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\title[P\'olya--Vinogradov with $\log\log q$]{The P\'olya--Vinogradov inequality with $\log\log q$}

\author[D. Bhattacharjee]{Deep Bhattacharjee}
\address{Formerly, Electro-Gravitational Space Propulsion Laboratory (EGSPL),
Bhubaneswar, Odisha 751030, India; Formerly, University of Calcutta, Senate House,
87/1 College Street, Kolkata 700073, West Bengal, India}
\email{itsdeep@live.com}
\email{d.bhattacharjee@erl-forschung.de}
\urladdr{https://orcid.org/0000-0003-0466-750X}
\thanks{Corresponding author: Deep Bhattacharjee. An earlier version of this paper is the
Hexagon preprint \texttt{2610.00256v1}, \url{https://hexagonmath.org/2610.00256v1}.}

\author[U. Bhattacharya]{Ushashi Bhattacharya}
\address{Formerly, National Taiwan University, Taipei 106319, Taiwan}
\email{rai.bhattacharya.in@gmail.com}
\urladdr{https://orcid.org/0009-0002-2254-3914}

\subjclass[2020]{Primary 11L40; Secondary 11M06, 11M20, 11L07, 11N13, 11N36, 11R42}
\keywords{P\'olya--Vinogradov inequality, character sums, zero-free regions, Selberg sieve,
Euler--Kronecker constants}

\begin{document}

\begin{abstract}
We prove that every primitive Dirichlet character $\chi$ modulo $q\ge3$ has character sums
$\sum_{n\le x}\chi(n)$ of size at most $C\sqrt q\log\log q$, with an absolute constant $C$.
Montgomery and Vaughan proved this bound under the generalised Riemann hypothesis, and Paley
showed that it is optimal. The only input from outside the paper is the zero-free half-plane
$\Re s>7/8$ for Dirichlet $L$-functions. On the major arcs this half-plane gives a power saving for
twisted prime sums; on the minor arcs we prove a weak form of the Montgomery--Vaughan bound for
exponential sums with multiplicative coefficients, with Selberg's sieve as the only arithmetic
input. The same half-plane gives $|L'/L(1,\chi)|\ll\log\log q$ for every nonprincipal $\chi$
and, for every prime $q$, the two-sided bound
$-C\log q\log\log q\le\gamma_q\le C\log q$ for the Euler--Kronecker constant of the cyclotomic
field $\mathbb Q(\zeta_q)$.
\end{abstract}

\maketitle

\section{Introduction}

Let $\chi$ be a primitive Dirichlet character modulo $q\ge3$. The P\'olya--Vinogradov inequality
states that $\bigl|\sum_{n\le x}\chi(n)\bigr|\ll\sqrt q\log q$ uniformly in
$x$~\cite[Chapter~23]{Dav00}.\footnote{The proof expands $\chi$ in additive characters through the
Gauss sum $\tau(\bar\chi)$, whose modulus is $\sqrt q$ for primitive $\chi$, and bounds the
resulting geometric series trivially; Lemma~\ref{lem:polya} below is the same argument with the
harmonic sums kept.} Paley proved that the factor $\log q$ cannot be replaced by anything smaller
than $\log\log q$: for infinitely many primitive quadratic characters the maximum over $x$ exceeds
$c\sqrt q\log\log q$~\cite{Pal32}. Montgomery and Vaughan~\cite{MV77} proved the upper bound
$\sqrt q\log\log q$ under the generalised Riemann hypothesis. Without hypothesis, Granville and
Soundararajan lowered the exponent of $\log q$ for characters of fixed odd
order~\cite{GS07}.\footnote{For a primitive character of odd order $g$ they obtain
$\sqrt q(\log q)^{1-\delta_g/2+o(1)}$ with $\delta_g=1-\frac g\pi\sin\frac\pi g$, and under the
generalised Riemann hypothesis the same exponent of $\log\log q$.} We prove the bound of Montgomery and Vaughan for every primitive character.

\begin{theorem}\label{thm:pv}
There is an absolute constant $C$ such that for every primitive character $\chi$ modulo $q\ge3$
and every $x\ge1$,
\[
  \Bigl|\sum_{n\le x}\chi(n)\Bigr|\le C\sqrt q\log\log q.
\]
\end{theorem}

By Paley's theorem, Theorem~\ref{thm:pv} is optimal up to the value of $C$. The input that
replaces the Riemann hypothesis is the following theorem.

\begin{theorem}[{\cite{OAI26}}]\label{thm:zerofree}
If $\chi$ is a nonprincipal Dirichlet character, then $L(s,\chi)\ne0$ for $\Re s>7/8$.
\end{theorem}

Theorem~\ref{thm:zerofree} is proved, and formalised in Lean~4, in~\cite{OAI26}. It is the only
result we use without proof; everything else is proved below from first principles, and the
constants are effective.

Montgomery and Vaughan use the Riemann hypothesis in one place: on the major arcs, for prime sums
of the characters $\chi\psi$ twisted by additive characters, where $\psi$ runs over the characters
of a small modulus $r$. A power saving in these sums is enough there, and
Theorem~\ref{thm:zerofree} provides one (Corollary~\ref{cor:prime}). On the minor arcs they use
their unconditional bound for exponential sums with multiplicative coefficients. A weaker form of
it suffices, and we prove that form in Section~\ref{sec:mv} with Selberg's sieve as the only
arithmetic input. Figure~\ref{fig:flow} shows how the parts fit together.

\begin{figure}[ht]
\centering
\includegraphics[width=\textwidth]{figures/fig_flow.png}
\caption{The structure of the proofs. The left branch proves Theorem~\ref{thm:pv}: the
Fourier expansion of a character sum reduces it to the sums
$\sum_{m\le H}\chi(m)e(m\beta)/m$, which are bounded on the major arcs by the half-plane and on
the minor arcs by bilinear forms and the sieve. The right branch proves
Theorem~\ref{thm:ek}.}
\label{fig:flow}
\end{figure}

The same half-plane controls the logarithmic derivative of $L(s,\chi)$ at $s=1$. For a number
field $K$ write $\zeta_K(s)=c_{-1}(s-1)^{-1}+c_0+O(s-1)$ near $s=1$; the Euler--Kronecker
constant of $K$ is $\gamma_K=c_0/c_{-1}$. Let $\gamma_q$ be the Euler--Kronecker constant of
$\mathbb Q(\zeta_q)$. For a prime $q$ the factorisation $\zeta_{\mathbb Q(\zeta_q)}(s)
=\zeta(s)\prod_{\chi\ne\chi_0}L(s,\chi)$, the product over the nonprincipal characters modulo
$q$~\cite[Theorem~4.3]{Was97},\footnote{All nonprincipal characters modulo a prime are
primitive, so no Euler factors have to be removed: the conductor of each nonprincipal $\chi$ is
$q$ itself, and $L(s,\chi)$ is the $L$-function of the corresponding primitive character.} gives
\begin{equation}\label{eq:gammaq}
  \gamma_q=\gamma+\sum_{\chi\ne\chi_0}\frac{L'}{L}(1,\chi),
\end{equation}
where $\gamma$ is Euler's constant.

\begin{theorem}\label{thm:ek}
There is an absolute constant $C$ with the following properties.
\begin{enumerate}
\item[(i)] For every nonprincipal character $\chi$ modulo $q$,
\[
  \Bigl|\frac{L'}{L}(1,\chi)\Bigr|\le C\log\log q.
\]
\item[(ii)] For every prime $q\ge3$,
$-C\log q\log\log q\le\gamma_q\le C\log q$.
\end{enumerate}
\end{theorem}

Under the generalised Riemann hypothesis the bound~(i) is classical, Ihara proved
$\gamma_q\le(2+o(1))\log q$ and Badzyan proved a lower bound of the order
of~(ii)~\cite{Iha06,Bad10}. Ford, Luca and Moree proved the two bounds of~(ii) without hypothesis
for a set of primes $q$ of density one~\cite{FLM14,Mor18}.\footnote{The difficulty for an
individual prime $q$ lies in the prime powers $n\equiv1\pmod q$, which enter $\gamma_q$ with the
weight $(q-1)\Lambda(n)/n$. Below $q^{17}$ we bound them for each prime $q$ separately by
Lemma~\ref{lem:progression}; beyond $q^{17}$ the zero-free half-plane makes their contribution
bounded.} For all primes the best unconditional bound we know is
$|\gamma_q|\ll_\varepsilon q^\varepsilon$, due to Ihara, Murty and Shimura~\cite{IMS09}. The
upper bound in~(ii) needs only Chebyshev's bound. The lower bound needs an upper bound for the
prime powers $n\le q^{17}$ with $n\equiv1\pmod q$, which we obtain from Selberg's sieve applied
in dyadic blocks (Lemma~\ref{lem:progression}).

