# Lower bounds for shifted moments of the Riemann zeta function

Michael J. Curran<EMAIL_ADDRESS>Mathematical Institute,
University of Oxford, Oxford, OX2 6GG, United Kingdom.

###### Abstract.

In previous work [8], the author gave upper bounds for the shifted moments of
the zeta function

$M_{\bm{\alpha},\bm{\beta}}(T)=\int_{T}^{2T}\prod_{k=1}^{m}|\zeta(\tfrac{1}{2}+i(t+\alpha_{k}))|^{2\beta_{k}}dt$

introduced by Chandee [5], where
$\bm{\alpha}=\bm{\alpha}(T)=(\alpha_{1},\ldots,\alpha_{m})$ and
$\bm{\beta}=(\beta_{1}\ldots,\beta_{m})$ satisfy $|\alpha_{k}|\leq T/2$ and
$\beta_{k}\geq 0$. Assuming the Riemann hypothesis, we shall prove the
corresponding lower bounds:

$M_{\bm{\alpha},\bm{\beta}}(T)\gg_{\bm{\beta}}T(\log
T)^{\beta_{1}^{2}+\cdots+\beta_{m}^{2}}\prod_{1\leq j<k\leq
m}|\zeta(1+i(\alpha_{j}-\alpha_{k})+1/\log T)|^{2\beta_{j}\beta_{k}}.$

##  1\. Introduction

This paper is concerned with the shifted moments

$M_{\bm{\alpha},\bm{\beta}}(T)=\int_{T}^{2T}\prod_{k=1}^{m}|\zeta(\tfrac{1}{2}+i(t+\alpha_{k}))|^{2\beta_{k}}dt,$
(1)

where $\bm{\alpha}=\bm{\alpha}(T)=(\alpha_{1},\ldots,\alpha_{m})$ and
$\bm{\beta}=(\beta_{1}\ldots,\beta_{m})$ satisfy $|\alpha_{k}|\leq T/2$ and
$\beta_{k}\geq 0$. These were first studied in general by Chandee [5]. In [8],
the author proved the upper bound

$M_{\bm{\alpha},\bm{\beta}}(T)\ll_{\bm{\beta}}T(\log
T)^{\beta_{1}^{2}+\cdots+\beta_{m}^{2}}\prod_{1\leq j<k\leq
m}|\zeta(1+i(\alpha_{j}-\alpha_{k})+1/\log T)|^{2\beta_{j}\beta_{k}}$

assuming the Riemann hypothesis. The goal of this paper is to obtain the
corresponding lower bound, showing that these bounds are of the correct order
or magnitude.

######  Theorem 1.1.

Assume the Riemann hypothesis. If $\beta_{k}\geq 0$ and $|\alpha_{k}|\leq T/2$
for $k=1,\ldots,m,$ then

$M_{\bm{\alpha},\bm{\beta}}(T)\gg_{\bm{\beta}}T(\log
T)^{\beta_{1}^{2}+\cdots+\beta_{m}^{2}}\prod_{1\leq j<k\leq
m}|\zeta(1+i(\alpha_{j}-\alpha_{k})+1/\log T)|^{2\beta_{j}\beta_{k}}$

for $T$ sufficiently large in terms of $\bm{\beta}$.

Note that the order of magnitude in this bound is consistent with the formula
for the shifted moments predicted by the famous work of Conrey Farmer Keating
Rubenstein and Snaith [7]. The reader may consult [8] for more background on
the shifted moments and a discussion of the previous estimates for
$M_{\bm{\alpha},\bm{\beta}}(T)$.

The method of proof is similar to a principle pioneered in the works of Heap
and Soundararajan [14] and Radziwiłł and Soundararajan [25]. These works
demonstrate that if one can asymptotically evaluate the twisted
$2k^{\text{th}}$ moment of a family of $L$-functions for some integer $k$,
then one can obtain sharp lower bounds for the $2\beta^{\text{th}}$ moments
for all $\beta\geq 0$. We will need a more elaborate argument in our setting
since there are more parameters for the shifted moments. To explain the
method, we will first introduce some notation. Set

$\beta_{\ast}:=\sum_{k\leq m}\max(1,\beta_{k}).$

We will choose a sequence of parameters $T_{j}=T^{c_{j}}$, where

$c_{0}=0\text{ and }c_{j}=\frac{e^{j}}{(\log_{2}T)^{2}}$

for $j>0$, where throughout this paper we will use $\log_{j}$ to denote the
$j$-fold iterated logarithm. Let $L$ be the largest integer such that
$T_{L}\leq T^{\delta}$ where $0<\delta<e^{-1000\beta_{\ast}}$ is some small
constant depending on $\bm{\beta}$ to be chosen later. For any $X$ let

$\mathcal{P}_{1,X}(s)=\sum_{p\leq T_{1}}\frac{1}{p^{s+1/\log X}}\frac{\log
X/p}{\log X}+\sum_{p\leq\log T}\frac{1}{2p^{2s}},$

and given any $2\leq j\leq L$ define

$\mathcal{P}_{j,X}(s)=\sum_{p\in(T_{j-1},T_{j}]}\frac{1}{p^{s+1/\log
X}}\frac{\log X/p}{\log X}.$

######  Remark.

In our proof, we could have also used the simpler Dirichlet polynomial

$\sum_{p\in(T_{j-1},T_{j}]}\frac{1}{p^{s}}$

in place of $\mathcal{P}_{j,X}(s)$ for all $1\leq j\leq L$. However the
$\mathcal{P}_{j,X}$ will appear when using the Riemann hypothesis, and using
the $\mathcal{P}_{j,X}$ from the onset will reduce the total number of mean
value calculations.

If $\mathcal{P}_{j,X}(s)$ is not too large, then (see lemma 2.2) we will be
able to efficiently approximate $\exp(\beta\mathcal{P}_{j,X}(s))$ with the
following Dirichlet polynomial

$\mathcal{N}_{j,X}(s;\beta):=\sum_{m\leq
100\beta_{\ast}^{2}K_{j}}\frac{\beta^{m}\mathcal{P}_{j,X}(s)^{m}}{m!},$

where $K_{j}=c_{j}^{-3/4}$ for $j\geq 1$. The set

$\mathcal{G}:=\left\\{t\in[T/2,5T/2]:|P_{j,T_{L}}(\tfrac{1}{2}+it)|\leq
K_{j}\text{ for all }1\leq j\leq L\right\\}.$

consists of the $t\in[T/2,5T/2]$ for which $\mathcal{N}_{j,T_{L}}(s;\beta)$ is
a good approximation to $\exp(\beta\mathcal{P}_{j,T_{L}}(s))$. We will be
computing moments over the set

$\mathcal{G}_{m}:=\left\\{t\in[T,2T]:t+\alpha_{k}\in\mathcal{G}\text{ for all
}1\leq k\leq m\right\\}.$

Fix a smooth function $w$ such that $1_{[5/4,7/4]}(t)\leq w(t)\leq 1_{[1,2]}$,
and suppose for a moment that we could compute the integral

$\displaystyle\mathcal{I}_{0}$
$\displaystyle=\int_{\mathcal{G}_{m}}\prod_{k\leq
m}\zeta(\tfrac{1}{2}+i(t+\alpha_{k}))$ $\displaystyle\times\prod_{j\leq
L}\exp\left((\beta_{k}-1)\mathcal{P}_{j,T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k}))+\beta_{k}\mathcal{P}_{j,T_{L}}(\tfrac{1}{2}-i(t+\alpha_{k}))\right)w(t/T)\
dt.$

Then Hölder’s inequality implies

$|\mathcal{I}_{0}|\leq
M_{\bm{\alpha},\bm{\beta}}(T)^{1/p}\times|\mathcal{J}|^{1/q}\times\prod_{k\leq
m}|\mathcal{I}_{k}|^{1/r_{k}},$

where

$\frac{1}{p}=\frac{1}{4\beta_{\ast}},\quad\frac{1}{r_{k}}=\frac{1}{2m}-\frac{\beta_{k}}{4m\beta_{\ast}},\quad\frac{1}{q}=1-\frac{1}{p}-\sum_{k\leq
m}\frac{1}{r_{k}}$

are conjugate exponents,

$\displaystyle\mathcal{I}_{k}=\int_{\mathcal{G}_{m}}|\zeta(\tfrac{1}{2}+i(t+\alpha_{k}))|^{2m}\prod_{j\leq
L}$ $\displaystyle\exp\left(2(\beta_{k}-m)\textup{Re
}\mathcal{P}_{j,T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k}))\right)$
$\displaystyle\times\prod_{\begin{subarray}{c}\ell\leq m\\\ \ell\neq
k\end{subarray}}\prod_{j\leq L}\exp\left(2\beta_{\ell}\textup{Re
}\mathcal{P}_{j,T_{L}}(\tfrac{1}{2}+i(t+\alpha_{\ell}))\right)w(t/T)dt,$

and

$\displaystyle\mathcal{J}=\int_{\mathcal{G}_{m}}\prod_{k\leq m}\prod_{j\leq
L}\exp\left(2\beta_{k}\textup{Re
}\mathcal{P}_{j,T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k}))\right)w(t/T)dt.$

Therefore, Theorem 1.1 will follow from the following three propositions.

