# Three conjectures about character sums

Andrew Granville  Départment de Mathématiques et Statistique, Université de
Montréal, CP 6128 succ Centre-Ville, Montréal, QC H3C 3J7, Canada; and
Department of Mathematics, University College London, Gower Street, London
WC1E 6BT, England<EMAIL_ADDRESS>and  Alexander P. Mangerel
Department of Mathematical Sciences, Durham University, Stockton Road, Durham,
DH1 3LE, England<EMAIL_ADDRESS>
###### Abstract.

We establish that three well-known and rather different looking conjectures
about Dirichlet characters and their (weighted) sums, (concerning the Pólya-
Vinogradov theorem for maximal character sums, the maximal admissible range in
Burgess’ estimate for short character sums, and upper bounds for $L(1,\chi)$
and $L(1+it,\chi)$) are more-or-less “equivalent”. We also obtain a new mean
value theorem for logarithmically weighted sums of 1-bounded multiplicative
functions.

We would like to thank Dimitris Koukoulopoulos and K. Soundararajan for
helpful discussions. A.G. is partially supported by grants from NSERC
(Canada). Most of this paper was completed while A.M. was a CRM-ISM
postdoctoral fellow at the Centre de Recherches Mathématiques.

##  1\. Introduction

###  1.1. Conjectures about weighted sums of Dirichlet characters

Let $\chi$ be a primitive character mod $q>1$ and

$S(\chi,N):=\sum_{n\leqslant N}\chi(n)\text{ for all }N\geqslant 1.$

The three most widely-used, unconditionally proved estimates about characters
sums are:

  * •

The _Pólya-Vinogradov_ theorem:

$M(\chi):=\max_{1\leqslant N\leqslant q}|S(\chi,N)|\leqslant c_{1}\sqrt{q}\log
q$

for some explicit $c_{1}>0$;

  * •

_Burgess’s theorem_ ([4], [17]): For $N\geqslant q^{c_{2}}$ (and $q$ _cube-
free_)

$|S(\chi,N)|=o(N)$

for any $c_{2}\geqslant\tfrac{1}{4}$; and

  * •

The Dirichlet $L$-function $L(s,\chi)$ at $s=1$ satisfies

$|L(1,\chi)|=\bigg{|}\sum_{n\geqslant 1}\frac{\chi(n)}{n}\bigg{|}\leqslant
c_{3}\log q$

for some explicit $c_{3}>0$. One can also show that for any fixed $T>0$, there
exists a constant $c_{3}(T)>0$ such that if $t\in[-T,T]$ then
$|L(1+it,\chi)|\leqslant c_{3}(T)\log q$.

The Riemann Hypothesis for $L(s,\chi)$ implies that one can take any
$c_{1},c_{2},c_{3}>0$ but this has resisted unconditional proof. One unlikely
but currently plausible ‘obstruction’ to establishing this unconditionally is
the possibility that $\chi(p)=1$ for all primes $p\leqslant q^{c}$, in which
case $c_{1},c_{2},c_{3}\gg c$, or indeed if $\chi$ is $1$-pretentious for the
primes up to $q$.111“Pretentiousness” will be defined in section 3.

Inspired by connections highlighted in [2, 7, 20] we show that improving any
one of these bounds will, more-or-less, improve the others.

######  Theorem 1.

The following statements are equivalent:

  * •

There exists $\kappa_{1}>0$ such that there are infinitely many primitive
characters $\chi\pmod{q}$ for which $M(\chi)\geqslant\kappa_{1}\sqrt{q}\log
q$;

  * •

There exists $\kappa_{3}>0$ such that there are infinitely many odd primitive
characters $\psi\pmod{r}$ for which $|L(1,\psi)|\geqslant\kappa_{3}\log r$.

This follows from a more precise connection:

######  Corollary 1.

Suppose that $\chi$ is a primitive character mod $q$. We have
$M(\chi)\gg\sqrt{q}\log q$ if and only if there exists a primitive character
$\xi\pmod{\ell}$ with $\xi(-1)=-\chi(-1)$ and $\ell\ll 1$ for which
$|L(1,\psi)|\gg\log q$, where $\psi$ is the primitive (odd) character that
induces $\chi\bar{\xi}$.

In other words we prove that if $M(\chi)\gg\sqrt{q}\log q$ then $\chi$ is
$\xi$-pretentious for some $\xi$ of bounded conductor, and we will also
establish a converse theorem.

Next we relate large $S(\chi,N)$-values with large $L(1+it,\chi)$-values:

######  Theorem 2.

The following statements are equivalent:

  * •

There exists $\kappa_{2}>0$ such that there are infinitely many primitive
characters $\chi\pmod{q}$ for which there is an integer
$N\in[q^{\kappa_{2}},q]$ such that $|S(\chi,N)|\geqslant\kappa_{2}N$;

  * •

There exists $\kappa_{3}>0$ and $T>0$ such that there are infinitely many
primitive characters $\chi\pmod{q}$ for which there exists $t\in[-T,T]$ such
that $|L(1+it,\chi)|\geqslant\kappa_{3}\log q$.

If we restrict attention here to characters of bounded order then one can take
$t=0$. The precise connection is given in the following result.

######  Proposition 1.

Fix $c>0$. Let $\chi$ be a primitive character mod $q$. There exists
$t\in\mathbb{R}$ with $|t|\ll 1$ for which $|L(1+it,\chi)|\geqslant c\log q$
if and only if there exist $\kappa=\kappa(c)>0$ and $x\in[q^{\kappa},q]$ for
which $|S(\chi,x)|\gg_{c}x$. If $\chi^{k}=\chi_{0}$ for some $k\ll 1$ then we
may take $t=0$.

In other words we prove that if $|S(\chi,N)|\gg N$ for some $N>q^{\kappa}$ and
$\kappa>0$ then $\chi$ is $n^{it}$-pretentious for some bounded real number
$t$, and we will also establish a converse theorem.

We can combine these results: If $M(\chi)\gg\sqrt{q}\log q$ and
$|S(\chi,N)|\gg N$ for some $N>q^{\kappa}$, then $\chi$ is both
$\xi$-pretentious and $n^{it}$-pretentious, which implies that $\xi$ is
$n^{it}$-pretentious, where $\xi$ is a primitive character of bounded
conductor. We will show that this implies $\xi=1$ and $t=0$, so that $\chi$ is
an odd character and $1$-pretentious for the primes up to $q$ (see Corollary
3). Therefore such a putative character is the only obstruction to improving
at least one of our three famous results unconditionally (that is, being able
to take any $c_{1}>0$ in the Pólya-Vinogradov theorem, or being able to take
any $c_{2}>0$ in Burgess’s theorem, or being able to take any $c_{3}>0$ in
bounds for $L(1,\chi)$).

Other new results on this topic will be discussed in section 2.

###  1.2. Logarithmic averages of multiplicative functions

We prove our results on sums of characters by viewing characters as examples
of multiplicative functions that take their values on the unit disk
$\mathbb{U}:=\\{z\in\mathbb{C}:|z|\leqslant 1\\}$. Halász’s theorem, which we
will discuss in detail below, bounds the mean value of $f(n)$ for $n$ up to
$x$ in terms of how “pretentious” $f$ is. In particular, if $f$ is real-valued
then Hall and Tenenbaum [16] showed that

$\displaystyle\sum_{n\leqslant x}f(n)\ll
x\,e^{-\tau\mathbb{D}(f,1;x)^{2}}\text{ where
}\mathbb{D}(f,1;x)^{2}=\sum_{p\leqslant x}\frac{1-\text{Re}(f(p))}{p},$ (1.1)

and

$\tau=0.3286\ldots=-\cos\theta\text{ where }\theta\in(0,\pi)\text{ satisfies
}\sin\theta-\theta\cos\theta=\frac{\pi}{2}.$

They gave an example where one attains equality in (1.1) (up to the inexplicit
constant).

We give an analogous result for logarithmic averages of the form
$\sum_{n\leqslant x}f(n)/n$ though, as discovered in [13], we do not need to
restrict attention to real-valued $f$. Here we let $\lambda\in\mathbb{R}$ be
such that

$\int_{0}^{1}|e(\theta)-\lambda|d\theta=2-\lambda,$

so that $\lambda=0.8221\dots$.

######  Proposition 2.

Let $f:\mathbb{N}\rightarrow\mathbb{U}$ be a multiplicative function.
(a) We have

$\sum_{n\leqslant x}\frac{f(n)}{n}\ll(\log
x)(1+\mathbb{D}(f,1;x)^{2})e^{-\lambda\,\mathbb{D}(f,1;x)^{2}}+\log\log x.$
(1.2)

(b) The exponent $\lambda$ is “best possible” in a result of this kind, since
there exists a multiplicative function $f:\mathbb{N}\rightarrow\mathbb{U}$
such that

$\bigg{|}\sum_{n\leqslant x}\frac{f(n)}{n}\bigg{|}\asymp(\log
x)e^{-\lambda\,\mathbb{D}(f,1;x)^{2}}.$ (1.3)

The $\lambda$ in the bound (1.2) improves on the $\tfrac{1}{2}$ in the bound
given in Lemma 4.3 of [13].

We deduce Proposition 2(b) from Theorem 3 which establishes asymptotics for
$\sum_{n\leqslant N}f(n)/n$ for a class of multiplicative functions
$f:\mathbb{N}\rightarrow\mathbb{U}$ for which $f(p)=g(\tau\log p)$ for each
prime $p$ for some fixed small real $\tau$, where $g(t)$ is a $1$-periodic
function with “well-behaved” Fourier coefficients.

More details, as well as other new results on this topic, will be discussed in
section 3.

##  2\. Connections between different sums of characters

###  2.1. Large character sums

Pólya gave the following Fourier expansion (see e.g., Lemma 1 of [21]) for
character sums: for $\alpha\in[0,1)$

$\displaystyle\sum_{n\leqslant\alpha q}\chi(n)=\frac{g(\chi)}{2\pi
i}\sum_{1\leqslant|n|\leqslant
q}\frac{\bar{\chi}(n)}{n}\bigg{(}1-e(-n\alpha)\bigg{)}+O(\log q),$ (2.1)

where $e(t):=e^{2\pi it}$ for $t\in\mathbb{R}$ and
$g(\chi):=\sum_{a\pmod{q}}\chi(a)e(\tfrac{a}{q})$ is the Gauss sum. When
$\chi$ is primitive we know that $|g(\chi)|=\sqrt{q}$ and

$\sum_{1\leqslant|n|\leqslant
q}\frac{\bar{\chi}(n)}{n}=(1-\chi(-1))L(1,\bar{\chi})+o(1),$ (2.2)

and so to estimate the left-hand side of (2.1) for any $\alpha$ we are left to
estimate the sums

$\sum_{n\leqslant q}\frac{\bar{\chi}(n)e(\pm n\alpha)}{n}.$

Note that (2.2) is large if and only if $\chi$ is an odd character and
$L(1,\chi)$ is large.

Fix $\tfrac{2}{\pi}<\Delta<1$ and let $R_{q}:=\exp(\tfrac{(\log
q)^{\Delta}}{\log\log q})$. For any $\alpha\in[0,1)$ we may obtain an
approximation $|\alpha-\tfrac{b}{m}|<\tfrac{1}{mR_{q}}$, with $(b,m)=1$ and
$m\leqslant R_{q}$, by Dirichlet’s theorem.

If $r_{q}:=(\log q)^{2-2\Delta}(\log\log q)^{4}<m\leqslant R_{q}$ then we say
that $\alpha$ is on a _minor arc_. By straightforward modifications to the
proof of Lemma 6.1 of [13], for such $\alpha$ we get

$\displaystyle\sum_{n\leqslant q}\frac{\chi(n)e(n\alpha)}{n}\ll\log\log
q+\frac{(\log r_{q})^{3/2}}{\sqrt{r_{q}}}\log q+\log R_{q}=o((\log
q)^{\Delta}).$ (2.3)

If $m\leqslant r_{q}$ then we say that $\alpha$ is on a _major arc_. Let
$N:=\min\\{q,\tfrac{1}{|m\alpha-b|}\\}$, so that $R_{q}\leqslant N\leqslant
q$. By Lemma 6.2 of [13],

$\displaystyle\sum_{1\leqslant|n|\leqslant
q}\frac{\chi(n)e(n\alpha)}{n}=\sum_{1\leqslant|n|\leqslant
N}\frac{\chi(n)e(n\tfrac{b}{m})}{n}+O(\log\log q).$ (2.4)

Therefore, if $M(\chi)\gg\sqrt{q}(\log q)^{\Delta}$ then either $\chi$ is odd
and $L(1,\chi)\gg(\log q)^{\Delta}$, or $M(\chi)=|S(\chi,\alpha q)|$ where
$\alpha$ lies on a major arc. The following proposition provides more detailed
information in these cases.

######  Proposition 3.

Fix $\tfrac{2}{\pi}<\Delta<1$ and let $\chi$ be a character mod $q$. We have

$M(\chi)\gg\sqrt{q}(\log q)^{\Delta}$

if and only if there is a primitive character $\xi\pmod{\ell}$ with
$\xi(-1)=-\chi(-1)$ and $\ell\leqslant(\log q)^{2-2\Delta}(\log\log q)^{4}$
such that

$\max_{1\leqslant N\leqslant q}\bigg{|}\sum_{n\leqslant
N}\frac{(\chi\bar{\xi})(n)}{n}\bigg{|}\gg\frac{\phi(\ell)}{\sqrt{\ell}}(\log
q)^{\Delta}.$

More precisely, in this case we have

$M(\chi)\sim\tau_{\chi,\xi}\cdot\frac{\sqrt{q\ell}}{\pi\phi(\ell)}\cdot\max_{1\leqslant
N\leqslant q}\bigg{|}\sum_{n\leqslant N}\frac{(\chi\bar{\xi})(n)}{n}\bigg{|}$
(2.5)

where $\tau_{\chi,1}\in[\tfrac{1}{2},3]$ and
$\tau_{\chi,\xi}=\max\\{1,|1-(\chi\bar{\xi})(2)|\\}$ if $\xi\neq 1$.

Throughout, $N_{q}$ denotes an integer value of $N$ that maximizes the right-
hand side of (2.5).

Using results from the next subsection, we will deduce Corollary 1 by showing
that if $M(\chi)\gg\sqrt{q}\log q$ then for $\psi=\chi\bar{\xi}$ we have

$\displaystyle|L(1,\psi)|\gg\log q\text{ if and only if
}\bigg{|}\sum_{n\leqslant N_{q}}\frac{\psi(n)}{n}\bigg{|}\gg\log q.$ (2.6)

However, there is not necessarily a correspondence between these two sums when
they are slightly smaller. For example, if $\psi(p)=1$ for all $p\leqslant
N:=\exp((\log q)^{\tau})$ and $\psi(p)=-1$ for all $\exp((\log
q)^{\tau})<p\leqslant q^{\delta}$ where $\tfrac{1}{2}<\tau<1$ and $\delta>0$
is some small fixed constant, then assuming $\psi$ is non-exceptional (see
(4.3)),

$\bigg{|}\sum_{n\leqslant N}\frac{\psi(n)}{n}\bigg{|}\asymp(\log
q)^{\tau}\text{ while }|L(1,\psi)|\asymp\prod_{p\leqslant
q}\bigg{|}1-\frac{\psi(p)}{p}\bigg{|}^{-1}\asymp(\log q)^{2\tau-1},$

which is much smaller. Moreover this (purported) example shows why _we cannot
assume_ that $N_{q}=q$ and that the largest sum is $\sim|L(1,\psi)|$.

Following an idea from [2], Proposition 3 has the following consequence for
quadratic non-residues:222It is worth recalling that the real primitive
characters are given by $1(\cdot)$ as well as $(\frac{\cdot}{n})$ if $n>1$,
$(\frac{2n}{\cdot})$ and $(\frac{4n}{\cdot})$ if $n\equiv 3\pmod{4}$, for each
odd squarefree integer $n$.

######  Corollary 2.

Fix $\Delta\in(\tfrac{2}{\pi},1)$. Let $n_{q}$ be the least quadratic non-
residue modulo a prime $q\equiv 3\pmod{4}$ and suppose that
$n_{q}\geqslant\exp((\log q)^{\Delta})$. For any odd, squarefree integer
$\ell\leqslant\big{(}\frac{\log n_{q}}{(\log q)^{\Delta}}\big{)}^{2}$ we have

$M((\tfrac{\cdot}{\ell q}))\geqslant(c\tau-o(1))\sqrt{q}\log n_{q}\text{ where
}c=\tfrac{2(\sqrt{e}-1)}{\pi}=0.41298\dots,$

$\tau=\frac{1}{2}$ if $\ell=1$, otherwise $\tau=1$ or $2$ depending on whether
$q\equiv 7$ or $3\pmod{8}$, respectively.

