# Quantitative bounds for products of simplices in subsets of the unit cube

Polona Durcik  Schmid College of Science and Technology, Chapman University,
One University Drive, Orange, CA 92866, USA<EMAIL_ADDRESS>and  Mario
Stipčić  Schmid College of Science and Technology, Chapman University, One
University Drive, Orange, CA 92866, USA<EMAIL_ADDRESS>
###### Abstract.

For each $1\leq i\leq n$, let $k_{i}\geq 1$ and let $\Delta_{i}$ be a set of
vertices of a non-degenerate simplex of $k_{i}+1$ points in
$\mathbb{R}^{k_{i}+1}$. If
$A\subseteq[0,1]^{k_{1}+1}\times\cdots\times[0,1]^{k_{n}+1}$ is a Lebesgue
measurable set of measure at least $\delta$, we show that there exists an
interval $I=I(\Delta_{1},\ldots,\Delta_{n},A)$ of length at least
$\exp(-\delta^{-C(\Delta_{1},\ldots,\Delta_{n})})$ such that for each
$\lambda\in I$, the set $A$ contains
$\Delta^{\prime}_{1}\times\cdots\times\Delta^{\prime}_{n}$, where each
$\Delta_{i}^{\prime}$ is an isometric copy of $\lambda\Delta_{i}$. This is a
quantitative improvement of a result by Lyall and Magyar. Our proof relies on
harmonic analysis. The main ingredient in the proof are cancellation estimates
for forms similar to multilinear singular integrals associated with
$n$-partite $n$-regular hypergraphs.

###### Key words and phrases:

Euclidean Ramsey theory, point configuration, singular integral.

######  Mathematics Subject Classification:

Primary 05D10; Secondary 42B20, 28A75

##  1\. Introduction

Finite point configurations in large but otherwise arbitrary subsets of
Euclidean spaces have been a topic of several recent investigations [1, 2, 3,
4, 8, 10, 12, 13, 14]. In this paper we study existence of isometric copies of
dilates of products of non-degenerate simplices in sets of positive Lebesgue
measure in the unit cube. The main focus is on a quantitative lower bound on
the size of the family of dilates whose isometric copies are detected in the
set, as stated in the following theorem.

######  Theorem 1.

Let $n\geq 1$. For each $1\leq i\leq n$, let $k_{i}\geq 1$ and let

$\Delta_{i}=\\{0,v_{i}^{1},\ldots,v_{i}^{k_{i}}\\},$

where $v_{i}^{1},\ldots,v_{i}^{k_{i}}\in\mathbb{R}^{k_{i}}$ are linearly
independent vectors. Let $\delta\in(0,1/2]$. There exists a constant
$C=C(\Delta_{1},\ldots,\Delta_{n})>0$ such that the following holds. For every
measurable set $A\subseteq[0,1]^{k_{1}+1}\times\cdots\times[0,1]^{k_{n}+1}$
with the Lebesgue measure $|A|\geq\delta$, there is an interval
$I=I(\Delta_{1},\ldots,\Delta_{n},A)\subseteq(0,\infty)$ of length at least

$(\exp(\delta^{-C(\Delta_{1},\ldots,\Delta_{n})}))^{-1}$

such that for every $\lambda\in I$ and $1\leq i\leq n$ one can find
$x_{i},y_{i}^{1},\ldots,y_{i}^{k_{i}}\in\mathbb{R}^{k_{i}+1}$ satisfying

$y_{i}^{l}\cdot y_{i}^{m}=\lambda v_{i}^{l}\cdot\lambda
v_{i}^{m}\quad\textup{for}\quad 1\leq l,m\leq k_{i}$

and $\Delta^{\prime}_{1}\times\cdots\times\Delta^{\prime}_{n}\subseteq A$,
where
$\Delta^{\prime}_{i}=\\{x_{i},x_{i}+y_{i}^{1},\ldots,x_{i}+y_{i}^{k_{i}}\\}$.
That is, $\Delta^{\prime}_{i}$ is an isometric copy of $\lambda\Delta_{i}$.

The case of a single simplex, i.e. case $n=1$ of Theorem 1, is a corollary of
a stronger quantitative finitary result by Bourgain [1, Proposition 3]. The
exact quantitative bound on the length of the interval is not written
explicitly in [1], but it can be tracked down from its proof in [1, Section
3]. Lyall and Magyar [14] proved a version of Theorem 1 for an arbitrary
finite product of simplices but with the lower bound on the length of the
interval $I$ being a reciprocal of a tower of exponentials

$(\exp(\exp(\cdots\exp(\delta^{-C(\Delta_{1},\ldots,\Delta_{n})}))\cdots)))^{-1}$

of length $n$. Their proof is based on a weak hypergraph regularity lemma
developed in the setting of Euclidean spaces. Kovač and the first author [3]
proved Theorem 1 when $k_{i}=1$ for all $1\leq i\leq n$, in which case each
simplex has only two points, i.e. their product is the vertex set of an
$n$-dimensional rectangular box. Theorem 1 is thus an extension of [3] and a
quantitative improvement of [14].

In analogy with the papers [1] and [14] we formulated Theorem 1 in the lowest
dimension possible, i.e. for sets
$A\subseteq[0,1]^{d_{1}}\times\cdots\times[0,1]^{d_{n}}$ where
$d_{i}=k_{i}+1$. As in these papers, our result extends to all higher
dimensions $d_{i}\geq k_{i}+1$ with mainly notational changes throughout the
proof.

Our approach builds on the methods from [2] and [3], which detect simplices
through certain counting forms. Following [2] and [3], in Section 2 we split
the counting forms into their structured, error, and uniform parts and analyze
each of the parts in the subsequent sections. The error part is handled
through a certain pigeonholing argument. The main novelty of this paper is
that to estimate the error part, we need to show cancellation in multilinear
singular integral forms associated with $n$-partite $n$-regular hypergraphs.

Certain dyadic models of the singular integral forms that we consider in this
paper are known to be bounded in an open range of Lebesgue spaces. This was
shown by Kovač [9], who proved $L^{p}$ estimates for multilinear forms
associated with graphs, and the second author [15], who extended this result
to hypergraphs. On the other hand, boundedness of the continuous variants of
the forms in [9] and their modified versions, which appear in this paper, is
still unknown in general. The only currently known bounds are in the special
case of hypercubes, which are hypergraphs having two vertices in each
partition class. This case was investigated by Slavikova, Thiele, and the
first author [6], who proved estimates for the associated forms in an open
range of Lebesgue exponents. However, to establish Theorem 1, one does not
need to show $L^{p}$ boundedness of the forms that arise in the proof since a
weaker statement suffices to deduce our combinatorial result. Namely, it is
enough to establish only non-trivial quantitative cancellation bounds for the
corresponding singular integral forms. That is, if the singular integral
kernel consists of finitely many consecutive scales, it suffices to show a
quantitative improvement over the trivial bound in the number of scales.
Moreover, to prove Theorem 1 it is enough to assume that all input functions
are supported in the unit cube and bounded by $1$. The proof of our
cancellation estimates uses an extension of the methods used to estimate
singular Brascamp-Lieb type integrals as in [3, 6, 9], to the hypergraph
setting.

A connection between cancellation estimates and Euclidean Ramsey theorems was
made explicit in [3], which deals with arithmetic progressions in the unit
cube. The paper [3] shows that if $n\geq 3$,
$p\in[1,\infty)\setminus\\{1,2,\ldots,n-1\\}$ and the dimension $d$ is large
enough, then there exists a constant $C=C(n,p,d)$ such that the following
holds. If $\delta\in(0,1/2]$ and $A\subseteq[0,1]^{d}$ has Lebesgue measure at
least $\delta$, then there is an interval $I$ whose length depends only on
$n,p,d$, and $\delta$, such that for all $\lambda\in I$, the set $A$ contains
arithmetic progression $x,x+y,\ldots,x+(n-1)y\in A$ with
$\|y\|_{\ell^{p}}=\lambda$. Here, the length of $I$ is a reciprocal of the
double exponential of a negative power of $\delta$ when $n=3$ or $4$ and the
triple exponential when $n\geq 5$, which reflects the best known bounds in
Szemerédi’s theorem. The case $n=3$ also follows along the lines of the proof
by Cook, Magyar, and Pramanik [2]. However, we point out that in contrast with
Theorem 1, the size of the gap $y$ in [3] cannot be measured in all $\ell^{p}$
norms, in particular not in $\ell^{2}$, as generalizations of Bourgain’s
counterexample [1] show. Moreover, the result in [3] is at present only known
in sufficiently large dimensions.