######  Proposition 1.2.

For large $T$

$\mathcal{J}\ll_{\bm{\beta}}T(\log
T)^{\beta_{1}^{2}+\cdots+\beta_{m}^{2}}\prod_{1\leq j<k\leq
m}|\zeta(1+i(\alpha_{j}-\alpha_{k})+1/\log T)|^{2\beta_{j}\beta_{k}}.$

######  Proposition 1.3.

Assuming the Riemann hypothesis, for large $T$

$|\mathcal{I}_{0}|\gg_{\bm{\beta}}T(\log
T)^{\beta_{1}^{2}+\cdots+\beta_{m}^{2}}\prod_{1\leq j<k\leq
m}|\zeta(1+i(\alpha_{j}-\alpha_{k})+1/\log T)|^{2\beta_{j}\beta_{k}}.$

######  Proposition 1.4.

Assuming the Riemann hypothesis, for large $T$ and $1\leq k\leq m$

$\mathcal{I}_{k}\ll_{\bm{\beta}}T(\log
T)^{\beta_{1}^{2}+\cdots+\beta_{m}^{2}}\prod_{1\leq j<k\leq
m}|\zeta(1+i(\alpha_{j}-\alpha_{k})+1/\log T)|^{2\beta_{j}\beta_{k}}.$

To prove Propositions 1.2 and 1.4, we will use the definition of
$\mathcal{G}_{m}$ to show that

$\mathcal{J}\ll\int_{T}^{2T}\prod_{k\leq
m}|\mathcal{N}_{T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\beta_{k})|^{2}dt$

and

$\mathcal{I}_{k}\ll\int_{\mathcal{G}_{m}}|\zeta(\tfrac{1}{2}+i(t+\alpha_{k}))|^{2m}|\mathcal{N}_{T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\beta_{k}-m)|^{2}\prod_{\begin{subarray}{c}\ell\leq
m\\\ \ell\neq
k\end{subarray}}|\mathcal{N}_{T_{L}}(\tfrac{1}{2}+i(t+\alpha_{\ell});\beta_{\ell})|^{2}dt,$

where

$\mathcal{N}_{X}(s;\beta):=\prod_{j\leq L}\mathcal{N}_{j,X}(s;\beta).$

Therefore $\mathcal{J}$ can be controlled by computing the mean value of a
Dirichlet polynomial, and the $\mathcal{I}_{k}$ can be bounded by computing
twisted $2m^{\text{th}}$ moments of the zeta function. Both of these
quantities can be controlled as $\mathcal{N}_{T_{L}}(s;\beta)$ is a short
Dirichlet polynomial. For the latter task, we will need to invoke the Riemann
hypothesis when $m>2$.

The proof of Proposition 1.3 is a bit more difficult. The first step is to
show that $\mathcal{I}_{0}$ is approximately equal to

$\displaystyle\int_{\mathcal{G}_{m}}\prod_{k\leq
m}\zeta(\tfrac{1}{2}+i(t+\alpha_{k}))\mathcal{N}_{T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}(\beta_{k}-1))^{2}\overline{\mathcal{N}_{T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}\beta_{k})}^{2}w(t/T)\
dt$

using the definition of $\mathcal{G}_{m}$ and the Riemann hypothesis. To
compute this quantity, we will write

$\int_{\mathcal{G}_{m}}=\int_{T}^{2T}-\int_{[T,2T]\setminus\mathcal{G}_{m}}.$

The integral over the entire interval $[T,2T]$ is a shifted pure
$m^{\text{th}}$ moment of zeta twisted by a Dirichlet polynomial, which we can
calculate unconditionally because $\mathcal{N}_{T_{L}}(s;\beta)$ is a short
Dirichlet polynomial. To bound the integral over
$[T,2T]\setminus\mathcal{G}_{m}$, we will first decompose this set according
to which subsum $\mathcal{P}_{j,X}(\tfrac{1}{2}+i(t+\alpha_{\ell});\beta)$ is
unusually large for some $\ell$, and then bound the contribution of such $t$
by

$\displaystyle\frac{1}{K_{j}^{2r}}\int_{T}^{2T}\prod_{k\leq
m}|\zeta(\tfrac{1}{2}+i(t+\alpha_{k}))|$
$\displaystyle\cdot|\mathcal{N}_{T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}(\beta_{k}-1))\mathcal{N}_{T_{L}}\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}\beta_{k})|^{2}$
$\displaystyle\quad\times|\mathcal{P}_{j,X}(\tfrac{1}{2}+i(t+\alpha_{\ell}))|^{2r}\
dt$

for some large integer $r$. To bound this quantity we are again forced to use
the Riemann hypothesis, at least when $m>2$. We can estimate this integral
using the same method in [8].

We now have all the necessary tools to begin our proof of Theorem 1.1. To help
simplify the notation going forward, we will omit subscripts depending on
$\bm{\beta}$ from all big-O or Vinogradov asymptotic notation. The reader
should keep in mind that all of the implicit constants in this notation can,
and usually will, depend on $\bm{\beta}$. We will also implicitly assume that
$T$ is sufficiently large in terms of $\bm{\beta}$. After covering some
preliminary results, we will prove Proposition 1.2, followed by Proposition
1.3, and we will conclude the proof by establishing Proposition 1.4.

## Acknowledgements

The author would like to thank Hung M. Bui and Stephen Lester for helpful
discussions.

##  2\. Preliminary Results

To control the size of zeta on the half line, we will use the following lemma
due to Soundararajan [27] and Harper [12].

######  Lemma 2.1.

Assume RH, let $t\in[T,2T]$, and $|\alpha|\leq T/2$. Then for $2\leq X\leq
T^{2}$

$\displaystyle\log|\zeta(\tfrac{1}{2}$
$\displaystyle+i(t+\alpha))|\leq\textup{Re }\sum_{p\leq
X}\frac{1}{p^{1/2+1/\log X+i(t+\alpha)}}\frac{\log X/p}{\log X}$
$\displaystyle+$ $\displaystyle\sum_{p\leq\min(\sqrt{X},\log
T)}\frac{1}{2p^{1+2i(t+\alpha)}}+\frac{\log T}{\log X}+O(1).$

We will often Taylor expand exponentials of the $\mathcal{P}_{j,X}$ in order
to turn moment generating function calculations into computing mean values of
short Dirichlet polynomials– a much simpler task. The following lemma will be
used frequently to accomplish this task.

######  Lemma 2.2.

If $\beta\leq\beta_{\ast}$ and $|\mathcal{P}_{j,X}(s)|\leq K_{j}$ for some
$1\leq j\leq L$, then

$\exp(\beta\mathcal{P}_{j,X}(s))=(1+O(e^{-50\beta_{\ast}^{2}K_{j}}))^{-1}\mathcal{N}_{j,X}(s;\beta).$

###### Proof.

Since $|\mathcal{P}_{j,X}(s)|\leq 2K_{j}$, Taylor expansion gives

$\mathcal{N}_{j,X}(s;\beta)=\exp(\beta\mathcal{P}_{j,X}(s))+O(e^{-100\beta_{\ast}^{2}K_{j}}).$

By assumption
$\exp(-2K_{j}\beta_{\ast})\leq|\exp(\beta\mathcal{P}_{j,X}(s))|\leq\exp(2K_{j}\beta_{\ast})$,
so the claim follows. ∎

Once we have reduced the proof to a number of mean value calculations, we will
use a few standard results to compute these averages. The first result is due
to Montgomery and Vaughan (see for example theorem 9.1 of [16]).

######  Lemma 2.3.

Given any complex numbers $a_{n}$

$\int_{T}^{2T}\left|\sum_{n\leq
N}\frac{a_{n}}{n^{it}}\right|^{2}dt=(T+O(N))\sum_{n\leq N}|a_{n}|^{2}.$

We will also make use of the property that Dirichlet polynomials supported on
distinct sets of primes are approximately independent in the mean square
sense. The precise formulation we will use is the following splitting lemma
which appears in equation (16) of [14].

