\documentclass[11pt]{amsart} \pdfoutput=1 \usepackage[usenames,dvipsnames]{xcolor} % \usepackage[utf8]{inputenc} \usepackage[T1]{fontenc} \usepackage[letterpaper,margin=1in]{geometry} \usepackage{lmodern} \usepackage{microtype} \usepackage{etoolbox} % \usepackage{lettrine} \usepackage{mathtools} \usepackage{amsmath,amssymb,amsfonts,amsxtra,bm} \usepackage{amsthm} \usepackage{mathrsfs} \usepackage{latexsym} \usepackage{nicefrac} % \usepackage{booktabs} \usepackage{makecell,multirow} % \usepackage{colortbl} \usepackage{graphicx} \usepackage{epstopdf} \usepackage{float} \floatstyle{plaintop} \restylefloat{table} \usepackage{footnote, tablefootnote} \usepackage{enumitem} \usepackage{algorithm} \usepackage{algorithmic} \usepackage{comment} \usepackage{xspace} \usepackage{url} \usepackage[most]{tcolorbox} \usepackage[round]{natbib} \usepackage{hyperref} \numberwithin{equation}{section} \makeatletter \definecolor{darkblue}{rgb}{0.0,0.0,0.65} \definecolor{darkred}{rgb}{0.68,0.05,0.0} \definecolor{darkgreen}{rgb}{0.0,0.29,0.29} \definecolor{darkpurple}{rgb}{0.47,0.09,0.29} \renewcommand{\topfraction}{0.85} \renewcommand{\bottomfraction}{0.4} \renewcommand{\textfraction}{0.1} \renewcommand{\floatpagefraction}{0.9} \patchcmd{\section}{\scshape}{\fontfamily{ppl}\bfseries\scshape\selectfont}{}{% \PackageWarning{aiproofs}{Could not bold the section font; amsart may have changed}} \newcommand{\aiauthor}{% \author[Suvrit Sra]{Suvrit Sra\textsuperscript{\normalfont*}}% \thanks{*\,TU Munich, Dept.~of Mathematics, School of CIT; Garching, Germany. Email: \texttt{s.sra@tum.de}.}} \makeatother \hypersetup{colorlinks=true,linkcolor=blue!45!black,citecolor=blue!45!black,urlcolor=red!65!black} \setlist{nosep,leftmargin=*} \renewcommand{\bibnumfmt}[1]{{\color{darkblue}\bf\textsc{#1}}.\hskip 6pt} \newenvironment{aiusage}{\section*{AI usage}}{} \newcommand{\aiacknowledgements}[1][]{% \section*{Acknowledgements} \if\relax\detokenize{#1}\relax\else #1\ \fi The author (SS) thanks the Alexander von Humboldt (AvH) Foundation for their generous support via a Humboldt Professorship in AI.% } \newcommand{\reals}{\mathbb{R}} \newcommand{\N}{\mathbb{N}} \newcommand{\pp}{\mathbb{P}} \newcommand{\E}{\mathbb{E}} \newcommand{\bmOne}{\bm{1}} \newcommand{\C}{\mathbb{C}} \newcommand{\Cc}{\mathcal{C}} \newcommand{\Dc}{\mathcal{D}} \newcommand{\Nc}{\mathcal{N}} \newcommand{\Xc}{\mathcal{X}} \newcommand{\Fc}{\mathcal{F}} \newcommand{\symm}{\mathfrak{S}} \newcommand{\ma}{\bm{A}} \newcommand{\mx}{\bm{X}} \newcommand{\my}{\bm{Y}} \newcommand{\mz}{\bm{Z}} \newcommand{\vx}{\bm{x}} \newcommand{\vy}{\bm{y}} \newcommand{\vv}{\bm{v}} \newcommand{\vz}{\bm{z}} \newcommand{\vr}{\bm{r}} \newcommand{\mb}{\bm{B}} \newcommand{\dpp}{\textsc{Dpp}} \newcommand{\etal}{et~al.\@\xspace} \newcommand{\half}{\tfrac{1}{2}} \newcommand{\kron}{\otimes} \newcommand{\inv}[1]{{#1}^{-1}} \newcommand{\nlsum}{\sum\nolimits} \newcommand{\nlprod}{\prod\nolimits} \newcommand{\ip}[2]{\langle {#1},\, {#2} \rangle} \newcommand{\iplr}[2]{\left\langle {#1},\, {#2}\right\rangle} \newcommand{\set}[1]{\left\lbrace#1\right\rbrace} \newcommand{\norm}[1]{\|{#1}\|} \newcommand{\bignorm}[1]{\left\lVert #1\right\rVert} \newcommand{\pnorm}[2]{\|#1\|_{#2}} \newcommand{\enorm}[1]{\bigl\|#1\bigr\|_{2}} \newcommand{\frob}[1]{\|#1\|_{\text{F}}} \newcommand{\fdel}[2]{\frac{\partial {#1}}{\partial {#2}}} \newcommand{\what}{\widehat} \newcommand{\ld}{\ell{}d} \newcommand{\gm}{\#} \newcommand{\nf}[2]{\nicefrac{#1}{#2}} \newcommand{\pfrac}[2]{\left(\tfrac{#1}{#2}\right)} \def\sint{\begingroup\textstyle\int\endgroup} % \DeclareMathOperator{\trace}{tr} \DeclareMathOperator{\diag}{diag} \DeclareMathOperator{\Diag}{Diag} \DeclareMathOperator{\real}{Re} \DeclareMathOperator{\imag}{Im} \DeclareMathOperator{\per}{per} \DeclareMathOperator{\Vol}{Vol} \DeclareMathOperator{\Imm}{Imm} \DeclareMathOperator{\mdet}{det} \DeclareMathOperator{\supp}{supp} \DeclareMathOperator{\qjsd}{QJSD} \DeclareMathOperator{\qjrd}{QJRD} \DeclareMathOperator*{\argmin}{argmin} \DeclareMathOperator*{\argmax}{argmax} \newtheorem{theorem}{Theorem}[section] \newtheorem{conj}[theorem]{Conjecture} \newtheorem{prop}[theorem]{Proposition} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corr}[theorem]{Corollary} \newtheorem{rmk}[theorem]{Remark} \newtheorem{example}[theorem]{Example} \newtheorem{claim}[theorem]{Claim} \theoremstyle{definition} \newtheorem{defn}[theorem]{Definition} \newtheorem{prob}[theorem]{Problem} \definecolor{cdarkred}{RGB}{180,70,30} % \definecolor{cdarkgreen}{RGB}{0,130,0} % \definecolor{azure}{RGB}{240,255,255} \definecolor{cmidgreen}{RGB}{201,219,216} \definecolor{clightgreen}{RGB}{200,255,200} \definecolor{clightgray}{RGB}{240,240,240} \definecolor{cmygreen}{RGB}{10,120,0} \definecolor{cmyblue}{RGB}{50,50,255} \definecolor{sblue}{RGB}{100,100,255} \definecolor{cmypurple}{RGB}{150,50,180} \definecolor{cmyred}{RGB}{238, 