\documentclass[12pt, notitlepage]{amsart} \usepackage{latexsym, amsfonts, amsmath, amssymb, amsthm, cite} \usepackage{enumerate} %%%%%%%%%% Start TeXmacs macros \newcommand{\mathbbm}{\mathbb} \newcommand{\tmem}{\textit} \newcommand{\tmmathbf}{\mathbf} \newcommand{\tmtextit}{\textit} \newcommand{\tmtextbf}{\textbf} \newcommand{\tmscript}[1]{\text{\scriptsize{$#1$}}} \newtheorem{lemma}{Lemma} \newtheorem{proposition}{Proposition} \newtheorem{theorem}{Theorem} \newtheorem*{nonumlemma}{Lemma} \newtheorem{corollary}{Corollary} \newtheorem{conjecture}{Conjecture} \newtheorem{remark}{Remark} %%%%%%%%%% End TeXmacs macros \newcommand{\assign}{:=} \newcommand{\mathcatalan}{C} \newcommand{\mathd}{\mathrm{d}} \newcommand{\upl}{+} %\newcommand{i}{\mathrm{i}} \newcommand{\tmop}{\text} \newcommand{\um}{-} \newcommand{\E}{\mathbb E} \pagestyle{headings} \oddsidemargin -0.25in \evensidemargin -0.25in \textwidth 6.5in \sloppy \flushbottom \parindent 1em \marginparwidth 48pt \marginparsep 10pt \columnsep 10mm \usepackage{graphicx} \usepackage{mathrsfs} %--------Meta Data: Fill in your info------ \title{Hilbert-type inequality with primes} \author{Kannan Soundararajan} \address{Department of Mathematics, Stanford University, Stanford, CA, USA} \email{ksound@stanford.edu} \thanks{The author was partially supported by NSF grant DMS-2100933.} \newcommand{\PP}{\mathbb P} \newcommand{\NN}{\mathbb N} \newcommand{\CC}{\mathbb C} \newcommand{\RR}{\mathbb R} \newcommand{\one}{\mathbf 1} \newcommand{\norm}[1]{\left\lVert #1\right\rVert} \newcommand{\abs}[1]{\left\lvert #1\right\rvert} \newcommand{\inner}[2]{\left\langle #1,#2\right\rangle} \DeclareMathOperator{\lcm}{lcm} \begin{document} \maketitle \vspace{-1.2em} \begin{abstract} Let $P$ be a finite set of primes, and let $K_P$ be the matrix whose off-diagonal entries are $|p-q|^{-1}$ and whose diagonal entries vanish. It is shown that the operator norm of $K_P$ on $\ell^2(P)$ is bounded by an absolute constant, uniformly in $P$. \end{abstract} \section{Introduction} \label{sec:statement} Answering Problem 22 in Montgomery's list \cite{Montgomery}, this note establishes the following Theorem. \begin{theorem} \label{thm:main} Let $P$ be a finite set of primes. Define an operator $K_P$ on the Hilbert space $\ell^2(P)$ by setting \begin{equation*} (K_P a)(p)=\sum_{\substack{q\in P\\q\ne p}}\frac{a(q)}{|p-q|}. \label{eq:kernel} \end{equation*} Then there is an absolute constant $C$ such that $\norm{K_P} \le C$. In other words, if $a(p)$ (for $p\in P$) are arbitrary complex numbers then \begin{equation} \Big|{\sum_{\substack{p,q\in P\\p\ne q}} \frac{a(p)\overline{a(q)}}{|p-q|}}\Big| \le \sum_{\substack{p,q\in P\\p\ne q}} \frac{|a(p)|\,|a(q)|}{|p-q|} \le C\sum_{p\in P}|a(p)|^2. \label{eq:main-form} \end{equation} \end{theorem} The proof given here is a rewriting of an argument supplied by ChatGPT 6. My initial interest had been Problem 21 in Montgomery's list, which asked for the optimal constant in a weighted version of Hilbert's inequality. This was a well-known open problem, with a guessed optimal constant of $\pi$. The LLM drew my attention to a very recent preprint of Brad Rodgers \cite{Rodgers}, which established that in fact the optimal answer cannot be $\pi$ and must be a little larger (although its true value is not yet determined). On a whim, I then prompted the LLM to consider Problem 22, and it quickly produced the solution explained here. For each integer $j\ge 0$, set $L_j = 2^{2^j}$, and define the