# Averages over the Gaussian Primes: Goldbach’s Conjecture and Improving
Estimates

Christina Giannitsi  School of Mathematics, Georgia Institute of Technology,
Atlanta GA 30332, USA<EMAIL_ADDRESS>,  Ben Krause  Department
of Mathematics, King’s College, London, WC2R 2LS, UK<EMAIL_ADDRESS>,
Michael T. Lacey  School of Mathematics, Georgia Institute of Technology,
Atlanta GA 30332, USA<EMAIL_ADDRESS>,  Hamed Mousavi  School of
Mathematics, Georgia Institute of Technology, Atlanta GA 30332, USA
<EMAIL_ADDRESS>and  Yaghoub Rahimi  School of Mathematics, Georgia
Institute of Technology, Atlanta GA 30332, USA<EMAIL_ADDRESS>
###### Abstract.

We prove versions of Goldbach conjectures for Gaussian Primes in arbitrary
sectors. Fix an interval $\omega\subset\mathbb{T}$. There is an integer
$N_{\omega}$, so that every odd integer $n$ with $\arg(n)\in\omega$ and
$N(n)>N_{\omega}$, is a sum of three Gaussian primes $n=p_{1}+p_{2}+p_{3}$,
with $\arg(p_{j})\in\omega$, for $j=1,2,3$. A density version of the binary
Goldbach conjecture is proved. Both follow from a High/Low decomposition of
the Fourier transform of averages over Gaussian primes, defined as follows.
Let $\Lambda(n)$ be the Von Mangoldt function for the Gaussian integers and
consider the norm function $N:\mathbb{Z}[i]\rightarrow\mathbb{Z}^{+}$,
$\alpha+i\beta\mapsto\alpha^{2}+\beta^{2}$. Define the averages

$A_{N}f(x)=\frac{1}{N}\sum_{N(n)<N}\Lambda(n)f(x-n).$

Our decomposition also proves the $\ell^{p}$ improving estimate

$\lVert A_{N}f\rVert_{\ell^{p^{\prime}}}\ll N^{1/p^{\prime}-1/p}\lVert
f\rVert_{\ell^{p}},\qquad 1<p\leq 2.,$

###### Contents

  1. 1 Introduction
  2. 2 Notation and Preliminaries
  3. 3 Inequalities involving Ramanujan Sums
  4. 4 The Vinogradov Inequality
  5. 5 Approximating the Kernel
  6. 6 Estimates for the High and Low Parts
  7. 7 Improving Inequalities
  8. 8 Goldbach Conjecture
    1. 8.1 The Binary Goldbach Conjecture
    2. 8.2 The Ternary Goldbach Conjecture

##  1\. Introduction

The principle goal of this paper is to establish a density version of the
strong Goldbach conjecture for Gaussian integers, restricted to sectors in the
complex plane.

Briefly, for integers $n\in\mathbb{Z}[i]$ we write $N(a+ib)=a^{2}+b^{2}$, and
if we express $n=\sqrt{N(n)}e^{i\theta}$, we set $\arg(n)=\theta$; the units
are $\pm 1$ and $\pm i$. The field $\mathbb{Z}[i]$ is a Euclidean domain,
inside of which an integer $p=a+ib$ is a prime if it has no integer factor $q$
with $N(q)<N(p)$. The Gaussian primes can take one of two forms. First if $a$
and $b$ are non-zero, then $N(p)=a^{2}+b^{2}$ is prime. Second, if $a$ or $b$
are zero, then it is of the form of a unit times a prime in $\mathbb{N}$
congruent to $3\mod 4.$

As in the case of $\mathbb{Z}$, the Goldbach conjecture over $\mathbb{Z}[i]$
requires a notion of evenness: a Gaussian integer $x=a+ib$ is _even_ if and
only if $N(x)=a^{2}+b^{2}$ is even. Evenness is equivalent to either condition
below.

  1. (1)

The integer $a+b$ is even.

  2. (2)

The Gaussian integer $1+i$ divides $x$.

An integer which is not even is _odd_.

The main result concerns Goldbach type results. We show that there are very
few even integers which are _not_ a sum of two primes. And, we do so even with
significant restrictions on the arguments of the integers involved. Similarly,
all sufficiently large odd integers are a sum of three primes. The only prior
results in this direction we could find in the literature correspond to the
entire complex plane. Below, we allow arbitrary sectors.

######  Theorem 1.1.

Fix an integer $B>10$ and interval $\omega\subset\mathbb{T}$. There exists an
$N_{\omega,B}>0$, such that for all integers $N>N_{\omega,B}$, the following
holds.

  1. (1)

Every odd integer $n$ with $N(n)>N_{\omega,B}$ and $\arg(n)\in\omega$ is a sum
of three Gaussian primes $n=p_{1}+p_{2}+p_{3}$, with $\arg(p_{j})\in\omega$
for $j=1,2,3$.

  2. (2)

We have $\lvert E_{2}(\omega,N)\rvert\ll\frac{N}{\log(N)^{B}}$, where
$E_{2}(\omega,N)$ is the set of _even_ Gaussian integers with $N(n)<N$ and
$\arg(n)\in\omega$, which _cannot_ be represented as sum of two Gaussian
primes $n=p_{1}+p_{2}$, with $\arg(p_{j})\in\omega$ for $j=1,2$.

We can further estimate the number of representations in both the binary and
ternary case.

Our proof derives from a detailed study of the analytic properties of averages
over the Gaussian primes. In particular, let
$f:\mathbb{Z}[i]\rightarrow\mathbb{C}$ be an arithmetic function, and for
$x\in\mathbb{Z}[i]$ define

$\displaystyle A_{N}f(x)=\frac{1}{N}\sum_{N(n)<N}\Lambda(n)f(x-n)$

where the von Mangoldt function on $\mathbb{Z}[i]$ is defined by

$\displaystyle\Lambda(n)=\begin{cases}\log(N(\rho))&\text{ if
}n=\rho^{\alpha}\text{ and }\rho\in\mathbb{Z}[i]\text{ is irreducible}\\\
0&\text{otherwise.}\end{cases}$

We prove an improving estimate for the averages above: The averages of
$\ell^{1}(\mathbb{Z}^{2})$ functions are nearly bounded.

######  Theorem 1.2.

For all $N$, and $1<p\leq 2$, we have,

$\displaystyle\langle A_{N}f,g\rangle\ll N^{1-2/p}\lVert f\rVert_{p}\lVert
g\rVert_{p}.$

Equivalently, for all $1<p\leq 2$, whenever $f$ is supported on a cube, $Q$,
of side length $\sqrt{N}$,

$\displaystyle\frac{\|A_{N}f\|_{p^{\prime}}}{|Q|^{1/p^{\prime}}}\ll\frac{\|f\|_{p}}{|Q|^{1/p}}$

To prove this, we need to develop many expected results, including Ramanujan
type identities, and the Vinogradov type estimates for the Fourier transform
of $A_{N}$. Not being able to identify clear cut references for many of these
estimates, we develop them below. We then develop a High/Low decomposition of
the Fourier transform of $A_{N}$. A particular innovation is our approach to
the major arc estimate in Lemma 5.4, which are typically proved by Abel
summation. But that method is is poorly adapted to the question at hand. We
develop a different method. Also noteworthy is that the Low term is defined in
terms of smooth numbers, a technique used in [210110401L]. Smoothness
facilitates the proof of Lemma 8.20. With this decomposition in hand, the
deduction of the Theorems is relatively straight forward.

Indeed, we develop this for the more specialized averages, over sectors of the
complex plane. For an interval $\omega\subset\mathbb{T}$, extend the
definition of the averages to this setting.

(1.3) $\displaystyle A_{N}^{\omega}f(x)=\frac{2\pi}{\lvert\omega\rvert
N}\sum_{\begin{subarray}{c}N(n)<N\\\
\arg(n)\in\omega\end{subarray}}\Lambda(n)f(x-n).$

To address the binary Goldbach question, note that

(1.4) $A_{N}^{\omega}\ast
A_{N}^{\omega}(m)=\frac{4\pi^{2}}{\lvert\omega\rvert^{2}N^{2}}\sum_{\begin{subarray}{c}N(n_{1}),N(n_{2})<N\\\
\arg(n_{1}),\arg(n_{2})\in\omega\\\
n_{1}+n_{2}=m\end{subarray}}\Lambda(n_{1})\Lambda(n_{2})$

The High/Low decomposition for the $A_{N}^{\omega}$ can be leveraged to study
the sum above. This is the path we follow to prove Theorem 1.1 in §8.

Our motivations come from number theory and analysis. In this setting, the
binary (and higher order) Goldbach conjectures has been addressed in the
number field setting. Mitsui [MR136590]*§11,12 addresses the binary and higher
order cases of Goldbach’s Conjecture in an arbitrary number field. Mitsui
finds that all sufficiently large totally positive odd integers are the sum of
odd totally positive primes. He does not address the sector case. Much earlier
work of Rademacher [MR3069422, MR3069434]. Rademacher [MR3069422] studied
representations of totally positive integers in a class of quadratic
extensions of $\mathbb{Z}$. These papers assume the Generalized Riemann
Hypothesis.

Holben and Jordan [MR238760] raised conjectures in the spirit of our Theorem.
Their Conjecture F states

######  Conjecture 1.5.

Each even Gaussian integer $n$ with $N(n)>10$ is the sum of two primes
$n=p_{1}+p_{2}$, with $\arg(n)-\arg(p_{j})\leq\pi/6$, for $j=1,2$.

As far as we know, there is no prior result in this direction, for neither the
binary nor the ternary version of the Goldbach conjecture. But, we also note
that our Theorem 1.1 provides a density version of a much stronger result. For
all $\delta>0$, most even Gaussian integers $n$ with $N(n)>N_{\delta}$, are
the sum of two primes $n=p_{1}+p_{2}$, with
$\lvert\arg(n)-\arg(p_{j})\rvert\leq\delta$, for $j=1,2$. Indeed, partition
the unit circle into intervals of length $\delta$, and apply our Theorem to
each interval.

The study of metric properties of the uniform distribution of $p\alpha$, for
irrational $\alpha\in\mathbb{C}$, and Gaussian prime $p$, is well developed by
Harman [MR2331072]*Chapters 11, [MR4045111], with effective results even small
sectors. See the extensions to certain quadratic number fields by Baier and
Technau [MR4206429]. The latter paper also addresses metric questions along
lines in the complex plane.

On the analytical side, our improving estimate above is part of Discrete
Harmonic Analysis, a subject invented by Bourgain [MR1019960]. The recent
textbook of one of us [MR4512201] serves as a comprehensive summary of the
subject. The study of the averages over the primes has been extensively
studied, beginning with the work of Bourgain-Wierdl [MR995574, BP1], and
continued by several [MR4434278, MR3370012, MR3299842, MR3646766, MR4072599].

##  2\. Notation and Preliminaries

Throughout the paper we are using $\|\beta\|$ to denote the distance of
$\beta\in\mathbb{C}$ to the closest point $n\in\mathbb{Z}[i]$, and for
$a,b,c\in\mathbb{Z}[i]$ we write $(a,b)=c$ to indicate that $c$ is the
greatest (in norm) common divisor of $a$ and $b$ up to a unitary element.

We will use the $\ell^{\infty}$ balls
$B_{\infty}(r)=\\{x=a+bi\in\mathbb{Z}[i]\,:\,-r\leq a,b<r\\}$ and
$B_{\infty}(c,r)=B_{\infty}(r)+c$. Notice that there is a small departure from
the traditional notation of a ball with respect to the $\infty$-norm, in the
sense that we are only including the lowest endpoint, however this variation
is useful as it allows us to obtain a tessellation of the complex plane. It is
obvious that translating and rotating the grid does not affect the
tessellation. For a $q=|q|e^{i\theta_{q}}\in\mathbb{Z}[i]$, we are interested
in the grid formed by the squares

$B_{q}:=e^{i\theta_{q}}B_{\infty}\left((1+i)\frac{|q|}{2},\frac{|q|}{2}\right)$

where we have rotated the squares by the argument of $q$, so that their sides
are parallel and orthogonal to $q$ respectively; see Figures 2 and 2. This
particular decomposition of lattice points coincides with all the _unique_
remainders modulo $q$ in $\mathbb{Z}[i]$ and simplifies our calculations of
complex exponentials. Indeed, it is known that for $q\in\mathbb{Z}[i]$, there
are $N(q)$ distinct remainders modulo $q$, which agrees with the number of
points inside $B_{q}$. It is straightforward to verify that any two points in
$B_{q}$ cannot be equivalent modulo $q$, which then proves our claim. We are
also going to need the following, more geometric, description of our boxes:

(2.1) $B_{q}=\\{r\in\mathbb{Z}[i]\mid 0\leq\langle r,q\rangle<N(q)\ \
\text{and}\ \ 0\leq\langle r,iq\rangle<N(q)\\}.$

$q$$iq$ Figure 1.
The lattice points in the box $B_{q}$, for $q=2+i$. There are exactly $N(q)=5$
points inside the box.

$q$$iq$$\frac{iq}{2}$$\frac{q}{2}$ Figure 2.
The box $B_{q}$ for $q=4+2i$. Note that $4+2i=2(2+i)$, hence $B_{4+2i}$ is
partitioned by $2^{2}$ Gaussian translations of $B_{2+i}$.

Let $e(x):=e^{2\pi ix}$. For a function $f:\mathbb{Z}[i]\to\mathbb{C}$, we can
define the discrete Fourier Transform over the box $B_{q}$ as

(2.2) $\displaystyle\mathcal{F}_{q}f\,(x):=\sum_{n\in
B_{q}}f(n)e\bigl{(}-\langle x,\frac{n}{q}\rangle\bigr{)},$

and the corresponding inverse discrete Fourier transform as

$\displaystyle\mathcal{F}_{q}^{-1}f\,(n):=\frac{1}{N(q)}\sum_{x\in
B_{q}}f(x)e\bigl{(}\langle n,\frac{x}{\bar{q}}\rangle\bigr{)}.$

The Euler totient function $\phi$ for Gaussian integers counts the number of
points in $B_{q}$ that are coprime to $z$. It is equal to

$\displaystyle\phi(z)$ $\displaystyle=N(z)\prod_{\begin{subarray}{c}p\divides
z\\\ p:\text{ prime in
}\mathbb{Z}[i]\end{subarray}}\left(1-\frac{1}{N(p)}\right).$

Clearly, $\phi(z)=\phi(\bar{z})$. Thus, $\lvert\mathbb{A}_{q}\rvert=\phi(q)$,
where

(2.3) $\mathbb{A}_{q}:=\\{r\in B_{q}\,:\,(r,q)=1\\}.$

We also need the arithmetic function $r_{2}(n)$, which is called _Sum of
Squares function_ , which counts the number of representations of an integer
$n$ as sum of two squares. Obviously $r_{2}(n)$ is the number of
$m\in\mathbb{Z}[i]$ such that $N(m)=n$. It is known that $r_{2}(n)=O(1)$ in an
average sense, namely

(2.4) $\displaystyle\sum_{n<N}r_{2}(n)$ $\displaystyle\simeq N$

Define the Möbius function as follows.

