# Variants of the Selberg sieve, and almost prime $k$-tuples

Paweł Lewulis Supported by NCN Sonatina 3, 2019/32/C/ST1/00341.

###### Abstract

Let $k\geq 2$ and $\mathcal{P}(n)=(A_{1}n+B_{1})\cdots(A_{k}n+B_{k})$ where
all the $A_{i},B_{i}$ are integers. Suppose that $\mathcal{P}(n)$ has no fixed
prime divisors. For each choice of $k$ it is known that there exists an
integer $\varrho_{k}$ such that $\mathcal{P}(n)$ has at most $\varrho_{k}$
prime factors infinitely often. We used a new weighted sieve set-up combined
with a device called an $\varepsilon$-trick to improve the possible values of
$\varrho_{k}$ for $k\geq 7$. As a by-product of our approach, we improve the
conditional possible values of $\varrho_{k}$ for $k\geq 4$, assuming the
generalized Elliott–Halberstam conjecture.

##  1 Introduction

### State of the art

Let us begin with recalling the following notion.

######  Definition (Admissible tuples).

Fix a positive integer $k$. For each $i=1,\dots,k$ fix integers $A_{i}$,
$B_{i}$, such that $A_{i}>0$, and let
$L_{i}\colon\mathbf{Z}^{+}\rightarrow\mathbf{Z}$ be a function given by the
formula $L_{i}(n):=A_{i}n+B_{i}$. For each positive integer $n$ put

$\mathcal{P}(n):=\prod_{i=1}^{k}L_{i}(n).$

We call $\mathcal{H}:=\\{L_{1},\dots,L_{k}\\}$ an admissible $k$–tuple, if for
every prime $p$ there is an integer $n_{p}$ such that none of the
$L_{i}(n_{p})$ is a multiple of $p$.

We are interested in the following problem being a vast generalization of the
twin primes conjecture.

######  Conjecture 1 (Dickson–Hardy–Littlewood).

Fix a positive integer $k$. Let $\\{L_{1},\dots,L_{k}\\}$ be an admissible
$k$–tuple. Then,

$\liminf_{n\rightarrow\infty}\Omega(\mathcal{P}(n))=k.$ (1)

One may reformulate the statement above into a general question about the
total number of prime factors contained within $\mathcal{P}(n)$. This creates
a way to ‘measure’ of how far we are from proving Conjecture 1.

######  Problem 2 ($DHL_{\Omega}$).

Fix positive integers $k$ and $\varrho_{k}\geq k$. Let
$\\{L_{1},\dots,L_{k}\\}$ be an admissible $k$–tuple. The task is to prove
that

$\liminf_{n\rightarrow\infty}\Omega(\mathcal{P}(n))\leq\varrho_{k}.$ (2)

From this point on, if inequality (2) is true for some precise choice of $k$
and $\varrho_{k}$, then we say that $DHL_{\Omega}[k;\varrho_{k}]$ holds. In
the case $k=1$, the classical Dirichlet’s theorem is equivalent to
$DHL_{\Omega}[1;1]$. This is also the only instance where the optimal possible
value of $\varrho_{k}$ is already known. For $k=2$ we have $DHL_{\Omega}[2;3]$
by Chen’s theorem proven in [Chen]. If $k\geq 3$, then the state of the art
and recent progress are described below.

Table A. State of the art – obtained values $\varrho_{k}$ for which
$DHL_{\Omega}[k;\varrho_{k}]$ holds.

Unconditional case.

$k$ |  3 |  4 |  5 |  6 |  7 |  8 |  9 |  10 |
---|---|---|---|---|---|---|---|---|---
Halberstam, Richert [Halberstam-Richert] |  10 |  15 |  19 |  24 |  29 |  34 |  39 |  45 |
Porter [Porter] |  8 |  |  |  |  |  |  |  |
Diamond, Halberstam [DH] |  |  12 |  16 |  20 |  25 |  29 |  34 |  39 |
Ho, Tsang [HT] |  |  |  |  |  24 |  28 |  33 |  38 |
Maynard [3-tuples, MaynardK] |  7 |  11 |  15 |  18 |  22 |  26 |  30 |  34 |
Lewulis [Lewulis] |  |  |  14 |  |  |  |  |  |
This work |  |  |  |  |  21 |  25 |  29 |  33 |

The $GEH$ case.

$k$ |  3 |  4 |  5 |  6 |  7 |  8 |  9 |  10 |
---|---|---|---|---|---|---|---|---|---
Sono [Sono] |  6 |  |  |  |  |  |  |  |
Lewulis [Lewulis] |  |  10 |  13 |  17 |  20 |  24 |  28 |  32 |
This work |  |  8 |  11 |  14 |  17 |  21 |  24 |  27 |

### Notation

The letter $p$ with possible indices always denotes a prime number and
${\log}$ denotes the natural logarithm. We use the notation
$\mathbf{N}=\\{1,2,3,\dots\\}$. We also use the following definitions listed
below:

  * •

$\varphi(n):=\\#\left(\mathbf{Z}/n\mathbf{Z}\right)^{\times}$ denotes Euler
totient function;

  * •

$\tau(n):=\sum_{d|n}1$ denotes the divisor function;

  * •

$\Omega(n)$ denotes the number of prime factors of $n$;

  * •

$\pi(x):=\\#\left\\{n\in\mathbf{N}:n\leq x,\leavevmode\nobreak\ n\text{ is
prime}\right\\}$;

  * •

$\pi(x;q,a):=\\#\left\\{n\in\mathbf{N}:n\leq x,\leavevmode\nobreak\ n\equiv
a\bmod q,\leavevmode\nobreak\ n\text{ is prime}\right\\}$;

  * •

$\log_{y}x:=\frac{\log x}{\log y}$ for $x,y>0$ and $y\not=1$;

  * •

By $(a,b)$ and $[a,b]$ we denote the greatest common divisor and the lowest
common multiple, respectively;

  * •

For a logical formula $\phi$ we define the indicator function
$\mathbf{1}_{\phi(x)}$ that equals $1$ when $\phi(x)$ is true and $0$
otherwise;

  * •

For a set $A$ we define the indicator function $\mathbf{1}_{A}$ that equals
$1$ when the argument belongs to $A$ and $0$ otherwise;

  * •

By $\text{gpf}(n)$ and $\text{lpf}(n)$ we denote the greatest and the lowest
prime divisor of $n$ respetively;

  * •

The condition $n\sim x$ means that $x<n\leq 2x$;

  * •

For a function $F$ being a map between some two abelian groups we define the
difference operator $\partial_{y}F(x):=F(x+y)-F(x)$;

  * •

We define an analogous operator for a function $F$ with $m$ variables, namely
$\partial_{y}^{(i)}F(x_{1},\dots,x_{m}):=F(x_{1},\dots,x_{i-1},x_{i}+y,x_{i+1},\dots,x_{m})-F(x_{1},\dots,x_{m})$;

  * •

For every compactly supported function
$F\colon[0,+\infty)\rightarrow\mathbf{R}$ we define

$S(F):=\sup\left(\\{x\in\mathbf{R}\colon F(x)\not=0\\}\cup\\{0\\}\right);$

  * •

We define a normalizing expression $B:=\frac{\varphi(W)\log x}{W}$ (cf. next
subsection);

  * •

Symmetric polynomials of degree $m$ and $k$ variables
$\sum_{j=1}^{k}t_{j}^{m}$ are denoted as $P_{m}$;

  * •

For a finitely supported arithmetic function
$f\colon\mathbf{N}\rightarrow\mathbf{C}$ we define a discrepancy

$\Delta\left(f;a\bmod q\right):=\sum_{n\equiv a\bmod
q}f(n)-\frac{1}{\varphi(q)}\sum_{(n,q)=1}f(n)\,;$

  * •

For any $f\colon\mathbf{R}\rightarrow\mathbf{R}$ we define a function related
to Selberg weights

$\lambda_{f}(n):=\sum_{d|n}\mu(d)f(\log_{x}d);$

  * •

We also make use of the ‘big $O$’, the ‘small $o$’, and the ‘$\ll$’ notation
in a standard way.

### The general set-up

Let us fix $k\in\mathbf{Z}^{+}$ and consider the expression

$\mathcal{S}:=\sum_{\begin{subarray}{c}n\sim x\\\ n\equiv b\bmod
W\end{subarray}}\left(\varrho_{k}-\Omega(\mathcal{P}(n))\right)\nu(n),$ (3)

where $\nu$ is some arbitrarily chosen sequence of non-negative weights. Put

$W:=\prod_{p<D_{0}}p$

for $D_{0}:=\log\log\log x$ and take some integer $b$ coprime to $W$. We
choose some residue class $b$ such that $\mathcal{P}(b)$ is coprime to $W$ and
then, we restrict our attention to $n\equiv b\bmod W$. This way we discard all
irregularities caused by very small prime numbers. Put
$A:=4\max\left\\{|A_{1}|,|B_{1}|,\dots,|A_{k}|,|B_{k}|\right\\}$. Assume that
$x>10\,000$ and $D_{0}>A$. Thus, our goal is to show that

$\mathcal{S}=\varrho_{k}\mathcal{S}_{0}-\mathcal{S}_{\Omega}>0,$ (4)

where

$\begin{split}\mathcal{S}_{0}&:=\sum_{\begin{subarray}{c}n\sim x\\\ n\equiv
b\bmod W\end{subarray}}\nu(n),\\\
\mathcal{S}_{\Omega}&:=\sum_{\begin{subarray}{c}n\sim x\\\ n\equiv b\bmod
W\end{subarray}}\Omega(\mathcal{P}(n))\nu(n).\\\ \end{split}$ (5)

The main difficulty is to calculate $\mathcal{S}_{\Omega}$ with sufficient
accuracy. One possible method and a good source of inspiration for new tools
is the following identity valid for square-free $n\leqslant x$:

$\Omega(n)=\sum_{p|n}1=\mathbf{1}_{\text{gpf}(n)>U}+\sum_{\begin{subarray}{c}p|n\\\
p\leq U\end{subarray}}1,$ (6)

where $U>x^{1/2}$ (usually, $U=x^{1/2+\epsilon}$ for some small $\epsilon>0$
has been considered). For instance, one can exploit the simple inequality

$\Omega(\mathcal{P}(n))\leavevmode\nobreak\ =\leavevmode\nobreak\
\sum_{i=1}^{k}\mathbf{1}_{\text{gpf}(L_{i}(n))>U}\leavevmode\nobreak\
+\sum_{\begin{subarray}{c}p|\mathcal{P}(n)\\\ p\leq
U\end{subarray}}1\leavevmode\nobreak\ \leq\leavevmode\nobreak\
k\leavevmode\nobreak\ +\sum_{\begin{subarray}{c}p|\mathcal{P}(n)\\\ p\leq
U\end{subarray}}1$ (7)

under the previous assumptions. This reasoning leads to results that are
nontrivial, but weaker than already existing in literature. However, the
interesting observation about this identity is that one does not need to rely
on any distributional claims about primes in arithmetic progressions in order
to exploit it.

In [MaynardK] and [Lewulis] the authors applied the following identity valid
for all square-free $n\sim x$:

$\Omega(n)=\frac{\log n}{\log T}+\sum_{p|n}\left(1-\frac{\log p}{\log
T}\right),$ (8)

where $T:=x^{l}$ for some exponent $l\in(0,1]$. This approach combined with
(6) gives some flexibility, because the expression in the parentheses in (8)
is negative for $p>T$. In such case, we can transform the task of seeking for
upper bounds for $\mathcal{S}_{\Omega}$ into problem of establishing lower
bounds. The idea was to apply the following partition of unity:

$1=\sum_{r}\mathbf{1}_{\Omega(n)=r}\geq\sum_{r\leq
H}\mathbf{1}_{\Omega(n)=r},$ (9)

valid for any $H>0$, and then, to calculate the contribution of $S_{\Omega}$
via (8) and (9), usually for $H=3,4$, depending on specific cases.

In this work we propose a different approach and we establish the asymptotic
behaviour of $\mathcal{S}_{\Omega}$. Such a result is sufficient to improve
the currently known values of $\varrho_{k}$ in the conditional case, that is
when $GEH$ (cf. Section ‘Preparing the sieve’ for definitions) is true. It
also greatly simplifies the unconditional results from [Lewulis] and explains
Conjecture 4.2 formulated there, which turns out to be slightly incorrect.

To tackle the uncondtitional case, we need to expand the sieve support beyond
the domain offered by the standard claims regarding primes in arithemetic
progressions (Theorem 5, in particular). Hence, we incorporate a device
invented in [Polymath8] called an $\varepsilon$-trick. In order to do so, we
have to apply (8). The reason for this is that the $\varepsilon$-trick is all
about bounding the sieve weights from below. In the same time, we wish to
apply this tool to $\mathcal{S}_{\Omega}$, which has to be estimated from
above. As we noticed before, (8) enables us to partially convert upper bounds
into lower bounds, at least until the prime factors are sufficiently large. On
the other hand, if they are small, we do not need to rely on any
distributional claim on primes in arithmetic progressions at all, so in this
case we can expand the sieve support almost freely.

To summarize, we propose a general set-up that is flexible enough to cover all
applications appearing in this work. We have the following criterion for our
main problem.

######  Lemma 3.

Let $k\geq 2$ and $\varrho\geq k$ be fixed integers. Suppose that for each
fixed admissible $k$–tuple $\\{L_{1},\dots,L_{k}\\}$ and each residue class
$b\bmod W$ such that $(L_{i}(b),W)=1$ for all $i=1,\dots,k$, one can find a
non-negative weight function $\nu\colon\mathbf{N}\rightarrow\mathbf{R}^{+}$
and fixed quantities $\alpha>0$, and $\beta_{1},\dots\beta_{k}\geq 0$, such
that one has the asymptotic lower bound

$\sum_{\begin{subarray}{c}n\sim x\\\ n\equiv b\bmod
W\end{subarray}}\nu(n)\geq\left(\alpha-o(1)\right)B^{-k}\frac{x}{W},$ (10)

and the asymptotic upper bounds

$\displaystyle\sum_{\begin{subarray}{c}n\sim x\\\ n\equiv b\bmod W\\\
\mathcal{P}(n)\textup{ sq-
free}\end{subarray}}\sum_{p|L_{i}(n)}\left(1-\ell\log_{x}p\right)\nu(n)$
$\displaystyle\leq(\beta_{i}+o(1))B^{-k}\frac{x}{W},$ (11)
$\displaystyle\sum_{\begin{subarray}{c}n\sim x\\\ n\equiv b\bmod W\\\
\mathcal{P}(n)\textup{ not sq-
free}\end{subarray}}\tau(\mathcal{P}(n))\left|\nu(n)\right|$
$\displaystyle\leq o(1)\times B^{-k}\frac{x}{W}$ (12)

for all $i=1,\dots,k$, and the key inequality

$\varrho>\frac{\beta_{1}+\dots+\beta_{k}}{\alpha}+\ell k.$

Then, $DHL_{\Omega}[k;\varrho]$ holds. Moreover, if one replaces inequalities
(10–11) with equalities, then the right-hand side of the key inequality above
is constant with respect to the $\ell$ variable.

