# Study guide for “On restriction projections to planes in $\mathbb{R}^{3}$”

Tainara Borges, Siddharth Mulherkar, and Tongou Yang

###### Abstract

This article is a study guide for “On restricted projections to planes in
$\mathbb{R}^{3}$” by Gan, Guo, Guth, Harris, Maldague and Wang [GGG+22]. We
first present the main problems and preliminaries related to restricted
projections in $\mathbb{R}^{3}$. Then we introduce the high-low method and
decoupling, which are the two central and novel ideas in their proofs. We hope
to provide as many details as possible so that this study guide is self-
contained, with the only exception of the Bourgain-Demeter decoupling
inequality for curves in the appendix.

###### Contents

  1. 1 Introduction
    1. 1.1 Marstrand’s and Mattila’s projection theorems
      1. 1.1.1 Dimension preserving property in $\mathbb{R}^{2}$
      2. 1.1.2 Exceptional set estimate in $\mathbb{R}^{2}$
      3. 1.1.3 Dimension preserving property in higher dimensions
    2. 1.2 Restricted projection theorems in $\mathbb{R}^{3}$
      1. 1.2.1 Background
      2. 1.2.2 Literature review
    3. 1.3 Main results
    4. 1.4 Reduction of Theorem 2 to Theorem 8
    5. 1.5 Preliminaries about $C^{2}$ curves
    6. 1.6 Notation
    7. 1.7 Acknowledgements
  2. 2 Some important definitions
  3. 3 Discretized Marstrand’s Theorem via high-low method
    1. 3.1 Statement of discretized Marstrand’s Theorem
    2. 3.2 Construction of wave packets
    3. 3.3 High-low decomposition
    4. 3.4 Analysis of high part
    5. 3.5 Analysis of low part
    6. 3.6 Choosing $K$ such that the high part dominates
    7. 3.7 Conclusion
    8. 3.8 Weight inequalities
      1. 3.8.1 Notation
      2. 3.8.2 Elementary inequalities
      3. 3.8.3 Proof of (3.9)
  4. 4 Discretization
    1. 4.1 Lemma 1 and Lemma 2
    2. 4.2 Discretization of Marstrand’s Theorem
    3. 4.3 Reduction of Theorem 1 to the discretized version in Theorem 4
  5. 5 Proof of Theorem 5
    1. 5.1 The decoupling machinery
    2. 5.2 Construction of wave packets
    3. 5.3 High-low decomposition
    4. 5.4 Analysis of high part
    5. 5.5 Analysis of low part
    6. 5.6 Choosing $K$ such that the low part is negligible
    7. 5.7 Analysis of $\delta^{1/2}$-mixed part
      1. 5.7.1 Construction of covering planks
      2. 5.7.2 Applying decoupling inequality
      3. 5.7.3 Analysis of each $f_{\omega}$
      4. 5.7.4 Comparison with the high part
    8. 5.8 Analysis of $\lambda$-mixed parts
      1. 5.8.1 Construction of covering planks
      2. 5.8.2 Applying decoupling inequalities
      3. 5.8.3 Analysis of each $f_{\tau}$
      4. 5.8.4 Comparison with the high part
    9. 5.9 Conclusion
    10. 5.10 Proof of dyadic pigeonholing
    11. 5.11 Proof of covering lemmas
  6. 6 Proof of Theorem 7
    1. 6.1 A bootstrap inequality
    2. 6.2 Construction of wave packets
    3. 6.3 High-low decomposition
    4. 6.4 Analysis of high and mixed parts
    5. 6.5 Analysis of low part
      1. 6.5.1 Maximally incomparable collection of thickened tubes
      2. 6.5.2 A pigeonholing argument
      3. 6.5.3 Analysis at new scale
    6. 6.6 Corollary 2
    7. 6.7 Corollary 3
  7. 7 Proof of Theorem 8
    1. 7.1 Construction of covering planks
    2. 7.2 Construction of wave packets
    3. 7.3 Good part and bad part
    4. 7.4 Controlling bad part
      1. 7.4.1 Far part
      2. 7.4.2 Near part
    5. 7.5 Controlling good part
      1. 7.5.1 Preparation
      2. 7.5.2 Estimate of $\mu_{g2}$
      3. 7.5.3 Estimate of $\mu_{g1}$
  8. 8 Refined Strichartz inequality
  9. 9 Appendix
    1. 9.1 Proof of flat decoupling
    2. 9.2 Proof of decoupling for $C^{2}$ curves
    3. 9.3 Proof of decoupling for $C^{2}$ cones
      1. 9.3.1 Preliminary reductions
      2. 9.3.2 Introducing intermediate scale
      3. 9.3.3 Approximation by cylinder
      4. 9.3.4 Applying decoupling for $C^{2}$ curves
      5. 9.3.5 Iteration

##  1 Introduction

###  1.1 Marstrand’s and Mattila’s projection theorems

####  1.1.1 Dimension preserving property in $\mathbb{R}^{2}$

In $\mathbb{R}^{2}$, given a direction $u\in\mathbb{S}^{1}$ on the unit
circle, consider the orthogonal projection $p_{u}(x):=x\cdot u$ for
$x\in\mathbb{R}^{2}$. For each Borel set $A\subseteq\mathbb{R}^{2}$, we
consider the size of the image $p_{u}(A)\subseteq\mathbb{R}$ relative to the
size of $A$. In particular, for each $u$ we ask if it is dimension preserving,
that is, whether we have111Here and throughout this article, $\dim$ will
always denote the Hausdorff dimension.

$\dim p_{u}(A)=\min\\{1,\dim A\\}.$ (1.1)

Note that $\leq$ in (1.1) is trivial since the range is a subset of
$\mathbb{R}$ and a projection cannot increase dimensions. The celebrated
Marstrand’s projection theorem [Mar54] states that $\sigma^{1}$-a.e.
$u\in\mathbb{S}^{1}$ is dimension preserving for $A$, where $\sigma^{1}$
denotes the standard surface measure on $\mathbb{S}^{1}$. For this reason, we
say a direction $u$ is exceptional if it is not dimension preserving, that is,
we have $<$ holds in (1.1). Lastly, we remark that the set of exceptional
directions naturally depend on the choice of $A$; for instance, if $A$ is a
straight line with direction $v\in\mathbb{S}^{1}$, then $u$ is dimension
preserving for $A$ if and only if $u\cdot v\neq 0$.

####  1.1.2 Exceptional set estimate in $\mathbb{R}^{2}$

In many cases, we can prove stronger results than dimension preserving
properties, known as exceptional set estimates. Given a Borel
$A\subseteq\mathbb{R}^{2}$. For each $0<s<\min\\{1,\dim A\\}$, we consider the
set of exceptional directions

$E_{s}:=\\{u\in\mathbb{S}^{1}:\dim p_{u}(A)<s\\}.$ (1.2)

If we can show that $\dim E_{s}<1$, then Marstrand’s theorem follows as a
corollary. In fact, Kaufman [Kau68] and Falconer [Fal82] established the bound
$\dim E_{s}\leq s$ and $\dim E_{s}\leq\max\\{0,1+s-\dim A\\}$, respectively.
For $\dim A<1$ Kaufman’s estimate is better, while for $\dim A>1$ Falconer’s
estimate is better. Recently, Ren and Wang [RW23] showed that we always have
$\dim E_{s}\leq\max\\{0,2s-\dim A\\}$, which is better than both Kaufman’s and
Falconer’s results. Moreover, a simple example given by Kaufman and Mattila
[KM75] shows that this is sharp.

####  1.1.3 Dimension preserving property in higher dimensions

Now we generalize to $\mathbb{R}^{d}$, $d\geq 2$. For each $1\leq k\leq d-1$,
consider the set $G(d,k)$ of all $k$-dimensional subspaces equipped with the
standard Haar measure $\sigma(d,k)$. Mattila [Mat75] generalized Marstrand’s
theorem [Mar54] to higher dimensions. More precisely, given $1\leq k\leq d-1$
and a subspace $S\in G(d,k)$, consider the projection $p_{S}$ from
$\mathbb{R}^{d}$ onto $S$. Then for every Borel set
$A\subseteq\mathbb{R}^{d}$, we have

$\dim p_{S}(A)=\min\\{k,\dim A\\},\quad\text{for $\sigma(d,k)$-a.e. $S\in
G(d,k)$}.$ (1.3)

One may also consider exceptional estimates in higher dimensions; however,
obtaining a sharp exceptional estimates will be much more difficult than in
$\mathbb{R}^{2}$, and so we will not discuss this topic here.

###  1.2 Restricted projection theorems in $\mathbb{R}^{3}$

####  1.2.1 Background

Now we consider $d=3$ and $k=1,2$. For $k=1$ we are considering projection to
lines, namely, for each $e\in\mathbb{S}^{2}$ we consider the function
$p_{e}(x):=x\cdot e$. For $k=2$ we are considering projection to planes,
namely, for each $e\in\mathbb{S}^{2}$ we consider the function
$\pi_{e}(x):=x-(x\cdot e)e$.

Denote by $\sigma^{2}$ the standard surface measure on $\mathbb{S}^{2}$.
Mattila’s theorem implies that given any Borel $A\subseteq\mathbb{R}^{3}$, for
$\sigma^{2}$-a.e. $e\in\mathbb{S}^{2}$ we have both

$\dim p_{e}(A)=\min\\{1,\dim A\\},\quad\dim\pi_{e}(A)=\min\\{2,\dim A\\}.$
(1.4)

However, this provides no information if we would like to study the
projections of $A$ along a subset $N\subseteq\mathbb{S}^{2}$ with zero
$\sigma^{2}$ measure, most typically when $N$ is the image of a continuous
curve on $\mathbb{S}^{2}$. Nevertheless, this problem is still interesting,
and we may ask the following main question: if
$\gamma(\theta):[0,1]\to\mathbb{S}^{2}$ is a continuous curve, and
$A\subseteq\mathbb{R}^{3}$ is a Borel set, is it true that for
$\mathcal{L}^{1}$-a.e. $\theta\in[0,1]$ we have either equality below:

$\dim p_{\gamma(\theta)}(A)=\min\\{1,\dim
A\\},\quad\dim\pi_{\gamma(\theta)}(A)=\min\\{2,\dim A\\}?$ (1.5)

The simple answer is no. In fact, if we take the image of $\gamma$ to be the
equator and $A$ to be the straight line passing through the north and south
poles, then the former equality of (1.5) fails to hold for any
$\theta\in[0,1]$. Similarly, if we take the image of $\gamma$ to be the
equator and $A$ a $2$-dimensional cube contained in the Equator plane, say
$A=[-1,1]^{2}\times\\{0\\}$ then the latter equality of (1.5) fails to hold
for any $\theta\in[0,1]$.

However, this triviality can be easily avoided once we require that the image
of $\gamma$ “escapes from being a great circle”. More precisely, we require
that $\gamma$ be $C^{2}$, and that the following non-degeneracy condition
hold:

$\det(\gamma,\gamma^{\prime},\gamma^{\prime\prime})(\theta)\neq
0,\quad\forall\,\theta\in[0,1].$ (1.6)

Geometrically, this means that $\gamma$ is the velocity of a curve with
nonzero torsion. A model case of $\gamma$ is a non-great circle which
generates the standard light cone, that is,

$\gamma(\theta)=\frac{1}{\sqrt{2}}\left(\cos\sqrt{2}\,\theta,\sin\sqrt{2}\,\theta,1\right).$
(1.7)

The factor $\sqrt{2}$ here is to normalize so that $|\gamma^{\prime}|=1$ as
well. We remark here that an interesting property of the standard light cone
is that $\gamma$ and $\gamma\times\gamma^{\prime}$ are two different
parametrizations of the same circle, since as one can easily check
$(\gamma\times\gamma^{\prime})(\theta)=\frac{1}{\sqrt{2}}(-\cos\sqrt{2}\theta,-\sin\sqrt{2}\theta,1))$.
This fails for a general non-degenerate curve $\gamma$.

####  1.2.2 Literature review

With the non-degeneracy condition (1.6), Fässler and Orponen [FO14] put
forward the following conjecture:

######  Conjecture 1.1.

Let $\gamma:[0,1]\to\mathbb{S}^{2}$ be $C^{2}$ and satisfies the non-
degeneracy condition (1.6). Then for every Borel set
$A\subseteq\mathbb{R}^{3}$, we have

$\dim p_{\gamma(\theta)}(A)=\min\\{1,\dim
A\\},\quad\dim\pi_{\gamma(\theta)}(A)=\min\\{2,\dim A\\}$ (1.8)

both hold for $\mathcal{L}^{1}$-a.e. $\theta\in[0,1]$.

In [FO14], the authors mentioned that the former equality in (1.8) is easy
when $\dim A\leq 1/2$, and the latter equality in (1.8) is easy when $\dim
A\leq 1$ (see [JJLL08]). In the same article, they were able to prove
something a little stronger. More precisely, if $\dim A>1/2$, they showed
there is some $\sigma>1/2$ depending on $\dim A$ such that $\dim
p_{\gamma(\theta)}>\sigma$ for a.e. $\theta$. If $\dim A>1$, then they showed
there is some $\sigma>1$ depending on $\dim A$ such that
$\dim\pi_{\gamma(\theta)}>\sigma$ for a.e. $\theta$.

Since [FO14], a dozens of partial results have been established towards the
main conjecture, see for instance [JJK14, OO15, OV18, KOV21, Har22]. The most
significant result among them is [KOV21], which is the first to establish the
former equality in (1.8) when $\gamma$ is given by a non-great circle such as
(1.7). Their main technique is to use Wolff’s [Wol00] incidence estimates on
tangency of circles, which relies in turn on the algebraic property of the
circle. Therefore, one major downside of this method is that it fails to work
in our case when $\gamma$ is merely $C^{2}$. We also remark here that [KOV21]
established the following exceptional set estimate: for every
$0<s<\min\\{1,\dim A\\}$, we have

$\dim E_{s}\leq\frac{s+1}{2},$ (1.9)

where the exceptional set $E_{s}$ is given by

$E_{s}:=\\{\theta\in[0,1]:\dim p_{\gamma(\theta)}(A)<s\\}.$ (1.10)

The full proof of Conjecture 1.1 is given by Pramanik-Yang-Zahl [PYZ22], Gan-
Guth-Maldague [GGM22] and Gan-Guo-Guth-Harris-Maldague-Wang [GGG+22]. Among
them, [PYZ22] and [GGM22] independently proved the former equality of (1.1),
using very different methods, and obtaining different exceptional set
estimates. More precisely, [PYZ22] proved a Kaufman type exceptional set
estimate, namely,

$\dim E_{s}\leq s,\quad\forall\,\,0<s<\min\\{1,\dim A\\}.$ (1.11)

In contrast, [GGM22] proved a Falconer type exceptional set estimate, namely,

$\dim E_{s}\leq\max\\{0,1+s-\dim A\\},\quad\forall\,\,0<s<\min\\{1,\dim A\\}.$
(1.12)

As mentioned right after (1.2), neither estimate is strictly better than the
other.

We remark that [Har23] also gives an analogue of Theorem 1.5 below in the case
of projection to lines in $\mathbb{R}^{3}$.

[GGG+22] proved the latter equality of (1.1) via a Falconer type exceptional
set estimate (see Theorem 1.2 below). This article is devoted to the
understanding of [GGG+22].

###  1.3 Main results

We now focus solely on the paper [GGG+22].

Let $\gamma:[0,1]\to\mathbb{S}^{2}$ be a $C^{2}$ curve. We assume it satisfies
the non-degeneracy condition in (1.6). We may also assume without loss of
generality that $|\gamma^{\prime}|=1$, whence

$\mathbf{e}_{1}:=\gamma,\quad\mathbf{e}_{2}:=\gamma^{\prime},\quad\mathbf{e}_{3}:=\gamma\times\gamma^{\prime}$
(1.13)

form an orthonormal basis for $\mathbb{R}^{3}$ known as a Frenet frame.

######  Theorem 1.2 (Theorem 1 in [GGG+22]).

Suppose $A\subseteq\mathbb{R}^{3}$ is a Borel set of Hausdorff dimension
$\alpha\in(0,3]$. For $0<s<\min\\{2,\alpha\\}$, define the exceptional set

$E_{s}=\\{\theta\in[0,1]\colon\textnormal{dim}(\pi_{\theta}(A))<s\\}$

Then we have

$\textnormal{dim}(E_{s})\leq\max\\{1+s-\alpha,0\\}$

The proof of Theorem 1.2 will be given in Section 4.3.

Theorem 1.2 gives the following corollary, which solves Conjecture 1.1.

######  Corollary 1.3 (Corollary 1 in [GGG+22]).

Suppose $A\subseteq\mathbb{R}^{3}$ is a Borel set of Hausdorff dimension
$\alpha$. Then we have

$\textnormal{dim}(\pi_{\theta}(A))=\min\\{2,\alpha\\},\,\text{for a.e.
}\theta\in[0,1]$

###### Proof.

There are two possible cases: $\textnormal{dim}(A)=\alpha\leq 2$ or
$\textnormal{dim}(A)=\alpha>2$. In the first case, take $s_{n}=\alpha-1/n$.
Theorem 1.2 gives $\textnormal{dim}(E_{s_{n}})\leq 1-1/n$. In particular,

$\mathcal{L}^{1}(E_{s_{n}})=0$

Since

$E_{\alpha}=\cup_{n=1}^{\infty}E_{s_{n}}$

it also follows that $\mathcal{L}^{1}(E_{\alpha})=0$. The case
$\textnormal{dim}(A)>2$ is similar, by taking $s_{n}=2-1/n$ and
$E_{2}=\cup_{n=1}^{\infty}E_{s_{n}}$. ∎

######  Theorem 1.4 (Theorem 2 in [GGG+22]).

Suppose $A\subseteq\mathbb{R}^{3}$ is a Borel set of Hausdorff dimension
greater than $2$. Then $\mathcal{H}^{2}(\pi_{\theta}(A))>0$ for a.e.
$\theta\in[0,1]$.