(2.5) $\displaystyle\mu(n)=\begin{cases}(-1)^{k}&\text{ if
}n=\epsilon\rho_{1}\rho_{2}\cdots\rho_{k}\\\ 1&\text{ if }n=\epsilon\\\
0&\text{otherwise.}\end{cases}$

where $\epsilon\in\\{\pm 1,\pm i\\}$ are units and $\rho_{i}$ are distinct
prime elements of $\mathbb{Z}[i]$.

We define the classical form of the average over a sector. For an interval
$\omega\subset\mathbb{T}$, set

(2.6) $M_{N}^{\omega}=M_{N}=\frac{\lvert\omega\rvert}{2\pi
N}\sum_{\begin{subarray}{c}N(n)<N\\\
\arg(n)\in\omega\end{subarray}}\delta_{n},$

where $\delta_{n}$ is a Dirac measure at $n$.

We emphasize that _we do not attempt to track constants that depend upon
$\omega$._ Frequently, we assume that we will assume that $N$ is large enough,
once $\omega$ is fixed.

######  Lemma 2.7.

Fix an interval $\omega\subset\mathbb{T}$, $0<|\omega|\leq 2\pi$. For
$N(\beta)<1$, and integers $N>N_{\omega}$,

(2.8) $\displaystyle\widehat{M}_{N}^{\omega}(\beta)$
$\displaystyle\coloneqq\frac{\lvert\omega\rvert}{2\pi
N}\sum_{\begin{subarray}{c}N(n)<N\\\ \arg(n)\in\omega\end{subarray}}e(-\langle
n,\beta\rangle)$ (2.9) $\displaystyle\ll_{\omega}\min\left(1,(N\cdot
N(\beta))^{-\frac{3}{4}}\right).+\frac{1}{\sqrt{N}}.$

The implied constant only depends on $\omega$.

###### Proof.

Let $I_{n}=B_{\infty}(n,1/2)$. For $n=0$, we have

(2.10) $\int_{I_{0}}e(-\langle
x,\beta\rangle)\;dx=\prod_{j=1}^{2}\frac{\sin(\pi\beta_{j})}{\pi\beta_{j}},\;\;\;\beta=(\beta_{1},\beta_{2}).$

This function is bounded above, and away from zero, since $N(\beta)<1$.
Suppress the dependence on $\omega$ in the notation. Modify the definition of

$S_{N}(\beta):=N\widehat{M}_{N}(\beta)$

to

(2.11) $\displaystyle T_{N}(\beta)$
$\displaystyle\coloneqq\sum_{\begin{subarray}{c}n\colon N(n)<N\\\
\arg(n)\in\omega\end{subarray}}\int_{I_{n}}e(-\langle x,\beta\rangle)\;dx$
(2.12)
$\displaystyle=\prod_{j=1}^{2}\frac{\sin(\beta_{j}/2)}{\beta_{j}/2}\sum_{\begin{subarray}{c}n\colon
N(n)<N\\\ \arg(n)\in\omega\end{subarray}}e(-\langle n,\beta\rangle)$ (2.13)
$\displaystyle=S_{N}(\beta)\prod_{j=1}^{2}\frac{\sin(\beta_{j}/2)}{\beta_{j}/2}.$

We see that it suffices to estimate $T_{N}(\beta)$.

The symmetric difference between the sector defined by $\omega$, and the area
of integration defining $T_{N}(\beta)$ is the set

(2.14) $\bigcup_{\begin{subarray}{c}n\colon N(n)<N\\\
\arg(n)\in\omega\end{subarray}}I_{n}\,\triangle\\{n\colon
N(n)<N,\arg(n)\in\omega\\}.$

It has measure at most $\ll\omega\sqrt{N}$, as the above set is supported in
an $O(1)$ neighborhood of the boundary $\\{n:N(n)<N,\ n\in\omega\\}$, which
has length $\omega\sqrt{N}$. Thus,

(2.15) $\displaystyle T_{N}(\beta)$
$\displaystyle\ll\int_{\begin{subarray}{c}N(x)<N\\\
\arg(x)\in\omega\end{subarray}}e(-\langle x,\beta\rangle)\;dx+\sqrt{N}$

By partitioning $\omega$ into smaller arcs and arguing as in [MONTGOMERY, Page
111-112], which addresses the case where $\omega=\mathbb{T}$, we may bound the
integral

$\frac{1}{N}\int_{\begin{subarray}{c}N(x)<N\\\
\arg(x)\in\omega\end{subarray}}e(-\langle
x,\beta\rangle)\;dx\ll_{\omega}\big{(}\frac{1}{N\cdot N(\beta)}\big{)}^{3/4}.$

∎

##  3\. Inequalities involving Ramanujan Sums

In this section, we prove two dimensional analogues of estimates and
identities already known for one dimensional Ramanujan sums, like the Cohen
Identity. This section is crucial to our High-Low decomposition. Some of these
properties are known, but we include details for completeness. We start with
standard facts about Fourier transform on $B_{q}$.

######  Lemma 3.1.

Consider the set $B_{q}$ and $n\in\mathbb{Z}[i]$. We have:

(3.2) $\displaystyle\sum_{r\in{B_{q}}}e(\langle
r,\frac{n}{\bar{q}}\rangle)=\begin{cases}N(q)&\textup{ if }\bar{q}\divides
n\\\ 0&\textup{ Otherwise.}\end{cases}$

Below, we divide by $\bar{d}$, where $d$ is a divisor of $q$.

######  Corollary 3.3.

For $d\divides q$ we have

(3.4) $\displaystyle\sum_{r\in B_{q}}e\left(\langle
r,\frac{n}{\bar{d}}\rangle\right)=\begin{cases}N(q)&\textup{ if
}\bar{d}\divides n\\\ 0&\textup{ Otherwise.}\end{cases}$

Another consequence of Lemma 3.1 is a form of Parseval’s Identity as stated
below.

######  Proposition 3.5.

For a function $f$ defined on $\mathbb{Z}[i]$ the Discrete Fourier Transform
$\mathcal{F}_{q}$ defined in (2.2) satisfies

$\displaystyle\sum_{n\in B_{q}}|f(n)|^{2}=\frac{1}{N(q)}\sum_{x\in
B_{\bar{q}}}|\mathcal{F}_{q}f(x)|^{2}.$

Above, on the left we have $B_{q}$, and on the right
$B_{\bar{q}}=\\{\bar{n}:n\in B_{q}\\}=\overline{B_{q}}$.

The analog of Ramanujan’s sum is

(3.6) $\displaystyle\tau_{q}(x)$
$\displaystyle:=\sum_{n\in\mathbb{A}_{q}}e(\langle x,\tfrac{n}{q}\rangle).$

######  Lemma 3.7.

Let $r\in\mathbb{Z}[i]$.

$\displaystyle\mathbf{1}_{\gcd(\bar{r},q)=1}=\frac{1}{N(q)}\sum_{k\in
B_{q}}e\left(\langle-r,\frac{k}{q}\rangle\right)\tau_{{\bar{q}}}(k).$

###### Proof.

Note from the definition in (3.6), that

(3.8)
$\tau_{{\bar{q}}}(k)=\bigl{\langle}\mathbf{1}_{\gcd(\bar{r},q)=1},e(\langle\cdot,k/\bar{q})\bigr{\rangle}.$

Then, the conclusion follows from general facts about orthogonal bases.

∎

######  Lemma 3.9.

For $x\in\mathbb{Z}[i]$ we have

(3.10) $\displaystyle\tau_{q}(x)$
$\displaystyle=\sum_{d\divides(q,\bar{x})}\mu(q/d)N(d).$

In particular, $\tau_{q}(q)=\phi(q)$ and $\tau_{q}(1)=\mu(q)$. In addition, if
$(q,\bar{x})=1$ then $\tau_{q}(x)=\mu(q)$.

###### Proof.

Note that thanks to Corollary 3.3 we have that if $d\divides q$ then

$\displaystyle\sum_{\begin{subarray}{c}k\in B_{q}\\\
d|k\end{subarray}}e\left(\langle x,\frac{k}{q}\rangle\right)$
$\displaystyle=\sum_{k\in B_{q}}e\left(\langle
x,\frac{k}{q}\rangle\right)\mathbf{1}_{d|k}$ $\displaystyle=\sum_{k\in
B_{q}}e\left(\langle x,\frac{k}{q}\rangle\right)\frac{1}{N(q)}\sum_{r\in
B_{q}}e\left(\langle-r,\frac{k}{d}\rangle\right)$
$\displaystyle=\frac{1}{N(q)}\sum_{r\in B_{q}}\sum_{k\in
B_{q}}e\left(\langle\frac{x-r\frac{\bar{q}}{\bar{d}}}{\bar{q}},k\rangle\right)$
$\displaystyle=\sum_{r\in
B_{q}}\mathbf{1}_{{\bar{q}}|x-r\frac{{\bar{q}}}{{\bar{d}}}}$
$\displaystyle=N(\frac{q}{d})\mathbf{1}_{\frac{q}{d}\divides\bar{x}}.$

The last equality comes from counting the number of $r$’s in $B_{q}$ for which
the indicator is non-zero. Let $\bar{x}=\frac{q}{d}x^{\prime}$. Then
$\bar{x}-\bar{r}\frac{q}{d}\equiv 0\mod
q\,\Leftrightarrow\,\frac{q}{d}(x^{\prime}-\bar{r})\equiv 0\mod\bar{q}$. This
means that $x^{\prime}\equiv\bar{r}\mod d$, which means that there exists a
unique such $\bar{r}\in B_{d}$ and $N(q/d)$ of them in $B_{q}$.

Hence, and with the assistance of the Inclusion-Exclusion principle, we
observe that

(3.11) $\displaystyle\tau_{q}(x)$ $\displaystyle=\sum_{\begin{subarray}{c}k\in
B_{q}\\\ (k,q)=1\end{subarray}}e(\langle x,\tfrac{k}{q}\rangle)$ (3.12)
$\displaystyle=\sum_{d\divides q}\mu(d)\sum_{\begin{subarray}{c}k\in B_{q}\\\
d|k\end{subarray}}e(\langle x,\tfrac{k}{q}\rangle)$ (3.13)
$\displaystyle=\sum_{d\divides
q}\mu(d)N(\tfrac{q}{d})\mathbf{1}_{\frac{q}{d}\divides\bar{x}}$ (3.14)
$\displaystyle=\sum_{d\divides q}\mu(q/d)N(d)\mathbf{1}_{d\divides\bar{x}}$
(3.15) $\displaystyle=\sum_{d\divides(q,\bar{x})}\mu(q/d)N(d).$

∎

######  Corollary 3.16.

The function $\tau_{q}(x)$ is multiplicative and we have that

(3.17) $\displaystyle\tau_{q}(x)$
$\displaystyle=\mu\Big{(}\frac{q}{(q,\bar{x})}\Big{)}\frac{\phi(q)}{\phi(\frac{q}{(q,\bar{x})})}.$

###### Proof.

Let $d=(q,\bar{x})$. A direct consequence of the Chinese Remainder Theorem
gives us

$\displaystyle\tau_{q}(x)$ $\displaystyle=\sum_{r\in\mathbb{A}_{q}}e(\langle
r,\tfrac{x}{\bar{q}}\rangle)$
$\displaystyle=\sum_{m\in\mathbb{A}_{q/d}}\sum_{k\in\mathbb{A}_{d}}e(\langle
k\frac{q}{d}+md,\frac{x}{\bar{q}}\rangle)$
$\displaystyle=\sum_{m\in\mathbb{A}_{q/d}}e(\langle
m,\tfrac{x}{\bar{q}/\bar{d}}\rangle)\sum_{k\in\mathbb{A}_{d}}e(\langle
k,\tfrac{x}{\bar{d}}\rangle)$ $\displaystyle=\tau_{q/d}(x)\,\phi(d)$

and the result follows immediately from the previous lemma and the
multiplicative properties of $\phi$. ∎

With these identities in mind, we prove Cohen’s identity.

######  Lemma 3.18.

For $x\in\mathbb{Z}[i]$, the following identity holds:

$\displaystyle\sum_{r\in\mathbb{A}_{\bar{q}}}\tau_{q}(x+r)=\mu(q)\tau_{q}(x).$

###### Proof.

We have

$\displaystyle\sum_{r\in\mathbb{A}_{\bar{q}}}\tau_{q}(x+r)$
$\displaystyle=\sum_{r\in B_{\bar{q}}}1_{(r,\bar{q})=1}\tau_{q}(x+r)$
$\displaystyle=\sum_{r\in B_{\bar{q}}}\frac{1}{N(q)}\sum_{k\in
B_{q}}e\left(\langle-r,\frac{k}{q}\rangle\right)\,\tau_{\bar{q}}(k)\,\tau_{q}(x+r)$
$\displaystyle=\frac{1}{N(q)}\sum_{k\in B_{q}}\tau_{\bar{q}}(k)\sum_{r\in
B_{\bar{q}}}e\left(\langle-r,\frac{k}{q}\rangle\right)\tau_{q}(x+r)$

Now we compute the inner sum.