###### Proof.

We have

$\sum_{\begin{subarray}{c}n\sim x\\\ n\equiv b\bmod
W\end{subarray}}\left(\varrho-\Omega(\mathcal{P}(n)\right)\nu(n)=\varrho\left(\sum_{\begin{subarray}{c}n\sim
x\\\ n\equiv b\bmod
W\end{subarray}}\nu(n)\right)-\left(\sum_{\begin{subarray}{c}n\sim x\\\
n\equiv b\bmod W\\\ \mathcal{P}(n)\textup{ sq-
free}\end{subarray}}\Omega(\mathcal{P}(n))\nu(n)\right)\\\
+O\left(\sum_{\begin{subarray}{c}n\sim x\\\ n\equiv b\bmod W\\\
\mathcal{P}(n)\textup{ not sq-
free}\end{subarray}}\tau(\mathcal{P}(n))\nu(n)\right).$ (13)

We also observe that

$\sum_{\begin{subarray}{c}n\sim x\\\ n\equiv b\bmod W\\\
\mathcal{P}(n)\textup{ sq-
free}\end{subarray}}\Omega(\mathcal{P}(n))\nu(n)=\left(\sum_{i=1}^{k}\sum_{\begin{subarray}{c}n\sim
x\\\ n\equiv b\bmod W\\\ \mathcal{P}(n)\textup{ sq-
free}\end{subarray}}\sum_{p|L_{i}(n)}\left(1-\ell\log_{x}p\right)\nu(n)\right)\\\
+(\ell k+o(1))\left(\sum_{\begin{subarray}{c}n\sim x\\\ n\equiv b\bmod
W\end{subarray}}\nu(n)\right).$ (14)

Combining (13–14) with the assumptions we arrive at

$\sum_{\begin{subarray}{c}n\sim x\\\ n\equiv b\bmod
W\end{subarray}}\left(\varrho-\Omega(\mathcal{P}(n)\right)\nu(n)\geq\left((\varrho+\ell
k)\,\alpha-\sum_{i=1}^{k}\beta_{i}-o(1)\right)B^{-k}\frac{x}{W}.$ (15)

Note that (15) becomes an equality, if one replaces inequalities (10–11) with
equalities – in such a case the left-hand side of (15) obviously does not
depend on the $\ell$ variable, so the same has to be true for the right-hand
side of (15). We conclude that the left-hand side of (15) is asymptotically
greater than $0$ if

$\varrho>\frac{\beta_{1}+\dots+\beta_{k}}{\alpha}+\ell k.$ (16)

∎

As mentioned in Table A, the main goal of this work is to prove the following
result.

######  Theorem 4 (Main Theorem).

$DHL_{\Omega}[k,\varrho_{k}]$ holds with the values $\varrho_{k}$ given in a
table below __

Table B.

$k$ |  2 |  3 |  4 |  5 |  6 |  7 |  8 |  9 |  10 |
---|---|---|---|---|---|---|---|---|---|---
Unconditionally |  4 |  7 |  11 |  14 |  18 |  21 |  25 |  29 |  33 |
Assuming $GEH$ |  3 |  6 |  8 |  11 |  14 |  17 |  21 |  24 |  27 |

_(bolded text indicates the novelties in the field).___

### Preparing the sieve

In this subsection we are focused on motivating our future choice of sieve
weights $\nu(n)$, so this discussion will be slightly informal. Our task is to
make the sum (3) greater than 0 for some fixed $\varrho_{k}$. That would be
sufficient to prove that $DHL_{\Omega}[k;\varrho_{k}]$ holds. Hence, the
weight $\nu$ has to be sensitive to almost prime $k$-tuples. We observe that
the von Mangoldt function satisfies

$\Lambda(n)=\left(\mu*\log\right)(n)=-\sum_{d|n}\mu(d)\log d,$

which for square-free $n\sim x$ gives

$\mathbf{1}_{n\text{ is
prime}}\approx\sum_{d|n}\mu(d)\left(1-\log_{x}d\right).$ (17)

That motivates the following construction of the Selberg sieve:

$\mathbf{1}_{n\text{ is prime}}\lessapprox
f(0)\left(\sum_{\begin{subarray}{c}d|n\end{subarray}}\mu(d)f(\log_{x}d)\right)^{2},$
(18)

where $f\colon[0,+\infty)\rightarrow\mathbf{R}$ is piecewise smooth and
supported on $[0,1)$. The problem is that the Bombieri–Vinogradov theorem
usually forces us to assume that $\mbox{supp}(f)\subset[0,\theta)$ for some
fixed positive $\theta$. The usual choice here is $\theta$ somewhat close to
$1/4$, or greater, if one assumes the Elliott–Halberstam conjecture.

In the multidimensional setting we have

$\mathbf{1}_{L_{1}(n),\dots,L_{k}(n)\text{ are all primes}}\lessapprox
f(0,\dots,0)\left(\sum_{\begin{subarray}{c}d_{1},\dots,d_{k}\\\ \forall
i\leavevmode\nobreak\
d_{i}|L_{i}(n)\end{subarray}}\left(\prod_{i=1}^{k}\mu(d_{i})\right)f\left(\log_{x}d_{1},\dots,\log_{x}d_{k}\right)\right)^{2}$
(19)

for some $f\colon[0,+\infty)^{k}\rightarrow\mathbf{R}$ being piecewise smooth
and compactly supported. In certain cases this approach can be more efficient
than (18), as was shown in [Maynard], where it was introduced. Dealing with
multivariate summations may be tedious at times, so we would like to transform
the right-hand side of $(\ref{MSS})$ a bit by replacing the function $f$ with
tensor products

$f_{1}(\log_{x}d_{1})\cdots f_{k}(\log_{x}d_{k}),$ (20)

where $f_{1},\dots,f_{k}\colon\mathbf{[}0,+\infty)\rightarrow\mathbf{R}$. By
the Stone–Weierstrass theorem we can approximate $f$ by a linear combination
of functions of such form, so essentially we lose nothing here. Our more
convenient sieve weights look as follows:

$\left(\sum_{j=1}^{J}c_{j}\prod_{i=1}^{k}\lambda_{f_{j,i}}(L_{i}(n))\right)^{2}$
(21)

with some real coefficients $c_{j}$, some smooth and compactly supported
functions $f_{i,j}$. Recall that

$\lambda_{f}(n):=\sum_{d|n}\mu(d)f(\log_{x}d).$

It is clear that such a weight can be decomposed into linear combination of
functions of the form

$n\mapsto\prod_{i=1}^{k}\lambda_{F_{i}}(L_{i}(n))\lambda_{G_{i}}(L_{i}(n)).$
(22)

In fact, (21) is exactly our choice in Section 4.

### Distributional claims concerning primes

In this work we refer to the generalised Elliott–Halberstam conjecture,
labeled further as $GEH[\vartheta]$ for some $0<\vartheta<1$. This broad
generalisation first appeared in [GEH]. Its precise formulation can be found
for example in [Polymath8]. The best known result in this direction is
currently proven by Motohashi [Motohashi].

######  Theorem 5.

$GEH[\vartheta]$ holds for every $\vartheta\in(0,1/2)$.

In this work we actually need only one specific corollary of $GEH$, which can
be perceived as an ‘Elliott–Halberstam conjecture for almost primes’.

######  Theorem 6.

Assume $GEH[\vartheta]$. Let $r\geq 1$, $\epsilon>0$, and $A\geq 1$ be fixed.
Let

$\Delta_{r,\epsilon}=\\{(t_{1},\dots,t_{r})\in[\epsilon,1]^{r}\colon\leavevmode\nobreak\
t_{1}\leq\dots\leq t_{r};\leavevmode\nobreak\ t_{1}+\dots+t_{r}=1\\},$

and let $F\colon\Delta_{r,\epsilon}\rightarrow{\bf R}$ be a fixed smooth
function. Let $f\colon{\bf N}\rightarrow{\bf R}$ be the function defined by
setting

$\displaystyle f(n)=F\left(\log_{n}p_{1},\dots,\log_{n}p_{r}\right)$

whenever $n=p_{1}\dots p_{k}$ is the product of $r$ distinct primes
$p_{1}<\dots<p_{r}$ with $p_{1}\geq x^{\epsilon}$ for some fixed $\epsilon>0$,
and $f(n)=0$ otherwise. Then for every $Q\ll x^{\vartheta}$, we have

$\displaystyle\sum_{q\leq
Q}\max_{\begin{subarray}{c}(a,q)=1\end{subarray}}\left|\Delta\left(\mathbf{1}_{[1,x]}f;a\bmod
q\right)\right|\ll x\log^{-A}x.$

##  2 Outline of the key ingredients

Let us start from presenting a minor variation of [Polymath8, Theorem 3.6].
The only change we impose is replacing the linear forms of the shape $n+h_{i}$
by slightly more general $L_{i}(n)$. This, however, does not affect the proof
in any way.

######  Proposition 7 (Non-$\Omega$ sums).

Let $k\geq 1$ be fixed, let $\\{L_{1},\dots,L_{k}\\}$ be a fixed admissible
k-tuple, and let $b\bmod W$ be such that $(L_{i}(b),W)=1$ for each
$i=1,\dots,k$. For each fixed $1\leq i\leq k$, let
$F_{i},\,G_{i}\colon[0,+\infty)\rightarrow\mathbf{R}$ be fixed smooth
compactly supported functions. Assume one of the following hypotheses:

  1. 1.

(Trivial case) One has

$\sum_{i=1}^{k}(S(F_{i})+S(G_{i}))<1.$

  2. 2.

(Generalized Elliott–Halberstam) There exist a fixed $0<\vartheta<1$ and
$i_{0}\in\\{1,\dots,k\\}$ such that $GEH[\vartheta]$ holds, and

$\sum_{\begin{subarray}{c}1\leq i\leq k\\\
i\not=i_{0}\end{subarray}}(S(F_{i})+S(G_{i}))<\vartheta.$

Then, we have

$\sum_{\begin{subarray}{c}n\sim x\\\ n\equiv b\bmod
W\end{subarray}}\prod_{i=1}^{k}\lambda_{F_{i}}(L_{i}(n))\lambda_{G_{i}}(L_{i}(n))=(c+o(1))B^{-k}\frac{x}{W},$

where

$c:=\prod_{i=1}^{k}\left(\int\limits_{0}^{1}F_{i}^{\prime}(t_{i})\,G_{i}^{\prime}(t_{i})\,dt_{i}\right).$

The next result is a crucial component of this work and is a novelty in the
topic. Together with Proposition 7 it creates a way to transform
$\mathcal{S}_{0}$ and $\mathcal{S}_{\Omega}$ into integrals, effectively
converting the main task of finding almost primes into an optimization
problem.

######  Proposition 8 (Sums containing $\Omega$ function).

Let $k\geq 1$ and $i_{0}\in\\{1,\dots,k\\}$ be fixed, let
$\\{L_{1},\dots,L_{k}\\}$ be a fixed admissible k-tuple, and let $b\bmod W$ be
such that $(L_{i}(b),W)=1$ for each $i=1,\dots,k$. For each fixed $1\leq i\leq
k$, let $F_{i},\,G_{i},\colon[0,+\infty)\rightarrow\mathbf{R}$ be fixed smooth
compactly supported functions, and let
$\Upsilon\colon[0,+\infty)\rightarrow\mathbf{R}$ be a bounded Riemann
integrable function continuous at $1$. Assume that there exist
$\vartheta,\,\vartheta_{0}\in(0,1)$ such that one of the following hypoteses
holds:

  1. 1.

(Trivial case) One has

$\sum_{i=1}^{k}(S(F_{i})+S(G_{i}))<1-\vartheta_{0}\leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
and\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ S(\Upsilon)<\vartheta_{0}.$

  2. 2.

(Generalized Elliott–Halberstam) Assume that $GEH[\vartheta]$ holds, and

$\sum_{\begin{subarray}{c}1\leq i\leq k\\\
i\not=i_{0}\end{subarray}}(S(F_{i})+S(G_{i}))<\vartheta.$

Then, we have

$\sum_{\begin{subarray}{c}n\sim x\\\ n\equiv b\bmod W\\\
\mathcal{P}(n)\leavevmode\nobreak\ \textup{sq-
free}\end{subarray}}\left(\sum_{p|L_{i_{0}}(n)}\Upsilon(\log_{x}p)\right)\prod_{i=1}^{k}\lambda_{F_{i}}(L_{i}(n))\lambda_{G_{i}}(L_{i}(n))=(c+o(1))B^{-k}\frac{x}{W},$
(23)

where

$\begin{split}c:=\left(\Upsilon(1)\,F_{i_{0}}(0)\,G_{i_{0}}(0)\leavevmode\nobreak\
+\leavevmode\nobreak\
\int\limits_{0}^{1}\frac{\Upsilon(y)}{y}\int\limits_{0}^{1-y}\partial_{y}F^{\prime}_{i_{0}}(t_{i_{0}})\,\partial_{y}G^{\prime}_{i_{0}}(t_{i_{0}})\,dt_{i_{0}}\,dy\right)\prod_{\begin{subarray}{c}1\leq
i\leq k\\\
i\not=i_{0}\end{subarray}}\left(\int\limits_{0}^{1}F_{i}^{\prime}(t_{i})\,G_{i}^{\prime}(t_{i})\,dt_{i}\right).\end{split}$

The first case of Proposition 8 is strongly related to [MaynardK, Proposition
5.1] and [Lewulis, Proposition 1.13]. It is worth mentioning that the
conditional results in the latter of these two cited papers relied only on
$GEH[2/3]$. It was not possible to invoke the full power of $GEH$ by methods
studied there due to certain technical obstacles. The second case of
Proposition 8 is strong enough to overcome them. It also paves a way to
conveniently apply a device called an $\varepsilon$-trick in the unconditional
setting.

The role of the last proposition in this section is to deal with the
contribution from $n$ such that $\mathcal{P}(n)$ is not square-free.

######  Proposition 9 (Sums with double prime factors).

Let $k\geq 1$ be fixed, let $\\{L_{1},\dots,L_{k}\\}$ be a fixed admissible
k-tuple, and let $b\bmod W$ be such that $(L_{i}(b),W)=1$ for each
$i=1,\dots,k$. For each fixed $1\leq i\leq k$, let
$F_{i},\,G_{i}\colon[0,+\infty)\rightarrow\mathbf{R}$ be fixed smooth
compactly supported functions. Then, we have

$\sum_{\begin{subarray}{c}n\sim x\\\ n\equiv b\bmod W\\\ \mathcal{P}(n)\emph{
not sq-
free}\end{subarray}}\tau(\mathcal{P}(n))\left|\prod_{i=1}^{k}\lambda_{F_{i}}(L_{i}(n))\lambda_{G_{i}}(L_{i}(n))\right|=o(1)\times
B^{-k}\frac{x}{W}.$

Now, we combine Propositions 7–9 to obtain Theorems 10, 12, and 13 giving us
criteria for the $DHL_{\Omega}$ problem. Theorem 10 refers to sieving on
standard simplex $\mathcal{R}_{k}$, which can be considered as a default range
for the multidimensional Selberg sieve. The next one, Theorem 12, deals with
the extended simplex $\mathcal{R}_{k}^{\prime}$, which was applied in
[Lewulis], where $DHL_{\Omega}[5;14]$ was proven. We also prove Theorem 13
being the most general of these three. It describes sieving on the epsilon-
enlarged simplex. In fact, Theorems 10 and 12 are corollaries from Theorem 13,
as noted in Remark 2.