###  1.4 Reduction of Theorem 2 to Theorem 8

We first state the following theorem, from which Theorem 1.4 follows as a
corollary.

######  Theorem 1.5 (Theorem 8 in [GGG+22]).

If $a\in(2,3]$ and $\mu$ is a compactly supported Borel measure on
$\mathbb{R}^{3}$ such that

$\rho_{a}(\mu):=\sup_{x\in\mathbb{R}^{3},r\in(0,1)}r^{-a}\mu(B^{3}(x,r))<\infty,$
(1.14)

then $\pi_{\theta\\#}\mu$ is absolutely continuous with respect to
$\mathcal{H}^{2}$, for a.e. $\theta\in[0,1]$. Here and throughout this
article, $\pi_{\theta\\#}\mu$ denotes the standard push-forward measure,
namely, for every Borel set $A\subseteq\pi_{\theta}(\mathbb{R}^{3})$,

$\pi_{\theta\\#}\mu(A):=\mu(\pi_{\theta}^{-1}(A)).$ (1.15)

The proof of Theorem 1.5 will be given in Section 7. We also remark that since
$\mu$ is compactly supported, we may actually regard $\mathcal{H}^{2}$ as
supported on the unit ball of $\pi_{\theta}(\mathbb{R}^{3})$, for all
$\theta\in[0,1]$. Also, Theorem 1.5 actually gives that $\pi_{\theta\\#}\mu$
can be identified with an integrable function on
$\pi_{\theta}(\mathbb{R}^{3})$ for a.e. $\theta$.

We now prove Theorem 1.4 assuming Theorem 1.5. For this purpose, we recall the
preliminary concept of Frostman measure.

######  Definition 1.6 (Frostman measure).

Let $A\subset\mathbb{R}^{d}$ be a Borel set. A finite nonzero measure $\mu$
supported in $A$ is said to be an $a$-Frostman measure in $A$ if
$\rho_{a}(\mu)<\infty$ (see (1.14) above).

We state Frostman’s Lemma below, where $\mathcal{H}^{a}$ stands for the
$a$-dimensional Hausdorff measure in $\mathbb{R}^{d}$.

######  Theorem 1.7 (Frostman’s lemma, Theorem 2.7 in [Mat15]).

Let $A\subset\mathbb{R}^{d}$ be a Borel set. Then $A$ admits an $a$-Frostman
measure if and only if $\mathcal{H}^{a}(A)\in(0,\infty]$.

###### Proof of Theorem 1.4 assuming Theorem 1.5.

We may of course assume $A\subseteq[-1,1]^{3}$. Since $\dim A\in(2,3]$, we
have $\mathcal{H}^{a}(A)>0$ where $a:=\frac{\dim A+2}{2}\in(2,\dim A)$
(indeed, since $\mathcal{H}^{a}(A)=\infty$ for such $a$). By Theorem 1.7, we
can find an $a$-Frostman measure $\mu$ supported in $A$. By Theorem 1.5, we
have $\pi_{\theta\\#}\mu$ is absolutely continuous with respect to
$\mathcal{H}^{2}$ for a.e. $\theta\in[0,1]$.

For every such $\theta$, we claim that $\mathcal{H}^{2}(\pi_{\theta}(A))>0$.
Otherwise, by definition of absolute continuity, we have
$\pi_{\theta\\#}\mu(\pi_{\theta}(A))=0$, that is,
$\mu(\pi_{\theta}^{-1}(\pi_{\theta}(A)))=0$. Since
$A\subseteq\pi_{\theta}^{-1}(\pi_{\theta}(A))$, this implies $\mu(A)=0$, a
contradiction. ∎

###  1.5 Preliminaries about $C^{2}$ curves

Define

$\mathbf{e}_{1}(\theta)=\gamma(\theta),\,\mathbf{e}_{2}(\theta)=\gamma^{\prime}(\theta),\,\mathbf{e}_{3}(\theta)=\mathbf{e}_{1}(\theta)\times\mathbf{e}_{2}(\theta).$
(1.16)

Then for any $\theta\in[0,1]$, $\\{\mathbf{e}_{i}(\theta)\colon i=1,2,3\\}$ is
an orthonormal basis of $\mathbb{R}^{3}$.

######  Lemma 1.8.

We have the following identities:

$\mathbf{e}_{1}^{\prime}=\mathbf{e}_{2},\quad\mathbf{e}_{2}^{\prime}=-\mathbf{e}_{1}+\uptau\mathbf{e}_{3},\quad\mathbf{e}_{3}^{\prime}=-\uptau\mathbf{e}_{2},$
(1.17)

where we denote $\uptau=\det(\gamma,\gamma^{\prime},\gamma^{\prime\prime})$.

###### Proof.

This is just a standard consequence of a Frenet frame with curvature $1$ and
torsion $\uptau$, but for completeness we also give a simple proof here. The
first relation is just the definition. To prove the second relation, we will
need the relation

$\gamma\cdot\gamma^{\prime\prime}=-1,$ (1.18)

which follows from differentiating $\gamma\cdot\gamma^{\prime}=0$ and using
$|\gamma^{\prime}|=1$. Now write

$\mathbf{e}_{2}^{\prime}=\gamma^{\prime\prime}:=a_{1}\gamma+a_{2}\gamma^{\prime}+a_{3}\gamma^{\prime}\times\gamma^{\prime},$
(1.19)

and using (1.18) and the fact that
$\\{\gamma,\gamma^{\prime},\gamma\times\gamma^{\prime}\\}$ forms an
orthonormal basis, we get $a_{1}=-1$, $a_{2}=0$, and

$a_{3}=\gamma^{\prime\prime}\cdot(\gamma\times\gamma^{\prime})=\det(\gamma,\gamma^{\prime},\gamma^{\prime\prime})=\uptau.$
(1.20)

Similarly, we note that

$\mathbf{e}_{3}^{\prime}=(\gamma\times\gamma^{\prime})^{\prime}=\gamma\times\gamma^{\prime\prime}.$
(1.21)

Write

$\gamma\times\gamma^{\prime\prime}=b_{1}\gamma+b_{2}\gamma^{\prime}+b_{3}\gamma\times\gamma^{\prime},$
(1.22)

and using the fact that
$\\{\gamma,\gamma^{\prime},\gamma\times\gamma^{\prime}\\}$ forms an
orthonormal basis, we get $b_{1}=0$ and $b_{2}=-\uptau$. To find $b_{3}$,
taking dot products with $\gamma^{\prime\prime}$ on both sides of (1.22) and
using $\gamma^{\prime}\cdot\gamma^{\prime\prime}=0$ which comes from
differentiating $|\gamma^{\prime}|^{2}=1$, we get $b_{3}\uptau=0$, and so
$b_{3}=0$ since $\uptau\neq 0$.

∎

The following slightly more generalized Taylor expansion theorem will be used
throughout the article to deal with low regularity issues. In words, with the
assumption below, we can pretend that $f$ was a $C^{2}$ function when we do
Taylor expansions.

######  Proposition 1.9.

Let $f:[a,b]\to\mathbb{R}^{3}$ be a $C^{1}$ function such that $f^{\prime}=gh$
where $g:[a,b]\to\mathbb{R}^{3}$ is $C^{1}$, $g(a)=0$ and
$h:[a,b]\to\mathbb{R}$ is bounded. Then we have the estimate

$|f(b)-f(a)|\leq\frac{1}{2}(b-a)^{2}\norm*{h}_{\infty}\norm*{g^{\prime}}_{\infty}.$
(1.23)

###### Proof.

By the fundamental theorem of calculus, we have

$\displaystyle|f(b)-f(a)|$
$\displaystyle=\left|\int_{0}^{1}(b-a)f^{\prime}((1-t)a+tb)dt\right|$
$\displaystyle\leq(b-a)\int_{0}^{1}\left|g((1-t)a+tb)h((1-t)a+tb)\right|dt$
$\displaystyle\leq(b-a)\norm*{h}_{\infty}\int_{0}^{1}\left|g((1-t)a+tb)\right|dt.$

For each $t\in[0,1]$, we can use the fundamental theorem of calculus again to
get

$\displaystyle|g((1-t)a+tb)|$ $\displaystyle=|g((1-t)a+tb)-g(a)|$
$\displaystyle=\left|\int_{0}^{1}t(b-a)g^{\prime}((1-s)a+s((1-t)a+tb))ds\right|$
$\displaystyle\leq\int_{0}^{1}\norm*{g^{\prime}}_{\infty}|t(b-a)|ds$
$\displaystyle=t(b-a)\norm*{g^{\prime}}_{\infty}.$

The result then follows. ∎

We also need the following elementary proposition.

######  Proposition 1.10.

If $|\theta-\theta^{\prime}|\leq\delta^{1/2}$, then we have

$\displaystyle|\mathbf{e}_{i}(\theta)\cdot\mathbf{e}_{i}(\theta^{\prime})-1|$
$\displaystyle\lesssim\delta,\quad\forall i=1,2,3,$ (1.24)
$\displaystyle|\mathbf{e}_{2}(\theta)\cdot\mathbf{e}_{j}(\theta^{\prime})|$
$\displaystyle\lesssim\delta^{1/2},\quad\forall j=1,3,$ (1.25)
$\displaystyle|\mathbf{e}_{1}(\theta)\cdot\mathbf{e}_{3}(\theta^{\prime})|$
$\displaystyle\lesssim\delta.$ (1.26)

###### Proof.

The estimate (1.25) is easy, since each $\mathbf{e}_{i}(\theta)$ is a
Lipschitz function, and $\\{\mathbf{e}_{i}(\theta):1\leq i\leq 3\\}$ is an
orthonormal basis for each $\theta$. The inequality (1.26) follows from Taylor
expanding $\mathbf{e}_{1}(\theta)$ with respect to the centre
$\theta^{\prime}$ and using that $\\{\mathbf{e}_{i}(\theta^{\prime}):1\leq
i\leq 3\\}$ is an orthonormal basis.

To prove (1.24) for $i=1$, we note that it follows directly from Lemma 1.8 and
Taylor expansion. For $i=3$ we need to be slightly more careful since
$\mathbf{e}_{3}$ is merely $C^{1}$. Take
$f(t)=\mathbf{e}_{3}(\theta)\cdot\mathbf{e}_{3}(t)$ and observe that by Lemma
1.8 one has
$f^{\prime}(t)=-\tau(t)\mathbf{e}_{3}(\theta)\cdot\mathbf{e}_{2}(t)=-\tau(t)g(t)$,
for $g(t)=\mathbf{e}_{3}(\theta)\cdot\mathbf{e}_{2}(t)$. Since $g(\theta)=0$
and $f(\theta)=1$, Proposition 1.9 gives
$|\mathbf{e}_{3}(\theta^{\prime})\cdot\mathbf{e}_{3}(\theta)-1|\lesssim|\theta-\theta^{\prime}|^{2}\lesssim\delta$.

It remains to prove the inequality for $i=2$. The idea is similar to the proof
of Proposition 1.9: we compute, using Lemma 1.8,

$\displaystyle|\mathbf{e}_{2}(\theta^{\prime})\cdot\mathbf{e}_{2}(\theta)-1|$
$\displaystyle=|\mathbf{e}_{2}(\theta^{\prime})\cdot\mathbf{e}_{2}(\theta)-\mathbf{e}_{2}(\theta)\cdot\mathbf{e}_{2}(\theta)|$
$\displaystyle=\left|\mathbf{e}_{2}(\theta)\cdot\int_{0}^{1}\mathbf{e}^{\prime}_{2}((1-t)\theta+t\theta^{\prime})(\theta^{\prime}-\theta)dt\right|$
$\displaystyle\leq|\theta^{\prime}-\theta|\int_{0}^{1}\left|\mathbf{e}_{2}(\theta)\cdot\mathbf{e}_{1}((1-t)\theta+t\theta^{\prime})\right|dt$
$\displaystyle+|\theta^{\prime}-\theta|\int_{0}^{1}\left|\uptau((1-t)\theta+t\theta^{\prime})\mathbf{e}_{2}(\theta)\cdot\mathbf{e}_{3}((1-t)\theta+t\theta^{\prime})\right|dt.$

Using $|\uptau|\sim 1$ and applying the already proved inequality (1.25) to
control the inner products left inside the integrals by $\delta^{1/2}$, we are
done. ∎

###  1.6 Notation

  1. 1.

$\lessapprox$ and $\gtrapprox$ will mean up to logarithmic losses. More
precisely, for two nonnegative functions $A(\delta),B(\delta)$, we use
$A(\delta)\lessapprox B(\delta)$ to mean the following: there exists some
constant $\alpha>0$ and $C>1$ such that

$A(\delta)\leq C|\log\delta|^{\alpha}B(\delta),\quad\forall\delta\in(0,1/2].$
(1.27)

We can define $\gtrapprox$ in a similar way.

  2. 2.

We say a collection $\mathcal{C}$ of sets is $A$-overlapping if
$\sum_{C\in\mathcal{C}}1_{C}\leq A$.

  3. 3.

Unless otherwise specified, all implicit constants in this article are allowed
to depend on the curve $\gamma$.

  4. 4.

For a family of subsets $\mathbb{T}$ of $\mathbb{R}^{n}$, we denote
$\cup\mathbb{T}:=\cup_{T\in\mathbb{T}}T$.

  5. 5.

$\text{dim}(A)$ stands for the Hausdorff dimension of $A$.

###  1.7 Acknowledgements

We are grateful to the organizers of the “Study Guide Writing Workshop 2023”
at the University of Pennsylvania for the opportunity of participating in this
very interesting project and to all participants of this workshop for many
fruitful discussions. We would like to give special thanks to our mentor,
Professor Yumeng Ou, for all her guidance and support.

##  2 Some important definitions

We first discuss the proof of Theorem 1.2. It is proved using a standard
technique commonly referred to as discretization. Namely, we first reduce
Theorem 1.2 to a discretized version which can be proved using Fourier
analysis. (see Theorem 4.5 below). We start with some preliminaries in
geometric measure theory.

We recall the definition of Hausdorff dimension in a metric space $X$. For
$A\subset X$,

$\textnormal{dim}(A)=\inf\\{s\geq 0\colon\mathcal{H}^{s}(A)=0\\}=\sup\\{s\geq
0\colon\mathcal{H}^{s}(A)=\infty\\}$

where $\mathcal{H}^{s}(A)=\lim_{\delta\rightarrow
0}\mathcal{H}^{s}_{\delta}(A)$ and for $0<\delta\leq\infty$,

$\mathcal{H}_{\delta}^{s}(A)=\inf\left\\{\sum_{i=0}^{\infty}(\textnormal{diam}\,U_{i})^{s}\colon
A\subset\cup_{i=0}^{\infty}\,U_{i},\,U_{i}\subset X\,\textnormal{ with
}\textnormal{diam}\,(U_{i})\leq\delta,\,\forall i\right\\}.$

######  Definition 2.1 (Definition 1 in [GGG+22]).

For a number $\delta$ and any set $X$, we use $|X|_{\delta}$ to denote the
maximal number of $\delta$-separated points in $X$.

######  Example 2.2.

Here are two typical cases:

  * •

If $X=B^{d}(0,r)$ and $\delta<r$, then $|X|_{\delta}\sim(r/\delta)^{d}$.

  * •

If $X=C_{n}\subseteq\mathbb{R}$, the $n$-th stage of Cantor middle third set,
then

    * –

if $\delta<3^{-n}$, we have $|X|_{\delta}\sim 2^{n}3^{-n}\delta^{-1}$.

    * –

if $\delta>3^{-n}$, say $\delta=3^{-m}$ where $m<n$, then $|X|_{\delta}\sim
2^{m}$. In particular, if $C$ is the Cantor middle third set, then
$|C|_{\delta}\sim 2^{m}=\delta^{-s}$ where $s=\text{dim}(C)=\log_{3}2$ (see
Figure 1).

$C_{n}$

$\delta$

Figure 1: Middle third Cantor set example.

The following proposition exhibits some uncertainty principle.

######  Proposition 2.3.

Given any finite set $X$, $|X|_{\delta}\sim|N_{\delta/4}(X)|_{\delta}$ where
$N_{\delta}(X)$ denotes the $\delta$-neighbourhood of $X$. In words, when
considering $|X|_{\delta}$, we can only see a $\delta/4$-blurred picture of
$X$.

###### Proof.

It suffices to show $|X|_{\delta}\gtrsim|N_{\delta/4}(X)|_{\delta}$. We need
to show that given any $\delta$-separated subset $A\subseteq N_{\delta/4}(X)$,
we have $\\#A\lesssim|X|_{\delta}$. For each point $a\in A$ choose $x_{a}\in
X$ with $|x_{a}-a|<\delta/4$. For any $u,v\in A$ with $u\neq v$ one has
$|u-v|>\delta$, and $|x_{u}-u|<\delta/4$, $|x_{v}-v|<\delta/4$. Thus
$|x_{u}-x_{v}|>\delta/2$. Thus we have obtained a $\delta/2$-separated subset
$B=\\{x_{a}\colon a\in A\\}\subseteq X$ with $\\#A=\\#B$. By dividing
$\mathbb{R}^{d}$ into a $\delta/2$-grid and a simple pigeonholing argument, we
can find a subset $B^{\prime}\subseteq B$ such that $B^{\prime}$ is
$\delta$-separated and $\\#B\sim\\#B^{\prime}$. Thus we have
$\\#A\sim\\#B^{\prime}\leq|X|_{\delta}$.

