qquad N_\lambda(s):=\#\{k:\lambda_k\leq s\}=s-o(s). \tag{10} $$

The second conclusion uses disjointness and unit length of the intervals; the endpoint discrepancy is at most one.

2.3. Reciprocal Bernstein-gamma zeros and coefficient signs

In this infinite paired situation, [1, Section 5.2.6] shows that

$$ F_\phi(z):=1/W_\phi(z) $$

extends entire, with precisely the simple zeros $-\lambda_k$. For completeness, its recurrence is

$$ F_\phi(z)=\phi(z)F_\phi(z+1). \tag{11} $$

Starting with its zero-free right half-plane [2, Theorem 4.1], continue left in strips of width one. Each zero $-\lambda_k$ of $\phi$ introduces a simple zero. The value $F_\phi(1-\lambda_k)$ is nonzero because two $\lambda$'s cannot differ by one. Each pole $-\lambda_k-1$ of $\phi$ cancels the simple zero of $F_\phi(z+1)$ at the same point. No other zeros or poles are introduced. This supplies the continuation argument used in [1] without relying on its cited working paper. A finite linear $\phi$ would instead produce a reciprocal-gamma lattice, which is why the finite case was removed before this step.

By (7), the real entire function

$$ f(x)=E(-x-1/2)=F_\phi(1/2-x) $$

has precisely the simple zeros $\lambda_k+1/2$. Their gaps exceed one, and their counting function satisfies

$$ N_f(x)=x-o(x). \tag{12} $$

Equation (8) says $\operatorname{sign}c_n=(-1)^n\operatorname{sign}f(n)$ and identifies zero coefficients with integer zeros of $f$. Let $c_m,c_n$ be consecutive nonzero coefficients, put $d=n-m$, and let $r$ count the zeros of $f$ in $(m,n)$. Endpoints are not zeros. Simplicity gives

$$ \operatorname{sign}(c_nc_m)=(-1)^{d+r}. $$

Separation gives $r\leq d$. A change of coefficient sign therefore consumes at least one unit of the nonnegative deficit $d-r$. Summing between consecutive nonzero coefficients from the first one to any later nonzero coefficient $c_n$ proves

$$ \#\{\text{coefficient sign changes through }n\} \leq n-N_f(n)+O(1)=o(n). \tag{13} $$

Zeros at intermediate integers are included in $r$. Thus no assumption that all coefficients are nonzero is needed. Since finite radius implies infinitely many nonzero coefficients, using the last nonzero index at most $N$ extends (13) to all $N$.

2.4. Fabry's theorem

For a set $S$ of indices, write $n_S(x)=\#(S\cap[0,x])$. The upper or maximum density appearing in the Fabry sign-change theorem is

$$ D_2(S)=\lim_{r\downarrow0}\limsup_{x\to\infty} \frac{n_S((1+r)x)-n_S(x)}{rx}. $$

Equation (13) implies $D_2(S)=0$: for each fixed $r>0$, the numerator is bounded by $o((1+r)x)$. The classical theorem, restated as Theorem A of Eremenko [4], says that a real power series of radius one whose coefficient sign changes have maximum density zero is singular at $+1$. This theorem does not impose a regularity condition on the magnitudes of its nonzero coefficients. The extra condition in Eremenko's stronger Theorem 1 is not being used.

Apply the theorem to $g(Rz)=\sum c_nR^nz^n$. It forces a singularity of $g$ at $+R$, contradicting the hypothesis of Theorem 2. This proves the general obstruction.

3. The xi density satisfies the obstruction

The standard Fourier density, up to a positive normalizing constant, is

$$ P(x)=\sum_{n\geq1} 2\pi n^2(2\pi n^2e^{2x}-3) \exp\{5x/2-\pi n^2e^{2x}\}. \tag{14} $$

Its normalization $p=P/\int_{\mathbb R}P$ has characteristic function $F_\xi$; see [1, (2.7)]. It is positive on $x\geq0$ term by term. The theta functional equation makes it even. It is holomorphic in $|\operatorname{Im}x|<\pi/4$ and has double exponential decay on the real line. We next give the precise boundary and moment arguments needed here.