######  Theorem 10 (Sieving on a standard simplex).

Suppose that there is an arbitrarily chosen fixed real parameter $\ell$ and a
fixed $\theta\in(0,\frac{1}{2})$ such that $GEH[2\theta]$ holds. Let $k\geq 2$
and $m\geq 1$ be fixed integers. For any fixed compactly supported square-
integrable function $F\colon[0,+\infty)^{k}\rightarrow\mathbf{R}$, define the
functionals

$\begin{split}I(F):=&\int_{[0,+\infty)^{k}}F(t_{1},\dots,t_{k})^{2}\,dt_{1}\dots
dt_{k},\\\ Q_{i}(F):=&\int_{0}^{\frac{1}{\theta}}\frac{1-\ell\theta
y}{y}\int_{[0,+\infty)^{k-1}}\left(\int_{0}^{\frac{1}{\theta}-y}\left(\partial_{y}^{(i)}F(t_{1},\dots,t_{k})\right)^{2}dt_{i}\right)\,dt_{1}\dots
dt_{i-1}\,dt_{i+1}\dots dt_{k}\,dy,\\\
J_{i}(F):=&\int_{[0,+\infty)^{k-1}}\left(\int_{0}^{\infty}F(t_{1},\dots,t_{k})\,dt_{i}\right)^{2}dt_{1}\dots
dt_{i-1}\,dt_{i+1}\dots dt_{k},\end{split}$ (24)

and let $\Omega_{k}$ be the infimum

$\Omega_{k}:=\inf_{F}\left(\frac{\sum_{i=1}^{k}\left(Q_{i}(F)+\theta(1-\ell)J_{i}(F)\right)}{I(F)}+\ell
k\right),$ (25)

over all square integrable functions $F$ that are supported on the simplex

$\mathcal{R}_{k}:=\\{(t_{1},\dots,t_{k})\in[0,+\infty)^{k}\colon
t_{1}+\dots+t_{k}\leq 1\\},$

and are not identically zero up to almost everywhere equivalence. If

$m>\Omega_{k},$

then $DHL_{\Omega}[k;m-1]$ holds.

######  Remark 1.

Due to the continuity of $\Omega_{k}$ we can replace the condition that
$GEH[2\theta]$ holds by a weaker one that $GEH[2\theta^{\prime}]$ holds for
all $\theta^{\prime}<\theta$. Therefore, we are also permitted to take
$\theta=1/4$ unconditionally and $\theta=1/2$ assuming $GEH$. The same remark
also applies to Theorems 11, 12, and 13.

The choice of parameter $\ell$ does not affect the value of $\Omega_{k}$.
Substituting

$F(t_{1},\dots,t_{k})=f(t_{1}+\dots+t_{k})$

for some $f\colon[0,+\infty)\rightarrow\mathbf{R}$ and fixing $\ell=1$ we get
the following result.

######  Theorem 11 (One-dimensional sieving).

Suppose that there is a fixed $\theta\in(0,\frac{1}{2})$ such that
$GEH[2\theta]$ holds. Let $k\geq 2$ and $m\geq 1$ be fixed integers. For any
fixed and locally square-integrable function
$f\colon[0,+\infty)\rightarrow\mathbf{R}$, define the functionals

$\begin{split}\bar{I}(f):=&\int\limits_{0}^{1}f(t)^{2}\,t^{k-1}\,dt,\\\
\bar{Q}^{(1)}(f):=&\int\limits_{0}^{1}\frac{1-\theta
y}{y}\int\limits_{0}^{1-y}\left(f(t)-f(t+y)\right)^{2}t^{k-1}\,dt\,dy,\\\
\bar{Q}^{(2)}(f):=&\left(\int\limits_{0}^{1}\int\limits_{1-y}^{1}\,+\,\int\limits_{1}^{\frac{1}{\theta}-1}\int\limits_{0}^{1}\,+\,\int\limits_{\frac{1}{\theta}-1}^{\frac{1}{\theta}}\int\limits_{0}^{\frac{1}{\theta}-y}\,\right)\frac{1-\theta
y}{y}\,f(t)^{2}\,t^{k-1}\,dt\,dy,\\\
\bar{Q}^{(3)}(f):=&\int\limits_{\frac{1}{\theta}-1}^{\frac{1}{\theta}}\frac{1-\theta
y}{y}\int\limits_{\frac{1}{\theta}-y}^{1}f(t)^{2}\,\left(t^{k-1}-\left(t+y-\frac{1}{\theta}\right)^{k-1}\right)dt\,dy,\end{split}$
(26)

and let $\bar{\Omega}_{k}$ be the infimum

$\bar{\Omega}_{k}:=\inf_{f}\left(\frac{\sum_{i=1}^{3}\bar{Q}^{(i)}(f)}{\bar{I}(f)}+1\right)\cdot
k,$

over all square integrable functions $f$ that are not identically zero up to
almost everywhere equivalence. If

$m>\bar{\Omega}_{k},$

then $DHL_{\Omega}[k;m-1]$ holds.

We obviously have $\bar{\Omega}_{k}\geq\Omega_{k}$ for every possible choice
of $k$. We may apply Theorem 11 to get some non-trivial improvements over the
current state of the art in the $GEH$ case. We perform optimization over
polynomials of the form $f(x)=a+b(1-x)+c(1-x)^{2}+d(1-x)^{3}$ for
$-1<a,b,c,d<1$. This choice transforms the functionals (26) into quadratic
forms depending on the parameters $a,b,c,d$. Details including close to
optimal polynomials (up to a constant factor) for each $k$ are covered in the
table below.

Table C. Upper bounds for $\Omega_{k}$.

$k$ |  $\theta=1/4$ |  $\theta=1/2$ |  $f(1-x)$ |
---|---|---|---|---
$2$ |  5.03947 |  3.84763 |  $3+25x-x^{2}+x^{3}$ |
$3$ |  8.15176 |  6.31954 |  $1+12x-2x^{2}+9x^{3}$ |
$4$ |  11.49211 |  9.00542 |  $1+15x-x^{2}+19x^{3}$ |
$5$ |  15.01292 |  11.86400 |  $1+16x+5x^{2}+32x^{3}$ |
$6$ |  18.68514 |  14.86781 |  $1+26x-8x^{2}+86x^{3}$ |
$7$ |  22.48318 |  17.99402 |  $1+24x+6x^{2}+110x^{3}$ |
$8$ |  26.39648 |  21.23219 |  $1+30x+x^{2}+200x^{3}$ |
$9$ |  30.40952 |  24.56817 |  $1+30x+3x^{2}+260x^{3}$ |
$10$ |  34.51469 |  27.99372 |  $1+36x-x^{2}+400x^{3}$ |

It turns out that close to optimal choices in the unconditional setting are
also close to optimal under $GEH$. These results are sufficient to prove the
conditional part of Theorem 4 in every case except for $k=4$. Unfortunately,
by this method we cannot provide any unconditional improvement over what is
already obtained in [MaynardK], as presented in Table C. Therefore, let us try
to expand the sieve support a bit.

######  Theorem 12 (Sieving on an extended simplex).

Suppose that there is a fixed $\theta\in(0,\frac{1}{2})$ such that
$GEH[2\theta]$ holds and an arbitrarily chosen fixed real parameter $\ell$.
Let $k\geq 2$ and $m\geq 1$ be fixed integers. Let $\Omega_{k}^{\emph{ext}}$
be defined as in (25), but where the supremum now ranges over all square-
integrable and non-zero up to almost everywhere equivalence $F$ supported on
the extended simplex

$\mathcal{R}^{\prime}_{k}:=\\{(t_{1},\dots,t_{k})\in[0,+\infty)^{k}\colon\forall_{i\in\\{1,\dots,k\\}}\leavevmode\nobreak\
t_{1}+\dots+t_{i-1}+t_{i+1}+\dots+t_{k}\leq 1\\}.$

If

$m>\Omega^{\emph{ext}}_{k},$

then $DHL_{\Omega}[k;m-1]$ holds.

It is difficult to propose a one-dimensional variation of Theorem 12 in a
compact form, because the precise shape of functionals analogous to (26)
varies depending on $k$. We deal with this problem in Subsection 5.2. Given
that, we apply Theorem 12 directly and perform optimization over polynomials
of the form $F(t_{1},\dots,t_{k})=a+b(1-P_{1})+c(1-P_{1})^{2}=:f(P_{1})$ for
$-1<a,b,c<1$. Our choice is motivated by the fact that the values of symmetric
polynomials generated only by $P_{1}$ depend only on the sum
$t_{1}+\dots+t_{k}$, so they behave ’one-dimensionally’, which makes all
necessary calculations much easier. Moreover, our numerical experiments
suggest that including $P_{2}$ does not provide much extra contribution. Some
good choices of polynomials (again, up to a constant factor) and the bounds
they produce are listed below.

Table D. Upper bounds for $\Omega^{\text{ext}}_{k}$.

$k$ |  $\theta=1/4$ |  $\theta=1/2$ |  $f(1-x)$ |
---|---|---|---|---
$2$ |  4.49560 |  3.35492 |  $6+8x+3x^{2}$ |
$3$ |  7.84666 |  6.03889 |  $2+7x+7x^{2}$ |
$4$ |  11.27711 |  8.80441 |  $1+6x+9x^{2}$ |
$5$ |  14.84534 |  11.70582 |  $1+7x+15x^{2}$ |
$6$ |  18.55409 |  14.74036 |  $1+9x+32x^{2}$ |
$7$ |  22.38208 |  17.89601 |  $1+10x+46x^{2}$ |
$8$ |  26.32546 |  21.16260 |  $1+10x+65x^{2}$ |
$9$ |  30.37012 |  24.52806 |  $1+10x+90x^{2}$ |
$10$ |  34.50669 |  27.98326 |  $1+11x+121x^{2}$ |

The results from the $\theta=1/4$ column in Table D predict the limitations of
methods developed in [Lewulis]. In the conditional case we also get a strong
enhancement over what is achievable by sieving on the standard simplex in the
$k=4$ case. In the $k=2$ case we observe a standard phenomenon that passing
through the constant $3$ seems impossible, most probably because of the parity
obstruction as mentioned in [Polymath8]. In this work we do not make any
attempt to break this notorious barrier, so we do not expect to outdo the
result of Chen – even assuming very strong distributional claims like $GEH$.

In order to push our results even more, we would like to apply a device called
an $\varepsilon$-trick, which made its debut in [Polymath8]. The idea is to
expand the sieve support even further than before, but at a cost of turning
certain asymptotics into lower bounds. This is also the place where the $\ell$
parameter starts to behave non-trivially.

######  Theorem 13 (Sieving on an epsilon-enlarged simplex).

Suppose that there is a fixed $\theta\in(0,\frac{1}{2})$ such that
$GEH[2\theta]$ holds, and arbitrarily chosen fixed real parameters $\ell>1$,
$\varepsilon\in[0,1)$, and $\eta\geq 1+\varepsilon$ subject to the constraint

$2\theta\eta+\frac{1}{\ell}\leq 1.$ (27)

Let $k\geq 2$ and $m\geq 1$ be fixed integers. For any fixed compactly
supported square-integrable function
$F\colon[0,+\infty)^{k}\rightarrow\mathbf{R}$, define the functionals

$\begin{split}J_{i,\varepsilon}(F):=&\int_{(1-\varepsilon)\cdot\mathcal{R}_{k-1}}\left(\int_{0}^{\infty}F(t_{1},\dots,t_{k})\,dt_{i}\right)^{2}dt_{1}\dots
dt_{i-1}\,dt_{i+1}\dots dt_{k},\\\
Q_{i,\varepsilon}(F):=&\int_{0}^{\frac{1}{\theta}}\frac{1-\ell\theta
y}{y}\int_{\Phi(y)\cdot\mathcal{R}_{k-1}}\left(\,\int_{0}^{\frac{1}{\theta}-y}\left(\partial_{y}^{(i)}F(t_{1},\dots,t_{k})\right)^{2}dt_{i}\right)dt_{1}\dots
dt_{i-1}\,dt_{i+1}\dots dt_{k}\,dy,\end{split}$ (28)

where $\Phi\colon[0,+\infty)\rightarrow\mathbf{R}$ is a function given by the
formula

$\Phi(y):=\begin{cases}1+\varepsilon,&\emph{for
}y\in\left[0,\frac{1}{\ell\theta}\right),\\\ 1-\varepsilon,&\emph{for
}y\in\left[\frac{1}{\ell\theta},\frac{1}{\theta}\right],\\\
0,&\emph{otherwise.}\end{cases}$ (29)

Let $\Omega_{k,\varepsilon}$ be the infimum

$\Omega_{k,\varepsilon}:=\inf_{\eta,F}\left(\frac{\sum_{i=1}^{k}\left(Q_{i,\varepsilon}(F)-\theta(\ell-1)J_{i,\varepsilon}(F)\right)}{I(F)}+\ell
k\right),$ (30)

over all square integrable functions $F$ that are supported on the region

$(1+\varepsilon)\cdot\mathcal{R}_{k}^{\prime}\,\cap\,\eta\cdot\mathcal{R}_{k},$

and are not identically zero up to almost everywhere equivalence. If

$m>\Omega_{k,\varepsilon},$

then $DHL_{\Omega}[k;m-1]$ holds. Moreover, if $\varepsilon=0$, then
constraint (27) can be discarded and the functional inside the parentheses in
(30) is constant with respect to the $\ell$ variable.

######  Remark 2.

Observe that Theorems 10 and 12 follow easily from Theorem 13. In the first
case we just consider $\varepsilon=0$ and $\eta=1$. To prove the latter, we
take the same $\varepsilon$ and any $\eta\geq k/(k-1)$.

Constraint (27) refers to the hypotheses mentioned in the ‘trivial case’ from
Proposition 8. Notice that we do not have to restrict the support of the
$Q_{i,\varepsilon}$ integrals for $y\in\left[0,\frac{1}{\ell\theta}\right)$,
because we do not apply any $EH$-like theorem/conjecture in this interval.
Below we present some upper bounds for $\Omega_{k,\varepsilon}$ obtained via
considering $\eta=1+\varepsilon$ and optimizing over polynomials of the form
$a+b(1-P_{1})+c(1-P_{1})^{2}$ for $-1<a,b,c<1$ supported on the simplex
$(1+\varepsilon)\cdot\mathcal{R}_{k}$:

Table E. Upper bounds for $\Omega_{k,\varepsilon}$.

$k$ |  $\varepsilon$ |  $\theta=1/4$ |
---|---|---|---
$2$ |  1/3 |  4.69949 |
$3$ |  1/4 |  7.75780 |
$4$ |  1/5 |  11.05320 |
$5$ |  1/6 |  14.54134 |
$6$ |  1/7 |  18.19060 |
$7$ |  1/9 |  21.99368 |
$8$ |  1/10 |  25.90287 |
$9$ |  1/10 |  29.90565 |
$10$ |  2/21 |  34.01755 |

We are also able to obtain the bound $33.93473$ for $k=10$ and the same
$\varepsilon$, if one optimizes over polynomials of the form
$a+b(1-P_{1})+c(1-P_{1})^{2}+d(1-P_{1})^{3}$ for $-1<a,b,c,d<1$. We observe
that results provided by the $\varepsilon$-trick are considerably stronger
than those listed in Table D for every $k\geq 3$. They surpass the currently
known value of $\varrho_{k}$ in (2) for $7\leq k\leq 10$. Let us also notice
that the bigger $k$ we take, the better improvement over Tables C and D we
obtain. The reason for this is that the region $\mathcal{R}^{\prime}_{k}$ is
much larger than simplex $\mathcal{R}_{k}$ for small $k$, but the difference
in size is far less spectacular for bigger values of $k$. In the same time,
the epsilon-enlarged simplex $(1+\varepsilon)\cdot\mathcal{R}_{k}$ does not
share this weakness.