Theorems~\ref{thm:pv} and~\ref{thm:ek} are not among the consequences of
Theorem~\ref{thm:zerofree} drawn in~\cite{OAI26}. The arguments work, with other constants, whenever $L(s,\chi)\ne0$ on $\Re s>\theta$ for all nonprincipal $\chi$
and some fixed $\theta<1$.
The paper is organised as follows. Section~\ref{sec:prelim} collects Chebyshev's bounds and
Selberg's sieve. Section~\ref{sec:riesz} bounds Riesz means of $\chi\Lambda$ by a contour shift.
Section~\ref{sec:mv} proves the bound for exponential sums with multiplicative coefficients,
Section~\ref{sec:pv} proves Theorem~\ref{thm:pv} and Section~\ref{sec:ek} proves
Theorem~\ref{thm:ek}.

\section{Preliminaries}\label{sec:prelim}

Throughout, $p$ and $\ell$ denote primes, $\Lambda$ is the von Mangoldt function,
$\theta(x)=\sum_{p\le x}\log p$, $\psi(x)=\sum_{n\le x}\Lambda(n)$, $e(x)=e^{2\pi ix}$, and
$\|x\|$ is the distance from $x$ to the nearest integer. We write $\omega(n)$ for the number of
prime factors of $n$, $\tau_3(n)$ for the number of ordered factorisations $n=abc$, and $P(n)$
for the largest prime factor of $n\ge2$, with $P(1)=1$.

\begin{lemma}\label{lem:chebyshev}
For $x\ge1$ and integers $N\ge2$,
\begin{enumerate}
\item[(i)] $\theta(x)\le x\log4$;
\item[(ii)] $\psi(x)-\theta(x)\le2\sqrt x\log x$ and $\psi(x)\le4x$;
\item[(iii)] $\sum_{p\le N}\frac{\log p}p\le\log N+\log4$;
\item[(iv)] $\sum_{p\le N}\frac1p\le\log\log N+10$.
\end{enumerate}
\end{lemma}

\begin{proof}
(i) It suffices to treat integers $m\ge1$, and we use induction on $m$. The cases $m=1,2$ are
clear, and for even $m>2$ we have $\theta(m)=\theta(m-1)$. Let $m=2n+1$. The primes in
$(n+1,2n+1]$ divide $\binom{2n+1}n$, and $2\binom{2n+1}n\le2^{2n+1}$ because $\binom{2n+1}n$ and
$\binom{2n+1}{n+1}$ are equal terms of the binomial expansion of $(1+1)^{2n+1}$. Hence
$\theta(2n+1)\le\theta(n+1)+n\log4\le(2n+1)\log4$.

(ii) We have $\psi(x)-\theta(x)=\sum_{k\ge2}\theta(x^{1/k})$, where only $k\le\log x/\log2$
contribute. Each term is at most $\sqrt x\log4$ by~(i), so the sum is at most
$(\log x/\log2)\sqrt x\log4=2\sqrt x\log x$. Since $\log x\le\sqrt x$, $\psi(x)\le(\log4+2)x\le4x$.

(iii) By Legendre's formula and~(i),
\[
  N\log N\ge\log N!\ge\sum_{p\le N}\Bigl\lfloor\frac Np\Bigr\rfloor\log p
  \ge N\sum_{p\le N}\frac{\log p}p-N\log4.
\]

(iv) Let $A(u)=\sum_{p\le u}\log p/p$. Partial summation and~(iii) give
\[
  \sum_{p\le N}\frac1p=\frac{A(N)}{\log N}+\int_2^N\frac{A(u)}{u\log^2u}\,du
  \le1+\frac{\log4}{\log2}+\int_2^N\Bigl(\frac1{u\log u}+\frac{\log4}{u\log^2u}\Bigr)du,
\]
and the integral is at most $\log\log N-\log\log2+2$.
\end{proof}

We use Selberg's upper bound sieve in the following form; for the method
see~\cite[Chapter~3]{HR74}, and we give the proof of the form we need.\footnote{The weights
$\lambda_d$ are those of Selberg's quadratic form, and the only feature used later is the bound
$|\lambda_d|\le1$, which makes the remainder term at most a sum of $3^{\omega(d)}|R_d|$ over
$d\le y$.}

\begin{lemma}\label{lem:selberg}
Let $a_1,\dots,a_X$ be integers, $P$ a squarefree integer, $\nu$ a multiplicative function with
$0<\nu(p)<1$ for $p\mid P$, and $y\ge1$. For $d\mid P$ put
$R_d=\#\{n\le X:d\mid a_n\}-\nu(d)X$, and for $l\mid P$ put
$g(l)=\prod_{p\mid l}\nu(p)/(1-\nu(p))$. Then
\[
  \#\{n\le X:(a_n,P)=1\}\le\frac XS+\sum_{\substack{d\mid P\\d\le y}}3^{\omega(d)}|R_d|,
  \qquad S=\sum_{\substack{l\mid P\\l\le\sqrt y}}g(l).
\]
\end{lemma}

\begin{proof}
Let $D$ be the set of divisors $d$ of $P$ with $d\le\sqrt y$, and let $(\lambda_d)_{d\in D}$ be
real numbers with $\lambda_1=1$. If $(a_n,P)=1$ then $\sum_{d\in D,\,d\mid a_n}\lambda_d=1$, and
otherwise this sum is real, so
\begin{align*}
  \#\{n:(a_n,P)=1\}&\le\sum_{n\le X}\Bigl(\sum_{\substack{d\in D\\d\mid a_n}}\lambda_d\Bigr)^2\\
  &=X\sum_{d_1,d_2\in D}\lambda_{d_1}\lambda_{d_2}\nu([d_1,d_2])
  +\sum_{d_1,d_2\in D}\lambda_{d_1}\lambda_{d_2}R_{[d_1,d_2]}.
\end{align*}
Write $Q$ for the double sum in the first term. Since $\nu$ is multiplicative and $d_1,d_2$ are
squarefree, $\nu([d_1,d_2])=\nu(d_1)\nu(d_2)/\nu((d_1,d_2))$. Let $h$ be the multiplicative
function on the divisors of $P$ with $h(p)=1/\nu(p)-1=1/g(p)$; then $1/\nu(m)=\sum_{l\mid m}h(l)$
for $m\mid P$, and
\[
  Q=\sum_{l\mid P}h(l)\Bigl(\sum_{\substack{d\in D\\l\mid d}}\lambda_d\nu(d)\Bigr)^2
  =\sum_{l\in D}\frac{y_l^2}{g(l)},\qquad y_l=\sum_{\substack{d\in D\\l\mid d}}\lambda_d\nu(d).
\]
M\"obius inversion gives $\lambda_d\nu(d)=\sum_{l\in D,\,d\mid l}\mu(l/d)y_l$, so the condition
$\lambda_1=1$ reads $\sum_{l\in D}\mu(l)y_l=1$. By Cauchy's inequality
\[
  1\le\Bigl(\sum_{l\in D}\frac{y_l^2}{g(l)}\Bigr)\Bigl(\sum_{l\in D}g(l)\Bigr)=QS,
\]
with equality for $y_l=\mu(l)g(l)/S$. We make this choice, so that $Q=1/S$. Then, writing
$l=de$,
\[
  \lambda_d=\frac{1}{\nu(d)S}\sum_{\substack{l\in D\\d\mid l}}\mu(l/d)\mu(l)g(l)
  =\frac{\mu(d)g(d)}{\nu(d)S}\sum_{\substack{e\in D,\ (e,d)=1\\de\le\sqrt y}}g(e).
\]
Since $g(d)/\nu(d)=\prod_{p\mid d}(1-\nu(p))^{-1}=\prod_{p\mid d}(1+g(p))=\sum_{f\mid d}g(f)$, we
get $|\lambda_d|\le S^{-1}\sum_{f\mid d}\sum_eg(fe)$, and the products $fe$ are distinct elements
of $D$; hence $|\lambda_d|\le1$. Finally $[d_1,d_2]\le d_1d_2\le y$, and every $d\mid P$ is the
least common multiple of at most $3^{\omega(d)}$ pairs $(d_1,d_2)$, so the second term is at most
$\sum_{d\mid P,\,d\le y}3^{\omega(d)}|R_d|$.
\end{proof}

\section{Riesz means of prime sums}\label{sec:riesz}

For an arithmetic function $a$ and $x\ge1$ put
\[
  S(a;x)=\sum_{n\le x}a(n)\Bigl(1-\frac nx\Bigr),\qquad
  S(\chi;x)=S(\chi\Lambda;x)=\sum_{n\le x}\chi(n)\Lambda(n)\Bigl(1-\frac nx\Bigr).
\]
The weight $1-n/x$ makes the Mellin kernel absolutely integrable, which keeps the contour shift
below free of truncation errors. The aim of this section is Proposition~\ref{prop:riesz}.

\begin{lemma}\label{lem:mellin}
Let $a\colon\mathbb N\to\C$ satisfy $\sum_n|a(n)|n^{-\sigma}<\infty$ for $\sigma>1$, and let
$F(s)=\sum_na(n)n^{-s}$ for $\Re s>1$.
\begin{enumerate}
\item[(i)] For $\Re s>1$, $\displaystyle\int_1^\infty S(a;x)\,x^{-s-1}\,dx=\frac{F(s)}{s(s+1)}$.
\item[(ii)] For $x>0$, $\displaystyle S(a;x)=\frac1{2\pi}\int_{-\infty}^{\infty}
F(2+it)\,\frac{x^{2+it}}{(2+it)(3+it)}\,dt$.
\end{enumerate}
In other words the following square commutes, and~(ii) inverts its right column on the line
$\Re s=2$:
\[
\begin{tikzcd}[column sep=6em,row sep=3em]
a \arrow[r,"\text{Riesz mean}"] \arrow[d,"\text{Dirichlet series}"'] &
S(a;\cdot) \arrow[d,"\text{Mellin transform}"] \\
F \arrow[r,"\times\,\frac1{s(s+1)}"'] & \dfrac{F(s)}{s(s+1)}
\end{tikzcd}
\]
\end{lemma}

\begin{proof}
(i) For $n\ge1$, $\int_n^\infty(1-n/x)x^{-s-1}dx=n^{-s}/s-n^{-s}/(s+1)=n^{-s}/(s(s+1))$. Since
$\sum_n|a(n)|\int_n^\infty x^{-\sigma-1}dx=\sigma^{-1}\sum_n|a(n)|n^{-\sigma}<\infty$, the sum
and the integral may be exchanged.