$\displaystyle\sum_{r\in
B_{\bar{q}}}e\left(\langle-r,\frac{k}{q}\rangle\right)\tau_{q}(x+r)$
$\displaystyle=\sum_{r\in
B_{\bar{q}}}e\left(\langle-r,\frac{k}{q}\rangle\right)\sum_{n\in\mathbb{A}_{q}}e\left(\langle\frac{n}{q},x+r\rangle\right)$
$\displaystyle=\sum_{n\in\mathbb{A}_{q}}e\left(\langle\frac{n}{q},x\rangle\right)\sum_{r\in
B_{\bar{q}}}e\left(\langle\frac{n-k}{q},r\rangle\right)$
$\displaystyle=\sum_{n\in\mathbb{A}_{q}}e\left(\langle\frac{n}{q},x\rangle\right)N(q)\mathbf{1}_{q\divides
n-k}$
$\displaystyle=N(q)e\left(\langle\frac{k}{q},x\rangle\right)\mathbf{1}_{(k,q)=1}.$

In addition, note that we have if $(k,q)=1$ then
$\tau_{\bar{q}}(k)=\tau_{\bar{q}}(1)$. Indeed, this means that $\bar{k}$ is a
unitary element modulo $\bar{q}$ so the set $\\{\bar{k}s\,:\,s\in
B_{\bar{q}}\\}$ is just a rearrangement of $B_{\bar{q}}$, and so

(3.19) $\displaystyle\tau_{\bar{q}}(k)$
$\displaystyle=\sum_{s\in\mathbb{A}_{\bar{q}}}e\left(\langle
k,\frac{s}{\bar{q}}\rangle\right)$ (3.20)
$\displaystyle=\sum_{s\in\mathbb{A}_{\bar{q}}}e\left(\langle
1,\frac{\bar{k}s}{\bar{q}}\rangle\right)$ (3.21)
$\displaystyle=\sum_{s\in\mathbb{A}_{\bar{q}}}e\left(\langle
1,\frac{s}{\bar{q}}\rangle\right)$ (3.22) $\displaystyle=\tau_{\bar{q}}(1).$

Therefore we have that

$\displaystyle\sum_{r\in\mathbb{A}_{\bar{q}}}\tau_{q}(x+r)$
$\displaystyle=\sum_{k\in
B_{q}}\tau_{\bar{q}}(k)e\left(\langle\frac{k}{q},x\rangle\right)\mathbf{1}_{(k,q)=1}$
$\displaystyle=\tau_{\bar{q}}(1)\sum_{k\in
B_{q}}e\left(\langle\frac{k}{q},x\rangle\right)\mathbf{1}_{(k,q)=1}$
$\displaystyle=\tau_{\bar{q}}(1)\tau_{q}(x).$

Finally, $\tau_{\bar{q}}(1)=\mu(q)$. ∎

######  Corollary 3.23.

For $x\in\mathbb{Z}[i]$ we have

(3.24)
$\displaystyle\bigg{|}\sum_{r\in\mathbb{A}_{\bar{q}}}\tau_{q}(x+r)\bigg{|}$
$\displaystyle\ll N(\gcd(x,q))\,N(q)^{\epsilon}.$

###### Proof.

By Lemma 3.18, it suffices to find a bound for $\tau_{q}(x)$. The inequality
follows immediately from Lemma 3.9 and Corollary 3.16

$\displaystyle|\tau_{q}(x)|\ll
N\big{(}\gcd(\bar{x},q)\big{)}{N(q)}^{\epsilon}.$

∎

Now we are ready to prove the Gaussian version of an inequality, originally
due to Bourgain, involving the Ramanujan sums. It assures us that while the
summands above can, in general, be as big as $q$, this happens infrequently as
$x$ varies.

######  Proposition 3.25.

For every $B,k>2$, integers $N>N_{k,B}$ and $Q\leq(\log N)^{B}$ we have

(3.26) $\displaystyle\frac{1}{N}\sum_{N(x)<N}\bigg{[}\sum_{q\colon
q<Q}\lvert\tau_{q}(x+r)\rvert\bigg{]}^{k}\ll Q^{k+\epsilon}.$

###### Proof.

The proof of this proposition is inspired from a proof due to Bourgain
[MR1209299]. In view of (3.24), it suffices to show that

(3.27)
$\frac{1}{N}\sum_{N(x)<N}\biggl{(}\sum_{N(q)<Q}N(\gcd(x,q))\biggr{)}^{k}\ll
Q^{k+\epsilon}.$

Expanding out the $k$th power, we have

$\displaystyle\sum_{N(q_{1}),N(q_{2}),\cdots,N(q_{k})<Q}\frac{1}{N}\sum_{N(x)<N}\prod_{i=1}^{k}N(\gcd(x,q_{i}))$

The function $x\to\prod_{i=1}^{k}N(\gcd(x,q_{i}))$ has period of
$\mathfrak{L}:=N(\operatorname{lcm}(q_{1},\cdots,q_{k}))$. We are free to
restrict attention to the case in which $N$ is much larger than
$\mathfrak{L}\leq Q^{k}$. Thus, the bound is reduced to establishing that

$\displaystyle\sum_{N(q_{1}),N(q_{2}),\cdots,N(q_{k})<Q}\frac{1}{\mathfrak{L}}\sum_{N(x)<\mathfrak{L}}\prod_{i=1}^{k}N(\gcd(x,q_{i}))\ll
Q^{k+\epsilon}.$

We will establish a bound for the inner sum, namely

(3.28) $\sum_{N(x)<\mathfrak{L}}\prod_{i=1}^{k}N(\gcd(x,q_{i}))\ll Q^{k}.$

The sum above is multiplicative in the variables $q_{1},\cdots,q_{k}$. So,
specializing to the case of $q_{i}=\rho^{r_{i}}$, we can estimate as follows.
Assume that $r=\max r_{i}$. Then $\mathfrak{L}=N(\rho^{r}).$ Let
$\rho^{a}\|x$. There are $\mathfrak{L}N(\rho^{-a})=N(\rho^{r-a})$ such $x$
with $N(x)<\mathfrak{L}$. Also
$N(\gcd(x,\rho^{r_{i}}))=N(\rho^{\min(r_{i},a)})$. So

(3.29)
$\displaystyle\sum_{N(x)<\mathfrak{L}}\prod_{i=1}^{k}N(\gcd(x,\rho^{r_{i}}))$
$\displaystyle\ll N(\rho^{r-a})\prod_{i=1}^{k}N(\rho^{\min(r_{i},a)})$ (3.30)
$\displaystyle\ll\prod_{i=1}^{k}N(\rho^{r_{i}}).$

Above, we can assume that $r=\max r_{i}=r_{1}$, and then estimate

(3.31) $N(\rho^{\min(r_{i},a)})\leq\begin{cases}N(\rho^{a})&i=1\\\
N(\rho^{r_{i}})&2\leq i\leq k\end{cases}.$

Using (3.29), the multiplicative property implies that

(3.32)
$\sum_{N(x)<\mathfrak{L}}\prod_{i=1}^{k}N(\gcd(x,q_{i}))\ll\prod_{i=1}^{k}N(q_{i})\leq
Q^{k}.$

Thus, to conclude the estimate, we note that

$\displaystyle\sum_{N(q_{1}),N(q_{2}),\cdots,N(q_{k})<Q}\frac{1}{\mathfrak{L}}\ll
Q^{\epsilon}$

This completes the proof. ∎

##  4\. The Vinogradov Inequality

We prove an analogue of the Vinogradov inequality for $\Lambda$ on
$\mathbb{Z}[i]$. That is, we control the Fourier transform of averages of the
von Mangoldt function over a sector (1.3), at a point which is close to a
rational with relatively large denominator.

######  Theorem 4.1.

Let $\alpha\in\mathbb{C}$ and $a,q\in\mathbb{Z}[i]$ with
$0<N(a)<N(q)<\sqrt{N}$ $\gcd(a,q)=1$ and
$N(\alpha-\frac{a}{q})<\frac{1}{N(q)^{2}}$. Fix $\omega\subset\mathbb{T}$. For
$N>N_{\omega}$, we have

$\displaystyle\sum_{\begin{subarray}{c}n\colon N(n)<N\\\
\arg(n)\in\omega\end{subarray}}\Lambda(n)e(\langle
n,\alpha\rangle)\ll\frac{N\log^{2}(N)}{N(q)^{1/2}}+N^{99/100}.$

The remainder of the section is devoted to the proof. We collect some
background material. The ring $\mathbb{Z}[i]$ is an Euclidean domain, which
means that for each $a,b\in\mathbb{Z}[i]$, there exists a unique pair of $q,r$
such that

$a=bq+r\textup{ where }{r\in B_{b}}$

Note that if $N(r)=\frac{N(q)}{2}$, then $r$ is on the edges of square
$\langle b,ib\rangle$, we pick the point with positive phase on the line that
contains origin, i.e. $0\leq\arg(r)<\pi$.

Similar to the Farey dissection, we can find an approximation of $\xi$ by
rational points in $\mathbb{Q}[i]$. In fact, we can say that for every
$N(\xi)<1$, there exists $\frac{a}{q}\in\mathbb{Q}[i]$ such that

$N(\xi-\tfrac{a}{q})<\frac{c(q)}{N(q)^{2}}$

where $c(q)\approx 1$ is a constant that depends on $q$ but is uniformly
bounded above and below. Thus, the assumption of the Theorem says that
$\alpha$ is well approximated by $a/q$, and we apply the result above with
$N(q)$ large.

Next, we recall the Prime Number Theorem for Gaussian integers. For any
choices of $\omega\subset\mathbb{T}$, integers $A$, and
$N(q)<\log^{\frac{A}{{2}}-1}N$,

(4.2) $\displaystyle\psi_{\Lambda}(x,q,r,\omega)$
$\displaystyle:=\sum_{\begin{subarray}{c}n\equiv r\mod q\\\ N(n)<x\\\
\arg(n)\in\omega\end{subarray}}\Lambda(n)=\frac{\lvert\omega\rvert}{2\pi}\frac{x}{\phi(q)}+O\left(\frac{x\,N(q)}{\log^{A}x}\right)$

the implied constants are absolute. The last two equations come from
[MR2061214]*Theorem 5.36 & following hint, which provides us with a uniform
property of the Prime Number Theorem.

The principal technical tool is this Lemma.

######  Lemma 4.3.

Let $N$ be a large integer and $m\in\mathbb{Z}[i]$ with $N(m)>N^{1/100}$ or
$m=0$. For large $T$, and $N(\alpha-\tfrac{a}{q})\leq\frac{1}{N(q)^{2}}$ where
$0<N(a)<N(q)$ and $(a,q)=1$

$\displaystyle B(T,N,\alpha,m)$ $\displaystyle:=\sum_{t\colon
0<N(t)<T}\min\left(\frac{N}{N(t)},\left(\frac{N}{N(t)}\right)^{\frac{1}{4}}\frac{1}{N(\|(\bar{t}+m)\alpha\|)^{\frac{3}{4}}}\right)$
$\displaystyle\ll\frac{N\log(T)}{N(q)}+N^{\frac{1}{4}}N(q)^{\frac{3}{4}}\log
N(q)+N(q)^{\frac{1}{4}}N^{\frac{1}{4}}T^{\frac{3}{4}}+N^{99/100}.$

###### Proof.

The proof proceeds by case analysis based on the values of $h$ and $r$
introduced here. Since $\mathbb{Z}[i]$ is a Euclidean domain, we know that
$t+\bar{m}=\bar{h}\bar{q}+\bar{r}$, with $r\in B_{q}$. Let
$\beta:=\alpha-\frac{a}{q}$. Then $N(\beta)<N(q)^{-2}$ and

$\|(\bar{t}+m)\alpha\|=\bigl{\lVert}hq\beta+r\beta+\tfrac{ra}{q}\bigr{\rVert}.$

So we can rewrite our sum in terms of $h$ and $r$ as follows.

$\displaystyle B(T,N,\alpha,m)=\sum_{N(h)<\frac{T}{N(q)}}\sum_{r\in
B_{q}}\min\left(\frac{N}{N(t)},\left(\frac{N}{N(t)}\right)^{\frac{1}{4}}\frac{1}{N(\|(\bar{t}+m)\alpha\|)^{\frac{3}{4}}}\right),$

where on the right we understand that $t=t_{h,r}$ is given by

(4.4) $t+\bar{m}=\bar{h}\bar{q}+\bar{r},$

Case 1: $\bar{t}+m=0$. Hence $\bar{t}=-m$, where $m$ is fixed. There is only
one term in the sum we are estimating. Since $N(t)>0$, we see that $m\neq 0$,
and so $N(t)>N^{1/100}$. Then we use the trivial bound, which gives the
contribution of at most $N^{99/100}$ and we see the inequality holds in the
case that $\bar{t}+m=0$.

Case 2: $h=0$, and $0<N(r)<N(q)/10$. By assumption
$N(r\beta)\leq(2N(q))^{-1}$, so $4N(\|(\bar{t}+m)\alpha\|)\geq
N(\|\frac{ra}{q}\|)-\frac{1}{2N(q)}$. Therefore

(4.5) $\displaystyle\sum_{0<N(r)\leq
N(q)/10}\left(\frac{N}{N(t)}\right)^{\frac{1}{4}}\frac{1}{N(\|(\bar{t}+m)\alpha\|)^{\frac{3}{4}}}$
$\displaystyle\ll\sum_{N(r)<N(q)}\frac{N^{\frac{1}{4}}}{N(r)^{\frac{1}{4}}\left(N(\|\frac{ra}{q}\|)-\frac{1}{4N(q)}\right)^{\frac{3}{4}}}$

Denote $d_{a}(r)\equiv ar\pmod{q}$, so that
$N(\|\frac{ra}{q}\|)=\frac{N(d_{a}(r))}{N(q)}$. We see that we can ignore
$\frac{1}{4N(q)}$ in the denominator of the right hand side. So

$\displaystyle\eqref{e:interestcase1}$ $\displaystyle\ll
N^{\frac{1}{4}}\sum_{N(r)<N(q)}\frac{1}{N(r)^{\frac{1}{4}}N(\|\frac{ra}{q}\|)^{\frac{3}{4}}}$

Now,the map $r\to ra$ is a permutation on $B_{q}$, so that

$\displaystyle\eqref{e:interestcase1}$ $\displaystyle\ll
N^{\frac{1}{4}}\sum_{N(r)<N(q)}\frac{1}{N(r)^{\frac{1}{4}}N(\frac{d_{a}(r)}{q})^{\frac{3}{4}}}$
$\displaystyle\ll
N^{\frac{1}{4}}N(q)^{3/4}\sum_{N(r)<N(q)}\frac{1}{N(r)^{\frac{1}{4}}N({d_{a}(r)\textbf{}})^{\frac{3}{4}}}$
$\displaystyle\ll N^{\frac{1}{4}}N(q)^{3/4}\sum_{N(r)<N(q)}\frac{1}{N(r)}\ll
N^{\frac{1}{4}}N(q)^{\frac{3}{4}}\log N(q).$

Above, we have used the $\ell^{4}$—$\ell^{4/3}$ Hölder inequality, and (2.4).
That completes Case 2.

Figure 3. The distances $\lVert\frac{ra}{q}+hq\beta+r\beta\rVert$ for a fixed
$h$ are essentially uniformly distributed in case 3.