######  Remark 3.

It is possible to consider other choices of $\eta$ than $1+\varepsilon$. One
of them is $(1+\varepsilon)k/(k-1)$, which gives an access to a larger domain
$(1+\varepsilon)\cdot\mathcal{R}_{k}^{\prime}$. However, expanding the sieve
support so far makes the constaint (27) more restrictive. As for this moment,
numerical experiments suggest that one loses more than wins by implementing
such a manouver. The author also tried excluding the fragment

$\\{(t_{1},\dots,t_{k})\in[0,+\infty)^{k}\colon\forall_{i\in\\{1,\dots,k\\}}\leavevmode\nobreak\
t_{1}+\dots+t_{i-1}+t_{i+1}+\dots+t_{k}>1-\varepsilon\\},$

motivated by the fact, that it contibutes neither to $J_{i,\varepsilon}(F)$,
nor the negative part of $Q_{i,\varepsilon}(F)$, and in the same time it
contributes to $I(F)$. Unfortunately, this technique did not generate any
substancial advantage.

## Lemmata

We have the following lemma enabling us to convert certain sums into
integrals.

######  Lemma 14.

Let $m\geq 1$ be a fixed integer and let
$f\colon(0,+\infty)^{m}\rightarrow\bf{C}$ be a fixed compactly supported,
Riemann integrable function. Then for $x>1$ we have

$\sum_{\begin{subarray}{c}p_{1},\dots,p_{m}\\\ p_{1}\cdots p_{m}\sim
x\end{subarray}}\,f\left(\log_{x}p_{1},\dots,\log_{x}p_{m}\right)=\left(c_{f}+o(1)\right)\frac{x}{\log
x},$

where

$c_{f}:=\int_{\begin{subarray}{c}t_{1}+\dots+t_{m}=1\end{subarray}}f(t_{1},\dots,t_{m})\frac{dt_{1}\dots
dt_{m-1}}{t_{1}\cdots t_{m}},$

where we lift Lebesgue measure $dt_{1}\dots dt_{m-1}$ up to the hyperplane
$t_{1}+\cdots+t_{m}=1$.

###### Proof.

Follows from prime number theorem combined with elementary properties of the
Riemann integral. ∎

We introduce an another useful lemma which helps us with discarding those
$n\sim x$ having low prime factors.

######  Lemma 15 (Almost primality).

Let $k\geq 1$ be fixed, let $(L_{1},\dots,L_{k})$ be a fixed admissible
$k$–tuple, and let $b\bmod W$ be such that $(L_{i}(b),W)=1$ for each
$i=1,\dots,k$. Let further
$F_{1},\dots,F_{k}\colon[0,+\infty)\rightarrow\mathbf{R}$ be fixed smooth
compactly supported functions, and let $m_{1},\dots,m_{k}\geq 0$ and
$a_{1},\dots,a_{k}\geq 1$ be fixed natural numbers. Then,

$\sum_{\begin{subarray}{c}n\sim x\\\ n\equiv b\bmod
W\end{subarray}}\prod_{j=1}^{k}\left(\left|\lambda_{F_{j}}(L_{j}(n))\right|^{a_{j}}\tau(L_{j}(n))^{m_{j}}\right)\ll
B^{-k}\frac{x}{W}.$

Furthermore, if $1\leq j_{0}\leq k$ is fixed and $p_{0}$ is a prime with
$p_{0}\leq x^{1/10k}$, then we have the variant

$\sum_{\begin{subarray}{c}n\sim x\\\ n\equiv b\bmod
W\end{subarray}}\prod_{j=1}^{k}\left(\left|\lambda_{F_{j}}(L_{j}(n))\right|^{a_{j}}\tau(L_{j}(n))^{m_{j}}\right)\mathbf{1}_{p_{0}|L_{j_{0}}(n)}\ll\frac{\log_{x}p_{0}}{p_{0}}B^{-k}\frac{x}{W}.$

As a consequence, we have

$\sum_{\begin{subarray}{c}n\sim x\\\ n\equiv b\bmod
W\end{subarray}}\prod_{j=1}^{k}\left(\left|\lambda_{F_{j}}(L_{j}(n))\right|^{a_{j}}\tau(L_{j}(n))^{m_{j}}\right)\mathbf{1}_{\textup{lpf}(L_{j_{0}}(n))\leq
x^{\epsilon}}\ll\epsilon B^{-k}\frac{x}{W},$

for any $\epsilon>0$.

###### Proof.

This is a trivial modification of [Polymath8, Proposition 4.2]. ∎

##  3 Proof of Propositions 8 and 9

Contraty to the numerical ordering, we tackle Proposition 9 first, because it
is going to be needed throughout the rest of this section.

### Propositon 9

###### Proof.

It suffices to show that

$\sum_{p}\,\sum_{\begin{subarray}{c}n\sim x\\\ n\equiv b\bmod W\\\
p^{2}|\mathcal{P}(n)\end{subarray}}\tau(\mathcal{P}(n))\left|\prod_{i=1}^{k}\lambda_{F_{i}}(L_{i}(n))\lambda_{G_{i}}(L_{i}(n))\,\right|\,=\,o(1)\times
B^{-k}\frac{x}{W}.$ (31)

Choose an $\epsilon>0$. We decompose the outer sum in (31) as follows:

$\sum_{p}\leavevmode\nobreak\ =\leavevmode\nobreak\ \sum_{p\leq
x^{\epsilon}}\leavevmode\nobreak\ +\leavevmode\nobreak\
\sum_{p>x^{\epsilon}}.$ (32)

We apply the divisor bound $\tau(n)\ll n^{o(1)}$, valid for all
$n\in\mathbf{N}$, to conclude that the second sum from the right-hand side of
(32) is

$\ll x^{o(1)}\sum_{p>x^{\epsilon}}\sum_{\begin{subarray}{c}n\sim x\\\
p^{2}|\mathcal{P}(n)\end{subarray}}1\ll x^{1-\epsilon+o(1)}.$ (33)

The first sum, by the third part of Lemma 15, can be easily estimated as being

$\ll\leavevmode\nobreak\ \epsilon B^{-k}\frac{x}{W}.$

To this end, we only have to send $\epsilon\rightarrow 0$ sufficently slowly.
∎

### The trivial case of Proposition 8

###### Proof.

We shall take $i_{0}=k$, as the other cases can be proven exactly the same
way. Proposition 9 implies that our task is equivalent to showing that

$\sum_{\begin{subarray}{c}n\sim x\\\ n\equiv b\bmod
W\end{subarray}}\sum_{p|L_{k}(n)}\Upsilon(\log_{x}p)\prod_{i=1}^{k}\lambda_{F_{i}}(L_{i}(n))\lambda_{G_{i}}(L_{i}(n))=(c+o(1))B^{-k}\frac{x}{W}.$
(34)

Interchanging the order of summation, we get that the left-hand side of (34)
equals

$\sum_{p}\Upsilon(\log_{x}p)\sum_{\begin{subarray}{c}d_{1},\dots,d_{k}\\\
e_{1},\dots,e_{k}\end{subarray}}\left(\prod_{i=1}^{k}\mu(d_{i})\mu(e_{i})F_{i}(\log_{x}d_{i})G_{i}(\log_{x}e_{i})\right)S_{p}(d_{1},\dots,d_{k},e_{1},\dots,e_{k}),$
(35)

where

$S_{p}(d_{1},\dots,d_{k},e_{1},\dots,e_{k}):=\sum_{\begin{subarray}{c}n\sim
x\\\ n\equiv b\bmod W\\\ \forall_{i}\,[d_{i},e_{i}]|L_{i}(n)\\\
p|L_{k}(n)\end{subarray}}1.$ (36)

By hypotheses, all the $L_{i}(n)$ are coprime to $W$. We also assumed that for
all distinct $i,\,j$ we have $|A_{i}B_{j}-A_{j}B_{i}|<D_{0}$. On the other
hand, if there exists a prime $p_{0}$ dividing both $[d_{i},e_{i}]$ and
$[d_{j},e_{j}]$, then $A_{i}B_{j}-A_{j}B_{i}\equiv 0\bmod p_{0}$, which forces
$p_{0}\leq D_{0}$. By this contradiction, we may further assume in this
subsection that $W,\,[d_{1},e_{1}],\dots,[d_{k},e_{k}]$ are pairwise coprime,
because otherwise $S_{p}$ vanishes. We mark this extra constraint by the ′
sign next to the sum (see (41) for an example). Under these assumptions, we
can can merge the congruences appearing under the sum in (36) into one:

$n\equiv a\bmod q,$ (37)

where

$q:=W\,[d_{k},e_{k},p]\prod_{i=1}^{k-1}[d_{i},e_{i}]$ (38)

and $(a,q)=1$. This gives

$S_{p}(d_{1},\dots,d_{k},e_{1},\dots,e_{k})=\sum_{\begin{subarray}{c}n\sim
x\\\ n\equiv a\bmod q\end{subarray}}1\,=\,\frac{x}{q}+O(1).$ (39)

The net contribution of the $O(1)$ error term to (35) is at most

$\ll\,\left(\sum_{d,e\leq
x}\frac{1}{[d,e]}\right)^{k-1}\sum_{\begin{subarray}{c}d,e,p\leq
x\end{subarray}}\frac{1}{[d,e,p]}\ll\left(\sum_{r\leq
x}\frac{\tau(r)^{O(1)}}{r}\right)^{k}\leq x^{o(1)}.$ (40)

Therefore, it suffices to show that

$\sum_{p}\frac{\Upsilon(\log_{x}p)}{p}\left(\prod_{i=1}^{k}\sideset{}{{}^{\prime}}{\sum}_{d_{i},e_{i}}\frac{\mu(d_{i})\mu(e_{i})F_{i}(\log_{x}d_{i})G_{i}(\log_{x}e_{i})}{\psi_{i}([d_{i},e_{i}])}\right)=(c+o(1))B^{-k},$
(41)

where

$\psi_{i}(n):=\begin{cases}n,&\text{for }i\in\\{1,\dots,k-1\\},\\\
[n,p]/p,&\text{for }i=k.\end{cases}$ (42)

By [Lewulis, Lemma 2.2 and Lemma 2.6] and the polarization argument we get

$\prod_{i=1}^{k}\sideset{}{{}^{\prime}}{\sum}_{d_{i},e_{i}}\frac{\mu(d_{i})\mu(e_{i})F_{i}(\log_{x}d_{i})G_{i}(\log_{x}e_{i})}{\psi_{i}([d_{i},e_{i}])}=(c^{\prime}c^{\prime\prime}+o(1))B^{-k},$
(43)

with

$\displaystyle c^{\prime}$
$\displaystyle:=\prod_{i=1}^{k-1}\int_{0}^{1}F^{\prime}_{i}(t)G^{\prime}_{i}(t)\,dt,$
(44) $\displaystyle c^{\prime\prime}$
$\displaystyle:=\int_{0}^{1-\log_{x}p}\partial_{y}F^{\prime}_{k}(t)\,\partial_{y}G^{\prime}_{k}(t)\,dt\,dy.$
(45)

######  Remark 4.

To justify this application, we need to consider (under the notation used
within the cited work)

$\lambda_{d_{1},\dots,d_{k}}:=\prod_{i=1}^{k}\mu(d_{i})\widetilde{F}_{i}(\log_{x}d_{i})$

in one case and

$\lambda_{d_{1},\dots,d_{k}}:=\prod_{i=1}^{k}\mu(d_{i})\widetilde{G}_{i}(\log_{x}d_{i})$

in the other – we are permitted to choose these weights arbitrarily due to
[Lewulis, Lemma 1.12]. The key relationship in that paper between
$\lambda_{d_{1},\dots,d_{k}}$ and $y_{r_{1},\dots r_{k}}$ may be established
via [Lewulis, (1.20) and Lemma 2.6]. Then, from a simple formula

$\widetilde{F}^{2}-\widetilde{G}^{2}=(\widetilde{F}-\widetilde{G})(\widetilde{F}+\widetilde{G})$

we deduce that after defining $\widetilde{F}$, $\widetilde{G}$ in such a way
that $F=\widetilde{F}-\widetilde{G}$ and $G=\widetilde{F}+\widetilde{G}$, and
comparing the two mentioned choices of $\lambda_{d_{1},\dots,d_{k}}$, our
argument is completed.

The expression $1-\log_{x}p$ in the upper limit of the integral may seem a bit
artificial. Its role is to unify this part of Proposition 8 with the second
one. Now, it suffices to show that

$\sum_{p}\frac{\Upsilon(\log_{x}p)}{p}\int_{0}^{1-\log_{x}p}\partial_{y}F^{\prime}_{k}(t)\,\partial_{y}G^{\prime}_{k}(t)\,dt=\int_{0}^{1}\frac{\Upsilon(y)}{y}\int_{0}^{1-y}\partial_{y}F^{\prime}_{k}(t_{k})\,\partial_{y}G^{\prime}_{k}(t_{k})\,dt_{k}\,dy.$
(46)

This is a direct application of Lemma 14. ∎

### The Elliott–Halberstam case of Proposition 8

###### Proof.

As in the previous subsection, we can take $i_{0}=k$ without loss of
generality. Again, by Proposition 9 we have to prove that

$\sum_{\begin{subarray}{c}n\sim x\\\ n\equiv b\bmod
W\end{subarray}}\sum_{p|L_{k}(n)}\Upsilon(\log_{x}p)\prod_{i=1}^{k}\lambda_{F_{i}}(L_{i}(n))\lambda_{G_{i}}(L_{i}(n))=(c+o(1))B^{-k}\frac{x}{W}.$
(47)

Take some $\epsilon>0$. We decompose the studied sum as follows:

$\sum_{\begin{subarray}{c}n\sim x\\\ n\equiv b\bmod
W\end{subarray}}=\sum_{\begin{subarray}{c}n\sim x\\\ n\equiv b\bmod W\\\
\text{lpf}(L_{k}(n))\leq
x^{\epsilon}\end{subarray}}+\sum_{\begin{subarray}{c}n\sim x\\\ n\equiv b\bmod
W\\\ \text{lpf}(L_{k}(n))>x^{\epsilon}\end{subarray}}.$ (48)

We show that the contribution of the first sum from the right-hand side of
(48) is $\ll\epsilon B^{-k}xW^{-1}$. To do so we bound

$\lambda_{F_{i}}(L_{i}(n))\lambda_{G_{i}}(L_{i}(n))\leq\frac{1}{2}\left(\lambda_{F_{i}}(L_{i}(n))^{2}+\lambda_{G_{i}}(L_{i}(n))^{2}\right)$
(49)

for each $i=1,\dots,k$. We also recall the trivial inequality

$\sum_{p|L_{k}(n)}\Upsilon(\log_{x}p)\ll\tau(L_{k}(n)).$ (50)

By (49) and (50) we can present the first sum from the right-hand side of (48)
as a linear combination of sums that can be threated straightforwardly by
Lemma 15.