######  Lemma 2.4.

Suppose for $1\leq j\leq J$ we have $j$ disjoint intervals $I_{j}$ and
Dirichlet polynomials $A_{j}(s)=\sum_{n}a_{j}(n)n^{-s}$ such that $a_{j}(n)$
vanishes unless $n$ is composed of primes in $I_{j}$. Then if $\prod_{j\leq
J}A_{j}(s)$ is a Dirichlet polynomial of length $N$

$\displaystyle\int_{T}^{2T}\prod_{j\leq
J}|A_{j}(\tfrac{1}{2}+it)|^{2}dt=(T+O(N))\prod_{j\leq
J}\left(\frac{1}{T}\int_{T}^{2T}|A_{j}(\tfrac{1}{2}+it)|^{2}dt\right)$

Finally, the following result due to Soundararajan [27, lemma 3] will simplify
calculations for high moments of Dirichlet polynomials supported on primes.

######  Lemma 2.5.

Let $r$ be a natural number and suppose $N^{r}\leq T/\log T$. Then given any
complex numbers $a_{p}$

$\int_{T}^{2T}\left|\sum_{p\leq N}\frac{a_{p}}{p^{it}}\right|^{2r}dt\ll
Tr!\left(\sum_{p\leq N}|a_{p}|^{2}\right)^{r}.$

To compute all the mean values that will arise, we will need two more
ingredients. The first is a special case of lemma 3.2 of [18], which will let
us estimate a sum over primes that will appear frequently throughout the
paper.

######  Lemma 2.6.

Given $\delta\in\mathbb{R}$ and $X\geq 2$

$\displaystyle\sum_{p\leq X}$ $\displaystyle\frac{\cos(\delta\log
p)}{p}=\log|\zeta(1+1/\log X+i\delta)|+O(1).$

The second and final ingredient we will need is a good estimate for the
coefficients of the Dirichlet polynomials

$\prod_{k\leq
m}\mathcal{N}_{j,X}(s+i\alpha_{k};\beta_{k}):=\sum_{n}\frac{b_{j,X,\bm{\alpha},\bm{\beta}}(n)}{n^{s}}.$

To this end, denote $a_{X}(p):=\log(X/p)p^{-1/\log X}/\log X$ and define
multiplicative functions $g_{X}$ and $h_{X}$ by

$g_{X}(p^{r};\beta):=\frac{\beta^{r}a_{X}(p)^{r}}{r!}$

and

$h_{X}(p^{r};\beta):=g_{X}(p^{r};\beta)+1_{p\leq\log
T}\sum_{t=1}^{r/2}\frac{\beta^{r-t}a_{X}(p)^{r-t}}{2^{t}t!(r-2t)!}.$

Next define $c_{1}(n)$ to be 1 if $n$ can be written as $n=n_{1}\cdots n_{r}$
where $r\leq 100\beta_{\ast}^{2}K_{1}$ and each $n_{i}$ is either a prime
$\leq T_{1}$ or a prime square $\leq\log T$. Finally for $2\leq j\leq L$ set
$c_{j}(n)$ to be 1 if $n$ is the product of at most $100\beta_{\ast}^{2}K_{j}$
not necessarily distinct primes in $(T_{j-1},T_{j}]$. In proposition 3.1 of
[8] it was shown that

######  Proposition 2.7.

For $2\leq j\leq L$

$\mathcal{N}_{j,X}(s;\beta)=\sum_{p\mid n\Rightarrow
p\in(T_{j-1},T_{j}]}\frac{g_{X}(n;\beta)c_{j}(n)}{n^{s}}.$

If

$\mathcal{N}_{1,X}(s;\beta)=\sum_{p\mid n\Rightarrow
p\in(T_{j-1},T_{j}]}\frac{f_{X}(n;\beta)}{n^{s}}$

then $f_{X}(n;\beta)\leq h_{X}(n;\beta)c_{1}(n)$ and
$f_{X}(p;\beta)=g_{X}(p;\beta)$.

Therefore $b_{1,X,\bm{\alpha},\bm{\beta}}(n)$ is the $m$-fold Dirichlet
convolution of $f_{X}(n;\beta_{k})n^{-i\alpha_{k}}$ and
$b_{j,X,\bm{\alpha},\bm{\beta}}(n)$ is the $m$-fold convolution of
$g_{X}(n;\beta_{k})c_{j}(n)n^{-i\alpha_{k}}$ for $2\leq j\leq L$. We will make
use of two more sets of coefficients. Let
$b^{\prime}_{j,X,\bm{\alpha},\bm{\beta}}(n)$ be the $m$-fold convolution of
$h_{X}(n;\beta_{k})n^{-i\alpha_{k}}1_{p|n\Rightarrow p\in(T_{0},T_{1}]}$ when
$j=1$ and the $m$-fold convolution of
$g_{X}(n;\beta_{k})n^{-i\alpha_{k}}1_{p|n\Rightarrow p\in(T_{j-1},T_{j}]}$
when $2\leq j\leq L$. Finally, let
$b^{\prime\prime}_{1,X,\bm{\alpha},\bm{\beta}}(n)$ be the $m$-fold convolution
of $h_{X}(n;\beta_{k})1_{p|n\Rightarrow p\in(T_{0},T_{1}]}$ when $j=1$ and the
$m$-fold convolution of $g_{X}(n;\beta_{k})1_{p|n\Rightarrow
p\in(T_{j-1},T_{j}]}$ when $2\leq j\leq L$. Note
$b^{\prime}_{j,X,\bm{\alpha},\bm{\beta}}$ and
$b^{\prime\prime}_{j,X,\bm{\alpha},\bm{\beta}}$ are multiplicative, and
$|b_{j,X,\bm{\alpha},\bm{\beta}}(n)|,|b^{\prime}_{j,X,\bm{\alpha},\bm{\beta}}(n)|\leq
b^{\prime\prime}_{j,X,\bm{\alpha},\bm{\beta}}(n)$. All the information we will
need about these coefficients is contained in lemma 3.2 of [8].

######  Lemma 2.8.

For $1\leq j\leq L$ and $p\in(T_{j-1},T_{j}]$

$b_{j,X,\bm{\alpha},\bm{\beta}}(p)=a_{X}(p)\sum_{k=1}^{m}\beta_{k}p^{-i\alpha_{k}},$

and $b^{\prime\prime}_{j,X,\bm{\alpha},\bm{\beta}}(p)\leq\beta_{\ast}$. If
$r\geq 2$

$b^{\prime\prime}_{j,X,\bm{\alpha},\bm{\beta}}(p^{r})\leq\frac{\beta_{\ast}^{r}m^{r}}{r!}$

holds whenever $2\leq j\leq L$ or $p>\log T$, and otherwise

$b^{\prime\prime}_{1,X,\bm{\alpha},\bm{\beta}}(p^{r})\leq
m\beta_{\ast}^{r}r^{2m}e^{-r\log(r/m)/2m+2r}.$

##  3\. Proof of Proposition 1.2

We will begin with the evaluation of

$\displaystyle\mathcal{J}=\int_{\mathcal{G}_{m}}\prod_{k\leq m}\prod_{j\leq
L}\exp\left(2\beta_{k}\textup{Re
}\mathcal{P}_{j,T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k}))\right)w(t/T)dt,$

which is the simplest computation. To compute this mean value, we apply lemma
2.2 and then extend the integral over the entire set $[T,2T]$ to find

$\displaystyle\mathcal{J}\ll\int_{T}^{2T}\prod_{k\leq
m}|\mathcal{N}_{T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\beta_{k})|^{2}dt.$

Here we have used that $\prod_{j\leq
L}(1+O(e^{-50\beta_{\ast}^{2}K_{j}}))^{2}=O(1)$. Note
$\mathcal{N}_{j,X}(s;\beta)$ is a Dirichlet polynomial of length $\leq
T_{j}^{100m\beta_{\ast}^{2}K_{j}}$ when $j>1$ and of length $\leq
T_{1}^{200m\beta_{\ast}^{2}K_{1}}$ when $j=1$, so

$\prod_{k\leq m}\mathcal{N}_{X}(s+i\alpha_{k};\beta_{k})$

has length at most
$T_{1}^{200m\beta_{\ast}^{2}K_{1}}T_{2}^{100m\beta_{\ast}^{2}K_{2}}\cdots
T_{L}^{100m\beta_{\ast}^{2}K_{L}}\leq T^{1/10}$ by choice of $T_{j},K_{j}$ and
$L$. Therefore using lemma 2.4 in tandem with proposition 3.3 of [8] we find