(ii) For $y>0$ let $k(y)=\frac1{2\pi i}\int_{(2)}y^s\,\frac{ds}{s(s+1)}$. If $y\ge1$, moving the
line of integration to $\Re s=-U$ and letting $U\to\infty$ picks up the residues $1$ at $s=0$ and
$-1/y$ at $s=-1$, so $k(y)=1-1/y$; if $y<1$, moving it to the right gives $k(y)=0$. As
$|s(s+1)|^{-1}$ is integrable on $\Re s=2$ and $\sum_n|a(n)|n^{-2}<\infty$, summation and
integration may be exchanged, and the right side equals $\sum_na(n)k(x/n)=S(a;x)$.
\end{proof}

\begin{lemma}\label{lem:size}
Let $\chi$ be nonprincipal modulo $q$. For $\Re s=\sigma>0$, $|L(s,\chi)|\le q|s|/\sigma$.
\end{lemma}

\begin{proof}
Put $A(t)=\sum_{n\le t}\chi(n)$; then $|A(t)|\le q$, and partial summation gives
$L(s,\chi)=s\int_1^\infty A(t)t^{-s-1}\,dt$ for $\sigma>0$. Hence
$|L(s,\chi)|\le q|s|\int_1^\infty t^{-\sigma-1}dt=q|s|/\sigma$.
\end{proof}

\begin{lemma}\label{lem:euler}
Let $A_0=\frac32\sum_\ell\ell^{-3/2}$. For every Dirichlet character $\chi$ the series
$E(s)=-\sum_\ell\log(1-\chi(\ell)\ell^{-s})$ is holomorphic on $\Re s>1$, $e^{E(s)}=L(s,\chi)$
there, and $|E(s)|\le A_0$ for $\Re s\ge3/2$.
\end{lemma}

\begin{proof}
For $\Re s\ge1$ we have $|\chi(\ell)\ell^{-s}|\le1/2$, and $|\log(1-w)|\le\sum_k|w|^k/k
\le\frac32|w|$ for $|w|\le1/2$. The series therefore converges locally uniformly on $\Re s>1$
and is bounded by $A_0$ on $\Re s\ge3/2$, and the Euler product gives $e^E=L$.
\end{proof}

\begin{lemma}[Borel--Carath\'eodory]\label{lem:bc}
Let $h$ be holomorphic on a neighbourhood of the disc $|z-c|\le R$ and let $0<r<R$. Then
\[
  \max_{|z-c|\le r}|h(z)-h(c)|\le\frac{2r}{R-r}\max_{|w-c|=R}\Re\bigl(h(w)-h(c)\bigr).
\]
\end{lemma}

\begin{proof}
Put $g(z)=h(c+z)-h(c)$ and $A=\max_{|w|=R}\Re g(w)$. The mean of $\Re g$ over the circle
$|w|=R$ is $\Re g(0)=0$, so $A\ge0$. If $A=0$, then $\Re g$ vanishes on the circle, hence on the
disc, and $g$ is an imaginary constant equal to $g(0)=0$. Let $A>0$. By the maximum principle
$\Re g\le A$ on $|z|\le R$, so $|2A-g|\ge A$ there and $\phi=g/(2A-g)$ is holomorphic. Since
$|g|^2-|2A-g|^2=4A(\Re g-A)\le0$, we have $|\phi|\le1$, and $\phi(0)=0$. Schwarz's lemma gives
$|\phi(z)|\le|z|/R$, and from $g=2A\phi/(1+\phi)$ we get $|g(z)|\le2A|z|/(R-|z|)$.
\end{proof}

\begin{lemma}\label{lem:logderiv}
Let $0<\delta<1/1000$, $1/2\le a\le1$, and let $\chi$ be nonprincipal modulo $q$ with
$L(s,\chi)\ne0$ for $\Re s>a+2\delta$. Put $K_1=(1+2A_0)/(2\delta^2)$. For
$a+8\delta\le\sigma\le2$ and $t\in\R$,
\[
  \Bigl|\frac{L'}{L}(\sigma+it,\chi)\Bigr|\le K_1\log\bigl(2q(3+|t|)^2\bigr).
\]
\end{lemma}

\begin{proof}
Let $c=2+it$, $R=2-a-3\delta$, $r=2-a-6\delta$ and $\Lambda_0=\log(2q(3+|t|)^2)\ge\log18>1$.
On the disc $|z-c|<2-a-2\delta$ we have $\Re z>a+2\delta$, so $L(z,\chi)$ has no zero there
and has a holomorphic logarithm $h$. On $|z-c|<1/2$ both $h$ and the function $E$ of
Lemma~\ref{lem:euler} are logarithms of $L$, so they differ by a constant in $2\pi i\Z$; we
choose $h=E$ there. Then $|h(c)|\le A_0$.

If $|z-c|\le R$, then $\Re z\ge a+3\delta\ge1/2$ and $|z|\le|c|+3/2\le4+|t|$, so
Lemma~\ref{lem:size} gives $\Re h(z)=\log|L(z,\chi)|\le\log(2q(4+|t|))\le\Lambda_0$.
Lemma~\ref{lem:bc} gives, for $|z-c|\le r$,
\[
  |h(z)-h(c)|\le\frac{2r}{R-r}\bigl(\Lambda_0+A_0\bigr)\le\frac{\Lambda_0+A_0}{\delta},
\]
as $r\le3/2$ and $R-r=3\delta$. Hence $|h(z)|\le A_0+(\Lambda_0+A_0)/\delta\le(1+2A_0)\Lambda_0/\delta$
on $|z-c|\le r$. For $w=\sigma+it$ with $a+8\delta\le\sigma\le2$ we have $|w-c|=2-\sigma\le r-2\delta$,
so the circle $|z-w|=2\delta$ lies in $|z-c|\le r$, and Cauchy's estimate gives
$|L'/L(w,\chi)|=|h'(w)|\le(1+2A_0)\Lambda_0/(2\delta^2)$.
\end{proof}

\begin{lemma}\label{lem:elementary}
For $q\ge1$, $t\in\R$ and $s=\sigma+it$ with $\sigma\ge1/2$,
\[
  \log\bigl(2q(3+|t|)^2\bigr)\le10\log(q+2)(1+|t|)^{1/2},\qquad |s(s+1)|\ge\tfrac18(1+|t|)^2.
\]
\end{lemma}

\begin{proof}
We have $\log(2q)\le2\log(q+2)$, and $\log y\le2y^{1/2}$ with $3+|t|\le4(1+|t|)$ gives
$\log(3+|t|)\le4(1+|t|)^{1/2}$; the first bound follows from $\log(q+2)\ge1$. For the second,
$|s|\ge\max(\sigma,|t|)\ge(1+|t|)/4$ and $|s+1|\ge\max(\sigma+1,|t|)\ge(1+|t|)/2$.
\end{proof}

\begin{proposition}\label{prop:riesz}
Let $0<\delta<1/1000$. There is $K=K(\delta)$ such that the following holds. Let $1/2\le a\le1$
and let $\chi$ be nonprincipal modulo $q$ with $L(s,\chi)\ne0$ for $\Re s>a+2\delta$. Then for
$x\ge1$,
\[
  |S(\chi;x)|\le K\,x^{a+8\delta}\log(q+2).
\]
\end{proposition}

\begin{proof}
Put $\sigma_0=a+8\delta$ and
\[
  \Phi(s)=-\frac{L'}{L}(s,\chi)\,\frac{x^s}{s(s+1)}.
\]
For $\Re s>1$, $-L'/L(s,\chi)=\sum_n\chi(n)\Lambda(n)n^{-s}$, so Lemma~\ref{lem:mellin}(ii) gives
$S(\chi;x)=\frac1{2\pi}\int\Phi(2+it)\,dt$. The function $\Phi$ is holomorphic on
$\sigma_0\le\Re s\le2$, and by Lemmas~\ref{lem:logderiv} and~\ref{lem:elementary},
\[
  |\Phi(\sigma+it)|\le80K_1\log(q+2)\,x^{\sigma}(1+|t|)^{-3/2}\qquad(\sigma_0\le\sigma\le2).
\]
This tends to $0$ as $|t|\to\infty$ uniformly in $\sigma$, so Cauchy's theorem on the rectangles
$[\sigma_0,2]\times[-T,T]$, $T\to\infty$, moves the integral to $\Re s=\sigma_0$
(Figure~\ref{fig:splane}). Since $\int_\R(1+|t|)^{-3/2}dt=4$, we get
$|S(\chi;x)|\le\frac{160K_1}{\pi}x^{\sigma_0}\log(q+2)$.
\end{proof}

\begin{figure}[ht]
\centering
\includegraphics[width=\textwidth]{figures/fig_splane.png}
\caption{The contour shift in the proof of Proposition~\ref{prop:riesz}. The zeros of
$L(s,\chi)$ lie in the strip $0\le\Re s\le7/8$ (dots), the kernel $x^s/(s(s+1))$ has poles at
$0$ and $-1$ (crosses), and the line of integration moves from $\Re s=2$ to
$\Re s=\sigma_0$ inside the zero-free half-plane.}
\label{fig:splane}
\end{figure}