64, 0} \definecolor{cdarkblue}{RGB}{30,75,170} \definecolor{csharpred}{RGB}{255,0,0} \newcommand{\sred}[1]{{\color{csharpred} #1}} \newcommand{\darkred}[1]{{\color{cdarkred} #1}} \newcommand{\bred}[1]{\textbf{\sred{#1}}} \newcommand{\darkblue}[1]{{\color{cdarkblue} #1}} \newcommand{\bblue}[1]{\textbf{\darkblue{#1}}} \newcommand{\darkbred}[1]{\textbf{\color{cdarkred} #1}} \newcommand{\darkgreen}[1]{{\color{cmygreen} #1}} \newtcolorbox{dashedbox}[1][]{ nobeforeafter, colback=yellow!10!white, boxrule=0pt, enhanced jigsaw, top=2pt, left=2pt, right=2pt, bottom=0pt, borderline horizontal={0.5pt}{0pt}{dashed}, borderline vertical={0.5pt}{0pt}{dashed}, #1 } \newtcbox{\tcbmathbox}[1][]{ nobeforeafter, colback=yellow!30!white, boxrule=0pt, enhanced jigsaw, top=0pt, left=1pt, right=1pt, bottom=0pt, borderline horizontal={0.5pt}{0pt}{dashed}, borderline vertical={0.5pt}{0pt}{dashed}, #1 } \setlength{\emergencystretch}{2em} \allowdisplaybreaks[2] \theoremstyle{plain} \newtheorem{assump}[theorem]{Assumption} \newcommand{\eps}{\varepsilon} \newcommand{\dd}{\,d} \newcommand{\RRreg}{\mathrm{RR}} \newcommand{\Kc}{\mathcal{K}} % \hypersetup{ pdftitle={Sharp order-four sign regularity of a Jacobi theta kernel}, pdfauthor={Suvrit Sra} } \makeatletter \newcommand{\preprintsfooter}[2]{% \def\ps@firstpage{\ps@plain \def\@oddfoot{\hfil\parbox[t]{\textwidth}{\centering\normalfont\scriptsize First public release: #2, as v1 in the author's preprint repository, \url{https://github.com/suvrit/preprints/tree/main/preprints/#1}; every version is kept there with its timestamp.\\[3pt]\thepage}\hfil \global\topskip\normaltopskip}% \let\@evenfoot\@oddfoot \def\@oddhead{\@serieslogo\hss}% \let\@evenhead\@oddhead}} \makeatother \begin{document} \title[Sharp order-four sign regularity of a Jacobi theta kernel]{Sharp order-four sign regularity of a Jacobi theta kernel} \aiauthor \begin{abstract} We prove that the Hankel kernel $\Phi(\sqrt{x+y})$, where $\Phi$ is the Jacobi theta kernel whose cosine transform is the Riemann $\Xi$-function, is strictly reverse-regular through order four. The order is sharp: an interval enclosure of the fifth Hankel--Wronskian at $t=10^{-2}$ shows failure at order five. The third-order condition is Csordas's conjecture that $s'^2-ss''$ is strictly log-concave for $s(t)=\Phi(\sqrt t)$. The proof reduces the kernel minors to Wronskian signs and establishes the signs by exact polynomial estimates for $t\ge1$ and interval arithmetic on $[0,1]$. The fourth-order sign also yields three positive iterates of $s\mapsto s'^2-ss''$, the triple Tur\'an inequalities for the normalized Taylor coefficients of $\Xi$, and quartic integral inequalities for $\Xi$. \end{abstract} \preprintsfooter{csordas}{10 October 2026} \maketitle \section{Introduction} The theta kernel \begin{equation}\label{eq:Phi-def} \Phi(x)=\sum_{n=1}^{\infty}\pi n^2(2\pi n^2e^{4x}-3)\exp(5x-\pi n^2e^{4x}) \end{equation} is an even, rapidly decreasing, positive kernel connected with the Riemann $\xi$-function. In the notation of \citet{csordas2015fourier}, it appears in the cosine transform representation \[ H(z)=\int_0^\infty \Phi(t)\cos(zt)\dd t. \] With the standard normalization \[ \xi(z)=\tfrac12z(z-1)\pi^{-z/2}\Gamma(z/2)\zeta(z), \qquad \Xi(z)=\xi(\tfrac12+iz), \] the exact relation is $H(z)=\Xi(z/2)/8$. Positivity, Tur\'an inequalities, generalized Laguerre inequalities, and sign regularity connect properties of $\Phi$ to the zero distribution of $\Xi$; see \citep{csordas1986,csordas2000conjectures,coffey2013logconcavity,csordas2015fourier,nuttall2013wronskians,karlin1968}. Csordas's Open Problem 4.14 \citep{csordas2015fourier}, which restates Problem 3.3 of \citet{csordas2000conjectures}, concerns the square-root composition \[ s(t)=\Phi(\sqrt t),\qquad t>0, \] and the Tur\'an expression \[ f(t)=s'(t)^2-s(t)s''(t). \] The problem asks whether \begin{equation}\label{eq:OP414} (\log f)''(t)<0\qquad (t>0). \end{equation} The positivity $f(t)>0$ follows from the strict logarithmic concavity of $s(t)$, proved by \citet[Theorem~2.1]{csordas1988moment}; see also \citep[Theorem~4.2(b)]{csordas2015fourier}. Thus \eqref{eq:OP414} asserts that the first Tur\'an expression attached to $s$ is itself strictly log-concave. \citet{csordas2000conjectures} showed that \eqref{eq:OP414} implies the double Tur\'an inequalities. \citet{planat2026second} have recently proved \eqref{eq:OP414}, and with it the third-order sign-regularity condition for the kernel $\Phi(\sqrt{x+y})$, by two computer-assisted arguments that share a local interval certificate near $t=0$: one continues with theta-tail estimates, and the other uses Jacobi's differential equation for $\theta_3$. The argument in this paper was developed independently\footnote{Our reconstructed chronology of this AI-assisted work records an explicit global proof claim for Csordas's conjecture on 25 February 2026. The first reported successful full run of a corrected compact verifier dates to 6 March 2026. A further verifier bug was repaired on 1 May, but the available record does not establish a completed run of that repair. On 23 May, compact certification was explicitly reported successful for the stronger $\RRreg_4$ theorem; the failure of $\RRreg_5$ was also obtained that day. These results were not publicly reported at the time. We record these milestones to document the paper's development and make no claim of priority.} and is also computer-assisted: exact polynomial estimates handle $t\ge1$, and interval arithmetic handles $0\le t\le1$. Inequality \eqref{eq:OP414} is not an isolated condition. Section~\ref{sec:determinant-reduction} shows that it is equivalent to the sign of a $3\times3$ Hankel determinant of derivatives of $s$, which is precisely the third infinitesimal sign-regularity condition for the Hankel kernel \[ K(x,y)=s(x+y)=\Phi(\sqrt{x+y}). \] The equivalence raises the question of how far the sign regularity of $K$ extends. For the related kernel $\Phi(x+y)$, \citet{nuttall2013wronskians} proved the third condition and reported numerical evidence that the fourth holds and the fifth fails. We prove that, for $K$, the fourth-order condition holds and the fifth fails. The third-order condition yields the double Tur\'an inequalities discussed by \citet{csordas2000conjectures}, as \citet{planat2026second} note, and the fourth-order condition also gives the triple Tur\'an inequalities and certain quartic Fourier inequalities. \subsection{Main results} For a smooth function $g$, define the Hankel--Wronskians \begin{equation}\label{eq:Wm-def} W_m[g](t)=\det\bigl(g^{(i+j-2)}(t)\bigr)_{i,j=1}^m. \end{equation} Let \[ \eps_m=(-1)^{m(m-1)/2}. \] The order-$m$ infinitesimal reverse-regularity sign is \[ \eps_m W_m[g](t)>0. \] A kernel $K$ on $[0,\infty)^2$ is \emph{strictly reverse-regular through order $q$} if \[ \eps_m\det[K(x_i,y_j)]_{i,j=1}^m>0 \quad(1\le m\le q) \] for all $0\le x_1<\cdots0\qquad (t\ge 0,\;m=1,2,3,4), \end{equation} where the value at $t=0$ is interpreted by analytic continuation of $s$. In particular, \[ W_1>0,\qquad W_2<0,\qquad W_3<0,\qquad W_4>0. \] Furthermore, the fifth condition fails: at $t=10^{-2}$ one has \begin{equation}\label{eq:W5-fail} W_5[s](10^{-2})<0, \end{equation} whereas $\eps_5=+1$. Consequently the Hankel kernel $K(x,y)=\Phi(\sqrt{x+y})$ is strictly reverse-regular through order four, but is not $\RRreg_5$. \end{theorem} The $m=3$ case is exactly Csordas's Open Problem 4.14. By Lemma~\ref{lem:ff-id}, \[ (\log f)''(t)=\frac{s(t)W_3[s](t)}{f(t)^2}. \] Since $s>0$ and $f>0$, the inequality $(\log f)''<0$ is equivalent to $W_3[s]<0$. Through the Desnanot--Jacobi identity, the signs for $m\le4$ also make the first three iterates of $g\mapsto g'^2-gg''$ positive at $s$ (Corollary~\ref{cor:differential-hierarchy}). Theorem~\ref{thm:main-full} has two consequences for the Riemann $\Xi$-function itself. The first concerns its Taylor coefficients. Let \begin{equation}\label{eq:H-beta-gamma} H(x)=\int_0^\infty \Phi(t)\cos(xt)\dd t =\sum_{k=0}^{\infty}(-1)^k\frac{\beta_k}{(2k)!}x^{2k}, \qquad \gamma_k=\frac{k!\beta_k}{(2k)!