In [3], bounds for the corresponding error part follow from cancellation
estimates for singular integrals similar to the multilinear Hilbert transform.
While non-trivial cancellation for the truncations of the multilinear Hilbert
transform to finitely many consecutive scales was first shown by Tao [17] and
later also by Zorin-Kranich [18], and Kovač, Thiele and the first author [5],
its boundedness on Lebesgue spaces is still one of the major open problems in
harmonic analysis. The only known bounds exist for the bilinear Hilbert
transform due to Lacey and Thiele [11]. In contrast to this, we expect that
boundedness of continuous variants of multilinear singular integrals
associated with hypergraphs, which arise in the proof of Theorem 1, is within
reach of current techniques. This is a problem for future investigation. In
view of this, we aim to keep the estimates for the error part and consequently
also the proof of Theorem 1 as short as possible, i.e. we do not aim to
optimize the bound in the number of scales for the singular integral as it
would not improve the quantitative bound in Theorem 1.

There are related questions on subsets of the Euclidean space of positive
upper Banach density, where one is concerned with finding all large dilates of
a fixed geometric configuration. The upper Banach density of a measurable set
$A\subseteq\mathbb{R}^{d}$ is defined as

$\overline{\delta}(A):=\limsup_{N\to\infty}\sup_{x\in\mathbb{R}^{d}}\frac{|A\cap(x+[0,N]^{d})|}{N^{d}}>0.$

The simplest example is a two-point pattern, for which it holds that whenever
$A\subseteq\mathbb{R}^{2}$ satisfies $\overline{\delta}(A)>0$, then there is a
threshold $\lambda_{0}=\lambda_{0}(A)$ such that for each
$\lambda\geq\lambda_{0}$, $A$ contains points $x,y$ with $|x-y|=\lambda$. This
was conjectured by Székely [16], the first proof was announced by Furstenberg,
Katznelson and Weiss [8], while in the meantime it was also independently
shown in [1] and by Falconer and Marstrand [7]. The paper [1] also generalizes
this result for a two-point pattern and shows existence of isometric copies of
all large dilates a fixed non-degenerate $(k+1)$-point simplex, in sets of
positive upper Banach density in $\mathbb{R}^{k+1}$. The paper [12] extends
this result to distance graphs, while the papers [13, 14] extend it also to
products of non-degenerate simplices. More precisely, in [14] it is shown that
if one is given $\Delta_{1},\ldots,\Delta_{n}$ as in Theorem 1 and
$A\subseteq\mathbb{R}^{k_{1}+1}\times\cdots\times\mathbb{R}^{k_{n}+1}$
satisfies $\overline{\delta}(A)>0$, then there exists
$\lambda_{0}=\lambda_{0}(A,\Delta_{1},\ldots,\Delta_{n})>0$ such that for all
$\lambda\geq\lambda_{0}$, the set $A$ contains
$\Delta_{1}^{\prime}\times\cdots\times\Delta_{n}^{\prime}$, where each
$\Delta_{i}^{\prime}$ an isometric copy of $\lambda\Delta_{i}$.

The corresponding result for all large dilates of rectangular boxes in sets of
positive upper Banach density was also obtained in [4] using harmonic analysis
techniques, but with some loss in the dimension of the ambient space. As
commented in [3], the arguments therein also give the corresponding result for
cubes in sets of positive upper Banach density and they extend to boxes as
well. Similarly, the techniques used in this paper can be extended to reprove
the full result from [14] in sets of positive upper Banach density, provided
one shows $L^{p}$ boundedness for the forms associated with hypergraphs, which
arise in the error part, or cancellation estimates to arbitrary, not
necessarily consecutive scales.

Recently, Kovač [10] initiated the study of density theorems for anisotropic
point configurations. The paper [10] established existence of all large
dilates of anisotropic simplices, anisotropic boxes, and an anisotropic
version of the result on distance trees, respectively, extending several
results contained in [1, 14], and [12] to the anisotropic setting. Proving
anisotropic versions of Theorem 1 and the corresponding results in sets of
positive density is yet another possible line of investigation.

Notation: For two non-negative quantities $A,B$ we write $A\lesssim B$ if
there exists a constant $C>0$ such that $A\leq CB$. We write $A\sim B$ if both
$A\lesssim B$ and $B\lesssim A$. In this paper, all constants are allowed to
depend on the dimensions $k_{i},n$, and simplices $\Delta_{i}$ and we will not
explicitly denote that in the notation.

##  2\. Counting forms and their decomposition

We begin this section by defining counting forms which detect isometric copies
of dilates of products of fixed non-degenerate simplices in a subset of the
unit cube.

Given $\Delta=\\{0,v^{1},\ldots,v^{d}\\}$, where $v^{1},\ldots,v^{d}$ are
linearly independent vectors in $\mathbb{R}^{d}$, we shall view them as
vertices of a simplex in $\mathbb{R}^{d+1}$ by identifying $\mathbb{R}^{d}$
with $\mathbb{R}^{d}\times\\{0\\}\subseteq\mathbb{R}^{d+1}$. First we recall
the definitions of suitable spherical measures associated with
$\lambda\Delta$, $\lambda>0$, as defined in [10] and previously used in [1,
13, 14].

For each $1\leq j\leq d$, let us write $v^{j}=|v^{j}|u^{j}$ where $u^{j}$ are
unit vectors. For $2\leq j\leq d$, the orthogonal projection of $u^{j}$ onto
$\textup{span}\\{u^{1},\ldots,u^{j-1}\\}$ is given by

$\beta_{j,1}u^{1}+\beta_{j,2}u^{2}+\ldots\beta_{j,j-1}u^{j-1}$

for some $\beta_{j,1},\ldots,\beta_{j,j-1}\in\mathbb{R}$. Let
$\widetilde{\sigma}$ denote the normalized surface measure on the unit sphere
$S^{d}$ in $\mathbb{R}^{d+1}$. For $2\leq j\leq d$ let

$\widetilde{\sigma}^{\Delta,x^{1},\ldots,x^{j-1}}$

be the normalized surface measure on the $(d-j+1)$-dimensional sphere in
$\mathbb{R}^{d+1}$ centered at

$\beta_{j,1}\frac{x^{1}}{|x^{1}|}+\beta_{j,2}\frac{x^{2}}{|x^{2}|}+\ldots\beta_{j,j-1}\frac{x^{j-1}}{|x^{j-1}|}$

with radius being equal to
$\textup{dist}(u^{j},\textup{span}\\{u^{1},\ldots,u^{j-1}\\}),$ which lies in
the $(d-j+2)$-dimensional plane in $\mathbb{R}^{d+1}$ that is orthogonal to
$\textup{span}\\{u^{1},\ldots,u^{j-1}\\}$.

For a Borel measure $\mu$ and a Borel set $A$ we define

$\mu_{\lambda}(A):=\mu(\lambda^{-1}A)$

and we write $\sigma^{\Delta}:=\widetilde{\sigma}_{|v^{1}|}$,
$\sigma^{\Delta,x^{1},\ldots,x^{j-1}}:=\widetilde{\sigma}^{\Delta,x^{1},\ldots,x^{j-1}}_{|v_{j}|}$.
We also define the measure $\sigma_{\Delta,\lambda}$ by

$\sigma_{\Delta,\lambda}(A):=\int_{\mathbb{R}^{d(d+1)}}\mathbbm{1}_{A}(y^{1},\ldots,y^{d})\,d\sigma_{\lambda}^{\Delta,y^{1},\ldots,y^{d-1}}(y^{d})\cdots
d\sigma_{\lambda}^{\Delta,y^{1}}(y^{2})\,d\sigma_{\lambda}^{\Delta}(y^{1})$

where $A$ is a Borel subset of $\mathbb{R}^{d(d+1)}$. The support of
$\sigma_{\Delta,\lambda}$ is the set of all points $y$ such that
$\\{0,y^{1},\ldots,y^{d}\\}$ is isometric to $\lambda\Delta$. Note that the
definition of $\sigma_{\Delta,\lambda}$ is symmetric in the choice of the
order of integration in $y^{1},\ldots,y^{d}$. This can be seen by rewriting
$\sigma_{\Delta,\lambda}$ using the Haar measure on $SO(d+1,\mathbb{R})$, see
for instance formulae (1) and (9) in [1]. By
$d\sigma_{\Delta,\lambda}(\cdot-c)$ we will denote integration with respect to
the measure $\sigma_{\Delta,\lambda}$ translated by $c\in\mathbb{R}^{d(d+1)}$.

Let us fix $n\geq 1$ and the dimensions $k_{i}\geq 1$ for $1\leq i\leq n$. We
also fix vertices of non-degenerate simplices of $k_{i}$ points $\Delta_{i}$
in $\mathbb{R}^{k_{i}+1},\,1\leq i\leq n$.