From now on fix $\delta=1/2000$ and $a=7/8-1/1000$, so that $a+2\delta=7/8$ and
\[
  \sigma_0=a+8\delta=\frac78+\frac3{1000},
\]
and let $K$ be the constant of Proposition~\ref{prop:riesz} for this $\delta$. By
Theorem~\ref{thm:zerofree}, Proposition~\ref{prop:riesz} applies to every nonprincipal character.

\begin{corollary}\label{cor:prime}
There is an absolute constant $C_3$ such that for every nonprincipal character $\xi$ modulo $k$:
\begin{enumerate}
\item[(i)] $|S(\xi;x)|\le Kx^{\sigma_0}\log(k+2)$ for $x\ge1$;
\item[(ii)] $\bigl|\sum_{p\le x}\xi(p)\log p\bigr|\le C_3x^{19/20}\log(k+2)$ for $x\ge1$;
\item[(iii)] for integers $2\le A\le B$ and $\gamma\in\R$,
\begin{equation}\label{eq:twistedprime}
  \Bigl|\sum_{A<p\le B}\xi(p)e(p\gamma)\Bigr|\le
  2C_3\,\frac{B^{19/20}\log(k+2)(2+2\pi|\gamma|B)}{\log A}.
\end{equation}
\end{enumerate}
\end{corollary}

\begin{proof}
(i) is Proposition~\ref{prop:riesz}. For (ii) put $T(y)=yS(\xi;y)$, that is,
$T(y)=\sum_{n\le y}\xi(n)\Lambda(n)(y-n)$, and let $0<h\le x$. Then
\[
  T(x+h)-T(x)=h\sum_{n\le x}\xi(n)\Lambda(n)+\sum_{x<n\le x+h}\xi(n)\Lambda(n)(x+h-n),
\]
and the last sum has at most $h+1$ terms, each of modulus at most $h\log(2x)$. With
$|T(y)|\le Ky^{1+\sigma_0}\log(k+2)$ this gives
\[
  \Bigl|\sum_{n\le x}\xi(n)\Lambda(n)\Bigr|\le\frac{2K(2x)^{1+\sigma_0}\log(k+2)}h+(h+1)\log(2x).
\]
Take $h=x^{47/50}$. Since $1+\sigma_0-47/50<19/20$ and $\log(2x)\le1+100x^{1/100}$, the right side
is at most $(8K+202)x^{19/20}\log(k+2)$. The prime powers contribute at most
$\psi(x)-\theta(x)\le2\sqrt x\log x\le200x^{19/20}$ by Lemma~\ref{lem:chebyshev}(ii), and~(ii)
follows with $C_3=8K+402$.

(iii) Let $\Pi(u)=\sum_{p\le u}\xi(p)\log p$ and $f(u)=e(u\gamma)/\log u$. Partial summation gives
\[
  \sum_{A<p\le B}\xi(p)e(p\gamma)=f(B)\Pi(B)-f(A)\Pi(A)-\int_A^B\Pi(u)f'(u)\,du,
\]
and $|f'(u)|\le2\pi|\gamma|/\log A+1/(u\log^2u)$ for $u\ge A$. By~(ii) the right side is at most
$C_3B^{19/20}\log(k+2)\bigl(3+2\pi|\gamma|B\bigr)/\log A$, which gives~\eqref{eq:twistedprime}.
\end{proof}

\section{Exponential sums with multiplicative coefficients}\label{sec:mv}

For $M>0$ put $w_M(\theta)=M$ if $\|\theta\|<1/M$ and $w_M(\theta)=1/\|\theta\|$ otherwise;
then $0\le w_M(\theta)\le M$ and $w_M(-\theta)=w_M(\theta)$.

\begin{theorem}\label{thm:mv}
There is an absolute constant $C_1$ with the following property. Let $f$ be completely
multiplicative with $|f(n)|\le1$, let $\alpha\in\R$, and let $a,r$ be coprime integers with
$r\ge1$, $|\alpha-a/r|\le r^{-2}$ and $2\le R\le r\le N/R$. Then
\[
  \Bigl|\sum_{n\le N}f(n)e(n\alpha)\Bigr|\le C_1\Bigl(\frac N{\log N}
  +\frac{N(\log N)^2}{R^{1/4}}\Bigr).
\]
\end{theorem}

Montgomery and Vaughan~\cite[Corollary~1]{MV77} proved this for multiplicative $f$ with
$N(\log R)^{3/2}R^{-1/2}$ in place of $N(\log N)^2R^{-1/4}$. The proof below follows their
approach, a bilinear decomposition together with the spacing of the multiples of $\alpha$ and an
upper bound for prime pairs, with cruder estimates, which suffice for Theorem~\ref{thm:pv}.

\begin{lemma}\label{lem:spacing}
Let $a,r$ be coprime with $r\ge2$ and $|\alpha-a/r|\le r^{-2}$, and let $H\ge0$ be an integer.
\begin{enumerate}
\item[(i)] $\|k\alpha\|\ge1/(2r)$ for every integer $k$ with $0<|k|\le r/2$.
\item[(ii)] $\#\{1\le h\le H:\|h\alpha\|<t\}\le(4H/r+1)(4rt+1)$ for every $t\ge0$.
\item[(iii)] For $M>0$ and every integer $K\ge0$ with $M\le2^K$,
\begin{align*}
  \sum_{h\le H}w_M(h\alpha)&\le\Bigl(\frac{4H}r+1\Bigr)(3M+4r+8rK),\\
  \sum_{h\le H}w_M(h\alpha)^2&\le\Bigl(\frac{4H}r+1\Bigr)(20rM+3M^2).
\end{align*}
\end{enumerate}
\end{lemma}

\begin{proof}
(i) Let $z$ be the integer nearest to $k\alpha$. Since $0<|k|<r$ and $(a,r)=1$, $r\nmid ka$, so
$|ka/r-z|\ge1/r$ and $\|k\alpha\|\ge1/r-|k|\,|\alpha-a/r|\ge1/(2r)$.

(ii) Let $L=\lfloor r/2\rfloor$, so $L\ge r/4$. For $k$ in a block of $L$ consecutive integers let
$y_k=k\alpha-z_k$ with $z_k$ the integer nearest to $k\alpha$. For $k\ne k'$ in the block,
$0<|k-k'|<L\le r/2$ and $y_k-y_{k'}\equiv(k-k')\alpha\pmod1$, so $|y_k-y_{k'}|\ge1/(2r)$ by~(i).
Points of $(-t,t)$ at mutual distances at least $1/(2r)$ number at most $4rt+1$, and $[1,H]$ is
covered by at most $H/L+1\le4H/r+1$ blocks.

(iii) If $\|\theta\|\ge1/M$, let $k\ge1$ be least with $\|\theta\|<2^k/M$; then $k\le K$ because
$\|\theta\|\le1/2<2^K/M$, and $\|\theta\|\ge2^{k-1}/M$, so $w_M(\theta)\le2^{1-k}M$. Hence
\[
  w_M(\theta)^j\le M^j\,[\|\theta\|<1/M]+\sum_{k=1}^K(2^{1-k}M)^j\,[\|\theta\|<2^k/M]\qquad(j=1,2).
\]
Summing over $h\le H$ with~(ii), the first sum is at most
$(4H/r+1)\bigl(4r+M+\sum_{k\le K}(8r+2^{1-k}M)\bigr)$ and the second at most
$(4H/r+1)\bigl(4rM+M^2+\sum_{k\ge1}(16rM2^{-k}+4M^24^{-k})\bigr)$.
\end{proof}

\begin{lemma}\label{lem:bilinear}
Let $\theta,\alpha\in\R$, $U>0$, and let $A,B,N$ be nonnegative integers with $B\le A+U$.
\begin{enumerate}
\item[(i)] $\bigl|\sum_{A<x\le B}e(x\theta)\bigr|\le w_U(\theta)$.
\item[(ii)] For a finite set $V$ of positive integers and $b:V\to\C$,
\[
  \sum_{A<u\le B}\Bigl|\sum_{\substack{v\in V\\uv\le N}}b(v)e(uv\alpha)\Bigr|^2\le
  \sum_{v,v'\in V}|b(v)|\,|b(v')|\,w_U\bigl((v-v')\alpha\bigr).
\]
\item[(iii)] For $M>0$ and an integer $H\ge0$,
$\sum_{1\le m,m'\le H}w_M((m-m')\alpha)\le H\bigl(M+2\sum_{h\le H}w_M(h\alpha)\bigr)$.
\end{enumerate}
\end{lemma}