$\displaystyle\mathcal{J}$ $\displaystyle\ll T\prod_{j\leq
L}\left(\prod_{p\in(T_{j-1},T_{j}]}\left(1+\frac{|b_{j,T_{L},\bm{\alpha},\bm{\beta}}(p)|^{2}}{p}+O\left(\frac{1}{p^{2}}\right)\right)+O(e^{-50\beta_{\ast}^{2}K_{j}})\right)$
$\displaystyle\ll T(\log T)^{\beta_{1}^{2}+\cdots+\beta_{m}^{2}}\prod_{1\leq
j<k\leq m}|\zeta(1+i(\alpha_{j}-\alpha_{k})+1/\log T)|^{2\beta_{j}\beta_{k}},$

where the latter bound follows from an application of lemma 2.6. This
concludes the proof of Proposition 1.2. ∎

##  4\. Proof of Proposition 1.3

Recall

$\displaystyle\mathcal{I}_{0}$
$\displaystyle=\int_{\mathcal{G}_{m}}\prod_{k\leq
m}\zeta(\tfrac{1}{2}+i(t+\alpha_{k}))$ $\displaystyle\times\prod_{j\leq
L}\exp\left((\beta_{k}-1)\mathcal{P}_{j,T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k}))+\beta_{k}\mathcal{P}_{j,T_{L}}(\tfrac{1}{2}-i(t+\alpha_{k}))\right)w(t/T)\
dt.$

To compute this, we will first need to replace the exponential by a Dirichlet
polynomial. To accomplish this, we will write $\mathcal{I}_{0}$ as a
telescoping sum and control the size of the intermediate increments. For
$0\leq J\leq L$ denote

$\displaystyle\qquad\qquad\qquad\mathcal{I}_{0}^{(J)}=\int_{\mathcal{G}_{m}}\prod_{k\leq
m}\zeta(\tfrac{1}{2}+i(t+\alpha_{k}))$ $\displaystyle\times\prod_{j\leq J}$
$\displaystyle\mathcal{N}_{j,T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}(\beta_{k}-1))^{2}\overline{\mathcal{N}_{j,T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}\beta_{k})}^{2}$
$\displaystyle\times\prod_{J<j\leq L}$
$\displaystyle\exp\left((\beta_{k}-1)\mathcal{P}_{j,T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k}))+\beta_{k}\mathcal{P}_{j,T_{L}}(\tfrac{1}{2}-i(t+\alpha_{k}))\right)w(t/T)\
dt.$

Since $\mathcal{I}_{0}=\mathcal{I}_{0}^{(0)}$, we may decompose

$\mathcal{I}_{0}=\mathcal{I}_{0}^{(L)}-\sum_{J\leq
L}\left(\mathcal{I}_{0}^{(J)}-\mathcal{I}_{0}^{(J-1)}\right)$

To prove Proposition 1.3 we will give a lower bound for $|I_{0}^{(L)}|$ and
then show that

$\sum_{J\leq
L}\big{|}\mathcal{I}_{0}^{(J)}-\mathcal{I}_{0}^{(J-1)}\big{|}\leq\frac{|I_{0}^{(L)}|}{2}.$

To control these differences, we will use the following consequence of lemma
2.2 and the definition of $\mathcal{G}_{m}$.

######  Lemma 4.1.

For $0\leq J\leq L$

$\displaystyle\big{|}\mathcal{I}_{0}^{(J)}-\mathcal{I}_{0}^{(J-1)}\big{|}$
$\displaystyle\ll$ $\displaystyle
e^{-50\beta_{\ast}^{2}K_{J}}\int_{\mathcal{G}_{m}}\prod_{k\leq
m}|\zeta(\tfrac{1}{2}+i(t+\alpha_{k}))|\cdot|\mathcal{N}_{T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}(\beta_{k}-1))\mathcal{N}_{T_{L}}\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}\beta_{k})|^{2}dt.$

To prove Proposition 1.3, we will now just need the following two estimates.

######  Proposition 4.2.

$|\mathcal{I}_{0}^{(L)}|\gg T\prod_{p\leq T_{L}}\left(1+\sum_{1\leq j,k\leq
m}\frac{\beta_{j}\beta_{k}}{p^{1+i(\alpha_{j}-\alpha_{k})}}\right).$

######  Proposition 4.3.

Assuming the Riemann hypothesis

$\displaystyle\int_{\mathcal{G}_{m}}\prod_{k\leq m}|\zeta(\tfrac{1}{2}+i(t+$
$\displaystyle\alpha_{k}))|\cdot|\mathcal{N}_{T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}(\beta_{k}-1))\mathcal{N}_{T_{L}}\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}\beta_{k})|^{2}dt$
$\displaystyle\ll T\prod_{p\leq T_{L}}\left(1+\sum_{1\leq j,k\leq
m}\frac{\beta_{j}\beta_{k}}{p^{i(\alpha_{j}-\alpha_{k})}}\right).$

Before proving these two statements, let us briefly see how they imply
Proposition 1.3.

###### Proof of Proposition 1.3.

Combining Propositions 4.3 and 4.2 with lemma 4.1, we see that

$\sum_{J\leq
L}\big{|}\mathcal{I}_{0}^{(J)}-\mathcal{I}_{0}^{(J-1)}\big{|}\ll|I_{0}^{(L)}|\sum_{J\leq
L}e^{-50\beta_{\ast}^{2}K_{J}}.$

By definition of $K_{J}$, the sum on the right hand side is equal to

$\sum_{J\leq
L}\exp\left(-50\beta_{\ast}^{2}\frac{(\log_{2}T)^{3/2}}{e^{3J/4}}\right).$

Because $T_{L}\leq T^{\delta}$, it also follows that $L\leq
2\log_{3}T+\log\delta$. Therefore by summing in reverse starting at $J=L$, we
see that this sum is bounded by

$\sum_{j\geq
1}\exp\left(-50\beta_{\ast}^{2}e^{-3\log\delta/4}e^{3j/4}\right)\leq\sum_{j\geq
1}\exp\left(-50\beta_{\ast}^{2}e^{-3\log\delta/4}j\right)\ll\exp\left(-50\beta_{\ast}^{2}e^{-3\log\delta/4}\right).$

Therefore if we choose $\delta>0$ sufficiently small in terms of $\bm{\beta}$
we may ensure that

$\sum_{J\leq
L}\big{|}\mathcal{I}_{0}^{(J)}-\mathcal{I}_{0}^{(J-1)}\big{|}\leq\frac{|I_{0}^{(L)}|}{2}.$

Therefore it follows that

$|I_{0}|\gg T\prod_{p\leq T_{L}}\left(1+\sum_{1\leq j,k\leq
m}\frac{\beta_{j}\beta_{k}}{p^{1+i(\alpha_{j}-\alpha_{k})}}\right).$

∎

###  4.1. Proof of Proposition 4.2

To estimate $\mathcal{I}_{0}^{(L)}$, we will first write
$\mathcal{I}_{0}^{(L)}=\mathcal{J}_{1}-\mathcal{J}_{2}$, where

$\mathcal{J}_{1}=\int_{T}^{2T}\prod_{k\leq
m}\zeta(\tfrac{1}{2}+i(t+\alpha_{k}))\mathcal{N}_{T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}(\beta_{k}-1))^{2}\overline{\mathcal{N}_{T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}\beta_{k})}^{2}w(t/T)\
dt$

and

$\mathcal{J}_{2}=\int_{[T,2T]\setminus\mathcal{G}_{m}}\prod_{k\leq
m}\zeta(\tfrac{1}{2}+i(t+\alpha_{k}))\mathcal{N}_{T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}(\beta_{k}-1))^{2}\overline{\mathcal{N}_{T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}\beta_{k})}^{2}w(t/T)\
dt.$

We will evaluate $\mathcal{J}_{1}$ first. We will make use of the following
approximate functional equation.

