A Fabry obstruction to complete Bernstein constructions of the Riemann xi function

1. Statement and relation to the KPS programme

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

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

Here $\phi$ is a Bernstein function, with the additional hypotheses specified in the respective results of [1]. A complete Bernstein function is a nonnegative Bernstein function whose Lévy measure has a completely monotone density. Equivalently, it has a holomorphic extension to the slit plane $\mathbb C\setminus(-\infty,0]$ which maps the upper half-plane into its closure [3]. Killing and drift terms are allowed.

Theorem 4.4 of [1] gives only real zeros under a stronger, separated meromorphic Pick hypothesis, denoted $B_{P1}$. Its simple zeros $-\lambda_k$ and poles $-\lambda_k-1$ satisfy $0\leq\lambda_1$ and $\lambda_{k+1}>\lambda_k+1$. That unit pairing is an additional assumption in [1, (4.14)]; it does not follow from the Pick property alone. Our result excludes every complete Bernstein factor, without assuming meromorphic continuation, unit pairing, or separation at the outset.

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 complete Bernstein function $\phi$ such that

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

This excludes the complete Bernstein subclass of the KPS construction for normalized xi, including the separated meromorphic Pick sufficient mechanism. It does not exclude the larger class $D_P$ posed in [1, Section 4.1], or its root-zero subclass generated by Bernstein functions with decreasing Lévy density, denoted $B_J$. A decreasing density need not be completely monotone. The conclusion gives neither an off-line zero nor a proof of RH.

We prove a general obstruction and then verify its hypotheses for the theta kernel. The new structural step is that entire continuation of the reciprocal Bernstein-gamma factor forces a complete Bernstein function to be meromorphic and its interlacing zero–pole pairs to have positive integer lengths. These pairs generate finite strings of simple Mellin-factor zeros with gaps at least one. Index-one regular variation forces density one of these strings. Mellin residues then force density-zero changes of sign in the Fourier density's Taylor coefficients, contradicting its regular positive Taylor endpoint by classical Fabry theory.

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 any complete Bernstein function $\phi$.

The conditions in (5) remove finite-pole complete Bernstein functions after meromorphic continuation is established. They hold for the xi density.

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. Entire continuation forces meromorphic integer pairing

Set $F(z)=E(z-1)$. By (7), it agrees with $1/W_\phi(z)$ first on $\operatorname{Re}z>1$, and then by analytic continuation on $\operatorname{Re}z>0$, where it is zero-free [2, Theorem 4.1]. It is entire by the Mellin continuation above. The recurrence gives

$$ F(z)=\phi(z)F(z+1),\qquad \phi(z)=\frac{F(z)}{F(z+1)}. \tag{9} $$

The quotient continues $\phi$ meromorphically to the whole plane. On the slit plane it agrees with the original complete Bernstein function by the identity theorem. It is real at every real point at which the quotient has no pole, because $E$ is real entire.

For completeness, the complete Bernstein representation is

$$ \phi(z)=a+bz+\int_{(0,\infty)}\frac{z}{z+t}\,\rho(dt), \quad a,b\geq0,\quad \int_{(0,\infty)}\frac{\rho(dt)}{1+t}<\infty. \tag{10} $$

Stieltjes inversion applied to $\phi(z)/z$ shows that $\rho$ vanishes on every real interval through which this real meromorphic function continues without a pole. Its support is therefore the locally finite set corresponding to negative poles. The measure is atomic there, and every pole of $\phi$ is simple with negative residue. This atomic representation and the real simple interlacing used below are standard meromorphic complete Bernstein facts; see, for example, [6, Lemma A.1] and its proof. There is no pole at zero: a Bernstein function has a finite nonnegative limit there, so a meromorphic singularity at zero is removable. Alternatively, $F(1)=1$ in (9) makes this immediate. In particular, the negative poles have no finite accumulation, including at zero.

A function with finitely many such poles is rational apart from the linear term $bz$. It is bounded as $s\to\infty$ when $b=0$, and has $\phi(s)/s\to b>0$ otherwise. Both possibilities contradict (5). Thus the pole sequence is infinite.