\begin{proof}
(i) If $\|\theta\|<1/U$, the sum has at most $U$ terms of modulus $1$. Otherwise
$\theta\notin\Z$, the sum has modulus at most $|e((B-A)\theta)-1|/|e(\theta)-1|$, and
$|e(\theta)-1|=2|\sin\pi\theta|\ge4\|\theta\|$, so the modulus is at most $1/(2\|\theta\|)$.
(ii) Expanding the square gives $\sum_{v,v'}b(v)\overline{b(v')}\sum_ue(u(v-v')\alpha)$, where
$u$ runs over the integers of $(A,\min(B,N/v,N/v')]$, an interval of length at most $U$; apply~(i).
(iii) For fixed $m$, the term $m'=m$ is $w_M(0)=M$, and each $h=|m-m'|\in[1,H]$ arises from at
most two $m'$.
\end{proof}

\begin{lemma}\label{lem:reduction}
Let $C_5=1+4\zeta(3/2)$. For every completely multiplicative $f$ with $|f(n)|\le1$, every
$\alpha\in\R$ and every integer $N\ge1$, the sum $S=\sum_{n\le N}f(n)e(n\alpha)$ satisfies
\[
  \Bigl|S\log N-\sum_{p\le N}\log p\,f(p)\sum_{m\le N/p}f(m)e(pm\alpha)\Bigr|\le C_5N.
\]
\end{lemma}

\begin{proof}
Write $S\log N=S_1+S_2$ with $S_1=\sum_{n\le N}f(n)e(n\alpha)\log n$ and
$S_2=\sum_{n\le N}f(n)e(n\alpha)\log(N/n)$. Then $|S_2|\le N\log N-\log N!\le N$, since
$N^N/N!\le e^N$. In $S_1$ write $\log n=\sum_{d\mid n}\Lambda(d)$ and $n=dm$; complete
multiplicativity gives $\sum_{d\le N}\Lambda(d)f(d)\sum_{m\le N/d}f(m)e(dm\alpha)$. The terms with
$d$ prime form the sum in the statement, and the others have modulus at most
$N\sum_p\sum_{k\ge2}\log p/p^k\le2N\sum_p\log p/p^2\le4\zeta(3/2)N$, since $\log p\le2\sqrt p$.
\end{proof}

Let $G(h)=\prod_{p\mid h}p/(p-1)$.

\begin{lemma}\label{lem:pairs}
\begin{enumerate}
\item[(i)] For every integer $X\ge2$ and every even $h\ge2$,
\[
  \#\{0\le n<X:n\text{ and }n+h\text{ prime}\}\le\frac{400\,G(h)^2X}{(\log X)^2}+3X^{3/5}.
\]
\item[(ii)] $\sum_{h\le X}G(h)^4\le e^{30}X$ for every $X\ge1$.
\end{enumerate}
\end{lemma}

\begin{proof}
(i) Apply Lemma~\ref{lem:selberg} to $a_n=n(n+h)$, $0\le n<X$, with $y=X^{1/5}$,
$z=\lfloor y\rfloor$, $P$ the product of the primes $p\le z$ with $p\nmid h$, and
$\nu(d)=2^{\omega(d)}/d$. Every $p\mid P$ is odd, so $\nu(p)=2/p<1$. For $d\mid P$ the residues $n$
with $d\mid n(n+h)$ form $2^{\omega(d)}$ classes modulo $d$ (namely $n\equiv0$ or $-h\pmod p$ for
each $p\mid d$, distinct as $p\nmid h$), and each class contains between $X/d-1$ and $X/d+1$ of
the integers $0\le n<X$; hence $|R_d|\le2^{\omega(d)}$. Since the prime factors of $d$ are at
least $3$, $3^{\omega(d)}|R_d|\le6^{\omega(d)}\le d^2\le y^2$, and the error term is at most
$y^3=X^{3/5}$.

Let $\mathcal M$ be the set of $m\le\sqrt y$ whose prime factors divide $P$. Since
$\nu(p)/(1-\nu(p))=\sum_{k\ge1}\nu(p)^k$ and $\nu(m)=2^{\Omega(m)}/m$ for these $m$, $g(l)$ is the
sum of $2^{\Omega(m)}/m$ over the $m$ whose set of prime factors is that of $l$. An $m\in\mathcal M$
is counted in $g(l)$ for $l$ the product of its prime factors, and $l\le m\le\sqrt y$. Hence,
with $\tau(m)\le2^{\Omega(m)}$ the number of divisors of $m$,
\[
  S\ge\sum_{m\in\mathcal M}\frac{2^{\Omega(m)}}m\ge\sum_{m\in\mathcal M}\frac{\tau(m)}m.
\]
Let $T=\lfloor y^{1/4}\rfloor$ and let $B$ be
the set of $b\le T$ that are prime to $h$; these $b$ are odd. For $b,b'\in B$ the product $bb'$
lies in $\mathcal M$, and a given $m$ is a product $bb'$ in at most $\tau(m)$ ways, so
$S\ge\bigl(\sum_{b\in B}1/b\bigr)^2$. If $E$ is a set of primes and $p\notin E$, every $b\le T$
prime to $E$ is uniquely $p^kb'$ with $b'$ prime to $E\cup\{p\}$, so the sum of $1/b$ over the
$b\le T$ prime to $E$ is at most $p/(p-1)$ times the same sum for $E\cup\{p\}$. Adding the prime
factors of $h$ one at a time gives
\[
  \sum_{b\in B}\frac1b\ge\frac1{G(h)}\sum_{b\le T}\frac1b\ge\frac{\log(T+1)}{G(h)}.
\]
As $T+1>y^{1/4}=X^{1/20}$, we get $S\ge(\log X)^2/(400G(h)^2)$, that is,
$X/S\le400G(h)^2X/(\log X)^2$. Finally, if $n<X$ and $n,n+h$ are prime, then $n\le z$ or
$(n(n+h),P)=1$, so the count is at most $z+1+X/S+y^3\le400G(h)^2X/(\log X)^2+3X^{3/5}$.

(ii) For $t=p-1\ge1$, $(t+1)^5\le(t+31)t^4$, that is, $(p/(p-1))^4\le1+30/p$. Hence
$G(h)^4\le\sum_d30^{\omega(d)}/d$ over the squarefree $d\mid h$, and
\begin{align*}
  \sum_{h\le X}G(h)^4&\le\sum_{d\le X}\mu(d)^2\frac{30^{\omega(d)}}d\cdot\frac Xd
  \le X\prod_p\Bigl(1+\frac{30}{p^2}\Bigr)\\
  &\le X\exp\Bigl(30\sum_{n\ge2}\frac1{n^2}\Bigr)\le e^{30}X.\qedhere
\end{align*}
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:mv}]
Let $S$ be the sum in the theorem, $L=\log N$ and $\ell=\lfloor\log_2N\rfloor$. Since
$N\ge rR\ge4$, we have $L\ge1$. If $\ell<64$, then $L<64\log2<45$ and $|S|\le N\le45N/L$. Let now
$\ell\ge64$, and put $I=\lfloor\log_2\ell\rfloor+1$ and $J=\ell+1-4I$. Then $16^I>\ell^4$, $1\le J\le\ell\le2L$,
$I\le\ell\le2L$, $2^I\le2\ell$, $N<2^{\ell+1}=2^J16^I$ and
$J^22^J\le(\ell+1)^22^{\ell+1}\ell^{-4}\le2N$. By Lemma~\ref{lem:reduction},
$|S|L\le C_5N+|T_1|+|T_2|$, where $T_1$ and $T_2$ are the parts of
$\sum_p\log p\,f(p)\sum_{m\le N/p}f(m)e(pm\alpha)$ with $p\le2^J$ and with $p>2^J$.

\emph{The small primes.} Let $0\le j<J$, $U=2^j$, $H=\lfloor N/U\rfloor$, and for $U<x\le2U$ let
$B(x)=\sum_{m\le H,\,xm\le N}f(m)e(xm\alpha)$; the inner sum for a prime $p\in(U,2U]$ is $B(p)$.
By Cauchy's inequality, Lemma~\ref{lem:bilinear}(ii), (iii) and Lemma~\ref{lem:spacing}(iii) with
$M=U$, $K=j$,
\[
  \sum_{U<p\le2U}\log p\,|B(p)|\le\bigl(U\log^2(2U)\bigr)^{1/2}Q_j^{1/2},
\]
where $Q_j=H\bigl(U+2(4H/r+1)(3U+4r+8rj)\bigr)$.
Since $HU\le N$, $UQ_j\le7NU+(8+16j)Nr+24N^2/r+(32+64j)N^2/U$, and $\log(2U)\le j+1$. Summing over
$j<J$ with $j+1\le J$, $\sum_{j<J}2^{j/2}\le3\cdot2^{J/2}$,
$\sqrt{(32+64j)2^{-j}}\le8(j+1)(3/4)^j$ and $\sum_{j\ge0}(j+1)^2(3/4)^j=112$,
\[
  |T_1|\le3\sqrt7J\sqrt{N2^J}+J^2N\sqrt{24/r}+896N+J^2\sqrt{24JNr}.
\]
Now $J^22^J\le2N$ gives $3\sqrt7J\sqrt{N2^J}\le12N$; $r\ge R\ge\sqrt R$ gives
$\sqrt{24/r}\le5R^{-1/4}$; and $r\sqrt R\le rR\le N$ gives $\sqrt{24JNr}\le5JNR^{-1/4}$. With
$J\le2L$ and $L\ge1$, $|T_1|\le908N+60L^3NR^{-1/4}$.