∎

To make the Definition 2 in [GGG+22] slightly more rigorous, we define

######  Definition 2.4 (Definition 2 in [GGG+22]).

Let $X\subseteq\mathbb{R}^{d}$ be a bounded set. Let $\delta>0$ and let $0\leq
s\leq d$. We say that $X$ is a $(\delta,s,C)$-set if for every ball
$B_{r}\subseteq\mathbb{R}^{d}$,

$|X\cap B_{r}|_{\delta}\leq C(r/\delta)^{s},\quad\forall\,\delta\leq r\leq 1.$
(2.1)

For simplicity, we say that $X$ is a $(\delta,s)$-set if $X$ is a
$(\delta,s,C)$-set for some $C$ depending on $s,d$ only.

By Proposition 2.3, we see that if $X$ is a $(\delta,s)$-set, then
$N_{\delta/4}(X)$ is also a $(\delta,s)$-set.

######  Example 2.5.

A typical case of a $(\delta,s)$-set is given by the union of $2^{n}$ disjoint
intervals of length $\delta$, which appears at the $n$-th stage of the
standard middle $(1-2\alpha)$-Cantor set $C$, where $\delta=\alpha^{n}$ and
$\dim C=\log_{\alpha^{-1}}(2)$.

To check such set $C_{n}$ is a $(\delta,s)$-set for $\delta=\alpha^{n}$ and
$s=\text{dim}(C)$ we observe that for any $0\leq m<n$ and $r=\alpha^{m}$ we
have $|C_{n}\cap B_{\alpha^{m}}|_{\alpha^{n}}\sim 2^{n-m}$ and

$(r/\delta)^{s}=\alpha^{-(n-m)s}=2^{n-m}\text{(see Figure \ref{fig:second
example})}.$ (2.2)

$C_{n}$

$C_{m}$

$3^{-m}$

$3^{-n}$

$2^{n-m}$ red intervals within each blue interval

Figure 2: $(3^{-n},\log_{3}(2))$-set example, with $n=3$ and $m=1$.

##  3 Discretized Marstrand’s Theorem via high-low method

To illustrate the high-low method in Fourier analysis, we pause the proof of
Theorem 1.2 and give an alternative proof of Marstrand’s projection theorem in
$\mathbb{R}^{2}$, resulting in a Falconer type exceptional set estimate (see
Section 1.1.2).

###  3.1 Statement of discretized Marstrand’s Theorem

We first slightly rephrase Proposition 1 of [GGG+22] as follows.

######  Proposition 3.1 (Proposition 1).

Fix $0<s<1$. Fix a small scale $\delta>0$ and a $\delta$-separated set
$\Theta\subseteq[0,\pi)$. For each $\theta\in\Theta$, define
$L_{\theta}:=\\{x\in\mathbb{R}^{2}:\arg(x)=\theta\\}\cup\\{(0,0)\\}$ and let
$p_{\theta}:\mathbb{R}^{2}\to L_{\theta}$ be the projection. Suppose

  * •

$A\subseteq B^{2}(0,1)$ is a union of disjoint $\delta$-balls with measure
$|A|=\delta^{2-a}$, or equivalently $|A|_{\delta}\sim\delta^{-a}$.

  * •

Assume for every $\theta\in\Theta$, $p_{\theta}(A)$ (which is a union of line
segments of length $\delta$ in $L_{\theta}$) satisfies the $s$-dimensional
condition: for each $r\in[\delta,1]$ and line segment $I_{r}\subseteq
L_{\theta}$ of length $r$, we have

$|p_{\theta}(A)\cap I_{r}|_{\delta}\lesssim(r/\delta)^{s}.$ (3.1)

Then

$\\#\Theta\lesssim_{s}\delta^{-1+a-s}.$ (3.2)

In particular, if $\\#\Theta\sim\delta^{-1}$, then the above becomes

$\delta^{-a}\lesssim_{s}\delta^{-s}.$ (3.3)

Under the uncertainty scale $\delta$, this form of $A$ is general enough. The
set of directions $\Theta^{\prime}$ can be thought of as containing almost all
directions since it has cardinality $\sim\delta^{-1}$. For simplicity, we may
just take $\Theta^{\prime}=\Theta$.

Intuitively, this proposition roughly implies the following. Assume $a=\dim
A$. Then for each uncertainty scale $\delta$, $A$ can be thought of as a
disjoint union of $\delta^{-a}$ disjoint $\delta$-balls. If the projections of
$A$ along almost all directions is a $(\delta,s)$-set, then $a\leq s$. This is
roughly the statement of Marstrand’s Projection Theorem. For more precise
details about the discretization step for Marstrand’s theorem we refer to
Subsection 4.2.

Now we start with the proof of Proposition 3.1, divided into the following
subsections.

###  3.2 Construction of wave packets

For each $\theta\in\Theta$, let $\mathbb{T}_{\theta}$ be a set of
$\delta\times 1$ tubes that cover $p_{\theta}^{-1}(p_{\theta}(A))\cap
B^{2}(0,1)$ and hence cover $A$. We may assume all tubes have the same
orientation, that is, perpendicular to $L_{\theta}$, and are all centred on
$L_{\theta}$. Note that $\\#\mathbb{T}_{\theta}\lesssim\delta^{-s}$ by the
$s$-dimensional condition.

Now we come to the Fourier side. Let $S_{\theta}$ be the dual rectangle to all
$T_{\theta}$’s, that is, $S_{\theta}$ is a $\delta^{-1}\times 1$ rectangle
centred at $0$ with orientation perpendicular to $T_{\theta}$.

For each $T_{\theta}\in\mathbb{T}_{\theta}$, define a Schwartz function
$\psi_{T_{\theta}}:\mathbb{R}^{2}\to\mathbb{R}$ Fourier supported on
$S_{\theta}$ such that $\psi_{T_{\theta}}\gtrsim 1$ on $T_{\theta}$ and decays
rapidly outside $T_{\theta}$.

For a concrete example, assume $L_{\theta}$ is the $x$-axis and
$T_{\theta}=[c-\delta,c+\delta]\times[-1,1]$. Then
$S_{\theta}=[-\delta^{-1},\delta^{-1}]\times[-1,1]$, and we may take

$\psi_{T_{\theta}}(x,y)=\eta(\delta^{-1}(x-c))\eta(y)$

where $\widehat{\eta}$ is a nonnegative even function supported on $(-1,1)$
and such that $\widehat{\eta}(0)=1$.

Lastly, we define

$f_{\theta}=\sum_{T_{\theta}\in\mathbb{T}_{\theta}}\psi_{T_{\theta}},\quad
f=\sum_{\theta\in\Theta}f_{\theta}.$ (3.4)

Thus $f_{\theta}$ is also Fourier supported on $S_{\theta}$, but physically
essentially supported on $\cup\mathbb{T}_{\theta}$. The function $f$ is
Fourier supported on $\cup_{\theta\in\Theta}S_{\theta}$, which is a union of
rectangles of the same size centered at $0$, but with $\delta$-separated
orientations.

###  3.3 High-low decomposition

The high-low (frequency) decomposition is based on the following simple fact
in plane geometry. This gives some sort of orthogonality when we deal with the
high part.

######  Lemma 3.2.

Let $\mathcal{S}$ be a collection of $\delta^{-1}\times 1$ rectangles centred
at the origin, with $\delta$-separated orientations. If $1\leq
K\leq\delta^{-1}$, then $\cup\mathcal{S}\backslash B(0,K^{-1}\delta^{-1})$ is
at most $O(K)$-overlapping.

###### Proof of lemma.

Let $T\in\mathcal{S}$ and let $l_{T}$ be the central axis along the longer
side of $T$. If $x\in T\backslash B(0,K^{-1}\delta^{-1})$, then the angle
between $Ox$ and $l_{T}$ is less than $O(K)\delta$ (see Figure 3). Thus if
$x\in(T\cap T^{\prime})\backslash B(0,K^{-1}\delta^{-1})$, by the triangle
inequality, the angle between $l_{T^{\prime}}$ and $l_{T}$ is less than
$O(K)\delta$. Since $\mathcal{S}$ is $\delta$-separated, this gives the claim.
∎

$\theta$$l_{T}$$x$$\delta^{-1}K^{-1}$$1$$\theta\sim\tan(\theta)\leq\frac{1}{\delta^{-1}K^{-1}}=\delta
K$

Figure 3: Illustration of Lemma 3.2.

Now given $K\in[1,\delta^{-1}]$ to be determined, and let $\eta_{l}$ be a
smooth bump function adapted to $B^{2}(0,K^{-1}\delta^{-1})$ in the frequency
space. Let $\eta_{h}=1-\eta_{l}$. We decompose

$f=f_{l}+f_{h}$ (3.5)

where $\widehat{f_{l}}=\widehat{f}\eta_{l}$,
$\widehat{f_{h}}=\widehat{f}\eta_{h}$.

###  3.4 Analysis of high part

In this part, we use Fourier analysis. By Plancherel’s identity and Lemma 3.2,
we have

$\begin{split}\int|f_{h}|^{2}=\int|\widehat{f_{h}}|^{2}&\leq\int|\sum_{\theta\in\Theta}\hat{f_{\theta}}\eta_{h}\cdot\mathbb{1}\text{supp}(\hat{f}_{\theta,h})|^{2}\\\
&\lesssim\int K\sum_{\theta\in\Theta}|\widehat{f_{\theta}}|^{2}\\\ &\lesssim
K\sum_{\theta\in\Theta}\int|f_{\theta}|^{2}.\end{split}$ (3.6)

By the rapid decay of $\psi_{T_{\theta}}$ (see (3.18) and Corollary 3.4 for
details), we have

$\int|f_{\theta}|^{2}\sim\delta\\#\mathbb{T}_{\theta}\lesssim\delta^{1-s}.$
(3.7)

Thus

$\int|f_{h}|^{2}\lesssim K(\\#\Theta)\delta^{1-s}.$ (3.8)

###  3.5 Analysis of low part

For the low part, we do not have orthogonality to aid us. Instead, we use the
$s$-dimensional condition (3.1) in the assumption of the proposition. We first
claim that for each $x\in A$ and $\theta\in\Theta$, we have

$|f_{\theta}*{\eta}_{l}^{\vee}(x)|\lesssim
K^{-1}\\#\\{T_{\theta}\in\mathbb{T}_{\theta}:T_{\theta}\cap
B^{2}(x,CK\delta)\neq\varnothing\\}.$ (3.9)

Intuitively, this holds by the uncertainty principle, since we now view the
rectangles $T_{\theta}$ blurred by scale $K\delta$.

We give a heuristic proof of (3.9) first. Assume for simplicity that
$f_{\theta}\approx\sum_{T_{\theta}\in\mathbb{T}_{\theta}}1_{T_{\theta}}$.
Assume also
$\eta_{l}^{\vee}\approx(K^{-1}\delta^{-1})^{2}1_{B^{2}(0,K\delta)}$. Then for
each $x\in\mathbb{R}^{2}$,

$1_{T_{\theta}}*1_{B^{2}(0,K\delta)}(x)\begin{cases}=0,&\text{ if
}T_{\theta}\cap B^{2}(x,K\delta)=\varnothing,\\\ \lesssim K\delta^{2},&\text{
if }T_{\theta}\cap B^{2}(x,K\delta)\neq\varnothing.\end{cases}$ (3.10)

The rigorous proof will be given later in Section 3.8 to avoid distraction
from the main idea of the proof.

Using (3.9) and the $s$-dimensional condition (3.1), for every $x\in A$ we
have

$|f_{l}(x)|\leq\sum_{\theta\in\Theta}|f_{\theta}*\eta_{l}^{\vee}(x)|\lesssim\\#\Theta
K^{-1}(K\delta/\delta)^{s}\sim\\#\Theta K^{s-1}.$ (3.11)

###  3.6 Choosing $K$ such that the high part dominates

For each $x\in A$, we have $f_{\theta}(x)\gtrsim 1$ by construction of
$\psi_{T_{\theta}}$. Thus $f(x)\gtrsim\\#\Theta$ for $x\in A$.

Since $s<1$, if $K=K(s)$ is chosen to be large enough, then
$|f_{l}(x)|<|f(x)|/2$ for each $x\in A$ by (3.11). Thus, by the triangle
inequality, the high part dominates for $x\in A$.

###  3.7 Conclusion

Since the high part dominates, by (3.8), we have

$\int_{A}f^{2}\lesssim K\\#\Theta\delta^{1-s}.$ (3.12)

On the other hand, since $f(x)\gtrsim\\#\Theta$ on $A$, we trivially have the
lower bound

$\int_{A}f^{2}\gtrsim(\\#\Theta)^{2}|A|\sim(\\#\Theta)^{2}\delta^{2-a}.$
(3.13)

Since $K$ depends on $s$ only, we have
$(\\#\Theta)^{2}\delta^{2-a}\lesssim_{s}\\#\Theta\delta^{1-s}$, as required.

###  3.8 Weight inequalities

We now give a rigorous proof of the heuristic argument used to prove (3.9), by
introducing weight functions. See also [BD17] for inequalities of weights.

####  3.8.1 Notation

Let $\psi:\mathbb{R}\to\mathbb{C}$ be a Schwartz function such that
$\psi\gtrsim 1$ in $(-1,1)$. Given an axis parallel rectangle
$T=[c_{1}-r_{1},c_{1}+r_{1}]\times[c_{2}-r_{2},c_{2}+r_{2}]$, we define the
(unnormalised) Schwartz function $\eta_{T}$ adapted to $T$ to be

$\eta_{T}(x_{1},x_{2})=\psi\left(\frac{x_{1}-c_{1}}{r_{1}}\right)\psi\left(\frac{x_{2}-c_{2}}{r_{2}}\right).$
(3.14)

For a general rectangle $T$, define $\eta_{T}$ in the most natural way with
rotations.

Let

$E=\max\left\\{100,1+\frac{2}{1-s}\right\\},$ (3.15)

where we recall $s\in(0,1)$ is given at the beginning of Proposition 3.1.

Similarly to (3.14), we define weight functions $w=w^{(E)}$

$w(x_{1},x_{2})=(1+|x_{1}|)^{-E}(1+|x_{2}|)^{-E}.$ (3.16)

In general, for a rectangle $T\subseteq\mathbb{R}^{d}$ with centre $c$, sides
$e_{i}$ and dimensions $r_{i}$, we define

$w_{T}(x)=w_{T,E}(x)=\prod_{i=1}^{d}\left(1+\frac{|(x-c)\cdot
e_{i}|}{r_{i}}\right)^{-E}.$ (3.17)

Similarly, for a ball $B=B^{2}(c,r)$, we define $\eta_{B}=\eta_{T}$ and
$w_{B}=w_{T}$ where $T$ is the square centred at $c$ of side length $r$.

It is easy to see that

$1_{T}\lesssim\eta_{T}\lesssim w_{T}.$ (3.18)

Also, if $T$ and $T^{\prime}$ are essentially the same, meaning
$C^{-1}T\subseteq T^{\prime}\subseteq CT$ for some absolute constant $C>1$,
then $w_{T}\sim w_{T^{\prime}}$. All implicit constants below are allowed to
depend on $E$. (Note that this property fails if $w$ is replaced by $\eta$.
Similar for Proposition 3.6 below.)

We also need the following proposition.

######  Proposition 3.3.

Suppose $\mathbb{I}$ is a tiling of $\mathbb{R}$ by intervals $I$ of the same
size. Let $C>1$, $x\in\mathbb{R}$ and $\mathbb{I}^{\prime}\subseteq\mathbb{I}$
be such that $\mathrm{dist}(x,I)\geq C|I|$ for every
$I\in\mathbb{I}^{\prime}$. Then

$\sum_{I\in\mathbb{I}^{\prime}}\left(1+\frac{|x-c_{I}|}{|I|}\right)^{-E}\lesssim
C^{-E+1},$ (3.19)

where $c_{I}$ denotes the centre of $I$.

###### Proof.

By scaling we may assume $|I|=1$, for all $I\in\mathbb{I}$. By translation we
may assume $x=0$. Then the left hand side is essentially bounded above by

$\int_{|y|\geq C}(1+|y|)^{-E}dy\sim C^{-E+1}.$ (3.20)

∎

######  Corollary 3.4.

If $\mathbb{T}$ is a tiling of $\mathbb{R}^{2}$ by rectangles $T$, then
$\sum_{T\in\mathbb{T}}w_{T}\sim 1$.

####  3.8.2 Elementary inequalities

######  Proposition 3.5.

Let $A,B\subseteq\mathbb{R}^{d}$ be Borel sets. Then

$1_{A}*1_{B}\leq\min\\{|A|,|B|\\}1_{A+B}.$ (3.21)

The proof is trivial. Using the case $d=1$ and $A,B$ being intervals entry-
wise, this can be upgraded to the following proposition:

######  Proposition 3.6.

Let $T,T^{\prime}$ be parallel (perpendicular) rectangles of dimensions
$a\times b$, $a^{\prime}\times b^{\prime}$, respectively. Then

$w_{T}*w_{T^{\prime}}\lesssim\min\\{a,a^{\prime}\\}\min\\{b,b^{\prime}\\}w_{T+T^{\prime}}.$
(3.22)

We leave the elementary proof to the reader.

####  3.8.3 Proof of (3.9)

For each $x$, denote
$\mathbb{T}_{\theta}(x):=\\{T_{\theta}\in\mathbb{T}_{\theta}:T_{\theta}\cap
B^{2}(x,CK\delta)\neq\varnothing\\}$.