operator $T_j$ on $\ell^2(P)$ by setting \begin{equation*} \label{1.3} (T_j a)(p) = \sum_{\substack{ q\in P \\ L_j \le |p-q| < L_{j+1}}} \frac{a(q)}{|p-q|}. \end{equation*} Note that only finitely many of the $T_j$ are non-zero, and that we may write \begin{equation*} \label{1.4} K_P = E + \sum_{j=0}^{\infty} T_j, \end{equation*} where the ``error'' $E$ accounts for the possibility that $2$ and $3$ both belong to $P$. Explicitly $E=0$ unless $2$ and $3$ are both in $P$, in which case $E$ sends $a(2)$ to $a(3)$ and vice versa, and sends all other $a(p)$ to $0$. Since $E$ has operator norm $\le 1$, to establish the boundedness of $K_P$ it is enough to show the boundedness of $\sum_{j=0}^{\infty} T_j$. Using Schur's inequality (see Lemma \ref{lem:schur}) together with the Brun-Titchmarsh inequality for the number of primes in an interval, one can show that $\norm{T_j} \le A$ for an absolute constant $A$ and all $j$. Further, by applying Schur's inequality with an upper bound sieve for counting pairs of primes $n$, $n+h$ with $n$ lying in an interval, one can establish that $\norm{T_jT_k} \le B 2^{-|j-k|/2}$ for an absolute constant $B$. This estimate implies that the operators $T_j$ and $T_k$ are nearly orthogonal when $j$ and $k$ are not close. The Cotlar-Stein Lemma (see Lemma \ref{lem:cotlar} below) gives a bound for the operator norm of a sum of nearly orthogonal operators, and enables one to deduce the boundedness of $K_P$. The immediate upper bound for $\norm{K_P}$ is $\ll \log \log (3+ |P|)$. In his notes to Problem 22, Montgomery mentions a construction of Ruzsa which indicates that the Brun-Titchmarsh bound alone is insufficient to go beyond this first bound. The new ingredient used in the proof here is a sieve bound for pairs of primes (see \eqref{eq:two-prime}) combined with the Cotlar-Stein Lemma. Indeed the proof would apply to any set of integers satisfying the sieve bounds recorded in Lemma \ref{lem:sieve}. Once $K_P$ is shown to be bounded, the inequality \eqref{eq:main-form} follows by Cauchy-Schwarz: setting $b(p) = |a(p)|$ we have $$ |\langle a, K_Pa \rangle | \le \langle b, K_P b\rangle \le \norm{b} \norm{K_P b} \le \norm{K_P} \norm{b}^2 = \norm{K_P} \norm{a}^2. $$ \section{Preliminary Lemmas} \label{sec:sieve-input} This section collects together three lemmas used in our proof. The first lemma records a simple bound for the operator norm of a matrix in terms of its absolute row and column sums. \begin{lemma}[Schur's inequality] \label{lem:schur} Let $M=(m_{uv})$ be a finite rectangular complex matrix. Suppose \[ \max_u\sum_v|m_{uv}|\le R, \qquad \max_v\sum_u|m_{uv}|\le C, \] so that the absolute row sums are bounded by $R$, and the absolute column sums by $C$. Then, the norm of $M$ as an operator from $\ell^2$ to $\ell^2$ satisfies \begin{equation*} \norm{M} \le\sqrt{RC}. \label{eq:schur} \end{equation*} %In particular, if $M$ is real symmetric and entrywise nonnegative, %$M\one\le A\one$ implies $\norm{M}\le A$. \end{lemma} \begin{proof} For every vector $x$, Cauchy--Schwarz gives \[ |(Mx)_u|^2 = \Big| \sum_{v} m_{uv} x_v \Big|^2 \le\Big(\sum_v|m_{uv}|\Big) \Big(\sum_v|m_{uv}|\,|x_v|^2\Big) \le R \sum_v|m_{uv}|\,|x_v|^2. \] Summing over $u$ and using the column bound yields $$ \norm{Mx}^2 \le R \sum_v|x_v|^2\sum_u|m_{uv}| \le RC\norm{x}^2. $$ \end{proof} Our next lemma is the key Cotlar-Stein estimate for the norm of nearly orthogonal operators (see Chapter VII \S 2 of \cite{S}). \begin{lemma}[Cotlar--Stein] \label{lem:cotlar} For $1\le j\le N$, let $S_j:\mathcal H\to\mathcal K$ be bounded linear operators between two complex Hilbert spaces. Put \begin{equation*} A=\max_{j}\sum_{k}\Vert{S_j^*S_k}\Vert^{\frac 12}, \qquad B=\max_{j}\sum_{k}\norm{S_jS_k^*}^{\frac 12}. \label{eq:cotlar-hypotheses} \end{equation*} Then \begin{equation*} \Big \Vert \sum_{j=1}^NS_j \Big \Vert \le\sqrt{AB}. \label{eq:cotlar-conclusion} \end{equation*} \end{lemma} \begin{proof} Write $S=\sum_{j}S_j$, and note that for any integer $m\ge 1$ \begin{equation} \norm{S}^{2m} = \Vert (S^*S)^{m} \Vert \le\sum_{i_1,\ldots,i_{2m} =1}^N \norm{S_{i_1}^*S_{i_2}S_{i_3}^*S_{i_4}\cdots S_{i_{2m-1}}^*S_{i_{2m}}}, \label{eq:cotlar-expansion} \end{equation} upon expanding out the power and using the triangle inequality. Observe that by sub-multiplicativity $$ \norm{S_{i_1}^*S_{i_2}S_{i_3}^*S_{i_4}\cdots S_{i_{2m-1}}^*S_{i_{2m}}} = \norm{\prod_{r=1}^{m} S_{i_{2r-1}}^*S_{i_{2r}}} \le \prod_{r=1}^{m} \norm{S_{i_{2r-1}}^* S_{i_{2r}}}. $$ On the other hand, by pairing terms differently, we also have $$ \norm{S_{i_1}^*S_{i_2}S_{i_3}^*S_{i_4}\cdots S_{i_{2m-1}}^*S_{i_{2m}}} = \norm{S_{i_1}^*\Big(\prod_{r=1}^{m-1} S_{i_{2r}}S_{i_{2r+1}}^*\Big)S_{i_{2m}}} \le B^2 \prod_{r=1}^{m-1} \norm{S_{i_{2r}} S_{i_{2r+1}}^*}, $$ where we used that $\Vert{S_{i_1}^*}\Vert = \Vert{S_{i_1}S_{i_1}^*}\Vert^{\frac 12} \le B$ and similarly $\Vert{S_{i_{2m}}}\Vert = \Vert{S_{i_{2m}}S_{i_{2m}}^*}\Vert^{\frac 12} \le B$. Using the geometric mean of both bounds, we conclude that \begin{equation} \label{eq:cotlar-chain} \norm{S_{i_1}^*S_{i_2}S_{i_3}^*S_{i_4}\cdots S_{i_{2m-1}}^*S_{i_{2m}}} \le B \Big( \prod_{r=1}^{m} \norm{S_{i_{2r-1}}^* S_{i_{2r}}}^{\frac 12}\Big) \Big( \prod_{r=1}^{m-1} \norm{S_{i_{2r}} S_{i_{2r+1}}^*}^{\frac 12} \Big). \end{equation} Substitute \eqref{eq:cotlar-chain} into \eqref{eq:cotlar-expansion} and sum first over $i_{2m}$. This index features only in one term in the products in \eqref{eq:cotlar-chain}, namely $\Vert S_{i_{2m-1}}^* S_{i_{2m}} \Vert^{\frac 12}$, so that we may sum over all choices for $i_{2m}$ and bound that contribution by the maximum row sum $A$. Sum now over $i_{2m-1}$, which features now only in the product $\Vert S_{i_{2m-2}}S_{i_{2m-1}}^*\Vert^{\frac 12}$ so that this sum may be bounded by $B$. Proceed in this manner, summing in order over the indices $i_{2m-2}$, $\ldots$, $i_2$, obtaining $m$ factors of $A$ and $m-1$ factors of $B$. We thus obtain $$ \Vert S\Vert^{2m} \le B A^m B^{m-1} \sum_{i_1=1}^{N} 1 = A^m B^{m} N. $$ Taking $2m$-th roots on both sides, and letting $m\to \infty$, the lemma follows. \end{proof} Finally our third lemma records classical sieve estimates for primes and pairs of primes in intervals. \begin{lemma} \label{lem:sieve} There are absolute constants $c_1,c_2$ such that, for every real interval $I$ of length $R\ge2$, \begin{equation} \# \{ n \in I: \ \ n\text{ prime } \} \le c_1\frac{R}{\log R}. \label{eq:one-prime} \end{equation} Moreover, for every positive integer $h$, \begin{equation} \#\{n\in I: n ,n+h \text { both prime } \} \le c_2\frac{h}{\phi(h)}\frac{R}{(\log R)^2}. \label{eq:two-prime} \end{equation} \end{lemma} \begin{proof} The bound \eqref{eq:one-prime} follows from the Brun-Titchmarsh inequality, see Corollary 3.4 of \cite{MV}. The bound \eqref{eq:two-prime} also follows from a straightforward application of the Selberg sieve. It may be found as Corollary 3.14 of \cite{MV}, where the right side is given as $$ \frac{8c(h)R}{(\log R)^2} \Big( 