Let us define a function

$\Omega^{\flat}(n):=\sum_{\begin{subarray}{c}p|n\\\
p>x^{\epsilon}\end{subarray}}\Upsilon(\log_{x}p)$

Now, it sufficies to show that for any $\epsilon>0$ we have

$\sum_{\begin{subarray}{c}n\sim x\\\ n\equiv b\bmod W\\\
\text{lpf}(L_{k}(n))>x^{\epsilon}\end{subarray}}\Omega^{\flat}(L_{k}(n))\prod_{i=1}^{k}\lambda_{F_{i}}(L_{i}(n))\lambda_{G_{i}}(L_{i}(n))=(c_{\epsilon}+o(1))B^{-k}\frac{x}{W},$
(51)

where $c_{\epsilon}\rightarrow c$ when $\epsilon\rightarrow 0$. After
expanding the $\lambda_{F_{i}},\,\lambda_{G_{i}}$ we conclude that the left-
hand side of (51) equals

$\sum_{\begin{subarray}{c}d_{1},\dots,d_{k-1}\\\
e_{1},\dots,e_{k-1}\end{subarray}}\left(\prod_{i=1}^{k-1}\mu(d_{i})\mu(e_{i})F_{i}(\log_{x}d_{i})G_{i}(\log_{x}e_{i})\right)S_{\epsilon}(d_{1},\dots,d_{k-1},e_{1},\dots,e_{k-1}),$
(52)

where

$S_{\epsilon}(d_{1},\dots,d_{k-1},e_{1},\dots,e_{k-1}):=\sum_{\begin{subarray}{c}n\sim
x\\\ n\equiv b\bmod W\\\ \text{lpf}(L_{k}(n))>x^{\epsilon}\\\
\forall_{i\not=k}\,[d_{i},e_{i}]|L_{i}(n)\end{subarray}}\Omega^{\flat}(L_{k}(n))\,\lambda_{F_{k}}(L_{k}(n))\,\lambda_{G_{k}}(L_{k}(n)).$
(53)

Notice that $n\equiv b\bmod W$ implies that all of the $L_{i}(n)$ are coprime
to $W$. We also assumed that for all distinct $i,\,j$ we have
$|A_{i}B_{j}-A_{j}B_{i}|<D_{0}$, so if there exists a prime $p_{0}$ dividing
both $[d_{i},e_{i}]$ and $[d_{j},e_{j}]$, then $A_{i}B_{j}-A_{j}B_{i}\equiv
0\bmod p_{0},$ which forces $p_{0}\leq D_{0}$. That is a contradiction.
Therefore, we may further assume in this subsection that
$W,\,[d_{1},e_{1}],\dots,[d_{k},e_{k}]$ are pairwise coprime and that
$\text{lpf}\,([d_{k},e_{k}])>x^{\epsilon}$, because otherwise $S_{\epsilon}$
vanishes. Under these assumptions we can merge all the congruences under the
sum (53) into two:

$n\equiv a\bmod q\,,\leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ L_{k}(n)\equiv
0\bmod[d_{k},e_{k},p],$ (54)

where we redefine $q$ and $a$ as

$q:=W\prod_{i=1}^{k-1}[d_{i},e_{i}],$ (55)

and $a$ being some residue class coprime to its modulus such that
$(L_{i}(a),W)=1$ for each possible choice of index $i$. This gives

$S_{\epsilon}(d_{1},\dots,d_{k-1},e_{1},\dots,e_{k-1})=\sum_{\begin{subarray}{c}n\sim
x\\\ \text{lpf}(L_{k}(n))>x^{\epsilon}\\\ n\equiv a\bmod
q\end{subarray}}\Omega^{\flat}(L_{k}(n))\,\lambda_{F_{k}}(L_{k}(n))\,\lambda_{G_{k}}(L_{k}(n)).$
(56)

We would like to perform a substitution $m:=L_{k}(n)$ in the sum from (56), so
we have to transform the congruence $n\equiv a\bmod q$ appropriately. In order
to do so, we split it into two: $n\equiv a\bmod[A_{k},q]/A_{k}$ and $n\equiv
a\bmod\mbox{rad}\,A_{k}$, where $\mbox{rad}\,A_{k}$ denotes the square-free
part of $A_{k}$. The former congruence is simply equivalent to $m\equiv
L_{k}(a)\bmod[A_{k},q]/A_{k}$. The latter is equivalent to $m\equiv
L_{k}(a)\bmod A_{k}\,\text{rad}\,A_{k}$ and it also implies $m\equiv
B_{k}\bmod A_{k}$, which has to be satisfied by our substitution. Note that

$(L_{k}(a),[A_{k},q]/A_{k})=(L_{k}(a),A_{k}\,\mbox{rad}A_{k})=1,$ (57)

so we can combine the two considered congruences into one $m\equiv
a^{\prime}\bmod[A_{k},q]\,\mbox{rad}A_{k}$. Hence,

$S_{\epsilon}(d_{1},\dots,d_{k-1},e_{1},\dots,e_{k-1})=\sum_{\begin{subarray}{c}A_{k}x+B_{k}<m\leq
2A_{k}x+B_{k}\\\ \text{lpf}(m)>x^{\epsilon}\\\ m\equiv a^{\prime}\bmod
q^{\prime}\end{subarray}}\Omega^{\flat}(m)\,\lambda_{F_{k}}(m)\,\lambda_{G_{k}}(m),$
(58)

where $q^{\prime}:=[A_{k},q]\,\mbox{rad}\,A_{k}=qA_{k}$ and $a^{\prime}$ is a
residue class $\bmod\,q$ coprime to its modulus. Thus, we have

$S_{\epsilon}(d_{1},\dots,d_{k-1},e_{1},\dots,e_{k-1})=\frac{1}{\varphi(q^{\prime})}\sum_{\begin{subarray}{c}A_{k}x+B_{k}<m\leq
2A_{k}x+B_{k}\\\
(m,q^{\prime})=1\end{subarray}}\Omega^{\flat}(m)\lambda_{F_{k}}(m)\lambda_{G_{k}}(m)\mathbf{1}_{\text{lpf}(m)>x^{\epsilon}}\,\\\
+\Delta\left(\Omega^{\flat}\lambda_{F_{k}}\lambda_{G_{k}}\mathbf{1}_{\text{lpf}(\cdot)>x^{\epsilon}}\mathbf{1}_{[A_{k}x+B_{k},2A_{k}x+B_{k}]};a^{\prime}\bmod
q^{\prime}\right).$ (59)

We split

$\sum_{p}S_{\epsilon}=S_{1}-S_{2}+S_{3},$ (60)

where

$\begin{split}S_{1}(d_{1},\dots,d_{k-1},e_{1},\dots,e_{k-1})&=\frac{1}{\varphi(q^{\prime})}\sum_{p}\Upsilon(\log_{x}p)\sum_{\begin{subarray}{c}A_{k}x+B_{k}<m\leq
2A_{k}x+B_{k}\\\
p|m\end{subarray}}\lambda_{F_{k}}(m)\lambda_{G_{k}}(m)\mathbf{1}_{\text{lpf}(m)>x^{\epsilon}},\\\
S_{2}(d_{1},\dots,d_{k-1},e_{1},\dots,e_{k-1})&=\frac{1}{\varphi(q^{\prime})}\sum_{\begin{subarray}{c}A_{k}x+B_{k}<m\leq
2A_{k}x+B_{k}\\\
(m,q^{\prime})>1\end{subarray}}\Omega^{\flat}(m)\lambda_{F_{k}}(m)\lambda_{G_{k}}(m)\mathbf{1}_{\text{lpf}(m)>x^{\epsilon}},\\\
S_{3}(d_{1},\dots,d_{k-1},e_{1},\dots,e_{k-1})&=\Delta\left(\Omega^{\flat}\lambda_{F_{k}}\lambda_{G_{k}}\mathbf{1}_{\text{lpf}(\cdot)>x^{\epsilon}}\mathbf{1}_{[A_{k}x+B_{k},2A_{k}x+B_{k}]};a^{\prime}\bmod
q^{\prime}\right).\end{split}$ (61)

For $j\in\\{1,2,3\\}$ we put

$\Sigma_{j}=\sum_{\begin{subarray}{c}d_{1},\dots,d_{k-1}\\\
e_{1},\dots,e_{k-1}\end{subarray}}\left(\prod_{i=1}^{k-1}\mu(d_{i})\mu(e_{i})F_{i}(\log_{x}d_{i})G_{i}(\log_{x}e_{i})\right)S_{j}(d_{1},\dots,d_{k-1},e_{1},\dots,e_{k-1}).$
(62)

Therefore, it suffices to derive the main term estimate

$\Sigma_{1}=(c_{\epsilon}+o(1))B^{-k}\frac{x}{W},\\\ $ (63)

the ‘correction’ error term estimate

$\Sigma_{2}\ll x^{1-\epsilon+o(1)},\\\ $ (64)

and the ‘GEH-type’ error term estimate

$\Sigma_{3}\ll x\log^{-A}x$ (65)

for any fixed $A>0$.

Let us begin with (64). We observe that since $\text{lpf}(m)>x^{\epsilon}$,
there exists a prime $x^{\epsilon}<p\leq x$ dividing both $m$ and one of
$d_{1},e_{1},\dots,d_{k-1},e_{k-1}$ (if $k=1$, then $\Sigma_{2}$ vanishes; we
also claim that $\epsilon$ tends to 0 slowly enough to ensure that
$D_{0}<x^{\epsilon}$). Thus, we may safely assume that $p|d_{1}$, for the
remaining $2k-3$ cases are analogous. Hence, we get

$\Sigma_{2}\ll x^{o(1)}\sum_{x^{\epsilon}<p\leq
x}\sum_{\begin{subarray}{c}d_{1},\dots,d_{k-1}\leq x\\\
e_{1},\dots,e_{k-1}\leq x\\\
p|d_{1}\end{subarray}}\prod_{i=1}^{k-1}\frac{1}{\varphi([d_{i},e_{i}])}\leavevmode\nobreak\
\sum_{\begin{subarray}{c}n\ll x\\\ p|n\end{subarray}}1\ll
x^{1+o(1)}\sum_{x^{\epsilon}<p\leq x}\frac{1}{p^{2}}\ll x^{1-\epsilon+o(1)}.$
(66)

To deal with (65) we just repeat the reasoning from [Polymath8, Subsection
‘The generalized Elliott-Halberstam case’, Eq (62)] combined with
$\Omega^{\flat}(m)=O(1/\epsilon)$.

Let us move to (63). We have

$\varphi(q^{\prime})=A_{k}\varphi\left(W\prod_{i=1}^{k-1}[d_{i},e_{i}]\right),$

so again by [Lewulis, Lemma 2.6] (or [Polymath8, Lemma 4.1] for an even more
direct application) we get

$\sideset{}{{}^{\prime}}{\sum}_{\begin{subarray}{c}d_{1},\dots,d_{k-1}\\\
e_{1},\dots,e_{k-1}\end{subarray}}\frac{\prod_{i=1}^{k-1}\mu(d_{i})\mu(e_{i})F_{i}(\log_{x}d_{i})G_{i}(\log_{x}e_{i})}{\varphi\left(q^{\prime}\right)}=\frac{A_{k}^{-1}}{\varphi(W)}(c^{\prime}+o(1))B^{1-k},$
(67)

where

$c^{\prime}:=\prod_{i=1}^{k-1}\int_{0}^{1}F^{\prime}_{i}(t)G^{\prime}_{i}(t)\,dt.$

By (61) it suffices to show that

$\sum_{p}\,\Upsilon(\log_{x}p)\sum_{\begin{subarray}{c}A_{k}x+B_{k}<m\leq
2A_{k}x+B_{k}\\\
p|m\end{subarray}}\lambda_{F_{k}}(m)\lambda_{G_{k}}(m)\mathbf{1}_{\text{lpf}(m)>x^{\epsilon}}=\left(c_{\epsilon}^{\prime\prime}+o(1)\right)\frac{A_{k}x}{\log
x},$ (68)

where $c_{\epsilon}^{\prime\prime}$ satisfies

$\lim_{\epsilon\rightarrow
0}c_{\epsilon}^{\prime\prime}=\Upsilon(1)\,F_{k}(0)\,G_{k}(0)\leavevmode\nobreak\
+\leavevmode\nobreak\
\int_{0}^{1}\frac{\Upsilon(y)}{y}\int_{0}^{1-y}\partial_{y}F^{\prime}_{k}(t)\,\partial_{y}G^{\prime}_{k}(t)\,dt\,dy.$
(69)

We simplify the restriction $A_{k}x+B_{k}<m\leq 2A_{k}x+B_{k}$ into $m\sim
A_{k}x$ at the cost of introducing to the left-hand side of (68) an error term
of size not greater than $x^{o(1)}$. We factorize $m=p_{1}\cdots p_{r}p$ for
some $x^{\epsilon}\leq p_{1}\leq\dots\leq p_{r}\leq 2A_{k}x$, $p\geq
x^{\epsilon}$, and $0\leq r\leq\frac{1}{\epsilon}$. The contribution of those
$m$ having repeated prime factors is readily $\ll x^{1-\epsilon}$, so we can
safely assume that $m$ is square-free. In such a case, we get

$\lambda_{F_{k}}(m)=(-1)^{r}\partial_{\,\log p_{1}}\dots\partial_{\,\log
p_{r}}(\partial_{\,\log p}F_{k}(0))$ (70)

and an analogous equation for $\lambda_{G_{k}}(m)$. Therefore, the left-hand
side of (68) equals

$\sum_{0\leq
r\leq\frac{1}{\epsilon}}\,\sum_{p}\Upsilon(\log_{x}p)\sum_{\begin{subarray}{c}x^{\epsilon}<p_{1}<\dots<p_{r}\\\
p_{1}\dots p_{r}p\,\sim A_{k}x\end{subarray}}\partial_{\,\log
p_{1}}\dots\partial_{\,\log p_{r}}(\partial_{\,\log
p}F_{k}(0))\,\cdot\,\partial_{\,\log p_{1}}\dots\partial_{\,\log
p_{r}}(\partial_{\,\log p}G_{k}(0)).$ (71)

Note that for the index $r=0$ the summand above equals

$(\Upsilon(1)+o(1))\sum_{p\sim A_{k}x}\,F_{k}(0)\,G_{k}(0).$ (72)

We apply Lemma 14 to (71–72) and obtain an asymptotic (68) with

$\begin{split}c_{\epsilon}^{\prime\prime}=\sum_{1\leq
r\leq\frac{1}{\epsilon}}\int_{0}^{1}\Upsilon(y)\int_{\begin{subarray}{c}\phantom{2}\\\
t_{1}+\dots+t_{r}=1-y\\\
\epsilon<t_{1}<\dots<t_{r}\end{subarray}}\partial_{t_{1}}\dots\partial_{t_{r}}(\partial_{y}F_{k}(0))\cdot\partial_{t_{1}}\dots\partial_{t_{r}}(\partial_{y}G_{k}(0))\frac{dy\,dt_{1}\dots
dt_{r-1}}{y\,t_{1}\cdots t_{r}}\\\ +\leavevmode\nobreak\
\Upsilon(1)\,F_{k}(0)\,G_{k}(0).\end{split}$ (73)

The first part of Lemma 15 gives us $c_{\epsilon}^{\prime\prime}\ll 1$ when
$\epsilon\rightarrow 0^{+}$. Now, consider any sequence of positive numbers
$(\epsilon_{1},\epsilon_{2},\dots)$ satisfying $\epsilon_{n}\rightarrow 0$ as
$n\rightarrow\infty$. In view of (68) and the last part of Lemma 15, we
conclude that
$\left(c_{\epsilon_{1}}^{\prime\prime},c_{\epsilon_{2}}^{\prime\prime}\dots\right)$
forms a Cauchy sequence, and hence it has a limit. Thus, by dominated
convergence theorem it suffices to establish for each $y\in[0,1]$ the
following equality