\emph{The large primes.} For $p>2^J$ we have $N/p<16^I$, so the $m$ in $T_2$ lie in
$(0,16^I-1]$, the union of the blocks $(16^i-1,16^{i+1}-1]$ with $0\le i<I$. Fix $i$, let
$A=16^i-1$, $U=15\cdot16^i$ and $X=\lfloor N/16^i\rfloor$, and assume $X\ge1$, the block being
empty otherwise. Only the primes $p\in(2^J,X]$ contribute, since $p\ge X+1$ and $m\ge16^i$ give
$pm>N$. Let $V$ be the set of these primes, which are odd, and
$B(m)=\sum_{p\in V,\,pm\le N}\log p\,f(p)e(pm\alpha)$. The block is
$\sum_{A<m\le A+U}f(m)B(m)$, and by Cauchy's inequality, Lemma~\ref{lem:bilinear}(ii) and
$\log p\le\lambda=\log(X+1)$, its square is at most $U\lambda^2W$, where
\[
  W=\sum_{p,p'\in V}w_U((p-p')\alpha)=|V|U+2\sum_{h=1}^Xw_U(h\alpha)\,\#\{p\in V:p+h\in V\}.
\]
The count vanishes for odd $h$, and for even $h$ Lemma~\ref{lem:pairs}(i), applied to $X+1$,
bounds it by $400G(h)^2(X+1)\lambda^{-2}+3(X+1)^{3/5}$. By Cauchy's inequality,
Lemma~\ref{lem:pairs}(ii) and Lemma~\ref{lem:spacing}(iii),
$\sum_{h\le X}G(h)^2w_U(h\alpha)\le e^{15}\sqrt XQ^{1/2}$ with $Q=(4X/r+1)(20rU+3U^2)$, and
$\sum_{h\le X}w_U(h\alpha)\le XU$. Hence $U\lambda^2W\le A_1+800e^{15}Z+A_3$, where
\[
  A_1=U^2\lambda^2X,\qquad Z=U(X+1)\sqrt XQ^{1/2},\qquad A_3=6U^2\lambda^2(X+1)^{3/5}X.
\]
Let $\kappa=\log(N+1)\ge\lambda$ and $T_i=4^i\kappa\,(N(2N)^{3/5})^{1/2}$. Since
$U^2X=225\cdot16^i(16^iX)\le225\cdot16^iN$ and $X+1\le2N$, we get $A_1\le225T_i^2$ and
$A_3\le1350T_i^2\le37^2T_i^2$. With $w=UX\le15N$, $X\le N/16^i$ and $R\le r\le N/R$,
\begin{align*}
  Z^2\le4U^2X^3Q&=320w^3X+\frac{48w^4}r+80rw^3+12Uw^3\\
  &\le\frac{320\cdot15^3N^4}{16^i}+\frac{48\cdot15^4N^4}R+\frac{80\cdot15^3N^4}R\\
  &\qquad+180\cdot15^3\cdot16^iN^3,
\end{align*}
and every coefficient is at most $40^4$, so $Z^{1/2}\le40(N2^{-i}+2NR^{-1/4}+2^iN^{3/4})$. The
block is therefore at most
\begin{align*}
  &15T_i+40\sqrt{800e^{15}}\bigl(N2^{-i}+2NR^{-1/4}+2^iN^{3/4}\bigr)+37T_i\\
  &\qquad\le K_b\bigl(N2^{-i}+NR^{-1/4}+2^iN^{3/4}+T_i\bigr),
\end{align*}
with $K_b=80\sqrt{800e^{15}}+52$. Summing over $i<I$,
\[
  |T_2|\le K_b\bigl(2N+INR^{-1/4}+2^IN^{3/4}+4^I\kappa(N(2N)^{3/5})^{1/2}\bigr).
\]
Here $I\le2L^3$; $2^I\le4L\le16N^{1/4}$; $4^I\le16L^2$ and $\kappa\le2L$; and
$\log x\le15x^{1/15}$ gives $L^3(N(2N)^{3/5})^{1/2}\le2\cdot15^3N$. Hence
$|T_2|\le K_b(216018N+2L^3NR^{-1/4})$.

Altogether $|S|L\le(C_5+908+216018K_b)N+(60+2K_b)L^3NR^{-1/4}$, and dividing by $L$ gives the
theorem with $C_1=45+C_5+908+216018K_b+60+2K_b$.
\end{proof}

\section{Proof of Theorem~\ref{thm:pv}}\label{sec:pv}

In this section $\chi$ is a primitive character modulo $q\ge3$.

\begin{lemma}\label{lem:polya}
Let $B\ge0$ be such that $\bigl|\sum_{m\le H}\bar\chi(m)e(m\beta)/m\bigr|\le B$ for every integer
$H\le q/2$ and every $\beta\in\R$. Then $\bigl|\sum_{n\le x}\chi(n)\bigr|\le\sqrt q\,(B+6)$ for
every $x\ge0$.
\end{lemma}

\begin{proof}
We may take $x=T$ to be an integer. Let $\tau=\sum_{a\bmod q}\bar\chi(a)e(a/q)$. Since
$\bar\chi$ is primitive, $\sum_a\bar\chi(a)e(an/q)=\chi(n)\tau$ for every $n$, and summing the
squares of the moduli of these identities over $n\bmod q$ gives $q\varphi(q)=|\tau|^2\varphi(q)$,
so $|\tau|=\sqrt q$. Summing over $1\le n\le T$,
\[
  \tau\sum_{n\le T}\chi(n)=\sum_{0<|a|\le q/2}\bar\chi(a)\,
  \frac{e(aT/q)-1}{1-e(-a/q)},
\]
where $a$ runs over the integers with $-q/2<a\le q/2$, $a\ne0$. For $0<|\theta|\le1/2$ put
$r(\theta)=(1-e(-\theta))^{-1}-(2\pi i\theta)^{-1}$; then $|1-e(-\theta)|=2|\sin\pi\theta|\ge4|\theta|$.
If $2\pi|\theta|\le1$, then $|e(-\theta)-1+2\pi i\theta|\le(2\pi\theta)^2$ and
$|r(\theta)|\le(2\pi\theta)^2/(4|\theta|\cdot2\pi|\theta|)<3$; otherwise
$|r(\theta)|\le1/(4|\theta|)+1/(2\pi|\theta|)<3$. The terms with $r(a/q)$ contribute at most
$6q$. Writing $a=\pm m$ with $1\le m\le q/2$, the remaining terms are $q/(2\pi i)$ times a
combination, with coefficients of modulus at most $1$, of four sums
$\sum_{m\le H}\bar\chi(m)e(m\beta)/m$ with $H\le q/2$, so they contribute at most $2qB/\pi$.
Dividing by $|\tau|$ gives the claim.
\end{proof}

\begin{lemma}\label{lem:smooth}
For $Y\ge2$ and $H\ge1$,
\[
  \sum_{\substack{n\le H\\P(n)\le Y}}\frac1n\le e^{11}\log Y.
\]
\end{lemma}

\begin{proof}
The sum is at most $\prod_{p\le Y}(1-1/p)^{-1}\le\exp\bigl(\sum_{p\le Y}\frac1{p-1}\bigr)$, and
$\sum_p\bigl(\frac1{p-1}-\frac1p\bigr)\le1$; apply Lemma~\ref{lem:chebyshev}(iv).
\end{proof}

\begin{lemma}\label{lem:major}
There is an absolute constant $C_2$ such that for $0<r<q$, $b\in\Z$, $\beta\in\R$, integers
$Y\ge\max(2,r)$ and $t\ge1$,
\[
  \Bigl|\sum_{\substack{n\le t\\P(n)>Y}}\chi(n)e\Bigl(n\Bigl(\frac br+\beta\Bigr)\Bigr)\Bigr|
  \le C_2\,\frac{r\log(qr+2)\,(2+2\pi|\beta|t)\,t}{Y^{1/20}\log Y}.
\]
\end{lemma}

\begin{proof}
Every $n\le t$ with $P(n)>Y$ is uniquely $n=pm$ with $p=P(n)$ prime, $Y<p\le t/m$ and
$P(m)\le p$, and then $m\le t/Y$. For such $p$ we have $p>r$, so $p\nmid r$, and on the units
modulo $r$ the function $u\mapsto e(umb/r)$ equals $\sum_\psi c_\psi\psi(u)$, the sum over the
characters $\psi$ modulo $r$, with $|c_\psi|\le1$.