###### Proof.

Let $\xi=(\tfrac{\cdot}{\ell})$ and $\chi=(\tfrac{\cdot}{\ell q})$, so that
$\chi\overline{\xi}=(\tfrac{\cdot}{q})1_{(\cdot,\ell)=1}$ and

$\sum_{n\leqslant
x}\frac{(\chi\bar{\xi})(n)}{n}=\sum_{\begin{subarray}{c}n\leqslant x\\\
(n,\ell)=1\end{subarray}}\frac{(\tfrac{n}{q})}{n}$

Let $y=n_{q}-1$ and $N=x=y^{w}$ where $w=e^{1/2}$. A _$y$ -smooth integer_ has
all of its prime factors $\leqslant y$, and any $y$-smooth integer here is a
quadratic residue mod $q$. Therefore

$\displaystyle\sum_{\begin{subarray}{c}n\leqslant x\\\
(n,\ell)=1\end{subarray}}\frac{(\tfrac{n}{q})}{n}\geqslant\sum_{\begin{subarray}{c}n\leqslant
x\\\ P(n)\leqslant y\\\
(n,\ell)=1\end{subarray}}\frac{1}{n}-\sum_{\begin{subarray}{c}n\leqslant x\\\
P(n)>y\\\
(n,\ell)=1\end{subarray}}\frac{1}{n}=2\sum_{\begin{subarray}{c}n\leqslant x\\\
P(n)\leqslant y\\\
(n,\ell)=1\end{subarray}}\frac{1}{n}-\sum_{\begin{subarray}{c}n\leqslant x\\\
(n,\ell)=1\end{subarray}}\frac{1}{n}$

where $P(n)$ is the largest prime factor of $n$. Therefore, since
$1_{(n,\ell)=1}=\sum_{d|(n,\ell)}\mu(d)$, and as $\ell\leqslant y$ we deduce
that

$\displaystyle\sum_{\begin{subarray}{c}n\leqslant x\\\
(n,\ell)=1\end{subarray}}\frac{(\tfrac{n}{q})}{n}\geqslant\sum_{d|\ell}\mu(d)\bigg{(}2\sum_{\begin{subarray}{c}n\leqslant
x\\\ d|n\\\ P(n)\leqslant
y\end{subarray}}\frac{1}{n}-\sum_{\begin{subarray}{c}n\leqslant x\\\
d|n\end{subarray}}\frac{1}{n}\bigg{)}=\sum_{d|\ell}\frac{\mu(d)}{d}\bigg{(}2\sum_{\begin{subarray}{c}m\leqslant
x/d\\\ P(n)\leqslant
y\end{subarray}}\frac{1}{m}-\sum_{\begin{subarray}{c}m\leqslant
x/d\end{subarray}}\frac{1}{m}\bigg{)}.$ (2.7)

Let $\psi(x,y)$ be the number of $y$-smooth integers $\leqslant x$. It is
well-known that $\psi(y^{u},y)=y^{u}\rho(u)(1+O(\frac{1}{\log y}))$ as
$y\to\infty$ for bounded $u$, with $\rho(u)=1$ for $0\leqslant u\leqslant 1$
and $\rho(u)=1-\log u$ for $1\leqslant u\leqslant 2$. Therefore by partial
summation we have

$\sum_{\begin{subarray}{c}n\leqslant x\\\ P(n)\leqslant
y\end{subarray}}\frac{1}{n}=\int_{t=1}^{x}\frac{\psi(t,y)}{t^{2}}dt+O(1)=\int_{u=0}^{w}\rho(u)du\cdot\log
y+O(1),$

and

$\int_{u=0}^{w}\rho(u)du=\int_{u=0}^{1}du+\int_{u=1}^{w}(1-\log
u)du=\frac{3}{2}w-1,$

so that

$2\sum_{\begin{subarray}{c}n\leqslant x\\\ P(n)\leqslant
y\end{subarray}}\frac{1}{n}-\sum_{\begin{subarray}{c}n\leqslant
x\end{subarray}}\frac{1}{n}=(2w-2)\log y+O(1).$

Since $d\leqslant\ell=y^{o(1)}$ we can use this in (2.7) with $x$ replaced by
$x/d$ to obtain

$\displaystyle\sum_{\begin{subarray}{c}n\leqslant x\\\
(n,\ell)=1\end{subarray}}\frac{(\tfrac{n}{q})}{n}$
$\displaystyle\geqslant\sum_{d|\ell}\frac{\mu(d)}{d}((2w-2)\log y+O(1))$
$\displaystyle=\frac{\phi(\ell)}{\ell}(2w-2)\log
y+O\bigg{(}\frac{\ell}{\phi(\ell)}\bigg{)}\sim\frac{\phi(\ell)}{\ell}(2\sqrt{e}-2)\log
n_{q}$ $\displaystyle\gg\frac{\phi(\ell)}{\sqrt{\ell}}(\log q)^{\Delta}$

using the hypothesis on the size of $\ell$. The hypothesis of the second part
of Proposition 3 is therefore satisfied (with $q$ replaced by $\ell q$), and
so by (2.5) and the last displayed equation we have

$M((\tfrac{\cdot}{\ell
q}))\sim\tau_{\chi,\xi}\cdot\frac{\sqrt{q}\ell}{\pi\phi(\ell)}\cdot\max_{1\leqslant
N\leqslant q}\bigg{|}\sum_{\begin{subarray}{c}n\leqslant N\\\
(n,\ell)=1\end{subarray}}\frac{(\tfrac{n}{q})}{n}\bigg{|}\gtrsim\frac{2(\sqrt{e}-1)\tau_{\chi,\xi}}{\pi}\cdot\sqrt{q}\log
n_{q}$

where $\tau_{\chi,1}\in[\tfrac{1}{2},3]$ and
$\tau_{\chi,\xi}=\max\\{1,|1-(\frac{2}{q})|\\}$ if $\xi\neq 1$. ∎

##  3\. Halász’s Theorem and beyond

For multiplicative functions $f,g:\mathbb{N}\rightarrow\mathbb{U}$ and
$x\geqslant 2$, we define the _pretentious distance_

$\mathbb{D}(f,g;x):=\bigg{(}\sum_{p\leqslant
x}\frac{1-\text{Re}(f(p)\bar{g}(p))}{p}\bigg{)}^{1/2}.$

It is well-known that $\mathbb{D}$ satisfies the triangle inequality:

$\displaystyle\mathbb{D}(f,h;x)\leqslant\mathbb{D}(f,g;x)+\mathbb{D}(g,h;x)\text{
for all }f,g,h:\mathbb{N}\rightarrow\mathbb{U}.$ (3.1)

With $2\leqslant y\leqslant x$ we also write $\mathbb{D}(f,g;y,x)$ to work
only with the primes in $(y,x]$. We say that $f$ is _$g$ -pretentious_ (for
the primes up to $x$) if $\mathbb{D}(f,g;x)\ll 1$; so if $f$ is
$g$-pretentious then $f(p)\approx g(p)$ frequently for $p\leqslant x$.

###  3.1. Halász’s theorem

For $T>0$, $x\geqslant 2$ and a multiplicative function
$f:\mathbb{N}\rightarrow\mathbb{U}$, we also define

$M(f;x,T):=\min_{|t|\leqslant T}\mathbb{D}(f,n^{it};x)^{2}.$

We let $t=t(f;x,T)$ be a real number in this range where we get equality.
Halász’s Theorem (see e.g., [12, Thm. 1]) states that if $1\leqslant
T\leqslant\log x$ then, for $M=M(f;x,T)$,

$\displaystyle\sum_{n\leqslant x}f(n)\ll(1+M)e^{-M}x+\frac{x}{T}.$ (3.2)

If $f(n)=n^{it}$ with $|t|\leqslant T$ then $M=0$, which reflects the fact
that $|\sum_{n\leqslant x}n^{it}|\sim x/|1+it|$.

Halász’s theorem shows that $\bigg{|}\sum_{n\leqslant x}f(n)\bigg{|}$ is
$o(x)$ if $f$ is not $n^{it}$-pretentious for any $t\in\mathbb{R}$. Elementary
estimates for $\zeta(s)$ to the right of the $1$-line imply that

$\mathbb{D}(1,n^{it};x)^{2}=\begin{cases}\log(1+|t|\log x)+O(1)&\text{ if
}|t|\leqslant 100;\\\ \log\log x+O(\log\log|t|)&\text{ if }|t|\geqslant
100.\end{cases}$ (3.3)

This shows that $1$ is _not_ $n^{it}$-pretentious unless $|t|\ll\tfrac{1}{\log
x}$. Therefore if $|t_{1}|,|t_{2}|\leqslant T:=(\log x)^{O(1)}$ then
$n^{it_{1}}$ cannot be $n^{it_{2}}$-pretentious unless $|t_{1}-t_{2}|\log x\ll
1$.

If $t(f;x,T)$ is not unique, say $t_{1}$ and $t_{2}$ both yield equality
above, then (3.1) implies that
$\mathbb{D}(n^{it_{1}},n^{it_{2}};x)\leqslant\mathbb{D}(f,n^{it_{1}};x)+\mathbb{D}(f,n^{it_{2}};x)=2\mathbb{D}(f,n^{it_{1}};x)$.
In particular if $f$ is $n^{it_{1}}$-pretentious then $f$ is not
$n^{it_{2}}$-pretentious unless $t_{2}=t_{1}+O(\tfrac{1}{\log x})$.

###  3.2. Halász-type bounds for logarithmically weighted sums

If $f$ is real-valued then we might expect that $t(f;x,T)=0$ but there are
examples where this is not so (which lead to the “best possible examples” in
Hall and Tenenbaum’s estimate (1.1)). The examples $f(n)=n^{it}$ show that
there cannot be an upper bound in terms of $\mathbb{D}(f,1;x)$ alone for
arbitrary $f:\mathbb{N}\rightarrow\mathbb{U}$, though such bounds are given in
[16] for $f$ belonging to certain restricted families of multiplicative
functions (most importantly those of bounded order).

In this article we will need bounds for the logarithmically weighted sums
$\sum_{n\leqslant x}f(n)/n$.

######  Proposition 4.

Let $x\geqslant 3$. Let $f:\mathbb{N}\rightarrow\mathbb{U}$ be a
multiplicative function with $M=M(f;x,1)$, and let $t\in[-1,1]$ minimize the
expression

$\tau\mapsto\sum_{p\leqslant x}\frac{2-\text{\rm
Re}((1+f(p))p^{-i\tau})}{p},\quad\tau\in[-1,1].$

If $|t|\leqslant\tfrac{1}{\log x}$ then

$\sum_{n\leqslant x}\frac{f(n)}{n}\ll(1+M)e^{-M}\log x+\log\log x.$ (3.4)

If $|t|\geqslant\tfrac{1}{\log x}$ then

$\sum_{n\leqslant x}\frac{f(n)}{n}\ll\frac{1}{|t|}(1+M+\log(|t|\log
x))e^{-M}+\log\log x.$ (3.5)

The bound (3.5) improves upon Theorem 2.4 in [8] and Theorem 1.4 in [19]
whenever $|t|\gg\tfrac{\log\log x}{\log x}$ (though the latter can be used to
replace $(1+M)e^{-M}$ by just $e^{-M}$).

###  3.3. Deductions

Using Proposition 4 we now prove Corollary 1.

###### Proof of more than (2.6).

Let $\psi=\chi\bar{\xi}$. By the definition of $N_{q}$,

$\mathcal{L}:=\bigg{|}\sum_{n\leqslant
N_{q}}\frac{\chi\bar{\xi}(n)}{n}\bigg{|}\geqslant\bigg{|}\sum_{n\leqslant
q}\frac{\chi\bar{\xi}(n)}{n}\bigg{|}=|L(1,\psi)|+O(1)$

(see (4.1) below), and so (2.6) follows if $|L(1,\psi)|\gg\log q$.

Conversely, by (1.2) of Proposition 2 ((which is a consequence of Proposition
4), we have

$\mathcal{L}:=\bigg{|}\sum_{n\leqslant
N_{q}}\frac{\chi\bar{\xi}(n)}{n}\bigg{|}\ll(\log
N_{q})\exp(-\\{\lambda+o(1)\\}\mathbb{D}(\psi,1;N_{q})^{2})$

so that $\exp(-\mathbb{D}(\psi,1;N_{q})^{2})\geqslant(\frac{\log
N_{q}}{\mathcal{L}})^{-1/\lambda+o(1)}\gg(\frac{\mathcal{L}}{\log
N_{q}})^{2}$. Now as $\psi$ is non-exceptional we have (see section 4) that

$\displaystyle|L(1,\psi)|$ $\displaystyle\asymp\log
q\,e^{-\mathbb{D}(\psi,1;q)^{2}}=\log
q\,e^{-\mathbb{D}(\psi,1;N_{q})^{2}-\mathbb{D}(\psi,1;N_{q},q)^{2}}$
$\displaystyle\gg\log q\,\bigg{(}\frac{\mathcal{L}}{\log
N_{q}}\bigg{)}^{2}\bigg{(}\frac{\log N_{q}}{\log
q}\bigg{)}^{2}=\frac{\mathcal{L}^{2}}{\log q}$

since $\mathbb{D}(\psi,1;N_{q},q)^{2}\leqslant 2\sum_{N_{q}<p\leqslant
q}\frac{1}{p}\leqslant 2\log(\frac{\log q}{\log N_{q}})+O(1)$. If
$\mathcal{L}\gg\log q$ then this establishes (2.6); if $\mathcal{L}\gg(\log
q)^{\tau}$ then this gives $|L(1,\psi)|\gg(\log q)^{2\tau-1}$ showing that the
example given after (2.6) is, in some sense, “best possible”. ∎

###### Proof of Corollary 1.

Suppose that $M(\chi)\gg\sqrt{q}\log q$. Proposition 3 shows that there is
$\xi$ primitive of conductor $\ell\leqslant\log q$ such that
$\xi(-1)=-\chi(-1)$ and

$\log
q\ll\frac{M(\chi)}{\sqrt{q}}\ll\frac{\sqrt{\ell}}{\phi(\ell)}\bigg{|}\sum_{n\leqslant
N_{q}}\frac{(\chi\bar{\xi})(n)}{n}\bigg{|}.$

The right-hand sum is $\ll\log q$, so $\ell\ll 1$. If $\chi\bar{\xi}$ is
induced by a primitive character $\psi\pmod{\ell^{\ast}}$ with
$\ell^{\ast}|\ell$ then by Lemma 4.4 of [13],

$\sum_{n\leqslant
N_{q}}\frac{\psi(n)}{n}=\prod_{p|\tfrac{\ell}{\ell^{\ast}}}\bigg{(}1-\frac{\psi(p)}{p}\bigg{)}^{-1}\sum_{n\leqslant
N_{q}}\frac{(\chi\bar{\xi})(n)}{n}+O(1),$ (3.6)

so $|\sum_{n\leqslant N_{q}}\tfrac{\psi(n)}{n}|\gg\log q$. By (2.6), we obtain
$|L(1,\psi)|\gg\log q$.

Conversely, suppose there is a primitive character $\xi$ of conductor $\ell\ll
1$ with $\xi(-1)=-\chi(-1)$ such that $|L(1,\psi)|\gg\log q$, where $\psi$ is
the primitive character that induces $\chi\bar{\xi}$. By (2.6) and (3.6) we
have $|\sum_{n\leqslant
N_{q}}\tfrac{(\chi\bar{\xi})(n)}{n}|\gg\frac{\phi(\ell)}{\sqrt{\ell}}\log q$,
and so Proposition 3 implies that

$M(\chi)\gg\frac{\sqrt{q\ell}}{\phi(\ell)}\bigg{|}\sum_{n\leqslant
N_{q}}\frac{(\chi\bar{\xi})(n)}{n}\bigg{|}\gg\sqrt{q}\log q,$

as claimed. ∎

###  3.4. A generalization of Halász’s theorem

Given a multiplicative function $f:\mathbb{N}\rightarrow\mathbb{C}$, let
$F(s):=\sum_{n\geqslant 1}f(n)/n^{s}$ denote its Dirichlet series, assumed to
be analytic and non-zero for $\text{Re}(s)>1$. For such $s$ we write
$-\tfrac{F^{\prime}}{F}(s)=\sum_{n\geqslant 1}\Lambda_{f}(n)/n^{s}$.

For fixed $\kappa\geqslant 1$ we restrict attention to those multiplicative
functions $f$ for which $|\Lambda_{f}(n)|\leqslant\kappa\Lambda(n)$, where
$\Lambda$ is the von Mangoldt function. A generalization of Halász’s Theorem
to such $f$ (Theorem 1.1 of [11]) states that

$\sum_{n\leqslant x}f(n)\ll(1+M)e^{-M}x(\log x)^{\kappa-1}+\frac{x}{\log
x}(\log\log x)^{\kappa},$ (3.7)

where $M$ is defined by

$e^{-M}(\log x)^{\kappa}=\max\big{\\{}|s^{-1}F(s)|:\ s=1+\tfrac{1}{\log
x}+it\text{ with }|t|\leqslant(\log x)^{\kappa}\big{\\}}.$

In section 5 we will apply this result (with $\kappa=2$) to the convolution
$1\ast f$, where $f:\mathbb{N}\rightarrow\mathbb{U}$ is multiplicative, in
order to prove Proposition 4.

##  4\. Large short character sums and large $L(1+it,\chi)$ values

In this section we will prove Proposition 1.