Case 3. $h\neq 0$, or $h=0$ and $N(r)>\frac{N(q)}{10}$. This is the principal
case. Observe that $N(t)\gg N(q)(N(h)+1)$. If $h\neq 0$ we can immediately see
that

$N(t)\gg N(h)\,N(q)-N(r)\geq N(h)\,N(q)-\frac{N(q)}{2}\gg N(q)\,(N(h)+1),$

or otherwise $N(t)=N(r)\geq{N(q)}/{10}$. Combining the two inequalities we see
that

$N(t)\gg N(q)(N(h)+1).$

The key claim comes in two parts: First, there are constants $0<C_{1}<C_{2}$
and $C_{3}>0$ so that for all $q$ and $h$ as above, and for all $r\in B_{q}$
there is a $d=d(r,h)\in B_{q}$ so that first,

(4.6)
$C_{1}N(\tfrac{d}{q})\leq\bigl{\lVert}\tfrac{ra}{q}+hq\beta+r\beta\bigr{\rVert}^{2}\leq
C_{2}N(\tfrac{d}{q}).$

and second, for all $d$, the cardinality of $\\{r\colon d(r,h)=d\\}$ is at
most $C_{3}$. (In this sense, $\|\frac{ra}{q}+hq\beta+r\beta\|$ runs over the
set $\|\frac{d}{q}\|$, as illustrated in Figure 3.)

###### Proof.

The middle term in (4.6) is at most one. We take $d=d(r,h)\in B_{q}$ to
minimize the distance between $d/q$ and the fractional part of
$\frac{ra}{q}+hq\beta+r\beta$. Then, the upper and lower bounds in (4.6) are
immediate.

The remainder of the argument concerns the second claim above. But that
follows, since we always have for $r_{1}\neq r_{2}\in B_{q}$

(4.7) $\displaystyle
N\left(\left\|\frac{r_{1}a}{q}+hq\beta+r_{1}\beta\right\|-\left\|\frac{r_{2}a}{q}+hq\beta+r_{2}\beta\right\|\right)$
$\displaystyle\gg N\left(\frac{(r_{1}-r_{2})a}{q}+(r_{1}-r_{2})\beta\right)$
(4.8)
$\displaystyle=N\left(\frac{r_{1}-r_{2}}{q}\alpha\right)\gg\frac{1}{N(q)}.$

∎

We can now turn to the sum $B(T,N,\alpha)$ in this case. Using the trivial
bound for the cases that $r=0$ and picking the nontrivial bound for the other
cases, we obtain

$\displaystyle B(T,N,\alpha)$
$\displaystyle\ll\sum_{N(h)<\frac{T}{N(q)}}\sum_{N(r)<N(q)/2}\min\left(\frac{N}{N(q)\,(N(h)+1)}\right.,$
$\displaystyle\hskip
142.26378pt\left.\frac{N^{1/4}}{N(q)^{1/4}\,(N(h)+1)^{1/4}\,N(\|hq\beta+\frac{ra}{q}+r\beta\|)^{3/4}}\right)$
$\displaystyle\ll\sum_{N(h)<\frac{T}{N(q)}}\Bigg{(}\Bigg{.}\frac{N}{N(q)\left(N(h)+1\right)}$
$\displaystyle\hskip
85.35826pt+\sum_{0<N(r)<N(q)/2}\frac{N^{\frac{1}{4}}}{N(q)^{\frac{1}{4}}(1+N(h))^{\frac{1}{4}}N(\|hq\beta+\frac{ra}{q}+r\beta\|)^{\frac{3}{4}}}\Bigg{.}\Bigg{)}$
$\displaystyle\ll\sum_{N(h)<\frac{T}{N(q)}}\left(\frac{N}{N(q)\left(N(h)+1\right)}+\sum_{0<N(d)<N(q)/2}\frac{N^{\frac{1}{4}}}{N(q)^{\frac{1}{4}}(1+N(h))^{\frac{1}{4}}N(\frac{d}{q})^{\frac{3}{4}}}\right)$
$\displaystyle\ll\sum_{N(h)<\frac{T}{N(q)}}\left(\frac{N}{N(q)\left(N(h)+1\right)}+\frac{\,N^{\frac{1}{4}}N(q)^{\frac{1}{2}}}{(1+N(h))^{\frac{1}{4}}}\sum_{k=1}^{N(q)}\frac{r_{2}(k)}{k^{\frac{3}{4}}}\right)$
$\displaystyle\ll\sum_{N(h)<\frac{T}{N(q)}}\left(\frac{N}{N(q)\left(N(h)+1\right)}+\frac{\,N^{\frac{1}{4}}N(q)^{\frac{3}{4}}}{(1+N(h))^{\frac{1}{4}}}\right)$
$\displaystyle\ll\sum_{0<\ell<\frac{T}{N(q)}}r_{2}(\ell)\left(\frac{N}{N(q)\ell}+\frac{N(q)^{\frac{3}{4}}N^{\frac{1}{4}}}{\ell^{\frac{1}{4}}}\right)$
$\displaystyle\ll\frac{N\log(T)}{N(q)}+N(q)^{\frac{1}{4}}N^{\frac{1}{4}}T^{\frac{3}{4}}$

where we have used the estimates in (2.4), and our proof is complete. ∎

Now are now ready to prove the main theorem of this section.

###### Proof of Theorem 4.1.

Vaughan’s identity, well-known in the one dimensional case, holds in two
dimensions as well. That is, let $|f|\leq 1$ be an arithmetic function and fix
$UV<N$. Then

(4.9) $\displaystyle\sum_{N(n)<N}f(n)\Lambda(n)\ll
U+\log(N)A_{1}(N,U,V)+N^{\frac{1}{2}}\log^{3}(N)A_{2}(N,U,V)$

where $A_{1}$ and $A_{2}$ are given by

(4.10) $\displaystyle A_{1}=$ $\displaystyle\sum_{N(t)\leq UV}\max_{1\leq
w\leq N}\biggl{\lvert}\sum_{w\leq N(r)\leq\frac{N}{N(t)}}f(rt)\biggr{\rvert}$
(4.11) $\displaystyle A_{2}=$ $\displaystyle\max_{U\leq M\leq N/V}\max_{V\leq
N(j)\leq N/M}\left(\sum_{V<N(k)\leq
N/M}\biggl{\lvert}\sum_{M<N(m)<\min(2M,\frac{N}{N(j)},\frac{N}{N(k)})}f(mj)\overline{f(mk)}\biggr{\rvert}\right)^{\frac{1}{2}}.$

The term we need to estimate is (4.9), with $f(x)=e(\langle x,\alpha\rangle)$.
The two auxiliary integers are $U=N^{\frac{1}{2}}$ and $V=N^{\frac{1}{4}}$.
The first term requires the exponential sum estimate (2.9), and Lemma 4.3 in
which we set $T=UV$. It follows that

(4.12) $\displaystyle A_{1}$
$\displaystyle=\sum_{N(t)<UV}\max_{w<\frac{N}{N(t)}}\biggl{|}\sum_{\begin{subarray}{c}w<N(r)<\frac{N}{N(t)}\\\
\arg(r)\in\omega\end{subarray}}e(\langle rt,\alpha\rangle)\biggr{|}$ (4.13)
$\displaystyle\ll\sum_{N(t)<UV}\min\biggl{(}\frac{N}{N(t)},\left(\frac{N}{N(t)}\right)^{\frac{1}{4}}\frac{1}{N(\|\bar{t}\alpha\|)^{\frac{3}{4}}}\biggr{)}$
(4.14)
$\displaystyle\ll\frac{N\log(N)}{N(q)}+N^{\frac{1}{4}}N(q)^{\frac{3}{4}}\log
N(q)+N^{\frac{1}{4}}[UV]^{\frac{3}{4}}N(q)^{\frac{1}{4}}+N^{99/100}$ (4.15)
$\displaystyle\ll\frac{N\log(N)}{N(q)}\Bigl{(}1+(N(q)/N)^{\frac{3}{4}}\log
N(q)+\frac{N(q)^{\frac{1}{4}}}{N^{\frac{3}{16}}}\Bigr{)}+N^{99/100}$ (4.16)
$\displaystyle\ll\frac{N\log^{2}(N)}{N(q)}+N^{99/100}$

where in the last inequality we have used Lemma 4.3 with the hypothesis that
$m=0$. This bound meets the claimed bound in Theorem 4.1.

The second term from Vaughan’s identity (4.9) is quadratic in nature.

$\displaystyle A_{2}^{2}$ $\displaystyle=\max_{U<M<\frac{N}{V}}\max_{V\leq
N(j)<N/M}\sum_{V<N(k)\leq N/M}\Biggl{|}\sum_{\begin{subarray}{c}M<N(m)<2M\\\
N(m)<N/N(k),N/N(j)\\\ \arg(m)\in\omega\end{subarray}}e\left(mj-
mk,\alpha\right)\Biggr{|}$ $\displaystyle\ll\max_{U<M<\frac{N}{V}}\max_{V\leq
N(j)<N/M}\sum_{V<N(k)\leq
N/M}\min\biggl{(}M,\left(\frac{N}{N(k)}\right)^{\frac{1}{4}}N(\|(j-k)\alpha\|)^{-\frac{3}{4}}\biggr{)}$

where we have used Lemma 2.7. Now we use Lemma 4.3 for $m=j$. The conclusion
of the Lemma applies since in the sum above, we have $M\leq\frac{N}{N(k)}$.
Therefore

(4.17) $\displaystyle A_{2}$
$\displaystyle\ll\max_{U<M<\frac{N}{V}}\max_{V\leq
N(j)<N/M}\left(M+\frac{N\log(N)}{N(q)}+\frac{N\,N(q)^{\frac{1}{4}}}{M^{3/4}}+\left(\frac{N}{M}\right)^{\frac{1}{4}}N(q)+N^{\frac{99}{100}}\right)^{\frac{1}{2}}$
(4.18) $\displaystyle\ll{\frac{N^{1/2}\log
N}{N(q)^{1/2}}}+\sqrt{\frac{N}{V}}+\frac{N^{\frac{1}{2}}N(q)^{\frac{1}{8}}}{U^{3/8}}+\left(\frac{N}{U}\right)^{\frac{1}{8}}N(q)^{\frac{1}{2}}+N^{\frac{99}{200}}.$

The last inequality follow from our choice of $U=N^{\frac{1}{2}}$ and
$V=N^{\frac{1}{4}}$, and completes the proof. ∎

##  5\. Approximating the Kernel

$q$$iq$ Figure 4.
The decomposition of the complex field.

Define the approximating multiplier $L_{N}^{a,q}$ as follows:

(5.1) $\displaystyle\widehat{L_{N}^{a,q}}(\xi)$
$\displaystyle:=\Phi(a,\bar{q})\widehat{M_{N}^{\omega}}(\xi-\frac{a}{q}).$

Above, we suppress the $\omega$ dependence in the already heavy notation, and
we use the notation

(5.2) $\displaystyle\Phi(a,q)$
$\displaystyle\coloneqq\frac{\tau_{q}(a)}{\phi(q)},$

In addition, recall that $M_{N}=M_{N}^{\omega}$ is an average over a sector
defined by a choice of interval $\omega\subset\mathbb{T}$, see (2.6). The
weighted variant is

(5.3) $A_{N}^{\omega}=A_{N}=\frac{2\pi}{\lvert\omega\rvert
N}\sum_{\begin{subarray}{c}N(n)<N\\\
\arg(n)\in\omega\end{subarray}}\Lambda(n)\delta_{n}.$

######  Lemma 5.4.

For $\omega\subset\mathbb{T}$, $\alpha\in\mathbb{C}$ with $N(\alpha)<1$,
assume that there are $0\leq N(a)<N(q)<Q$ such that
$N(\alpha-\frac{a}{q})<\frac{Q}{{N\cdot N(q)}}$ and $a\in\mathbb{A}_{q}$).
Then we have, for $A>1$, and $N>N_{\omega,A}$,

(5.5)
$\bigl{\lvert}\widehat{A_{N}^{\omega}}(\alpha)-\widehat{L_{N}^{a,q}}(\alpha)\bigl{\lvert}\ll\frac{Q^{4}}{\log^{A}(N)}.$

The usual one dimensional approach to these estimates uses the Prime Number
Theorem, and Abel summation. Implementing that argument in the two dimensional
case engages a number of complications. After all, the two dimensional Prime
Number Theorem is adapted to annular sectors, whereas Abel summation is most
powerful on rectangles in the plane. We avoid these technicalities below.
(Mitsui [MR136590] sums over rectangular regions.)

###### Proof.

The quantifications of the Prime Number Theorem are decisive. Write

$\displaystyle\widehat{A_{N}^{\omega}}(\alpha)$
$\displaystyle=\frac{2\pi}{\lvert\omega\rvert
N}\sum_{\begin{subarray}{c}N(n)<N\\\
\arg(n)\in\omega\end{subarray}}\Lambda_{\mathbb{Z}[i]}(n)e(\left\langle
n,\alpha\right\rangle)$ $\displaystyle=\frac{2\pi}{\lvert\omega\rvert
N}\sum_{\begin{subarray}{c}r\in B_{\bar{q}}\\\
(r,\bar{q})=1\end{subarray}}\;\sum_{\begin{subarray}{c}n\equiv r\mod\bar{q}\\\
N(n)<N\\\ \arg(n)\in\omega\end{subarray}}\Lambda(n)e(\left\langle
n,\beta+a/q\right\rangle)$ $\displaystyle=\frac{2\pi}{\lvert\omega\rvert
N}\sum_{\begin{subarray}{c}r\in B_{\bar{q}}\\\
(r,\bar{q})=1\end{subarray}}e(\left\langle
r,a/q\right\rangle)\sum_{\begin{subarray}{c}n\equiv r\mod\bar{q}\\\ N(n)<N\\\
\arg(n)\in\omega\end{subarray}}\Lambda(n)e(\left\langle n,\beta\right\rangle)$
$\displaystyle=\frac{1}{\phi(q)}\sum_{\begin{subarray}{c}r\in B_{\bar{q}}\\\
(r,\bar{q})=1\end{subarray}}e(\left\langle r,a/q\right\rangle)B_{N}(r,\beta),$

where we define $B_{N}(r,\beta)$, and a closely related quantity by

(5.6) $\displaystyle B_{N}(r,\beta)$
$\displaystyle\coloneqq\frac{2\pi\phi(q)}{\lvert\omega\rvert
N}\sum_{\begin{subarray}{c}n\equiv r\mod\bar{q}\\\ N(n)<N\\\
\arg(n)\in\omega\end{subarray}}\Lambda(n)e(\left\langle
n,\beta\right\rangle),$ (5.7) $\displaystyle B^{\prime}_{N}(r,\beta)$
$\displaystyle\coloneqq\frac{2\pi\phi(q)}{\lvert\omega\rvert
N}\sum_{\begin{subarray}{c}n\equiv r\mod\bar{q}\\\ N(n)<N\\\
\arg(n)\in\omega\end{subarray}}e(\left\langle n,\beta\right\rangle).$

$\quad{(\log N)^{-10A}}$$N_{j}\;$$N_{j+1}=\rho N_{j}$ Figure 5.
figure description