$\sum_{r\geq 1}\int_{\begin{subarray}{c}\phantom{2}\\\
t_{1}+\dots+t_{r}=1-y\\\
0<t_{1}<\dots<t_{r}\end{subarray}}\partial_{t_{1}}\dots\partial_{t_{r}}(\partial_{y}F_{k}(0))\cdot\partial_{t_{1}}\dots\partial_{t_{r}}(\partial_{y}G_{k}(0))\frac{dt_{1}\dots
dt_{r-1}}{t_{1}\cdots t_{r}}\\\
=\int_{0}^{1-y}\partial_{y}F_{k}^{\prime}(t)\,\partial_{y}G_{k}^{\prime}(t)\,dt.$
(74)

By depolarization argument it suffices to show that for each $y\in[0,1]$, we
have

$\sum_{r\geq 1}\int_{\begin{subarray}{c}\phantom{2}\\\
t_{1}+\dots+t_{r}=1-y\\\
0<t_{1}<\dots<t_{r}\end{subarray}}\left|\partial_{t_{1}}\dots\partial_{t_{r}}(\partial_{y}F(0))\right|^{2}\frac{\,dt_{1}\dots
dt_{r-1}}{t_{1}\cdots
t_{r}}=\int_{0}^{1-y}\left|\partial_{y}F^{\prime}(t)\right|^{2}\,dt$ (75)

for any smooth $F\colon[0,\infty)\rightarrow\mathbf{R}$. For the sake of
clarity, we relabel $\partial_{y}F(x)$ as $H(x)$. We substitute $u:=t/(1-y)$
and $u_{i}:=t_{i}/(1-y)$ for all possible choices of $i$. With these settings
(75) is equivalent to

$\sum_{r\geq 1}\int_{\begin{subarray}{c}\phantom{2}\\\ u_{1}+\dots+u_{r}=1\\\
0<u_{1}<\dots<u_{r}\end{subarray}}\left|\partial_{(1-y)u_{1}}\dots\partial_{(1-y)u_{r}}H(0)\right|^{2}\frac{\,du_{1}\dots
du_{r-1}}{u_{1}\cdots
u_{r}}=(1-y)^{2}\int_{0}^{1}\left|H^{\prime}(u(1-y))\right|^{2}\,du.$ (76)

Note that one of the $(1-y)$ appeared from transforming $t_{r}\mapsto u_{r}$.
Put $\widetilde{H}(x):=H(x(1-y))$. We get

$\partial_{(1-y)u_{1}}\dots\partial_{(1-y)u_{r}}H(0)=\partial_{u_{1}}\dots\partial_{u_{r}}\widetilde{H}(0),$

and $\widetilde{H}^{\prime}(x)=(1-y)H^{\prime}(x(1-y))$ by the chain rule.
Thus, it suffices to show that

$\sum_{r\geq 1}\int_{\begin{subarray}{c}\phantom{2}\\\ u_{1}+\dots+u_{r}=1\\\
0<u_{1}<\dots<u_{r}\end{subarray}}\left|\partial_{u_{1}}\dots\partial_{u_{r}}\widetilde{H}(0)\right|^{2}\frac{\,du_{1}\dots
du_{r-1}}{u_{1}\cdots
u_{r}}=\int_{0}^{1}\left|\widetilde{H}^{\prime}(u)\right|^{2}\,du.$ (77)

To this end, we apply the key combinatorial identity [Polymath8, (67)]. ∎

##  4 Proof of Theorem 13

###### Proof.

Let $k,m,\varepsilon,\theta,\ell$ be as in Theorem 13. Let us assume that we
have a non-zero square integrable function
$F\colon[0,+\infty)^{k}\rightarrow\mathbf{R}$ supported on
$(1+\varepsilon)\cdot\mathcal{R}_{k}^{\prime}\cap\,\eta\cdot\mathcal{R}_{k}$
and satisfying

$\frac{\sum_{i=1}^{k}\left(Q_{i,\varepsilon}(F)-\theta(\ell-1)J_{i,\varepsilon}(F)\right)}{I(F)}+\ell
k<m.$ (78)

Now, we perform an analogous sequence of simplifications as in [Polymath8,
(72–84)] and eventually arrive at a non-zero smooth function
$f\colon\mathbf{R}^{k}\rightarrow\mathbf{R}$ being the linear combination of
tensor products – namely

$f(t_{1},\dots,t_{k})=\sum_{j=1}^{J}c_{j}f_{1,j}(t_{1})\cdots f_{k,j}(t_{j})$
(79)

with $J$, $c_{j}$, $f_{i,j}$ fixed, for which all the components
$f_{1,j}(t_{1}),\dots,f_{k,j}(t_{k})$ are supported on the region

$\left\\{(t_{1},\dots,t_{k})\in\mathbf{R}^{k}\colon\sum_{i=1}^{k}\max\left(t_{i},\delta\right)\leq\theta\eta-\delta\right\\}\\\
\cap\left\\{(t_{1},\dots,t_{k})\in\mathbf{R}^{k}\colon\forall_{1\leq i_{0}\leq
k}\sum_{\begin{subarray}{c}1\leq i\leq k\\\
i\not=i_{0}\end{subarray}}\max\left(t_{i},\delta\right)\leq(1+\varepsilon)\theta-\delta\right\\}$
(80)

for some sufficently small $\delta>0$ – that obeys

$\frac{\sum_{i=1}^{k}\left(\widetilde{Q}_{i,\varepsilon}(f)-(\ell-1)\widetilde{J}_{i,\varepsilon}(f)\right)}{\widetilde{I}(f)}+\ell
k<m,$ (81)

where

$\displaystyle\widetilde{I}(f):=$
$\displaystyle\int\limits_{[0,+\infty)^{k}}\left|\frac{\partial^{k}}{\partial
t_{1}\dots\partial t_{k}}f(t_{1},\dots,t_{k})\right|^{2}dt_{1}\dots dt_{k},$
(82) $\displaystyle\widetilde{J}_{i,\varepsilon}(f):=$
$\displaystyle\int\limits_{(1-\varepsilon)\theta\cdot\mathcal{R}_{k-1}}\left|\frac{\partial^{k-1}}{\partial
t_{1}\dots\partial t_{i-1}\partial t_{i+1}\dots\partial
t_{k}}f(t_{1},\dots,t_{i-1},0,t_{i+1},\dots,t_{k})\right|^{2}dt_{1}\dots
dt_{i-1}dt_{i+1}\dots dt_{k},$
$\displaystyle\widetilde{Q}_{i,\varepsilon}(f):=$
$\displaystyle\int\limits_{0}^{1}\frac{1-\ell
y}{y}\int\limits_{{\Psi}(y)\cdot\mathcal{R}_{k-1}}\left(\int\limits_{0}^{1-y}\left|\partial_{y}^{(i)}\frac{\partial^{k}}{\partial
t_{1}\dots\partial
t_{k}}f(t_{1},\dots,t_{k})\right|^{2}dt_{i}\right)dt_{1}\dots
dt_{i-1}\,dt_{i+1}\dots dt_{k}\,dy,$

with $\Psi\colon[0,+\infty)\rightarrow\mathbf{R}$ being a function given as

$\Psi(y):=\begin{cases}1+\varepsilon,&\text{for
}y\in\left[0,\frac{1}{\ell}\right),\\\ 1-\varepsilon,&\text{for
}y\in\left[\frac{1}{\ell},1\right],\\\ 0,&\text{otherwise.}\end{cases}$ (83)

We construct a non-negative sieve weight
$\nu\colon\mathbf{N}\rightarrow\mathbf{Z}$ by the formula

$\nu(n):=\left(\sum_{j=1}^{J}c_{j}\lambda_{f_{1,j}}(L_{1}(n))\cdots\lambda_{f_{k,j}}(L_{k}(n))\right)^{2}.$
(84)

Notice that if $\varepsilon>0$, then for any $1\leq j,j^{\prime}\leq J$ we
have

$\sum_{i=1}^{k}(S(f_{i,j})+S(f_{i,j^{\prime}}))<2\theta\eta<1$ (85)

from the $2\theta\eta+\frac{1}{\ell}\leq 1$ assertion. On the flip side, if
$\varepsilon=0$, then $\text{supp}(F)\subset\mathcal{R}_{k}^{\prime}$ and
consequently for every $1\leq i_{0}\leq k$ we have

$\sum_{\begin{subarray}{c}1\leq i\leq k\\\
i\not=i_{0}\end{subarray}}(S(f_{i,j})+S(f_{i,j^{\prime}}))<2\theta.$ (86)

Applying results from [Polymath8, Subsection ‘Proof of Theorem 3.12’], we get

$\sum_{\begin{subarray}{c}n\sim x\\\ n\equiv b\bmod
W\end{subarray}}\nu(n)=\left(\alpha+o(1)\right)B^{-k}\frac{x}{W},$ (87)

where

$\alpha=\widetilde{I}(f).$

Now, let us consider the sum

$\sum_{\begin{subarray}{c}n\sim x\\\ n\equiv b\bmod W\\\
\mathcal{P}(n)\textup{ sq-
free}\end{subarray}}\nu(n)\sum_{p|L_{k}(n)}\left(1-\ell\log_{x}p\right).$ (88)

We can expand the sum above as a linear combination of expressions

$\sum_{\begin{subarray}{c}n\sim x\\\ n\equiv b\bmod W\\\
\mathcal{P}(n)\textup{ sq-
free}\end{subarray}}\sum_{p|L_{k}(n)}\left(1-\ell\log_{x}p\right)\prod_{i=1}^{k}\lambda_{f_{i,j}}(L_{i}(n))\lambda_{f_{i,j^{\prime}}}(L_{i}(n))$
(89)

for various $1\leq j,j^{\prime}\leq J$. We seek for the upper bound of the sum
(89). We can achieve this goal by applying Proposition 8. We also observe that
the first part of this result should be more effective for smaller values of
$p$, and the second part for larger values of $p$. Therefore, we perform a
decomposition of the expression (89) as follows:

$\sum_{\begin{subarray}{c}n\sim x\\\ n\equiv b\bmod W\\\
\mathcal{P}(n)\textup{ sq-
free}\end{subarray}}\sum_{p|L_{k}(n)}=\sum_{\begin{subarray}{c}n\sim x\\\
n\equiv b\bmod W\\\ \mathcal{P}(n)\textup{ sq-
free}\end{subarray}}\left(\sum_{\begin{subarray}{c}p|L_{k}(n)\\\ p\leq
x^{1/\ell}\end{subarray}}+\sum_{\begin{subarray}{c}p|L_{k}(n)\\\
p>x^{1/\ell}\end{subarray}}\right).$ (90)

For the $p\leq x^{\ell}$ sum we apply the trivial case of Proposition 8 with
$\vartheta_{0}=1/\ell$ and

$\Upsilon(y)=(1-\ell y)\mathbf{1}_{y\leq 1/\ell}.$

Under these assumptions we have

$\displaystyle\sum_{i=1}^{k}\left(S(f_{i,j})+S(f_{i,j^{\prime}})\right)$
$\displaystyle<2\theta(1+\varepsilon)\leq 1-\frac{1}{\ell},$ (91)
$\displaystyle S(\Upsilon)$ $\displaystyle\leq\frac{1}{\ell},$ (92)

so the necessary hypotheses from the ‘trivial case’ of Proposition 8 are
indeed satisfied. Observe that under $\varepsilon=0$ the inequality (91)
satisfies the second case of Proposition 8, so in these circumstances we do
not have to rely on the constraint (27) any longer. Thus, we get

$\sum_{\begin{subarray}{c}n\sim x\\\ n\equiv b\bmod W\\\
\mathcal{P}(n)\textup{ sq-
free}\end{subarray}}\nu(n)\sum_{\begin{subarray}{c}p|L_{i}(n)\\\ p\leq
x^{1/\ell}\end{subarray}}\left(1-\ell\log_{x}p\right)=\left(\beta_{k}^{(1)}+\,o(1)\right)B^{-k}\frac{x}{W},$
(93)

where

$\begin{split}\beta_{k}^{(1)}=\sum_{j,j^{\prime}=1}^{J}c_{j}c_{j^{\prime}}\left(\int\limits_{0}^{1}\frac{\Upsilon(y)}{y}\int\limits_{0}^{1-y}\partial_{y}f_{k,j}^{\prime}(t_{k})\,\partial_{y}f_{k,j^{\prime}}^{\prime}(t_{k})\,dt_{k}\,dy\right)\prod_{i=1}^{k-1}\left(\int\limits_{0}^{1}f_{k,j}^{\prime}(t_{i})\,f_{k,j^{\prime}}^{\prime}(t_{i})\,dt_{i}\right).\end{split}$
(94)

From (84) we see that $\beta_{k}^{(1)}$ factorizes as

$\beta_{k}^{(1)}=\int\limits_{0}^{1/\ell}\frac{1-\ell
y}{y}\int\limits_{(1+\varepsilon)\theta\cdot\mathcal{R}^{k-1}}\int\limits_{0}^{1-y}\left|\partial_{y}^{(k)}\frac{\partial^{k}}{\partial
t_{1}\dots\partial t_{k}}f(t_{1},\dots,t_{k})\right|^{2}dt_{i}\,dt_{1}\dots
dt_{k-1}\,dy.$ (95)

Now we deal with the $p>x^{\ell}$ case. We apply the $GEH$ case of Proposition
8 with $\vartheta=1/2$ and

$\Upsilon(y)=(1-\ell y)\mathbf{1}_{y>1/\ell}.$

We decompose $\\{1,\dots,J\\}$ into $\mathcal{J}_{1}\cup\mathcal{J}_{2}$,
where $\mathcal{J}_{1}$ consists of those indices $j\in\\{1,\dots,J\\}$
satisfying

$\sum_{i=1}^{k-1}S(f_{i,j})<(1-\varepsilon)\theta,$ (96)

and $\mathcal{J}_{2}$ is the complement. As in [Polymath8] we apply the
elementary inequality

$(x_{1}+x_{2})^{2}\geq(x_{1}+2x_{2})x_{1}$

to obtain the pointwise lower bound

$\begin{split}\nu(n)\geq\left(\left(\sum_{j\in\mathcal{J}_{1}}+\leavevmode\nobreak\
2\sum_{j\in\mathcal{J}_{2}}\right)c_{j}\lambda_{f_{1,j}}(L_{1}(n))\cdots\lambda_{f_{k,j}}(L_{k}(n))\right)\left(\sum_{j^{\prime}\in\mathcal{J}_{1}}c_{j^{\prime}}\lambda_{f_{1,j^{\prime}}}(L_{1}(n))\cdots\lambda_{f_{k,j^{\prime}}}(L_{k}(n))\right).\end{split}$
(97)

Therefore, if $j\in\mathcal{J}_{1}\cup\mathcal{J}_{2}$ and
$j^{\prime}\in\mathcal{J}_{1}$, then from (96) one has

$\sum_{i=1}^{k-1}\left(S(f_{i,j})+S(f_{i,j^{\prime}})\right)<2\theta,$

so the hypothesis from the ‘Generalised Elliott–Halberstam’ case of
Proposition 8 is indeed satisfied. Thus, by Proposition 8 and (97) we get