###  4.1. Truncations of $L(1,\chi)$

Let $t\in\mathbb{R}$. By partial summation we have

$\bigg{|}\sum_{n>N}\frac{\chi(n)}{n^{1+it}}\bigg{|}\leqslant\frac{|S(\chi,N)|}{N}+|1+it|\int_{N}^{\infty}\frac{|S(\chi,y)|}{y^{2}}dy\leqslant(2+|t|)\frac{M(\chi)}{N},$

so by the Pólya-Vinogradov theorem,

$L(1+it,\chi)=\sum_{n\leqslant
N}\frac{\chi(n)}{n^{1+it}}+O\bigg{(}(2+|t|)\frac{\sqrt{q}\log q}{N}\bigg{)}.$
(4.1)

We wish to also truncate the Euler product for $L(1+it,\chi)$ at $q$ when
$|t|\ll 1$, losing at most a constant multiple. The prime number theorem in
arithmetic progressions tells us that there exist constants $A,c>0$ such that
if $L(s,\chi)$ has no exceptional zero then

$\psi(x,\chi)=\sum_{n\leqslant x}\chi(n)\Lambda(n)\ll
x\exp\bigg{(}-c\frac{\log x}{\log q}\bigg{)}+\frac{x}{\log x}$ (4.2)

for all $x\geqslant q^{A}$. By partial summation we deduce that if $B>A$ then

$\sum_{p>q^{B}}\frac{\chi(p)}{p^{1+it}}=-(1+it)\int_{q^{B}}^{\infty}\frac{\psi(u,\chi)}{u^{2+it}\log
u}du+O(1)\ll(1+|t|)\frac{e^{-cB}}{B}+1.$

Now let $B=\max\\{2A,(1/c)\log(1+|t|)\\}$ so that since $\sum_{q<p\leqslant
q^{B}}\chi(p)/p^{1+it}\leqslant\sum_{q<p\leqslant q^{B}}1/p\ll\log B$, we have

$\sum_{p>q}\frac{\chi(p)}{p^{1+it}}\ll\log B+(1+|t|)\frac{e^{-cB}}{B}+1\ll
1+\log\log(1+|t|).$

Hence, if $\chi$ is non-exceptional then for all $t\in\mathbb{R}$, $|t|\ll 1$,

$\displaystyle|L(1+it,\chi)|\asymp\bigg{|}\prod_{p\leqslant
q}\bigg{(}1-\frac{\chi(p)}{p^{1+it}}\bigg{)}^{-1}\bigg{|}\asymp(\log
q)e^{-\mathbb{D}(\chi,n^{it};q)^{2}}.$ (4.3)

Taking $N=q$ in (4.1) and assuming $|t|\ll 1$, we have

$\displaystyle L(1+it,\chi)=\sum_{n\leqslant
q}\frac{\chi(n)}{n^{1+it}}+o(1)=({1+it})\int_{1}^{q}\frac{S(\chi,u)}{u^{2+it}}du+o(1)\ll\int_{1}^{q}\frac{|S(\chi,u)|}{u^{2}}du+o(1),$

and so, for any $c>0$, we have

$|L(1+it,\chi)|\ll\bigg{(}c+\max_{q^{c}\leqslant x\leqslant
q}\frac{1}{x}|S(\chi,x)|\bigg{)}\log q$ (4.4)

using the bound $|S(\chi,u)|\leqslant u$ for $u\leqslant q^{c}$.

###  4.2. Exceptional characters

Landau proved that there exists an absolute constant $c>0$ such that for any
$Q$ sufficiently large there is _at most_ one $q\leqslant Q$, one primitive
real character $\chi\pmod{q}$ and one real number $\beta\in(0,1)$ such that

$L(\beta,\chi)=0\text{ and }\beta\geqslant 1-\frac{c}{\log Q}.$

If such a triple $(q,\chi,\beta)$ exists then we call $q$ an _exceptional
modulus_ , $\beta$ an _exceptional zero_ and $\chi$ an _exceptional
character_.

If exceptional zeros exist there must be infinitely many of them (otherwise we
can decrease $c$ as needed). If $\\{\beta_{j}\\}_{j}$ is a sequence of
exceptional zeros and $\\{q_{j}\\}_{j}$ is the corresponding set of
exceptional moduli then

$(1-\beta_{j})\log q_{j}\rightarrow 0\text{ as $j\rightarrow\infty$.}$

It is an important open problem to obtain effective lower bounds for
$1-\beta$. Siegel’s theorem (see e.g., [18, Thm. 5.28]) states that if $\beta$
is the largest real zero of $L(s,\chi)$ then
$1-\beta\gg_{\epsilon}q^{-\epsilon}$ for any $\epsilon>0$, but the implicit
constant is ineffective unless $\epsilon\geqslant 1/2$.

If $L(s,\chi)$ has an exceptional zero then $\chi(p)=-1$ for many “small”
primes. This suggests (but does not directly imply) the following result:

######  Lemma 1.

Suppose that $\chi$ is an exceptional character modulo $q$. Then:
a) $|L(1+it,\chi)|=o(\log q)$ when $|t|\ll 1$, and
b) for fixed $c>0$ we have $|S(\chi,x)|=o_{q\to\infty}(x)$ for all $x\geqslant
q^{c}$.

###### Proof.

a) ($t=0$): As $\chi$ is exceptional it must be real, and there is a
$\beta\in(0,1)$ such that $L(\beta,\chi)=0$ with $\eta:=(1-\beta)\log q=o(1)$.
By the truncation argument in (4.1),

$\displaystyle L(\beta,\chi)$ $\displaystyle=\sum_{n\leqslant
q}\frac{\chi(n)}{n}n^{1-\beta}+O\bigg{(}\frac{M(\chi)}{q^{\beta}}\bigg{)}=\sum_{n\leqslant
q}\frac{\chi(n)}{n}\bigg{(}1+O(\eta)\bigg{)}+O(q^{\tfrac{1}{2}-\beta}\log q)$
$\displaystyle=\sum_{n\leqslant q}\frac{\chi(n)}{n}+O\bigg{(}\eta\log
q\bigg{)}$

since $\eta\gg q^{-o(1)}$. By (4.1) we deduce that for any $\epsilon>0$,

$\displaystyle|L(1,\chi)|=\bigg{|}\sum_{n\leqslant
q}\frac{\chi(n)}{n}\bigg{|}+O(1/q^{\epsilon})\ll\eta\log q$ (4.5)

since $L(\beta,\chi)=0$, which implies a).

b) We use the above to observe that

$\frac{1}{q}\sum_{n\leqslant q}(1\ast\chi)(n)=\frac{1}{q}\sum_{n\leqslant
q}\chi(n)\left\lfloor\frac{q}{n}\right\rfloor=\sum_{n\leqslant
q}\frac{\chi(n)}{n}+O(1)\ll\eta\log q+1;$

on the other hand we have

$\frac{1}{q}\sum_{n\leqslant q}(1\ast\chi)(n)\gg e^{-ue^{u/2}}\log
q+O(1)\text{ where }u=\mathbb{D}(\chi,1;q)^{2}$

by [15, (3.5)], so that
$\mathbb{D}(\chi,1;q)^{2}=u\geqslant\log\log(\tfrac{1}{\theta})+O(1)$ where
$\theta:=\max\\{\eta,\tfrac{1}{\log q}\\}$.333One can obtain the better lower
bound $\mathbb{D}(\chi,1;q)^{2}\geqslant\\{2+o(1)\\}\log\log(\tfrac{1}{\eta})$
by summing [23, (3.23)] over all $m\ll\sqrt{\log 1/\eta}$ (note that their
$\eta$ is our $1/\eta$). If $x\in[q^{c},q]$ that
$\mathbb{D}(\chi,1;x)^{2}=\mathbb{D}(\chi,1;q)^{2}+O(1)\geqslant\log\log(\tfrac{1}{\theta})+O(1)$.
Therefore since $\chi$ is real, Hall and Tenenbaum’s estimate (1.1) yields

$|S(\chi,x)|\ll
xe^{-\tau\mathbb{D}(\chi,1;x)^{2}}\ll\frac{x}{(\log(1/\theta))^{\tau}}=o(x).$

a) ($|t|\ll 1$): We insert the bound from (b) into (4.4), and let $c\to 0$ to
deduce our result. ∎

###### Proof of Proposition 1.

We may assume that $\chi$ is an unexceptional character, since the result
follows vacuously when $\chi$ is exceptional by Proposition 1.

Now if $|L(1+it,\chi)|\gg\log q$ for some $|t|\ll 1$ and if $c>0$ is
sufficiently small there exists $x\in[q^{c},q]$ for which $|S(\chi,x)|\gg x$
by (4.4).

Now suppose that $|S(\chi,x)|\gg x$ for some $x\in[q^{\kappa},q]$ for some
fixed $\kappa\in(0,1]$. For a sufficiently large constant $T$, Halász’s
Theorem (3.2) implies that $\mathbb{D}(\chi,n^{it};x)\ll 1$ for some
$|t|\leqslant T$, and so

$\displaystyle\mathbb{D}(\chi,n^{it};q)^{2}=\mathbb{D}(\chi,n^{it};x)^{2}+\sum_{x<p\leqslant
q}\frac{1-\text{Re}(\chi(p)p^{-it})}{p}\ll 1+2\log\bigg{(}\frac{\log q}{\log
x}\bigg{)}\ll_{\kappa}1.$ (4.6)

Then (4.3) implies that $|L(1+it,\chi)|\gg_{\kappa}\log q$ as $\chi$ is non-
exceptional.

Suppose now that $\chi^{k}=\chi_{0}$ with $k\ll 1$. As $x>q^{\kappa}$ we note
then that

$x\ll|S(\chi,x)|\leqslant\\#\\{n\leqslant x:(n,q)=1\\}\ll\frac{\phi(q)}{q}x,$

which implies that $\sum_{p|q}\tfrac{1}{p}\ll 1$. Next, repeatedly using the
triangle inequality (3.1) for $\mathbb{D}$ together with (4.6),

$\mathbb{D}(1,n^{ikt};q)=\mathbb{D}(\chi^{k},n^{ikt};q)+O\bigg{(}1+\sum_{p|q}\frac{1}{p}\bigg{)}\leqslant
k\,\mathbb{D}(\chi,n^{it};q)+O(1)\ll_{\kappa}1.$

By (3.3) we deduce that $\log(1+k|t|\log q)\ll_{\kappa}1$, so that
$|t|\ll_{\kappa}\tfrac{1}{\log q}$. It follows then that
$\mathbb{D}(\chi,1;q)^{2}=\mathbb{D}(\chi,n^{it},q)^{2}+O_{\kappa}(1)\ll 1$,
and so $|L(1,\chi)|\asymp\log q\,e^{-\mathbb{D}(\chi,1;q)^{2}}\gg_{\kappa}\log
q$ by (4.3). ∎

We would like to deduce that $|L(1,\chi)|\gg\log q$ from
$|L(1+it,\chi)|\gg\log q$ with $|t|\ll 1$, for characters of higher order.
This is not necessarily guaranteed, though we can prove the following.

######  Lemma 2.

Let $\chi$ be a complex character mod $q$ and fix $T\geqslant 1$. If
$|L(1+it,\chi)|\gg\log q$ with $|t|\leqslant T$ then
$|L(1+it_{0},\chi)|\gg\log q$ where $t_{0}=t(\chi;q,T)$ and
$|t-t_{0}|\ll\tfrac{1}{\log q}$.

###### Proof.

Since $\chi$ is not real, it is non-exceptional. By (4.3) we see that

$|L(1+it,\chi)|\gg\log q\text{ if and only if }\mathbb{D}(\chi,n^{it};q)\ll
1.$

Let $t_{0}=t(\chi;q,T)$ so that
$\mathbb{D}(\chi,n^{it_{0}};q)\leqslant\mathbb{D}(\chi,n^{it};q)\ll 1$, and
therefore $|L(1+it_{0},\chi)|\gg\log q$ by (4.3). Moreover

$\mathbb{D}(1,n^{i(t-t_{0})};q)=\mathbb{D}(n^{it_{0}},n^{it};q)\leqslant\mathbb{D}(\chi,n^{it};q)+\mathbb{D}(\chi,n^{it_{0}};q)\ll
1$

by (3.1), and so we deduce that $|t-t_{0}|\ll\tfrac{1}{\log q}$ by (3.3). ∎

##  5\. A variant of Halász’s Theorem

In this section we will prove Propositions 4 and 2(a), our various upper
bounds for logarithmic averages.

###### Proof of Proposition 4.

As in the proof of Lemma 1,

$\sum_{m\leqslant x}(1*f)(m)=\sum_{n\leqslant
x}f(n)\left\lfloor\frac{x}{n}\right\rfloor=x\sum_{n\leqslant
x}\frac{f(n)}{n}+O(x).$

Applying (3.7) to the mean value of $1*f$ with $\kappa=2$, we then obtain

$\sum_{n\leqslant x}\frac{f(n)}{n}=\frac{1}{x}\sum_{m\leqslant
x}(1*f)(m)+O(1)\ll(1+M)e^{-M}\log x+1\ $

where $e^{-M}(\log x)^{2}=|\tfrac{1}{s}\zeta(s)F(s)|$ with $s=1+1/\log x+it$,
for some real $t,|t|\leqslant(\log x)^{2}$. We have
$|\zeta(s)|\leqslant\zeta(1+\frac{1}{\log x})\ll\log x$,

$|F(s)|\asymp\zeta(1+\tfrac{1}{\log
x})\exp(-\mathbb{D}(f,n^{it};x)^{2})\asymp(\log
x)\exp(-\mathbb{D}(f,n^{it};x)^{2})\leqslant\log x;$ (5.1)

and $M\ll\log\log x$. If $|t|\geqslant 1$ then $|\zeta(s)|\ll\log(2+|t|)$, and
so

$\bigg{|}\sum_{n\leqslant x}\frac{f(n)}{n}\bigg{|}\ll(1+M)\frac{|F(s)|}{\log
x}\frac{\log(2+|t|)}{1+|t|}\ll\log\log x.$

We henceforth assume that $|t|\leqslant 1$. We have

$|\zeta(s)/s|\asymp\log y_{t}\text{ where }y_{t}=\min\\{e^{1/|t|},x\\},$

and

$e^{-M}=\frac{|\zeta(s)F(s)/s|}{(\log
x)^{2}}\asymp\exp\bigg{(}-\min_{|\tau|\leqslant
1}(\mathbb{D}(1,n^{i\tau};x)^{2}+\mathbb{D}(f,n^{i\tau};x)^{2})\bigg{)}.$

By (3.3), $\mathbb{D}(1,n^{it};x)^{2}=\log(\frac{\log x}{\log y_{t}})+O(1)$,
so we deduce that

$\displaystyle\sum_{n\leqslant x}\frac{f(n)}{n}$ $\displaystyle\ll\log
y_{t}\,\exp(-\mathbb{D}(f,n^{it};x)^{2})\cdot\mathcal{L}$ (5.2)
$\displaystyle\asymp\mathcal{L}\,e^{-\mathcal{L}}\log x,$ (5.3)

where we have set

$\mathcal{L}:=1+\mathbb{D}(f,n^{it};x)^{2}+\log(\tfrac{\log x}{\log
y_{t}})\ll\log\log x.$

From (5.2) and $\mathbb{D}(f,n^{it};x)^{2}\geqslant M(f;x,1)$ we obtain (3.4)
when $|t|\leqslant\tfrac{1}{\log x}$ and (3.5) when $\tfrac{1}{\log
x}<|t|\leqslant 1$. ∎

The next proof continues on using the results in the previous proof:

###### Proof of Proposition 2(a).

If $|t|\leqslant\tfrac{1}{\log x}$ then
$\mathbb{D}(f,1;x)=\mathbb{D}(f,n^{it};x)+O(1)$, and in this case we also
obtain (1.2) (for the previous proof) with the better constant $1$ in place of
$\lambda$.

For $\tfrac{1}{\log x}\leqslant|t|\leqslant 1$, we now prove the lower bound
$\mathcal{L}\geqslant\lambda\,\mathbb{D}(f,1;x)^{2}+O(1)$. When we substitute
this into (5.3), we obtain (1.2) since $y\mapsto ye^{-y}$ is a decreasing
function for $y\geqslant 0$.

First, as $p^{-it}=1+O(|t|\log p)$ when $p\leqslant y_{t}$, we obtain

$\mathbb{D}(f,n^{it};y_{t})^{2}-\mathbb{D}(f,1;y_{t})^{2}=\sum_{p\leqslant
y_{t}}\frac{\text{Re}(f(p)(1-p^{-it}))}{p}\ll|t|\sum_{p\leqslant
y_{t}}\frac{\log p}{p}\ll 1.$

The prime number theorem implies that

$\displaystyle\sum_{y_{t}<p\leqslant
x}\frac{\text{Re}(f(p)(p^{-it}-\lambda))}{p}\leqslant\sum_{y_{t}<p\leqslant
x}\frac{|p^{-it}-\lambda|}{p}$
$\displaystyle=\bigg{(}\int_{0}^{1}|e(u)-\lambda|du\bigg{)}\log\bigg{(}\frac{\log
x}{\log y_{t}}\bigg{)}+O(1)$ $\displaystyle=(2-\lambda)\sum_{y_{t}<p\leqslant
x}\frac{1}{p}+O(1),$ (5.4)

using the definition of $\lambda$. Re-organised, this gives

$\mathbb{D}(f,n^{it};y_{t},x)^{2}+\log\bigg{(}\frac{\log x}{\log
y_{t}}\bigg{)}+O(1)\geqslant\lambda\,\mathbb{D}(f,1;y_{t},x)^{2}+O(1),$

so that

$\displaystyle\mathcal{L}$
$\displaystyle=(\mathbb{D}(f,n^{it};y_{t},x)^{2}+\log(\tfrac{\log x}{\log
y_{t}}))+\mathbb{D}(f,n^{it};y_{t})^{2}+1$
$\displaystyle\geqslant\lambda\,\mathbb{D}(f,1;y_{t},x)^{2}+\mathbb{D}(f,1;y_{t})^{2}+O(1)\geqslant\lambda\,\mathbb{D}(f,1;x)^{2}+O(1).\qed$

##  6\. Large character sums

###  6.1. Consequences of repulsion

Suppose that we are given a primitive character $\chi\pmod{q}$. Fix $A>0$. For
each primitive character $\psi\pmod{\ell}$ with $\ell\leqslant(\log q)^{A}$
select $|t|\leq 1$ for which $\mathbb{D}(\chi,\psi\,n^{it};q)$ is minimized.
Index the pairs $(\psi,t)$ so that $(\psi_{j},t_{j})$ is the pair that gives
the $j$-th smallest distance $\mathbb{D}(\chi,\psi_{j}\,n^{it_{j}};q)$
(breaking ties arbitrarily if needed). A simple modification of [1, Lem. 3.1]
shows that for each $k\geqslant 2$ we have

$\mathbb{D}(\chi,\psi_{k}\,n^{it_{k}};q)^{2}\geq(c_{k}+o(1))\log\log q$

where $c_{k}\geqslant 1-\tfrac{1}{\sqrt{k}}$. As any $1\leqslant n\leqslant
N\leqslant q$ has $P(n)\leqslant q$, [19, Thm. 6.4] yields

$\sum_{n\leqslant
N}\frac{(\chi\bar{\psi_{k}})(n)}{n}=\sum_{\begin{subarray}{c}n\leqslant N\\\
P(n)\leqslant q\end{subarray}}\frac{(\chi\bar{\psi_{k}})(n)}{n}\ll(\log
q)e^{-\mathbb{D}(\chi,\psi_{k}\,n^{it_{k}};q)^{2}}+1\ll(\log
q)^{1-c_{k}+o(1)}.$ (6.1)

Under the additional hypothesis that $\psi_{1}\psi_{2}$ is an even character
(which follows if we restrict attention to $\psi$ with $(\chi\psi)(-1)=-1$) we
can take any $c_{2}>1-\tfrac{2}{\pi}$ as we will see below in Lemma 3.