1+ O\Big(\frac{\log \log R}{\log R}\Big)\Big), $$ uniformly in $h$, where $c(h) = c \prod_{p|h, p>2} \frac{p-1}{p-2}$ with $c= 2 \prod_{p>2} (1-1/(p-1)^2)$ being the twin-prime constant. In Corollary 3.14 it is assumed that $h$ is even and $R\ge 4$, but these are plainly unimportant, and the stated version follows immediately. \end{proof} \section{Proof of the main theorem} \label{sec:main-proof} All operators in this section act on $\ell^2(P)$, and $\one$ denotes the all-ones vector. We begin by showing that the operator norms of $T_j$ are bounded (uniformly for all $j$). \begin{lemma} \label{lem:block-row} There is an absolute constant $A_0\ge1$ such that for all $j\ge 0$ we have $T_j\one\le A_0\one$, and therefore $\norm{T_j}\le A_0$. \end{lemma} \begin{proof} If $p \in P$, then by definition $$ (T_j\one)(p) = \sum_{\substack{ q\in P \\ L_j \le |p-q| < L_{j+1}}} \frac{1}{|p-q|} \le \sum_{m=2^{j}}^{2^{j+1}-1} \frac{1}{2^m} \sum_{\substack{ q\in P \\ 2^m \le |p-q| <2^{m+1}}} 1, $$ upon splitting the interval $[L_j,L_{j+1})$ into dyadic blocks. Using the Brun-Titchmarsh inequality \eqref{eq:one-prime}, the above is $$ \ll \sum_{m=2^{j}}^{2^{j+1}-1} \frac{1}{2^m} \frac{2^m}{m} \ll 1. $$ This bound shows that the row sums of the matrix defining $T_j$ (in the obvious basis for $\ell^2(P)$) are bounded. Since the operator is symmetric, the same holds for the column sums. Therefore, by Schur's inequality (Lemma \ref{lem:schur}) we conclude that $\norm{T_j}\ll 1$. \end{proof} The next lemma is the key step and bounds the operator norm of $T_jT_k$, showing that this norm tends rapidly to zero as $|j-k|$ increases. \begin{lemma} \label{lem:mixed-row} There is an absolute constant $B_0$ such that, for $0\le jk$: $\norm{T_kT_j} = \norm{(T_kT_j)^*} = \norm{T_j^*T_k^*} = \norm{T_jT_k}$. Finally the case $j=k$ is already covered by Lemma \ref{lem:block-row}. This completes our proof, and moreover a permissible value for $D$ is $\max( A_0^2, A_0 \sqrt{B_0})$. \end{proof} The proof of our main theorem is now at hand. Since the operators $T_j$ are self-adjoint, and only finitely many of them are non-zero, we may apply the Cotlar-Stein Lemma together with the bound of Lemma \ref{lem:mixed-row}. It follows that $$ \max_j \sum_{k} \norm{T_j^* T_k}^{\frac 12} \le \max_j \sum_k \sqrt{D} 2^{-|j-k|/4} \le 2\sqrt{D} \sum_{\ell= 0}^{\infty} 2^{-\ell/4} \le 20 \sqrt{D}, $$ and the same bound applies to $\max_j \sum_k \norm{T_j T_k^*}^{\frac 12}$. We conclude that $$ \norm{K_P} \le \norm{E} + \Big \Vert \sum_{j} T_j \Big\Vert \le 1+ 20\sqrt{D}, $$ so that $\norm{K_P}$ is uniformly bounded by an absolute constant, as desired. \begin{thebibliography}{9} \bibitem{Montgomery} H.~L.~Montgomery, \emph{Ten Lectures on the Interface Between Analytic Number Theory and Harmonic Analysis}, CBMS Regional Conference Series in Mathematics, vol.~84, American Mathematical Society, Providence, RI, 1994. \bibitem{MV} H.~L.~Montgomery and R.~C.~Vaughan, \emph{Multiplicative Number Theory I. Classical Theory}, Cambridge Studies in Advanced Mathematics, vol.~97, Cambridge University Press, 2007. \bibitem{Rodgers} Brad Rodgers, \emph{On the optimal constant in the Montgomery-Vaughan weighted Hilbert inequality}, preprint, {\tt arXiv:2608.12315v1}, (2026). \bibitem{S} E.~M.~Stein, \emph{Harmonic Analysis: Real-variable methods, Orthogonality, and Oscillatory Integrals}. Princeton Mathematical Series, vol.~43, Princeton University Press, 1993. \end{thebibliography} \end{document}