For any tuple of integers $L=(l_{1},\ldots,l_{n})$ satisfying $1\leq l_{i}\leq
k_{i}$ for each $1\leq i\leq n$, let $\mathcal{H}_{L}$ be the set of all
functions $h\colon~{}\\{1,\ldots,n\\}\rightarrow\mathbbm{N}\cup\\{0\\}$
satisfying $h(i)\leq l_{i}$ for each $1\leq i\leq n$. We will write
$L+1:=L+(1,\ldots,1)$. We will also denote

$K:=(k_{1},\ldots,k_{n}),\quad
k:=k_{1}+\cdots+k_{n},\quad\textup{and}\quad\mathcal{H}:=\mathcal{H}_{K}.$

For $h\in\mathcal{H}_{L}$ and

$x=(x^{0}_{1},\ldots,x^{l_{1}}_{1},x_{2}^{0},\ldots
x_{2}^{l_{2}},\ldots,x_{n}^{0},\ldots,x_{n}^{l_{n}})\in\mathbb{R}^{(K+1)\cdot(L+1)}$

we define $\Pi_{h}:\mathbb{R}^{(K+1)\cdot(L+1)}\to\mathbb{R}^{k+n}$ by

$\Pi_{h}{x}:=(x^{h(1)}_{1},\ldots,x^{h(n)}_{n}).$

Now we are ready to define the counting forms which detect isometric copies of
dilates of our simplices. For $\lambda>0$ and a measurable set
$A\subseteq\mathbb{R}^{k+n}$ we define the form

$\displaystyle\mathcal{N}^{0}_{\lambda}(A)=\int_{\mathbb{R}^{(K+1)\cdot(K+1)}}\Big{(}\prod_{h\in\mathcal{H}}\mathbbm{1}_{A}(\Pi_{h}x)\Big{)}\prod_{i=1}^{n}d\sigma_{\Delta_{i},\lambda}((x_{i}^{r}-x_{i}^{0})_{r=1}^{k_{i}})\,dx_{i}^{0}.$

Note that if $\mathcal{N}^{0}_{\lambda}(A)>0,$ then $A$ contains
$\Delta_{1}^{\prime}\times\cdots\times\Delta_{n}^{\prime}$, where each
$\Delta_{i}^{\prime}$ is an isometric copy of the dilated simplex
$\lambda\Delta_{i}$.

If $\varphi$ is a function on $\mathbb{R}^{m}$ and $t\in\mathbb{R}$, we denote
$\varphi_{t}(x):=t^{-m}\varphi(t^{-1}x)$. Let $g$ be the Gaussian

$g(x):=e^{-\pi|x|^{2}}.$

The dimension of Gaussians will always be understood from the context and will
not be part of the notation. For $\varepsilon>0$ we define a smoothed form

$\mathcal{N}^{\varepsilon}_{\lambda}(A):=\int_{\mathbb{R}^{(K+1)\cdot(K+1)}}\Big{(}\prod_{h\in\mathcal{H}}\mathbbm{1}_{A}(\Pi_{h}x)\Big{)}\Big{(}\prod_{i=1}^{n}{\sigma}_{\Delta_{i},\lambda}\ast{g}_{\varepsilon\lambda}((x_{i}^{r}-x_{i}^{0})_{r=1}^{k_{i}})\Big{)}\,dx.$

In the limit as $\varepsilon\to 0$ it recovers $\mathcal{N}^{0}_{\lambda}(A)$.
This can be seen similarly as in [3, Section 3].

Theorem 1 will be deduced from the following three lemmas.

######  Lemma 2.

There exists a constant $C>0$ such that for any $\lambda\in(0,1]$,
$\delta\in(0,1/2]$, and a measurable set
$A\subseteq[0,1]^{k_{1}+1}\times\cdots\times[0,1]^{k_{n}+1}$ with
$|A|\geq\delta$,

$\mathcal{N}^{1}_{\lambda}(A)\geq
C\delta^{(k_{1}+1)(k_{2}+1)\cdots(k_{n}+1)}.$ (2.1)

######  Lemma 3.

There exists a constant $C>0$ such that for any $\varepsilon\in(0,1]$,
positive integer $J$, any $\lambda_{j}\in(2^{-j},2^{-j+1}]$, $1\leq j\leq J$,
and any measurable set
$A\subseteq[0,1]^{k_{1}+1}\times\cdots\times[0,1]^{k_{n}+1}$,

$\sum_{j=1}^{J}\big{|}\mathcal{N}^{\varepsilon}_{\lambda_{j}}(A)-\mathcal{N}^{1}_{\lambda_{j}}(A)\big{|}\leq
CJ^{1-2^{-n}}\varepsilon^{-1-\sum_{i=1}^{n}k_{i}(k_{i}+2)}.$ (2.2)

######  Lemma 4.

There exists a constant $C>0$ such that for any
$\lambda,\,\varepsilon\in(0,1]$ and any measurable set
$A\subseteq[0,1]^{k_{1}+1}\times\cdots\times[0,1]^{k_{n}+1}$,

$\big{|}\mathcal{N}^{0}_{\lambda}(A)-\mathcal{N}^{\varepsilon}_{\lambda}(A)\big{|}\leq
C\varepsilon^{\frac{1}{2}}.$ (2.3)

Let us first show how to deduce Theorem 1 from these three lemmas. We
decompose

$\mathcal{N}^{0}_{\lambda}(A)=\mathcal{N}^{1}_{\lambda}(A)+(\mathcal{N}^{\varepsilon}_{\lambda}(A)-\mathcal{N}^{1}_{\lambda}(A))+(\mathcal{N}^{0}_{\lambda}(A)-\mathcal{N}^{\varepsilon}_{\lambda}(A)).$

Let $C_{1}$-$C_{3}$ be the constants that appear in Lemmas 2-4, respectively.
We may assume $C_{1}\in(0,1],C_{2},C_{3}\in[1,\infty)$. Let
$\delta\in(0,1/2]$, $\kappa:=(k_{1}+1)\cdots(k_{n}+1)$, and
$\varrho:=\sum_{i=1}^{n}k_{i}(k_{i}+2)$. Take

$\varepsilon:=(C_{1}\delta^{\kappa})^{2}(3C_{3})^{-2},\quad
J:=\lfloor(3C_{2}C_{1}^{-1}\varepsilon^{-\varrho-1}\delta^{-\kappa})^{2^{n}}\rfloor+1.$

Note that $J\lesssim\delta^{-\kappa 2^{n}(2\varrho+3)},$ which gives

$2^{-J}\geq 2^{-C\delta^{-\kappa
2^{n}(2\varrho+3)}}=e^{-C^{\prime}\delta^{-\kappa 2^{n}(2\varrho+3)}}$

for some constants $C$ and $C^{\prime}$.

By (2.2), there exists $1\leq j\leq J$ such that for each
$\lambda\in(2^{-j},2^{-j+1}]$,

$\big{|}\mathcal{N}^{\varepsilon}_{\lambda}(A)-\mathcal{N}^{1}_{\lambda}(A)\big{|}\leq
C_{2}J^{-2^{-n}}\varepsilon^{-\varrho-1}\leq\frac{C_{1}}{3}\delta^{\kappa}.$

Combined with (2.1) and (2.3), we obtain the desired estimate

$\mathcal{N}^{0}_{\lambda}(A)\geq\mathcal{N}^{1}_{\lambda}(A)-\big{|}\mathcal{N}^{\varepsilon}_{\lambda}(A)-\mathcal{N}^{1}_{\lambda}(A)\big{|}-\big{|}\mathcal{N}^{0}_{\lambda}(A)-\mathcal{N}^{\varepsilon}_{\lambda}(A)\big{|}$
$\geq
C_{1}\delta^{\kappa}-\frac{C_{1}}{3}\delta^{\kappa}-\frac{C_{1}}{3}\delta^{\kappa}=\frac{C_{1}}{3}\delta^{\kappa}>0.$

##  3\. The structured part: Proof of lemma 2

Let $\delta\in(0,1/2]$ and
$A\subseteq[0,1]^{k_{1}+1}\times\cdots\times[0,1]^{k_{n}+1}$ with
$|A|\geq\delta$. Let $m$ be the unique positive integer such that
$\lambda\in(2^{-m},2^{-m+1}]$. For each $1\leq i\leq n$, let
$\mathcal{Q}_{m}^{k_{i}}$ be a collection of all dyadic cubes in
$[0,1]^{k_{i}+1}$ with side length $2^{-m}$. We write
$\mathcal{Q}_{m}:=\mathcal{Q}_{m}^{k_{1}}\times\cdots\times\mathcal{Q}_{m}^{k_{n}}$.

We partition the domain of integration of each $x_{i}^{r_{i}}$ into congruent
cubes of side length $2^{-m}$ and restrict the domain of integration so that
for each $i$, all $x_{i}^{r_{i}}$, $1\leq r_{i}\leq k_{i}$, lie in the same
cube from $\mathcal{Q}_{m}$. This yields

$\mathcal{N}^{1}_{\lambda}(A)\geq
2^{-D}\sum_{(Q_{1},\ldots,Q_{n})\in\mathcal{Q}_{m}}\fint_{Q_{1}^{k_{1}+1}\times\cdots\times
Q_{n}^{k_{n}+1}}\Big{(}\prod_{h\in\mathcal{H}}\mathbbm{1}_{A}(\Pi_{h}x)\Big{)}\prod_{i=1}^{n}\sigma_{\Delta_{i},\lambda}*g_{\lambda}((x_{i}^{r}-x_{i}^{0})_{r=1}^{k_{i}})\,dx$

where $D=m(K+1)\cdot(K+1)$ and $Q^{k_{i}}_{i}$ denotes the $k_{i}$-fold
Cartesian product of $Q_{i}$.