Compare $B_{N,r}$ to $B^{\prime}_{N,r}$, as follows. Using the trivial
estimate for $N(n)\leq\sqrt{N}$

(5.8) $\displaystyle B_{N,r}(\beta)-B_{N}^{\prime}(r,\beta)\ll$ $\displaystyle
N^{-1/2}+\frac{2\pi\phi(q)}{\lvert\omega\rvert
N}\sum_{\begin{subarray}{c}n\colon\sqrt{N}<N(n)<N\\\ n\equiv r\mod\bar{q}\\\
\arg(n)\in\omega\end{subarray}}\bigl{(}\Lambda(n)-\tfrac{q}{{\pi}\phi(q)}\bigr{)}e(\langle
n,\beta\rangle).$

We continue with the last sum above. It is divided into annular rectangles, as
follows. Let $\mathcal{P}$ be a partition of the arc
$[0,\omega]\subset\mathbb{T}$ into intervals of length approximately $(\log
N)^{-10A}$. Set $\rho=1+(\log N)^{-10A}$. For integers $j$ with

(5.9) $N^{1/{2}}\leq N_{j}=\rho^{j}{\sqrt{N}}<N,$

and an interval $P\in\mathcal{P}$, set

(5.10) $R(j,P)=\\{n\colon N_{j}\leq N(n)<N_{j+1},\ \textup{arg}(n)\in P,\
n\equiv r\mod\bar{q}\\}.$

See Figure 5. The set $R(j,P)$ is the symmetric difference of four sets to
which the prime counting function estimate (4.2) applies. From it, we see that

(5.11) $\displaystyle D(j,P)$ $\displaystyle=\sum_{n\in
R(j,P)}\bigl{(}\Lambda(n)-\tfrac{q}{\phi(q)}\bigr{)}e(\langle n,\beta\rangle)$
(5.12) $\displaystyle\leq\sup_{n,m\in R(j,P)}\lvert 1-e(\langle
n-m,\beta\rangle)\rvert\sum_{n\in R(j,P)}\Lambda(n)+\tfrac{q}{\phi(q)}$ (5.13)
$\displaystyle\qquad+\Bigl{\lvert}\sum_{n\in
R(j,P)}\Lambda(n)-\tfrac{q}{\phi(q)}\Bigr{\rvert}$ (5.14)
$\displaystyle\ll\Bigl{[}\frac{Q}{N}\cdot\frac{N}{\log^{10}N}\Bigr{]}^{1/2}\lvert
R(j,P)\rvert+\frac{N_{j+1}-N_{j}}{(\log N)^{10A}}$ (5.15)
$\displaystyle\ll\sqrt{Q}\frac{N_{j+1}-N_{j}}{(\log N)^{10A}}.$

The bound for the first term comes from the condition that
$N(\beta)\leq\frac{Q}{N}$, and for the second from (4.2).

Control of the absolute value of the $D(j,P)$ is sufficient, since

(5.16) $\displaystyle B_{N,r}(\beta)-B_{N,r}^{\prime}(\beta)$
$\displaystyle\ll
N^{-1/2}+\frac{\phi(q)}{N}\sum_{P\in\mathcal{P}}\sum_{j\colon\rho^{j}\leq\sqrt{N}}\lvert
D(j,P)\rvert$ (5.17) $\displaystyle\ll
N^{-1/2}+\frac{\phi(q)}{N}\sum_{P\in\mathcal{P}}\sum_{j\colon\rho^{j}\leq\sqrt{N}}\frac{N_{j+1}-N_{j}}{(\log
N)^{10A}}$ (5.18) $\displaystyle\ll\frac{Q(\log N)^{A}}{(\log N)^{10A}}$

as there are only $\ll(\log N)^{A}$ choices of the interval $P$. We are free
to choose $A$ as large as we want.

This holds for all $r\in B_{\bar{q}}$, so that we have

(5.19)
$\widehat{A_{N}^{\omega}}(\alpha)-\frac{1}{\phi(q)}\sum_{\begin{subarray}{c}r\in
B_{\bar{q}}\\\ (r,\bar{q})=1\end{subarray}}e(\left\langle
r,a/q\right\rangle)B_{N}^{\prime}(r,\beta)\ll\frac{Q}{(\log N)^{A}}.$

Then, observe the elementary inequality that for $r,s\in B_{\bar{q}}$,

(5.20) $\displaystyle B_{N}^{\prime}(r,\beta)-B_{N}^{\prime}(s,\beta)$
$\displaystyle\ll\lvert r-s\rvert\cdot\lvert\beta\rvert\ll
Q\Bigl{[}\frac{Q}{N}\Bigr{]}^{1/2}$

which just depends upon the Lipschitz bound on exponentials, and the upper
bound on $\beta$.

The conclusion of the argument is then clear. Up to an error term of magnitude
$Q^{3/2}(\log N)^{A}$ we can write

(5.21) $\displaystyle\widehat{A_{N}^{\omega}}(\alpha)$
$\displaystyle=\frac{1}{\phi(q)}\sum_{\begin{subarray}{c}r\in B_{\bar{q}}\\\
(r,\bar{q})=1\end{subarray}}e(\left\langle
r,a/q\right\rangle)B_{N}^{\prime}(0,\beta)$ (5.22)
$\displaystyle=\Phi(a,\bar{q})B_{N}^{\prime}(0,\beta)$ (5.23)
$\displaystyle=\Phi(a,\bar{q})\frac{1}{\lvert B_{\bar{q}}\rvert}\sum_{r\in
B_{\bar{q}}}B_{N}^{\prime}(r,\beta)$ (5.24)
$\displaystyle=\Phi(a,\bar{q})\widehat{M_{N}^{\omega}}(\beta).$

That is the conclusion of Lemma 5.4. ∎

Consider the following dyadic decomposition of rationals

(5.25) $\mathcal{R}_{s}=\left\\{\frac{a}{q}\,:\,2^{s}\leq
N(q)<2^{s+1},\,a\in\mathbb{A}_{q}\right\\}.$

Let $\Delta$ be be a continuous function on $\mathbb{C}$, a tensor product of
piecewise linear functions, with

(5.26) $\eta(\xi)=\begin{cases}1&\mbox{if }\xi=(0,0)\\\ 0&\mbox{if
}N(\xi)\geq\lVert\xi\rVert_{\infty}\geq 1\end{cases}$

and let $\Delta_{s}(\xi):=\eta(16^{s}\xi)$. Here, we remark that this
definition is different from many related papers in the literature. With this
defintion, the function $\check{\eta}$ is a tensor product Féjer kernels. In
particular, they are non-negative averages. Imposing this choice here will
simplify considerations in the analysis of the Goldbach conjectures.

Recalling definitions of $L_{N}^{a,q}$ in (5.1), further define

$\widehat{B^{\omega}_{N}}(\xi):=\sum_{s\geq
0}\sum_{a/q\in\mathcal{R}_{s}}\widehat{L_{N}^{a,q}}\left(\xi\right)\Delta_{s}\left(\xi-\frac{a}{q}\right).$

We remind the reader that the $\omega$ dependence is suppressed in the
notation on the right.

######  Theorem 5.27.

Fix an integer $A>10$ and $\omega\subset\mathbb{T}$. Then, there is an
$N_{\omega}$ to that for all $N>N_{\omega}$,

(5.28) $\|\widehat{A^{\omega}_{N}}-\widehat{B^{\omega}_{N}}\|_{\infty}\ll(\log
N)^{-A}.$

The implied constant is independent of $\omega$.

###### Proof.

A useful and familiar fact we will reference below is that for each $s$, the
functions below are disjointly supported.

(5.29)
$\widehat{L_{N}^{a,q}}\Bigl{(}\xi\Bigr{)}\Delta_{s}\Bigl{(}\xi-\frac{a}{q}\Bigr{)},\qquad
a/q\in\mathcal{R}_{s}$

Fix $\xi\in\mathbb{T}^{2}$. By the higher-dimensional Dirichlet’s Theorem
there are relatively prime numbers $a$ and $q$ that satisfy $1\leq N(a)\leq
N(q)\leq N^{1/4}$ such that

$N\bigl{(}\xi-\tfrac{a}{q}\bigr{)}\leq\frac{1}{N(q)\cdot N^{1/4}}.$

To prove the theorem we need to consider two cases based on the value of
$N(q)$.

Case 1: Suppose $1\leq N(q)\leq(\log N)^{4A}$. For
$\frac{a^{\prime}}{q^{\prime}}\neq\frac{a}{q}$ and $N(q^{\prime})\leq(\log
N)^{2A}$, we have

$\displaystyle N\left(\xi-\frac{a^{\prime}}{q^{\prime}}\right)$
$\displaystyle\gg
N\left(\frac{a^{\prime}}{q^{\prime}}-\frac{a}{q}\right)-N\left(\xi-\frac{a}{q}\right)$
$\displaystyle\geq\frac{1}{N(q^{\prime})N(q)}-\frac{1}{N(q)}\frac{1}{N^{1/4}}\gg(\log
N)^{-6A}.$

Using this, and the decay estimate for $\widehat{M}_{N}$ in (2.9) to see that

$\left|\widehat{L_{N}^{a^{\prime},q^{\prime}}}(\xi)\Delta_{s}\left(\xi-\frac{a^{\prime}}{q^{\prime}}\right)\right|\ll\frac{1}{\sqrt{N(q^{\prime})}}\left(NN\left(\xi-\frac{a^{\prime}}{q^{\prime}}\right)\right)^{-3/4}\ll
N^{-1/2}.$

Appeal to the disjointness property (5.29). We have

$\Biggl{\lvert}\sum_{s\colon 2^{s}\leq(\log
N)^{2A}}\sum_{\frac{a^{\prime}}{q^{\prime}}\in\mathcal{R}_{s},\frac{a^{\prime}}{q^{\prime}}\neq\frac{a}{q}}\widehat{L_{N}^{a^{\prime},q^{\prime}}}(\xi)\Delta_{s}\left(\xi-\frac{a^{\prime}}{q^{\prime}}\right)\Biggr{\rvert}\ll
N^{-1/2}\sum_{s\colon 2^{s}\leq(\log N)^{2A}}2^{-s}\ll N^{-1/2}.$

For $2^{s}>(\log N)^{2A}$ we use the trivial bound for
$\widehat{M_{N}^{\omega}}$, the estimate for the Gauss sums in (5.2), as well
as the support property (5.29). This yields the estimate

(5.30) $\sum_{s\colon 2^{s}>(\log
N)^{2A}}\sum_{\begin{subarray}{c}\frac{a^{\prime}}{q^{\prime}}\in\mathcal{R}_{s}\\\
\frac{a^{\prime}}{q^{\prime}}\neq\frac{a}{q}\end{subarray}}\widehat{L_{N}^{a^{\prime},q^{\prime}}}(\xi)\Delta_{s}\Bigl{(}\xi-\frac{a^{\prime}}{q^{\prime}}\Bigr{)}\ll\sum_{s\colon
2^{s}>(\log N)^{2A}}2^{-3s/4}\ll(\log N)^{-A}.$

Above, the sums exclude the case of $a^{\prime}/q^{\prime}=a/q$. That is the
central case, the one Lemma 5.4 was designed for.

We turn to the case of $a^{\prime}/q^{\prime}=a/q$ here. With appropriate
choice of $Q$ and $A$ in that Lemma, we obtain from (5.28),

$\displaystyle\Bigl{\lvert}\widehat{A_{N}^{\omega}}(\xi)-\widehat{L_{N}^{a,q}}(\xi)\Delta_{s}\Bigl{(}\xi-\frac{a}{q}\Bigr{)}\Bigr{\rvert}$
$\displaystyle\leq\Bigl{\lvert}\widehat{A_{N}^{\omega}}(\xi)-\widehat{L_{N}^{a,q}}(\xi)\Bigr{\rvert}+\Bigl{\lvert}\widehat{L_{N}^{a,q}}(\xi)\Bigl{(}1-\Delta_{s}\Bigl{(}\xi-\frac{a}{q}\Bigr{)}\Bigr{)}\Bigr{\rvert}$
$\displaystyle\ll(\log N)^{-A}+\frac{N(q)^{2}}{N(\xi-a/q)^{1/2}}\ll(\log
N)^{-A}.$

This holds by choice of $a/q$. That completes this case.

Case 2: Suppose $(\log N)^{4A}\leq N(q)$. Both terms are small. By the
Vinogradov inequality in Theorem 4.1 we have

$\displaystyle\lvert\widehat{A_{N}^{\omega}}(\xi)\lvert\ll(\log N)^{-A}.$

It remains to show that $\widehat{B}_{N}(\xi)$ is also small. That function is
a sum over integers $s\geq 0$. For $2^{s}>(\log N)^{2A}$, we only need to use
the estimate (5.30). Thus, our focus turns to the case of $2^{s}\leq(\log
N)^{2A}$.

For $2^{s}\leq(\log N)^{2A}$ we have for
$\frac{a^{\prime}}{q^{\prime}}\in\mathcal{R}_{s}$

$\displaystyle N\Bigl{(}\xi-\frac{a^{\prime}}{q^{\prime}}\Bigr{)}$
$\displaystyle\geq
N\Bigl{(}\frac{a^{\prime}}{q^{\prime}}-\frac{a}{q}\Bigr{)}-N\Bigl{(}\xi-\frac{a}{q}\Bigr{)}$
$\displaystyle\geq\frac{1}{N(q^{\prime})N(q)}-\frac{1}{N(q)}\frac{1}{N^{1/4}}$
$\displaystyle\geq\frac{2^{-s-1}}{N(q)}\gg N^{-1/8}.$

From the decay estimate in (2.9), we have

$\widehat{L_{N}^{a^{\prime},q^{\prime}}}(\xi)\Delta_{s}\Bigl{(}\xi-\frac{a^{\prime}}{q^{\prime}}\Bigr{)}\ll\left(NN\left(\xi-\frac{a^{\prime}}{q^{\prime}}\right)\right)^{-3/4}\ll
N^{-3/32}$

Using the disjointness property (5.29), it then easy to see that

(5.31)
$\sum_{2^{s}\leq\sqrt{Q}}\sum_{\frac{a^{\prime}}{q^{\prime}}\in\mathcal{R}_{s},\frac{a^{\prime}}{q^{\prime}}\neq\frac{a}{q}}\widehat{L_{N}^{a^{\prime},q^{\prime}}}(\xi)\Delta_{s}\left(\xi-\frac{a^{\prime}}{q^{\prime}}\right)\ll(\log
N)^{-A}.$

That completes the second case, and hence the proof of our Theorem. ∎

##  6\. Estimates for the High and Low Parts

Our High and Low decomposition of the multiplier incorporates a notion of
smooth numbers. For integer $Q=2^{q_{0}}\ll(\log N)^{B}$, we say that a
Gaussian integer is $Q$-smooth if $q$ is square free and the product of primes
$\rho$ with $N(\rho)\leq Q$. Here, $B$ will be a fixed integer.