For each integer $k\geqslant 1$ we define

$\gamma_{k}:=\frac{1}{k}\sum_{a=0}^{k-1}|\cos(\pi
a/k)|=\frac{1}{k}\sum_{a=0}^{k-1}\big{|}1+e\big{(}a/k\big{)}\big{|}.$

Using the Fourier expansion

$\displaystyle|1+e(\alpha)|=\frac{4}{\pi}\bigg{(}1-\sum_{d\neq
0}\frac{(-1)^{d}}{4d^{2}-1}e(d\alpha)\bigg{)},\quad\alpha\in\mathbb{R}/\mathbb{Z},$
(6.2)

we easily deduce that

$\gamma_{k}=\frac{2}{\pi}\Bigg{(}1-2\sum_{\begin{subarray}{c}r\geqslant 1\\\
k|r\end{subarray}}\frac{(-1)^{r}}{4r^{2}-1}\Bigg{)}=\begin{cases}\frac{\text{\rm
cosec}(\pi/2k)}{k}&\text{if}\ k\ \text{is odd},\\\
\frac{\cot(\pi/2k)}{k}&\text{if}\ k\ \text{is even};\end{cases}$ (6.3)

(see also Lemma 5.2 of [10]). Therefore
$\gamma_{k}=\tfrac{2}{\pi}+O(\tfrac{1}{k^{2}})$, and the $\gamma_{k}$, with
$k$ even, increase towards $\tfrac{2}{\pi}=0.6366\ldots$:

$\gamma_{2}=\tfrac{1}{2}<\gamma_{4}=0.6035\ldots<\gamma_{6}=0.6220\ldots<\gamma_{8}=0.6284\ldots$

whereas the $\gamma_{k}$, with $k$ odd, decrease towards $\tfrac{2}{\pi}$:

$\gamma_{1}=1>\gamma_{3}=\tfrac{2}{3}>\gamma_{5}=0.6472\ldots>\gamma_{7}=0.6419\ldots>\gamma_{9}=0.6398\ldots$

For $k>1$, we have $\gamma_{k}\leqslant\tfrac{2}{3}$.

######  Lemma 3.

Fix $C>0$ and let $m\leqslant(\log q)^{C}$. Let $t_{1},t_{2}\in\mathbb{R}$ be
chosen such that $t:=t_{2}-t_{1}$ satisfies $|t|\ll 1$. Let $\chi_{1}$ and
$\chi_{2}=\chi_{1}\xi$ be characters with modulus in $[q,q^{2}]$, where
$\xi\pmod{m}$ is an odd character. Suppose that
$\mathbb{D}(\chi_{2},n^{it_{2}};q)\geqslant\mathbb{D}(\chi_{1},n^{it_{1}};q)$.
Then for any $\epsilon>0$,

$\mathbb{D}(\chi_{2},n^{it_{2}};q)^{2}\geqslant(1-\tfrac{2}{\pi}-\epsilon\theta)\log\log
q+O(\log\log\log q),$

where $\theta:=1$ if $\xi^{j}$ is exceptional for some $1\leqslant
j\leqslant\min\\{\phi(m),\log\log q\\}$, and $\theta:=0$ otherwise.

###### Proof.

Suppose that $\xi\pmod{m}$ has order $k$. We note that

$\displaystyle\mathbb{D}(\chi_{2},n^{it_{2}};q)^{2}$
$\displaystyle\geqslant\frac{1}{2}\bigg{(}\mathbb{D}(\chi_{2},n^{it_{2}};q)^{2}+\mathbb{D}(\chi_{1},n^{it_{1}};q)^{2}\bigg{)}$
$\displaystyle=\log\log q-\frac{1}{2}\text{Re}\bigg{(}\sum_{p\leqslant
q}\frac{\chi_{1}(p)p^{-it_{1}}+\chi_{2}(p)p^{-it_{2}}}{p}\bigg{)}+O(1)$
$\displaystyle=\log\log q-\frac{1}{2}\text{Re}\bigg{(}\sum_{p\leqslant
q}\frac{\chi_{1}(p)p^{-it_{1}}(1+\xi(p)p^{-it})}{p}\bigg{)}+O(1)$
$\displaystyle\geqslant\log\log q-\frac{1}{2}\sum_{p\leqslant
q}\frac{|1+\xi(p)p^{-it}|}{p}+O(1)$ (6.4)

Using (6.2) we deduce that

$\displaystyle\sum_{p\leqslant
q}\frac{|1+\xi(p)p^{-it}|}{p}=\frac{4}{\pi}\bigg{(}\log\log
q-\sum_{1\leqslant|d|\leqslant D}\frac{(-1)^{d}}{4d^{2}-1}S_{d}\bigg{)}+O(1),$

where $D:=\log\log q$ and

$S_{d}:=\sum_{p\leqslant q}\frac{\xi(p)^{d}}{p^{1+idt}}.$

If $k$ divides $d$ then $\xi(p)^{d}=1_{p\nmid m}$, and so

$\displaystyle S_{d}$ $\displaystyle=\sum_{p\leqslant
q}\frac{1}{p^{1+idt}}+O\bigg{(}\sum_{p|m}\frac{1}{p}\bigg{)}=\sum_{p\leqslant
q}\frac{1}{p^{1+idt}}+O(\log\log\log\log q).$

The sum over $p\leqslant z:=\min\\{q,e^{2\pi/|dt|}\\}$ equals

$\displaystyle\sum_{p\leqslant z}\frac{1}{p}+O\bigg{(}|dt|\sum_{p\leqslant
z}\frac{\log p}{p}\bigg{)}$ $\displaystyle=\log(\min\\{\log
q,\tfrac{2\pi}{|dt|}\\})+O(1)$ $\displaystyle=\log(\min\\{\log
q,\tfrac{1}{|t|}\\})+O(\log\log\log q).$

The remaining set of primes $z<p\leqslant q$ is non-empty only if
$z=e^{2\pi/|dt|}$, which happens when $|dt|>\frac{2\pi}{\log q}$. Let
$\sigma:=\tfrac{dt}{|dt|}\in\\{-1,+1\\}$. Applying the prime number theorem,

$\displaystyle\sum_{z<p\leqslant
q}\frac{1}{p^{1+idt}}=\int_{e^{2\pi/|dt|}}^{q}u^{-idt}\frac{du}{u\log
u}+O(1)=\int_{1}^{X}e(-\sigma v)\frac{dv}{v}+O(1)=O(1),$

putting $v=\tfrac{|dt|\log u}{2\pi}$ and $X=\tfrac{\log q}{\log z}$
($\geqslant 1$) and then integrating by parts. We deduce that if $k$ divides
$d$ with $d\leqslant D$ then $S_{d}=\log(\min\\{\log
q,\tfrac{1}{|t|}\\})+O(\log\log\log q)$.

If $k$ does not divide $d$ let $\alpha:=\xi^{d}$, a non-principal character
mod $m$, and let $T=dt$ so that $|T|\ll D$. For $\epsilon>0$ small, let

$Y:=\begin{cases}\exp((\log m)^{10}+T)&\text{ if $\alpha$ is non-
exceptional,}\\\ \exp((\log q)^{\epsilon})&\text{ if $\alpha$ is
exceptional.}\end{cases}$

Trivially bounding the primes $p\leqslant Y$ we deduce that

$S_{d}=\sum_{Y<p\leqslant q}\frac{\alpha(p)}{p^{1+iT}}+O(\log\log Y).$

Since $\psi(y,\alpha)\ll\tfrac{y}{\log y}$ for $y\geqslant Y$ by the Siegel-
Walfisz theorem when $\alpha$ is exceptional, and using (4.2) otherwise, this
is

$\displaystyle\int_{Y}^{q}\frac{d\psi(y,\alpha)}{y^{1+iT}\log
y}+O\bigg{(}\frac{1}{Y}\bigg{)}$
$\displaystyle=\left[\frac{\psi(y,\alpha)}{y^{1+iT}\log
y}\right]_{Y}^{q}+(1+iT)\int_{Y}^{q}\frac{\psi(y,\alpha)}{y^{2+iT}\log
y}dy+O\bigg{(}\frac{1}{(\log Y)^{2}}\bigg{)}$ $\displaystyle\ll\frac{1}{(\log
Y)^{2}}+T\int_{Y}^{q}\frac{1}{y(\log y)^{2}}dy\ll 1.$

We conclude that $S_{d}\ll\log\log Y$ whenever $k\nmid d$.

Putting these estimates into (6.3), we find that

$\displaystyle\sum_{p\leqslant q}\frac{|1+\xi(p)p^{-it}|}{p}$
$\displaystyle=\frac{4}{\pi}\log\log
q+2\bigg{(}\gamma_{k}-\frac{2}{\pi}\bigg{)}\log(\min\\{\log
q,\tfrac{1}{|t|}\\})$ $\displaystyle+O(\log\log\log q+\log\log Y),$

and $\log\log Y\ll\epsilon\theta\log\log q+\log\log\log q$. Changing
$\epsilon$ by a (possibly ineffective) constant factor if needed, we deduce
from (6.4) that

$\displaystyle\mathbb{D}(\chi_{2},n^{it_{2}};q)^{2}\geqslant(1-\tfrac{2}{\pi}-\epsilon\theta)\log\log
q+(\tfrac{2}{\pi}-\gamma_{k})\log(\min\\{\log
q,\tfrac{1}{|t|}\\})+O(\log\log\log q).$

Now since $\xi$ is odd, its order $k$ must be even and therefore
$\gamma_{k}\leqslant\tfrac{2}{\pi}$; the result follows since the middle term
is $\gg-O(1)$. ∎

For any $\epsilon>0$ let $K>1/\epsilon^{2}$ so that $1-c_{K}<\epsilon$. As a
consequence of (6.1) and Lemma 3, established just below, we have

$\max_{1\leqslant N\leqslant q}\bigg{|}\sum_{n\leqslant
N}\frac{(\chi\bar{\psi})(n)}{n}\bigg{|}=\begin{cases}O((\log
q)^{\epsilon})&\text{ if }\psi\neq\psi_{k}\text{ for all }k<K,\\\
O_{\epsilon}((\log q)^{\tfrac{2}{\pi}+\epsilon})&\text{ if
}\psi=\psi_{k}\text{ for some }k,1<k<K.\end{cases}$ (6.5)

The implicit constant in the second case is effective as long as $\psi^{j}$ is
non-exceptional for all $j\geqslant 1$, and in this case
$\tfrac{2}{\pi}+\epsilon$ can be replaced by $\tfrac{2}{\pi}+o(1)$.

###  6.2. Working with large character sums

In this subsection we set the stage for the proof of Proposition 3. We recall
that $\Delta\in(\tfrac{2}{\pi},1)$ and $q\geqslant 3$ are given, and

$R_{q}:=\exp(\tfrac{(\log q)^{\Delta}}{\log\log q}),\quad r_{q}:=(\log
q)^{2-2\Delta}(\log\log q)^{4}.$

######  Proposition 5.

For any given primitive character $\chi\pmod{q}$ there exists a primitive
character $\xi\pmod{\ell}$ with $\ell\leqslant r_{q}$ and $(\chi\xi)(-1)=-1$
such that if $|\alpha-\tfrac{b}{m}|\leqslant\frac{1}{mR_{q}}$ with $m\leqslant
r_{q}$ and $N:=\min\\{q,\frac{1}{|m\alpha-b|}\\}$ then

$\sum_{1\leqslant|n|\leqslant
q}\frac{\chi(n)e(n\alpha)}{n}=1_{\ell|m}\frac{2\eta
g(\xi)}{\phi(m)}\prod_{p|\tfrac{m}{\ell}}(1-(\bar{\chi}\xi)(p))\sum_{1\leqslant
n\leqslant N}\frac{(\chi\bar{\xi})(n)}{n}+o((\log q)^{\Delta}),$

where, whenever $\ell|m$ we write $\ell_{q}$ to denote the largest divisor of
$m/\ell$ that is coprime to $q$, $m_{q}\ell_{q}=m/\ell$ and

$\eta:=\mu(m_{q})\chi(\ell_{q})\bar{\xi}(b)\xi(m_{q})\in S^{1}\cup\\{0\\}.$

###### Proof.

By (2.4) we have

$\displaystyle\sum_{1\leqslant|n|\leqslant
q}\frac{\chi(n)e(n\alpha)}{n}=\sum_{1\leqslant|n|\leqslant
N}\frac{\chi(n)e(n\tfrac{b}{m})}{n}+O(\log\log q).$ (6.6)

With the intent of replacing exponentials by Dirichlet characters on the
right-hand side, we split the $n$th summand according to the common factors of
$n$ with $m$. Therefore if $(n,m)=d$ with $m=cd$ and $n=rd$ we have

$\displaystyle\sum_{1\leqslant|n|\leqslant
N}\frac{\chi(n)}{n}e(n\tfrac{b}{m})$
$\displaystyle=\sum_{cd=m}\frac{\chi(d)}{d}\sum_{1\leqslant|r|\leqslant
N/d\atop(r,c)=1}\frac{\chi(r)}{r}e(r\tfrac{b}{c})$
$\displaystyle=\sum_{cd=m}\frac{\chi(d)}{d\phi(c)}\sum_{\psi\pmod{c}}\bar{\psi}(b)g(\psi)\sum_{1\leqslant|n|\leqslant
N/d}\frac{\chi\bar{\psi}(n)}{n}$
$\displaystyle=2\sum_{cd=m}\frac{\chi(d)}{d\phi(c)}\sum_{\psi\pmod{c}\atop\chi\psi(-1)=-1}\bar{\psi}(b)g(\psi)\sum_{1\leqslant
n\leqslant N/d}\frac{\chi\bar{\psi}(n)}{n},$

since if $(k,c)=1$ then
$e(\tfrac{k}{c})=\frac{1}{\phi(c)}\sum_{\psi\pmod{c}}\bar{\psi}(k)g(\psi)$,
and then noting the $n$ and $-n$ terms cancel if $\chi\psi(-1)=1$.

To control the size of $g(\psi)$ we split the sum over characters modulo $c$
according to the primitive characters that induce them. Since each $\psi$
factors as $\psi^{\ast}\psi_{0}^{(f)}$, where $\psi^{\ast}$ is primitive
modulo $e$ and $\psi_{0}^{(f)}$ is principal modulo $f$ with $ef=c$, the
right-hand side of the above sum is

$\displaystyle
2\sum_{efd=m}\frac{\chi(d)}{d\phi(ef)}\sideset{}{{}^{\ast}}{\sum}_{\psi^{\ast}\pmod{e}\atop\chi\psi^{\ast}(-1)=-1}\bar{\psi}^{\ast}(b)g(\psi^{\ast}\psi_{0}^{(f)})\sum_{1\leqslant
n\leqslant N/d\atop(n,f)=1}\frac{\chi\bar{\psi}^{\ast}(n)}{n}$
$\displaystyle=2\sum_{efd=m\atop(e,f)=1}\frac{\mu(f)\chi(d)}{d\phi(ef)}\sideset{}{{}^{\ast}}{\sum}_{\psi^{\ast}\pmod{e}\atop\chi\psi^{\ast}(-1)=-1}\psi^{\ast}(f\bar{b})g(\psi^{\ast})\sum_{1\leqslant
n\leqslant N/d\atop(n,f)=1}\frac{\chi\bar{\psi}^{\ast}(n)}{n},$ (6.7)

since $g(\psi^{\ast}\psi_{0}^{(f)})=\psi^{\ast}(f)\mu(f)g(\psi^{\ast})$.