}. \end{equation} Then the infinite Hankel matrix $(\gamma_{i+j})_{i,j\ge0}$ inherits strict reverse regularity through order four. The same transfer works for real indices: for $a\ge0$, set \[ \beta_a=\int_0^\infty\Phi(r)r^{2a}\dd r, \qquad \gamma_a=\frac{\sqrt\pi\,\beta_a}{4^a\Gamma(a+1/2)}. \] At integer indices, $\gamma_a$ agrees with \eqref{eq:H-beta-gamma}. \begin{theorem}[Coefficient reverse regularity]\label{thm:coef-RR4} Under Assumption~\ref{assump:interval}, for $m=1,2,3,4$, and for all strictly increasing nonnegative real sequences \[ r_0<\cdots0. \end{equation} In particular, the consecutive Hankel minors satisfy \[ \eps_m\det(\gamma_{n+i+j})_{i,j=0}^{m-1}>0 \qquad(n\ge0,\;m\le4). \] \end{theorem} At integer indices, determinant condensation gives the first three iterates of the Tur\'an operator. \begin{corr}[Tur\'an, double Tur\'an, and triple Tur\'an]\label{cor:triple-turan} Define \[ T_k=\gamma_k^2-\gamma_{k-1}\gamma_{k+1}\quad(k\ge1), \] \[ E_k=T_k^2-T_{k-1}T_{k+1}\quad(k\ge2), \] and \[ Q_k=E_k^2-E_{k-1}E_{k+1}\quad(k\ge3). \] Under Assumption~\ref{assump:interval}, \[ T_k>0\quad(k\ge1),\qquad E_k>0\quad(k\ge2),\qquad Q_k>0\quad(k\ge3). \] \end{corr} The second consequence is a Fourier-side family of multilinear integral inequalities for the Riemann $\Xi$-function. Define \begin{equation}\label{eq:XiPhi} \Xi_\Phi(\lambda)=2\int_0^\infty \Phi(r)\cos(\lambda r)\dd r. \end{equation} Thus $\Xi_\Phi(\lambda)=\Xi(\lambda/2)/4=2H(\lambda)$. Fourier inversion gives \[ \Phi(r)=\frac1\pi\int_0^\infty \Xi_\Phi(\lambda)\cos(r\lambda)\dd\lambda. \] For $r>0$, put \begin{equation}\label{eq:Theta-def} \Theta_j(r,\lambda)=\left(\frac{1}{2r}\frac{\partial}{\partial r}\right)^j\cos(r\lambda). \end{equation} \begin{theorem}[Fourier-side determinant inequalities]\label{thm:Fourier-side} Under Assumption~\ref{assump:interval}, for every $r>0$ and every $m=1,2,3,4$, \begin{equation}\label{eq:Fourier-ineq} \eps_m \int_{(0,\infty)^m} \left(\prod_{\ell=1}^m\Xi_\Phi(\lambda_\ell)\right) \det\bigl(\Theta_{i+j-2}(r,\lambda_j)\bigr)_{i,j=1}^m \dd\lambda_1\cdots\dd\lambda_m>0. \end{equation} For $m=5$, the left-hand side is negative at $r=1/10$. \end{theorem} The $m=3$ case is the Fourier-inverted form of Csordas's conjecture; the $m=4$ case is a quartic constraint on $\Xi_\Phi$. \subsection{Strategy and computer assistance} By Lemma~\ref{lem:Wronskian-RR}, Theorem~\ref{thm:main-full} reduces to the signs of the Wronskians $W_m[s]$. For $m=3,4$ the proof has an analytic tail component and a compact computer-assisted component. For $t\ge1$ the first term of the theta series dominates, and the signs of $W_3$ and $W_4$ reduce to exact polynomial and integer inequalities. For $0\le t\le1$ we verify the signs in interval arithmetic, and a single interval enclosure at $t=10^{-2}$ gives the failure at order five. The coefficient, differential, and Fourier consequences inherit the same arithmetic assumption. \begin{assump}[Interval-arithmetic trust model]\label{assump:interval} The computer-assisted statements use exact Python integer arithmetic, exact symbolic polynomial operations in SymPy, and the outward interval operations of \texttt{mpmath.iv}. We assume these implementations are correct. \end{assump} All error bounds and sign tests run in interval arithmetic with 90 decimal digits, with integer factorials and rational partition endpoints; the overlap of the $t$- and $u$-ranges is checked exactly. \paragraph{Limitations.} The signs on the compact range rest on Assumption~\ref{assump:interval}; an independent implementation in another interval package would provide an additional check. The sharp order-four boundary concerns the specific Hankel kernel $\Phi(\sqrt{x+y})$. The order-five failure rules out reverse regularity of all orders for that kernel, but not higher positivity orders after a different transform: the coefficient integration in Section~\ref{sec:coeff-consequences} need not preserve a failure of sign, so the fifth-order witness does not prove failure of the fifth coefficient Hankel condition or of later Tur\'an iterates. Finally, the coefficient and Fourier inequalities are not pointwise Laguerre inequalities for $\Xi$; in particular, they do not by themselves prove $\Xi'^2-\Xi\Xi''\ge0$ on $\mathbb R$. For orders two and three, the consecutive coefficient inequalities are necessary conditions for the Riemann hypothesis \citep{csordas2000conjectures}; for order four this is not known, since Karlin's conjecture fails in general. \subsection{Related work}\label{sec:related} The theta kernel $\Phi$ and its Fourier transform have long been used to formulate necessary conditions for the Riemann hypothesis. \citet{csordas1986} proved P\'olya's Tur\'an inequalities for the Taylor coefficients of $\xi$; their key step is the strict log-concavity of $\int_t^\infty\Phi(\sqrt u)\dd u$, that is, reverse regularity of order two for the corresponding Hankel kernel. For the kernel $\Phi(x+y)$, \citet{nuttall2013wronskians} worked through the Wronskians $\det[\Phi^{(i+j-2)}(t)]$ and proposed a cumulant-based route past the obstruction at order five. Whether the regularity orders of $\Phi(x+y)$ and $\Phi(\sqrt{x+y})$ are