######  Lemma 4.4.

Let

$V(x,t)=\frac{1}{2\pi
i}\int_{(1)}\frac{e^{s^{2}}}{s}\left(\frac{t^{3m}}{x}\right)^{s}ds$

and $\tau_{\bm{\alpha}}(n)=\sum_{n_{1}\cdots
n_{m}=n}n_{1}^{-i\alpha_{1}}\cdots n_{m}^{-i\alpha_{m}}$. For $\alpha_{j}\leq
T/2$ and $t\in[T,2T]$

$\prod_{k\leq
m}\zeta(\tfrac{1}{2}+it+i\alpha_{k})=\sum_{n}\frac{\tau_{\bm{\alpha}}(n)}{n^{1/2+it}}V(n,t)+O(1/T).$

###### Proof.

We will evaluate the integral

$I=\frac{1}{2\pi i}\int_{(1)}\frac{e^{s^{2}}}{s}t^{3ms}\prod_{k\leq
m}\zeta(\tfrac{1}{2}+it+i\alpha_{k}+s)ds$

in two ways. Expanding $\prod_{k\leq m}\zeta(\tfrac{1}{2}+it+i\alpha_{k}+s)$
into its Dirichlet series and simplifying shows

$I=\sum_{n}\frac{\tau_{\bm{\alpha}}(n)}{n^{1/2+it}}V(n,t).$

Alternatively, by shifting the contour to $\textup{Re }s=-1$, we pass over
poles at $s=0$ and $s=-\tfrac{1}{2}-it-i\alpha_{k}$. Only the residue at $s=0$
contributes because of the rapid decay of $e^{s^{2}}$ and because
$|t+\alpha_{k}|\geq T/2$. Therefore

$I=\prod_{k\leq m}\zeta(\tfrac{1}{2}+it+i\alpha_{k})+\frac{1}{2\pi
i}\int_{(-1)}\frac{e^{s^{2}}}{s}t^{3ms}\prod_{k\leq
m}\zeta(\tfrac{1}{2}+it+i\alpha_{k}+s)ds+O_{A}(T^{-A}).$

To conclude, by the standard estimate $\zeta(-\tfrac{1}{2}+it)\ll 1+|t|$, we
may bound this final integral by

$\ll\int_{\mathbb{R}}\frac{e^{-y^{2}}}{y+1}T^{-3m}\prod_{j\leq
s}(1+|t+\alpha_{j}+y|)dy.$

The integral over the region $|y|\leq T$ can be bounded by

$\ll\int_{0}^{T}\frac{e^{-y^{2}}}{y+1}T^{-3m}\times T^{m}\ dy\ll T^{-1},$

and the integral over the region $|y|\geq T$ is

$\ll\int_{T}^{\infty}\frac{e^{-y^{2}}}{y}T^{-2m}\times y^{m}\ dy\ll T^{-1}.$

∎

Next we will need to understand the coefficients of the Dirichlet polynomials

$\prod_{k\leq
m}\mathcal{N}_{j,T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}(\beta_{k}-1))^{2}=\sum_{n}\frac{q_{j}(n)}{n^{1/2+it}},$
$\prod_{k\leq
m}\mathcal{N}_{T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}(\beta_{k}-1))^{2}=\sum_{n}\frac{q(n)}{n^{1/2+it}},$

and

$\prod_{k\leq
m}\mathcal{N}_{j,T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}\beta_{k})^{2}=\sum_{n}\frac{r_{j}(n)}{n^{1/2+it}},$
$\prod_{k\leq
m}\mathcal{N}_{T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}\beta_{k})^{2}=\sum_{n}\frac{r(n)}{n^{1/2+it}}.$

Notice that each $q_{j}(n)$ is the twofold Dirichlet convolution of
$b_{j,T_{L},\bm{\alpha},\tfrac{1}{2}(\bm{\beta}-1)}(n)$ with itself and each
$r_{j}(n)$ is the twofold Dirichlet convolution of
$b_{j,T_{L},\bm{\alpha},\tfrac{1}{2}\bm{\beta}}(n)$ with itself, where
$\tfrac{1}{2}(\bm{\beta}-1)$ and $\tfrac{1}{2}\bm{\beta}$ denote the vectors
with $j^{\text{th}}$ element $\tfrac{1}{2}(\beta_{j}-1)$ and
$\tfrac{1}{2}\beta_{j}$ respectively. As before, we will define two more sets
of coefficients: Let $q^{\prime}_{j}(n)$ be the twofold Dirichlet convolution
of $b^{\prime}_{j,T_{L},\bm{\alpha},\tfrac{1}{2}(\bm{\beta}-1)}(n)$ with
itself and let $r^{\prime}_{j}(n)$ be the twofold Dirichlet convolution of
$b^{\prime}_{j,T_{L},\bm{\alpha},\tfrac{1}{2}\bm{\beta}}(n)$ with itself.
Finally define $q^{\prime\prime}_{j}(n)$ and $r^{\prime\prime}_{j}(n)$ in an
analogous fashion. As before, the key point is that $q_{j}^{\prime}$ and
$r^{\prime}_{j}$ are multiplicative approximations of $q_{j}$ and $r_{j}$ and
that $q^{\prime\prime}_{j}$ and $r^{\prime\prime}_{j}$ are non-negative
multiplicative coefficients satisfying $|q_{j}|,|q^{\prime}_{j}|\leq
q^{\prime\prime}$ and $|r_{j}|,|r^{\prime}_{j}|\leq r^{\prime\prime}$. Before
proceeding further, we will need the following estimate.

######  Lemma 4.5.

For $p\in(T_{j-1},T_{j}]$

$q_{j}(p^{l}),r_{j}(p^{l})\ll
m^{2}\beta_{\ast}^{l}l^{4m}e^{-l\log(l/2m)/4m+2l}.$

###### Proof.

The bound for $q_{j}$ follows by combining the formula

$q_{j}(p^{l})=\sum_{x+y=l}b_{j,T_{L},\bm{\alpha},\tfrac{1}{2}(\bm{\beta}-1)}(p^{x})b_{j,T_{L},\bm{\alpha},\tfrac{1}{2}(\bm{\beta}-1)}(p^{y})$

with the bound for $b_{j,T_{L},\bm{\alpha},\tfrac{1}{2}(\bm{\beta}-1)}(p^{x})$
given in lemma 2.8 and noting that either $x\geq l/2$ or $y\geq l/2$. The
argument for $r_{j}$ is the same. ∎

We can now compute

$\displaystyle\mathcal{J}_{1}$ $\displaystyle=\frac{1}{2\pi
i}\sum_{n}\sum_{h,k<T^{1/2}}\frac{\tau_{\bm{\alpha}}(n)q(h)\overline{r(k)}}{n^{1/2}h^{1/2}k^{1/2}}\int_{(1)}\frac{e^{s^{2}}}{s}\int_{T}^{2T}\left(\frac{nh}{k}\right)^{-it}\frac{t^{3ms}}{n^{s}}w(t/T)\
dt\ ds$ $\displaystyle+O\left(\frac{1}{T}\int_{T}^{2T}\prod_{k\leq
m}|\mathcal{N}_{T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}(\beta_{k}-1))\mathcal{N}_{T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}\beta_{k})|^{2}dt\right).$

Because

$\prod_{k\leq
m}\mathcal{N}_{T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}(\beta_{k}-1))\mathcal{N}_{T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}\beta_{k})$

is a short Dirichlet polynomial whose coefficients $c(n)$ satisfy
$c(n)\ll_{\varepsilon}n^{\varepsilon}$, it follows that the error term is
$\ll_{\varepsilon}T^{\varepsilon}$ by lemma 2.3. Returning now to the main
term, the next step is to discard the terms where $nh\neq k$. An exercise in
contour integration shows that

$t^{j}\frac{\partial^{j}}{\partial
t^{j}}V(x,t)\ll_{A,j}\left(1+|x/t^{3m}|\right)^{-A}.$

Therefore repeated integration by parts implies that

$\int_{T}^{2T}w(t/T)\left(\frac{nh}{k}\right)^{-it}V(x,t)\
dt\ll_{j,A}\frac{(1+n/T^{3m})^{-A}}{|\log(nh/k)|^{j}T^{j}}.$

Because $h$ and $k$ are at most $T^{1/2}$, it follows that $\log(nh/k)\gg
T^{-1/2}$ if $nh\neq k$. Therefore the contribution to the sum of terms with
$hn\neq k$ is $O_{A}(T^{-A})$ because $h,k<T^{1/2}$.