Fix $ef|m$ and $\psi^{\ast}\pmod{e}$. We extend the inner sum in (6.7) to all
$n\leqslant N$ as

$\displaystyle\sum_{1\leqslant n\leqslant
N/d\atop(n,f)=1}\frac{\chi\bar{\psi}^{\ast}(n)}{n}=\sum_{1\leqslant n\leqslant
N\atop(n,f)=1}\frac{\chi\bar{\psi}^{\ast}(n)}{n}+O(\log(2d)).$ (6.8)

By Lemma 4.4 of [13],

$\displaystyle\sum_{1\leqslant n\leqslant
N\atop(n,f)=1}\frac{\chi\bar{\psi}^{\ast}(n)}{n}=\prod_{p|f}\bigg{(}1-\frac{\chi\bar{\psi}^{\ast}(p)}{p}\bigg{)}\sum_{1\leqslant
n\leqslant
N}\frac{\chi\bar{\psi}^{\ast}(n)}{n}+O\bigg{(}(\log\log(2+f))^{2}\bigg{)}.$
(6.9)

We next observe the identity

$\sum_{fd=m/e\atop(e,f)=1}\frac{\mu(f)\chi(d)}{d\phi(ef)}\psi^{\ast}(f)\prod_{p|f}\bigg{(}1-\frac{\chi\bar{\psi}^{\ast}(p)}{p}\bigg{)}=\frac{\chi(e_{q})\psi^{\ast}(m_{q})\mu(m_{q})}{\phi(m)}\prod_{p|\tfrac{m}{e}}(1-(\bar{\chi}\psi^{\ast})(p)),$

where now $e_{q}$ is the largest divisor of $m/e$ which is coprime to $q$, and
$m_{q}e_{q}=m/e$. The main terms from (6.9) thus contribute

$\displaystyle\frac{2}{\phi(m)}\sum_{e|m}\mu(m_{q})\chi(e_{q})\sideset{}{{}^{\ast}}{\sum}_{\psi^{\ast}\pmod{e}\atop\chi\psi^{\ast}(-1)=-1}g(\psi^{\ast})\psi^{\ast}(m_{q}\bar{b})\prod_{p|\tfrac{m}{e}}(1-(\bar{\chi}\psi^{\ast})(p))\sum_{1\leqslant
n\leqslant N}\frac{\chi\bar{\psi}^{\ast}(n)}{n}$ (6.10)

in (6.7). By noting that $|g(\psi^{\ast})|=\sqrt{e}=\sqrt{m/df}$ the
contribution of the error terms from (6.8) and (6.9) in (6.7) is bounded by

$\displaystyle\ll\sqrt{m}\sum_{fd|m}\frac{(\log 2f)(\log
2d)}{d^{3/2}\phi(f)f^{1/2}}\ll\sqrt{m}\leqslant\sqrt{r_{q}}=o((\log
q)^{\Delta}).$ (6.11)

We now apply (6.5) with $N=\min\\{q,\tfrac{1}{|m\alpha-b|}\\}$. Set
$\xi:=\psi_{1}$, whose contribution,

$\frac{2\mu(m_{q})\chi(\ell_{q})\xi(\bar{b}m_{q})g(\xi)}{\phi(m)}\prod_{p|\tfrac{m}{\ell}}(1-(\bar{\chi}\xi)(p))\sum_{1\leqslant
n\leqslant N}\frac{\chi\bar{\xi}(n)}{n}$

only appears in (6.10) if the conductor $\ell$ of $\xi$ divides $m$. By (6.5)
the contributions to (6.10) from all the characters $\psi\neq\psi_{k}$ for all
$k<K$ is, for $\epsilon$ sufficiently small,

$\ll(\log
q)^{\epsilon}\sum_{e|m}\frac{\sqrt{e}\,\tau(m/e)\phi(e)}{\phi(m)}\ll\sqrt{m}(\log
q)^{2\epsilon}\leqslant\sqrt{r_{q}}(\log q)^{2\epsilon}\ll(\log
q)^{1-\Delta+3\epsilon}=o((\log q)^{\Delta}).$

Since the coefficient in front of each individual sum over $n$ in (6.10) is
bounded, again by (6.5) the contribution of the main terms from all of the
characters $\psi_{k}$ with $1<k<K$ is $\ll_{\epsilon}K\cdot(\log
q)^{\tfrac{2}{\pi}+\epsilon}=o((\log q)^{\Delta})$, if $\epsilon$ is
sufficiently small. We insert these estimates into (6.6) to obtain the result.
∎

The proof of Proposition 5 is very similar to the proof of the main results in
[3].

###### Proof of Proposition 3.

Let $\alpha\in[0,1)$ be chosen so that $M(\chi)=|S(\chi,\alpha q)|$. Applying
(2.1), we have

$\frac{M(\chi)}{\sqrt{q}}=\frac{1}{2\pi}\bigg{|}\sum_{1\leqslant|n|\leqslant
q}\frac{\chi(n)}{n}-\sum_{1\leqslant|n|\leqslant
q}\frac{\chi(n)e(n\alpha)}{n}\bigg{|}+O(1).$

Let $\xi:=\psi_{1}$ once again, and let $\ell$ be its conductor. The proof is
split up according to whether $\ell>1$ or $\ell=1$.

Case 1: Assume $\ell>1$, so $\xi$ is non-trivial. Suppose first that
$|M(\chi)|\gg\sqrt{q}(\log q)^{\Delta}$. In light of (2.3), $\alpha$ is on a
major arc, so there is $\tfrac{b}{m}$ such that $m\leqslant r_{q}$ and
$|\alpha-\tfrac{b}{m}|\leqslant\tfrac{1}{mR_{q}}$, with $\ell|m$ by
Proposition 5. Note that if we vary $\alpha$ in the interval
$\left[\tfrac{b}{m}-\tfrac{1}{mR_{q}},\tfrac{b}{m}+\tfrac{1}{mR_{q}}\right]$
then $N=N(\alpha)=\min\\{q,\tfrac{1}{|m\alpha-b|}\\}$ varies in the range
$R_{q}\leqslant N\leqslant q$. As $\ell>1$, Proposition 5 also shows that
$\sum_{1\leqslant|n|\leqslant q}\frac{\chi(n)}{n}=o((\log q)^{\Delta})$, and
moreover, writing $m_{q}\ell_{q}=m/\ell$ as before,

$\frac{M(\chi)}{\sqrt{q}}=\frac{\sqrt{\ell}}{\pi\phi(m)}\prod_{p|\tfrac{m}{\ell}}|1-(\bar{\chi}\xi)(p)|\cdot\bigg{|}\sum_{1\leqslant
n\leqslant N}\frac{(\chi\bar{\xi})(n)}{n}\bigg{|}+o((\log q)^{\Delta})$

provided $\mu(m_{q})\chi(\ell_{q})\xi(m_{q})\neq 0$. Evidently, to maximize
the right-hand side we must have $N(\alpha)=N_{q}$.

Next, we find $m=d\ell$, given $\xi$ and $N_{q}$, that maximizes

$s_{d}:=\frac{1}{\phi(d\ell)}\prod_{p|d}|1-(\bar{\chi}\xi)(p)|.$

Suppose that $p^{e}\|d$ and $D=dp^{e}$. If $p|\ell$ then
$s_{d}=s_{D}/p^{e}<s_{D}$ so we may assume that $p\nmid\ell$. In that case
$s_{d}\leqslant 2s_{D}/\phi(p^{e})\leqslant s_{D}$ unless $p^{e}=2$. Hence
$d=1$ or $2$ and $\phi(d\ell)=\phi(\ell)$, and so

$\frac{M(\chi)}{\sqrt{q}}=\frac{\sqrt{\ell}}{\pi\phi(\ell)}\max\\{1,|1-(\bar{\chi}\xi)(2)|\\}\bigg{|}\sum_{1\leqslant
n\leqslant N_{q}}\frac{(\chi\bar{\xi})(n)}{n}\bigg{|}+o((\log q)^{\Delta}),$

which proves (2.5) when $\ell>1$, and also that $|\sum_{n\leqslant
N_{q}}\frac{(\chi\bar{\xi})(n)}{n}|\gg\frac{\phi(\ell)}{\sqrt{\ell}}(\log
q)^{\Delta}$.

Conversely, assume that $\bigg{|}\sum_{n\leqslant
N_{q}}\frac{(\chi\bar{\psi})(n)}{n}\bigg{|}\gg\frac{\phi(r)}{\sqrt{r}}(\log
q)^{\Delta}$ for some primitive character $\psi$ of conductor $r\leqslant
r_{q}$ with $\psi(-1)=-\chi(-1)$. In view of (6.5), it follows that $\psi=\xi$
and $r=\ell$. The assumption also implies that $\log N_{q}+O(1)\geqslant(\log
q)^{\Delta}$, so $N_{q}\geqslant R_{q}$.

Selecting $\beta\in[\tfrac{1}{\ell}-\tfrac{1}{\ell
R_{q}},\tfrac{1}{\ell}+\tfrac{1}{\ell R_{q}}]$ so that
$N(\beta)=N_{q}\in[R_{q},q]$ and applying Proposition 5,

$\displaystyle\bigg{|}\sum_{1\leqslant|n|\leqslant
q}\frac{\chi(n)e(n\beta)}{n}\bigg{|}=\frac{2\sqrt{\ell}}{\phi(\ell)}\bigg{|}\sum_{1\leqslant
n\leqslant N_{q}}\frac{\chi\bar{\xi}(n)}{n}\bigg{|}+o((\log
q)^{\Delta})\gg(\log q)^{\Delta}.$ (6.12)

Combining (2.1) with (6.12) and a second application of Proposition 5, we get

$M(\chi)\geqslant|S(\chi,\beta
q)|\geqslant\frac{\sqrt{q}}{2\pi}\bigg{|}\sum_{1\leqslant|n|\leqslant
q}\frac{\chi(n)e(n\beta)}{n}\bigg{|}-\frac{\sqrt{q}}{2\pi}\bigg{|}\sum_{1\leqslant|n|\leqslant
q}\frac{\chi(n)}{n}\bigg{|}+O(\log q)\gg\sqrt{q}(\log q)^{\Delta},$

as required.

Case 2: Assume now that $\ell=1$ and $\xi$ is trivial so that $\chi$ is odd.
If $\alpha$ is on a minor arc then from (2.1) and (2.3) we get

$\displaystyle\frac{M(\chi)}{\sqrt{q}}=\frac{1-\chi(-1)}{2\pi}\bigg{|}\sum_{n\leqslant
q}\frac{\chi(n)}{n}\bigg{|}+o((\log q)^{\Delta}).$ (6.13)

On the other hand, if $\alpha$ is on a major arc then by Proposition 5,

$\frac{M(\chi)}{\sqrt{q}}=\frac{1-\chi(-1)}{2\pi}\max_{\begin{subarray}{c}R_{q}\leqslant
N\leqslant q\\\ 1\leqslant m\leqslant
r_{q}\end{subarray}}\bigg{|}\sum_{n\leqslant
q}\frac{\chi(n)}{n}-\frac{\chi(\ell_{q})\mu(m_{q})}{\phi(m)}\prod_{p|m}(1-\bar{\chi}(p))\sum_{1\leqslant
n\leqslant N}\frac{\chi(n)}{n}\bigg{|}+o((\log q)^{\Delta}).$

The coefficient of the sum up to $N$ is $\leqslant 2$ (which is attained if
$m=2$ and $\chi(2)=-1$), so that by the triangle inequality

$\displaystyle\frac{M(\chi)}{\sqrt{q}}\leqslant\frac{3}{\pi}\bigg{|}\sum_{n\leqslant
N_{q}}\frac{\chi(n)}{n}\bigg{|}+o((\log q)^{\Delta}).$ (6.14)

We obtain this as an equality when $\chi(2)=-1$ with $N_{q}=q$ and $m=2$.

By (6.13) and then by taking $m=1$ and $N=N_{q}$ above, we obtain

$\displaystyle\frac{M(\chi)}{\sqrt{q}}$
$\displaystyle\geqslant\frac{1}{\pi}\max\bigg{\\{}\bigg{|}\sum_{n\leqslant
q}\frac{\chi(n)}{n}\bigg{|},\bigg{|}\sum_{n\leqslant
q}\frac{\chi(n)}{n}-\sum_{n\leqslant
N_{q}}\frac{\chi(n)}{n}\bigg{|}\bigg{\\}}+o((\log q)^{\Delta})$
$\displaystyle\geqslant\frac{1}{2\pi}\bigg{|}\sum_{n\leqslant
N_{q}}\frac{\chi(n)}{n}\bigg{|}+o((\log q)^{\Delta}).$ (6.15)

(6.14) and (6.2) imply (2.5). Together these bounds also yield that
$M(\chi)\gg\sqrt{q}(\log q)^{\Delta}$ if and only if $|\sum_{n\leqslant
N_{q}}\tfrac{\chi(n)}{n}|\gg(\log q)^{\Delta}$. ∎

######  Corollary 3.

Let $\chi$ be a primitive character modulo $q$. Assume $M(\chi)\geqslant
c_{1}\sqrt{q}\log q$ and $|S(\chi,N)|\gg N$ for some $N\in[q^{c_{2}},q]$, with
$c_{1},c_{2}\gg 1$. Then $|L(1,\chi)|\gg\log q$ , $\mathbb{D}(\chi,1;q)\ll 1$
and $\chi$ is odd.

###### Proof.

Assume that $M(\chi)\geqslant c_{1}\sqrt{q}\log q$ and $|S(\chi,N)|\gg N$ for
some $N\in[q^{c_{2}},q]$. By Corollary 1 and Proposition 1, there is $|t|\ll
1$ and $\ell\ll 1$ such that, simultaneously,

$|L(1+it,\chi)|\gg\log q\text{ and }|L(1,\chi\bar{\xi})|\gg|L(1,\psi)|\gg\log
q,$

where $\xi$ is primitive modulo $\ell$ and $\xi(-1)=-\chi(-1)$, and $\psi$ is
the primitive character that induces $\chi\bar{\xi}$. By (4.1),

$\bigg{|}\sum_{n\leqslant
q}\frac{\chi(n)n^{-it}}{n}\bigg{|},\bigg{|}\sum_{n\leqslant
q}\frac{\chi(n)\bar{\xi}(n)}{n}\bigg{|}\gg\log q.$

Applying (1.2) of Proposition 4 we obtain
$\mathbb{D}(\chi,n^{it};q),\mathbb{D}(\chi,\xi;q)\ll 1$, so by (3.1),

$\mathbb{D}(\xi,n^{it};q)\leqslant\mathbb{D}(\chi,\xi;q)+\mathbb{D}(\chi,n^{it};q)\ll
1.$

Therefore $\ell=1$, else we let $Y:=\exp((\log(2\ell))^{10}+|t|)\ll 1$, and
apply (4.2) and partial summation as in the proof of Lemma 3 to get

$\sum_{p\leqslant q}\frac{\xi(p)p^{-it}}{p}=\sum_{Y<p\leqslant
q}\frac{\xi(p)p^{-it}}{p}+O(\log\log Y)\ll 1,$

which implies that $\mathbb{D}(\xi,n^{it};q)^{2}=\log\log q+O(1)$, a
contradiction.

We deduce that $\xi$ is trivial so that $\chi(-1)=\chi\xi(-1)=-1$ and that
$|L(1,\chi)|\gg\log q$. By Lemma 1, $\chi$ must be non-exceptional, so (4.3)
gives $|L(1,\chi)|\asymp\log q\,e^{-\mathbb{D}(\chi,1;q)^{2}}$ and we deduce
that $\mathbb{D}(\chi,1;q)\ll 1$. ∎

##  7\. A class of examples

###  7.1. The set-up

Let $g:\mathbb{R}\rightarrow\mathbb{U}$ be a $1$-periodic function with
$g(0)=1$ and Fourier expansion

$g(t)=\sum_{n\in\mathbb{Z}}g_{n}e(nt)$

so that

$g_{n}:=\hat{g}(n):=\int_{0}^{1}g(u)e(-nu)du\text{ for all integers }n,$

and therefore $|g_{n}|\leqslant\int_{0}^{1}|g(u)|du\leqslant 1$ for all $n$.
We will assume that $|g_{n}|\ll|n|^{-3}$ for all integers $n\neq 0$ (so that
$\\{g_{n}\\}_{n}$ is absolutely summable).444The proof works provided
$|g_{n}|\ll 1/|n|^{2+\epsilon}$ for all integers $n\neq 0$.

Write $\gamma_{0}=g_{0}+1$ and $\gamma_{n}=g_{n}$ for all integers $n\neq 0$.
Then
$\sum_{n\in\mathbb{Z}}\text{Re}(\gamma_{n})=\sum_{n\in\mathbb{Z}}\text{Re}(g_{n})+1=\text{Re}(g(0))+1=2$
so that $\mu:=\max_{n}\text{Re}(\gamma_{n})>0$. Let
$\mathcal{L}:=\\{\ell\in\mathbb{Z}:\text{Re}(\gamma_{\ell})=\mu\\}$, which is
a non-empty set, and finite as $|g_{n}|\ll|n|^{-3}$. Moreover there exists
$\delta>0$ such that $\text{Re}(g_{n})\leqslant\mu-\delta$ for all
$n\not\in\mathcal{L}$.

Fix $t\in(0,1]$. We define a multiplicative function
$f=f_{t}:\mathbb{N}\rightarrow\mathbb{U}$ at primes $p$ by

$f_{t}(p):=g\bigg{(}\frac{t\log p}{2\pi}\bigg{)}\in\mathbb{U},$ (7.1)

and inductively on prime powers $p^{m}$, $m\geqslant 2$, via the convolution
formula

$\displaystyle f_{t}(p^{m}):=\frac{1}{m}\sum_{1\leqslant j\leqslant
m}f_{t}(p^{m-j})g\bigg{(}\frac{t\log p^{j}}{2\pi}\bigg{)}.$ (7.2)

Under these assumptions we will prove the following estimate:

######  Theorem 3.

Let $t\in[-1,1]$ be such that $|t|$ is small but $|t|\gg(\log X)^{-\epsilon}$
for all $\epsilon>0$. Then

$\sum_{n\leqslant
X}\frac{f_{t}(n)}{n}=(1+O(|t|))\sum_{\ell\in\mathcal{L}}\frac{X^{i\ell
t}}{i\ell^{\prime}t}\frac{C_{\ell}(it\log
X)^{\gamma_{\ell}-1}}{\Gamma(\gamma_{\ell})}$

where $C_{\ell}:=\prod_{k\neq 0}k^{-g_{\ell-k}}$, and $\ell^{\prime}=1$ if
$\ell=0$ and $\ell^{\prime}=\ell$ otherwise.