On every real pole-free interval, the derivative obtained from (10) is

$$ \phi'(x)=b+\int_{(0,\infty)}\frac{t}{(x+t)^2}\,\rho(dt)>0. $$

The integral is locally convergent there; the representing measure has finite mass on bounded intervals and satisfies the stated tail integrability. At a pole, $\phi$ tends to $-\infty$ on its right and to $+\infty$ on its left. There is exactly one simple zero between consecutive poles, and one between the first pole and zero, allowing a zero at zero. Nonconstant Pick functions have no nonreal zeros, and positivity excludes positive zeros. Write all zeros and poles as

$$ 0\leq a_1<b_1<a_2<b_2<\cdots,\qquad \phi(-a_j)=0,\quad -b_j\text{ a pole}. \tag{11} $$

We next show that each $L_j=b_j-a_j$ is a positive integer. For $t\geq0$, let $m(t)$ be the multiplicity of the zero of $F$ at $-t$, taking $m(t)=0$ at a nonzero value. Set $m(t)=0$ for $t<0$, as $F$ is zero-free on the right half-plane. Orders in (9) give the exact identity

$$ m(t)-m(t-1)=\nu(t),\qquad \nu(t)=\sum_j\bigl(\mathbf1_{\{t=a_j\}}-\mathbf1_{\{t=b_j\}}\bigr). \tag{12} $$

Iterating to the right yields a finite sum for each $m(t)$. Before $b_1$ the only source is $a_1$. At $t=b_1$, nonnegativity of $m$ would fail unless $b_1-a_1\in\mathbb N$. When this difference is the positive integer $L_1$, the pair's total contribution is exactly one at

$$ a_1,a_1+1,\ldots,b_1-1 $$

and zero everywhere else. In particular, it has no remaining tail at or beyond $b_1$. Inductively all earlier pairs have canceled before $a_j$. The same argument at $b_j$ forces $L_j=b_j-a_j\in\mathbb N$ and leaves precisely the string

$$ a_j,a_j+1,\ldots,a_j+L_j-1. \tag{13} $$

This induction accounts for every source in (12). There are no additional zeros of $F$: an order at any real point is determined by its finitely iterated recurrence, while a nonreal zero would shift by positive integers to the zero-free right half-plane, since $\phi$ has no nonreal zeros or poles. Thus all zeros of $F$ are precisely the simple negatives of (13). Within a string the gap is one. Between strings it is $a_{j+1}-(b_j-1)>1$.

2.3. Density of the strings and coefficient signs

The negative-sign set of $\phi(-t)$ is

$$ U=\bigcup_{j\geq1}(a_j,b_j). $$

In the upper half-plane, the principal logarithm $\log\phi$ has bounded harmonic imaginary part between zero and $\pi$. Its boundary imaginary part equals $\pi$ precisely on $-U$. The Poisson representation of a bounded harmonic function has this bounded boundary function as density, with no linear harmonic term or singular boundary measure. Differentiating its holomorphic representation gives

$$ \frac{\phi'(s)}{\phi(s)} =\int_0^\infty\frac{\mathbf1_U(t)}{(s+t)^2}\,dt, \qquad s>0. \tag{14} $$

This is also the logarithmic Pick representation [3]. Bernstein concavity and index-one regular variation 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. Subtracting (14) from $\int_0^\infty(s+t)^{-2}\,dt=1/s$ yields

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

The weight on $[0,s]$ is at least $1/(4s)$, hence $|[0,s]\setminus U|=o(s)$. Each complete interval $(a_j,b_j)$ has integer length $L_j$ and contributes exactly $L_j$ zeros from (13). At most one interval can be partially cut by the endpoint $s$, and its length and number of included string points differ by at most one. Consequently,

$$ \#\{t\in[0,s]:F(-t)=0\}=s-o(s). \tag{15} $$

The real entire function $f(x)=E(-x-1/2)=F(1/2-x)$ therefore has precisely simple zeros, gaps at least one, and counting function $N_f(x)=x-o(x)$.