Now by shifting the contour to $\textup{Re }s=-1/4$ and simplifying the
diagonal terms, we find that

$\mathcal{J}_{1}=T\|w\|_{1}\sum_{\begin{subarray}{c}h,k\leq T^{1/2}\\\
h|k\end{subarray}}\frac{\tau_{\bm{\alpha}}(k/h)q(h)\overline{r(k)}}{k}+O(T^{1-\varepsilon}).$

By multiplicativity, we may factor the inner sum as

$\sum_{\begin{subarray}{c}h,k\leq T^{1/2}\\\
h|k\end{subarray}}\frac{\tau_{\bm{\alpha}}(k/h)q(h)\overline{r(k)}}{k}=\prod_{j\leq
L}\sum_{\begin{subarray}{c}p|h,k\Rightarrow p\in(T_{j-1},T_{j}]\\\
h|k\end{subarray}}\frac{\tau_{\bm{\alpha}}(k/h)q_{j}(h)\overline{r_{j}(k)}}{k}.$

Next notice if $q_{j}(h)\neq q^{\prime}_{j}(h)$, then it must be that
$\Omega(h)\geq 100\beta_{\ast}^{2}K_{j}$. So replacing $q_{j}$ and $r_{j}$
with $q^{\prime}_{j}$ and $r^{\prime}_{j}$ respectively incurs an error of
order

$\displaystyle
e^{-100\beta_{\ast}^{2}K_{j}}\sum_{\begin{subarray}{c}p|h,k\Rightarrow
p\in(T_{j-1},T_{j}]\\\
h|k\end{subarray}}\frac{e^{\Omega(h)}|\tau_{\bm{\alpha}}(k/h)|q^{\prime\prime}_{j}(h)r^{\prime\prime}_{j}(k)}{k}$
$\displaystyle\ll
e^{-100\beta_{\ast}^{2}K_{j}}\prod_{p\in(T_{j-1},T_{j}]}\left(1+\frac{2e\beta_{\ast}^{2}}{p}+O\left(\frac{1}{p^{2}}\right)\right)\ll
e^{-50\beta_{\ast}^{2}K_{j}}.$

for each $j$. Here we have used the divisor bound
$|\tau_{\bm{\alpha}}(p^{y})|\ll_{\varepsilon}p^{\varepsilon y}$ and lemma 4.5
to bound the contribution of the terms of order smaller than $1/p^{2}$.
Therefore, suppressing terms of order $T^{1-\varepsilon}$, we see that

$\displaystyle\mathcal{J}_{1}$ $\displaystyle=T\|w\|_{1}\prod_{j\leq
L}\left(\prod_{p\in(T_{j-1},T_{j}]}\left(\sum_{0\leq x\leq
y}\frac{\tau_{\bm{\alpha}}(p^{y-x})q_{j}(p^{x})\overline{r_{j}(p^{y})}}{p^{y}}\right)+O(e^{-50\beta_{\ast}^{2}K_{j}})\right)$
$\displaystyle=T\|w\|_{1}\prod_{j\leq
L}\left(\prod_{p\in(T_{j-1},T_{j}]}\left(1+\frac{q_{j}(p)\overline{r_{j}(p)}+\tau_{\bm{\alpha}}(p)\overline{r_{j}(p)}}{p}+O\left(\frac{1}{p^{2}}\right)\right)+O(e^{-50\beta_{\ast}^{2}K_{j}})\right).$

A quick computation shows that

$q_{j}(p)+\tau_{\bm{}{\alpha}}(p)=a_{T_{L}}(p)\sum_{k\leq
m}\beta_{k}p^{-i\alpha_{k}}=r_{j}(p).$

Therefore, recalling $a_{T_{L}}(p)=p^{-1/\log T_{L}}(1-\log p/\log T_{L})$, we
deduce

$\displaystyle\mathcal{J}_{1}=T\|w\|_{1}\prod_{j\leq
L}\left(\prod_{p\in(T_{j-1},T_{j}]}\left(1+\frac{1}{p}\Bigg{|}\sum_{k\leq
m}\beta_{k}p^{-i\alpha_{k}}\Bigg{|}^{2}+O\left(\frac{\log p}{p\log
T_{L}}+\frac{1}{p^{2}}\right)\right)+O(e^{-50\beta_{\ast}^{2}K_{j}})\right),$

and we may readily conclude

######  Proposition 4.6.

$\mathcal{J}_{1}\gg T\prod_{p\leq T_{L}}\left(1+\sum_{1\leq j,k\leq
m}\frac{\beta_{j}\beta_{k}}{p^{1+i(\alpha_{j}-\alpha_{k})}}\right).$

We will now show, for $\delta$ sufficiently small, that
$|\mathcal{J}_{2}|\leq|\mathcal{J}_{1}|/2$. To do this we will need to
decompose the set $[T,2T]\setminus\mathcal{G}_{m}$ into many pieces. First
recall the definition

$\mathcal{G}:=\left\\{t\in[T/2,5T/2]:|P_{j,T_{L}}(\tfrac{1}{2}+it)|\leq
K_{j}\text{ for all }1\leq j\leq L\right\\}.$

Now given a subset $A$ of $[m]:=\\{1,\ldots,m\\}$, we will define

$\mathcal{G}_{A}:=\left\\{t\in[T,2T]:t+\alpha_{k}\in\mathcal{G}\text{ if and
only if }k\in A\right\\},$

and for each $1\leq j\leq L$ let

$\displaystyle\mathcal{B}_{j}:=\\{t\in[T/2$
$\displaystyle,5T/2]:|\mathcal{P}_{r,T_{s}}(\tfrac{1}{2}+it)|\leq K_{j}\text{
for all }1\leq r<j\text{ and }r\leq s\leq L$ $\displaystyle\text{ but
}|\mathcal{P}_{j,T_{s}}(\tfrac{1}{2}+it)|>K_{j}\text{ for some }j\leq s\leq
L\\}.$

To evaluate $\mathcal{J}_{2}$, we must evaluate an integral over all of the
$\mathcal{G}_{A}$ with $A$ ranging over all proper subsets of $[m]$. Without
loss of generality we will write $A=[m]\setminus[a]$. For each
$t\in\mathcal{G}_{A}$, there is a function $F_{t}:[a]\rightarrow[L]$ such that
$t+\alpha_{j}\in\mathcal{B}_{f(j)}$. We will further partition
$\mathcal{G}_{A}$ into the sets

$\mathcal{B}_{A,n}=\\{t\in\mathcal{G}_{A}:\min_{j\in[a]}F_{t}(j)=n\\}.$

With this new notation, we may write

$[T,2T]\setminus\mathcal{G}_{m}=\bigsqcup_{n\leq
L}\bigsqcup_{A\subsetneq[m]}\mathcal{B}_{A,n}.$

When $n>1$, we may proceed in a similar manner to [8]. Using Lemma 2.1 with
$X=T_{n-1}$ and the definition of $\mathcal{B}_{A,n}$ we find

$\displaystyle\int_{\mathcal{B}_{A,n}}$ $\displaystyle\prod_{k\leq
m}|\zeta(\tfrac{1}{2}+i(t+\alpha_{k}))||\mathcal{N}_{T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}(\beta_{k}-1))\mathcal{N}_{T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}\beta_{k})|^{2}w(t/T)\
dt$
$\displaystyle\ll\int_{\mathcal{B}_{A,n}}\prod_{k=1}^{m}\exp\Bigg{(}\textup{Re
}\Bigg{(}\sum_{j<n}\mathcal{P}_{j,T_{n-1}}(\tfrac{1}{2}+i(t+\alpha_{k}))+1/c_{n-1}\Bigg{)}$
$\displaystyle\qquad\qquad\qquad\qquad\times|\mathcal{N}_{T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}(\beta_{k}-1))\mathcal{N}_{T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}\beta_{k})|^{2}\
dt$
$\displaystyle\ll~{}e^{1/c_{n-1}}\int_{\mathcal{B}_{A,n}}\prod_{k=1}^{m}\prod_{j<n}\exp\left(\textup{Re
}\mathcal{P}_{j,T_{n-1}}(\tfrac{1}{2}+i(t+\alpha_{k}))\right)$
$\displaystyle\qquad\qquad\qquad\qquad\times|\mathcal{N}_{T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}(\beta_{k}-1))\mathcal{N}_{T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}\beta_{k})|^{2}\
dt$ $\displaystyle\ll~{}e^{1/c_{n-1}}\max_{\begin{subarray}{c}\ell\in[a]\\\
s\in[L]\end{subarray}}\int_{T}^{2T}|\mathcal{P}_{n,T_{s}}(\tfrac{1}{2}+i(t+\alpha_{\ell}))/K_{n}|^{2\lceil
1/20c_{n}\rceil}\prod_{k=1}^{m}\prod_{j<n}|\mathcal{N}_{j,T_{n-1}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2})|^{2}$
$\displaystyle\qquad\qquad\qquad\qquad\times|\mathcal{N}_{T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}(\beta_{k}-1))\mathcal{N}_{T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}\beta_{k})|^{2}\
dt.$

So by lemma 2.4 we need the following mean value calculations.