One can make the weaker assumption that $|g_{n}|\ll 1/|n|^{1+\epsilon}$ for
all integers $n\neq 0$, and obtain the weaker, but satisfactory, error term
$O(|t|^{\epsilon/2})$ in place of $O(|t|)$.

Henceforth fix $t$ and use $f=f_{t}$. By (7.2) and induction on $m\geqslant 1$
we have

$|f(p^{m})|\leqslant\frac{1}{m}\sum_{1\leqslant j\leqslant
m}|f(p^{m-j})|\leqslant 1,$

so that $f$ indeed takes values in $\mathbb{U}$. If $F(s)$ is the Dirichlet
series of $f$ for $\text{Re}(s)>1$ then $F(s)$ is analytic and non-vanishing
in that half-plane, and so $-\tfrac{F^{\prime}}{F}(s)$ is also analytic there.
The convolution identity (7.2) implies that

$-\frac{F^{\prime}}{F}(s)=\sum_{n\geqslant 1}g\bigg{(}\frac{t\log
n}{2\pi}\bigg{)}\frac{\Lambda(n)}{n^{s}}.$

Integrating $-\tfrac{F^{\prime}}{F}(s)$ termwise, we see that when
$\text{Re}(s)>1$,

$\log F(s)=\sum_{n\geqslant 1}\frac{\Lambda_{f}(n)}{n^{s}\log
n}=\sum_{p^{k}}\frac{g(\tfrac{t\log
p^{k}}{2\pi})}{kp^{ks}}=\sum_{m\in\mathbb{Z}}g_{m}\sum_{p^{k}}\frac{p^{ikmt}}{kp^{ks}}=\sum_{m\in\mathbb{Z}}g_{m}\log\zeta(s-imt),$

swapping orders of summation using the absolute summability of
$\\{g_{m}\\}_{m}$. For $\text{Re}(s)>1$, we may thus write

$F(s)=\prod_{m\in\mathbb{Z}}\zeta(s-imt)^{g_{m}}.$

We will work with the finite truncations of this product,

$F_{N}(s):=\prod_{|m|\leqslant 2N}\zeta(s-imt)^{g_{m}}.$

The proof of Theorem 3 relies on a technical contour integration argument
complicated by the possibility that the zeros and poles of $\zeta(s-imt)$
might contribute essential singularities whenever $g_{m}\neq 0$. The following
key technical lemma will be proved in section A.2.

For given $\tau\in\mathbb{R}$ we define

$\sigma(\tau):=\frac{c}{\log(2+|\tau|)},$

where $c>0$ is chosen sufficiently small so that $\zeta(\sigma+i\tau)\neq 0$
whenever $\sigma\geqslant 1-\sigma(\tau)$.

######  Lemma 4.

Let $t\in[-1,1]$ be such that $|t|$ is small but $|t|\gg(\log X)^{-\epsilon}$
for all $\epsilon>0$. Fix $A\geqslant 2$, let $N:=\lceil\tfrac{(\log
X)^{A}}{|t|}\rceil$ and $T:=(N+\tfrac{1}{2})|t|$. Also let
$r_{0}:=\tfrac{1}{4}\min\\{\sigma(3T),|t|\\}$.

(a) If $s=\sigma+i\tau$ with $\sigma\geqslant\frac{1}{\log X}$ and
$|\tau|\leqslant T$ then

$F(s+1)=F_{N}(s+1)+O((\log X)^{-2}).$

(b) We have

$\max_{|\tau|\leqslant T}|F_{N}(1-r_{0}+i\tau)|\ll_{\epsilon}(\log
X)^{\epsilon}.$

(c) Let $\eta\in\\{-1,+1\\}$. Then

$\max_{-r_{0}\leqslant\sigma\leqslant r_{0}}|F_{N}(1+\sigma+i\eta
T)|\ll_{\epsilon}(\log X)^{\epsilon}.$

(d) If $|t|$ is sufficiently small then for any $\ell\in\mathbb{Z}$,

$\prod_{\begin{subarray}{c}|k|\leqslant 2N\\\
k\neq\ell\end{subarray}}\zeta(1-i(k-\ell)t)^{g_{k}}=(1+O(|t|))C_{\ell}(it)^{g_{\ell}-1}.$

More generally when $|n|\leqslant N$ and $|s|\leqslant 2r_{0}$,

$\prod_{\begin{subarray}{c}|k|\leqslant 2N\\\ k\neq
n\end{subarray}}|\zeta(1+s-i(k-n)t)^{g_{k}}|\ll_{\epsilon}(\log
X)^{\epsilon}.$

###### Proof of Theorem 3.

Let $c_{0}:=\tfrac{1}{\log X},A=2$ and $N$ and $T$ be as in Lemma 4 so that
$T\geqslant(\log X)^{2}$. By a quantitative form of Perron’s formula [24, Cor.
2.4], we have

$\sum_{n\leqslant X}\frac{f(n)}{n}=\frac{1}{2\pi
i}\int_{(c_{0})}F(s+1)\frac{X^{s}}{s}ds=\frac{1}{2\pi
i}\int_{c_{0}-iT}^{c_{0}+iT}F(s+1)\frac{X^{s}}{s}ds+O\bigg{(}\frac{1}{\log
X}\bigg{)}.$

By Lemma 4(a),

$\displaystyle\sum_{n\leqslant X}\frac{f(n)}{n}=\frac{1}{2\pi
i}\int_{c_{0}-iT}^{c_{0}+iT}F_{N}(s+1)\frac{X^{s}}{s}ds+O\bigg{(}\frac{1}{\log
X}\bigg{)}.$ (7.3)

We now deform the path $[c_{0}-iT,c_{0}+iT]$ into a contour intersecting with
the critical strip within the common zero- and pole-free regions of
$\\{\zeta(s-int)\\}_{|n|\leqslant 2N}$. Since $|\text{Im}(s)-nt|\leqslant 3T$
we see that $\zeta(s+1-int)\neq 0$ for all $|n|\leqslant 2N$ and
$|\text{Im}(s)|\leqslant T$ whenever $\text{Re}(s)\geqslant-\sigma(3T)$.

Let $\mathcal{H}$ denote the Hankel contour555That is, for $r:=\frac{1}{\log
X}$ the contour consisting of the circular segment
$\\{s\in\mathbb{C}:|s|=r,\text{arg}(s)\in(-\pi,\pi)\\}$ (omitting the point
$s=-r$) together with the lines $\text{Re}(s)\leqslant-r$ covered twice, once
at argument $+\pi$ and once at argument $-\pi$, traversed counterclockwise.
[24, p. 179] of radius $\tfrac{1}{\log X}$, and let
$r_{0}:=\tfrac{1}{4}\min\\{|t|,\sigma(3T)\\}$. For each $|n|\leqslant N$ we
write

$\mathcal{H}_{n}:=(\mathcal{H}+int)\cap\\{\sigma+i\tau\in\mathbb{C}:\sigma\geqslant-
r_{0}\\}.$

We glue the paths $\\{\mathcal{H}_{n}\\}_{|n|\leqslant N}$ together and to the
horizontal lines $[-r_{0}+iT,c_{0}+iT]$ and $[-r_{0}-iT,c_{0}-iT]$ using the
line segments

$\displaystyle L_{n}$ $\displaystyle:=(-r_{0}+in|t|,-r_{0}+i((n+1)|t|))\text{
for }-N\leqslant n\leqslant N-1,$ $\displaystyle B_{1}$
$\displaystyle:=[-r_{0}-iT,-r_{0}-iN|t|),\quad\quad
B_{2}:=(-r_{0}+iN|t|,-r_{0}+iT]$

Denote this concatenated path by $\Gamma_{N}$ and define the contour

$\Gamma:=[c_{0}-iT,c_{0}+iT]\cup[c_{0}+iT,-r_{0}+iT]\cup\Gamma_{N}\cup[-r_{0}-iT,c_{0}-iT],$

traversed counterclockwise. Since $F_{N}(s+1)/s$ is analytic in the interior
of the component cut out by $\Gamma$, the residue theorem implies that

$\displaystyle\frac{1}{2\pi
i}\int_{c_{0}-iT}^{c_{0}+iT}F_{N}(s+1)\frac{X^{s}}{s}ds=\mathcal{M}+\mathcal{R},$
(7.4)

where $\mathcal{M}:=\frac{1}{2\pi i}\sum_{|n|\leqslant
N}\int_{\mathcal{H}_{n}}\frac{F_{N}(s+1)}{s}X^{s}ds$ is the contribution from
the Hankel contours, and

$\displaystyle\mathcal{R}$ $\displaystyle:=\frac{1}{2\pi
i}\bigg{(}\int_{B_{1}}\frac{F_{N}(s+1)}{s}X^{s}ds+\int_{B_{2}}\frac{F_{N}(s+1)}{s}X^{s}ds+\sum_{-N\leqslant
n\leqslant N-1}\int_{L_{n}}\frac{F_{N}(s+1)}{s}X^{s}ds\bigg{)}$
$\displaystyle-\frac{1}{2\pi
i}\bigg{(}\int_{-r_{0}-iT}^{c_{0}-iT}\frac{F_{N}(s+1)}{s}X^{s}ds-\int_{-r_{0}+iT}^{c_{0}+iT}\frac{F_{N}(s+1)}{s}X^{s}ds\bigg{)}.$

Along the segments $L_{n}$ and $B_{j}$, where $\text{Re}(s+1)=1-r_{0}$, we
apply Lemma 4(b) to obtain

$\displaystyle\sum_{-N\leqslant n\leqslant
N-1}\bigg{|}\int_{L_{n}}\frac{F_{N}(s+1)}{s}X^{s}ds\bigg{|}+\sum_{j=1,2}\bigg{|}\int_{B_{j}}\frac{F_{N}(s+1)}{s}X^{s}ds\bigg{|}$
$\displaystyle\ll_{\epsilon}X^{-r_{0}}(\log
X)^{\epsilon}\bigg{(}\sum_{|n|\leqslant
N}\frac{1}{r_{0}+|nt|}+\frac{1}{T}\bigg{)}\ll\frac{1}{\log X}.$

Along the horizontal segments we use Lemma 4(c) to give

$\displaystyle\bigg{|}\int_{-r_{0}\pm iT}^{c_{0}\pm
iT}\frac{F_{N}(s+1)}{s}X^{s}ds\bigg{|}\ll_{\epsilon}\frac{(\log
X)^{\epsilon}}{T}\ll\frac{1}{\log X}.$

Thus, $\mathcal{R}\ll\tfrac{1}{\log X}$, and it remains to treat
$\mathcal{M}$. For each $|n|\leqslant N$ note that
$\mathcal{H}_{n}=\mathcal{H}_{0}+int$, and so by a change of variables,

$\mathcal{M}_{n}:=\frac{1}{2\pi
i}\int_{\mathcal{H}_{n}}\frac{F_{N}(s+1)}{s}X^{s}ds=\frac{X^{int}}{2\pi
i}\int_{\mathcal{H}_{0}}\frac{G_{n}(s)}{s+int}\frac{X^{s}}{s^{g_{n}}}ds,$

where we set

$G_{n}(s):=\left[s\zeta(s+1)\right]^{g_{n}}\prod_{\begin{subarray}{c}|m|\leqslant
2N\\\ m\neq n\end{subarray}}\zeta(s+1-i(m-n)t)^{g_{m}}.$

$G_{n}$ is analytic near $0$, and when
$|s|\leqslant\tfrac{1}{2}\min\\{|t|,\sigma(3T)\\}=2r_{0}$ we can write

$G_{0}(s)=\sum_{j\geqslant
0}\mu_{0,j}(t)s^{j},\quad\quad\frac{G_{n}(s)}{s+int}=\sum_{j\geqslant
0}\mu_{n,j}(t)s^{j}\ \ \text{ if }n\neq 0.$

The functions $\mu_{n,j}(t)$ are determined by Cauchy’s integral formula as

$\mu_{n,j}(t)=\frac{1}{2\pi i}\int_{|s|=r}\frac{G_{n}(s)}{(s+int)^{1_{n\neq
0}}}\frac{ds}{s^{j+1}},\quad 0<r\leqslant 2r_{0}.$ (7.5)

Note in particular that

$\mu_{0,0}(t)=\prod_{\begin{subarray}{c}|m|\leqslant 2N\\\ m\neq
0\end{subarray}}\zeta(1-imt)^{g_{m}},\quad\mu_{n,0}(t)=\frac{1}{int}\prod_{\begin{subarray}{c}|m|\leqslant
2N\\\ m\neq n\end{subarray}}\zeta(1-i(m-n)t)^{g_{m}}\text{ if }n\neq 0,$

while for $j\geqslant 1$ we take $r=2r_{0}$ and apply Lemma 4(d) in (7.5) to
get666 Here and below, we repeatedly use the fact that if
$z\in\mathbb{C}\backslash\\{0\\}$ then, choosing any appropriate branch of
complex argument, we have
$|z^{g_{n}}|=|z|^{\text{Re}(g_{n})}\exp(-\text{Im}(g_{n})\cdot\text{arg}(z))\asymp|z|^{\text{Re}(g_{n})}$
for all $n\in\mathbb{Z}$, as $\sup_{n\in\mathbb{Z}}|g_{n}|\leqslant 1$ and
$|\text{arg}(z)|$ is uniformly bounded. for all $|n|\leqslant N$,

$\displaystyle|\mu_{n,j}(t)|$ $\displaystyle\leqslant
r^{-j}\max_{|s|=r}\frac{\prod_{\begin{subarray}{c}|k|\leqslant 2N\\\ k\neq
n\end{subarray}}|\zeta(s+1-i(k-n)t)^{g_{k}}|}{|s+int|^{1_{n\neq
0}}}\ll_{\epsilon}\frac{2^{j}(\log
X)^{\epsilon}}{\min\\{|t|,\sigma(3T)\\}^{j}(1_{n=0}+|nt|)}.$ (7.6)

Integrating over $\mathcal{H}_{0}$ (noting that $|s|\leqslant r/2$ for all
$s\in\mathcal{H}_{0}$) and applying [24, Cor. 0.18], when $n\neq 0$ we obtain

$\displaystyle\mathcal{M}_{n}$ $\displaystyle=\frac{X^{int}\mu_{n,0}(t)}{2\pi
i}\int_{\mathcal{H}_{0}}X^{s}s^{-g_{n}}ds+O_{\epsilon}\bigg{(}\frac{(\log
X)^{\epsilon}}{|nt|\min\\{|t|,\sigma(3T)\\}}\int_{\mathcal{H}_{0}}X^{\text{Re}(s)}|s|^{1-\text{Re}(g_{n})}|ds|\bigg{)}$
$\displaystyle=\frac{X^{int}\mu_{n,0}(t)}{\Gamma(g_{n})}\bigg{(}(\log
X)^{g_{n}-1}+O(X^{-r_{0}})\bigg{)}$
$\displaystyle+O_{\epsilon}\bigg{(}\frac{(\log
X)^{2\epsilon}}{|n|t}\bigg{(}\int_{-r_{0}}^{-\tfrac{1}{\log
X}}|\sigma|^{1-\text{Re}(g_{n})}X^{\sigma}d\sigma+(\log
X)^{\text{Re}(g_{n})-2}\bigg{)}\bigg{)}$
$\displaystyle=\frac{X^{int}}{int}\frac{(\log
X)^{\gamma_{n}-1}}{\Gamma(\gamma_{n})}\bigg{(}\prod_{\begin{subarray}{c}|k|\leqslant
2N\\\ k\neq
n\end{subarray}}\zeta(1-i(k-n)t)^{g_{k}}+O_{\epsilon}\bigg{(}\frac{1}{(\log
X)^{1-\epsilon}}\bigg{)}\bigg{)}$

and similarly

$\mathcal{M}_{0}=\frac{(\log
X)^{\gamma_{0}-1}}{\Gamma(\gamma_{0})}\bigg{(}\prod_{\begin{subarray}{c}|k|\leqslant
2N\\\ k\neq
0\end{subarray}}\zeta(1-ikt)^{g_{k}}+O_{\epsilon}\bigg{(}\frac{1}{(\log
X)^{1-\epsilon}}\bigg{)}\bigg{)}.$

We next focus on the products of $\zeta$-values. When $n=\ell\in\mathcal{L}$,
Lemma 4(d) gives

$\prod_{\begin{subarray}{c}|k|\leqslant 2N\\\
k\neq\ell\end{subarray}}\zeta(1-i(k-\ell)t)^{g_{k}}=(1+O(|t|))C_{\ell}(it)^{g_{\ell}-1}.$

We saw above that $\text{Re}(\gamma_{n})\leqslant\mu-\delta$ for all
$n\not\in\mathcal{L}$. Combining this with Lemma 4(d) and the estimates
$|t|^{\text{Re}(g_{\ell})-1}\geqslant 1$ (since $|g_{\ell}|\leqslant 1$ for
all $\ell$) and $1/\Gamma(g_{n})\ll 1$ uniformly (since $1/\Gamma$ is entire),
when $\ell\notin\mathcal{L}$ we obtain