We write
$A_{N}^{\omega}=\textup{Lo}_{Q,N}^{\omega}+\textup{Hi}_{Q,N}^{\omega}$, where

(6.1) $\displaystyle\widehat{\operatorname{Lo}}_{Q,N}^{\omega}(\xi)$
$\displaystyle=\sum_{q\colon
N(q)<Q}\sum_{a\in\mathbb{A}_{q}}\Phi(a,\bar{q})\widehat{M_{N}^{\omega}}(\xi-\frac{a}{q})\Delta_{q_{0}}(\xi-\frac{a}{q}).$

Here, we recall that $\omega\subset\mathbb{T}$ is an interval, and
$N>N_{\omega}$ is sufficiently large. The average $M^{\omega}_{N}$ is defined
in (2.6), the Gauss sum $\Phi(a,\bar{q})$ in (5.2). (Note that the Gauss sum
will be zero if $q$ contains a square.) This definition is inspired by Theorem
5.27. But, the definition above incorporates not only smoothness, but the
cutoff function $\Delta_{q_{0}}$ is a function of $Q$. Both changes are useful
in the next section. There are two key properties of these terms. The first,
is that the ‘High’ part has small $\ell^{2}$ norm.

######  Lemma 6.2.

For any $\epsilon>0$, $\omega\subset\mathbb{T}$, there in an $N_{\omega}$ so
that for all $N>N_{\omega}$,

(6.3)
$\displaystyle\|\operatorname{Hi}_{Q,N}^{\omega}\|_{\ell_{2}\rightarrow\ell_{2}}\ll
Q^{-1+\varepsilon}.$

###### Proof.

The $\ell^{2}$ norm is estimated on the frequency side. By Theorem 5.27, the
High term is a sum of three terms. They are, suppressing the dependence on
$\omega$,

(6.4) $\displaystyle\widehat{\operatorname{Hi}_{Q,N}^{1}}(\xi)$
$\displaystyle=\sum_{Q<2^{s+1}}\sum_{\begin{subarray}{c}2^{s}\leq
N(q)<2^{s-1}\\\ N(q)\geq
Q\end{subarray}}\sum_{\frac{a}{q}\in\mathcal{R}_{s}}\Phi(a,\bar{q})\widehat{M_{N}^{\omega}}(\xi-\frac{a}{q})\Delta_{s}(\xi-\frac{a}{q})$
(6.5) $\displaystyle\widehat{\operatorname{Hi}_{Q,N}^{2}}(\xi)$
$\displaystyle=\sum_{s}\sum_{\begin{subarray}{c}2^{s}\leq N(q)<2^{s-1}\\\
N(q)<Q\end{subarray}}\sum_{\frac{a}{q}\in\mathcal{R}_{s}}\Phi(a,\bar{q})\widehat{M_{N}^{\omega}}(\xi-\frac{a}{q})\\{\Delta_{q_{0}}(\xi-\frac{a}{q})-\Delta_{s}(\xi-\frac{a}{q})\\}$
(6.6) $\displaystyle\widehat{\operatorname{Hi}_{Q,N}^{3}}(\xi)$
$\displaystyle\coloneqq\widehat{A^{\omega}_{N}}-\widehat{B^{\omega}_{N}}.$

We address them in reverse order. The last term is controlled by Theorem 5.27.
It clearly satisfies our claim (6.3).

In the $\widehat{\operatorname{Hi}_{Q,N}^{2}}$, the key term is the last
difference between $\Delta_{q_{0}}$ and $\Delta_{s}$. In particular, if
$\Delta_{q_{0}}(\xi)-\Delta_{s}(\xi)\neq 0$, we have $N(\xi)\gg 2^{-q_{0}}$.
It follows that $\widehat{M_{N}^{\omega}}(\xi)$ is relatively small. This
allows us to estimate

(6.7)
$\displaystyle\lVert\widehat{\operatorname{Hi}_{Q,N}^{2}}(\xi)\rVert_{\infty}$
$\displaystyle\leq\sum_{s}\sum_{\begin{subarray}{c}2^{s}\leq q<2^{s-1}\\\
N(q)<Q\end{subarray}}\max_{\frac{a}{q}\in\mathcal{R}_{s}}\bigl{\lVert}\Phi(a,\bar{q})\widehat{M_{N}^{\omega}}(\xi-\frac{a}{q})\\{\Delta_{q_{0}}(\xi-\frac{a}{q})-\Delta_{s}(\xi-\frac{a}{q})\\}\bigr{\rVert}_{\infty}$
(6.8)
$\displaystyle\ll\sum_{s}2^{-3s/4}\min\\{1,(N2^{-q_{0}})^{-3/4}+N^{-1/2}\\}\ll
Q^{-1}.$

Here, we have used the disjointness of support for the different functions,
and the exponential sum estimate Lemma 2.7.

It remains to bound the term $\widehat{\operatorname{Hi}_{Q,N}^{1}}$. But the
smallest denominator $q$ that we sum over satisfies at least $N(q)>Q$, so that
a similar argument leads to

(6.9)
$\displaystyle\lVert\widehat{\operatorname{Hi}_{Q,N}^{1}}\rVert_{\infty}$
$\displaystyle\ll\sum_{Q<2^{s}<N^{1/4}}\sum_{\begin{subarray}{c}2^{s}\leq
N(q)<2^{s-1}\\\ N(q)\geq
Q\end{subarray}}\max_{\frac{a}{q}\in\mathcal{R}_{s}}\lvert\Phi(a,\bar{q})\rvert$
(6.10) $\displaystyle=\sum_{Q<2^{s}<N^{1/4}}2^{-(1-\epsilon)s}\ll
Q^{-1+\epsilon}.$

∎

We turn to the Low term. It has an explicit form in spatial variables.

######  Lemma 6.11.

For $x\in\mathbb{Z}[i]$ we have the equality below, in which recall that
$2^{q_{0}}=Q<(\log N)^{B}$.

(6.12)
$\textup{Lo}_{N,Q}^{\omega}(x)=\left(M_{N}^{\omega}\ast\widecheck{\Delta_{q_{0}}}\right)(x)\sum_{q\colon
N(q)<Q}\frac{\mu(q)\tau_{q}(x)}{\phi(q)}$

And, moreover, for all $\epsilon>0$, and non-negative $f$

(6.13) $\displaystyle\operatorname{Lo}_{Q,N}^{\omega}f(x)\ll
Q^{\epsilon}\bigl{[}(M_{N}^{\omega}\ast\widecheck{\Delta_{q_{0}}})\ast
f^{1+\epsilon}(x)\bigr{]}^{1/(1+\epsilon)}.$

###### Proof.

We have for fixed $q$,

(6.14)
$\displaystyle\sum_{a\in\mathbb{A}_{q}}\Phi(a,\bar{q})\int_{D}\widehat{M_{N}^{\omega}}(\xi-\frac{a}{q})\Delta_{q_{0}}(\xi-\frac{a}{q})e(\langle
x,\xi\rangle)d\xi$
$\displaystyle=\sum_{a\in\mathbb{A}_{q}}\Phi(a,\bar{q})e(\langle
x,\frac{a}{q}\rangle)\left(M_{N}\ast\widecheck{\Delta_{q_{0}}}\right)(x)$
(6.15)
$\displaystyle=\left(M_{N}^{\omega}\ast\widecheck{\Delta_{q_{0}}}\right)(x)\frac{1}{\phi(\bar{q})}\sum_{a\in\mathbb{A}_{q}}\tau_{\bar{q}}(a)e(\langle
x,\frac{a}{q}\rangle)$ (6.16)
$\displaystyle=\left(M_{N}^{\omega}\ast\widecheck{\Delta_{q_{0}}}\right)(x)\frac{1}{\phi(q)}\sum_{r\in\mathbb{A}_{\bar{q}}}\tau_{q}(x+r).$

We then apply Lemma 3.18, and sum over $Q$-smooth denominators $q$ to conclude
the first claim (6.12).

For the second claim, we use (3.26) in the standard way. Fix $2^{s-1}\leq Q$,
and consider the operator $A$ with kernel

(6.17)
$A_{s}(x)=\left(M_{N}^{\omega}\ast\widecheck{\Delta_{q_{0}}}\right)(x)\sum_{q\colon
2^{s}\leq q<2^{s}}\frac{\lvert\tau_{q}(x)\rvert}{\phi(q)}.$

For an integer $k>2\epsilon^{-1}$, and non-negative $f\in\ell^{1+\epsilon}$,
we have

(6.18) $\displaystyle A_{s}f(x)$
$\displaystyle=\sum_{y}\left(M_{N}^{\omega}\ast\widecheck{\Delta_{q_{0}}}\right)(y)\sum_{q\colon
2^{s}\leq q<2^{s}}\frac{\lvert\tau_{q}(y)\rvert}{\phi(q)}f(x-y)$ (6.19)
$\displaystyle\leq\Bigl{[}\sum_{y}\left(M_{N}^{\omega}\ast\widecheck{\Delta_{q_{0}}}\right)(y)f(x-y)^{k/(k-1)}\Bigr{]}^{(k-1)/k}$
(6.20)
$\displaystyle\times\Bigl{[}\sum_{y}\left(M_{N}^{\omega}\ast\widecheck{\Delta_{q_{0}}}\right)(y)\sum_{q\colon
2^{s}\leq
q<2^{s}}\Bigl{[}\frac{\lvert\tau_{q}(y)\rvert}{\phi(q)}\Bigr{]}^{k}\Bigr{]}^{1/k}$
(6.21) $\displaystyle\ll 2^{\epsilon
s}\Bigl{[}\sum_{y}\left(M_{N}^{\omega}\ast\widecheck{\Delta_{q_{0}}}\right)(y)f(x-y)^{k/(k-1)}\Bigr{]}^{(k-1)/k}.$

We sum this over $2^{s-1}<Q$ to complete the proof of (6.13).

∎

##  7\. Improving Inequalities

In this brief section, we establish the improving inequalities, namely Theorem
1.2. And, list some additional results that we could establish. For the
convenience of the reader, we restate the improving inequality here, in a
slightly more convenient form for our subsequent discussion. And, there is no
loss of generality to reduce to the trivial case of a sector.

######  Theorem 7.1.

For all $N$, and $1<p\leq 2$, we have, for $\omega=\mathbb{T}$, and functions
$f,g$ supported on $[0,\sqrt{N}]^{2}$,

$\displaystyle N^{-1}\langle A_{N}^{\mathbb{T}}f,g\rangle\ll N^{-2/p}\lVert
f\rVert_{p}\lVert g\rVert_{p}.$

###### Proof.

As the angle $\omega=\mathbb{T}$, we suppress it in the notation. We must
prove the inequalities over the open range of $1<p<2$. So, it suffices to
consider the case that $f=\mathbf{1}_{F}$ and $g=\mathbf{1}_{G}$, for
$F,G\subset\\{n\colon N(n)\leq N\\}$. Interpolation proves them as stated.

Dominating the von Mangoldt function by $\log N$, we always have

$\displaystyle\langle A_{N}f,g\rangle\ll N(\log N)\lvert F\rvert\cdot\lvert
G\rvert.$

We can then immediately deduce the inequality if

(7.2) $N^{-2}\lvert F\rvert\cdot\lvert G\rvert\ll(\log N)^{-2p^{\prime}}$

So, we assume that this inequality fails, which will allow us to use our High
Low decomposition. Namely, for $0<\epsilon<1/2$ sufficiently small, set

(7.3) $Q^{\frac{2(1+\epsilon)}{1-\epsilon}}\simeq\frac{N^{2}}{\lvert
F\rvert\cdot\lvert G\rvert}\ll(\log N)^{2p^{\prime}}.$

Write $A_{N}=\textup{Hi}_{N,Q}+\textup{Lo}_{N,Q}$. Appealing to (6.3) for the
High term, and (6.13) for the Low term, we have

(7.4) $\displaystyle\langle\textup{Hi}_{N,Q}f,g\rangle$ $\displaystyle\ll
Q^{-1+\epsilon}\lVert f\rVert_{2}\lVert g\rVert_{2}\ll Q^{-1+\epsilon}[\lVert
f\rVert_{1}\lVert g\rVert_{1}]^{1/2},$ (7.5)
$\displaystyle\langle\textup{Lo}_{N,Q}f,g\rangle$ $\displaystyle\ll
NQ^{\epsilon}\lVert f\rVert_{1+\epsilon}\lVert g\rVert_{1+\epsilon}.$

By choice of $Q$, the two upper bounds nearly agree and are at most

(7.6) $\langle\textup{Lo}_{N,Q}f,g\rangle\simeq N^{-1+2\epsilon}\bigl{[}\lvert
F\rvert\cdot\lvert G\rvert]^{1-2\epsilon}.$

That is the desired inequality, for $p^{\prime}=\frac{1+2\epsilon}{\epsilon}$.
And so completes the proof.

∎

The techniques developed to establish the improving inequality can be
elaborated on to prove additional results. We briefly describe them here.

  1. (1)

An $\ell^{p}\to\ell^{p}$, for $1<p<\infty$ inequality for the maximal function
$\sup_{N}\lvert A_{N}f\rvert$. Compare to [MR995574].

  2. (2)

A $(p,p)$, $1<p<2$, sparse bound for the maximal function. Here we use the
terminology of [MR4072599], for instance. The interest in the sparse bound is
that it immediately implies a range of weighted inequalities.

  3. (3)

One can establish pointwise convergence of ergodic averages. Let
$(T_{1},T_{2})$ be commuting invertible measure preserving transformations of
a probability space $(X,\mu)$. For all $1<p<\infty$, and $f\in L^{p}(X)$, the
limit

(7.7) $\lim_{N}\frac{1}{N}\sum_{N(n)<N}\Lambda(n)f(T^{n}x)$

exists for a.e.$(x)$. Here, $T^{a+ib}=T_{1}^{a}T_{2}^{b}$. Compare to
[MR3646766].

We have given references particular to the primes (in $\mathbb{Z}$).

##  8\. Goldbach Conjecture

The purpose of this section is to prove analogues of the Goldbach Conjecture
on the Gaussian setting. We recall some elementary facts about Gaussian
primes. We address a binary and ternary form of the Goldbach conjecture. The
binary form is addressed in density form. Namely, most even integers are the
sum of two primes. On the other hand all sufficiently large odd integers are
the sum of three primes. We further restrict the arguments of the integers to
be in a fixed interval.