Since the measure $\sigma_{\Delta_{i},\lambda}$ is supported in the ball
$B(0,\lambda C_{\Delta_{i}})$ in $\mathbb{R}^{k_{i}(k_{i}+1)}$ of radius
$\lambda C_{\Delta_{i}}$ centered at the origin, where $C_{\Delta_{i}}$ is the
diameter of $\Delta_{i}$, for each $1\leq i\leq n$ and $y\in B(0,\lambda R)$,

$\sigma_{\Delta_{i},\lambda}*g_{\lambda}(y)=\lambda^{-k_{i}(k_{i}+1)}\int_{\mathbb{R}^{k_{i}(k_{i}+1)}}e^{-\pi|\lambda^{-1}(y-u)|^{2}}d\sigma_{\Delta_{i},\lambda}(u)$
$\geq\lambda^{-k_{i}(k_{i}+1)}e^{-\pi(C_{\Delta_{i}}+R)^{2}}\int_{\mathbb{R}^{k_{i}(k_{i}+1)}}d\sigma_{\Delta_{i},\lambda}(u)=\lambda^{-k_{i}(k_{i}+1)}e^{-\pi(C_{\Delta_{i}}+R)^{2}}$

The last equality follows after integrating in $u^{k},u^{k-1}$, all the way to
$u^{1}$, where $u=(u^{1},\ldots,u^{k})$, using that the spherical measures are
normalized. Therefore, with $R^{2}=k_{i}(k_{i}+1)$ we obtain

$\sigma_{\Delta_{i},\lambda}\ast
g_{\lambda}\gtrsim\lambda^{-k_{i}(k_{i}+1)}\mathbbm{1}_{[-\lambda,\lambda]^{k_{i}(k_{i}+1)}}\gtrsim
2^{mk_{i}(k_{i}+1)}\mathbbm{1}_{[-2^{-m},2^{-m}]^{k_{i}(k_{i}+1)}}.$

Estimating $\sigma_{\Delta_{i},\lambda}*g_{\lambda}$ from below for each $i$
we obtain

$\mathcal{N}^{1}_{\lambda}(A)\gtrsim
2^{-m(k+n)}\sum_{(Q_{1},\ldots,Q_{n})\in\mathcal{Q}_{m}}\fint_{Q_{1}^{k_{1}+1}\times\cdots\times
Q_{n}^{k_{n}+1}}\Big{(}\prod_{h\in\mathcal{H}}\mathbbm{1}_{A}(\Pi_{h}x)\Big{)}$
$\times\prod_{i=1}^{n}\mathbbm{1}_{[-2^{-m},2^{-m}]^{k_{i}(k_{i}+1)}}((x_{i}^{r}-x_{i}^{0})_{r=1}^{k_{i}})\,dx.$

Since for $1\leq r_{i}\leq k_{i}$ both $x_{i}^{r_{i}}$ and $x_{i}^{0}$ belong
to the same cube $Q_{i}$, this gives

$\mathcal{N}^{1}_{\lambda}(A)\gtrsim
2^{-m(k+n)}\sum_{(Q_{1},\ldots,Q_{n})\in\mathcal{Q}_{m}}\fint_{Q_{1}^{k_{1}+1}\times\cdots\times
Q_{n}^{k_{n}+1}}\prod_{h\in\mathcal{H}}\mathbbm{1}_{A}(\Pi_{h}x)\,dx.$

To complete the proof of the estimate (2.1), we claim that

$\fint_{Q_{1}^{k_{1}+1}\times\cdots\times
Q_{n}^{k_{n}+1}}\prod_{h\in\mathcal{H}}\mathbbm{1}_{A}(\Pi_{h}x)\,dx\geq\bigg{(}\fint_{Q_{1}\times\cdots\times
Q_{n}}\mathbbm{1}_{A}\bigg{)}^{\kappa}$ (3.1)

where $\kappa=(k_{1}+1)\cdots(k_{n}+1)$. This claim then implies

$\mathcal{N}^{1}_{\lambda}(A)\gtrsim
2^{-m(k+n)}\sum_{(Q_{1},\ldots,Q_{n})\in\mathcal{Q}_{m}}\bigg{(}\fint_{Q_{1}\times\cdots\times
Q_{n}}\mathbbm{1}_{A}\bigg{)}^{\kappa}$
$\geq\bigg{(}2^{-m(k+n)}\sum_{(Q_{1},\ldots,Q_{n})\in\mathcal{Q}_{m}}\fint_{Q_{1}\times\cdots\times
Q_{n}}\mathbbm{1}_{A}\bigg{)}^{\kappa}=\bigg{(}\int_{[0,1]^{k+n}}\mathbbm{1}_{A}\bigg{)}^{\kappa}\geq\delta^{\kappa}.$

The second inequality follows by applying Jensen’s inequality for the
normalized counting measure over the set of cardinality $2^{m(k+n)}$. This
finishes the proof the lemma, up to verification of the bound (3.1).

The estimate (3.1) can be proven by induction on $n$. We rewrite

$\fint_{Q_{1}^{k_{1}+1}\times\cdots\times
Q_{n}^{k_{n}+1}}\prod_{h\in\mathcal{H}}\mathbbm{1}_{A}(\Pi_{h}x)\,dx=\fint_{Q_{1}^{k_{1}+1}\times\cdots\times
Q_{n}^{k_{n}+1}}\prod_{a=0}^{k_{n}}\prod_{h\in\mathcal{H}:h(n)=a}\mathbbm{1}_{A}(\Pi_{h}x)\,dx$
$=\fint_{Q_{1}^{k_{1}+1}\times\cdots\times
Q_{n-1}^{k_{n-1}+1}}\Big{(}\fint_{Q_{n}}\prod_{h\in\mathcal{H}:h(n)=0}\mathbbm{1}_{A}(\Pi_{h}x)dx_{n}^{0}\Big{)}^{k_{n}+1}dx^{\prime}$
(3.2)

where $x^{\prime}=(x^{0}_{1},\ldots,x^{k_{1}}_{1},x_{2}^{0},\ldots
x_{2}^{k_{2}},\ldots,x_{n-1}^{0},\ldots,x_{n-1}^{k_{n-1}})$. From here we see
that the claim holds for $n=1$ since in this case, (3.1) is an equality. For
the induction step we use Fubini and Jensen’s inequality for the integral over
$x^{\prime}$, which bounds (3.2) from below by

$\Big{(}\fint_{Q_{n}}\fint_{Q_{1}^{k_{1}+1}\times\cdots\times
Q_{n-1}^{k_{n-1}+1}}\prod_{h\in\mathcal{H}:h(n)=0}\mathbbm{1}_{A}(\Pi_{h}x)\,dx^{\prime}\,dx_{n}^{0}\Big{)}^{k_{n}+1}.$

By the induction hypothesis applied to the set

$\\{y\in\mathbb{R}^{k+n-k_{n}-1}:(y,x_{n}^{0})\in A\\}$

for a fixed $x_{n}^{0}$, we estimate the last expression from below by

$\bigg{(}\fint_{Q_{n}}\bigg{(}\fint_{Q_{1}\times\cdots\times
Q_{n-1}}\mathbbm{1}_{A}(y)dy\bigg{)}^{(k_{1}+1)\cdots(k_{n-1}+1)}dx_{n}^{0}\bigg{)}^{(k_{n}+1)}.$

Another application of Jensen’s inequality, this time for the integral over
$x_{n}^{0}$, gives (3.1). This completes the proof of Lemma 3.

##  4\. The error part: Proof of lemma 3

The key ingredient in the proof of bounds for the error part are cancellation
estimates for forms similar to multilinear singular integrals associates with
$n$-regular $n$-partite hypergraphs. These are the content of Lemma 5.