######  Proposition 4.7.

For $j<n$

$\displaystyle\int_{T}^{2T}\prod_{k=1}^{m}$
$\displaystyle|\mathcal{N}_{j,T_{n-1}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2})\mathcal{N}_{j,T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}(\beta_{k}-1))\mathcal{N}_{j,T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}\beta_{k})|^{2}dt$
$\displaystyle\leq\ $ $\displaystyle
T\prod_{p\in(T_{j-1},T_{j}]}\left(1+\frac{1}{p}\sum_{1\leq j,k\leq
m}\frac{\beta_{j}\beta_{k}}{p^{i(\alpha_{j}-\alpha_{k})}}+O\left(\frac{\log
p}{p\log
T_{n-1}}+\frac{1}{p^{2}}\right)\right)+O(e^{-50\beta_{\ast}^{2}K_{j}}).$

###### Proof.

All of the following calculations are similar to earlier computations, so we
will just sketch their proofs. The coefficients of the Dirichlet polynomial

$\prod_{k=1}^{m}\mathcal{N}_{j,T_{n-1}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2})\mathcal{N}_{j,T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}(\beta_{k}-1))\mathcal{N}_{j,T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}\beta_{k})$

are given by the triple convolution

$a(n):=b_{j,T_{n-1},\bm{\alpha},\bm{1/2}}\ast
b_{j,T_{L},\bm{\alpha},\tfrac{1}{2}(\bm{\beta}-1)}\ast
b_{j,T_{L},\bm{\alpha},\tfrac{1}{2}\bm{\beta}}(n),$

where $\bm{1/2}$ is a vector of $m$ copies of $1/2$. Using Rankin’s trick, one
can replace these with the multiplicative coefficients

$a^{\prime}(n):=b^{\prime}_{j,T_{n-1},\bm{\alpha},\bm{1/2}}\ast
b^{\prime}_{j,T_{L},\bm{\alpha},\tfrac{1}{2}(\bm{\beta}-1)}\ast
b^{\prime}_{j,T_{L},\bm{\alpha},\tfrac{1}{2}\bm{\beta}}(n)$

at a cost of $O(e^{-50\beta_{\ast}^{2}K_{j}})$. Using lemma 2.8, one may then
show that

$|a^{\prime}(p)|^{2}=\sum_{1\leq j,k\leq
m}\frac{\beta_{j}\beta_{k}}{p^{i(\alpha_{j}-\alpha_{k})}}+O\left(\frac{\log
p}{\log T_{n-1}}\right)$

and

$\sum_{r\geq 2}\frac{|a(p^{r})|^{2}}{p^{r}}\ll\frac{1}{p^{2}}.$

The claim now follows by multiplicativity and lemma 2.3. ∎

######  Proposition 4.8.

For $1<n\leq L$

$\displaystyle\int_{T}^{2T}|\mathcal{P}_{n,T_{s}}(\tfrac{1}{2}+i(t+\alpha_{\ell}))/K_{n}|^{2\lceil
1/20c_{n}\rceil}$
$\displaystyle\prod_{k=1}^{m}|\mathcal{N}_{n,T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}(\beta_{k}-1))\mathcal{N}_{n,T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}\beta_{k})|^{2}\
dt$ $\displaystyle\ll Te^{-\log(1/c_{n})/40c_{n}}.$

###### Proof.

By Cauchy Schwarz, the relevant mean value is at most

$\left(\int_{T}^{2T}|\mathcal{P}_{n,T_{s}}(\tfrac{1}{2}+i(t+\alpha_{\ell}))/K_{n}|^{2\lceil
1/20c_{n}\rceil}\ dt\right)^{1/2}$
$\times\left(\int_{T}^{2T}\prod_{k=1}^{m}|\mathcal{N}_{T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}(\beta_{k}-1))\mathcal{N}_{T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}\beta_{k})|^{2}\
dt\right)^{1/2}.$

By Proposition 4.2 of [8], the first integral is $\ll
Te^{-\log(1/c_{n})/20c_{n}}$. Using the same reasoning in the previous proof,
one may show that the second integral is

$\ll\prod_{p\in{(T_{n-1},T_{n}]}}\left(1+O\left(\frac{1}{p}\right)\right)\ll
1$

because $\log T_{n}/\log T_{n-1}=e$ for $n>1$. The claim now follows.

∎

######  Proposition 4.9.

For $j>n$

$\displaystyle\int_{T}^{2T}\prod_{k=1}^{m}|\mathcal{N}_{j,T_{L}}$
$\displaystyle(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}(\beta_{k}-1))\mathcal{N}_{j,T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}\beta_{k})|^{2}dt$
$\displaystyle\leq
T\prod_{p\in(T_{j-1},T_{j}]}\left(1+O\left(\frac{1}{p}\right)\right)+O(e^{-50\beta_{\ast}^{2}K_{j}}).$

###### Proof.

The proof is a simpler version of the proof of Proposition 4.7. One now has a
double convolution instead of a triple convolution, and uses a cruder bound
for the Dirichlet polynomial coefficients. The details are omitted. ∎

The previous three propositions and Mertens’ estimate now imply that

$\displaystyle\int_{\mathcal{B}_{A,n}}\prod_{k\leq
m}|\zeta(\tfrac{1}{2}+i(t+\alpha_{k}))||\mathcal{N}_{T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}(\beta_{k}-1))\mathcal{N}_{T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}\beta_{k})|^{2}w(t/T)\
dt$ $\displaystyle\ll
T\prod_{j<n}\left(\prod_{p\in(T_{j-1},T_{j}]}\left(1+\frac{1}{p}\sum_{1\leq
j,k\leq
m}\frac{\beta_{j}\beta_{k}}{p^{i(\alpha_{j}-\alpha_{k})}}+O\left(\frac{\log
p}{p\log
T_{n-1}}+\frac{1}{p^{2}}\right)\right)+O(e^{-50\beta_{\ast}^{2}K_{j}})\right)$
$\displaystyle\qquad\qquad\times
e^{1/c_{n-1}-\log(1/c_{n})/40c_{n}}\prod_{n<j\leq
L}\left(\prod_{p\in(T_{j-1},T_{j}]}\left(1+O\left(\frac{1}{p}\right)\right)+O(e^{-50\beta_{\ast}^{2}K_{j}})\right)$
$\displaystyle\ll
T\exp\left(1/c_{n-1}-\log(1/c_{n})/40c_{n}+O\left((\log_{2}T)^{2}e^{-n}\right)\right)\prod_{p\leq
T_{n-1}}\left(1+\frac{1}{p}\sum_{1\leq j,k\leq
m}\frac{\beta_{j}\beta_{k}}{p^{i(\alpha_{j}-\alpha_{k})}}\right)$
$\displaystyle\ll
T\exp\left(1/c_{n-1}-\log(1/c_{n})/40c_{n}+O\left((\log_{2}T)^{2}e^{-n}+L-n\right)\right)\prod_{p\leq
T_{L}}\left(1+\frac{1}{p}\sum_{1\leq j,k\leq
m}\frac{\beta_{j}\beta_{k}}{p^{i(\alpha_{j}-\alpha_{k})}}\right).$

When $n=1$, we will instead use the estimate

######  Proposition 4.10.

Assuming the Riemann hypothesis, for any $A\subsetneq[m]$

$\displaystyle\int_{\mathcal{B}_{A,1}}\prod_{k\leq
m}\zeta(\tfrac{1}{2}+i(t+\alpha_{k}))\mathcal{N}_{T_{L}}(\tfrac{1}{2}+i(t$
$\displaystyle+\alpha_{k});\tfrac{1}{2}(\beta_{k}-1))^{2}\overline{\mathcal{N}_{T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}\beta_{k})}^{2}w(t/T)\
dt$ $\displaystyle\ll_{A}T(\log T)^{-A}.$

###### Proof.

By Hölder’s inequality, the integral over $\mathcal{B}_{A,1}$ is at most

$(\text{meas }\mathcal{B}_{A,1})^{1/3}\left(\int_{T}^{2T}\prod_{k\leq
m}|\zeta(\tfrac{1}{2}+i(t+\alpha_{k}))|^{3}\ dt\right)^{1/3}$
$\times\left(\int_{T}^{2T}|\mathcal{N}_{T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}(\beta_{k}-1))\mathcal{N}_{T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}\beta_{k})|^{6}\
dt\right)^{1/3}$

By the same reasoning used in the proof of Proposition 4.7, one may show the
final integral is $\ll T(\log T)^{O(1)}$. By the main theorem in [8], the
first integral is also $\ll T(\log T)^{O(1)}$. The claim now follows by using
lemma 2.7 of [8] to bound $\text{meas }\mathcal{B}_{A,1}.$ ∎

We now have all the necessary tools to show

######  Proposition 4.11.

Assuming the Riemann hypothesis, if $\delta$ is sufficiently small in terms of
$\bm{\beta}$