$\sum_{\begin{subarray}{c}n\sim x\\\ n\equiv b\bmod W\\\
\mathcal{P}(n)\textup{ sq-
free}\end{subarray}}\nu(n)\sum_{\begin{subarray}{c}p|L_{i}(n)\\\
p>x^{1/\ell}\end{subarray}}\left(1-\ell\log_{x}p\right)\leq\left(\beta_{k}^{(2)}+\,o(1)\right)B^{-k}\frac{x}{W},$
(98)

where

$\begin{split}\beta_{k}^{(2)}=\left(\sum_{j\in\mathcal{J}_{1}}+\leavevmode\nobreak\
2\sum_{j\in\mathcal{J}_{2}}\right)\sum_{j^{\prime}\in\mathcal{J}_{1}}c_{j}c_{j^{\prime}}\left(\Upsilon(1)\,f_{k,j}(0)\,f_{k,j^{\prime}}(0)+\int\limits_{0}^{1}\frac{\Upsilon(y)}{y}\int\limits_{0}^{1-y}\partial_{y}f_{k,j}^{\prime}(t_{k})\,\partial_{y}f_{k,j^{\prime}}^{\prime}(t_{k})\,dt_{k}\,dy\right)\\\
\times\,\prod_{i=1}^{k-1}\left(\int\limits_{0}^{1}f_{k,j}^{\prime}(t_{i})\,f_{k,j^{\prime}}^{\prime}(t_{i})\,dt_{i}\right).\end{split}$
(99)

For $s=1,2$ let us define

$f_{s}(t_{1},\dots,t_{k}):=\sum_{j\in\mathcal{J}_{s}}c_{j}f_{1,j}(t_{1})\cdots
f_{k,j}(t_{k}).$

From (84) we observe that $\beta_{k}^{(2)}$ can be factorized as

$\beta_{k}^{(2)}=\beta_{k}^{(2,1)}+\beta_{k}^{(2,2)},$ (100)

where

$\beta_{k}^{(2,1)}:=\int\limits_{1/\ell}^{1}\frac{1-\ell
y}{y}\int\limits_{(1-\varepsilon)\theta\cdot\mathcal{R}^{k-1}}\int\limits_{0}^{1-y}\left(\partial_{y}^{(k)}\frac{\partial^{k}}{\partial
t_{1}\dots\partial
t_{k}}f_{1}(t_{1},\dots,t_{k})+2\partial_{y}^{(k)}\frac{\partial^{k}}{\partial
t_{1}\dots\partial t_{k}}f_{2}(t_{1},\dots,t_{k})\right)\\\
\times\partial_{y}^{(k)}\frac{\partial^{k}}{\partial t_{1}\dots\partial
t_{k}}f_{1}(t_{1},\dots,t_{k})\,dt_{i}\,dt_{1}\dots dt_{k-1}\,dy$

and

$\beta_{k}^{(2,2)}:=(1-\ell)\int\limits_{(1-\varepsilon)\theta\cdot\mathcal{R}_{k-1}}\left(\frac{\partial^{k-1}}{\partial
t_{1}\dots\partial
t_{k-1}}f_{1}(t_{1},\dots,t_{k-1},0)+2\frac{\partial^{k-1}}{\partial
t_{1}\dots\partial t_{k-1}}f_{2}(t_{1},\dots,t_{k-1},0)\right)\\\
\times\frac{\partial^{k-1}}{\partial t_{1}\dots\partial
t_{k-1}}f_{1}(t_{1},\dots,t_{k-1},0)\,dt_{1}\dots dt_{k-1}.$

Let $\delta_{1}>0$ be a sufficiently small fixed quantity. By a smooth
partitioning, we may assume without loss of generality that all of the
$f_{i,j}$ are supported on intervals of length at most $\delta_{1}$, while
keeping the sum

$\sum_{j=1}^{J}|c_{j}||f_{1,j}(t_{1})|\cdots|f_{k,j}(t_{k})|$

bounded uniformly in $t_{1},\dots,t_{k}$ and in $\delta_{1}$. Therefore, the
supports of $f_{1}$ and $f_{2}$ overlap only on some set of measure at most
$O(\delta_{1})$. Hence, we conclude that

$\beta_{k}:=\beta_{k}^{(1)}+\beta_{k}^{(2)}=\,\widetilde{J}_{k,\varepsilon}(f)+\widetilde{Q}_{k,\varepsilon}(f)+O(\delta_{1}),$
(101)

which implies

$\sum_{\begin{subarray}{c}n\sim x\\\ n\equiv b\bmod W\\\
\mathcal{P}(n)\textup{ sq-
free}\end{subarray}}\nu(n)\sum_{p|L_{k}(n)}\left(1-\ell\log_{x}p\right)\leq\left(\beta_{k}+o(1)\right)B^{-k}\frac{x}{W}.$
(102)

A similar argument provides results analogous to (102) for all remaining
indices $1\leq i\leq k-1$. If we set $\delta_{1}$ to be small enough, then the
claim $DHL_{\Omega}[k;\varrho_{k}]$ follows from Lemma 3 and (81). We also
note that if $\varepsilon=0$, then (102) becomes an equality, because in this
case we have $\mathcal{J}_{2}=\emptyset$. ∎

##  5 Solving variational problems

In this Section we focus on applying Theorems 10, 12, and 13 to prove Theorem
4.

###  5.1 Proof of Theorem 11

###### Proof.

This is a direct application of Theorem 10. We choose
$F(t_{1},\dots,t_{k})=\bar{f}(t_{1}+\dots+t_{k})$ for a function
$\bar{f}\colon[0,+\infty)\rightarrow\mathbf{R}$ defined as

$\bar{f}(x):=\begin{cases}f(x),&\text{for }x\in[0,1],\\\
0,&\text{otherwise.}\end{cases}$ (103)

We also set $\ell=1$, so the contribution from $J_{i}(F)$ vanishes for each
possible choice of index $i$.

First, we calculate $I(F)$. We substitute $t_{1}+\dots+t_{k}\mapsto t$ and
leave $t_{j}$ the same for $j=2,\dots,k$. We get

$I(F)=\int\limits_{0}^{1}f(t)^{2}\left(\int\limits_{t\cdot\mathcal{R}_{k-1}}dt_{2}\dots
dt_{k}\right)dt\leavevmode\nobreak\ =\leavevmode\nobreak\
\frac{1}{(k-1)!}\,\int\limits_{0}^{1}f(t)^{2}\,t^{k-1}dt\leavevmode\nobreak\
=\leavevmode\nobreak\ \bar{I}(f).$ (104)

Let us move on to the $Q_{i}(F)$ integral. For the sake of convenience let us
choose $i=k$. By the same substitution as before we arrive at

$Q_{k}(F)=\int\limits_{0}^{\frac{1}{\theta}}\frac{1-\theta
y}{y}\int\limits_{0}^{1}\left(\bar{f}(t)-\bar{f}(t+y)\right)^{2}\int\limits_{t\cdot\mathcal{R}_{k-1}}\mathbf{1}_{t_{k}\leq\frac{1}{\theta}-y}\,dt_{2}\dots
dt_{k}\,dt\,dy.$ (105)

We wish to replace $\bar{f}$ with $f$ and discard the indicator function. The
latter can be simply performed by calculating the inner integral. Note that it
may be geometrically intepreted as a volume of a ‘bitten’ simplex. We define

$H_{y,t}:=\left\\{(t_{2},\dots,t_{k})\in\mathbf{R}^{k-1}\colon
t_{2}+\dots+t_{k}\leq t\leavevmode\nobreak\ \leavevmode\nobreak\
\text{and}\leavevmode\nobreak\ \leavevmode\nobreak\
t_{k}>1/\theta-y\right\\}.$

Observe that $H_{y,t}$ is just a translated simplex
$(t-1/\theta+y)\cdot\mathcal{R}_{k-1}$ for $1/\theta-t<y\leq 1/\theta$ and an
empty set for $y\leq 1/\theta-t$ . Thus, we obtain

$\frac{1}{(k-1)!}\int\limits_{t\cdot\mathcal{R}_{k-1}}\mathbf{1}_{t_{k}\leq\frac{1}{\theta}-y}\,dt_{2}\dots
dt_{k}\\\
=\text{Vol}(t\cdot\mathcal{R}_{k-1})-\text{Vol}(H_{y,t})=\begin{cases}t^{k-1},&\text{for
}y\in[0,\frac{1}{\theta}-t],\\\ t^{k-1}-(t-1/\theta+y)^{k-1},&\text{for
}y\in(\frac{1}{\theta}-t,\frac{1}{\theta}].\end{cases}$ (106)

For $0\leq y\leq 1$ we also have

$\bar{f}(t)-\bar{f}(t+y)=\begin{cases}f(t)-f(t+y),&\text{for }t\in[0,1-y],\\\
f(t),&\text{for }t\in(1-y,1],\end{cases}$ (107)

and simply $\bar{f}(t)-\bar{f}(t+y)=\bar{f}(t)$ for greater $y$. We decompose
the domain of integration

$D:=\\{(y,t)\in\mathbf{R}^{2}\colon 0<t<1\leavevmode\nobreak\
\text{and}\leavevmode\nobreak\ 0<y<1/\theta\\}$

into

$D=D_{1}\cup D_{2}\cup D_{3}\cup D_{4}\cup D_{5}\cup\left(\text{some set of
Lebesgue measure 0}\right),$ (108)

where

$\displaystyle D_{1}$ $\displaystyle:=\\{(y,t)\in\mathbf{R}^{2}\colon
0<y<1\leavevmode\nobreak\ \text{and}\leavevmode\nobreak\ 0<t<1-y\\},$
$\displaystyle D_{2}$ $\displaystyle:=\\{(y,t)\in\mathbf{R}^{2}\colon
0<y<1\leavevmode\nobreak\ \text{and}\leavevmode\nobreak\ 1-y<t<1\\},$
$\displaystyle D_{3}$ $\displaystyle:=\\{(y,t)\in\mathbf{R}^{2}\colon
1<y<1/\theta-1\leavevmode\nobreak\ \text{and}\leavevmode\nobreak\ 0<t<1\\},$
$\displaystyle D_{4}$ $\displaystyle:=\\{(y,t)\in\mathbf{R}^{2}\colon
1/\theta-1<y<1/\theta\leavevmode\nobreak\ \text{and}\leavevmode\nobreak\
0<t<1/\theta-y\\},$ $\displaystyle D_{5}$
$\displaystyle:=\\{(y,t)\in\mathbf{R}^{2}\colon
1/\theta-1<y<1/\theta\leavevmode\nobreak\ \text{and}\leavevmode\nobreak\
1/\theta-y<t<1\\}.$

Therefore, from (105–108) we get

$Q_{k}(F)=\iint\limits_{D_{1}}\frac{1-\theta
y}{y}\,(f(t)-f(t+y))^{2}\,t^{k-1}\,dt\,dy\leavevmode\nobreak\ \\\
+\iint\limits_{D_{2}\cup D_{3}\cup D_{4}}\frac{1-\theta
y}{y}\,f(t)^{2}\,t^{k-1}\,dt\,dy\leavevmode\nobreak\ +\leavevmode\nobreak\
\iint\limits_{D_{5}}\frac{1-\theta
y}{y}\,f(t)^{2}\,\left(t^{k-1}-\left(t+y-1/\theta\right)^{k-1}\right)\,dt\,dy.$
(109)

The same reasoning applies to $Q_{i}(F)$ for $i=1,\dots,k-1$.

###  5.2 Collapse of Theorem 12 into one dimension and bounds for
$\Omega_{k}^{\textmd{ext}}$

We wish to transform Theorem 12 into its one-dimensional analogue in a similar
manner as we did in Subsection 5.1. For the sake of convenience, let us assume
in this subsection that $k\geq 3$. In the $k=2$ case, Theorem 12 can be
applied directly without any intermediate simplifications – it also does not
provide anything beyond what is already known anyway, as presented in Table D.
We take

$F(t_{1},\dots,t_{k})=f(t_{1}+\dots+t_{k})\mathbf{1}_{(t_{1},\dots,t_{k})\in\mathcal{R}_{k}^{\prime}},$

where $f\colon[0,+\infty)\rightarrow\mathbf{R}$ is some locally square-
integrable function. We also put $\ell=1$, so the contribution from $J_{i}(F)$
vanishes for each possible choice of index $i$. Let us begin with $I(F)$
integral. This time we substitute

$\begin{cases}t_{1}+\dots+t_{k}&\longmapsto\leavevmode\nobreak\
\leavevmode\nobreak\ x,\\\ t_{1}+\dots+t_{k-1}&\longmapsto\leavevmode\nobreak\
\leavevmode\nobreak\ t,\\\ t_{1}&\longmapsto\leavevmode\nobreak\
\leavevmode\nobreak\ t_{1},\\\ &\vdots\\\
t_{k-2}&\longmapsto\leavevmode\nobreak\ \leavevmode\nobreak\
t_{k-2}.\end{cases}$ (110)

We also relabel $t_{k-1}$ as $s$. It is calculated in [Lewulis, Subsubsection
‘Calculating J’] that

$I(F)=\int\limits_{\mathcal{R}_{k}^{\prime}}F(t_{1},\dots,t_{k})^{2}\,dt_{1}\dots
dt_{k}\leavevmode\nobreak\ =\leavevmode\nobreak\
\frac{1}{(k-3)!}\int\limits_{0}^{1}\int\limits_{0}^{t}\int\limits_{t}^{1+\frac{s}{k-1}}f(x)^{2}\,(t-s)^{k-3}\,dx\,ds\,dt.$
(111)

Let us focus on the $Q_{i}(F)$ integral. Again, for the sake of convenience we
choose $i=k$. We have

$Q_{k}(F)=\int\limits_{0}^{\frac{1}{\theta}}\frac{1-\theta
y}{y}\int\limits_{\mathcal{R}_{k-1}}\left(\int\limits_{0}^{\rho(t_{1},\dots,t_{k-1})}\left(\partial_{y}\bar{f}(t_{1}+\dots+t_{k})\right)^{2}\mathbf{1}_{t_{k}\leq\frac{1}{\theta}-y}\,dt_{k}\right)\,dt_{1}\dots
dt_{k-1}\,dy,$ (112)

where

$\rho(t_{1},\dots,t_{k}):=\sup\\{t_{k}\in\mathbf{R}\colon(t_{1},\dots,t_{k})\in\mathcal{R}_{k}^{\prime}\\}.$

We observe that any permutation of the variables $t_{1},\dots,t_{k-1}$ does
not change the integrand. We also notice that if we consider an extra
assertion $0<t_{1}<\dots<t_{k-1}$, then

$\rho(t_{1},\dots,t_{k-1})=1-t_{2}-\dots-t_{k-1}.$

Therefore, $Q_{k}(F)$ equals

$(k-1)!\,\int\limits_{0}^{\frac{1}{\theta}}\frac{1-\theta
y}{y}\int\limits_{\begin{subarray}{c}\mathcal{R}_{k-1}\\\
0<t_{1}<\dots<t_{k-1}\end{subarray}}\left(\int\limits_{0}^{1-t_{2}-\dots-
t_{k-1}}\left(\partial_{y}\bar{f}(t_{1}+\dots+t_{k})\right)^{2}\mathbf{1}_{t_{k}\leq\frac{1}{\theta}-y}\,dt_{k}\right)\,dt_{1}\dots
dt_{k-1}\,dy,$ (113)

In order to calculate the inner integral, we perform the same substitution as
described (110). This way we obtain

$\int\limits_{\begin{subarray}{c}\mathcal{R}_{k-1}\\\
0<t_{1}<\dots<t_{k-1}\end{subarray}}\left(\int\limits_{0}^{1-t_{2}-\dots-
t_{k-1}}\left(\partial_{y}\bar{f}(t_{1}+\dots+t_{k})\right)^{2}\mathbf{1}_{t_{k}\leq\frac{1}{\theta}-y}\,dt_{k}\right)\,dt_{1}\dots
dt_{k-1}\\\ =\leavevmode\nobreak\
\int\limits_{0}^{1}\int\limits_{\begin{subarray}{c}0<t_{1}<\dots<t_{k-2}<t-\sum_{i=1}^{k-2}t_{i}\end{subarray}}\left(\int\limits_{t}^{1+t_{1}}\left(\partial_{y}\bar{f}(x)\right)^{2}\mathbf{1}_{x-t\leq\frac{1}{\theta}-y}\,dx\right)\,dt_{1}\dots
dt_{k-2}\,dt$ (114)