###  8.1. The Binary Goldbach Conjecture

The Goldbach Conjecture states that every even Gaussian integer can be written
as the sum of two primes. We prove a density version of this result. It uses
the High/Low decomposition. Observe that

(8.1) $A_{N}^{\omega}\ast
A_{N}^{\omega}(n)=\frac{2\pi}{\lvert\omega\rvert^{2}N^{2}}\sum_{\begin{subarray}{c}N(m_{1}),N(m_{2})<N\\\
m_{1}+m_{2}=n\\\
\arg(m_{1}),\arg(m_{2})\in\omega\end{subarray}}\Lambda(m_{1})\Lambda(m_{2}).$

If the sum is non-zero, $n$ can be represented as the sum of two numbers in
the support of the von Mangoldt function $\Lambda$ intersected with
$S^{\omega}_{N}$, where we define

(8.2) $S^{\omega}_{N}=\\{n\colon N(n)<N,\arg(n)\in\omega\\}.$

The von Mangoldt function is supported on the Gaussian primes, and their
powers. The powers have density less than $\ll\sqrt{N}$. Thus, it suffices to
establish that

(8.3) $\lvert\\{n\in S^{\omega}_{N}\colon\textup{$n$ even \&
}A_{N}^{\omega}\ast A_{N}^{\omega}(n)=0\\}\rvert\ll\frac{N}{(\log N)^{B}}.$

Recall from (6.1), that we can write

(8.4) $A_{N}^{\omega}={\textup{Hi}}+{\textup{Lo}}.$

This depends upon a choice of $Q=(\log N)^{B}$ for some sufficiently large
power of $B$, and we suppress the dependence on $Q$, $N$ and $\omega$ in the
notation. Thus, we write

(8.5) $A_{N}\ast A_{N}=\textup{Hi}\ast
A_{N}+\textup{Lo}\ast\textup{Hi}+\textup{Lo}\ast\textup{Lo}.$

On the right, the last term that is crucial. We further write it as

(8.6) $\textup{Lo}\ast\textup{Lo}=\textup{Main}+\textup{Error}.$

Aside from the main term, everything is small. Our Theorem easily follows from
the Lemma below, in which we collect the required estimates.

######  Lemma 8.7.

We have the estimates below, valid for all choices of $B>1$.

(8.8) $\displaystyle N^{-1}$ $\displaystyle\ll\min_{\begin{subarray}{c}n\in
S^{\omega}_{N}\\\ n\ \textup{even}\end{subarray}}\textup{Main}(n),$ (8.9)
$\displaystyle\lvert\\{0<N(n)<N\colon\lvert\textup{Error}(n)\rvert$
$\displaystyle>N^{-1}(\log N)^{-(B-1)/2}\rvert\ll N(\log N)^{-(B-1)2},$ (8.10)
$\displaystyle\lVert\textup{Lo}\ast\textup{Hi}(n)\rVert_{\ell^{2}}$
$\displaystyle\ll(N(\log N)^{B-1})^{-1/2},$ (8.11)
$\displaystyle\lVert\textup{Hi}\ast A_{N}(n)\rVert_{\ell^{2}}$
$\displaystyle\ll(N(\log N)^{B-1})^{-1/2}.$

We focus on the first estimate above. The Main term is

(8.12) $\textup{Main}(x)\coloneqq\sum_{\textup{$q$ is
$Q$-smooth}}\frac{1}{\phi(q)^{2}}\sum_{a\in\mathbb{A}_{q}}\tau_{q}(a)^{2}e\bigl{(}\langle\tfrac{a}{q},x\rangle\bigr{)}\int_{\mathbb{T}^{2}}\widehat{M_{N}^{\omega}}(\xi)^{2}e\bigl{(}\langle\xi,x\rangle\bigr{)}\;d\xi$

Above, we say that $q$ is _$Q$ -smooth_ if $q$ is square free and all prime
factors $\rho$ of $q$ satisfy $N(q)<Q$. The expression above can be calculated
explicitly, using the Ramanujan like sum (3.6).

######  Lemma 8.13.

Recall that $2^{q_{0}}=Q<(\log N)^{B}$. For every $N(x)<N$ we have

(8.14) $\displaystyle\textup{Main}(x)=M_{N}^{\omega}\ast
M_{N}^{\omega}(x)\sum_{\textup{$q$ is
$Q$-smooth}}\frac{|\mu(q)|}{\phi(q)^{2}}\tau_{q}(x),$

###### Proof.

The term in (8.12) is

$\displaystyle\textup{Main}(x)$ $\displaystyle=M_{N}(x)^{\omega}\ast
M_{N}^{\omega}(x)\sum_{\textup{$q$ is
$Q$-smooth}}\frac{1}{\phi(q)^{2}}\sum_{a\in\mathbb{A}_{q}}\tau_{q}(a)^{2}e\bigl{(}\langle\tfrac{a}{q},x\rangle\bigr{)}$

In the arithmetic term above, we fix $q$, expand the Ramanujan sums, and we
use Lemma 3.18. This gives us

(8.15)
$\displaystyle\frac{1}{\phi(q)^{2}}\sum_{a\in\mathbb{A}_{q}}\tau_{q}(a)^{2}e\bigl{(}\langle\tfrac{a}{q},x\rangle\bigr{)}$
$\displaystyle=\frac{1}{\phi(q)^{2}}\sum_{a\in\mathbb{A}_{q}}\Bigl{[}\sum_{r\in\mathbb{A}_{q}}e\bigl{(}\langle\tfrac{a}{q},r\rangle\bigr{)}\Bigr{]}^{2}e\bigl{(}\langle\tfrac{a}{q},x\rangle\bigr{)}$
(8.16)
$\displaystyle=\frac{1}{\phi(q)^{2}}\sum_{r_{1}\in\mathbb{A}_{q}}\sum_{r_{2}\in\mathbb{A}_{q}}\tau_{q}(x+r_{1}+r_{2})$
(8.17)
$\displaystyle=\frac{1}{\phi(q)^{2}}\sum_{r_{1}\in\mathbb{A}_{q}}\tau_{q}(x+r_{1})\tau_{\bar{q}}(1)$
(8.18) $\displaystyle=\frac{1}{\phi(q)^{2}}\tau_{q}(x)\tau_{\bar{q}}(1)^{2},$

where in the last line we have used Lemma 3.18 again. Finally
$\tau_{\bar{q}}(1)^{2}=\lvert\mu(q)\rvert$. ∎

On the right in our equality for the Low-Low expression (8.14), the first
convolution satisfies

(8.19) $\min_{n\in S^{\omega}_{N}}M_{N}^{\omega}\ast M_{N}^{\omega}(n)\gg
N^{-1}.$

The implied constant is a function of $\omega$, but at no point have we sought
to track this dependence.

We analyze the arithmetic part here, and the Lemma below completes the proof
of (8.8). Indeed, it is exactly this Lemma and its proof that motivates the
use of the smooth numbers.

######  Lemma 8.20.

We have for all even $x$,

(8.21) $\displaystyle\sum_{\textup{$q$ is
$Q$-smooth}}\frac{|\mu(q)|}{\phi(q)^{2}}\tau_{q}(x)\gg 1$

###### Proof.

Exploit the multiplicative structure of the sum. One sees that it is a product
over primes. For any integer $x$,

(8.22) $\displaystyle\sum_{\textup{$q$ is
$Q$-smooth}}\frac{|\mu(q)|}{\phi(q)^{2}}\tau_{q}(x)$
$\displaystyle=\prod_{\rho\colon
N(\rho)<Q}1+\frac{\tau_{\rho}(x)}{\phi(\rho)^{2}}$

Above, the product is over primes $\rho$ with $N(\rho)<Q$, up to
multiplication by units. And recall that if $\rho\mid x$, we have
$\tau_{\rho}(x)=\phi(\rho)$.

It is important to single out the prime $1+i$. This is the unique prime with
$\phi(1+i)=1$, and

(8.23) $1+\frac{\tau_{1+i}(x)}{\phi(1+i)^{2}}=\begin{cases}2&1+i\mid x\\\
0&1+i\nmid x\end{cases}$

Thus, if $1+i\nmid x$, the sum in (8.22) is zero. But, evenness of $x$ is
equivalent to $1+i\mid x$. Thus, for even $x$, single out the case of
$\rho=1+i$.

(8.24) $\displaystyle\prod_{\rho\colon N(\rho)\leq
Q}1+\frac{\tau_{\rho}(x)}{\phi(\rho)^{2}}$
$\displaystyle=2\prod_{\begin{subarray}{c}\rho|x,\ N(\rho)\leq Q\\\ \rho\neq
1+i\end{subarray}}\left(1+\frac{1}{\phi(\rho)}\right)\left(1-\frac{1}{\phi(\rho)^{2}}\right)^{-1}\times\prod_{\begin{subarray}{c}\rho,\
N(\rho)\leq Q\\\ \rho\neq 1+i\end{subarray}}1-\frac{1}{\phi(\rho)^{2}}$ (8.25)
$\displaystyle=2\prod_{\begin{subarray}{c}\rho|x,\ N(\rho)\leq Q\\\ \rho\neq
1+i\end{subarray}}\frac{\phi(\rho)}{\phi(\rho)-1}\times\prod_{\begin{subarray}{c}\rho,\
N(\rho)\leq Q\\\ \rho\neq 1+i\end{subarray}}1-\frac{1}{\phi(\rho)^{2}}$ (8.26)
$\displaystyle\coloneqq h(x)\mathcal{G}.$

In the last line, we have written the product as a ‘local term’ $h(x)$ and a
‘global term,’ $\mathcal{G}$. If there is no $Q$-smooth prime that divides
$x$, we understand that the local term is $2$. Thus, $h(x)$ is always at least
2 for even $x$. The global term is finite:

(8.27) $\mathcal{G}\geq\prod_{\rho}1-\frac{1}{\phi(\rho)^{2}},$

and the infinite sum is convergent to a positive number, since

(8.28) $\displaystyle\sum_{\rho\ \textup{prime}}\frac{1}{\phi(\rho)^{2}}$
$\displaystyle\ll\sum_{\rho\
\textup{prime}}N(p)^{-2}\ll\sum_{k=1}^{\infty}k2^{-k}<\infty.$

That completes our proof. ∎

###### Proof of (8.9).

We control the difference between the Low term and the Main term. This is the
term Error, and we only seek a distributional estimate on it. We begin by
writing out this term explicitly. Recall that

(8.29) $\widehat{\operatorname{Lo}}(\xi)=\sum_{q\colon
N(q)<Q}\sum_{a\in\mathbb{A}_{q}}\Phi(a,\bar{q})\widehat{M_{N}^{\omega}}(\xi-\frac{a}{q})\Delta_{q_{0}}(\xi-\frac{a}{q}).$

By the disjointness of the supports of $\Delta_{q_{0}}(\cdot-\frac{a}{q})$,
for $N(q)<Q$, and the definition of the Main term in (8.12), we see that

(8.30)
$\displaystyle\widehat{\operatorname{Lo}}(\xi)\cdot\widehat{\operatorname{Lo}}(\xi)$
$\displaystyle=\sum_{N(q)\leq
Q}\sum_{a\in\mathbb{A}_{q}}\tau_{q}(a)^{2}\Delta_{q_{0}}(\xi-a/q)^{2}\widehat{M_{N}^{\omega}}(\xi-a/q)^{2}$

We have done the work to invert this Fourier transform. In particular from
(8.18), we have

(8.31) $\displaystyle{\operatorname{Lo}}\ast{\operatorname{Lo}}(x)$
$\displaystyle=M_{N}^{\omega}\ast\check{\Delta}_{q_{0}}\ast
M_{N}^{\omega}\ast\check{\Delta}_{q_{0}}(x)\sum_{q\colon
N(q)<N}\frac{\lvert\mu(q)\rvert}{\phi(q)^{2}}\tau_{q}(x).$

We can then explicitly write $\textup{Error}(x)$ as the sum of these three
terms.

(8.32) $\displaystyle E_{1}(x)$
$\displaystyle\coloneqq\int(1-\Delta_{q_{0}}(\xi)^{2})\widehat{M_{N}^{\omega}}(\xi)^{2}e(\langle\xi,x\rangle)\;d\xi\sum_{q\colon
N(q)<Q}\frac{\lvert\mu(q)\rvert}{\phi(q)^{2}}\tau_{q}(x),$ (8.33)
$\displaystyle E_{2}(x)$ $\displaystyle\coloneqq M_{N}^{\omega}\ast
M_{N}^{\omega}(x)\sum_{q\colon Q\leq
N(q)<N^{1/8}}\frac{\lvert\mu(q)\rvert}{\phi(q)^{2}}\tau_{q}(x),$ (8.34)
$\displaystyle E_{3}(x)$ $\displaystyle\coloneqq M_{N}^{\omega}\ast
M_{N}^{\omega}(x)\sum_{q\colon N(q)\geq
N^{1/8}}\frac{\lvert\mu(q)\rvert}{\phi(q)^{2}}\tau_{q}(x).$

We address them in order.

For the control of $E_{1}$ defined in (8.32), an easily accessible $\ell^{2}$
estimate applies. We recall that $Q\ll(\log N)^{B}$. By selection of
$\Delta_{q_{0}}$ as a scaled version of a Fejer kernel, we have
$1-\Delta_{q_{0}}(\xi)^{2}=O(Q^{-3})$ if $N(\xi)<Q^{-4}$. So in this range we
have

(8.35)
$\displaystyle\bigl{\lVert}\int_{N(\xi)<Q^{-4}}(1-\Delta_{q_{0}}(\xi)^{2})\widehat{M_{N}^{\omega}}(\xi)^{2}e(\langle\xi,x\rangle)\;d\xi\bigr{\rVert}_{\ell^{2}}\ll
Q^{-3}\int|\widehat{M_{N}^{\omega}}(\xi)|^{2}d\xi\ll N^{-1}Q^{-3}.$

But then it follows that

(8.36)
$\bigl{\lVert}\int(1-\Delta_{q_{0}}(\xi)^{2})\widehat{M_{N}^{\omega}}(\xi)^{2}e(\langle\xi,x\rangle)\;d\xi\bigr{\rVert}_{\ell^{2}}\ll
N^{-1}Q^{-3}.$

On the other hand, it is easy to see that

(8.37) $\sum_{q\colon
N(q)<Q}\frac{\lvert\mu(q)\rvert}{\phi(q)^{2}}\tau_{q}(x)\ll Q.$

These two estimates prove the control required in (8.9). In particular we have

(8.38) $\lVert E_{1}\rVert_{\ell^{2}}\ll(N^{2}(\log N)^{B-1})^{-1/2}.$

This estimate is stronger than than the required distributional estimate.