$|\mathcal{J}_{2}|\leq|\mathcal{J}_{1}|/2.$

###### Proof.

The preceding calculations in tandem with Proposition 4.6 imply that for $n>1$

$\displaystyle\int_{\mathcal{B}_{A,n}}\prod_{k\leq m}|\zeta(\tfrac{1}{2}$
$\displaystyle+i(t+\alpha_{k}))||\mathcal{N}_{T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}(\beta_{k}-1))\mathcal{N}_{T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}\beta_{k})|^{2}w(t/T)\
dt$
$\displaystyle\ll\exp\left(1/c_{n-1}-\log(1/c_{n})/40c_{n}+O\left((\log_{2}T)^{2}e^{-n}+L-n\right)\right)|\mathcal{J}_{1}|.$

Therefore by summing over all $A\subsetneq[m]$ and applying Proposition 4.10
we see that

$|\mathcal{J}_{2}|\ll|\mathcal{J}_{1}|\sum_{2\leq n\leq
L}\exp\left(1/c_{n-1}-\log(1/c_{n})/40c_{n}+O\left((\log_{2}T)^{2}e^{-n}+L-n\right)\right)+O_{A}\left(T(\log
T)^{-A}\right).$ (2)

The $T(\log T)^{-A}$ term is negligible. To bound the sum over $n$, we use the
definition of the $c_{n}$ to find

$\displaystyle\sum_{2\leq n\leq
L}\exp\left(1/c_{n-1}-\log(1/c_{n})/40c_{n}+O\left((\log_{2}T)^{2}e^{-n}+L-n\right)\right)$
$\displaystyle=\sum_{2\leq n\leq
L}\exp\left(e^{-n}(\log_{2}T)^{2}(O(1)+\tfrac{1}{40}n-\tfrac{1}{20}\log_{3}T)\right)+O(L-n)).$

Because $T_{L}\leq T^{\delta}$ it follows that $L\leq 2\log_{3}T+\log\delta$
so the sum is at most

$\displaystyle\sum_{2\leq n\leq
L}\exp\left(e^{-n}(\log_{2}T)^{2}(O(1)+\tfrac{1}{40}\log\delta)\right)+O(L-n)).$

By summing in reverse, we may bound this sum by

$\sum_{j\geq
0}\exp\left(e^{j}(O(1)+\tfrac{1}{40}\log\delta)+O(j)\right)\leq\sum_{j\geq
1}\exp\left((O(1)+\tfrac{1}{40}\log\delta)j\right)\leq\exp\left(O(1)+\tfrac{1}{40}\log\delta\right).$

Therefore by taking $\delta$ sufficiently small in terms of $\bm{\beta}$, we
can ensure that this sum times the implicit constant in (2) is at most $1/3$,
so the claim follows. ∎

Combining this with Proposition 4.6 completes the proof of Proposition 4.2 ∎

###  4.2. Proof of Proposition 4.3

Now only Proposition 4.3 is needed to complete the proof of Proposition 1.3.
To accomplish this, we will now use Lemma 2.1 with $X=T_{L}$ and the
definition of $\mathcal{G}_{m}$ to deduce

$\displaystyle\int_{\mathcal{G}_{m}}$ $\displaystyle\prod_{k\leq
m}|\zeta(\tfrac{1}{2}+i(t+\alpha_{k}))||\mathcal{N}_{T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}(\beta_{k}-1))\mathcal{N}_{T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}\beta_{k})|^{2}w(t/T)\
dt$ $\displaystyle\ll\int_{\mathcal{G}_{m}}\prod_{k=1}^{m}\prod_{j\leq
L}\exp\left(\textup{Re
}\mathcal{P}_{j,T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k}))\right)$
$\displaystyle\qquad\qquad\qquad\times|\mathcal{N}_{T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}(\beta_{k}-1))\mathcal{N}_{T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}\beta_{k})|^{2}\
dt$
$\displaystyle\ll\int_{T}^{2T}\prod_{k=1}^{m}|\mathcal{N}_{T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2})|^{2}$
$\displaystyle\qquad\qquad\qquad\times|\mathcal{N}_{T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}(\beta_{k}-1))\mathcal{N}_{T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}\beta_{k})|^{2}\
dt.$

So all that remains to is compute the mean square of the Dirichlet polynomial

$\prod_{k=1}^{m}\mathcal{N}_{T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2})\mathcal{N}_{T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}(\beta_{k}-1))\mathcal{N}_{T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\tfrac{1}{2}\beta_{k}).$

Proposition 4.7 with $n=L+1$ and lemma 2.4 imply that this mean value is

$\ll T\prod_{p\leq T_{L}}\left(1+\frac{1}{p}\sum_{1\leq j,k\leq
m}\frac{\beta_{j}\beta_{k}}{p^{i(\alpha_{j}-\alpha_{k})}}\right).$

This now concludes the proof of Proposition 4.3, and therefore also the proof
of Proposition 1.3.

∎

##  5\. Proof of Proposition 1.4

All that remains is to bound the quantities

$\displaystyle\mathcal{I}_{k}=\int_{\mathcal{G}_{m}}|\zeta(\tfrac{1}{2}+i(t+\alpha_{k}))|^{2m}\prod_{j\leq
L}$ $\displaystyle\exp\left(2(\beta_{k}-m)\textup{Re
}\mathcal{P}_{j,T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k}))\right)$
$\displaystyle\times\prod_{\begin{subarray}{c}\ell\leq m\\\ \ell\neq
k\end{subarray}}\prod_{j\leq L}\exp\left(2\beta_{\ell}\textup{Re
}\mathcal{P}_{j,T_{L}}(\tfrac{1}{2}+i(t+\alpha_{\ell}))\right)w(t/T)dt.$

This is very similar to the proof of Proposition 4.3. By applying lemma 2.1
with $X=T_{L}$ and lemma 2.2 it follows that

$\displaystyle\mathcal{I}_{k}\ll\int_{T}^{2T}|\mathcal{N}_{T_{L}}(\tfrac{1}{2}$
$\displaystyle+i(t+\alpha_{k});m)\mathcal{N}_{T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\beta_{k}-m)|^{2}$
$\displaystyle\times\prod_{\begin{subarray}{c}\ell\leq m\\\ \ell\neq
k\end{subarray}}|\mathcal{N}_{T_{L}}(\tfrac{1}{2}+i(t+\alpha_{\ell});\beta_{\ell})|^{2}\
dt.$

Before proceeding it may be helpful to review the definitions made preceding
Proposition 2.7. When $j>1$, the coefficients $a_{j}(n)$ of the Dirichlet
polynomial

$\mathcal{N}_{j,T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});m)\mathcal{N}_{j,T_{L}}(\tfrac{1}{2}+i(t+\alpha_{k});\beta_{k}-m)\prod_{\begin{subarray}{c}\ell\leq
m\\\ \ell\neq
k\end{subarray}}\mathcal{N}_{j,T_{L}}(\tfrac{1}{2}+i(t+\alpha_{\ell});\beta_{\ell})$

are given by the $m+1$ fold Dirichlet convolution of
$g_{T_{L}}(n;m)n^{-i\alpha_{k}}c_{j}(n)$ and
$g_{T_{L}}(n;\beta_{k}-m)n^{-i\alpha_{k}}c_{j}(n)$ with
$g_{T_{L}}(n;\beta_{\ell})n^{-i\alpha_{\ell}}c_{j}(n)$ for all $\ell\leq m$
not equal to $k$. When $j=1$ a similar formula holds with $f_{T_{L}}$ in place
of $g_{T_{L}}$. As before, one may replace $c_{j}(n)$ with $1_{p|n\Rightarrow
p\in(T_{j-1},T_{j}]}$ at a cost of $O(e^{-50\beta_{\ast}^{2}K_{j}})$ to obtain
multiplicative coefficients $a_{j}^{\prime}(n)$ which

$|a^{\prime}_{j}(p)|^{2}=\sum_{1\leq j,k\leq
m}\frac{\beta_{j}\beta_{k}}{p^{i(\alpha_{j}-\alpha_{k})}}+O\left(\frac{\log
p}{\log T_{L}}\right)$

and

$\sum_{r\geq 2}\frac{|a(^{\prime}p^{r})|^{2}}{p^{r}}\ll\frac{1}{p}$

for $T_{j-1}<p\leq T_{j}$. As before, this allows us to conclude that

$\mathcal{I}_{k}\ll T\prod_{p\leq T_{L}}\left(1+\sum_{1\leq j,k\leq
m}\frac{\beta_{j}\beta_{k}}{p^{i(\alpha_{j}-\alpha_{k})}}\frac{1}{p}\right).$

Proposition 1.4, and therefore Theorem 1.1, readily follows.

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