related is not addressed here. \citet{csordas2015fourier} lists further inequalities and open problems for $\Phi$, and the general theory of totally positive and sign-regular kernels is due to \citet{karlin1968}. \section{From logarithmic concavity to Hankel--Wronskians}\label{sec:determinant-reduction} Two reductions turn Open Problem 4.14 and its higher-order analogues into sign conditions on Hankel--Wronskians. Lemma~\ref{lem:ff-id} identifies \eqref{eq:OP414} with the sign of $W_3[s]$, and Lemma~\ref{lem:Wronskian-RR} shows that Wronskian signs give the signs of all ordered kernel minors. Let $s$ be a $C^4$ function and put \[ f=s'^2-ss''. \] Define \[ W_3[s](t)=\det\begin{pmatrix} s&s'&s''\\ s'&s''&s^{(3)}\\ s''&s^{(3)}&s^{(4)} \end{pmatrix}(t). \] \begin{lemma}\label{lem:ff-id} For every $C^4$ function $s$, \begin{equation}\label{eq:ff-id} ff''-(f')^2=sW_3[s]. \end{equation} Thus, wherever $s>0$ and $f>0$, \begin{equation}\label{eq:logf-W3} (\log f)''=\frac{sW_3[s]}{f^2}. \end{equation} \end{lemma} \begin{proof} Differentiating $f=s'^2-ss''$ gives \[ f'=s's''-ss^{(3)},\qquad f''=(s'')^2-ss^{(4)}. \] Expanding $ff''-(f')^2$ gives \[ s\{ss''s^{(4)}-s(s^{(3)})^2-s'^2s^{(4)}+2s's''s^{(3)}-(s'')^3\}, \] which is $sW_3[s]$. Dividing by $f^2$ proves \eqref{eq:logf-W3}. \end{proof} For $s(t)=\Phi(\sqrt t)$, the Csordas--Varga theorem gives $s>0$ and $f>0$ on $(0,\infty)$. Hence Open Problem 4.14 is equivalent to \begin{equation}\label{eq:W3-target} W_3[s](t)<0\qquad(t>0). \end{equation} The second reduction passes from Wronskians to the ordered minors in the definition of reverse regularity. We include the classical Chebyshev-system argument \citep{karlin1968}, with the derivative signs appropriate to reverse regularity, because Section~\ref{sec:rr5-failure} also needs its converse. \begin{lemma}[Wronskians and ordered minors]\label{lem:Wronskian-RR} Let $s\in C^{2q-2}([0,\infty))$. If $\eps_jW_j[s](t)>0$ for $t\ge0$ and $1\le j\le q$, then $s(x+y)$ is strictly reverse-regular through order $q$. Conversely, reverse regularity through order $q$ implies $\eps_jW_j[s](t)\ge0$ for $t>0$. \end{lemma} \begin{proof} We first recall the evaluation criterion for a Chebyshev system. If the initial Wronskians of $f_1,\ldots,f_k$ are positive on an interval, then $\det[f_j(x_i)]_{i,j=1}^k>0$ at increasing nodes. For completeness, divide the $i$th row by $f_1(x_i)>0$ and put $g_j=(f_{j+1}/f_1)'$. Their initial Wronskians satisfy \[ W(g_1,\ldots,g_j)=\frac{W(f_1,\ldots,f_{j+1})}{f_1^{j+1}}>0. \] Subtract consecutive normalized rows, from bottom to top, and expand in the first column. The remaining determinant is the integral of $\det[g_j(t_i)]$ over $\prod_{i=1}^{k-1}[x_i,x_{i+1}]$. Induction makes its integrand positive in the interior, proving the criterion, including endpoints. Fix $z\le0$ and apply this criterion in $x\ge0$ to $(-1)^js^{(j)}(x-z)$, $0\le j0$. For fixed increasing $x_i$, the resulting evaluation determinants are the transposes of the initial Wronskians in $z$ of $s(x_i-z)$. A second application gives $\det[s(x_i-z_j)]>0$ for increasing $z_j\le0$. Set $z_j=-y_{k+1-j}$ and reverse the columns; their reversal has sign $\eps_k$. For the converse, coalesce increasing $x$- and $y$-nodes around positive points whose sum is $t$. Divide the determinant by the two positive Vandermonde products and pass to the limit. The limit is $W_j[s](t)/\prod_{i=0}^{j-1}(i!)^2$, proving the necessary sign. \end{proof} \section{Proof of Theorem~\ref{thm:main-full}}\label{sec:proof-architecture} By Lemma~\ref{lem:Wronskian-RR}, Theorem~\ref{thm:main-full} reduces to the signs of $W_m[s]$. To read these signs off the theta series, we first write each $W_m[s]$ as a positive factor times a polynomial in normalized theta sums; the tail, compact, and order-five arguments all work with this polynomial. \subsection{Derivative polynomials and normalized determinants}\label{sec:normalized-determinants} Set \[ t=u^2, \qquad y=\pi e^{4u}. \] Define polynomials $q_k$ by \begin{equation}\label{eq:q-rec} q_0(y)=2y-3, \qquad q_{k+1}(y)=4yq_k'(y)+(5-4y)q_k(y). \end{equation} Then \begin{equation}\label{eq:Phi-deriv-Uk} \Phi^{(k)}(u)=\pi e^{5u-y}U_k(u), \end{equation} where \begin{equation}\label{eq:Uk} U_k(u)=\sum_{n=1}^{\infty} n^2q_k(n^2y)e^{-(n^2-1)y}. \end{equation} Induction proves this identity: differentiating a summand applies the operator $4y\partial_y+(5-4y)$ to the polynomial factor. The substitution $t=u^2$ introduces powers of $u^{-1}$, so we remove them before evaluating the determinants. Define