The residue identity (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

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

Gaps at least one give $r\leq d$: an open interval of integer length $d$ with nonzero endpoints cannot contain more than $d$ such zeros. 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

$$ \#\{\text{coefficient sign changes through }n\} \leq n-N_f(n)+O(1)=o(n). \tag{16} $$

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 (16) 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 (16) implies $D_2(S)=0$: for each fixed $r>0$, the numerator is bounded by $o((1+r)x)$. Here a change of sign means a change between consecutive nonzero coefficients, ignoring intervening zero coefficients. 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

$$ P(x)=\sum_{n\geq1} 2\pi n^2(2\pi n^2e^{2x}-3) \exp\{5x/2-\pi n^2e^{2x}\}. \tag{17} $$

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

$$ P(x)=t^{1/4}\bigl(2t^2\Theta''(t)+3t\Theta'(t)\bigr). \tag{18} $$

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

$$ \Theta(a+i)=\sum_{n\in\mathbb Z}(-1)^ne^{-\pi n^2a} =a^{-1/2}\sum_{m\in\mathbb Z} \exp\{-\pi(m+1/2)^2/a\}. \tag{19} $$

The function in (19), 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, (18) implies

$$ P\bigl(\tfrac12\log(a+i)\bigr)=o(a^N) \quad\text{for every }N>0. \tag{20} $$

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, (20) 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

$$ R=\pi^2/16. \tag{21} $$

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 (17) gives

$$ P(x)\sim4\pi^2\exp\{9x/2-\pi e^{2x}\} \quad(x\to\infty). \tag{22} $$

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 (22). 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 (22) gives

$$ \log M(s)\geq2s\log x_*-O\bigl(s/(\log s)^2\bigr)-O(\log s) =2s\log L-o(s). \tag{23} $$

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, (23) 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 $s$,

$$ \int_b^\infty x^{2s}p(x)\,dx \leq\frac{C b^{2s}e^{-ce^{2b}}}{2ce^{2b}-2s/b}. \tag{24} $$

Here $e^{2b}=s^{1+\varepsilon}$. The bound in (24), 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

$$ \frac{M(s-1)}{M(s)}\sim\frac4{(\log s)^2},\qquad \phi_p(s)\sim\frac{16s}{(\log s)^2}. \tag{25} $$

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 $B_{P1}$, or even that it is complete Bernstein. Our result rules out the complete Bernstein subclass, including proposed constructions by a positive mixture of exponential Lévy densities. A possible broader $B_J$ factor must have a decreasing Lévy density that is not completely monotone. This restriction does not settle whether such a factor exists.

The imported ingredients are the KPS construction and product law [1], Bernstein-gamma theory [2], complete Bernstein and logarithmic Pick theory [3], and classical Fabry theory [4]. The meromorphic atomic representation and interlacing are also explicitly recorded in [6, Lemma A.1]; neither interlacing alone nor its proof is a novelty claim here. The contribution is their combination into Theorem 2, including the entire-quotient and integer-string argument, and the resulting unconditional exclusion for the actual theta kernel. Real-moment asymptotics for xi coefficients are well studied, for example in [5]; our concentration estimate supplies only what the obstruction needs.

Version 1 treated only $B_{P1}$ with unit zero–pole pairing. The present version removes those assumptions. Meromorphic continuation and positive integer pairing are instead consequences of the full probability-law identity, the kernel's analytic continuation at zero, and the complete Bernstein property. This is a broader class exclusion, while remaining strictly weaker than exclusion of ordinary $B_J$ membership.

No zero of $\xi$ has been located or excluded by this argument. 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 and Mellin residue identity; the passage from the entire quotient to an atomic Pick representation; the multiplicity recurrence and its integer-string induction; the bounded-harmonic logarithmic representation; the coefficient sign deficit including zero coefficients; and the theta cusp and inverse-square moment bounds. These are 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

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