Let $x\in A$. Then $\mathbb{T}_{\theta}(x)\neq\varnothing$. Write

$|f_{\theta}*\eta_{l}^{\vee}(x)|\leq\sum_{T_{\theta}\in\mathbb{T}_{\theta}(x)}|\psi_{T_{\theta}}*\eta_{l}^{\vee}(x)|+\sum_{T_{\theta}\notin\mathbb{T}_{\theta}(x)}|\psi_{T_{\theta}}*\eta_{l}^{\vee}(x)|:=I_{1}(x)+I_{2}(x).$
(3.23)

Using (3.18) and Proposition 3.6, we bound

$|\psi_{T_{\theta}}*\eta_{l}^{\vee}(x)|\lesssim K^{-2}\delta^{-2}\cdot
CK\delta\cdot\delta
w_{T^{\prime}_{\theta}}(x)=CK^{-1}w_{T^{\prime}_{\theta}}(x),$ (3.24)

where $T^{\prime}_{\theta}:=T_{\theta}+B^{2}(0,K\delta)$ is a rectangle with
the same orientation and centre as $T_{\theta}$ and with dimensions $1\times
CK\delta$.

Using this and $w_{T^{\prime}_{\theta}}\lesssim 1$, we trivially bound

$I_{1}(x)\lesssim CK^{-1}\\#\mathbb{T}_{\theta}(x).$ (3.25)

To bound $I_{2}(x)$, we note that $\\{T^{\prime}_{\theta}\\}$ are at most
$O(K)$ overlapping. Since all centres of rectangles $T_{\theta}$ are on the
same straight line, using Proposition 3.3, we have

$I_{2}(x)\lesssim CK^{-1}\cdot KC^{-E+1}.$ (3.26)

Since $\\#\mathbb{T}_{\theta}(x)\geq 1$ for each $x\in A$, choosing
$C=K^{\frac{1}{E-1}}$ gives that

$|f_{\theta}*\eta^{\vee}_{l}(x)|\lesssim K^{-1+\frac{1}{E-1}},$ (3.27)

which is actually slightly weaker than (3.9), but it it turns out to be
sufficient for our purpose. Indeed, note that (3.11) becomes

$|f_{l}(x)|\lesssim C\\#\Theta K^{s-1}\lesssim\\#\Theta K^{\frac{s-1}{2}},$
(3.28)

in view of our choice of $E$ in (3.15). Thus the low part is still dominated
by the high part.

##  4 Discretization

In order to make some steps of the discretization of Theorem 1.2 more precise,
we will make use of Frostman measures in place of the Hausdorff content, since
$\mathcal{H}^{a}_{\infty}$ have the disadvantage of only being an outer
measure. See Definition 1.6 for the definition of Frostman measure.

By Frostman’s Lemma (see Theorem 1.7) we know that $A$ admits an $s$-Frostman
measure if and only if $\mathcal{H}^{s}(A)\in(0,\infty]$. Hence, $A$ admits an
$s$-Frostman measure $\nu_{s}$ for any $0<s<\text{dim}(A)$.

###  4.1 Lemma 1 and Lemma 2

######  Lemma 4.1 (Lemma 1 in [GGG+22]).

Let $\delta\in(0,1)$, $s\in(0,d]$ and $B\subseteq[0,1]^{d}$. Assume there is
an $s$-Frostman measure $\nu$ such that $\nu(B)=\kappa>0$. Then, there exists
a $\delta$-separated finite $(\delta,s)$-set $P\subseteq B$ with cardinality
$\\#P\gtrsim\rho_{s}(\nu)^{-1}\kappa\delta^{-s}$.

###### Proof of Lemma 4.1.

By losing at most a dimensional constant in $\gtrsim$, we may assume
$\delta=2^{-n}$ for some $n\in\mathbb{N}$. For each $0\leq m\leq n$, denote by
$\mathcal{D}_{m}$ the tiling of $[0,1]^{d}$ with axis parallel cubes of side
length $2^{-m}$ that intersect $B$. We may also assume the cubes are closed on
the left and open on the right for each coordinate; thus the tiling becomes
strictly disjoint.

We say a finite set $P\subseteq[0,1]^{d}$ satisfies the $s$-dimensional
condition at scale $r=2^{-m}$ if

$|P\cap Q|_{\delta}\leq 2(r/\delta)^{s},\quad\forall Q\in\mathcal{D}_{m}.$
(4.1)

The factor $2$ here is introduced just for technical treatment of rounding
errors. Note that if $P$ is assumed to be $\delta$-separated, then the above
is equivalent to

$\\#(P\cap Q)\leq 2(r/\delta)^{s},\quad\forall Q\in\mathcal{D}_{m}.$ (4.2)

Thus a $\delta$-separated set $P\subseteq[0,1]^{d}$ is a $(\delta,s)$-set if
and only if it satisfies the $s$-dimensional condition (4.2) above at every
dyadic scale $r\in[\delta,1]$.

We start at scale $2^{-n}$. Define $P_{n}$ by choosing one point from each
member of $\mathcal{D}_{n}$;

The set $P_{n}$ clearly satisfies the $s$-dimensional condition at scale
$2^{-n}$. Note $B\subseteq\cup\mathcal{D}_{n}$. Denote
$\mathcal{C}_{n}:=\mathcal{D}_{n}$.

We now consider the scale $2^{1-n}$. For each $Q_{n-1}\in\mathcal{D}_{n-1}$,
we say it is bad (at this level) if $Q_{n-1}\cap P_{n}$ does not satisfy the
$s$-dimensional condition at scale $2^{1-n}$. In this case, we remove points
from $Q_{n-1}\cap P_{n}$ until it just becomes good. Meanwhile, we remove from
$\mathcal{C}_{n}$ all cubes $Q_{n}\subseteq Q_{n-1}$, and add $Q_{n-1}$ to it.
Perform this for each bad cube $Q_{n-1}$, and call the resulting point set
$P_{n-1}\subseteq P_{n}$ and the resulting collection of cubes
$\mathcal{C}_{n-1}$. Note that $P_{n-1}$ satisfies the $s$-dimensional
condition at scales $2^{-n}$ and $2^{1-n}$, and also
$B\subseteq\cup\mathcal{C}_{n-1}$.

We now consider the scale $2^{2-n}$. For each $Q_{n-2}\in\mathcal{D}_{n-2}$,
if $Q_{n-2}\cap P_{n-1}$ does not satisfy the $s$-dimensional condition at
scale $2^{2-n}$, we remove points from $Q_{n-2}\cap P_{n-1}$ until it does so.
Also, replace all cubes $Q_{n-1}\subseteq Q_{n-2}$ from $\mathcal{C}_{n-1}$
and replace them with $Q_{n-2}$. Perform this for each bad cube $Q_{n-2}$, and
call the resulting point set $P_{n-2}\subseteq P_{n-1}$ and collection of
cubes $\mathcal{C}_{n-2}$. Note that $P_{n-2}$ satisfies the $s$-dimensional
condition at scales $2^{-n}$, $2^{1-n}$ and $2^{2-n}$. Also,
$B\subseteq\cup\mathcal{C}_{n-2}$.

We run the above algorithm exactly $n$ times until we arrive at a point set
$P=P_{0}$ and a collection of cubes $\mathcal{C}_{0}$. By construction, $P$ is
a $(\delta,s)$-set. Also, $B\subseteq\cup\mathcal{C}_{0}$.

It remains to prove that $\\#P\gtrsim\rho_{s}(\nu)^{-1}\kappa\delta^{-s}$. For
each $0\leq m\leq n$, we denote by $\mathcal{C}^{m}$ the collection of cubes
in $\mathcal{C}_{0}$ that has length $2^{-m}$. Using the fact that $\nu$ is an
$s$-Frostman measure, we have

$\sum_{m=0}^{n}2^{-ms}\rho_{s}(\nu)\\#(\mathcal{C}^{m})\gtrsim\sum_{m=0}^{n}\nu(\cup\mathcal{C}^{m})\geq\nu(\cup\mathcal{C}_{0})\geq\nu(B)=\kappa.$
(4.3)

On the other hand, recall that for each bad cube $Q_{m}$ at each scale
$2^{-m}$, $0\leq m\leq n-1$, we removed points from $P_{m+1}\cap Q_{m}$ until
$Q_{m}$ just becomes good. Also, if $Q_{m}\in\mathcal{C}^{m}$, this means that
$Q_{m}$ is left untouched by the algorithm at subsequent larger scales. Thus

$\\#(P\cap Q_{m})=\\#(P_{m}\cap Q_{m})>2(2^{-m}/2^{-n})^{s}-1\geq 2^{(n-m)s}.$
(4.4)

For $Q_{n}\in\mathcal{C}^{n}$, by construction we have $\\#(P\cap
Q_{n})=1=2^{(n-n)s}$. Thus we have

$\begin{split}\\#P=\sum_{m=0}^{n}\\#(P\cap
Q_{m})\\#(\mathcal{C}^{m})\geq&\sum_{m=0}^{n}2^{(n-m)s}\\#(\mathcal{C}^{m})\\\
\gtrsim&2^{ns}\rho_{s}(\nu)^{-1}\kappa=\rho_{s}(\nu)^{-1}\kappa\delta^{-s},\end{split}$
(4.5)

where we have used (4.3). ∎

We will now move to the proof of Lemma 2 in [GGG+22]. We introduce the
following notation: let $\mathcal{D}$ be the collection of dyadic squares in
$[0,1]^{2}$ and let $\mathcal{D}_{2^{-j}}$ denote the collection of dyadic
squares with side-lengths $2^{-j}$.

######  Lemma 4.2 (Lemma 2 in [GGG+22]).

Fix $\epsilon>0$ and suppose $\dim(X)<s$. Then there exists a family of
collections of dyadic squares $\\{\mathcal{C}_{2^{-k}}\\}_{k=1}^{\infty}$ such
that $\mathcal{C}_{2^{-k}}\subset\mathcal{D}_{2^{-k}}$ such that the following
properties hold.

  1. 1.

$\sum_{k\geq 0}\sum_{D\in\mathcal{C}_{2^{-k}}}r(D)^{s}<\epsilon$

  2. 2.

$X\subseteq\bigcup_{k\geq 0}\bigcup_{D\in\mathcal{C}_{2^{-k}}}D$

  3. 3.

For any $l<k$ and any $D\in\mathcal{D}_{2^{-l}}$ we have that

$\\#\\{D^{\prime}\in\mathcal{C}_{2^{-k}}:D^{\prime}\subset D\\}\leq
2^{(k-l)s}.$

This is called the $s$-dimensional condition.

Remark: The first two conditions can be deduced immediately from the condition
that $\dim(X)<s$. The third condition corresponds to the construction of a
$(\delta,s)$ set (see Lemma 1) where in this particular case $\delta=2^{-k}$.

###### Proof.

We consider the collection $\Pi$ of all coverings $\mathcal{C}$ of $X$ by
dyadic squares which also satisfy conditions $(1)$ and $(2)$ in Lemma 4.2. For
each such covering, $\mathcal{C}$, we may assume (by the structure of dyadic
squares) that the dyadic squares in $\mathcal{C}$ are disjoint. We define a
partial order $\prec$ over the collection of all such coverings as follows: If
$\mathcal{C}=\mathcal{C}^{\prime}$ we declare that
$\mathcal{C}\prec\mathcal{C}^{\prime}$ and also that
$\mathcal{C}^{\prime}\prec\mathcal{C}$. If
$\mathcal{C}\neq\mathcal{C}^{\prime}$, we say
$\mathcal{C}\prec\mathcal{C}^{\prime}$ when
$\mathcal{C}\neq\mathcal{C}^{\prime}$ if the following two conditions hold:

  1. 1.

There exists (a unique) $k_{0}>0$ such that
$\mathcal{C}\cap\mathcal{D}_{2^{-k}}=\mathcal{C}^{\prime}\cap\mathcal{D}_{2^{-k}}$
for $k<k_{0}$ and
$\mathcal{C}\cap\mathcal{D}_{2^{-k_{0}}}\subsetneq\mathcal{C}^{\prime}\cap\mathcal{D}_{2^{-k_{0}}}$.

  2. 2.

For each $x\in X$, the dyadic square contained in $\mathcal{C}$ that contains
$x$ must be a subset of of the dyadic square in $\mathcal{C}^{\prime}$ that
contains $x$.

We check that $\prec$ determines a partial order on $\Pi$. It is easy to see
that $\prec$ is reflexive and antisymmetric. We now that that it is
transitive. Suppose, $\mathcal{C}\prec\mathcal{C}^{\prime}$ and
$\mathcal{C}^{\prime}\prec\mathcal{C}^{\prime\prime}$. We must show that
$\mathcal{C}\prec\mathcal{C}^{\prime\prime}$. Note that if either
$\mathcal{C}=\mathcal{C}^{\prime}$ or
$\mathcal{C}^{\prime}=\mathcal{C}^{\prime\prime}$ then this is immediate.
Thus, let use assume that
$\mathcal{C}\neq\mathcal{C}^{\prime}\neq\mathcal{C}^{\prime\prime}$. By
condition (a) we see that there exists $k_{1}>1$ such that
$\mathcal{C}\cap\mathcal{D}_{2^{-k}}=\mathcal{C}\cap\mathcal{D}_{2^{-k}}$ for
$k<k_{1}$ and
$\mathcal{C}\cap\mathcal{D}_{2^{-k_{1}}}\subsetneq\mathcal{C}\cap\mathcal{D}_{2^{-k_{1}}}$
and similarly there exists $k_{2}>0$ such that
$\mathcal{C}\cap\mathcal{D}_{2^{-k}}=\mathcal{C}^{\prime\prime}\cap\mathcal{D}_{2^{-k}}$
for $k<k_{2}$ and
$\mathcal{C}\cap\mathcal{D}_{2^{-k_{2}}}\subsetneq\mathcal{C}\cap\mathcal{D}_{2^{-k_{2}}}$.
Let $k_{0}=\min\\{k_{1},k_{2}\\}$. Note that for $k<k_{0}$ we have that
$\mathcal{C}\cap\mathcal{D}_{2^{-k}}=\mathcal{C}^{\prime\prime}\cap\mathcal{D}_{2^{-k}}$.
Moreover we have that either
$\mathcal{C}\cap\mathcal{D}_{2^{-k_{0}}}\subsetneq\mathcal{C}^{\prime}\cap\mathcal{D}_{2^{-k_{0}}}$
or
$\mathcal{C}^{\prime}\cap\mathcal{D}_{2^{-k_{0}}}\subsetneq\mathcal{C}^{\prime\prime}\cap\mathcal{D}_{2^{-k_{0}}}$.
AS a consequence we have that
$\mathcal{C}\cap\mathcal{D}_{2}^{-k_{0}}\subsetneq\mathcal{C}^{\prime\prime}\cap\mathcal{D}_{2^{-k_{0}}}$.
This proves condition (a). If $x\in X$, then since the
$\mathcal{C}\prec\mathcal{C}^{\prime}\prec\mathcal{C}^{\prime\prime}$, the
dyadic cube that contains $x$ in $\mathcal{C}^{\prime\prime}$ contains the
dyadic cube that contains $x$ in $\mathcal{C}^{\prime}$ which in turn contains
the dyadic cube that contains $x$ in $\mathcal{C}$. Thus condition (b) is also
satisfied and thus $\mathcal{C}\prec\mathcal{C}^{\prime\prime}$.

Suppose $\mathcal{C}$ is a covering that is maximal with respect to the
ordering $\prec$. We claim that such a covering must satisfy condition (3) in
Lemma 4.2. Indeed, suppose in order to obtain a contradiction that (3) does
not hold: then there exist $l$ and $k$ with $l<k$ such that for some square
$D\in\mathcal{D}_{2^{-l}}$ we have the following:

$\\#\\{D^{\prime}\in\mathcal{C}_{2^{-k}}:D^{\prime}\subset D\\}>2^{(k-l)s}$
(4.6)

We construct a collection $\mathcal{C}^{\prime}\in\Pi$ such that
$\mathcal{C}\prec\mathcal{C}^{\prime}$. We construct $\mathcal{C^{\prime}}$ by
removing all elements in
$\\{D^{\prime}\in\mathcal{C}_{2^{-k}}:D^{\prime}\subset D\\}$ from
$\mathcal{C}$ and replace this with $D$ in $\mathcal{C}^{\prime}$. It is easy
to see that $\mathcal{C}\prec\mathcal{C^{\prime}}$. Moreover if
$\mathcal{C^{\prime}}$ satisfies condition (2) then
$\mathcal{C}\prec\mathcal{C}^{\prime}$. Thus we show that

$\sum_{D\in\mathcal{C}^{\prime}}r(D)^{s}\leq\sum_{D\in\mathcal{C}}r(D)^{s}\leq\epsilon$

After cancelling like terms in the previous equation and using the condition
in (4.6), this is reduced to showing that

$r(D)^{s}\leq\sum_{D^{\prime}\in\mathcal{C}:D^{\prime}\subset
D}r(D^{\prime})^{s}\iff\sum 2^{-2ls}\leq 2^{(k-l)s}2^{-2ls}$

which is true since $l<k$.

Thus in order to prove this lemma it suffices to find such a maximal element.
By Zorn’s lemma, this is reduced to the task of showing that any totally
ordered ascending chain has a maximal element. To that end let
$(\mathcal{C}_{j})_{j\in J}$ be such a chain. We claim that

$\mathcal{C}=\bigcap_{j\in
J}\bigcup_{\mathcal{C}_{i}\succeq\mathcal{C}_{j}}\mathcal{C}_{i}$

is our required maximal element. First we note that $\mathcal{C}$ covers $X$.
Indeed, if $x\in X$, let $D$ be the largest dyadic cube that contains $x$ in
$\bigcup_{j\in J}\mathcal{C}_{j}$. If $D\in\mathcal{C}_{k}$ and if
$\mathcal{C}_{k}\prec\mathcal{C}_{l}$, then since the dyadic square in
$\mathcal{C}_{l}$ contains that dyadic square in $\mathcal{C}_{k}$ which
contains $x$ (which is $D$). Thus $D\in\mathcal{C}$ since
$D\in\bigcup_{\mathcal{C}_{i}\succeq\mathcal{C}_{j}}\mathcal{C}_{i}$ for all
$j\in J$.