For $1\leq v\leq n$, parameters $\lambda,a,b>0$, and
$A\subseteq[0,1]^{k_{1}+1}\times\cdots\times[0,1]^{k_{n}+1}$ we introduce

$N(v,\lambda,a,b,A):=\int_{a}^{b}\int_{\mathbb{R}^{(K+1)\cdot(K+1)}}\Big{(}\prod_{h\in\mathcal{H}}\mathbbm{1}_{A}(\Pi_{h}x)\Big{)}$
$\times{\sigma}_{\Delta_{v},\lambda}\ast{(\Delta
g)}_{t\lambda}((x_{v}^{r}-x_{v}^{0})_{r=1}^{k_{v}})\prod_{i=1,i\neq
v}^{n}{\sigma}_{\Delta_{i},\lambda}\ast{g}_{t\lambda}((x_{i}^{r}-x_{i}^{0})_{r=1}^{k_{i}})\,dx\,\frac{dt}{t}.$
(4.1)

Here, $\Delta g$ is the Laplacian of $g$ in $\mathbb{R}^{{k_{v}}(k_{v}+1)}$
dimensions. If we further write

$\Delta g((x_{v}^{r}-x_{v}^{0})_{r=1}^{k_{v}})=\sum_{w=1}^{k_{v}}\Delta
g(x_{v}^{w}-x_{v}^{0})\,g((x_{v}^{r}-x_{v}^{0})_{r=1,r\neq w}^{k_{v}}),$ (4.2)

and expand out the convolutions, we obtain

$N(v,\lambda,a,b,A)=\sum_{w=1}^{k_{v}}\int_{a}^{b}\int_{\mathbb{R}^{(K+1)\cdot(K+1)+k_{v}(k_{v}+1)}}\Big{(}\prod_{h\in\mathcal{H}}\mathbbm{1}_{A}(\Pi_{h}x)\Big{)}$
$\times(\Delta
g)_{t\lambda}(x_{v}^{w}-x_{v}^{0}-p_{v}^{w})\,g_{t\lambda}((x_{v}^{r}-x_{v}^{0}-p_{v}^{r})_{r=1,r\neq
w}^{k_{v}})$
$\times\Big{(}\prod_{r=1}^{k_{v}}d\sigma_{\lambda}^{\Delta_{v},p_{v}^{1},\ldots
p_{v}^{k_{v}-r}}(p_{v}^{k_{v}+1-r})\Big{)}\prod_{i=1,i\neq
v}^{n}{\sigma}_{\Delta_{i},\lambda}\ast{g}_{t\lambda}((x_{i}^{r}-x_{i}^{0})_{r=1}^{k_{i}})\,dx\,\frac{dt}{t}.$
(4.3)

Using the fundamental theorem of calculus in $t$, the Leibniz rule and the
heat equation

$(\Delta g)_{t\lambda}=2\pi t\partial_{t}(g_{t\lambda})$ (4.4)

we thus obtain the identity

$\sum_{j=1}^{J}(\mathcal{N}^{1}_{\lambda_{j}}(A)-\mathcal{N}^{\varepsilon}_{\lambda_{j}}(A))=\frac{1}{2\pi}\sum_{v=1}^{n}\sum_{j=1}^{J}N(v,\lambda_{j},\varepsilon,1,A).$

Due to symmetry with respect to the simplices and the definitions of the
measures $\sigma_{\Delta_{i},\lambda}$, it suffices to prove the corresponding
bound from Lemma 2.2 for $v=1$ and $w=k_{1}$ in (4.3).

Now, for $j\in\mathbb{N}$, $t\in(0,\infty)$ and $s\in[2^{-j-5}t,2^{-j-4}t]$
let

$r_{j}(s,t):=\sqrt{t^{2}\lambda_{j}^{2}-s^{2}},\quad
c_{j}(s,t):=\frac{t^{2}\lambda_{j}^{2}}{sr_{j}(s,t)}.$

We can see that $s\sim 2^{-j}t\sim t\lambda_{j}\sim r_{j}(s,t)$ and thus
$c_{j}(s,t)\sim 1$. Also, for a $d$-dimensional Gaussian $g$ we have

$(\Delta
g)_{t\lambda_{j}}=c_{j}(s,t)\sum_{m=1}^{d}(\partial_{m}g)_{r_{j}(s,t)}\ast(\partial_{m}g)_{s}.$
(4.5)

We will also split and estimate

$\prod_{h\in\mathcal{H}}\mathbbm{1}_{A}(\Pi_{h}x)=\prod_{a=0}^{k_{1}}\prod_{h\in\mathcal{H}:h(1)=a}\mathbbm{1}_{A}(\Pi_{h}x)\leq\prod_{h\in\mathcal{H}:h(1)=k_{1}}\mathbbm{1}_{A}(\Pi_{h}x)$
(4.6)

We start by multiplying the integrand (4.3) by

$(\log 2)^{-1}\int_{2^{-j-5}}^{2^{-j-4}}\frac{ds}{s}=1.$

Applying the triangle inequality and using (4.5) and (4.6), we estimate
$\sum_{j=1}^{J}|N(1,\lambda_{j},\varepsilon,1,A)|$ up to an absolute constant
by

$\sum_{j=1}^{J}\sum_{m=1}^{k_{1}+1}\int_{\varepsilon}^{1}\int_{2^{-j-5}t}^{2^{-j-4}t}\int_{\mathbb{R}^{D}}\mathbbm{1}_{[0,1]^{k+n}}(x_{1}^{0},\ldots,x_{n}^{0})\,\Big{|}\int_{\mathbb{R}^{k_{1}+1}}\Big{(}\prod_{h\in\mathcal{H}:h(1)=k_{1}}\mathbbm{1}_{A}(\Pi_{h}x)\Big{)}$
$(\partial_{m}g)_{s}(x_{1}^{k_{1}}-q)\,dx_{1}^{k_{1}}\Big{|}\times(\sigma_{\lambda_{j}}^{\Delta_{1},p_{1}^{1},\ldots,p_{1}^{k_{1}-1}}\ast|\partial_{m}g|_{r_{j}(s,t)})(x_{1}^{0}-q)\,{g}_{t\lambda_{j}}((x_{1}^{r}-x_{1}^{0}-p_{1}^{r})_{r=1}^{k_{1}-1})$
$\times\Big{(}\prod_{r=1}^{k_{v}}d\sigma_{\lambda}^{\Delta_{v},p_{v}^{1},\ldots
p_{v}^{k_{v}-r}}(p_{v}^{k_{v}+1-r})\Big{)}\,d((x_{1}^{r})_{r=0}^{k_{1}-1})\,dq$
$\times\Big{(}\prod_{i=2}^{n}\sigma_{\Delta_{i},\lambda_{j}}*{g}_{t\lambda_{j}}((x_{i}^{r}-x_{i}^{0})_{r=1}^{k_{i}})d((x_{i}^{r})_{r=0}^{k_{i}})\Big{)}\,\frac{ds}{s}\,\frac{dt}{t}$
(4.7)

where now $D=(K+1)\cdot(K+1)+(k_{1}-1)(k_{1}+1)$.

If $\mu$ is a probability measure on $\mathbb{R}^{d}$, $d\geq 2$, whose
support is contained in a sphere in $\mathbb{R}^{d}$ and whose radius depends
only on the simplices $\Delta_{i}$, and $\phi$ a Schwartz function on
$\mathbb{R}^{d}$, for $t\in[\varepsilon,1]$ we can estimate

$(\mu_{t^{-1}}\ast\phi)(y)\lesssim\int_{\mathbb{R}^{d}}(1+|y-ut^{-1}|)^{-d-1}d\mu(u)\lesssim\varepsilon^{-d-1}(1+|y|)^{-d-1}$
(4.8)

with the implicit constant depending on $\phi$. The second inequality follows
from

$1+|y|\leq 1+|y-ut^{-1}|+|u|t^{-1}\lesssim\varepsilon^{-1}(1+|y-ut^{-1}|).$

We can rescale inequality (4.8) by $t\lambda_{j}s^{-1}\sim 1$ in order to get

$(\mu_{\lambda_{j}s^{-1}}\ast\phi_{t\lambda_{j}s^{-1}})(y)\lesssim\varepsilon^{-d-1}(1+|y|)^{-d-1}.$

Analogously we obtain

$(\mu_{\lambda_{j}s^{-1}}\ast\phi_{r_{j}(s,t)s^{-1}})(y)\lesssim\varepsilon^{-d-1}(1+|y|)^{-d-1}.$

Dominating the right-hand side by a superposition of Gaussians in
$\mathbb{R}^{d}$

$\int_{1}^{\infty}g_{\beta}(y)\,\frac{d\beta}{\beta^{2}}\sim(1+|y|)^{-d-1}$

and rescaling by $s$ gives

$(\mu_{\lambda_{j}}\ast\phi_{t\lambda_{j}})(y)\lesssim\varepsilon^{-d-1}\int_{1}^{\infty}{g}_{\beta
s}(y)\,\frac{d\beta}{\beta^{2}},\quad(\mu_{\lambda_{j}}\ast\phi_{r_{j}(s,t)})(y)\lesssim\varepsilon^{-d-1}\int_{1}^{\infty}{g}_{\beta
s}(y)\,\frac{d\beta}{\beta^{2}}.$ (4.9)

Now we apply the second inequality in (4.9) to the convolution

$\sigma_{\lambda_{j}}^{\Delta_{1},p_{1}^{1},\ldots,p_{1}^{k_{1}-1}}\ast|\partial_{m}g|_{r_{j}(s,t)}(x_{1}^{0}-q)$

in the expression (4.7). After that we can recognize another convolution, this
time after integrating in the variable $p_{1}^{k_{1}-1}$, i.e.