$\displaystyle\sum_{\begin{subarray}{c}|n|\leqslant N\\\
n\neq\ell\end{subarray}}\prod_{\begin{subarray}{c}|k|\leqslant 2N\\\ k\neq
n\end{subarray}}|\zeta(1-i(k-n)t)^{g_{k}}|\frac{(\log
X)^{\text{Re}(g_{n})}}{|\Gamma(g_{n}+1_{n=0})|(1_{n=0}+|nt|\log X)}$
$\displaystyle\ll_{\epsilon}\frac{(\log X)^{\mu-1-\delta+\epsilon}}{|t|}$
$\displaystyle\ll\frac{(|t|\log X)^{\mu-1-\delta+\epsilon}}{|t|}.$

Accounting for the error term for $\mathcal{M}_{\ell}$ and using $\delta<1$,
it follows that

$\displaystyle\mathcal{M}$
$\displaystyle=\bigg{(}1+O(|t|)\bigg{)}\sum_{\ell\in\mathcal{L}}\frac{X^{i\ell
t}}{i\ell^{\prime}t}\frac{C_{\ell}(it\log
X)^{\gamma_{\ell}-1}}{\Gamma(\gamma_{\ell})}+O_{\epsilon}\bigg{(}\frac{(|t|\log
X)^{\mu-1+\epsilon}}{|t|}\cdot\frac{1}{(\log X)^{\delta}}\bigg{)}.$

Setting $\delta^{\prime}:=\tfrac{1}{2}\delta$ and taking
$\epsilon<\delta^{\prime}$ we obtain

$\displaystyle\mathcal{M}$
$\displaystyle=\bigg{(}1+O\bigg{(}|t|+\frac{1}{(\log
X)^{\delta^{\prime}}}\bigg{)}\bigg{)}\sum_{\ell\in\mathcal{L}}\frac{X^{i\ell
t}}{i\ell^{\prime}t}\frac{C_{\ell}(it\log
X)^{\gamma_{\ell}-1}}{\Gamma(\gamma_{\ell})}.$

The proof is completed upon combining this estimate with our bound for
$\mathcal{R}$ in (7.4), and then using (7.3). ∎

The values of $f_{t}(p)$ at primes $p$ are crucial in obtaining the shape of
the asymptotic formula in Theorem 3, not the values $f_{t}(p^{m})$ with
$m\geqslant 2$ at prime powers, as the following Corollary shows:

######  Corollary 4.

Assume the hypotheses of Theorem 3. Let $f:\mathbb{N}\rightarrow\mathbb{U}$ be
a multiplicative function such that $f(p)=f_{t}(p)$ for all primes $p$, and
define a multiplicative function $h$ so that $f:=f_{t}\ast h$. Then

$\sum_{n\leqslant X}\frac{f(n)}{n}=\sum_{\ell\in\mathcal{L}}\frac{X^{i\ell
t}}{i\ell^{\prime}t}\frac{C_{\ell}(it\log
X)^{\gamma_{\ell}-1}}{\Gamma(\gamma_{\ell})}(H(1+i\ell t)+O(|t|)+o(1))\text{
where }H(s):=\sum_{n\geqslant 1}\frac{h(n)}{n^{s}};$

moreover, for each $\ell\in\mathcal{L}$ we have $H(1+i\ell t)\neq 0$ unless
$f(2^{k})=-2^{ik\ell t}$ for all $k\geqslant 1$.

One expects that $\mathcal{L}$ typically contains just one element,
$\\{\ell\\}$, and so an asymptotic is given by this formula for all large $X$
if $H(1+i\ell t)\neq 0$ (that is, if $f(2^{k})\neq-2^{ik\ell t}$ for some
$k\geqslant 1$). If $\mathcal{L}$ contains more than one element first note
that $H(1+i\ell t)=0$ for at most one value of $\ell$, so we have a sum of
main terms of similar magnitude. For $X\in[Z,Z^{1+o(1)}]$ we get a formula of
the form $(\sum_{\ell\in\mathcal{L}}c_{\ell}X^{i\ell t}+o(1))(\log Z)^{\mu-1}$
(where the $c_{\ell}$ depend on $Z$ but not $X$) and such a finite length
trigonometric polynomial will have size $o(1)$ for a logarithmic measure $0$
set of $X$-values (that is for $X\in[Z,YZ]$ where $|t|\log Y\to\infty$ with
$\log Y=o(\log Z)$).

###### Proof.

Each $h(p)=0$ so that $h(n)=0$ unless $n$ is powerful. Since each
$|f_{t}(p^{k})|,|f(p^{k})|\leqslant 1$, we deduce by induction that
$|h(p^{k})|\leqslant 2^{k-1}$ for each $k\geqslant 2$. We begin by assuming
that each $h(2^{k})=h(3^{k})=0$, and so if $(n,6)=1$ then $|h(n)|\leqslant
n^{\kappa}$ where $\kappa=\frac{\log 2}{\log 5}$($<\frac{1}{2}$).

As $h(n)=0$ unless $n$ is powerful and $(n,6)=1$, we have

$\sum_{n>N}\frac{|h(n)|}{n}\ll N^{\kappa-\frac{1}{2}}\text{ and
}\sum_{b\geqslant 1}\frac{|h(b)|\log b}{b}<\infty.$

Select $A>0$ so that $A(\frac{1}{2}-\kappa)>2-\mu$. Then with $M:=(\log
X)^{A}$ we have $\sum_{b>M}\frac{|h(b)|}{b}=o((\log X)^{\mu-2})$, and by
Theorem 3,

$\displaystyle\sum_{n\leqslant X}\frac{f(n)}{n}$
$\displaystyle=\sum_{\begin{subarray}{c}ab\leqslant X\\\ b\leqslant
M\end{subarray}}\frac{f_{t}(a)h(b)}{ab}+\sum_{\begin{subarray}{c}ab\leqslant
X\\\ b>M\end{subarray}}\frac{f_{t}(a)h(b)}{ab}$
$\displaystyle=\sum_{b\leqslant M}\frac{h(b)}{b}\sum_{a\leqslant
X/b}\frac{f_{t}(a)}{a}+O\bigg{(}\sum_{a\leqslant
X}\frac{1}{a}\sum_{M<b\leqslant X/a}\frac{|h(b)|}{b}\bigg{)}$
$\displaystyle=(1+O(|t|))\sum_{\ell\in\mathcal{L}}\frac{X^{i\ell
t}}{i\ell^{\prime}t}\frac{C_{\ell}(it\log
X)^{\gamma_{\ell}-1}}{\Gamma(\gamma_{\ell})}\sum_{b\leqslant
M}\frac{h(b)}{b^{1+i\ell t}}\bigg{(}1-\frac{\log b}{\log
X}\bigg{)}^{\gamma_{\ell}-1}$ $\displaystyle+o((\log X)^{\mu-1}).$

The claimed formula follows since

$\sum_{b\leqslant M}\frac{h(b)}{b^{1+i\ell t}}\bigg{(}1-\frac{\log b}{\log
X}\bigg{)}^{\gamma_{\ell}-1}=H(1+i\ell
t)+O\bigg{(}\sum_{b>M}\frac{|h(b)|}{b}+\frac{1}{\log X}\sum_{b\leqslant
M}\frac{|h(b)|\log b}{b}\bigg{)}=H(1+i\ell t)+o(1).$

Now suppose that $h(3^{k})$ is not necessarily $0$. The key issue is

$\sum_{n>N}\frac{|h(n)|}{n}=\sum_{k\geqslant
0}\frac{|h(3^{k})|}{3^{k}}\sum_{\begin{subarray}{c}n>N/3^{k}\\\
(n,6)=1\end{subarray}}\frac{|h(n)|}{n}\ll\sum_{k\geqslant
0}\frac{|h(3^{k})|}{3^{k}}(N/3^{k})^{\kappa-\frac{1}{2}}\leqslant
N^{\kappa-\frac{1}{2}}\sum_{k\geqslant
0}\frac{2^{k-1}}{(3^{\kappa+\frac{1}{2}})^{k}}\ll N^{\kappa-\frac{1}{2}}$

since $3^{\kappa+\frac{1}{2}}>2$.

Our assumptions guarantee that the sum for $H(s)$ converges on the $1$-line.
Is $H(1+i\tau)\neq 0$ for $\tau\in\mathbb{R}$? We see that the Euler factors
converge on the $1$-line and indeed

$\bigg{|}\sum_{k\geqslant 0}\frac{h(p^{k})}{p^{k(1+i\tau)}}\bigg{|}\geqslant
1-\sum_{k\geqslant 1}\frac{|h(p^{k})|}{p^{k}}\geqslant 1-\sum_{k\geqslant
2}\frac{2^{k-1}}{p^{k}}=1-\frac{2}{p(p-2)}>0$

for each prime $p\geqslant 3$.

Now suppose that $h(2^{k})$ is not necessarily $0$. The analogous argument
works for any $h(.)$ for which there exists $\epsilon>0$ such that
$|h(2^{k})|\ll(2^{k})^{1-\epsilon}$. To establish this we first assume that
each $|f_{t}(2^{k})|=1$ so write $f_{t}(2^{k})=e(\theta_{k})$ and
$g(\frac{t\log 2^{j}}{2\pi})=r_{j}e(\gamma_{j})$ with $0\leqslant
r_{j}\leqslant 1$. Then (7.2) becomes $m\,e(\theta_{m})=\sum_{1\leqslant
j\leqslant m}r_{j}e(\theta_{m-j}+\gamma_{j})$. This implies that $r_{j}=1$ and
$\theta_{m}=\theta_{m-j}+\gamma_{j}\ \pmod{1}$ for $1\leqslant j\leqslant m$.
Now $\theta_{0}=\gamma_{0}=0$ and so $\theta_{m}=\gamma_{m}=m\gamma_{1}\
\pmod{1}$. But then $\sum_{k\geqslant 0}f_{t}(2^{k})/2^{ks}=\sum_{k\geqslant
0}(e(\gamma_{1})/2^{s})^{k}=(1-e(\gamma_{1})/2^{s})^{-1}$ and so if
$k\geqslant 1$ then $h(2^{k})=f(2^{k})-e(\gamma_{1})f(2^{k-1})$ and so each
$|h(2^{k})|\leqslant 2$.

Otherwise there exists a minimal $k\geqslant 1$ such that $|f_{t}(2^{k})|<1$;
let $\delta=1-|f_{t}(2^{k})|\in(0,1]$. Now select $\alpha>0$ for which
$\delta(\alpha-1)=\alpha^{k}(2-\alpha)$ so that $1<\alpha<2$. We claim that
$|h(2^{m})|\leqslant\kappa\alpha^{m}$ for all $m\geqslant 0$, where
$\kappa:=\max_{0\leqslant m\leqslant k}|h(2^{m})|/\alpha^{m}$. This is
trivially true for $m\leqslant k$; otherwise for $m>k$ we have (as
$h(1)=1,h(2)=0$)

$\displaystyle|h(2^{m})|$
$\displaystyle=\bigg{|}f(2^{m})-\sum_{j=0}^{m-1}f_{t}(2^{m-j})h(2^{j})\bigg{|}\leqslant
2+\sum_{\begin{subarray}{c}j=2\\\ j\neq
m-k\end{subarray}}^{m-1}|h(2^{j})|+(1-\delta)|h(2^{m-k})|$
$\displaystyle\leqslant\sum_{\begin{subarray}{c}j=0\\\ j\neq
m-k\end{subarray}}^{m-1}\kappa\alpha^{j}+(1-\delta)\kappa\alpha^{m-k}<\kappa\alpha^{m}\bigg{(}\sum_{i\geqslant
1}\alpha^{-i}-\delta\alpha^{-k}\bigg{)}=\kappa\alpha^{m}$

as $2\leqslant\kappa+\kappa\alpha$, by induction, using the definition of
$\alpha$.

Finally we wish to determine whether the Euler factor of $H(1+i\ell t)$ at $2$
equals $0$. This equals the Euler factor for $f$ at $1$ divided by the Euler
factor for $f_{t}$ at $1$. Since each $|f_{t}(2^{k})|\leqslant 1$ the
denominator is bounded; since $|f(2^{k})|\leqslant 1$ we have
$\sum_{k\geqslant 0}\frac{f(2^{k})}{2^{k(1+i\ell t)}}=0$ if and only if
$f(2^{k})=-2^{ik\ell t}$ for all $k\geqslant 1$.

∎

###  7.2. Our specific example

We now use Theorem 3 to construct a multiplicative function $f$ satisfying the
conclusion of Proposition 2(b). We will use the auxiliary 1-periodic function

$g(u)=\frac{e(u)-\lambda}{|e(u)-\lambda|}=\frac{|e(u)-\lambda|}{e(-u)-\lambda}=\sum_{n\in\mathbb{Z}}g_{n}e(nu).$

We see that $g$ takes values on $S^{1}$ with $g(0)=1$. We will verify the
following properties of $\\{g_{n}\\}_{n}$ in the appendix (section A.1), which
shows that $g$ satisfies the assumptions required to apply Theorem 3.

######  Lemma 5.

For the $\\{g_{n}\\}_{n}$ defined just above we have:

  1. (a)

$g_{n}\in\mathbb{R}$ for all $n$,

  2. (b)

$|g_{n}|\ll\ (\tfrac{2\lambda}{1+\lambda^{2}})^{|n|}\leqslant 0.99^{|n|}$ for
all $n\in\mathbb{Z}$,

  3. (c)

$g_{-n}<g_{n}$ for all $n\geqslant 1$,

  4. (d)

$g_{1}=0.7994\dots$, and there is $\delta>0$ such that $g_{n}\leq
g_{1}-\delta$ or all $n\neq 0,1$, and

  5. (e)

$g_{0}<g_{1}-1$.

###### Deduction of Proposition 2(b).

Let $x$ be large. Let $t\in[\tfrac{1}{\log\log x},1]$ be small, and set
$y_{t}:=e^{\tfrac{1}{t}}$ and $f=f_{t}$. For small enough $t$, Theorem 3
yields

$\bigg{|}\sum_{n\leqslant x}\frac{f(n)}{n}\bigg{|}\asymp t^{-1}(t\log
x)^{g_{1}-1}\asymp\log x\exp\bigg{(}(g_{1}-2)\sum_{y_{t}<p\leqslant
x}\frac{1}{p}\bigg{)}.$ (7.7)

Using the definition of $\lambda$,

$\displaystyle
2-\lambda=\int_{0}^{1}\frac{|e(u)-\lambda|}{e(-u)-\lambda}(e(-u)-\lambda)du=\int_{0}^{1}g(u)e(-u)du-\lambda\int_{0}^{1}g(u)du=g_{1}-\lambda
g_{0},$

so that $g_{1}-2=\lambda(g_{0}-1)$. By partial summation and the prime number
theorem we have

$\displaystyle\mathbb{D}(f,1;y_{t},x)^{2}=\sum_{y_{t}<p\leqslant
x}\frac{1-\text{Re}(g(\tfrac{t}{2\pi}\log p))}{p}$
$\displaystyle=\text{Re}\bigg{(}\int_{0}^{1}\bigg{(}1-g(u)\bigg{)}du\bigg{)}\log\bigg{(}\frac{\log
x}{\log y_{t}}\bigg{)}+O(1)$ $\displaystyle=(1-g_{0})\sum_{y_{t}<p\leqslant
x}\frac{1}{p}+O(1).$

Combining these last few observations we deduce that

$\bigg{|}\sum_{n\leqslant x}\frac{f(n)}{n}\bigg{|}\asymp\log
x\exp\bigg{(}\lambda(g_{0}-1)\sum_{y_{t}<p\leqslant
x}\frac{1}{p}\bigg{)}\asymp\log x\exp\bigg{(}-\lambda\
\mathbb{D}(f,1;y_{t},x)^{2}\bigg{)}.$

Finally, since $g$ is Lipschitz and $g(0)=1$,

$|g(\tfrac{t}{2\pi}\log p)-1|\ll t\log p\text{ for all }p\leqslant y_{t},$

and so by Mertens’ theorem,

$\mathbb{D}(f,1;y_{t})^{2}=\sum_{p\leqslant
y_{t}}\frac{1-\text{Re}(g(\tfrac{t}{2\pi}\log p))}{p}\ll t\sum_{p\leqslant
y_{t}}\frac{\log p}{p}\ll 1.$

Combining these last two estimates, (1.3) follows. ∎

##  Appendix A Auxiliary results towards Proposition 2(b)

We establish the technical lemmas 5 and 4, used in the proofs of Proposition
2(b) and Theorem 3.

###  A.1. On the Fourier Coefficients of $g$

###### Proof of Lemma 5.

(a) Since $g(u)=\bar{g}(-u)$, a term-by-term comparison of the Fourier series
of each shows that $g_{n}=\bar{g}_{n}$ and thus $g_{n}\in\mathbb{R}$ for each
$n\in\mathbb{Z}$.