The character $\chi\psi$ modulo $qr$ is nonprincipal. Otherwise $\chi(n)=\bar\psi(n)$ for every
$n$ prime to $qr$. Let $d=(q,r)$ and let $n$ be prime to $q$ with $n\equiv1\pmod d$. By the
Chinese remainder theorem there is $n'$ prime to $qr$ with $n'\equiv n\pmod q$ and
$n'\equiv1\pmod r$, and then $\chi(n)=\chi(n')=\bar\psi(n')=1$. So $\chi$ is trivial on the
units congruent to $1$ modulo $d$, and is induced by a character modulo $d\le r<q$, which
contradicts primitivity.

Applying~\eqref{eq:twistedprime} to $\xi=\chi\psi$ with $\gamma=m\beta$,
$A=\max(Y,P(m)-1)$ and $B=\lfloor t/m\rfloor$ (the inner sum is empty if $A\ge B$) gives
\[
  \Bigl|\sum_{\substack{n\le t\\P(n)>Y}}\chi(n)e\Bigl(n\Bigl(\frac br+\beta\Bigr)\Bigr)\Bigr|
  \le2C_3r\,\frac{\log(qr+2)(2+2\pi|\beta|t)}{\log Y}\sum_{m\le t/Y}\Bigl(\frac tm\Bigr)^{19/20},
\]
and $\sum_{m\le M}m^{-19/20}\le20M^{1/20}$. Hence $C_2=40C_3$ will do.
\end{proof}

\begin{proposition}\label{prop:hsum}
There is an absolute constant $C_4$ such that if $\log q\ge200$, then
$\bigl|\sum_{m\le H}\chi(m)e(m\beta)/m\bigr|\le C_4\log\log q$ for every integer $H\le q$ and
every $\beta\in\R$.
\end{proposition}

\begin{proof}
Let $L=\log q$, $M=\lceil L\rceil$, $R=M^{16}$ and $Y=M^{700}$. Since
$e^{L/16}\ge(L/16)^4/24>L+1\ge M$, we have $2\le R<q$ and $R\le Y$. Split the sum according to
$P(m)\le Y$ or $P(m)>Y$. By Lemma~\ref{lem:smooth} the first part is at most $e^{11}\log Y$. For
the second put $A(t)=\sum_{n\le t,\,P(n)>Y}\chi(n)e(n\beta)$ and $w(t)=1/t-1/(t+1)$. Partial
summation and $|A(H)|\le H$ give
\[
  \Bigl|\sum_{\substack{m\le H\\P(m)>Y}}\frac{\chi(m)e(m\beta)}m\Bigr|\le1+\sum_{1\le t<H}|A(t)|\,w(t).
\]
If $t\le Y$, then $A(t)=0$. Let $t>Y$ and $Q=\lfloor t/R\rfloor\ge1$. By Dirichlet's theorem
there are coprime $a,r$ with $1\le r\le Q$ and $|\beta-a/r|\le1/(r(Q+1))$; then
$|\beta-a/r|\,t\le R$ (Figure~\ref{fig:arcs}). If $r\le R$, then Lemma~\ref{lem:major} with $b=a$
gives $|A(t)|\le C_2kt$, where
\[
  k=\frac{R\log(qR+2)(2+2\pi R)}{Y^{1/20}\log Y}.
\]
If $r>R$, then $R\le r\le t/R$ and $|\beta-a/r|\le r^{-2}$, so Theorem~\ref{thm:mv} for $f=\chi$
bounds the full sum $\sum_{n\le t}\chi(n)e(n\beta)$, and since $\log t\le L$, removing the
$\Psi(t)=\#\{n\le t:P(n)\le Y\}$ smooth terms gives
\[
  |A(t)|\le C_1\Bigl(\frac t{\log t}+t\rho\Bigr)+\Psi(t),\qquad
  \rho=\frac{L^2}{R^{1/4}}=\frac{L^2}{M^4}.
\]
Since $tw(t)\le1/t$ and $w(t)\le1/t^2$,
\[
  \sum_{1\le t<H}|A(t)|\,w(t)\le C_1\sum_{2\le t\le H}\frac1{t\log t}
  +(C_1\rho+C_2k)\sum_{t\le H}\frac1t+\sum_{1\le t<H}\Psi(t)w(t).
\]
The first sum is at most $3+\log L$, the second at most $1+L$, and by partial summation the third
is at most $\sum_{n\le H,\,P(n)\le Y}1/n\le e^{11}\log Y$. Finally $L\le M\le2L$ gives
$\rho(1+L)\le2/M\le1$; $\log(qR+2)\le18M$, $2+2\pi R\le8M^{16}$, $Y^{1/20}=M^{35}$ and
$\log Y=700\log M\ge700$ give $k(1+L)\le288M^{34}/(700M^{35})\le1$; and
$\log Y=700\log M\le1400\log L$. Collecting terms, the sum is at most
$1+4C_1+C_2+(2800e^{11}+C_1)\log L\le(2800e^{11}+1+5C_1+C_2)\log L$.
\end{proof}

\begin{figure}[ht]
\centering
\includegraphics[width=\textwidth]{figures/fig_arcs.png}
\caption{The dissection of $\mathbb R/\mathbb Z$ in the proof of
Proposition~\ref{prop:hsum}, drawn for $R=4$. Near a fraction $a/r$ with $r\le R$ (major arcs)
the sum $A(t)$ is controlled by the twisted prime sums of Corollary~\ref{cor:prime}; elsewhere
(minor arcs) by Theorem~\ref{thm:mv}. The heights only indicate that the major arcs carry the
large values.}
\label{fig:arcs}
\end{figure}

\begin{proof}[Proof of Theorem~\ref{thm:pv}]
The character $\bar\chi$ is primitive modulo $q$. If $\log q\ge200$, then
Proposition~\ref{prop:hsum} for $\bar\chi$ and Lemma~\ref{lem:polya} give
\[
  \Bigl|\sum_{n\le x}\chi(n)\Bigr|\le\sqrt q\,(C_4\log\log q+6)\le(C_4+6)\sqrt q\log\log q.
\]
If $\log q<200$, then every sum $\sum_{m\le H}\bar\chi(m)e(m\beta)/m$ with $H\le q$ has modulus at
most $1+\log q<201$, and Lemma~\ref{lem:polya} gives
\[
  \Bigl|\sum_{n\le x}\chi(n)\Bigr|\le207\sqrt q\le\frac{207}{\log\log3}\sqrt q\log\log q.\qedhere
\]
\end{proof}

\section{The logarithmic derivative at \texorpdfstring{$s=1$}{s=1}}\label{sec:ek}

Recall that $\sigma_0=7/8+3/1000$ and that $K$ is the constant of Corollary~\ref{cor:prime}.

\begin{lemma}\label{lem:atone}
For every nonprincipal character $\chi$ modulo $q$,
\[
  -\frac{L'}{L}(1,\chi)=2\int_1^\infty\frac{S(\chi;t)}{t^2}\,dt,
\]
and the integral converges absolutely.
\end{lemma}

\begin{proof}
By Lemma~\ref{lem:mellin}(i) with $a=\chi\Lambda$, for real $s>1$,
\[
  \int_1^\infty S(\chi;t)\,t^{-s-1}\,dt=-\frac{L'}{L}(s,\chi)\,\frac1{s(s+1)}.
\]
As $s\to1^+$ the left side tends to $\int_1^\infty S(\chi;t)t^{-2}dt$ by dominated convergence,
since $|S(\chi;t)|\le Kt^{\sigma_0}\log(q+2)$ and $\sigma_0<1$. The function $L(s,\chi)$ is
entire and does not vanish at $s=1$ by Theorem~\ref{thm:zerofree}, so the right side tends to
$-\frac12\frac{L'}{L}(1,\chi)$.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:ek}(i)]
Let $\chi$ be nonprincipal modulo $q$, so $q\ge3$, and put
$Y=\bigl(\log(q+2)\bigr)^{1/(1-\sigma_0)}\ge1$. Since $|S(\chi;t)|\le\psi(t)\le4t$ by
Lemma~\ref{lem:chebyshev}(ii), Lemma~\ref{lem:atone} gives
\[
  \Bigl|\frac{L'}{L}(1,\chi)\Bigr|\le2\int_1^Y\frac{4\,dt}t
  +2K\log(q+2)\int_Y^\infty t^{\sigma_0-2}\,dt
  =\frac{8\log\log(q+2)+2K}{1-\sigma_0}.
\]
As $q+2\le q^2$, $\log\log(q+2)\le\log2+\log\log q$, and since $\log\log q\ge\log\log3>0$ the
right side is at most $C\log\log q$.
\end{proof}

For the lower bound in Theorem~\ref{thm:ek}(ii) we need to control the prime powers $n\le q^{17}$
with $n\equiv1\pmod q$. We count them in dyadic blocks with Selberg's sieve.