We turn to the term $E_{2}$ defined in (8.33). For this and $E_{3}$, the
leading term $M_{N}^{\omega}\ast M_{N}^{\omega}(x)\ll N^{-1}$, so our focus is
on the arithmetic terms. The definition $E_{2}$ requires that the denominators
$q$ are at least $Q$, and less than $N^{1/8}$. The point is that the Ramanujan
function $\tau_{q}$ is rarely more than $1$, as quantified by (3.26). For
integers $s$ with $Q/2<2^{s}\leq N^{1/8}$, we have

(8.39) $\Bigl{\lvert}\Bigl{\\{}N(x)<N\colon\sum_{2^{s}<N(q)\leq
2^{s+1}}\lvert\tau_{q}(x)\rvert>2^{3s/2}\Bigr{\\}}\Bigr{\rvert}\ll N2^{-3s}.$

This follows from (3.26), with $k=8$, and trivial bounds on the totient
function. It is clear that this can be summed over these values of $s$ to
complete the proof of (8.33) in this case. Indeed, we have

(8.40) $\displaystyle\Bigl{\lvert}\Bigl{\\{}N(x)<N\colon\sum_{2^{s}<N(q)\leq
2^{s+1}}\lvert\tau_{q}(x)\rvert>2^{3s/2}\Bigr{\\}}\Bigr{\rvert}$
$\displaystyle\ll 2^{-12s}\sum_{x\colon N(x)<N}\Bigr{[}\sum_{2^{s}<N(q)\leq
2^{s+1}}\lvert\tau_{q}(x)\rvert\Bigr{]}^{8}$ (8.41) $\displaystyle\ll
2^{-3s}N.$

Using a trivial bound lower bound on the Totient function will complete this
case.

We turn to the term $E_{3}$ defined in (8.34). In this case, we require
$N^{1/8}<N(q)$. And $N(q)$ can be as large and $e^{Q}=e^{c(\log N)^{B}}$. This
is too large to directly apply the previous argument. Instead, we will
specialize the proof of (3.26) to this setting. For an integer $s$ with
$N^{1/8}<2^{s+1}$, estimate

(8.42) $\displaystyle\sum_{\begin{subarray}{c}q\colon 2^{s}\leq
N(q)<2^{s+1}\\\ q\textup{\
$Q$-smooth}\end{subarray}}\frac{\lvert\mu(q)\rvert}{\phi(q)^{2}}\tau_{q}(x)$
$\displaystyle\ll s^{2}2^{-2s}\sum_{\begin{subarray}{c}q\colon 2^{s}\leq
N(q)<2^{s+1}\\\ q\textup{\ $Q$-smooth}\end{subarray}}N((x,q)).$

Above, we have used the familiar upper bound $\tau_{q}(x)\ll N((x,q))$. Write
$(x,q)=d$ and $q=q^{\prime}d$. Continue

(8.43) $\displaystyle\eqref{e:E31}$ $\displaystyle\ll
s^{2}2^{-2s}\sum_{d\textup{\ $Q$-smooth}}\mathbf{1}_{d\mid
x}N(d)\sum_{\begin{subarray}{c}q^{\prime}\colon 2^{s}\leq
N(q^{\prime})N(d)<2^{s+1}\\\ q^{\prime}\textup{\
$Q$-smooth}\end{subarray}}\mathbf{1}$ (8.44) $\displaystyle\ll
s^{2}2^{-s}\sum_{d\textup{\ $Q$-smooth}}\mathbf{1}_{d\mid x}\ll 2^{-s/2}.$

In the last line, we have $0<N(x)<N$, and $2^{s+1}>N^{1/8}$, so that we can
use a favorable estimate on the divisor function. This estimate is summable in
$s$. It follows that for $0<N(x)<N$, that we have

(8.45) $E_{3}(x)\ll N^{-17/16}.$

This completes the analysis of the Error term.

∎

###### Proof of (8.10).

We have from (6.3) and (6.13),

(8.46) $\displaystyle\lVert\textup{Lo}\ast\textup{Hi}\rVert_{\ell^{2}}^{2}$
$\displaystyle=\int_{\mathbb{T}^{2}}\lvert\widehat{\textup{Lo}}\cdot\widehat{\textup{Hi}}\rvert^{2}\;d\xi$
(8.47) $\displaystyle\ll
Q^{-1+\epsilon}\int_{\mathbb{T}^{2}}\lvert\widehat{\textup{Lo}}\rvert^{2}\;d\xi$
(8.48) $\displaystyle\ll\frac{Q^{-1+2\epsilon}}{N}.$

∎

The previous argument immediately implies the third and final estimate (8.11).
That completes the proof of Lemma 8.7, and hence the proof of our binary
Goldbach Theorem.

###  8.2. The Ternary Goldbach Conjecture

We turn to the ternary Goldbach conjecture, using the same notation. We will
show that for all intervals $\omega\subset\mathbb{T}$, there is an
$N_{\omega}$, so that for $N>N_{\omega}$, we have

(8.49) $\inf_{\begin{subarray}{c}n\in S^{\omega}_{N}\\\ \textup{$n$
odd}\end{subarray}}A_{N}^{\omega}\ast A_{N}^{\omega}\ast A_{N}^{\omega}(n)\gg
N^{-1}.$

That is, every odd integer in $S^{\omega}_{N}$ has many representations as a
sum of three primes, each of which is in the sector $\omega$. Indeed, there
are as many as $N^{2}/(\log N)^{2}$ representations. This holds for all large
$N$, and we can assume that the sector is small, so that completes the proof
of ternary Goldbach Theorem.

It remains to establish (8.49). We turn to the High/Low decomposition as in
(8.4), and write

(8.50) $\displaystyle A_{N}^{\omega}\ast A_{N}^{\omega}\ast A_{N}^{\omega}$
$\displaystyle=\textup{Hi}\ast A_{N}^{\omega}\ast
A_{N}^{\omega}+\textup{Lo}\ast A_{N}^{\omega}\ast A_{N}^{\omega}$ (8.51)
$\displaystyle=\textup{Hi}\ast A_{N}^{\omega}\ast
A_{N}^{\omega}+\textup{Lo}\ast\textup{Hi}\ast
A_{N}^{\omega}+\textup{Lo}\ast\textup{Lo}\ast A_{N}^{\omega}$ (8.52)
$\displaystyle=\textup{Hi}\ast A_{N}^{\omega}\ast
A_{N}^{\omega}+\textup{Lo}\ast\textup{Hi}\ast
A_{N}^{\omega}+\textup{Lo}\ast\textup{Lo}\ast\textup{Hi}+\textup{Lo}\ast\textup{Lo}\ast\textup{Lo}$
(8.53)
$\displaystyle\eqqcolon\textup{Err}+\textup{Lo}\ast\textup{Lo}\ast\textup{Lo}.$

As before, in (8.12) we will write

(8.54)
$\textup{Lo}\ast\textup{Lo}\ast\textup{Lo}=\textup{Main}+\textup{Error}.$

Thus, our focus is on the Lemma below. It easily completes the proof.

######  Lemma 8.55.

We have the estimates

(8.56) $\displaystyle N^{-1}$ $\displaystyle\ll\min_{\begin{subarray}{c}x\in
S^{\omega}_{N}\\\ \textup{$x$ odd}\end{subarray}}\textup{Main}(x),$ (8.57)
$\displaystyle\lVert\textup{Error}\rVert_{\ell^{\infty}}+\lVert\textup{Err}\rVert_{\ell^{\infty}}$
$\displaystyle\ll[N(\log N)^{B-3}]^{-1}.$

Recalling the definition of $Q$-smooth from (8.12), we set

(8.58) $\textup{Main}(x)\coloneqq\tilde{M}(x)\sum_{\textup{$q$ is $Q$
smooth}}\frac{\mu(q)}{\phi(q)^{3}}\tau_{q}(x),$

where

(8.59) $\tilde{M}(x)\coloneqq M_{N}^{\omega}\ast M_{N}^{\omega}\ast
M_{N}^{\omega}\ast\widecheck{\Delta_{q_{0}}}\ast\widecheck{\Delta_{q_{0}}}\ast\widecheck{\Delta_{q_{0}}}.$

###### Proof of (8.56).

The details here are very close to those of the binary case, so we will be a
little brief. Following the arguments of Lemma 8.13, in the definition of the
Main term, we have $\inf_{x\in S^{\omega}_{N}}\tilde{M}(x)\gg N^{-1}$. In the
arithmetic part of (8.58), we have the Möbius function $\mu(q)$, instead of
$\lvert\mu(q)\rvert$ as in the binary case, and a third power of the totient
function. Using the multiplicative properties, we have

(8.60) $\displaystyle\sum_{\textup{$q$ is $Q$
smooth}}\frac{\mu(q)}{\phi(q)^{3}}\tau_{q}(x)$
$\displaystyle=\prod_{N(p)<Q,p|x}\left(1-\frac{1}{\phi^{2}(p)}\right)\prod_{N(p)<Q,(p,x)=1}\left(1+\frac{1}{\phi^{3}(p)}\right)$
(8.61) $\displaystyle\coloneqq h_{3}(x)\mathcal{G}_{3}$

Parity is again crucial. We have

(8.62) $1-\frac{\tau_{1+i}(x)}{\phi^{3}(1+i)}=\begin{cases}0&\textup{$x$
even}\\\ 2&\textup{$x$ odd}\end{cases}$

That is, $h_{3}(x)$ is positive for odd $x$. Note that $\mathcal{G}_{3}=O(1)$,
because

(8.63)
$\displaystyle\mathcal{G}_{3}=\prod_{N(p)<Q,(p,x)=1}\left(1+\frac{1}{\phi^{3}(p)}\right)<\sum_{n}\frac{|\mu(n)|}{N(n)^{2}}=O(1).$

This completes the proof of (8.56). ∎

###### Proof of (8.57).

There are two estimates to prove. The first is to bound the $\ell^{\infty}$
norm of

(8.64) Error
$\displaystyle\coloneqq\textup{Lo}\ast\textup{Lo}\ast\textup{Lo}-\textup{Main}.$

Switch to Fourier variables. We have

(8.65)
$\displaystyle\widehat{\textup{Lo}}(\xi)^{3}=\sum_{q<Q}\sum_{a\in\mathbb{A}_{q}}\widehat{\tilde{M}}(\xi-a/q)\frac{\tau_{q}(a)^{3}}{\phi(q)^{3}}.$

It follows that

(8.66)
$\displaystyle\widehat{\textup{Error}}(\xi)=\sum_{\begin{subarray}{c}q\textup{\
$Q$-smooth}\\\ q\geq
Q\end{subarray}}\sum_{a\in\mathbb{A}_{q}}\widehat{\tilde{M}}(\xi-a/q)\frac{\tau_{q}(a)^{3}}{\phi(q)^{3}}.$

So, the $\ell^{\infty}$ norm of Error is at most the $L^{1}(\mathbb{T}^{2})$
norm of the expression above. That is at most

(8.67) $\displaystyle\sum_{\begin{subarray}{c}q\textup{\ $Q$-smooth}\\\ q\geq
Q\end{subarray}}\Bigl{\lVert}\sum_{a\in\mathbb{A}_{q}}\widehat{\tilde{M}}(\xi-a/q)$
$\displaystyle\tau_{q}(a)^{3}\Bigr{\rVert}_{L^{1}(\mathbb{T}^{2})}$ (8.68)
$\displaystyle\ll N^{-1}\sum_{\begin{subarray}{c}q\geq
Q\end{subarray}}\phi(q)^{-2+\epsilon}\ll N^{-1}Q^{-1+\epsilon}.$

By our choice of $Q=(\log N)^{B}$, this estimate meets our requirements.

The second estimate concerns the term Err. The term Err is a sum of three
terms of the form $\phi_{1}\ast\phi_{2}\ast\phi_{3}$, where
$\phi_{j}\in\\{A_{N},\textup{Hi},\textup{Lo}\\}$. And, at least one of the
terms is a High term. See (8.52). We control each term. To fix ideas, consider

(8.69) $\displaystyle\lVert\operatorname{Hi}\ast A_{N}^{\omega}\ast
A_{N}^{\omega}\rVert_{\infty}$ $\displaystyle\leq\lVert\operatorname{Hi}\ast
A_{N}^{\omega}\rVert_{2}\lVert A_{N}^{\omega}\rVert_{2}$ (8.70)
$\displaystyle\leq\lVert\widehat{\operatorname{Hi}}\widehat{A_{N}^{\omega}}\rVert_{2}\lVert\widehat{A_{N}^{\omega}}\rVert_{2}$
(8.71)
$\displaystyle\leq\lVert\widehat{\operatorname{Hi}}\rVert_{\infty}\lVert\widehat{A_{N}^{\omega}}\rVert_{2}^{2}$

where the last inequality is trivial. Now,
$\lVert\widehat{\operatorname{Hi}}\rVert_{\infty}\ll Q^{-1+\epsilon}$, by
(6.3). Recall that $Q=(\log N)^{B}$, for an integer $B>10$. And,

(8.72) $\displaystyle\lVert\widehat{A_{N}^{\omega}}\rVert_{2}^{2}$
$\displaystyle=\lVert{A_{N}^{\omega}}\rVert_{2}^{2}$ (8.73) $\displaystyle\ll
N^{-2}\sum_{N(n)<N,\arg(n)\in\omega}\Lambda(n)^{2}$ (8.74) $\displaystyle\ll
N^{-1}(\log N)^{2}$

We see that $\lVert\operatorname{Hi}\ast A_{N}^{\omega}\ast
A_{N}^{\omega}\rVert_{\infty}\ll N^{-1}(\log N)^{-(1-\epsilon)B-2}$. That
completes this case.

The second term to control is

(8.75)
$\displaystyle\lVert\operatorname{Hi}\ast\operatorname{Lo}\ast\operatorname{Lo}\rVert_{\infty}$
$\displaystyle\leq\lVert\widehat{\operatorname{Hi}}\rVert_{\infty}\lVert\operatorname{Lo}\rVert_{2}^{2}.$

To estimate this last term $\lVert\operatorname{Lo}\rVert_{2}^{2}$, use
(6.13), which gives a better estimate than the first term. The third term to
control is

(8.76) $\displaystyle\lVert\operatorname{Hi}\ast
A_{N}^{\omega}\ast\operatorname{Lo}\rVert_{\infty}$
$\displaystyle\leq\lVert\widehat{\operatorname{Hi}}\rVert_{\infty}\lVert
A_{N}^{\omega}\rVert_{2}\lVert\operatorname{Lo}\rVert_{2}.$

But the right hand side is the geometric mean of the other two terms. That
completes the proof.

∎

## References