3.1. Finite Taylor radius, regular positive endpoint

Put $t=e^{2x}$ and $\Theta(t)=\sum_{n\in\mathbb Z}e^{-\pi n^2t}$. Direct differentiation gives

$$ P(x)=t^{1/4}\bigl(2t^2\Theta''(t)+3t\Theta'(t)\bigr). \tag{15} $$

The right side is holomorphic when $\operatorname{Re}t>0$, with the branch $t^{1/4}=e^{x/2}$. Evenness also follows by writing it as one half of $(d^2/dx^2-1/4)$ applied to $e^{x/2}\Theta(e^{2x})$, which is even by $\Theta(t)=t^{-1/2}\Theta(1/t)$.

At $t=a+i$, for real $a>0$, Poisson summation gives

$$ \Theta(a+i)=\sum_{n\in\mathbb Z}(-1)^ne^{-\pi n^2a} =a^{-1/2}\sum_{m\in\mathbb Z} \exp\{-\pi(m+1/2)^2/a\}. \tag{16} $$

The function in (16), and each fixed derivative with respect to $a$, is bounded by a power of $a^{-1}$ times $e^{-\pi/(4a)}$ as $a\downarrow0$. Since differentiation in $a$ is differentiation in $t$ on this path, (15) implies

$$ P\bigl(\tfrac12\log(a+i)\bigr)=o(a^N) \quad\text{for every }N>0. \tag{17} $$

The map $a\mapsto\tfrac12\log(a+i)$ is locally holomorphic at zero, with nonzero derivative, and takes zero to $i\pi/4$. If $P$ extended holomorphically at that point, (17) would force the composition to vanish identically near zero. The identity theorem would then force $P$ to vanish on its strip, contradicting its positive real values. Thus $i\pi/4$ is a genuine obstruction to holomorphic continuation.

Strip holomorphy and this obstruction show that the Taylor radius of $P$ at zero is exactly $\pi/4$. Because $P$ is even, the radius for $g(y)=p(\sqrt y)$ is exactly

$$ R=\pi^2/16. \tag{18} $$

Nevertheless, $g$ is holomorphic near $+R$: choose the local square-root branch near $R$, whose value at $R$ is the real point $\pi/4$, and use strip holomorphy near that real point. This is local continuation near $+R$, not a larger disk about zero.

3.2. Real-moment concentration and the forced factor

Separating $n=1$ in (14) gives

$$ P(x)\sim4\pi^2\exp\{9x/2-\pi e^{2x}\} \quad(x\to\infty). \tag{19} $$

The remaining terms are exponentially smaller. Consequently, for suitable positive constants and all sufficiently large $x$, $p(x)$ lies between two constant multiples of the exponential in (19). Globally on $x\geq0$ we can also use $p(x)\leq C\exp(-ce^{2x})$.

Let $L=(\log s)/2$ and $x_*=L-\log\log s$. Integrating on $[x_*,x_*+1]$ and using the lower estimate in (19) gives

$$ \log M(s)\geq2s\log x_*-O\bigl(s/(\log s)^2\bigr)-O(\log s) =2s\log L-o(s). \tag{20} $$

Fix $0<\varepsilon<1$, and put $a=(1-\varepsilon)L$, $b=(1+\varepsilon)L$. For $s\geq1$,

$$ \int_0^a x^{2s}p(x)\,dx\leq\tfrac12a^{2s},\qquad \int_0^a x^{2s-2}p(x)\,dx\leq\tfrac12a^{2s-2}. $$

Upon division by $M(s)$ and $M(s)/L^2$, respectively, (20) makes both bounds tend to zero exponentially in $s$. The second estimate controls the inverse-square weight at zero, so concentration in probability alone is not being substituted for a moment estimate.