\begin{lemma}\label{lem:block}
Let $q$ be prime, let $M\ge1$ be an integer and let $1<z<q$. The number of $m$ with
$M\le m<2M$ for which $1+mq$ is a prime power is at most
\[
  \frac{2M}{\log z}+z(1+\log z)^2+z.
\]
\end{lemma}

\begin{proof}
If $1+mq=p^k$ with $p\le z$, then $p^k\in(Mq,2Mq)$, and for each prime $p$ at most one power of
$p$ lies in this interval, since $p^{k+1}\ge2p^k$. These $m$ number at most $z$. If $p>z$, then
$1+mq$ has no prime factor up to $z$, and we apply Lemma~\ref{lem:selberg} to the $M$ integers
$a_m=1+mq$, $M\le m<2M$, with $P$ the product of the primes up to $z$, $\nu(d)=1/d$ and $y=z$.
Every prime $p\le z$ differs from $q$, so for $d\mid P$ the condition $d\mid1+mq$ fixes $m$ in one
residue class modulo $d$, and $|R_d|\le1$. Here $g(l)=\prod_{p\mid l}(p-1)^{-1}$, which is the sum
of $1/n$ over the $n$ whose set of prime factors is that of $l$. Every $n\le\sqrt z$ has all its
prime factors at most $z$ and the product of its distinct prime factors is at most $\sqrt z$, so
\[
  S=\sum_{\substack{l\mid P\\l\le\sqrt z}}g(l)\ge\sum_{n\le\sqrt z}\frac1n\ge\frac{\log z}2.
\]
For squarefree $d$, $3^{\omega(d)}=\tau_3(d)$, and
$\sum_{d\le z}\tau_3(d)=\#\{(a,b,c):abc\le z\}\le z\bigl(\sum_{a\le z}1/a\bigr)^2\le z(1+\log z)^2$.
Lemma~\ref{lem:selberg} therefore bounds the number of these $m$ by $2M/\log z+z(1+\log z)^2$.
\end{proof}

\begin{lemma}\label{lem:progression}
There is an absolute constant $C_0$ such that for every prime $q\ge3$,
\[
  \sum_{\substack{1<n\le q^{17}\\n\equiv1\ (\mathrm{mod}\ q)}}\frac{\Lambda(n)}n
  \le C_0\,\frac{\log q\log\log q}q.
\]
\end{lemma}

\begin{proof}
Write $n=1+mq$ with $1\le m<q^{16}$, and let $J=\lfloor16\log_2q\rfloor+1$, so that $2^J>q^{16}$
and $J\le25\log q$. We split the range of $m$ into the blocks $2^j\le m<2^{j+1}$, $0\le j<J$
(Figure~\ref{fig:blocks}), and let $N_j$ be the number of $m$ in the $j$-th block with $1+mq$ a
prime power. We claim $N_j\le393\cdot2^j/(j+1)$. For $j<32$ this holds trivially. For $j\ge32$
apply Lemma~\ref{lem:block} with $M=2^j$ and $z=2^u$, $u=j/32\ge1$. Since
$2^j<2^J\le2q^{16}$, we have $z<2^{1/32}q^{1/2}<q$. Then $2M/\log z=64\cdot2^j/(j\log2)\le
96\cdot2^j/(j+1)$, and since $\log z\le u$,
\[
  z(1+\log z)^2+z\le2^u(2+u)^2\le\frac{297\cdot2^{4u}}{32u+1}\le\frac{297\cdot2^j}{j+1},
\]
because $(2+u)^2(32u+1)\le9u^2\cdot33u=297u^3$ and $u^3\le2^{3u}$. This proves the claim.

A term of the $j$-th block has $\Lambda(n)\le\log(1+2^{j+1}q)\le(j+2)\log2+\log q$ and
$n>2^jq$, so the block contributes at most
\[
  \frac{393}{(j+1)q}\bigl((j+2)\log2+\log q\bigr)\le\frac{393}q\Bigl(2\log2+\frac{\log q}{j+1}\Bigr).
\]
Summing over $j<J$ and using $\sum_{j<J}1/(j+1)\le1+\log J$, $J\le25\log q$ and
$\log J\le4+\log\log q$, the sum in the lemma is at most
$\frac{393}q\bigl(35\log q+5\log q+\log q\log\log q\bigr)$. Since $\log\log q\ge\log\log3>1/11$,
this is at most $C_0\log q\log\log q/q$ with $C_0=393\cdot441$.
\end{proof}

\begin{figure}[ht]
\centering
\includegraphics[width=0.85\textwidth]{figures/fig_blocks.png}
\caption{The dyadic blocks in the proof of Lemma~\ref{lem:progression}. The block
$2^j\le m<2^{j+1}$ contributes a constant part, of total size $\ll J\ll\log q$, and a part of
size $\log q/(j+1)$, whose total is $\ll\log q\log J$. The second part produces the factor
$\log\log q$.}
\label{fig:blocks}
\end{figure}

\begin{proof}[Proof of Theorem~\ref{thm:ek}(ii)]
Let $q\ge3$ be prime and put $c(n)=\sum_{\chi\ne\chi_0}\chi(n)$, the sum over the nonprincipal
characters modulo $q$. By orthogonality $c(n)=(q-1)\mathbf 1[n\equiv1\ (\mathrm{mod}\ q)]
-\mathbf 1[q\nmid n]$. Put
\[
  T(t)=\sum_{\chi\ne\chi_0}S(\chi;t)=\sum_{n\le t}c(n)\Lambda(n)\Bigl(1-\frac nt\Bigr),\qquad
  I=\int_1^\infty\frac{T(t)}{t^2}\,dt.
\]
By Lemma~\ref{lem:atone}, $\sum_{\chi\ne\chi_0}\frac{L'}{L}(1,\chi)=-2I$, so by~\eqref{eq:gammaq}
it suffices to show $-C\log q\le I\le C\log q\log\log q$.

By Corollary~\ref{cor:prime}(i), $|T(t)|\le qK\log(q+2)\,t^{\sigma_0}$. Let $Y=q^{17}$. Since
$17(\sigma_0-1)<-2$ and $\log(q+2)\le q$,
\[
  \Bigl|\int_Y^\infty\frac{T(t)}{t^2}\,dt\Bigr|\le\frac{qK\log(q+2)\,Y^{\sigma_0-1}}{1-\sigma_0}
  \le\frac{K\log(q+2)}{q(1-\sigma_0)}\le\frac K{1-\sigma_0}.
\]
For the lower bound, $c(n)\ge-1$ gives $-T(t)\le\psi(t)\le4t$, so
\[
  -I\le\int_1^Y\frac{4\,dt}t+\frac K{1-\sigma_0}=68\log q+\frac K{1-\sigma_0}.
\]
For the upper bound, $c(n)\le(q-1)\mathbf 1[n\equiv1\ (\mathrm{mod}\ q)]$ gives
\begin{align*}
  \int_1^Y\frac{T(t)}{t^2}\,dt&\le(q-1)\sum_{\substack{n\le Y\\n\equiv1\ (\mathrm{mod}\ q)}}
  \Lambda(n)\int_n^\infty\Bigl(1-\frac nt\Bigr)\frac{dt}{t^2}\\
  &=\frac{q-1}2\sum_{\substack{1<n\le q^{17}\\n\equiv1\ (\mathrm{mod}\ q)}}\frac{\Lambda(n)}n,
\end{align*}
which is at most $\frac{C_0}2\log q\log\log q$ by Lemma~\ref{lem:progression}. Hence
\[
  I\le\frac{C_0}2\log q\log\log q+\frac K{1-\sigma_0},
\]
and both bounds follow since $\log q>1$ and $\log\log q>1/11$.
\end{proof}

\subsection*{Data and code availability}
The Lean~4 formalisation of every proof in this paper, compiled together with the unchanged Lean
sources of the zero-free half-plane~\cite{OAI26}, and the programs that check the numerical
steps are in the repository \url{\repoURL}, release v1.1.2, \url{\releaseURL}~\cite{BB26code}.
The main theorems are \texttt{Vinogradov.PV.polya\_vinogradov\_loglog} (Theorem~\ref{thm:pv}),
\texttt{Vinogradov.logDeriv\_one\_le} and \texttt{Vinogradov.sum\_logDeriv\_one}
(Theorem~\ref{thm:ek}) and \texttt{Vinogradov.MV.weak\_mv} (Theorem~\ref{thm:mv}); they depend only
on the axioms \texttt{propext}, \texttt{Classical.choice} and \texttt{Quot.sound}.

\subsection*{Use of AI tools}
The proofs, the Lean~4 formalisation, the checking programs and the text of this paper were
drafted with Claude (Anthropic), used under the direction of the authors, who checked every
statement and take full responsibility for the content of the paper, including its correctness
and its citations. The ideas that came from this work are: replacing the generalised Riemann
hypothesis on the major arcs by the half-plane of Theorem~\ref{thm:zerofree}
(Section~\ref{sec:riesz}, Corollary~\ref{cor:prime} and Lemma~\ref{lem:major}); the weak form of
the Montgomery--Vaughan bound with Selberg's sieve as its only arithmetic input
(Section~\ref{sec:mv}); and the dyadic application of Selberg's sieve to prime powers in
progressions (Lemma~\ref{lem:progression}). Theorem~\ref{thm:zerofree} is the work of
OpenAI~\cite{OAI26} and is not claimed here.

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\end{document}