$\sigma_{\lambda_{j}}^{\Delta_{1},p_{1}^{1},\ldots
p_{1}^{k_{1}-2}}*g_{t\lambda_{j}}(x_{1}^{k_{1}-1}-x_{1}^{0}).$

We apply the first inequality in (4.9) to this convolution. We continue the
same procedure in the variables $p_{1}^{k_{1}-2},\ldots,p_{1}^{1}$ until we
have estimated all convolutions in these variables one by one. Then we also
use (4.9) on the remaining convolutions
$\sigma_{\Delta_{i},\lambda_{j}}*g_{t\lambda_{j}}$. If we denote
$\beta=(\beta_{1},\ldots,\beta_{n})$ and
$\beta_{i}=(\beta_{i}^{0},\beta_{i}^{1},\ldots,\beta_{i}^{k_{i}})\in(1,\infty)^{k_{i}+1}$
for $1\leq i\leq n$, we obtain an estimate for (4.7) by

$\varepsilon^{-\varrho}\,\sum_{j=1}^{J}\sum_{m=1}^{k_{1}+1}\int_{(1,\infty)^{k+n}}\int_{\varepsilon}^{1}\int_{2^{-j-5}t}^{2^{-j-4}t}\int_{\mathbb{R}^{(K+1)\cdot(K+1)}}\mathbbm{1}_{[0,1]^{k+n}}(x_{1}^{0},\ldots,x_{n}^{0})$
$\times\Big{|}\int_{\mathbb{R}^{k_{1}+1}}\Big{(}\prod_{h\in\mathcal{H}:h(1)=k_{1}}\mathbbm{1}_{A}(\Pi_{h}x)\Big{)}(\partial_{m}g)_{s}(x_{1}^{k_{1}}-q)dx_{1}^{k_{1}}\Big{|}\,d\mu^{(1,k_{1})}_{K,g,\beta}\,\frac{ds}{s}\,\frac{dt}{t}\,\mathcal{D}(\beta)^{-2}\,d\beta$
(4.10)

where $\varrho$ is as in Lemma 3 and $\mathcal{D}(\beta)$ is just the product
of all $\beta_{i}^{r}$ over $1\leq i\leq n$ and $0\leq r\leq k_{i}$. Here, for
$L=(l_{1},\ldots,l_{n})$, a tuple $\alpha=(\alpha_{1},\ldots,\alpha_{n})$,
$\alpha_{i}=(\alpha_{i}^{0},\alpha_{i}^{1},\ldots,\alpha_{i}^{l_{i}})\in(1,\infty)^{l_{i}+1}$,
a Schwartz function $\varphi$, $1\leq v\leq n$, and $1\leq w\leq l_{v}$ we set

$d{\mu}^{(v,w)}_{L,\varphi,\alpha}=\varphi_{s\alpha_{v}^{0}}(x_{v}^{0}-q)\Big{(}\prod_{r=1,r\neq
w}^{l_{v}}g_{s\alpha_{v}^{r}}(x_{v}^{r}-x_{v}^{0})dx_{v}^{r}\Big{)}\Big{(}\prod_{i=1,i\neq
v}^{n}g_{s\alpha_{i}}((x_{i}^{r}-x_{i}^{0})_{r=1}^{l_{i}})dx_{i}\Big{)}\,dq,$

and we set $x_{i}=(x_{i}^{0},\ldots,x_{i}^{l_{i}})$. Note that the integrand
in (4.10) is constant in $\beta_{1}^{k_{1}}$ and $\beta_{i}^{0}$, $i\neq v$,
but we added an integral in these variables that equals to one for uniform
notation.

Now, for $1\leq v\leq n$, $1\leq m\leq k_{v}+1$, any tuple
$\alpha=(\alpha_{1},\ldots,\alpha_{n})$ as above, any real numbers $a,b$ with
$b/a\geq 2$, any $L=(l_{1},\ldots,l_{n})$ with $1\leq l_{i}\leq k_{i}$, and
$1\leq w\leq l_{v}$, we define

$\widetilde{\Theta}^{(v,w,m)}_{L,\varphi,a,b,\alpha}(A):=\int_{a}^{b}\int_{\mathbb{R}^{(K+1)\cdot(L+1)}}\mathbbm{1}_{[0,1]^{k+n}}(x_{1}^{0},\ldots,x_{n}^{0})$
$\times\Big{|}\int_{\mathbb{R}^{k_{v}+1}}\Big{(}\prod_{h\in\mathcal{H}_{L}:h(v)=w}\mathbbm{1}_{A}(\Pi_{h}x)\Big{)}(\partial_{m}g)_{s\alpha_{v}^{w}}(x_{v}^{w}-q)dx_{v}^{w}\Big{|}\,d\mu^{(v,w)}_{L,\varphi,\alpha}\,\frac{ds}{s}.$
(4.11)

Then, after summing in $j$, (4.10) can be recognized as

$\varepsilon^{-\varrho}\,\sum_{m=1}^{k_{1}+1}\int_{(1,\infty)^{k+n}}\int_{\varepsilon}^{1}\widetilde{\Theta}_{K,g,2^{-J-5}t,2^{-5}t,\widetilde{\beta}}^{(1,k_{1},m)}(A)\,\frac{dt}{t}\,\mathcal{D}(\beta)^{-2}\,d\beta$
(4.12)

where $\widetilde{\beta}_{1}^{k_{1}}=1$, $\widetilde{\beta}_{i}^{0}=1$ for all
$i\neq v$, and $\widetilde{\beta}$ agrees with $\beta$ in all other
coordinates. Thus it suffices to bound the form inside the inner integral for
each fixed $t$ and $\beta_{i}^{k_{i}}$.

In order to do that, for the same range of parameters $v,L,a,b,\alpha$ as in
the definition of (4.11) we define yet another form

$\Theta^{(v)}_{L,a,b,\alpha}(A):=-\int_{a}^{b}\int_{\mathbb{R}^{(K+1)\cdot(L+1)}}\Big{(}\prod_{h\in\mathcal{H}_{L}}\mathbbm{1}_{A}(\Pi_{h}x)\Big{)}$
$\times(\Delta
g)_{s\alpha_{v}}((x_{v}^{r}-x_{v}^{0})_{r=1}^{l_{v}})\prod_{i=1,i\neq
v}^{n}g_{s\alpha_{i}}((x_{i}^{r}-x_{i}^{0})_{r=1}^{l_{i}})\,dx\,\frac{ds}{s}.$
(4.13)

Note that as opposed to (4.10), this form is not necessarily non-negative. The
bound for (4.10) will follow from the following crucial lemma, which
establishes a cancellation estimate for both (4.11) and (4.13). In the proof
we inductively simplify the kernels of the singular forms using repeated
applications of Cauchy-Schwarz, partial integration, and positivity arguments.

######  Lemma 5.

For any $L,\varphi,a,b,\alpha,v,w,m$ as above,

$|\Theta^{(v)}_{L,a,b,\alpha}(A)|,\widetilde{\Theta}^{(v,w,m)}_{L,\varphi,a,b,\alpha}(A)\lesssim(\log(b/a))^{1-2^{-n}}.$
(4.14)

Applying Lemma 5 specified to
$(L,\varphi,a,b,\alpha,v,w,m)=(K,g,2^{-J-5}t,2^{-5}t,\widetilde{\beta},1,k_{1},m)$
gives the desired estimate for the form (4.10). Indeed, we obtain a bound for
(4.12) by

$J^{1-2^{-n}}\varepsilon^{-\varrho}\sum_{m=1}^{k_{1}+1}\int_{(1,\infty)^{k+n}}\int_{\varepsilon}^{1}\,\frac{dt}{t}\,\mathcal{D}(\beta)^{-2}\,d\beta$

In the end it remains to integrate in $t$ and $\beta$ and sum in $m$, which
yields the conclusion of Lemma 3.

###### Proof of Lemma 5.

This lemma will be proven by downward induction on a parameter $0\leq\tau\leq
n$. Let $P(\tau)$ be the statement that the estimates

$|\Theta^{(v)}_{L,a,b,\alpha}(A)|,\widetilde{\Theta}^{(v,w,m)}_{L,\varphi,a,b,\alpha}(A)\lesssim(\log(b/a))^{1-2^{-n+\tau}}$

hold for all tuples $L$ satisfying $l_{i}=1$ and
$\alpha_{i}^{0}=\alpha_{i}^{1}$ for $1\leq i\leq\tau$. The estimate on (4.14)
will follow from $P(0)$. The base case is $P(n)$. Pick $0\leq\tau\leq n$. We
may inductively assume that $P(\tau^{\prime})$ holds for all
$\tau<\tau^{\prime}\leq n$.