We note the following: For each $k\in\mathbb{N}$ we can choose
$\mathcal{C}_{j}$ for some $j\in J$ such that
$\mathcal{C}_{i}\cap\mathcal{D}_{2^{-k}}=\mathcal{C}_{j}\cap\mathcal{D}_{2^{-k}}$
whenever $\mathcal{C}_{i}\succeq\mathcal{C}_{j}$. Indeed suppose this is not
true, then we can find a sequence $\\{j_{1},j_{2},...\\}\in J$ such that
$\mathcal{C}_{j_{a}}\prec\mathcal{C}_{j_{b}}$ and
$\mathcal{C}_{j_{a}}\cap\mathcal{D}_{2^{-k}}\subset\mathcal{C}_{j_{b}}\cap\mathcal{D}_{2^{-k}}$
whenever $a<b$ and where the containment $\subset$ is a strict containment.
However since $|\mathcal{D}_{2^{-k}}|<\infty$, it is impossible for these
containments to always be strict and thus we can find $j\in J$ such that
$\mathcal{C}_{i}\cap\mathcal{D}_{2^{-k}}=\mathcal{C}_{j}\cap\mathcal{D}_{2^{-k}}$
whenever $\mathcal{C}_{i}\succeq\mathcal{C}_{j}$. We can similarly show (as a
consequence of the fact that $\\{\mathcal{C}_{i}\\}_{i\in J}$ is totally
ordered that given any $K\in\mathbb{N}$ we can find $j\in J$ such that
$\mathcal{C}_{i}\cap\mathcal{D}_{2^{-k}}=\mathcal{C}_{j}\cap\mathcal{D}_{2^{-k}}$
for $0\leq k\leq K$ whenever $\mathcal{C}_{i}\succeq\mathcal{C}_{j}$. Fix this
$j$. Then we have that (since
$\mathcal{C}\subseteq\bigcup_{\mathcal{C}_{i}\succeq\mathcal{C}_{j}}\mathcal{C}_{i}$)

$\sum_{k=0}^{K}\sum_{D\in\mathcal{C}\cap\mathcal{D}_{2^{-k}}}r(D)^{s}\leq\sum_{k=0}^{K}\sum_{D\in\mathcal{C}_{j}\cap\mathcal{D}_{2^{-k}}}r(D)^{s}\leq\epsilon$

Sending $K\rightarrow\infty$ shows that $\mathcal{C}$ satisfies condition (3).
We finally check that $\mathcal{C}_{i}\prec\mathcal{C}$ for $\mathcal{C}_{i}$.
For each $x\in X$, notice that if the dyadic square in $\mathcal{C}$ that
contains $x$ is the largest square containing $x$ in
$\bigcup_{j}\mathcal{C}_{j}$, thus the dyadic square in $\mathcal{C}$ that
contains $x$ must contain the dyadic square in $\mathcal{C}_{i}$ that contains
$x$ for any $i$. This verifies condition (b). For condition (a) we proceed to
prove this by contradiction. Suppose $\mathcal{C}$ and $\mathcal{C}_{i}$ are
distinct collections. If we have that $\mathcal{C}_{i}\not\prec\mathcal{C}$,
then we must have that
$\mathcal{D}_{{2}^{-k}}\cap\mathcal{C}_{i}=\mathcal{D}_{2^{-k}}\cap\mathcal{C}$
for $k<k_{0}$ and
$\mathcal{C}_{i}\cap\mathcal{D}_{2}^{-k_{0}}\not\subseteq\mathcal{C}\cap\mathcal{D}_{2}^{-k_{0}}$
for some $k_{0}>0$. Choose $D$ such that
$D\in\mathcal{C}_{i}\cap\mathcal{D}_{2^{-k_{0}}}$ but
$D\not\in\mathcal{C}\cap\mathcal{D}_{2^{-k_{0}}}$. Let $x\in X$ be such that
$x\in D$. Then $x\in\tilde{D}$ where $\tilde{D}\in\mathcal{C}\cap D_{2^{-k}}$
for some $k<k_{0}$ (by property (b) just established). However, this means
that $\tilde{D}\in\mathcal{C}$ (since the intersections of $\mathcal{C}$ and
$\mathcal{C}^{\prime}$ with dyadic squares upto scale $2^{-k_{0}}$ are the
same.), which is impossible since $D\cap\tilde{D}\neq\emptyset$. ∎

Remark: Lemma 3 in the paper requires a small extra condition to this lemma as
presently stated, where we require our collection to be of the form
$\\{\mathcal{C}_{2^{-k}}\\}_{k=k_{0}}^{\infty}$ for some fixed $k_{0}$ instead
of $\\{\mathcal{C}_{2^{-k}}\\}_{k=1}^{\infty}$. This can clearly be done, by
repeating the above procedure and noting that since $\dim(X)<s$ we have that
$\mathcal{H}_{\delta}^{s}(X)=0$ for arbitrarily small $\delta$ (where we
choose $\delta=2^{-k_{0}}$).

###  4.2 Discretization of Marstrand’s Theorem

We first restate Marstrand’s projection theorem.

######  Theorem 4.3 (Marstrand’s projection theorem).

For each $\theta\in[0,\pi]$, define
$L_{\theta}=\\{x\in\mathbb{R}^{2}\colon\textnormal{arg}(x)=\theta\\}$ and let
$p_{\theta}:\mathbb{R}^{2}\rightarrow L_{\theta}$ be the projection. Suppose
$A\subset\mathbb{R}^{2}$ is a Borel set, then we have
$\text{dim}(p_{\theta}(A))=\min\\{1,\textnormal{dim}(A)\\}$ for a.e.
$\theta\in[0,\pi]$.

The following easy lemma will be useful for this and the next subsection.

######  Lemma 4.4.

Let $A\subset\mathbb{R}^{n}$ and $p:A\rightarrow\mathbb{R}^{m}$ be a Lipschitz
function. Then $\textnormal{dim}(p(A))\leq\textnormal{dim}(A)$.

###### Proof.

It enough to show that for any $s>\textnormal{dim}(A)$, one has
$\mathcal{H}^{s}(p(A))=0$. Fix such an $s$. Since $\mathcal{H}^{s}(A)=0$ and
$\mathcal{H}^{s}_{\delta}(A)$ is monotone decreasing in $\delta$ , one has
$\mathcal{H}^{s}_{\delta}(A)=0$ for all $\delta$. Then for every
$\varepsilon>0$ and every $\delta>0$, there exists a cover
$\\{U_{i}\\}_{i\in\mathbb{N}}$ for $A$ with
$\sum_{i}(\textnormal{diam}(U_{i}))^{s}<\varepsilon$, and
$\textnormal{diam}(U_{i})\leq\delta$. That gives a cover
$p(A)\subset\cup_{i}p(U_{i})$ satisfying that $\textnormal{diam}(p(U_{i}))\leq
M\textnormal{diam}(U_{i})$ and $\sum_{i}(\textnormal{diam}\,p(U_{i}))^{s}\leq
M^{s}\varepsilon$ where $M$ is the Lipchitz constant of $p$. That implies
$\mathcal{H}_{M\delta}^{s}(p(A))\leq M^{s}\varepsilon$. Since $\varepsilon>0$
is arbitrary, $\mathcal{H}_{M\delta}^{s}(p(A))=0$. We are done after sending
$\delta$ to $0$.

∎

From the lemma above and since $p_{\theta}(A)\subset L_{\theta}$,
$\textnormal{dim}(L_{\theta})=1$, we know that for every $\theta\in[0,\pi]$,
$\textnormal{dim}(p_{\theta}(A))\leq\min\\{1,\textnormal{dim}(A)\\}$.

In the sequel we explain how one can go about reducing Marstrand Theorem to
the discretized version stated in Proposition 3.1.

###### Proof that Proposition 3.1 implies Theorem 4.3.

Let $\alpha=\textnormal{dim}(A)$ and $s_{A}=\min\\{1,\alpha\\}$. The case
$s_{A}=0$ is trivial, so we assume $s_{A}>0$.

Because for any $\theta\in[0,1]$ it holds that
$\textnormal{dim}(p_{\theta}(A))\leq s_{A}$, if we show that the exceptional
set

$E_{s_{A}}=\\{\theta\in[0,1]\colon\textnormal{dim}(p_{\theta}(A))<s_{A}\\}$
(4.7)

has zero Lebesgue measure, then $\textnormal{dim}(p_{\theta}(A))=s_{A}$ for
a.e. $\theta\in[0,1]$, and we are done.

Moreover, since for $s_{n}:=s_{A}-s_{A}/n$,
$E_{s_{A}}=\cup_{n=1}^{\infty}E_{s_{n}}$, it is enough to show that

$\mathcal{L}^{1}(E_{s_{n}})=0,\text{ for all }n\in\mathbb{N}.$

Assume by contradiction that $\mathcal{L}^{1}(E_{s_{n}})>0$, then
$\mathcal{L}^{1}$ is an $1$-Frostman measure in $E_{s_{n}}$ and from Lemma
4.1, for any $\delta\in(0,1)$ there exists a $\delta$-separated
$(\delta,1)$-set $P\subset E_{s_{n}}$ of cardinality $\\#P\gtrsim\delta^{-1}$.
For $a<s_{A}$, one can reduce things to the discretization as stated in
Proposition 3.1 with $\\#\Theta\gtrsim_{\varepsilon}\delta^{-1+\varepsilon}$
by following the same argument in the later discretization (see Section 4.3).
After the discretization, the application of Proposition 3.1 will give for
every $\varepsilon_{0}>0$ that

$\delta^{-1+\varepsilon}\lesssim_{\varepsilon,s}\delta^{-1+a-s_{n}}$

for some $j$ with $\delta=2^{-j}\leq\varepsilon_{0}$. Send $\varepsilon_{0}$
to zero to conclude that $a-s_{n}\leq\varepsilon$, for any $\varepsilon$. So,
$a\leq s_{n}<s_{A}$ which gives a contradiction when we take $a$ sufficiently
close to $s_{A}$ ($a>s_{n}$ for example). ∎

###  4.3 Reduction of Theorem 1 to the discretized version in Theorem 4

Now we return to the discretization of the main Theorem 1.2. We first recall
the statement of Theorem 4 in [GGG+22].

######  Theorem 4.5 (Theorem 4 in [GGG+22]).

Fix $0<s<2$. For each $\varepsilon>0$, there exists $C_{s,\varepsilon}>0$ so
that the following holds. Let $\delta>0$. Let $H\subseteq B^{3}(0,1)$ be a
union of $h$ many disjoint $\delta$-balls. Let $\Theta$ be a
$\delta$-separated $(\delta,t)$-subset of $[0,1]$ of cardinality
$\\#\Theta\gtrapprox\delta^{-t}$, such that for each $\theta\in\Theta$, we
have a collection of $\delta\times\delta\times 1$-tubes $\mathbb{T}_{\theta}$
pointing at the direction $\gamma(\theta)$ and satisfying the following
conditions:

  1. 1.

$\mathbb{T}_{\theta}$ satisfies the $s$-dimensional condition: for any
$r\in[\delta,1]$ and any ball $B_{r}\subseteq\mathbb{R}^{3}$ of radius $r$, we
have

$\\#\\{T\in\mathbb{T}_{\theta}:T\cap
B_{r}\neq\varnothing\\}\lesssim(r/\delta)^{s}.$ (4.8)

In particular, $\\#\mathbb{T}_{\theta}\lesssim\delta^{-s}$.

  2. 2.

Each $\delta$-ball contained in $H$ intersects $\gtrapprox\\#\Theta$ many
tubes from $\cup_{\theta\in\Theta}\mathbb{T}_{\theta}$.

Then

$\delta^{-t}h\leq C_{s,\varepsilon}\delta^{-1-s-\varepsilon}.$ (4.9)

Let us now discuss the details of how Theorem 4.5 implies Theorem 1.2. Assume
$A\subset B^{3}(0,1)$ and let $0<s<\min\\{2,\alpha\\}$ where
$\alpha=\textnormal{dim}(A)$. Recall the definition of the exceptional set

$E_{s}=\\{\theta\in[0,1]\colon\textnormal{dim}(\pi_{\theta}(A))<s\\}.$

Observe that can reduce ourselves to checking that Theorem 4.5 implies Theorem
1.2 for $s>\max\\{0,\alpha-1\\}$ because for the remaining case $\alpha>1$ and
$s\leq\alpha-1$ what Theorem 1.2 claims is that $\text{dim}(E_{s})=0$, and
that will follow from
$\text{dim}(E_{s})\leq\text{dim}(E_{\alpha-1+\varepsilon})\leq\varepsilon$,
for all $\varepsilon>0$.

Fix $a<\text{dim}(A)$, and $t<\text{dim}(E_{s})$. We need to check that

$a+t\leq 1+s$ (4.10)

then sending $a\rightarrow\text{dim}(A)$ and $t\rightarrow\text{dim}(E_{s})$
we get Theorem 1.2.

Let $\nu_{A}$ be an $a$-Frostman measure in $A$ and $\nu_{E_{s}}$ a
$t$-Frostman measure in $E_{s}$.

For any $\theta$ in the exceptional set $E_{s}$, Lemma 4.2 guarantees one can
find a nice cover $\mathbb{D}_{\theta}=\\{D\\}$ for
$\pi_{\theta}(A)\subset\gamma(\theta)^{\perp}\cong\mathbb{R}^{2}$ comprised of
dyadic cubes with small associated $s$-dimensional content, in the sense that
$\sum_{D\in\mathbb{D}_{\theta}}r(D)^{s}<1$, and the extra $s$-dimensional
condition for each generation. That is, for any $j\in\mathbb{N}$ and any ball
$B_{r}^{2}\subset\gamma(\theta)^{\perp}$, with $2^{-j}\leq r\leq 1$,

$\\#\\{D\in\mathbb{D}_{\theta,j}\colon D\subset
B_{r}^{2}\\}\lesssim\left(\frac{r}{2^{-j}}\right)^{s}$ (4.11)

where $\mathbb{D}_{\theta,j}=\\{D\in\mathbb{D}_{\theta}\colon r(D)=2^{-j}\\}$.

Let $\mathbb{T}_{\theta}:=\\{T=\pi_{\theta}^{-1}(D)\cap B^{3}(0,1)\colon
D\in\mathbb{D}_{\theta}\\}$. Split $\mathbb{T}_{\theta}$ in generations
$\mathbb{T}_{\theta}=\cup_{j}\mathbb{T}_{\theta,j}$, where for each $j$,
$\mathbb{T}_{\theta,j}$ corresponds to the pre-images by ${\pi}_{\theta}$ of
dyadic cubes $D\in\mathbb{D}_{\theta,j}$. Similar condition is then inherited
by the tubes in the collection $\mathbb{T}_{\theta,j}$. Namely,

$\\#\\{T\in\mathbb{T}_{\theta,j}\colon T\cap
B^{3}(x,r)\neq\emptyset\\}\lesssim\left(\frac{r}{2^{-j}}\right)^{s}\text{ for
}2^{-j}\leq r\leq 1.$ (4.12)

Given $\varepsilon_{0}>0$, as remarked right after Lemma 4.2 we can assume
that we only use sufficiently small cubes, that is, $2^{-j}<\varepsilon_{0}$.
Observe that for each $j>|\log_{2}(\varepsilon_{0})|$ the collection
$\mathbb{D}_{\theta,j}=\\{D\in\mathbb{D}_{\theta}\colon r(D)=2^{-j}\\}$ is
pairwise disjoint and the same is then true for the $2^{-j}\times 2^{-j}\times
1$ tubes in $\mathbb{T}_{\theta,j}$. Also, by construction the tubes in
$\mathbb{T}_{\theta}$ point in the direction $\gamma(\theta)$ and
$A\subset\cup_{j>|\log_{2}(\varepsilon_{0})|}\cup_{T\in\mathbb{T}_{\theta,j}}T$.

Here is an overview of the discretization argument: Recall we fixed $\nu_{A}$
$a$-Frostman measure in $A$ and $\nu_{E_{s}}$ a $t$-Frostman measure in
$E_{s}$.

The key will be to find a suitable generation $j$ (among the $j$ such that
$2^{-j}<\varepsilon_{0}$) for which there is a positive $a$-dimensional
content worth of points in $A$, say $A_{g}\subset A$ with
$\nu_{A}(A_{g})\gtrsim\frac{1}{j^{2}}>0$, which are covered by
$\cup_{T\in\mathbb{T}_{\theta,j}}T$ for a big fraction of the angles $\theta$
in a suitable $\delta$-separated set $\Theta$ of angles in $E_{s}$. $\Theta$
will show up by application of Lemma 4.1 for a subset $E_{s,j}\subset E_{s}$
to be defined later satisfying $\nu_{E_{s}}(E_{s,j})\gtrsim\frac{1}{j^{2}}>0$,
and we will know that $\\#\Theta\gtrsim\frac{1}{j^{2}}2^{jt}$. By applying
Lemma 4.1 to the set $A_{g}$ we get a $2^{-j}$-separated set of cardinality
$h$ with $h\gtrsim\nu_{A}(A_{g})(2^{-j})^{-a}\gtrsim\frac{1}{j^{2}}2^{aj}$,
and we use these collection of $2^{-j}$-separated points as centers of the
$2^{-j}$ balls that will comprise the $H$ that we need for application of
Theorem 4.5. That theorem will then give

$(2^{-j})^{-t}h\leq C_{s,\varepsilon}(2^{-j})^{-1-s-\varepsilon}$ (4.13)

which will turn into $\frac{1}{j^{2}}2^{(a+t)j}\leq
C_{s,\varepsilon}2^{(s+1+\varepsilon)j}$.