integers $a_{r,k}$ by $a_{0,0}=1$, zero outside $0\le k\le r$, and \[ a_{r+1,k}=(k-2r)a_{r,k}+a_{r,k-1}. \] Repeated application of $(2u)^{-1}\partial_u$ gives \[ \frac{s^{(r)}(u^2)}{\pi e^{5u-y}} =\frac{R_r(U;u)}{2^r u^{2r}},\qquad R_r(U;u)=\sum_{k=0}^r a_{r,k}u^kU_k. \] For $r\ge1$, $a_{r,0}=0$, so $R_r$ is divisible by $u$. Every term in $\det[R_{i+j}]_{i,j=0}^{m-1}$ contains at least $m-1$ such factors. Thus \[ P_m(U;u)=u^{1-m}\det[R_{i+j}(U;u)]_{i,j=0}^{m-1} \] is a polynomial, homogeneous of degree $m$ in $U$, and \begin{equation}\label{eq:scaled-Wm} W_m[s](u^2)= \frac{(\pi e^{5u-y})^m}{2^{m(m-1)}u^{(m-1)(2m-1)}}P_m(U;u), \qquad u>0. \end{equation} In particular, the denominators for $m=3,4$ are $64u^{10}$ and $4096u^{21}$. The recurrence determines every polynomial using exact integer arithmetic. Near $u=0$ we instead use the Taylor series in $t$, which avoids division by small powers of $u$. The compact verification of $W_3$ in Section~\ref{sec:compact} works with the cubic polynomial explicitly: \begin{align}\label{eq:N-W3} N(u)=&6U_0U_1^2-12uU_0U_1U_2-u^3U_0U_1U_4 +6u^2U_0U_2^2+u^4U_0U_2U_4-u^4U_0U_3^2\notag\\ &+10uU_1^3-6u^2U_1^2U_2+4u^3U_1^2U_3-u^4U_1^2U_4 -3u^3U_1U_2^2+2u^4U_1U_2U_3-u^4U_2^3, \end{align} and \[ W_3[s](u^2)=\frac{(\pi e^{5u-y})^3}{64u^{10}}N(u). \] \subsection{Analytic tail estimates}\label{sec:tail} For $t\ge1$ the first theta summand dominates every $U_k$, and exact estimates fix the signs of $W_3$ and $W_4$. \begin{prop}[Tail signs]\label{prop:tail-signs} For $t\ge1$, \[ W_3[s](t)<0, \qquad W_4[s](t)>0. \] \end{prop} \begin{proof} For $u\ge1$, put $y=\pi e^{4u}>159$ and write $U_k=A_k+\eta_k$, where $A_k=q_k(y)$ is the $n=1$ contribution. We bound $P_m(A;u)$ away from zero, then show that the $n\ge2$ tail cannot change its sign. \emph{\textbf{The first summand.}} For $m=3$, write $P_3(A;u)=\sum_{j=0}^6a_j(u)y^j$. The leading coefficient is $a_6(u)=-1024u(4u-1)^3\le-27648u^4$. The absolute coefficient sums of $a_0,\ldots,a_5$ are \[ (S_0,\ldots,S_5)=(10800,268200,1450800,3197280,2223936,776192). \] Each $a_j$ has degree at most four. Since $\sum_{j=0}^5S_j/159^{6-j}<5000$, it follows that \[ P_3(A;u)<-22648u^4y^6\le-10000u^4y^6. \] For $m=4$, write $P_4(A;u)=\sum_{j=0}^{10}b_j(u)y^j$. Here $b_{10}(u)=786432u^3(4u-1)^6\ge573308928u^9$. For $j=0,\ldots,9$, the constants \[ (C_0,\ldots,C_9)=(1,1,3,9,4,66,60,53,23,6) \] satisfy $|b_j(u)|\le C_jb_{10}(u)$: substituting $u=v+1$ makes every coefficient of $C_jb_{10}\pm b_j$ nonnegative. The rational inequality $\sum_{j=0}^9C_j/159^{10-j}<1/25$ therefore gives \[ P_4(A;u)\ge\frac{24}{25}\,573308928u^9y^{10} >500000000u^9y^{10}. \] The recurrences of Section~\ref{sec:normalized-determinants} generate the coefficients in these finite checks, which we verify in exact integer arithmetic. \emph{\textbf{The remaining summands.}} Let $P_*$ be the largest absolute coefficient sum among the required $q_k$. Since $y\ge159$, \[ |A_k|\le P_*y^d=:M,\qquad |\eta_k|\le P_*y^d2^{2d+3}e^{-3y}=:\delta, \] where $(d,P_*)=(5,151269)$ for $m=3$ and $(d,P_*)=(7,51450613)$ for $m=4$. Indeed, $|q_k(n^2y)|\le P_*(n^2y)^d$, and the ratio of successive terms $n^{2d+2}e^{-(n^2-1)y}$ is at most $(3/2)^{2d+2}e^{-5y}<1/2$ for $n\ge2$. Summing from $n=2$ gives the displayed bound; also $\delta450000000u^9y^{10}$. The prefactors in \eqref{eq:scaled-Wm} are positive, which proves the proposition. \end{proof} \subsection{Compact verification}\label{sec:compact} On $0\le t\le1$ no theta summand dominates, and we verify the signs in interval arithmetic: with Taylor polynomials in $t$ near zero, and with the theta series in $u$ elsewhere. \begin{prop}[Compact signs]\label{prop:compact-signs} Under Assumption~\ref{assump:interval}, \[ W_3[s](t)<0, \qquad W_4[s](t)>0, \qquad 0\le t\le1. \] \end{prop} \begin{proof} \emph{\textbf{Small $t$.}} The evenness and analyticity of $\Phi$ at zero give \[ s(t)=\sum_{j\ge0}a_jt^j,\qquad a_j=\frac{\Phi^{(2j)}(0)}{(2j)!