For the sake of clarity, we relabel $t_{1}$ as $s$. Thus, the expression from
(114) equals

$\int\limits_{0}^{1}\int\limits_{0}^{\frac{t}{k-1}}\left(\int\limits_{t}^{1+s}\left(\partial_{y}\bar{f}(x)\right)^{2}\mathbf{1}_{x-t\leq\frac{1}{\theta}-y}\,dx\right)\left(\int\limits_{s}^{\frac{t-s}{k-2}}\int\limits_{t_{2}}^{\frac{t-s-
t_{2}}{k-3}}\cdots\int\limits_{t_{k-3}}^{\frac{t-s-t_{2}-\dots-
t_{k-3}}{2}}\,dt_{k-2}\dots dt_{2}\right)\,ds\,dt.$ (115)

If $k=3$, then the inner integral simplifies to $1$. For $0\leq s\leq t$ let
us define

$\mathscr{L}(k;t,s):=\int\limits_{s}^{\frac{t-s}{k-2}}\int\limits_{t_{2}}^{\frac{t-s-
t_{2}}{k-3}}\cdots\int\limits_{t_{k-3}}^{\frac{t-s-t_{2}-\dots-
t_{k-3}}{2}}\,dt_{k-2}\dots dt_{2}.$

We apply the induction over $k$ to show that

$\mathscr{L}(k;t,s)=\frac{(t-(k-1)s)^{k-3}}{(k-2)!(k-3)!}.$ (116)

Our claim is obviously true for $k=3$. For every $k\geq 3$ we observe the
identity

$\mathscr{L}(k+1;t,s)=\int\limits_{s}^{\frac{t-s}{k-1}}\mathscr{L}(k;t-s,u)\,du.$
(117)

To finish the proof of the claim one has to put (116) into (117) and
substitute

$t-s-(k-1)u\mapsto z.$

Combining (113–115) with the claim discussed above we conclude that $Q_{k}(F)$
equals

$\frac{(k-1)!}{(k-2)!(k-3)!}\,\int\limits_{0}^{\frac{1}{\theta}}\frac{1-\theta
y}{y}\int\limits_{0}^{1}\int\limits_{0}^{\frac{t}{k-1}}\left(\int\limits_{t}^{1+s}\left(\partial_{y}\bar{f}(x)\right)^{2}\mathbf{1}_{x-t\leq\frac{1}{\theta}-y}\,dx\right)(t-(k-1)s)^{k-3}\,ds\,dt\,dy.$
(118)

Let us relabel $s$ as $s/(k-1)$ to simplify the expression above. We arrive at

$\displaystyle Q_{k}(F)$
$\displaystyle=\frac{1}{(k-3)!}\,\int\limits_{0}^{\frac{1}{\theta}}\frac{1-\theta
y}{y}\int\limits_{0}^{1}\int\limits_{0}^{t}\left(\int\limits_{t}^{1+\frac{s}{k-1}}\left(\partial_{y}\bar{f}(x)\right)^{2}\mathbf{1}_{x-t\leq\frac{1}{\theta}-y}\,dx\right)(t-s)^{k-3}\,ds\,dt\,dy$
$\displaystyle=\frac{1}{(k-3)!}\int\limits_{E}\frac{1-\theta
y}{y}\left(\partial_{y}\bar{f}(x)\right)^{2}(t-s)^{k-3}\,dx\,dt\,ds\,dy,$
(119)

where

$E:=\left\\{(y,s,t,x)\in\mathbf{R}^{4}\colon
0<y<\frac{1}{\theta},\leavevmode\nobreak\ 0<t<1,\leavevmode\nobreak\
0<s<t,\leavevmode\nobreak\ t<x<1+\frac{s}{k-1},\leavevmode\nobreak\
x-t<\frac{1}{\theta}-y\right\\}.$

We wish to drop the bar from $\bar{f}$. Hence, we decompose

$E=E_{1}\cup E_{2},$

where

$\displaystyle E_{1}$ $\displaystyle:=\left\\{(y,s,t,x)\in E\colon x+y\leq
1+\frac{s}{k-1}\right\\},$ $\displaystyle E_{2}$
$\displaystyle:=\left\\{(y,s,t,x)\in E\colon x+y>1+\frac{s}{k-1}\right\\}.$

From (5.2) we have that $Q_{k}(F)$ equals $1/(k-3)!$ times

$\int\limits_{E_{1}}\frac{1-\theta
y}{y}\left(f(x)-f(x+y)\right)^{2}(t-s)^{k-3}\,dx\,dt\,ds\,dy\leavevmode\nobreak\
+\leavevmode\nobreak\ \int\limits_{E_{2}}\frac{1-\theta
y}{y}f(x)^{2}(t-s)^{k-3}\,dx\,dt\,ds\,dy.$ (120)

Now, we would like to convert two integrals above into a finite sum of
integrals with explicitely given limits, just like in (109). If we choose the
order of integration

$y\rightarrow x\rightarrow t\rightarrow s,$

then we get

$\displaystyle\int\limits_{E_{1}}\boxtimes\leavevmode\nobreak\ $
$\displaystyle=\leavevmode\nobreak\
\int\limits_{0}^{1}\int\limits_{s}^{1}\int\limits_{t}^{1+\frac{s}{k-1}}\leavevmode\nobreak\
\int\limits_{0}^{1+\frac{s}{k-1}-x}\boxtimes\leavevmode\nobreak\
dy\,dx\,dt\,ds,$ (121)
$\displaystyle\int\limits_{E_{2}}\boxtimes\leavevmode\nobreak\ $
$\displaystyle=\leavevmode\nobreak\
\int\limits_{0}^{1}\int\limits_{s}^{1}\int\limits_{t}^{1+\frac{s}{k-1}}\int\limits_{1+\frac{s}{k-1}-x}^{\frac{1}{\theta}+t-x}\boxtimes\leavevmode\nobreak\
dy\,dx\,dt\,ds,$ (122)

where $\boxtimes$ denotes an arbitrary integrable function.

######  Remark 5.

From the computational point of view, the variable $y$ should be integrated in
the last order, because it engages a non-polynomial function. The author found
the following order of integration as the most computationally convenient:

$x\rightarrow t\rightarrow s\rightarrow y.$

Unfortunately, in this case there is no decompsition of $E$ similar to (108)
that is common for all possible choices of $k$ and $\theta$. In the
$k=4,\,\theta=1/2$ case, which accordingly to Tables C and D is the only one,
where we can expect a qualitative improvement over Theorem 11, we are able to
convert the integral over $E$ into 15 integrals with explicitely given limits.
Such a conversion is a straightforward operation (quite complicated to perform
without a computer program, though). We do not present the precise shape of
these integrals here.

Let us set $k=4$, $\theta=1/2$, and

$f(x)=12+63x+100x^{2}.$

Combining (111) with (121–122), and performing the calculations on a computer,
we get

$\displaystyle I(F)=\leavevmode\nobreak\ $
$\displaystyle\frac{2977019}{51030}>58.3386047422.$ $\displaystyle
Q_{k}(F)=\leavevmode\nobreak\ $
$\displaystyle\frac{132461570733345\log\frac{5}{3}-997242435\log
3-49178701703144}{4629441600}$
$\displaystyle+\frac{6144554}{105}\log\frac{6}{5}-\frac{15996989}{280}\,\text{arcoth}\,4<70.0214943902.$

This combined with Theorem 12 gives

$\Omega_{4}^{\text{ext}}\left(\theta=\frac{1}{2}\right)<8.80105,$ (123)

which proves the $k=4$ case of the conditional part of Theorem 4.

###  5.3 bounds for $\Omega_{k,\varepsilon}$

We apply Theorem 13 with $\eta=1+\varepsilon$ and some $\varepsilon$, $\ell$
satisfying

$2\theta(1+\varepsilon)+\frac{1}{\ell}=1.$

We choose $F(t_{1},\dots,t_{k})=\bar{f}(t_{1}+\dots+t_{k})$ for a function
$\bar{f}\colon[0,+\infty)\rightarrow\mathbf{R}$ satisfying

$\bar{f}(x):=\begin{cases}f(x),&\text{for }x\in[0,1+\varepsilon],\\\
0,&\text{otherwise.}\end{cases}$ (124)

First, we calculate $I(F)$. We proceed just like in (104) and get

$I(F)\,=\
\int\limits_{0}^{1+\varepsilon}f(t)^{2}\left(\,\int\limits_{t\cdot\mathcal{R}_{k-1}}dt_{2}\dots
dt_{k}\right)dt\leavevmode\nobreak\ =\leavevmode\nobreak\
\frac{1}{(k-1)!}\,\int\limits_{0}^{1+\varepsilon}f(t)^{2}\,t^{k-1}\,dt.$ (125)

Next, let us consider $J_{i,\varepsilon}(F)$. As before, let us put $i=k$. We
have

$J_{k,\varepsilon}(F)\,=\,\int\limits_{(1-\varepsilon)\cdot\mathcal{R}_{k-1}}\left(\int\limits_{0}^{1+\varepsilon-
t_{1}-\dots-t_{k-1}}f(t_{1}+\dots+t_{k})\,dt_{k}\right)^{2}dt_{1}\dots
dt_{k-1}.$ (126)

We perform the same substitution as in (104). We get that
$J_{k,\varepsilon}(F)$ equals

$\begin{gathered}\,\int\limits_{0}^{1-\varepsilon}\left(\int\limits_{0}^{1+\varepsilon-t}f(t+t_{k})\,dt_{k}\right)^{2}\int\limits_{t\cdot\mathcal{R}_{k-1}}\,dt_{1}\dots
dt_{k-2}\,dt\,=\,\int\limits_{0}^{1-\varepsilon}\left(\int\limits_{t}^{1+\varepsilon}f(x)\,dx\right)^{2}\frac{t^{k-2}}{(k-2)!}\,dt.\end{gathered}$
(127)

We perform analogous calculations for $i=1,\dots,k-1$.

Let us move to $Q_{i,\varepsilon}(F)$. Put

$\begin{cases}t_{1}+\dots+t_{k-1}&\longmapsto\leavevmode\nobreak\
\leavevmode\nobreak\ t,\\\ t_{2}&\longmapsto\leavevmode\nobreak\
\leavevmode\nobreak\ t_{2},\\\ &\vdots\\\
t_{k}&\longmapsto\leavevmode\nobreak\ \leavevmode\nobreak\ t_{k}.\end{cases}$
(128)

and split

$Q_{k,\varepsilon}(F)=Q_{(1)}(f)+Q_{(2)}(f),$ (129)

where

$\displaystyle Q_{(1)}(f)$
$\displaystyle:=\frac{1}{(k-2)!}\int\limits_{0}^{\frac{1}{\ell\theta}}\frac{1-\ell\theta
y}{y}\int\limits_{0}^{1+\varepsilon}\left(\,\int\limits_{0}^{\frac{1}{\theta}-y}\left(\bar{f}(t+t_{k})-\bar{f}(t+t_{k}+y)\right)^{2}\,dt_{k}\right)t^{k-2}\,dt\,dy,$
$\displaystyle Q_{(2)}(f)$
$\displaystyle:=\frac{1}{(k-2)!}\int\limits_{\frac{1}{\ell\theta}}^{\frac{1}{\theta}}\frac{1-\ell\theta
y}{y}\int\limits_{0}^{1-\varepsilon}\left(\,\int\limits_{0}^{\frac{1}{\theta}-y}\left(\bar{f}(t+t_{k})-\bar{f}(t+t_{k}+y)\right)^{2}\,dt_{k}\right)t^{k-2}\,dt\,dy,$
(130)

Therefore, we put $t_{k}+t\mapsto x$ and decompose

$(k-2)!\left(Q_{(1)}(f)+Q_{(2)}(f)\right)=\\\\[4.30554pt]
\int\limits_{H_{1}\cup H_{3}}\frac{1-\ell\theta
y}{y}f(x)^{2}\,t^{k-2}\,dx\,dt\,dy\,+\,\int\limits_{H_{2}\cup
H_{4}}\frac{1-\ell\theta
y}{y}\left(f(x)-f(x+y)\right)^{2}t^{k-2}\,dx\,dt\,dy,$ (131)

where

$H:=\\{(y,t,x)\in\mathbf{R}^{3}\colon 0<y<1/\theta,\leavevmode\nobreak\
0<t<x<1+\varepsilon,\leavevmode\nobreak\ x-t<1/\theta-y\\},$

and

$\displaystyle H_{1}$ $\displaystyle:=\\{(y,t,x)\in H\colon 0<y\leq
1/(\ell\theta)\leavevmode\nobreak\ \text{and}\leavevmode\nobreak\
x+y<1+\varepsilon\\},$ $\displaystyle H_{2}$ $\displaystyle:=\\{(y,t,x)\in
H\colon 0<y\leq 1/(\ell\theta)\leavevmode\nobreak\
\text{and}\leavevmode\nobreak\ x+y>1+\varepsilon\\},$ $\displaystyle H_{3}$
$\displaystyle:=\\{(y,t,x)\in H\colon
1/(\ell\theta)<y<1/\theta\leavevmode\nobreak\ \text{and}\leavevmode\nobreak\
0<t<1-\varepsilon\leavevmode\nobreak\ \text{and}\leavevmode\nobreak\
x+y<1+\varepsilon\\},$ $\displaystyle H_{4}$ $\displaystyle:=\\{(y,t,x)\in
H\colon 1/(\ell\theta)<y<1/\theta\leavevmode\nobreak\
\text{and}\leavevmode\nobreak\ 0<t<1-\varepsilon\leavevmode\nobreak\
\text{and}\leavevmode\nobreak\ x+y>1+\varepsilon\\},$

Unfortunately, with varying $k$, $\varepsilon$, $\theta$ there is no uniform
way to decompose $H_{1},\dots,H_{4}$ further into integrals with explicitely
given limits. In the unconditional setting, namely with $\theta=1/4$ fixed,
every choice of parameters described in Table E provides less than 10
different integrals to calculate. For these choices we present close to
optimal polynomials minimalizing the $\Omega_{k,\varepsilon}$ functional.

Table G. Upper bounds for $\Omega_{k,\varepsilon}$.

$k$ |  $\varepsilon$ |  $f(1+\varepsilon-x)$ |  bounds for $\Omega_{k}$ |
---|---|---|---|---
$2$ |  1/3 |  $1+5x+3x^{2}$ |  4.6997 |
$3$ |  1/4 |  $1+7x+10x^{2}$ |  7.7584 |
$4$ |  1/5 |  $1+7x+19x^{2}$ |  11.0533 |
$5$ |  1/6 |  $1+7x+33x^{2}$ |  14.5415 |
$6$ |  1/7 |  $1+7x+51x^{2}$ |  18.1907 |
$7$ |  1/9 |  $1+8x+70x^{2}$ |  21.9939 |
$8$ |  1/10 |  $1+8x+102x^{2}$ |  25.9038 |
$9$ |  1/10 |  $1+5x+132x^{2}$ |  29.9059 |
$10$ |  2/21 |  $1+35x+30x^{2}+470x^{3}$ |  33.9384 |

These bounds are sufficient to prove the unconditional part of Theorem 4.

∎

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