Sending $\varepsilon_{0}$ to zero (so $j$ to infinity) we get $a+t\leq
1+s+\varepsilon$, for all $\varepsilon>0$. Therefore, $a+t\leq 1+s$.

We now give the details for the ingredients used in the overview above. For
each $\theta$, since
$A\subset\cup_{j>|\log_{2}\varepsilon_{0}|}(\cup_{T\in\mathbb{T}_{\theta,j}}T)$,
by pigeonholing, and by sub-additivity of the measure $\nu_{A}$, there exists
a $j(\theta)>|\log_{2}(\varepsilon_{0})|$, such that

$\nu_{A}(A\cap(\cup_{T\in\mathbb{T}_{\theta,j(\theta)}}T))\geq\frac{1}{10j(\theta)^{2}}\nu_{A}(A)=\frac{1}{10j(\theta)^{2}}.$
(4.14)

Define for each $j>|\log_{2}(\varepsilon_{0})|$

$E_{s,j}=\\{\theta\in E_{s}\colon j(\theta)=j\\}.$

This gives a partition $E_{s}=\sqcup_{j>|\log_{2}\varepsilon_{0}|}E_{s,j}$.
Again by pigeonholing there exists $j>|\log_{2}\varepsilon_{0}|$ with

$\nu_{E_{s}}(E_{s,j})\gtrsim\frac{1}{j^{2}}.$

Set $\delta:=2^{-j}$ for this suitable fixed $j$. An application of Lemma 4.1
guarantees the existence of a $\delta$-separated set of angles $\Theta\subset
E_{s,j}$ such that $\Theta$ is a $(\delta,t)$-set and
$\\#\Theta\gtrsim(\log\delta^{-1})^{-2}\delta^{-t}$.

Consider

$S:=\\{(x,\theta)\in A\times\Theta\colon
x\in\cup_{T\in\mathbb{T}_{\theta,j}}T\\}$ (4.15)

and the sections

$S_{x}=\\{\theta\in\Theta\colon(x,\theta)\in S\\},\,S_{\theta}=\\{x\in
A\colon(x,\theta)\in S\\}.$

In words, $S_{x}$ is comprised of all the angles $\theta\in\Theta$ such that
$x$ is covered by $\cup_{T\in\mathbb{T}_{\theta,j}}T$, while $S_{\theta}$ is
comprised of the points in $A$ covered by union of tubes in
$\mathbb{T}_{\theta,j}$.

Let $\mu$ be the counting measure in $\Theta$.

By inequality (4.14) and since $\Theta\subset E_{s,j}$, we know that

$\begin{split}(\nu_{A}\times\mu)(S)=&\int_{\Theta}\left(\int_{A}\chi_{S}(x,\theta)d\nu_{A}(x)\right)d\mu(\theta)\\\
=&\int_{\Theta}\nu_{A}(A\cap\cup_{T\in\mathbb{T}_{\theta,j}}T)d\mu(\theta)\\\
\geq&\int_{\Theta}\frac{1}{10j^{2}}d\mu(\theta)=\frac{1}{10j^{2}}\mu(\Theta).\end{split}$
(4.16)

Observe that by Fubini,

$\begin{split}(\nu_{A}\times\mu)&(\\{(x,\theta)\in
S\colon\mu(S_{x})\leq\frac{1}{20j^{2}}\mu(\Theta)\\})\\\
=&\int_{A}\chi_{\mu(S_{x})\leq\frac{1}{20j^{2}}\mu(\Theta)}(x)\left(\int_{\Theta}\chi_{S}(x,\theta)d\mu(\theta)\right)d\nu_{A}(x)\\\
=&\int_{A}\chi_{\mu(S_{x})\leq\frac{1}{20j^{2}}\mu(\Theta)}(x)\mu(S_{x})d\nu_{A}(x)\\\
&\leq\frac{1}{20j^{2}}\nu_{A}(A)\mu(\Theta)=\frac{1}{20j^{2}}\mu(\Theta).\end{split}$
(4.17)

Therefore,

$\begin{split}(\nu_{A}\times\mu)&(\\{(x,\theta)\in
S\colon\mu(S_{x})>\frac{1}{20j^{2}}\mu(\Theta)\\})\\\
\geq&(\nu_{A}\times\mu)(S)-(\nu_{A}\times\mu)(\\{(x,\theta)\in
S\colon\mu(S_{x})\leq\frac{1}{20j^{2}}\mu(\Theta)\\})\\\
\geq&\left(\frac{1}{10j^{2}}-\frac{1}{20j^{2}}\right)\mu(\Theta)=\frac{1}{20j^{2}}\mu(\Theta).\end{split}$
(4.18)

Take $A_{g}=\\{x\in A\colon\mu(S_{x})>\frac{1}{20j^{2}}\mu(\Theta)\\}$, and
notice that

$\displaystyle(\nu_{A}\times\mu)(\\{(x,\theta)\in
S\colon\mu(S_{x})>\frac{1}{20j^{2}}\mu(\Theta)\\})=$
$\displaystyle(\nu_{A}\times\mu)(\\{(x,\theta)\in S\colon x\in A_{g}\\})$
$\displaystyle\leq$ $\displaystyle\mu(\Theta)\nu_{A}(A_{g}).$

Hence, inequality (4.18) implies

$\nu_{A}(A_{g})\geq\frac{1}{20j^{2}}.$

Then, since

$\nu_{A}(A_{g})\gtrsim\frac{1}{j^{2}}=(\log\delta^{-1})^{-2},$

an application of Lemma 4.1 gives a $\delta$-separated set $C_{g}\subset
A_{g}$ which is a $(\delta,a)$-set and cardinality
$\gtrsim(\log\delta^{-1})^{-2}\delta^{-a}$.

Define $H$ as the disjoint union of the $\delta$ balls centered at the points
in $C_{g}$.

Observe that $H$ and $\Theta$ will satisfy the hypothesis of Theorem 4.5.
Indeed, the $s$-dimensional condition was obtained by construction in (4.12)
and the center of which $2^{-j}$-ball contained in $H$ lives in $A_{g}$, so it
intersects at least $\frac{1}{20j^{2}}\\#\Theta$ many tubes in
$\cup_{\theta\in\Theta}\mathbb{T}_{\theta,j}$. Applying Theorem 4.5, we thus
arrived at the desired inequality (4.13) stated before.

##  5 Proof of Theorem 5

###  5.1 The decoupling machinery

Let $\gamma$ be a non-degenerate curve as discussed in the introduction. We
first state in full detail the main decoupling theorem we will be using here.
For completeness, the proof will be given in Section 9.3.

######  Theorem 5.1 (Bourgain-Demeter decoupling for the cone).

Let $\delta\in 4^{-\mathbb{N}}$ and $\Omega$ be the $\delta^{1/2}$-lattice in
$[0,1]$. Given $C,K\geq 1$, consider the collection of planks
$\\{A_{\omega,\delta^{1/2}}:\omega\in\Omega\\}$, where

$A_{\omega,\delta^{1/2}}:=\left\\{\sum_{i=1}^{3}\xi_{i}\mathbf{e}_{i}(\omega):|\xi_{1}|\leq
C\delta,|\xi_{2}|\leq C\delta^{1/2},C^{-1}K^{-1}\leq\xi_{3}\leq C\right\\}.$
(5.1)

Then we have

  1. 1.

If $C$ is large enough (depending on $\gamma$ only), the collection
$\\{A_{\omega,\delta^{1/2}}:\theta\in\Omega\\}$ covers the
$\delta$-neighbourhood of the truncated cone

$\Gamma_{K^{-1}}:=\\{r\mathbf{e}_{3}(\omega):\omega\in[0,1],r\in[K^{-1},1]\\}.$
(5.2)

Moreover, the collection $\\{A_{\omega,\delta^{1/2}}:\omega\in\Omega\\}$ is at
most $C^{\prime}K^{2}$-overlapping, where $C^{\prime}$ depends on $\gamma,C$
but not $K,\delta$.

  2. 2.

The following Fourier decoupling inequality holds: for each $2\leq p\leq 6$,
every family of Schwartz functions $\\{f_{\omega}:\omega\in\Omega\\}$ where
each $f_{\omega}$ has Fourier support in $A_{\omega,\delta^{1/2}}$, we have

$\left\|\sum_{\omega\in\Omega}f_{\omega}\right\|_{L^{p}(\mathbb{R}^{3})}\leq
C_{\varepsilon}K^{10}\delta^{-\varepsilon}\norm*{\norm*{f_{\omega}}_{L^{p}(\mathbb{R}^{3})}}_{\ell^{2}(\omega\in\Omega)},\quad\forall\varepsilon>0,$
(5.3)

where the constant $C_{\varepsilon}$ depends on $\varepsilon,C,p$ but not
$K,\delta$ or the test functions $f_{\omega}$.

We will also use the following flat decoupling inequality. To avoid technical
issues, we just state a version that is sufficient for our use here. We
include in the appendix a proof of this inequality.

######  Theorem 5.2 (Flat decoupling).

In $\mathbb{R}^{3}$, let $\mathbb{T}$ be a finite collection of tubes with
possibly different dimensions and orientations. Let $C\geq 1$ and assume the
collection of $C$-concentrically enlarged tubes $\\{CT:T\in\mathbb{T}\\}$ is
at most $K$-overlapping. Then we have the following flat decoupling
inequality: for every $2\leq p\leq\infty$, every family of Schwartz functions
$\\{f_{T}:T\in\mathbb{T}\\}$ where each $f_{T}$ has Fourier support in $T$, we
have

$\left\|\sum_{T\in\mathbb{T}}f_{T}\right\|_{L^{p}(\mathbb{R}^{3})}\leq
C^{\prime}K(\\#\mathbb{T})^{\frac{1}{2}-\frac{1}{p}}\norm*{\norm*{f_{T}}_{L^{p}(\mathbb{R}^{3})}}_{\ell^{2}(T\in\mathbb{T})},$
(5.4)

where $C^{\prime}$ depends on $C$ but not $K$, $\\#\mathbb{T}$, the dimensions
of the tubes, or the test functions $f_{T}$.

Remark. Flat decoupling is sharp when $\mathbb{T}$ is a tiling of a larger
tube by congruent parallel smaller tubes. See, for instance, Section [Yan21]
for a proof.

Now we restate Theorem 5 in [GGG+22], which is a rescaled version of Theorem
4.5, more convenient for the application of the Decoupling theorems later.

######  Theorem 5.3 (Theorem 5).

Fix $0<s<2$. For each $\varepsilon>0$, there exists $C_{s,\varepsilon}>0$ so
that the following holds. Let $\delta\in(0,1]$.

Let $H\subseteq B^{3}(0,\delta^{-1})$ be a union of $\delta^{-a}$ many
disjoint unit balls so that $H$ has measure $|H|\sim\delta^{-a}$. Let $\Theta$
be a $\delta$-separated $(\delta,t)$-subset of $[0,1]$ of cardinality
$\\#\Theta\gtrapprox\delta^{-t}$, such that for each $\theta\in\Theta$, we
have a collection of $1\times 1\times\delta^{-1}$-tubes $\mathbb{T}_{\theta}$
pointing at the direction $\gamma(\theta)$ and satisfying the following
conditions:

  1. 1.

$\mathbb{T}_{\theta}$ satisfies the $s$-dimensional condition: for any
$r\in[1,\delta^{-1}]$ and any ball $B_{r}\subseteq\mathbb{R}^{3}$ of radius
$r$, we have

$\\#\\{T\in\mathbb{T}_{\theta}:T\cap B_{r}\neq\varnothing\\}\lesssim r^{s}.$
(5.5)

In particular, $\\#\mathbb{T}_{\theta}\lesssim\delta^{-s}$.

  2. 2.

Each unit ball contained in $H$ intersects $\gtrapprox\\#\Theta$ many tubes
from $\cup_{\theta}\mathbb{T}_{\theta}$.

Then

$\delta^{-t-a}\leq C_{s,\varepsilon}\delta^{-1-s-\varepsilon}.$ (5.6)

The proof structure will be very similar to that of Proposition 3.1.

We remark that the decoupling theorem follows for every scale
$\delta\in(0,1]$. For $\delta\notin 4^{-\mathbb{N}}$, we should replace the
tiling of $[0,1]$ by $\delta^{1/2}$ intervals by a slightly more general
subset, namely, a maximally $\delta^{1/2}$-separated subset of $[0,1]$. Once
we can show decoupling for all $\delta\in 4^{-\mathbb{N}}$, the slightly more
general result follows from a trivial tiling argument, triangle and Cauchy-
Schwarz, where we lose at most an absolute constant.

###  5.2 Construction of wave packets

Fix $\theta\in\Theta$. We define a frequency slab $P_{\theta}$ centred at the
origin that has dimensions $1\times 1\times\delta$, and its shortest direction
is parallel to $\mathbf{e}_{1}(\theta)$. That is, $P_{\theta}$ is dual to
every $T_{\theta}\in\mathbb{T}_{\theta}$. We may also assume without loss of
generality that $\mathbb{T}_{\theta}$ is a subcollection of the lattice that
tiles $[-10\delta^{-1},10\delta^{-1}]^{3}$.

For each $T_{\theta}\in\mathbb{T}_{\theta}$, define a Schwartz function
$\psi_{T_{\theta}}:\mathbb{R}^{3}\to\mathbb{R}$ Fourier supported on
$P_{\theta}$ such that $\psi_{T_{\theta}}\gtrsim 1$ on $T_{\theta}$ and decays
rapidly outside $T_{\theta}$. With this, we define

$f_{\theta}=\sum_{T_{\theta}\in\mathbb{T}_{\theta}}\psi_{T_{\theta}},\quad
f=\sum_{\theta\in\Theta}f_{\theta}.$ (5.7)

Thus $f_{\theta}$ is also Fourier supported on $P_{\theta}$, but physically
essentially supported on $\cup\mathbb{T}_{\theta}$. The function $f$ is
Fourier supported on $\cup_{\theta\in\Theta}P_{\theta}$, which is a union of
slabs of the same size centered at $0$, but with normal orientations
prescribed by $\mathbf{e}_{1}(\theta)$.

Thus by definition, for every $x\in H$ we have

$\\#\Theta\lessapprox f(x)\lesssim\\#\Theta.$ (5.8)

The first inequality is true by Hypothesis 2 for $\mathbb{T}_{\theta}$ in
Theorem 5.3 and the second inequality follows from $f_{\theta}\lesssim 1$
(essentially by the disjointedness of the tubes in $\mathbb{T}_{\theta}$). In
particular, for every $p<\infty$,

$\int_{H}|f(x)|^{p}\gtrapprox|H|(\\#\Theta)^{p}\sim\delta^{-a}(\\#\Theta)^{p}.$
(5.9)

###  5.3 High-low decomposition

Similar to the proof of Proposition 3.1, the high-low decomposition is based
on an elementary fact in Euclidean geometry, which requires the nondenegenacy
condition on $\mathbf{e}_{1}(\theta)=\gamma(\theta)$. Before we state the
geometric fact, let us first make the following decomposition. Let
$K\in[1,\delta^{-1}]$ to be determined (see (5.32) below). Decompose
$P_{\theta}$ into:

  1. 1.

The high part:

$P_{\theta,\mathrm{high}}:=\left\\{\sum_{i=1}^{3}\xi_{i}\mathbf{e}_{i}(\theta):|\xi_{1}|\leq\delta,K^{-1}\leq|\xi_{2}|\leq
1,|\xi_{3}|\leq 1\right\\}.$ (5.10)

  2. 2.

The low part:

$P_{\theta,\mathrm{low}}:=\left\\{\sum_{i=1}^{3}\xi_{i}\mathbf{e}_{i}(\theta):|\xi_{1}|\leq\delta,|\xi_{2}|\leq
K^{-1},|\xi_{3}|\leq K^{-1}\right\\}.$ (5.11)

  3. 3.

The $\delta^{1/2}$-mixed part:

$P_{\theta,\delta^{1/2}}:=\left\\{\sum_{i=1}^{3}\xi_{i}\mathbf{e}_{i}(\theta):|\xi_{1}|\leq\delta,|\xi_{2}|\leq\delta^{1/2},K^{-1}\leq|\xi_{3}|\leq
1\right\\}.$ (5.12)

  4. 4.

For dyadic numbers $\lambda\in(\delta^{1/2},K^{-1}]$, the $\lambda$-mixed
part:

$P_{\theta,\lambda}:=\left\\{\sum_{i=1}^{3}\xi_{i}\mathbf{e}_{i}(\theta):|\xi_{1}|\leq\delta,\lambda/2\leq|\xi_{2}|\leq\lambda,K^{-1}\leq|\xi_{3}|\leq
1\right\\}.$ (5.13)

Thus for each $\theta$,

$P_{\theta}=P_{\theta,\mathrm{high}}\sqcup
P_{\theta,\mathrm{low}}\sqcup(\sqcup_{\text{dyadic
}\lambda\in[\delta^{1/2},K^{-1}]}P_{\theta,\lambda}).$ (5.14)

We choose a smooth partition of unity adapted to this partition which we
denote by $\eta_{\theta,\mathrm{high}}$, $\eta_{\theta,\mathrm{low}}$,
$\eta_{\theta,\mathrm{\lambda}}$, respectively, so that

$\eta_{\theta,\mathrm{high}}+\eta_{\theta,\mathrm{low}}+\sum_{\text{dyadic
}\lambda\in[\delta^{1/2},K^{-1}]}\eta_{\theta,\lambda}=1$ (5.15)

on $P_{\theta}$. Since $\mathrm{supp}\widehat{f_{\theta}}\subseteq
P_{\theta}$, we also obtain a decomposition of $f_{\theta}$

$f_{\theta}=f_{\theta,\mathrm{high}}+f_{\theta,\mathrm{low}}+\sum_{\text{dyadic
}\lambda\in[\delta^{1/2},K^{-1}]}f_{\theta,\lambda},$ (5.16)

where
$\widehat{f}_{\theta,\mathrm{high}}=\eta_{\theta,\mathrm{high}}\widehat{f_{\theta}}$,
$\widehat{f}_{\theta,\mathrm{low}}=\eta_{\theta,\mathrm{low}}\widehat{f_{\theta}}$,
and $\widehat{f}_{\theta,\lambda}=\eta_{\theta,\lambda}\widehat{f_{\theta}}$.
Similarly, we have a decomposition of $f$

$f=f_{\mathrm{high}}+f_{\mathrm{low}}+\sum_{\text{dyadic
}\lambda\in[\delta^{1/2},K^{-1}]}f_{\lambda},$ (5.17)

where $f_{\mathrm{high}}=\sum_{\theta\in\Theta}f_{\theta,\mathrm{high}}$,
$f_{\mathrm{low}}=\sum_{\theta\in\Theta}f_{\theta,\mathrm{low}}$ and
$f_{\lambda}=\sum_{\theta\in\Theta}f_{\theta,\lambda}$.