}. \] For $0<\rho<\pi/8$, the real part of $e^{4u}$ is at least $e^{-4\rho}\cos(4\rho)>0$ on $|u|\le\rho$, so the theta series gives the explicit Cauchy bound \[ |\Phi(u)|\le M_\rho:= \sum_{n\ge1}\pi n^2(2\pi n^2e^{4\rho}+3) e^{5\rho-\pi n^2e^{-4\rho}\cos(4\rho)}. \] Thus $|a_j|\le M_\rho\rho^{-2j}$. If $p_M(t)=\sum_{j=0}^Ma_jt^j$ and $q=t_0/\rho^2<1$, then, for $0\le t\le t_0$, \begin{equation}\label{eq:Taylor-error} |s^{(r)}(t)-p_M^{(r)}(t)| \le M_\rho t_0^{-r} \frac{(M+1)^rq^{M+1}} {1-((M+2)/(M+1))^rq}, \end{equation} provided the denominator is positive. To prove \eqref{eq:Taylor-error}, bound the falling factorial by $j^r$ and sum the resulting geometric majorant. The coefficients $a_j$ and $M_\rho$ are themselves enclosed by theta series with geometric tail bounds. For $W_3$, we use $(t_0,\rho,M)=(1/100,3/10,30)$ and 500 equal rational subintervals. For $W_4$, we use $(t_0,\rho,M)=(3/200,19/50,50)$ and 750 subintervals. We expand $\det[p_M^{(i+j)}]$ as a polynomial before evaluating it on each interval. If its entries have modulus at most $B$, and every derivative error is at most $\delta$, the determinant error is at most $m!m(B+\delta)^{m-1}\delta$. The determinant bound follows by telescoping the $m$ factors in each of the $m!$ determinant terms. \emph{\textbf{Away from zero.}} For the cubic, we use the mean-value bound \[ N(u)\le N(c)+h\sup_{v\in[c-h,c+h]}|N'(v)| \] on a subdivision of $99/1000\le u\le1$ into 28175 rational intervals. The derivative is obtained from $U_k'=U_{k+1}-(5-4y)U_k$. Each $U_k$ is enclosed by its first four theta summands and the geometric tail bound of Section~\ref{sec:rr5-failure} with $N=4$. For the quartic, substitute the first four theta summands into $P_4$, producing a finite function $F_4(u)$. On each interval $[c-h,c+h]$, use \[ F_4(u)\ge F_4(c)-\sum_{j=1}^6\frac{|F_4^{(j)}(c)|h^j}{j!} -\frac{h^7}{7!}\sup_{v\in[c-h,c+h]}|F_4^{(7)}(v)|. \] An additional bound $4C(B+\delta)^3\delta$ controls the omitted $n\ge5$ summands; here $B$ bounds the four-summand values of the $U_k$ and $\delta$ their tails on the interval, and $C$ is the sum of the absolute coefficient bounds of $P_4$ on the interval. There are 71 rational subintervals covering $153/1250\le u\le1$, with widths at most $1/200$ below $1/5$, $1/100$ up to $1/2$, and $1/50$ thereafter. The strict rational inequalities \[ (99/1000)^2<1/100,\qquad (153/1250)^2<3/200 \] ensure overlap with the small-$t$ regions. Evaluated in interval arithmetic, these bounds give a positive signed lower bound on every subinterval. \end{proof} \subsection{Failure at order five}\label{sec:rr5-failure} The fifth-order witness is not needed for Open Problem 4.14; it shows that order four in Theorem~\ref{thm:main-full} is sharp. \begin{prop}[Failure of $\RRreg_5$]\label{prop:rr5-fails} Under Assumption~\ref{assump:interval}, the kernel $K(x,y)=\Phi(\sqrt{x+y})$ is not $\RRreg_5$. \end{prop} \begin{proof} For $m=5$, $\eps_5=+1$. If $K$ were $\RRreg_5$, coalescing five increasing nodes in each variable would force $W_5[s](t)\ge0$ for every $t>0$ (Lemma~\ref{lem:Wronskian-RR}). We show that this sign fails at $t=10^{-2}$. Set $u=1/10$ and $y=\pi e^{4u}$, and define the positively scaled determinant \[ \widetilde W_5(u)=\frac{W_5[s](u^2)}{(\pi e^{5u-y})^5}. \] Interval arithmetic gives \begin{equation}\label{eq:W5-enclosure} -1.106\cdot10^{29}<\widetilde W_5(1/10)<-1.105\cdot10^{29}. \end{equation} The tail bound used for each $U_k$, $0\le k\le8$, is \[ \left|\sum_{n>N}n^2q_k(n^2y)e^{-(n^2-1)y}\right| \le S_k y^{k+1} \frac{(N+1)^{2k+4}e^{-((N+1)^2-1)y}} {1-((N+2)/(N+1))^{2k+4}e^{-(2N+3)y}}, \] where $S_k$ is the absolute coefficient sum of $q_k$ and $N=12$. The denominator is positive. Its geometric ratio bounds all subsequent ratios because both factors decrease with $n$. We insert these enclosures into the exact chain-rule recurrence and evaluate all $5!$ signed determinant terms in interval arithmetic. Thus \eqref{eq:W5-enclosure} encloses both series truncation and arithmetic error, and $W_5[s](10^{-2})<0$ contradicts the necessary sign. \end{proof} \begin{proof}[Proof of Theorem~\ref{thm:main-full}] Propositions~\ref{prop:tail-signs} and~\ref{prop:compact-signs} prove the $m=3,4$ signs. Positivity gives $m=1$, and the strict log-concavity of $s$ \citep[Theorem~2.1]{csordas1988moment} gives $m=2$ for $t>0$. At zero, an interval enclosure of the theta series gives $-38.0500,\qquad \mathcal L^2(s)=s\tau_3>0,\qquad \mathcal L^3(s)=\tau_2\bigl(\tau_3^2+s^2\tau_4\bigr)>0. \end{equation} In contrast, $(\log W_4[s])''(10^{-2})>0$. \end{corr} \begin{proof} Theorem~\ref{thm:main-full} and \eqref{eq:log-Am} give the three concavity assertions. Applying $\mathcal L(g)=-g^2(\log g)''$ first to $\tau_2$ and then to $s\tau_3$ gives \eqref{eq:differential-iterates}. Finally, $(\log W_4)''=W_3W_5/W_4^2>0$ at the certified witness. \end{proof} \subsection{Coefficient inequalities}\label{sec:coeff-consequences} The numbers $\beta_a$ are moments of $s$, so integrating the kernel $s(x+y)$ against powers in each variable transfers its minors to the matrix $(\gamma_{r_i+c_j})$. The transfer