(b) We will prove a numerically sharper bound, which will be used in part (d).
Note that

$g(u)=\frac{e(u)-\lambda}{\sqrt{1+\lambda^{2}}}\cdot\bigg{(}1-\frac{\lambda}{1+\lambda^{2}}(e(u)+e(-u))\bigg{)}^{-1/2}.$

As $2\lambda<0.99(1+\lambda^{2})$, Taylor expanding the bracketed expression
gives

$\displaystyle g(u)$
$\displaystyle=\frac{e(u)-\lambda}{\sqrt{1+\lambda^{2}}}\sum_{j\geqslant
0}\binom{-1/2}{j}(-1)^{j}\bigg{(}\frac{\lambda}{1+\lambda^{2}}\bigg{)}^{j}\sum_{0\leqslant
i\leqslant j}\binom{j}{i}e((2i-j)u)$
$\displaystyle=\frac{e(u)-\lambda}{\sqrt{1+\lambda^{2}}}\sum_{n\in\mathbb{Z}}e(nu)\sum_{\begin{subarray}{c}j\geqslant|n|\\\
2|(n+j)\end{subarray}}\binom{j}{(j+n)/2}2^{-2j}\binom{2j}{j}\bigg{(}\frac{\lambda}{1+\lambda^{2}}\bigg{)}^{j}$
$\displaystyle=\sum_{m\in\mathbb{Z}}e(mu)\frac{h_{m-1}-\lambda
h_{m}}{\sqrt{1+\lambda^{2}}},$

where we have set

$\displaystyle h_{n}:=\sum_{\begin{subarray}{c}j\geqslant|n|\\\
2|(j+n)\end{subarray}}2^{-2j}\binom{2j}{j}\binom{j}{(j+|n|)/2}\bigg{(}\frac{\lambda}{1+\lambda^{2}}\bigg{)}^{j}=\sum_{l\geqslant
0}\binom{|n|+2l}{|n|+l}\binom{2(|n|+2l)}{|n|+2l}\bigg{(}\frac{\lambda}{4(1+\lambda^{2})}\bigg{)}^{|n|+2l}.$
(A.1)

It follows that

$g_{n}=\frac{h_{n-1}-\lambda h_{n}}{\sqrt{1+\lambda^{2}}}\text{ for all
}n\in\mathbb{Z}.$

We use Stirling’s approximation in the form $\sqrt{2\pi n}(n/e)^{n}\leqslant
n!\leqslant e^{\tfrac{1}{12}}\sqrt{2\pi n}(n/e)^{n}$ for $n\in\mathbb{N}$ (see
[22]), and bound $\binom{j}{(j+|n|)/2}$ by a central binomial coefficient as
$\binom{j}{(j+|n|)/2}\leqslant\frac{1}{2^{\nu}}\binom{j+\nu}{(j+\nu)/2}$,
where $\nu=1_{2\nmid j}$. When $n\neq 0$ this gives

$2^{-2j}\binom{2j}{j}\binom{j}{(j+|n|)/2}\leqslant
2^{-2j}\bigg{(}\tfrac{e^{\tfrac{1}{12}}}{\sqrt{\pi
j}}2^{2j}\bigg{)}\bigg{(}\tfrac{e^{\tfrac{1}{12}}}{2^{\nu}\sqrt{\pi(j+\nu)/2}}2^{j+\nu}\bigg{)}\leqslant\frac{e^{\tfrac{1}{6}}\sqrt{2}}{\pi
j}2^{j},\text{ if }2|(j+n).$

Setting $c:=\tfrac{2\lambda}{1+\lambda^{2}}$, we find using the first
expression for $h_{n}$ in (A.1) that

$0\leqslant h_{m}\leqslant\sum_{\begin{subarray}{c}j\geqslant|m|\\\
2|(j+m)\end{subarray}}\frac{e^{\tfrac{1}{6}}\sqrt{2}}{\pi
j}\,2^{j}\bigg{(}\frac{\lambda}{1+\lambda^{2}}\bigg{)}^{j}\leqslant\frac{e^{\tfrac{1}{6}}\sqrt{2}}{(1-c^{2})\pi}\,\frac{c^{|m|}}{|m|},\quad
m\neq 0.$

Using the second expression for $h_{n}$ in (A.1), when $|n|\geqslant 1$ we get

$\displaystyle h_{n-1}-\lambda h_{n}$ $\displaystyle=\sum_{l\geqslant
0}\binom{|n|-1+2l}{|n|-1+l}\binom{2(|n|-1+2l)}{|n|-1+2l}\bigg{(}\frac{\lambda}{4(1+\lambda^{2})}\bigg{)}^{|n|-1+2l}\bigg{(}1-\frac{2|n|-1+4l}{2(|n|+l)}\frac{\lambda^{2}}{1+\lambda^{2}}\bigg{)}$
$\displaystyle\leqslant\bigg{(}1-\frac{2|n|-1}{2|n|}\frac{\lambda^{2}}{1+\lambda^{2}}\bigg{)}h_{n-1}.$

Together with the previous bound, it follows that when $|n|\geqslant 2$,

$\displaystyle|g_{n}|\leqslant\frac{e^{\tfrac{1}{6}}\sqrt{2}}{(1-c^{2})\pi\sqrt{1+\lambda^{2}}}\bigg{(}1-\frac{2|n|-1}{2|n|}\frac{\lambda^{2}}{1+\lambda^{2}}\bigg{)}\frac{c^{|n|-1}}{|n|-1},$
(A.2)

and the bound $|g_{n}|\ll c^{|n|}$ immediately follows.

(c) We deduce from (A.1) that $h_{n}=h_{-n}$ for any $n\geqslant 1$. Thus,

$\sqrt{1+\lambda^{2}}g_{n}=\begin{cases}h_{n-1}-\lambda h_{n}&\text{ if
}n\geqslant 1,\\\ h_{|n|+1}-\lambda h_{|n|}&\text{ if }n\leqslant
0.\end{cases}$ (A.3)

To prove $g_{-n}<g_{n}$ for all $n\geqslant 1$ we need only show that
$h_{n+1}<h_{n-1}$ for all $n\geqslant 1$.
To see this, note that

$\displaystyle h_{n+1}$
$\displaystyle=\bigg{(}\frac{\lambda}{2(1+\lambda^{2})}\bigg{)}^{2}\sum_{l\geqslant
0}\binom{n-1+2l}{n-1+l}\binom{2(n-1+2l)}{n-1+2l}\bigg{(}\frac{\lambda}{4(1+\lambda^{2})}\bigg{)}^{n-1+2l}\frac{(2n+4l)^{2}-1}{(n+l+1)(n+l)}$
$\displaystyle\leqslant
16\bigg{(}\frac{\lambda}{2(1+\lambda^{2})}\bigg{)}^{2}\sum_{l\geqslant
0}\binom{n-1+2l}{n-1+l}\binom{2(n-1+2l)}{n-1+2l}\bigg{(}\frac{\lambda}{4(1+\lambda^{2})}\bigg{)}^{n-1+2l},$

and thus
$h_{n+1}\leqslant\bigg{(}\frac{2\lambda}{1+\lambda^{2}}\bigg{)}^{2}h_{n-1}<h_{n-1}$,
as required.

(d) Since $g_{-1}<g_{1}$, and $g_{-n}<g_{n}$ for $n\geqslant 2$ by (c), it is
enough to show that there is $\delta>0$ such that $g_{n}\leqslant
g_{1}-\delta$ for all $n\geqslant 2$.

The upper bound (A.2) is decreasing with $n$, and one may verify that it gives
$\leqslant 0.7<g_{1}-\tfrac{1}{20}$ for $n=10$. Thus, we clearly have
$g_{n}\leqslant g_{1}-\tfrac{1}{20}$ for all $n\geqslant 10$.

On the other hand, a computer-assisted calculation shows that $g_{n}\leqslant
g_{1}-\tfrac{1}{4}$, say, for all $2\leqslant n\leqslant 9$:

$\displaystyle g_{1}=0.7994...,\quad g_{2}=0.2848...,\quad
g_{3}=0.1659...\quad g_{4}=0.1102...,\quad g_{5}=0.0778...,\quad$
$\displaystyle g_{6}=0.0568...,\quad g_{7}=0.0423...,\quad
g_{8}=0.0321...,\quad g_{9}=0.0246....$

The claim thus follows with $\delta=\tfrac{1}{20}$.

(e) Though this may be verified by computer, we give a proof. From (A.3),

$g_{1}-1-g_{0}=\frac{1}{\sqrt{1+\lambda^{2}}}(h_{0}-\lambda
h_{1}-(h_{1}-\lambda
h_{0}))-1=\frac{1+\lambda}{\sqrt{1+\lambda^{2}}}(h_{0}-h_{1})-1.$

It is enough to show that
$h_{0}-h_{1}>\frac{\sqrt{1+\lambda^{2}}}{1+\lambda}$. To see this, we write

$h_{0}-h_{1}=\sum_{l\geqslant
0}2^{-2l}\binom{2l}{l}\binom{4l}{2l}\bigg{(}\frac{\lambda}{1+\lambda^{2}}\bigg{)}^{l}\bigg{(}1-\frac{\lambda}{2(1+\lambda^{2})}\frac{4l+1}{l+1}\bigg{)}\geqslant
1-\frac{\lambda}{2(1+\lambda^{2})},$

using
$\tfrac{\lambda}{2(1+\lambda^{2})}\frac{4l+1}{l+1}\leqslant\tfrac{2\lambda}{1+\lambda^{2}}<1$
and positivity to restrict to the term $l=0$. But we see that

$1-\frac{\lambda}{2(1+\lambda^{2})}>\bigg{(}1-\frac{2\lambda}{(1+\lambda)^{2}}\bigg{)}^{1/2}=\frac{\sqrt{1+\lambda^{2}}}{1+\lambda},$

since, setting $t=\frac{2\lambda}{(1+\lambda)^{2}}$, we have

$1-\frac{\lambda}{2(1+\lambda^{2})}=1-\frac{t}{4}\cdot\bigg{(}1+\frac{2\lambda}{1+\lambda^{2}}\bigg{)}\geqslant
1-\frac{t}{2}>(1-t)^{1/2},$

as required. ∎

###  A.2. On products of shifted zeta functions

###### Proof of Lemma 4.

Throughout, set $G:=\sum_{n\in\mathbb{Z}}|g_{n}|<\infty$ (since we assumed
that $|g_{n}|\ll 1/(1+|n|)^{3}$) and $F_{N}(s):=\prod_{|n|\leqslant
2N}\zeta(s-int)^{g_{n}}$.

(a) Let $\sigma\geqslant\tfrac{1}{\log X}$. Since $|f(n)|\leqslant 1$,

$|F(1+\sigma+i\tau)|\leqslant\zeta(1+\sigma)\ll\log X\text{ for all
}|\tau|\leqslant T.$

When $|n|>2N$, $|\tau-nt|\geqslant(2N+1)|t|-T\geqslant T$, so that also

$\zeta(1+\sigma+i(\tau-nt))\ll\log(2+|\tau-nt|)\ll\log(2|n|)\text{ for all
}|\tau|\leqslant T.$

Now $|\zeta(2+2\sigma)/\zeta(1+\sigma)|<|\zeta(1+\sigma+i(\tau-
nt))|<|\zeta(1+\sigma)|$ and so if $|g_{n}|<\frac{1}{\log\log X}$ then

$\displaystyle\zeta(1+\sigma+i(\tau-nt))^{-g_{n}}=1+O(g_{n}\log\log X).$ (A.4)

This holds for $|n|>2N$ as we assumed that $|g_{n}|\ll|n|^{-3}$. Since
$N\geqslant(\log X)^{A}$ for some $A\geqslant 2$ it follows that

$\displaystyle\max_{|\tau|\leqslant
T}|F_{N}(1+\sigma+i\tau)-F(1+\sigma+i\tau)|$
$\displaystyle\ll\max_{|\tau|\leqslant
T}\bigg{(}|F(1+\sigma+i\tau)|\cdot\bigg{|}\exp\bigg{(}\sum_{|n|>2N}\log\bigg{|}\zeta(1+\sigma+i(\tau-
nt))^{-g_{n}}\bigg{|}\bigg{)}-1\bigg{|}\bigg{)}$ $\displaystyle\ll(\log
X)\sum_{n>2N}|n|^{-3}\log\log(2n)\ll_{\epsilon}(\log X)N^{-2+\epsilon}\ll(\log
X)^{-2},$ (A.5)

as required.

(b) Let $|\tau|\leqslant T$ and $|n|\leqslant 2N$, and put $\sigma:=1-r_{0}$.
If $|\sigma-1+i(\tau-nt)|\leqslant 1$ then

$|\zeta(\sigma+i(\tau-nt))|^{\pm 1}\ll|\sigma-1+i(\tau-nt)|^{-1}\leqslant
r_{0}^{-1}.$

Otherwise, $|\zeta(\sigma+i(\tau-nt))|^{\pm 1}\ll\log(2+|\tau-nt|)\ll\log T$.
Thus, for any $|\tau|\leqslant T$,

$\displaystyle|F_{N}(\sigma+i\tau)|\ll\prod_{\begin{subarray}{c}|n|\leqslant
2N\\\ |-r_{0}+i(\tau-nt)|\leqslant
1\end{subarray}}r_{0}^{-|g_{n}|}\cdot\prod_{\begin{subarray}{c}|n|\leqslant
2N\\\ |-r_{0}+i(\tau-nt)|>1\end{subarray}}(\log T)^{|g_{n}|}\ll(r_{0}^{-1}\log
T)^{G}.$

Since $\min\\{\sigma(3T),|t|\\}^{-1}\log T\ll_{\epsilon}(\log X)^{\epsilon}$,
the claim follows.

(c) Assume $\eta=+1$; the claim with $\eta=-1$ is completely analogous. Note
that $|T-kt|\geqslant|t|/2$ for all $|k|\leqslant 2N$. If
$|\sigma+i(T-kt)|\leqslant 1$ then $|kt|\geqslant T/2\geqslant N|t|/2$, i.e.,
$|k|\geqslant N/2$, and then

$|\zeta(1+\sigma+i(T-kt))^{g_{k}}|\ll|T-kt|^{-|g_{k}|}\ll|t|^{-|g_{k}|}.$

Splitting the product as in (b) and using $|g_{n}|\ll 1/(1+|n|)^{3}$ for each
$|\sigma|\leqslant r_{0}$ we get

$\displaystyle|F_{N}(1+\sigma+iT)|$
$\displaystyle\ll\prod_{|n|\geqslant\tfrac{N}{2}}|t|^{-|g_{n}|}\cdot\prod_{\begin{subarray}{c}|n|\leqslant
2N\\\ |\sigma+i(T-nt)|>1\end{subarray}}(\log T)^{|g_{n}|}$
$\displaystyle\ll\exp\bigg{(}\log(\tfrac{1}{|t|})\sum_{|n|\geqslant\tfrac{N}{2}}|n|^{-3}\bigg{)}(\log
T)^{G}\ll(\log\log x)^{G},$

and the claim follows.

(d) Observe that if $|m|\leqslant|t|^{-1}$ then $imt\zeta(1+imt)=1+O(|mt|)$;
otherwise, if $|mt|\geqslant 1$ then when $|t|$ is small enough $m$ is large
and we may Taylor expand

$[imt\zeta(1+imt)]^{g_{\ell-m}}=1+O\bigg{(}|g_{\ell-m}|\log(2+|mt|)\bigg{)},$

similarly to (A.4). Thus, since $|g_{\ell-m}|\ll(1+|m-\ell|)^{-3}$ we have

$\displaystyle\bigg{|}\prod_{m\neq 0}[imt\zeta(1+imt)]^{g_{\ell-m}}-1\bigg{|}$
$\displaystyle\ll|t|\sum_{|m|\leqslant|t|^{-1}}|g_{\ell-m}||m|+\sum_{|m|>|t|^{-1}}|g_{\ell-m}|\log(2+|mt|)$
$\displaystyle\ll|t|\sum_{|m|\leqslant|t|^{-1}}(1+|m-\ell|)^{-2}+\sum_{|m|>|t|^{-1}}\frac{\log(2+|m|)}{(1+|m-\ell|)^{3}}\ll|t|.$

Since also $|t|\geqslant\tfrac{1}{\log X},$ it follows that (handling the
range $|k|>2N$ as in (a))

$\displaystyle\prod_{\begin{subarray}{c}|k|\leqslant 2N\\\
k\neq\ell\end{subarray}}\zeta(1-i(k-\ell)t)^{g_{k}}$
$\displaystyle=\bigg{(}1+O\bigg{(}\frac{1}{(\log
X)^{2}}\bigg{)}\bigg{)}\prod_{m\neq 0}(imt)^{-g_{\ell-m}}\prod_{n\neq
0}[int\zeta(1+int)]^{g_{\ell-n}}$
$\displaystyle=\bigg{(}1+O(|t|)\bigg{)}C(it)^{g_{\ell}-1}$

since $\sum_{m\in\mathbb{Z}}g_{m}=g(0)=1$, and the first claim is proved.
Suppose more generally that $|n|\leqslant N$, and that
$|s|\leqslant\tfrac{1}{2}\min\\{|t|,\sigma(3T)\\}=2r_{0}$. Whenever $k\neq n$
we have $|s-i(k-n)t|\geqslant|t|/2$. Thus, arguing as in (c),

$\prod_{\begin{subarray}{c}|k|\leqslant 2N\\\ k\neq
n\end{subarray}}|\zeta(s+1-i(k-n)t)^{g_{k}}|\ll\prod_{\begin{subarray}{c}|k|\leqslant
2N\\\ k\neq n\\\ |s-i(k-n)t|\leqslant
1\end{subarray}}|t|^{-|g_{k}|}\prod_{\begin{subarray}{c}|k|\leqslant 2N\\\
k\neq n\\\ |s-i(k-n)t|>1\end{subarray}}(\log T)^{|g_{k}|}\ll(|t|^{-1}\log
T)^{G},$

and since $|t|\gg(\log X)^{-\epsilon}$ for any $\epsilon>0$ the claim follows.
∎

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