Let $L$ be a tuple satisfying $l_{i}=1$ for $1\leq i\leq\tau$. Let $1\leq
v\leq n$. First we reduce the bound for (4.13) to the bound for (4.11).
Applying the convolution identity

$-\frac{1}{2}(\Delta
g)_{\sqrt{2}s}(x_{v}^{w}-x_{v}^{0})=\sum_{m=1}^{k_{v}+1}\int_{\mathbb{R}^{k_{v}+1}}(\partial_{m}g)_{s}(x_{v}^{0}-q)(\partial_{m}g)_{s}(x_{v}^{w}-q)\,dq,$
(4.15)

rewriting the integral in $x_{v}^{w}$ as the innermost, localizing
$(x_{1}^{0},\ldots,x_{n}^{0})$ to the cube $[0,1]^{k+n}$, estimating all
characteristic functions $\mathbbm{1}_{A}$ which do not depend on the variable
$x_{v}^{w}$ by $1$ and using the triangle inequality gives

$|\Theta^{(v)}_{L,a,b,\alpha}(A)|\leq\sum_{w=1}^{l_{v}}\sum_{m=1}^{k_{v}+1}\widetilde{\Theta}^{(v,w,m)}_{L,|\partial_{m}g|,a,b,\widetilde{\alpha}}(A),$
(4.16)

where
$\widetilde{\alpha}_{v}^{0}=\widetilde{\alpha}_{v}^{w}=2^{-\frac{1}{2}}\alpha_{v}^{w}$
and $\widetilde{\alpha}$ agrees with $\alpha$ in all other coordinates. Thus,
it suffices to bound

$\sum_{m=1}^{k_{v}+1}\widetilde{\Theta}^{(v,w,m)}_{L,\varphi,a,b,\alpha}(A)$
(4.17)

for each fixed $w$ and any $L,\varphi,a,b,\alpha$ as above.

We apply the Cauchy-Schwarz inequality in all variables but $x_{v}^{w}$ and
the summation in $m$, which estimates (4.17) by the geometric mean of

$\sum_{m=1}^{k_{v}+1}\int_{a}^{b}\int_{\mathbb{R}^{(K+1)\cdot(L+1)}}\Big{|}\int_{\mathbb{R}^{k_{v}+1}}\Big{(}\prod_{h\in\mathcal{H}_{L}:h(v)=w}\mathbbm{1}_{A}(\Pi_{h}x)\Big{)}(\partial_{m}g)_{s\alpha_{v}^{w}}(x_{v}^{w}-q)dx_{v}^{w}\Big{|}^{2}\,d{\mu}^{(v,w)}_{L,\varphi,\alpha}\,\frac{ds}{s}$
(4.18)

and

$\sum_{m=1}^{k_{v}+1}\int_{a}^{b}\int_{\mathbb{R}^{(K+1)\cdot(L+1)}}\mathbbm{1}_{[0,1]^{k+n}}(x_{1}^{0},\ldots,x_{n}^{0})\,d{\mu}^{(v,w)}_{L,\varphi,\alpha}\,\frac{ds}{s}.$
(4.19)

Let us focus on the second term (4.19), which we estimate trivially. Expanding
out the measure, we first integrate in $q$. Then for each $1\leq i\leq n$, we
integrate in all $x_{i}^{r}$ for $r\neq 0$ that appear in $\mu$ and finally in
$x_{1}^{0},\ldots,x_{n}^{0}$. In the end we integrate in $s$ and sum in $m$,
which yields a bound for (4.19) by an absolute constant times

$\log(b/a).$ (4.20)

It remains to estimate (4.18). We integrate in $x_{v}^{r}$ for $r\neq w$ and
finally in $x_{v}^{0}$. Expanding out also the square of the integrand we see
that (4.18) is up to a constant equal to

$\sum_{m=1}^{k_{v}+1}\int_{a}^{b}\int_{\mathbb{R}^{(K+1)\cdot(L+1)-l_{v}(k_{v}+1)}}\Big{(}\int_{\mathbb{R}^{k_{v}+1}}\Big{(}\prod_{h\in\mathcal{H}_{L}:h(v)=0}\mathbbm{1}_{A}(\Pi_{h}x)\Big{)}(\partial_{m}g)_{s\alpha_{v}^{w}}(x_{v}^{0}-q)\,dx_{v}^{0}\Big{)}$
$\times\Big{(}\int_{\mathbb{R}^{k_{v}+1}}\Big{(}\prod_{h\in\mathcal{H}_{L}:h(v)=1}\mathbbm{1}_{A}(\Pi_{h}x)\Big{)}(\partial_{m}g)_{s\alpha_{v}^{w}}(x_{v}^{1}-q)\,dx_{v}^{1}\Big{)}$
$\times\Big{(}\prod_{i=1,i\neq
v}^{n}g_{s\alpha_{i}}((x_{i}^{r}-x_{i}^{0})_{r=1}^{l_{i}})\,dx_{i}\Big{)}\,dq\,\frac{ds}{s}.$

Here we have also relabelled $x_{v}^{w}$ to $x_{v}^{0}$ and denoted by
$x_{v}^{1}$ its copy after the square expansion.

We can rewrite

$\Big{(}\prod_{h\in\mathcal{H}_{L}:h(v)=0}\mathbbm{1}_{A}(\Pi_{h}x)\Big{)}\Big{(}\prod_{h\in\mathcal{H}_{L}:h(v)=1}\mathbbm{1}_{A}(\Pi_{h}x)\Big{)}=\prod_{h\in\mathcal{H}_{\widetilde{L}}}\mathbbm{1}_{A}(\Pi_{h}x),$

where $\widetilde{L}$ agrees with $L$ in all coordinates except possibly at
$v$-th, where $\widetilde{l}_{v}=1$. Using (4.15) again we recognize this as a
positive constant multiple of

$\Theta^{(z)}_{\widetilde{L},a,b,\widetilde{\alpha}}$ (4.21)

specified to $z=v$, where $\widetilde{\alpha}$ agrees with $\alpha$ in all
slots except possibly at $v$-th, where
$\widetilde{\alpha}_{v}=(\alpha_{v}^{w},\alpha_{v}^{w})$.

We distinguish two cases. If $v>\tau$, then by symmetry we may assume
$v=\tau+1$. The induction hypothesis, i.e. the statement $P(\tau+1)$, gives

$\Theta^{(v)}_{\widetilde{L},a,b,\widetilde{\alpha}}\lesssim(\log(b/a))^{1-2^{-n+\tau+1}}.$

Taking the geometric mean of this bound and (4.20) yields the desired estimate
for (4.16).

It remains to consider the case $v\leq\tau$, which will also prove the base
case. If $v\leq\tau$, then $l_{v}=1$. First we observe that for any $1\leq
z\leq n$ and any $L,a,b,\alpha$ as in (4.13), the form
$\Theta^{(z)}_{{L},a,b,{\alpha}}$ is non-negative if $l_{z}=1$ and
$\alpha_{z}^{0}=\alpha_{z}^{1}$. Indeed, using (4.15) we can write it up to a
positive constant as

$\sum_{m=1}^{k_{z}+1}\int_{a}^{b}\int_{\mathbb{R}^{(K+1)\cdot(L+1)}}\Big{(}\int_{\mathbb{R}^{k_{z}+1}}\Big{(}\prod_{h\in\mathcal{H}_{L}:h(z)=0}\mathbbm{1}_{A}(\Pi_{h}x)\Big{)}(\partial_{m}g)_{s\alpha_{z}^{1}}(x_{r}^{0}-q)\,dx_{r}^{0}\Big{)}^{2}$
$\times\Big{(}\prod_{i=1,i\neq
z}^{n}g_{s\alpha_{i}}((x_{i}^{r}-x_{i}^{0})_{r=1}^{l_{i}})\,dx_{i}\Big{)}\,dq\,\frac{ds}{s}.$

Furthermore, by the fundamental theorem of calculus in $s$, the Leibniz rule,
the heat equation (4.4), and the identity (4.2), we have for any
$L,a,b,\alpha$ as above

$\sum_{z=1}^{n}\Theta^{(z)}_{{L},a,b,{\alpha}}=2\pi\big{(}\Xi_{L,a,\alpha}-\Xi_{L,b,\alpha}\big{)},$
(4.22)

where for any $L,a,\alpha$ as above we have denoted

$\Xi_{L,a,\alpha}(A):=\int_{\mathbb{R}^{(K+1)\cdot(L+1)}}\Big{(}\prod_{h\in\mathcal{H}_{L}}\mathbbm{1}_{A}(\Pi_{h}x)\Big{)}\prod_{i=1}^{n}g_{a\alpha_{i}}((x_{i}^{r}-x_{i}^{0})_{r=1}^{l_{i}})\,dx.$

We also have the estimate

$\Xi_{L,a,\alpha}(A)\lesssim 1,$ (4.23)

can be seen by localizing $(x_{1}^{0},\ldots,x_{n}^{0})$ to the cube
$[0,1]^{k+n}$, bounding all functions $\mathbbm{1}_{A}$ by $1$, integrating
first in $x_{i}^{r}$ for all $r\neq 0$ and finally in $x_{i}^{0}$.

Recall that to finish the proof of the lemma it remains to estimate (4.21) for
$z=v$ where $v\leq\tau$. By non-negativity of (4.21) for each $1\leq
z\leq\tau$, it then suffices to estimate

$\sum_{z=1}^{\tau}\Theta^{(z)}_{\widetilde{L},a,b,\widetilde{\alpha}}.$

In order to bound this sum for $1\leq\tau\leq n$, by (4.22) and (4.23) applied
to $\widetilde{L},a,b,\widetilde{\alpha}$ it suffices to estimate the form
(4.21) for each $z>\tau$. However this has already been achieved. ∎

##  5\. The uniform part: Proof of lemma 4

In this section we proceed similarly as in [13] and [10], with the necessary
modifications arising from the multilinear setting.