applies the continuous Cauchy--Binet (Andr\'eief) formula twice and handles the power determinants as Chebyshev systems. \begin{proof}[Proof of Theorem~\ref{thm:coef-RR4}] Since \[ \beta_k=\int_0^\infty \Phi(r)r^{2k}\dd r =\frac12\int_0^\infty s(u)u^{k-1/2}\dd u, \] consider, for increasing nonnegative real sequences $r_i,c_j$, the matrix \[ I_{ij}=\int_0^\infty\int_0^\infty x^{r_i-3/4}y^{c_j-3/4}s(x+y)\dd x\dd y. \] All integrals converge: the exponents at zero exceed $-1$, and $s$ decays faster than any power at infinity. Applying the continuous Cauchy--Binet formula twice gives \[ \det I=\int_{x_1<\cdots0$, with strictness on the whole integration domain. Changing variables $w=x+y$, $v=x/(x+y)$, gives \[ I_{ij}=B\left(r_i+\frac14,c_j+\frac14\right) \int_0^\infty s(w)w^{r_i+c_j-1/2}\dd w =2B\left(r_i+\frac14,c_j+\frac14\right)\beta_{r_i+c_j}. \] Using \[ \gamma_a=\frac{\sqrt\pi}{4^a\Gamma(a+1/2)}\beta_a, \] we may write \[ I_{ij}=\frac2{\sqrt\pi} \bigl[4^{r_i}\Gamma(r_i+1/4)\bigr] \bigl[4^{c_j}\Gamma(c_j+1/4)\bigr]\gamma_{r_i+c_j}. \] All displayed row and column factors are positive. Hence $\det(I_{ij})$ and $\det(\gamma_{r_i+c_j})$ have the same sign, which proves \eqref{eq:gamma-RR4}. \end{proof} \begin{proof}[Proof of Corollary~\ref{cor:triple-turan}] Let \[ \Delta_m(n)=\det(\gamma_{n+i+j})_{i,j=0}^{m-1}. \] Theorem~\ref{thm:coef-RR4} gives $\Delta_2(n)<0$, $\Delta_3(n)<0$, and $\Delta_4(n)>0$. Since \[ \Delta_2(n)=\gamma_n\gamma_{n+2}-\gamma_{n+1}^2=-T_{n+1}, \] we get $T_k>0$. Desnanot--Jacobi condensation gives \[ \Delta_3(n)\gamma_{n+2}=\Delta_2(n)\Delta_2(n+2)-\Delta_2(n+1)^2 =-E_{n+2}. \] Since $\Delta_3(n)<0$ and $\gamma_{n+2}>0$, the identity gives $E_k>0$. A second condensation identity gives \[ \Delta_4(n)\Delta_2(n+2)=\Delta_3(n)\Delta_3(n+2)-\Delta_3(n+1)^2. \] Because $\Delta_4(n)>0$ and $\Delta_2(n+2)<0$, the sequence $D_n=-\Delta_3(n)>0$ is strictly log-concave: \[ D_{n+1}^2>D_nD_{n+2}. \] Since $E_{n+2}=\gamma_{n+2}D_n$, the same identities give the more explicit factorization \begin{equation}\label{eq:Q-factorization} Q_k=T_k\bigl(D_{k-2}^2+ \gamma_{k-1}\gamma_{k+1}\Delta_4(k-3)\bigr)>0, \qquad k\ge3. \end{equation} \end{proof} \subsection{Fourier-side inequalities}\label{sec:Fourier-side} Fourier inversion writes each derivative of $s$ as an integral of $\Xi_\Phi$ against an explicit kernel, so each Wronskian sign becomes a multilinear integral inequality for $\Xi_\Phi$. The theta kernel and all its derivatives decay rapidly on the real line, so $\Phi$ and $\Xi_\Phi$ are Schwartz functions. For fixed $r>0$, each $\Theta_j(r,\lambda)$ has at most polynomial growth in $\lambda$. Thus differentiation under the Fourier integral and the applications of Fubini below are justified by absolute convergence. Fourier inversion gives \[ s(t)=\Phi(\sqrt t) =\frac1\pi\int_0^\infty \Xi_\Phi(\lambda)\cos(\lambda\sqrt t)\dd\lambda. \] Writing $r=\sqrt t$, we have \[ s^{(j)}(t)=\frac1\pi\int_0^\infty \Xi_\Phi(\lambda)\Theta_j(r,\lambda)\dd\lambda, \] where $\Theta_j$ is defined by \eqref{eq:Theta-def}. Multilinearity of the determinant in its columns gives \[ W_m[s](r^2)=\frac1{\pi^m} \int_{(0,\infty)^m} \left(\prod_{\ell=1}^m\Xi_\Phi(\lambda_\ell)\right) \det(\Theta_{i+j-2}(r,\lambda_j))_{i,j=1}^m \dd\lambda_1\cdots\dd\lambda_m. \] Multiplying by $\eps_m$ and using Theorem~\ref{thm:main-full} gives \eqref{eq:Fourier-ineq} for $m\le4$. The failure for $m=5$ follows from \eqref{eq:W5-fail}, completing the proof of Theorem~\ref{thm:Fourier-side}. The first four kernels are \[ \Theta_0(r,\lambda)=\cos(r\lambda), \] \[ \Theta_1(r,\lambda)=-\frac{\lambda}{2r}\sin(r\lambda), \] \[ \Theta_2(r,\lambda)=\frac{\lambda}{4r^3}\sin(r\lambda)-\frac{\lambda^2}{4r^2}\cos(r\lambda), \] \[ \Theta_3(r,\lambda)=\frac{r^2\lambda^3\sin(r\lambda)+3r\lambda^2\cos(r\lambda)-3\lambda\sin(r\lambda)}{8r^5}. \] For $m=2$, the inequality is the quadratic integral constraint \[ \int_0^\infty\int_0^\infty \Xi_\Phi(\lambda)\Xi_\Phi(\mu)\Kc_2(r;\lambda,\mu) \dd\lambda\dd\mu>0, \] where symmetrizing in $\lambda,\mu$ gives \begin{align*} \Kc_2(r;\lambda,\mu)=& \frac{\lambda\mu}{4r^2}\sin(r\lambda)\sin(r\mu) +\frac{\lambda^2+\mu^2}{8r^2}\cos(r\lambda)\cos(r\mu)\\ &-\frac{\lambda\sin(r\lambda)\cos(r\mu)+\mu\cos(r\lambda)\sin(r\mu)}{8r^3}. \end{align*} The $m=3$ and $m=4$ cases are the corresponding cubic and quartic determinant inequalities. \begin{aiusage} % The results in this paper were developed with an AI assistant (OpenAI's GPT~Pro). 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