Lastly, in view of (5.9) and using triangle inequality, we have

$\int_{H}|f_{\mathrm{high}}|^{p}+\int_{H}|f_{\mathrm{low}}|^{p}+\sum_{\text{dyadic
}\lambda\in[\delta^{1/2},K^{-1}]}\int_{H}|f_{\lambda}|^{p}\gtrapprox\delta^{-a}(\\#\Theta)^{p}.$
(5.18)

###  5.4 Analysis of high part

The geometric fact that gives orthogonality for the high part is as follows.

######  Lemma 5.4.

The collection of all high parts $\\{P_{\theta,high}\\}_{\theta\in\Theta}$ is
at most $O(K^{2})$-overlapping.

###### Proof of lemma.

We first perform a preliminary reduction. Partition $\Theta$ into intervals of
length $\ll K^{-1}$. By losing a factor of order $O(K)$, it suffices to assume
$\Theta$ is contained in such an interval of length $\ll K^{-1}$.

We may ignore the thin direction $\mathbf{e}_{1}$. More precisely, consider
the plane

$\Pi_{\theta}:=\left\\{\xi_{2}\mathbf{e}_{2}(\theta)+\xi_{3}\mathbf{e}_{3}(\theta):K^{-1}\leq|\xi_{2}|\leq
1,|\xi_{3}|\leq 1\right\\},$ (5.19)

whose $\delta$ neighbourhood contains $P_{\theta,\mathrm{high}}$. We claim the
following: for any $\theta\in[0,1]$ and any
$\theta^{\prime}=\theta+\Delta\in[0,1]$ with $CK\delta\leq\Delta\leq
C^{-1}K^{-1}$ ($C$ to be determined at the end of the proof of this lemma), if
$\xi\in\Pi_{\theta}$, then

$\mathrm{dist}(\xi,\Pi_{\theta^{\prime}})>10\delta.$ (5.20)

Observe that if this holds then $P_{\theta,\mathrm{high}}\cap
P_{\theta^{\prime},\mathrm{high}}=\varnothing$ for
$\Delta\in[CK\delta,C^{-1}K^{-1}]$. Indeed, if $\xi\in
P_{\theta,\mathrm{high}}\cap P_{\theta^{\prime},\mathrm{high}}$, then there
exists $\xi_{1}\in\Pi_{\theta}$ and $\xi_{2}\in\Pi_{\theta^{\prime}}$ such
that $|\xi-\xi_{1}|\leq\delta$, $|\xi-\xi_{2}|\leq\delta$. Since
$\xi_{1}\in\Pi_{\theta}$, we have
$\mathrm{dist}(\xi_{1},\Pi_{\theta^{\prime}})>10\delta$. Thus
$\mathrm{dist}(\xi,\Pi_{\theta^{\prime}})>9\delta$. But
$|\xi-\xi_{2}|\leq\delta$, so
$\mathrm{dist}(\xi_{2},\Pi_{\theta^{\prime}})>8\delta$, which contradicts the
fact that $\xi_{2}\in\Pi_{\theta^{\prime}}$.

Thus if $P_{\theta,\mathrm{high}}\cap
P_{\theta^{\prime},\mathrm{high}}\neq\varnothing$ then
$\Delta\notin[CK\delta,C^{-1}K^{-1}]$. By the preliminary reduction at the
beginning we see that if $P_{\theta,\mathrm{high}}$ and
$P_{\theta+\Delta,\mathrm{high}}$ overlap then we must have $\Delta\leq
CK\delta$. Since $\Theta$ is $\delta$-separated, we see that
$\\{P_{\theta,\mathrm{high}}\\}_{\theta\in\Theta}$ is at most $O(K)$
overlapping. Thus it remains to prove (5.20).

Let $\xi\in\Pi_{\theta}$. Write

$\xi=a\mathbf{e}_{2}(\theta)+b\mathbf{e}_{3}(\theta)=a\gamma^{\prime}(\theta)+b\gamma(\theta)\times\gamma^{\prime}(\theta),$
(5.21)

where $|a|\in[K^{-1},1]$ and $|b|\leq 1$. Since the normal direction of
$\Pi_{\theta+\Delta}$ is $\gamma(\theta+\Delta)$, we have
$\mathrm{dist}(\xi,\Pi_{\theta^{\prime}})=\xi\cdot\gamma(\theta^{\prime})$,
and so we need to show

$\left|\gamma(\theta+\Delta)\cdot\left(a\gamma^{\prime}(\theta)+b\gamma(\theta)\times\gamma^{\prime}(\theta)\right)\right|>10\delta.$
(5.22)

By Taylor expansion, we have
$\gamma(\theta+\Delta)=\gamma(\theta)+\Delta\gamma^{\prime}(\theta)+O(\Delta^{2})$.
Using $|a|\geq K^{-1}$, we see the left hand side of (5.22) is

$|a|\Delta+O(\Delta^{2})\geq(|a|-O(\Delta))\Delta\geq(|a|-O(C^{-1}K^{-1}))\Delta>10\delta,$
(5.23)

if $C$ is large enough. ∎

Now we are ready to estimate $\int_{H}|f_{\mathrm{high}}|^{p}$. By
Plancherel’s identity and Lemma 5.4, we have

$\int|f_{\mathrm{high}}|^{2}=\int|\widehat{f}_{\mathrm{high}}|^{2}\lesssim
K^{2}\sum_{\theta\in\Theta}\int|\widehat{f_{\theta}}|^{2}=K^{2}\sum_{\theta\in\Theta}\int|f_{\theta}|^{2}.$
(5.24)

By the rapid decay of $\psi_{T_{\theta}}$, we have

$\int|f_{\theta}|^{2}\sim\delta^{-1}\\#\mathbb{T}_{\theta}.$ (5.25)

Thus

$\int|f_{\mathrm{high}}|^{2}\lesssim
K^{2}\delta^{-1}\sum_{\theta\in\Theta}\\#\mathbb{T}_{\theta}.$ (5.26)

Thus, for $p\in[2,\infty)$, we have

$\int_{H}|f_{\mathrm{high}}|^{p}\leq\int_{H}|f_{\mathrm{high}}|^{2}(\\#\Theta)^{p-2}\lesssim(\\#\Theta)^{p-2}K^{2}\delta^{-1}\sum_{\theta\in\Theta}\\#\mathbb{T}_{\theta}.$
(5.27)

Using $\\#\mathbb{T}_{\theta}\lesssim\delta^{-s}$, we have

$\int_{H}|f_{\mathrm{high}}|^{p}\lesssim(\\#\Theta)^{p-1}K^{2}\delta^{-1-s}.$
(5.28)

###  5.5 Analysis of low part

For the low part, we do not have orthogonality to aid us. Instead, we use the
$s$-dimensional condition (5.5). Heuristically, for each $x\in H$,
$\theta\in\Theta$ and $T_{\theta}\in\mathbb{T}_{\theta}$, if we ignore the
tails of all Schwartz functions, then

$|\psi_{T_{\theta}}*\eta_{\theta,\mathrm{low}}^{\vee}(x)|\lesssim\begin{cases}\delta
K^{-2}\cdot\delta^{-1},&\text{if }T_{\theta}\cap
B^{3}(x,CK)\neq\varnothing,\\\ 0,&\text{if }T_{\theta}\cap
B^{3}(x,CK)=\varnothing.\end{cases}$ (5.29)

As a result,

$|f_{\theta,\mathrm{low}}(x)|\lesssim
K^{-2}\\#\\{T_{\theta}\in\mathbb{T}_{\theta}:T_{\theta}\cap
B^{3}(x,CK)\neq\varnothing\\}.$ (5.30)

Intuitively, this follows from the uncertainty principle, since we now view
the tubes $T_{\theta}$ blurred by scale $K$. For heuristic purposes, one can
pretend
$\check{\eta}_{\theta,low}=\frac{1}{\delta^{-1}K^{2}}\mathbb{1}_{\tilde{T}}$
where $\tilde{T}$ is a $\delta^{-1}\times K\times K$ tube with same
orientation as $T_{\theta}$, and the convolution
$\psi_{T_{\theta}}*\mathbb{1}_{\tilde{T}}(x)$ can capture at most the volume
of a $\delta^{-1}\times 1\times 1$ tube. The rigorous proof is left as an
exercise, since it is similar to the argument in Section 3.5, except that the
treatment of weights in 2D should be generalised to 3D in the natural way. (In
particular, note that the $K^{s-2}$ in (5.31) below would be similarly
replaced by $K^{\frac{s-2}{2}}$ with a suitably chosen exponent $E=E(s)$ for
the weights. This will also slightly affect our choice of $K$ in (5.32)
below.)

Using (5.30) and the $s$-dimensional condition (5.5), for every $x\in H$ we
have

$|f_{\mathrm{low}}(x)|\leq\sum_{\theta\in\Theta}|f_{\theta,\mathrm{low}}(x)|\lesssim\\#\Theta
K^{-2}K^{s}\sim\\#\Theta K^{s-2}.$ (5.31)

###  5.6 Choosing $K$ such that the low part is negligible

Recall from (5.8) that $|f(x)|\gtrapprox\\#\Theta$, that is,
$|f(x)|\geq|\log\delta|^{-\alpha}\\#\Theta$ for some $\alpha=O(1)$. We now
choose

$K:=|\log\delta|^{\frac{\alpha+1}{2-s}},$ (5.32)

so that for $\delta$ small enough we have for every $x\in H$ that
$|f_{\mathrm{low}}(x)|<|f(x)|/2$. Thus we need only consider the contributions
from the high part and the mixed parts.

###  5.7 Analysis of $\delta^{1/2}$-mixed part

The main tool for the analysis of mixed parts is the Bourgain-Demeter
$\ell^{2}(L^{6})$ decoupling for the cone, stated as in Theorem 5.1. We
further divide this section into a few subsections.

####  5.7.1 Construction of covering planks

To use Theorem 5.1, the idea is to construct a cover of the truncated cone

$\Gamma_{K^{-1}}:=\\{r\mathbf{e}_{3}(\theta):K^{-1}\leq r\leq
1,0\leq\theta\leq 1\\}$ (5.33)

by essentially disjoint slabs $P^{+}_{\theta,\delta^{1/2}}$, where
$P^{+}_{\theta,\delta^{1/2}}:=P_{\theta,\delta^{1/2}}\cap\\{\xi_{3}>0\\}$, and
we define $P^{-}_{\theta,\delta^{1/2}}$ in a similar way. In the following we
make this precise.

For each $C\geq 1$, denote the slightly enlarged plank

$CP^{+}_{\theta,\delta^{1/2}}:=\left\\{\sum_{i=1}^{3}\xi_{i}\mathbf{e}_{i}(\theta):|\xi_{1}|\leq
C\delta,\,\,|\xi_{2}|\leq C\delta^{1/2},\,\,C^{-1}K^{-1}\leq\xi_{3}\leq
C\right\\}.$ (5.34)

Remark on notation. For simplicity, from now on we abuse notation and simply
denote $CP^{+}_{\theta,\delta^{1/2}}$ by $CP_{\theta,\delta^{1/2}}$.

Strictly speaking, the object we need to cover is not $\Gamma_{K^{-1}}$, but
rather $\cup_{\theta\in\Theta}P_{\theta,\delta^{1/2}}$. For this purpose, we
need the following lemma.

######  Lemma 5.5.

We have the following observations.

  1. 1.

If $|\theta-\theta^{\prime}|\leq\delta^{1/2}$, then $P_{\theta,\delta^{1/2}}$
and $P_{\theta^{\prime},\delta^{1/2}}$ are essentially the same in the sense
that $P_{\theta,\delta^{1/2}}\subseteq C_{1}P_{\theta^{\prime},\delta^{1/2}}$
for some suitable constant $C_{1}$.

  2. 2.

If $C_{2}K\delta^{1/2}\leq|\theta-\theta^{\prime}|\leq C_{2}^{-1}$ where
$C_{2}=C_{2}(C_{1})$ is large enough, then $C_{1}P_{\theta,\delta^{1/2}}$ and
$C_{1}P_{\theta^{\prime},\delta^{1/2}}$ are disjoint.

The proof is given in Section 5.11 after the main proof.

Let $\Omega$ be the $\delta^{1/2}$-lattice contained in $[0,1]$. Lemma 5.5
immediately implies the following.

######  Lemma 5.6.

The enlarged planks $\\{C_{1}P_{\omega,\delta^{1/2}}:\omega\in\Omega\\}$ cover
the union of planks $\cup_{\theta\in\Theta}P_{\theta,\delta^{1/2}}$. More
precisely, for each $\theta\in\Theta$, take $\omega\in\Omega$ to be the
closest to $\theta$ (so that $|\omega-\theta|\leq\delta^{1/2}$). Then we have

$P_{\theta,\delta^{1/2}}\subseteq C_{1}P_{\omega,\delta^{1/2}}.$ (5.35)

Moreover, the planks $\\{C_{1}P_{\omega,\delta^{1/2}}:\omega\in\Omega\\}$ are
$O(K^{2})$ overlapping.

####  5.7.2 Applying decoupling inequality

First, we may assume each $f_{\theta,\delta^{1/2}}$ has Fourier support on
$P_{\theta,\delta^{1/2}}$ only. With a loss of a constant factor, we may
assume without loss of generality that all $\Theta$ lie in an interval of
length less than $C_{2}^{-1}$ where we recall the constant $C_{2}$ as in Lemma
5.5.

For $\omega\in\Omega$ and $\theta\in\Theta$, we introduce a notation:
$\theta\prec\omega$ if $\theta-\omega\in(-\delta^{1/2}/2,\delta^{1/2}/2]$. We
also refer to this as “$\theta$ is attached to $\omega$”. Each
$\theta\in\Theta$ can be attached to a unique $\omega\in\Omega$. With this, we
define

$f_{\omega}=\sum_{\theta\prec\omega}f_{\theta,\delta^{1/2}}.$ (5.36)

Thus $f_{\delta^{1/2}}=\sum_{\omega\in\Omega}f_{\omega}$. For each
$\omega\in\Omega$, the Fourier support of $f_{\omega}$ is equal to
$\cup_{\theta\prec\omega}P_{\theta,\delta^{1/2}}$, which is contained in
$C_{1}P_{\omega,\delta^{1/2}}$ by Lemma 5.6. Since
$C_{1}P_{\omega,\delta^{1/2}}$ are $K^{O(1)}$ overlapping and
$K\lesssim_{\varepsilon}\delta^{-\varepsilon}$ by (5.32), we may apply
Bourgain-Demeter’s decoupling Theorem 5.1222The case $p=6$ is sharp for
decoupling, but we shall apply this argument to a general $2\leq p\leq 6$,
which allows us to see what would happen if we did not have sharp decoupling.
to get

$\norm*{f_{\delta^{1/2}}}_{p}\lesssim_{\varepsilon}\delta^{-\varepsilon}\norm*{\norm*{f_{\omega}}_{p}}_{\ell^{2}(\omega\in\Omega)}.$
(5.37)

We actually only need a slightly weaker $\ell^{p}(L^{p})$ decoupling
inequality, which follows from the above via Hölder’s inequality:

$\norm*{f_{\delta^{1/2}}}_{p}\lesssim_{\varepsilon}\delta^{-\varepsilon}(\\#\\{\omega\in\Omega:f_{\omega}\neq
0\\})^{1/2-1/p}\norm*{\norm*{f_{\omega}}_{p}}_{\ell^{p}(\omega\in\Omega)}.$
(5.38)

Note that a priori, each $\omega$ may have a different number of
$\theta\in\Theta$ for which $\theta\prec\omega$. However, after a standard
process which we call dyadic pigeonholing, we may assume there is a number $N$
such that for each $\omega$, the number of $\theta$’s for which
$\theta\prec\omega$ is between $N$ and $2N$. In this process, we only pay the
price of a logarithmic factor $O(|\log\delta|)$. The details are given in
Section 5.10 below.