For the high tail write $x=b+y$. The inequalities $\log(b+y)\leq\log b+y/b$ and $e^{2y}\geq1+2y$ yield, for large $s$,

$$ \int_b^\infty x^{2s}p(x)\,dx \leq\frac{C b^{2s}e^{-ce^{2b}}}{2ce^{2b}-2s/b}. \tag{21} $$

Here $e^{2b}=s^{1+\varepsilon}$. The bound in (21), divided by $M(s)$, tends to zero even after multiplication by a fixed power of $L$. On this high region $x^{-2}\leq b^{-2}$.

The tilted probability $x^{2s}p(x)\,dx/M(s)$ therefore has mass tending to one on $[a,b]$, with negligible inverse-square contribution outside. It follows that

$$ (1+\varepsilon)^{-2} \leq\liminf_{s\to\infty}\frac{L^2M(s-1)}{M(s)} \leq\limsup_{s\to\infty}\frac{L^2M(s-1)}{M(s)} \leq(1-\varepsilon)^{-2}. $$

Letting $\varepsilon$ decrease to zero proves

$$ \frac{M(s-1)}{M(s)}\sim\frac4{(\log s)^2},\qquad \phi_p(s)\sim\frac{16s}{(\log s)^2}. \tag{22} $$

Thus all hypotheses of Theorem 2 hold, proving Theorem 1.

4. What the exclusion changes

The xi membership question in [1, Section 4.1] concerns the larger class $D_P$. It was not stated there as a claim that the factor belongs to $B_{P1}$. The present result distinguishes the broad membership programme from this particular sufficient condition for real zeros. An attempted proof through that sufficient condition must fail at the separated meromorphic Pick hypothesis, even if broader membership is eventually established.

The ingredients imported here are the KPS construction and product law [1], Bernstein-gamma theory [2], logarithmic Pick theory [3], and the classical Fabry theorem in the form recorded in [4]. The contribution is their combination into Theorem 2 and the resulting unconditional exclusion for the actual theta kernel. Real-moment asymptotics for xi coefficients are well studied, for example in [5]; (20)--(22) give only the elementary concentration estimate needed for the exclusion. They are not advanced coefficient asymptotics or a separate novelty claim.

No zero of $\xi$ has been located or excluded by this argument. No conclusion about broader Bernstein membership follows. The Riemann hypothesis remains unresolved.

5. Verification and acknowledgments

The result is analytic and does not depend on finite positivity tables, numerical Mellin zeros, or computations. Its checks are the product normalization in (6), the residue identity (8), the absence of additional terms in (9), the zero-pole cancellation in (11), the treatment of zero coefficients in (13), and the cusp and moment bounds (16)--(22). Each is included in the proof.

An early proof sketch was discussed in r/riemann_hypothesis/post/18. We thank halasz_resonator for discussion of the cusp and the scope of the Pick hypothesis. That discussion is not used as a substitute for the proof or as an independent verification of every premise.

References

[1] T. Konstantopoulos, P. Patie and R. Sarkar. A new class of solutions to the van Dantzig problem, the Lee--Yang property, and the Riemann hypothesis. Annales de l'Institut Fourier 74 (2024), no. 1, 377--421. https://doi.org/10.5802/aif.3600

[2] P. Patie and M. Savov. Bernstein-gamma functions and exponential functionals of Lévy processes. Electronic Journal of Probability 23 (2018). https://arxiv.org/abs/1604.05960

[3] R. L. Schilling, R. Song and Z. Vondraček. Bernstein Functions: Theory and Applications, second edition. De Gruyter, 2012. https://doi.org/10.1515/9783110269338

[4] A. Eremenko. A version of Fabry's theorem for power series with regularly varying coefficients. Proceedings of the American Mathematical Society 136 (2008), no. 12, 4389--4394. Theorem A. https://doi.org/10.1090/S0002-9939-08-09652-4 ; https://arxiv.org/abs/0710.5894

[5] D. Romik. Orthogonal polynomial expansions for the Riemann xi function. arXiv:1902.06330 (2019), Section 6.1. https://arxiv.org/abs/1902.06330