We again use the fundamental theorem of calculus, the Leibniz rule and the
heat equation (4.4), which yields

$\mathcal{N}^{\varepsilon}_{\lambda}(A)-\mathcal{N}^{\vartheta}_{\lambda}(A)=\frac{1}{2\pi}\sum_{v=1}^{n}N(v,\lambda,\vartheta,\varepsilon,A),$
(5.1)

where $N(v,\lambda,\vartheta,\varepsilon,A)$ has been defined in (4.1). We
write out each $N(v,\lambda,\vartheta,\varepsilon,A)$ as in (4.3). By symmetry
it again suffices to consider the case $v=1$ and the summand for $w=k_{1}$.

In the summand for $w=k_{1}$ in $N(1,\lambda,\vartheta,\varepsilon,A)$ we
consider the integrand in $p_{1}^{1},\ldots,p_{1}^{k_{1}-1}$ and
$x_{1}=(x_{1}^{0},\ldots,x_{1}^{k_{1}}),$ i.e.

$\int_{\mathbb{R}^{2k_{1}(k_{1}+1)}}\Big{(}\prod_{h\in\mathcal{H}}\mathbbm{1}_{A}(\Pi_{h}x)\Big{)}\sigma_{\lambda}^{\Delta_{1},p_{1}^{1},\ldots,p_{1}^{k_{1}-1}}*(\Delta
g)_{t\lambda}(x_{1}^{k_{1}}-x_{1}^{0})\,g_{t\lambda}\big{(}(x_{1}^{r}-x_{1}^{0}-p_{1}^{r})_{r=1}^{k_{1}-1}\big{)}\,d\nu\,dx_{1}$
(5.2)

where

$d\nu=d\sigma_{\lambda}^{\Delta_{1},p_{1}^{1},\ldots
p_{1}^{k_{1}-2}}(p_{1}^{k_{1}-1})\cdots
d\sigma_{\lambda}^{\Delta_{1}}(p_{1}^{1}).$

Integrating in $x_{1}^{k_{1}}$ we rewrite it as

$\int_{\mathbb{R}^{(2k_{1}-1)(k_{1}+1)}}\Big{(}\prod_{h\in\mathcal{H}:h(1)\neq
k_{1}}\mathbbm{1}_{A}(\Pi_{h}x)\Big{)}\mathcal{F}\ast{\sigma}_{\lambda}^{\Delta_{1},p_{1}^{1},\ldots,p_{1}^{k_{1}-1}}\ast(\Delta
g)_{t\lambda}(x_{1}^{0})$ $\times
g_{t\lambda}\big{(}(x_{1}^{r}-x_{1}^{0}-p_{1}^{r})_{r=1}^{k_{1}-1}\big{)}\,d\nu\,dx_{1}^{0}\cdots
dx_{1}^{k_{1}-1},$ (5.3)

where $\mathcal{F}$ denotes the function

$x_{1}^{k_{1}}\mapsto\prod_{h\in\mathcal{H}:h(1)=k_{1}}\mathbbm{1}_{A}(\Pi_{h}x).$

By Jensen’s inequality applied to the probabilistic measure

$g_{t\lambda}\big{(}(x_{1}^{r}-x_{1}^{0}-p_{1}^{r})_{r=1}^{k_{1}-1}\big{)}\mathbbm{1}_{[0,1]^{k_{1}+1}}(x_{1}^{0})\,d\nu\,dx_{1}^{0}\cdots
dx_{1}^{k_{1}-1}$

for fixed $p_{1}^{r}$, the square of (5.3) is bounded by

$\int_{\mathbb{R}^{(2k_{1}-1)(k_{1}+1)}}\Big{(}\prod_{h\in\mathcal{H}:h(1)\neq
k_{1}}\mathbbm{1}_{A}(\Pi_{h}x)\Big{)}|\mathcal{F}\ast{\sigma}_{\lambda}^{\Delta_{1},p_{1}^{1},\ldots,p_{1}^{k_{1}-1}}\ast(\Delta
g)_{t\lambda}(x_{1}^{0})|^{2}$ $\times
g_{t\lambda}\big{(}(x_{1}^{r}-x_{1}^{0}-p_{1}^{r})_{r=1}^{k_{1}-1}\big{)}\,d\mu\,dx_{1}^{0}\cdots
dx_{1}^{k_{1}-1}.$

Bounding all characteristic functions that do not make up $\mathcal{F}$ by one
and integrating in $x_{1}^{1},\ldots,x_{1}^{k_{1}-1}$, we estimate the last
display by

$\int_{\mathbb{R}^{k_{1}(k_{1}+1)}}|\mathcal{F}\ast{\sigma}_{\lambda}^{\Delta_{1},p_{1}^{1},\ldots,p_{1}^{k_{1}-1}}\ast(\Delta
g)_{t\lambda}(x_{1}^{0})|^{2}\,dx_{1}^{0}\,d\nu.$

By Plancherel’s theorem in $x_{1}^{0}$, this equals

$\int_{\mathbb{R}^{k_{1}(k_{1}+1)}}|\widehat{\mathcal{F}}(\xi)|^{2}|\widehat{{\sigma}}_{\lambda}^{\Delta_{1},p_{1}^{1},\ldots,p_{1}^{k_{1}-1}}(\xi)|^{2}|\widehat{(\Delta
g)_{t\lambda}}(\xi)|^{2}\,d\xi\,d\nu.$ (5.4)

We will use the following estimate for the spherical measure that can be found
in [1] and [13]:

$|\widehat{{\sigma}}^{\Delta_{1},p_{1}^{1},\ldots,p_{1}^{k_{1}-1}}(\xi)|\lesssim\textnormal{dist}(\xi,\textnormal{span}\\{p_{1}^{1},\ldots,p_{1}^{k_{1}-1}\\})^{-\frac{1}{2}}$

for each $\xi\in\mathbb{R}^{k_{1}+1}$. Together with

$\widehat{\Delta g}(\xi)=-4\pi^{2}|\xi|^{2}e^{-\pi|\xi|^{2}},$

this allows us to estimate (5.4) by

$\int_{\mathbb{R}^{k_{1}(k_{1}+1)}}\
|\widehat{\mathcal{F}}(\xi)|^{2}\textnormal{dist}(\lambda\xi,\textnormal{span}\\{p_{1}^{1},\ldots,p_{1}^{k_{1}-1}\\})^{-1}|t\lambda\xi|^{4}e^{-2\pi|t\lambda\xi|^{2}}\,d\xi\,d\nu.$
(5.5)

Next, we have

$\int_{\mathbb{R}^{(k_{1}+1)(k_{1}-1)}}\textnormal{dist}(\lambda\xi,\textnormal{span}\\{p_{1}^{1},\ldots,p_{1}^{k_{1}-1}\\})^{-1}\,d\nu\lesssim|\lambda\xi|^{-1}$

This estimate is proven in Section 3.3 of [10]. Applying this to (5.5) yields
a bound by

$\int_{\mathbb{R}^{k_{1}+1}}|\widehat{\mathcal{F}}(\xi)|^{2}|\lambda\xi|^{-1}|t\lambda\xi|^{4}e^{-2\pi|t\lambda\xi|^{2}}\,d\xi\lesssim
t\int_{\mathbb{R}^{k_{1}+1}}|\widehat{\mathcal{F}}(\xi)|^{2}\,d\xi\leq t,$

For the second inequality we used that $y\mapsto y^{3}e^{-2\pi y^{2}}$ is
uniformly bounded, while for the last inequality we applied Plancherel one
more time. This gives an estimate for (5.2) by $t^{\frac{1}{2}}.$ Finally,
applying this estimate first to (5.2) and then integrating in (5.1) in the
remaining variables, we see that

$|\mathcal{N}^{\vartheta}_{\lambda}(A)-\mathcal{N}^{\varepsilon}_{\lambda}(A)|\lesssim\int_{\vartheta}^{\varepsilon}\int_{\mathbb{R}^{k+n-k_{1}-1}}t^{\frac{1}{2}}\mathbbm{1}_{[0,1]^{k+n-k_{1}-1}}(x_{2}^{0},\ldots,x_{n}^{0})\,dx_{2}^{0}\cdots
dx_{n}^{0}\frac{dt}{t}\lesssim\varepsilon^{\frac{1}{2}}.$

Letting $\vartheta\to 0$ yields (2.3).

### Acknowledgments.

M. Stipčić was supported in part by a fellowship through the Grand Challenges
Initiative at Chapman University and the Croatian Science Foundation under the
project UIP-2017-05-4129 (MUNHANAP). The authors thank Vjekoslav Kovač for
valuable discussions and comments.

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