For simplicity, we will denote $\\#\omega=\\#\\{\omega\in\Omega:f_{\omega}\neq
0\\}$, and so we have

$\\#\Theta\sim N\\#\omega.$ (5.39)

####  5.7.3 Analysis of each $f_{\omega}$

Fix $\omega$. By the triangle and Hölder’s inequalities, we have

$\norm*{f_{\omega}}_{p}\leq\sum_{\theta\in\Theta:\theta\prec\omega}\norm*{f_{\theta,\delta^{1/2}}}_{p}\leq\left(\sum_{\theta\in\Theta:\theta\prec\omega}\norm*{f_{\theta,\delta^{1/2}}}^{p}_{p}\right)^{1/p}N^{1-1/p}.$
(5.40)

We remark that this is the best we can do in general, since all planks
$P_{\theta,\delta^{1/2}}$ where $\theta\prec\omega$ are essentially the same,
in the sense similar to Lemma 5.5. Thus, combined with (5.38) and (5.39), we
arrive at

$\begin{split}\norm*{f_{\delta^{1/2}}}_{p}&\lesssim_{\varepsilon}\delta^{-\varepsilon}(\\#\omega)^{\frac{1}{2}-\frac{1}{p}}\norm*{\norm*{f_{\theta,\delta^{1/2}}}_{p}}_{\ell^{p}(\theta\in\Theta)}N^{1-\frac{1}{p}}\\\
&\sim\delta^{-\varepsilon}(\\#\Theta)^{\frac{1}{2}-\frac{1}{p}}\norm*{\norm*{f_{\theta,\delta^{1/2}}}_{p}}_{\ell^{p}(\theta\in\Theta)}N^{\frac{1}{2}}.\end{split}$
(5.41)

Now we use the assumption that $\Theta$ is a $\delta$-separated
$(\delta,t)$-set to get

$N\lesssim\\#\left\\{\theta\in\Theta\colon\theta\in\left(\omega-\frac{\delta^{1/2}}{2},\omega+\frac{\delta^{1/2}}{2}\right]\right\\}\lesssim\delta^{-t/2}.$
(5.42)

Also, for each $f_{\theta,\delta^{1/2}}$, by the elementary property of
Fourier cutoff and rapid decay of $\psi_{T_{\theta}}$, we have

$\norm*{f_{\theta,\delta^{1/2}}}_{p}\lesssim\norm*{f_{\theta}}_{p}\sim((\\#\mathbb{T}_{\theta})\delta^{-1})^{1/p},$
(5.43)

and by plugging these back in inequality (5.41),

$\norm*{f_{\delta^{1/2}}}_{p}\lesssim_{\varepsilon}\delta^{-\varepsilon-\frac{t}{4}}(\\#\Theta)^{\frac{1}{2}-\frac{1}{p}}\left(\sum_{\theta\in\Theta}(\\#\mathbb{T}_{\theta})\delta^{-1}\right)^{\frac{1}{p}}.$
(5.44)

Using $\\#\mathbb{T}_{\theta}\lesssim\delta^{-s}$, we have

$\norm*{f_{\delta^{1/2}}}_{p}\lesssim_{\varepsilon}\delta^{-\varepsilon-\frac{s+1}{p}-\frac{t}{4}}(\\#\Theta)^{\frac{1}{2}}.$
(5.45)

####  5.7.4 Comparison with the high part

We now make a comparison between the bounds
$\norm*{f_{\mathrm{high}}}_{L^{p}(H)}$ given by (5.28) and
$\norm*{f_{\delta^{1/2}}}_{p}$. If $p>4$, using the lower bound
$\\#\Theta\gtrapprox\delta^{-t}$, we see that for small enough $\delta$,

$(\\#\Theta)^{\frac{p-1}{p}}K^{\frac{2}{p}}\delta^{-\frac{s+1}{p}}\geq\delta^{-\varepsilon-\frac{s+1}{p}-\frac{t}{4}}(\\#\Theta)^{\frac{1}{2}},$
(5.46)

which means that the upper bound for the $L^{p}$ norm of the
$\delta^{1/2}$-mixed part is dominated by the upper bound for the $L^{p}$ norm
of the high part of $f$ in $H$.

###  5.8 Analysis of $\lambda$-mixed parts

The main tool for this part is still decoupling, but more complicated than the
previous step. Fix a dyadic scale $\lambda\in[\delta^{1/2},K^{-1}]$.

For technical reasons that arise in the proof of Lemma 5.7 in Section 5.11
below, we first perform a preliminary reduction. Let $C_{0}$ be a large dyadic
constant to be determined. If $\lambda\in[\delta^{1/2},C_{0}\delta^{1/2}]$,
then the same method of Section 5.7 can be used. Thus we assume
$\lambda>C_{0}\delta^{1/2}$ throughout the rest of this subsection.

####  5.8.1 Construction of covering planks

First, since $P_{\theta,\lambda}$ consists of four similar parts, we may
assume without loss of generality that each $f_{\theta,\lambda}$ is Fourier
supported on one part:

$Q_{\theta,\lambda}:=\left\\{\sum_{i=1}^{3}\xi_{i}\mathbf{e}_{i}(\theta):|\xi_{1}|\leq\delta,\lambda/2\leq\xi_{2}\leq\lambda,K^{-1}\leq\xi_{3}\leq
1\right\\}.$ (5.47)

Similarly, for $C\geq 1$, we define the slightly enlarged planks

$CQ_{\theta,\lambda}:=\left\\{\sum_{i=1}^{3}\xi_{i}\mathbf{e}_{i}(\theta):|\xi_{1}|\leq
C\delta,(2C)^{-1}\lambda\leq\xi_{2}\leq C\lambda,C^{-1}K^{-1}\leq\xi_{3}\leq
C\right\\}.$ (5.48)

Note that this is the same as the $S_{\tau}$ defined on Page 15 of [GGG+22].

Define $\Sigma$ to be the $\lambda$-lattice contained in $[0,1]$. For each
$\sigma\in\Sigma$, consider the following plank at scale $\lambda$:

$P_{\sigma,\lambda}=\left\\{\sum_{i=1}^{3}\xi_{i}\mathbf{e}_{i}(\sigma):|\xi_{1}|\leq\lambda^{2},|\xi_{2}|\leq\lambda,K^{-1}\leq\xi_{3}\leq
1\right\\}.$ (5.49)

Also recall that for $C\geq 1$,

$CP_{\sigma,\lambda}=\left\\{\sum_{i=1}^{3}\xi_{i}\mathbf{e}_{i}(\sigma):|\xi_{1}|\leq
C\lambda^{2},|\xi_{2}|\leq C\lambda,C^{-1}K^{-1}\leq\xi_{3}\leq C\right\\}.$
(5.50)

Since we have an extra intermediate scale $\lambda$, we will need the
following lemma, which is rephrased from Lemma 3 of [GGG+22].

######  Lemma 5.7.

By choosing $C_{1}$ and then $C_{2}$ to be large enough, we have

  1. 1.

If $|\theta-\theta^{\prime}|\leq\lambda^{-1}\delta$, then $Q_{\theta,\lambda}$
and $Q_{\theta^{\prime},\lambda}$ are essentially the same in the sense that
$Q_{\theta,\lambda}\subseteq C_{1}Q_{\theta^{\prime},\lambda}$.

  2. 2.

If $C_{2}\lambda^{-1}\delta\leq|\theta-\theta^{\prime}|\leq
C_{2}^{-1}\lambda$, then $C_{1}Q_{\theta,\lambda}$ and
$C_{1}Q_{\theta^{\prime},\lambda}$ are disjoint.

The proof is slightly different from Lemma 5.5, so we include it in Section
5.11. The following lemma is almost identical to Lemma 5.5, so we omit the
proof.

######  Lemma 5.8.

The following statements are true.

  1. 1.

If $|\theta-\theta^{\prime}|\leq\lambda$, then $C_{1}P_{\theta,\lambda}$ and
$C_{1}P_{\theta^{\prime},\lambda}$ are essentially the same in the sense that
$C_{1}P_{\theta,\lambda}\subseteq C_{3}P_{\theta^{\prime},\lambda}$ where
$C_{3}=C_{3}(C_{1})$ is large enough.

  2. 2.

If $C_{4}K\lambda\leq|\theta-\theta^{\prime}|\leq C_{4}^{-1}K^{-1}$ where
$C_{4}=C_{4}(C_{3})$ is large enough, then $C_{3}P_{\theta,\lambda}$ and
$C_{3}P_{\theta^{\prime},\lambda}$ are disjoint.

The key difference from the $\delta^{1/2}$-mixed case is that the natural
separation in this case is $\lambda^{-1}\delta$ which is smaller than
$\lambda$, the aperture of $Q_{\theta,\lambda}$. Moreover, in terms of the
canonical scale of decoupling, the thickness $\delta$ of $Q_{\theta,\lambda}$
does not match its aperture $\lambda$ which was supposed to be $\delta^{1/2}$.
When $\lambda=\delta^{1/2}$ as in the previous case, we had
$\lambda^{-1}\delta=\delta^{1/2}=\lambda$ which made the scenario simpler.

Motivated by Lemma 5.7, we define $\mathcal{T}$ to be the
$\lambda^{-1}\delta$-lattice in $[0,1]$. Now we have three subsets of $[0,1]$:

$\Theta\subseteq\delta\mathbb{Z}\cap[0,1],\quad\mathcal{T}=(\lambda^{-1}\delta)\mathbb{Z}\cap[0,1],\quad\Sigma=\lambda\mathbb{Z}\cap[0,1].$
(5.51)

We will define relationships between their elements. For each
$\theta\in\Theta$, we attach it to a $\tau\in\mathcal{T}$ if
$\theta-\tau\in(\lambda^{-1}\delta/2,\lambda^{-1}\delta/2]$, which we denote
by $\theta\prec\tau$. Thus each $\theta$ is attached to a unique $\tau$. With
this, we define

$f_{\tau}=\sum_{\theta\prec\tau}f_{\theta,\lambda}.$ (5.52)

Similarly, for each $\tau\in\mathcal{T}$, we attach it to a $\sigma\in\Sigma$
if $\tau-\sigma\in(\lambda/2,\lambda/2]$, which we denote by
$\tau\prec\sigma$. Thus each $\tau$ is attached to a unique $\sigma$. With
this, we define

$f_{\sigma}=\sum_{\tau\prec\sigma}f_{\tau}.$ (5.53)

We also write $\theta\prec\sigma$ if there is a $\tau$ such that
$\theta\prec\tau$ and $\tau\prec\sigma$. Thus each $\theta$ is attached to a
unique $\sigma$. Thus we have

$f_{\lambda}=\sum_{\sigma\in\Sigma}f_{\sigma}=\sum_{\sigma\in\Sigma}\sum_{\tau\prec\sigma}f_{\tau}=\sum_{\sigma\in\Sigma}\sum_{\tau\prec\sigma}\sum_{\theta\prec\tau}f_{\theta,\lambda}.$
(5.54)

####  5.8.2 Applying decoupling inequalities

Before estimating $\int|f_{\lambda}|^{p}$, we may first apply a similar
preliminary reduction to that of the $\delta^{1/2}$-mixed part to reduce to
the case that $\mathrm{diam}(\Theta)\leq C_{4}^{-1}K^{-1}$. Next, we may apply
a dyadic pigeonholing argument to subsets of $\Theta,\mathcal{T},\Sigma$
(still denoted by $\Theta,\mathcal{T},\Sigma$ by abuse of notation), so that
there are numbers $M,N$ such that

$\\#\\{\theta\in\Theta:\theta\prec\tau\\}\sim
M,\quad\\#\\{\tau\in\mathcal{T}:\tau\prec\sigma\\}\sim N.$ (5.55)

We only pay the price of logarithmic losses here; the detail is similar to the
proof in Section 5.10, which we ask the reader to verify.

We first apply Theorem 5.1 to decouple $f_{\lambda}$ into different
$f_{\sigma}$’s. The detail is as follows. First, the Fourier support of each
$f_{\tau}$ is equal to $\cup_{\theta\prec\tau}Q_{\theta,\lambda}$ and in turn
contained in $C_{1}Q_{\tau,\lambda}$ by Part 1 of Lemma 5.7. Thus the Fourier
support of each $f_{\sigma}$ is contained in
$\cup_{\tau\prec\sigma}C_{1}Q_{\tau,\lambda}$, which is contained in
$C_{3}P_{\sigma,\lambda}$ by Part 1 of Lemma 5.8 and the trivial fact that
$C_{1}Q_{\tau,\lambda}\subseteq C_{1}P_{\tau,\lambda}$. Also, since
$\text{diam}(\Theta)\leq C_{4}^{-1}K^{-1}$, by Part 2 of Lemma 5.8,
$\\{C_{3}P_{\sigma,\lambda}:\sigma\in\Sigma\\}$ has bounded overlap.

We thus apply Theorem 5.1 with respect to the planks $C_{3}P_{\sigma,\lambda}$
at scale $\lambda^{2}$ in place of $\delta$, followed by Hölder’s inequality
to get

$\norm*{f_{\lambda}}_{p}\lesssim_{\varepsilon}\lambda^{-\varepsilon}\norm*{\norm*{f_{\sigma}}_{p}}_{\ell^{2}(\sigma\in\Sigma)}\leq\lambda^{-\varepsilon}(\\#\sigma)^{\frac{1}{2}-\frac{1}{p}}\norm*{\norm*{f_{\sigma}}_{p}}_{\ell^{p}(\sigma\in\Sigma)},$
(5.56)

where, for simplicity, we denoted
$\\#\sigma:=\\#\\{\sigma\in\Sigma:f_{\sigma}\neq 0\\}$.

Now we need to analyse each $f_{\sigma}$, whose Fourier transform has support
contained in $\cup_{\tau\prec\sigma}C_{1}Q_{\tau,\lambda}$. By Part 2 of Lemma
5.7, the collection $\\{C_{1}Q_{\tau,\lambda}:\tau\prec\sigma\\}$ has bounded
overlap. Thus we can apply the flat decoupling Theorem 5.2 to get (recall
(5.55)):

$\norm*{f_{\sigma}}_{p}\lessapprox
N^{1-\frac{2}{p}}\norm*{\norm*{f_{\tau}}_{p}}_{\ell^{p}(\tau\prec\sigma)}.$
(5.57)

####  5.8.3 Analysis of each $f_{\tau}$

Fix $\tau$. We use triangle and Hölder’s inequalities (recall (5.55)):

$\norm*{f_{\tau}}_{p}\lesssim
M^{1-\frac{1}{p}}\norm*{\norm*{f_{\theta,\lambda}}_{p}}_{\ell^{p}(\theta\prec\tau)}.$
(5.58)

This is the best we can do in general since for each $\theta\prec\tau$, the
planks $Q_{\theta,\lambda}$ are essentially the same.

Combining the above inequalities (5.56) through (5.58) gives

$\norm*{f_{\lambda}}_{p}\lesssim_{\varepsilon}\delta^{-\varepsilon}(\\#\sigma)^{\frac{1}{2}-\frac{1}{p}}M^{1-\frac{1}{p}}N^{1-\frac{2}{p}}\norm*{\norm*{f_{\theta,\lambda}}_{p}}_{\ell^{p}(\theta\in\Theta)}.$
(5.59)

Now for each $\theta$, by the elementary property of Fourier cutoff and rapid
decay of each $\psi_{T_{\theta}}$, we have

$\norm*{f_{\theta,\lambda}}_{p}\lesssim\norm*{f_{\theta}}_{p}\sim(\\#\mathbb{T}_{\theta}\delta^{-1})^{1/p}.$
(5.60)

Now by bounded overlap, we observe that

$MN\sim\\#\\{\theta\in\Theta:\theta\prec\sigma\\}\lesssim(\lambda/\delta)^{t},\quad\\#\Theta\sim(\\#\sigma)MN.$
(5.61)

where we have used the fact that $\Theta$ is a $\delta$-separated
$(\delta,t)$-set in the first relation. We also have $M\lesssim\lambda^{-t}$
by the $(\delta,t)$-set condition. Using this, noting $p\geq 2$ and (5.59),
(5.60) and (5.61), we thus have

$\displaystyle\norm*{f_{\lambda}}_{p}$
$\displaystyle\lesssim_{\varepsilon}\delta^{-\varepsilon}(\\#\Theta)^{\frac{1}{2}-\frac{1}{p}}(MN)^{\frac{1}{2}-\frac{1}{p}}M^{\frac{1}{p}}\delta^{-\frac{1}{p}}\left(\sum_{\theta\in\Theta}\\#\mathbb{T}_{\theta}\right)^{\frac{1}{p}}$
$\displaystyle\lesssim\delta^{-\varepsilon-\frac{1}{p}}\left(\sum_{\theta\in\Theta}\\#\mathbb{T}_{\theta}\right)^{\frac{1}{p}}(\\#\Theta)^{\frac{1}{2}-\frac{1}{p}}(\lambda/\delta)^{t(\frac{1}{2}-\frac{1}{p})}\lambda^{-\frac{t}{p}}$
$\displaystyle=\delta^{-\varepsilon-\frac{1}{p}-t(\frac{1}{2}-\frac{1}{p})}\left(\sum_{\theta\in\Theta}\\#\mathbb{T}_{\theta}\right)^{\frac{1}{p}}(\\#\Theta)^{\frac{1}{2}-\frac{1}{p}}\lambda^{t(\frac{1}{2}-\frac{2}{p})}.$
(5.62)

We remark that when $\lambda\sim\delta^{1/2}$, this reduces to the same
expression in (5.45) in the $\delta^{1/2}$-mixed part. Using
$\\#\mathbb{T}_{\theta}\lesssim\delta^{-s}$, we have

$\norm*{f_{\lambda}}_{p}\lesssim_{\varepsilon}\delta^{-\varepsilon-\frac{s+1}{p}-t(\frac{1}{2}-\frac{1}{p})}(\\#\Theta)^{\frac{1}{2}}\lambda^{t(\frac{1}{2}-\frac{2}{p})}.$
(5.63)

####  5.8.4 Comparison with the high part

We now make a comparison between the bounds to
$\norm*{f_{\mathrm{high}}}_{p}^{p}$ given by (5.28) and the bounds for

$\sum_{\lambda\text{
dyadic};\lambda\in(\delta^{1/2},K^{-1}]}\norm*{f_{\lambda}}_{p}^{p}$

obtained from the previous step.

If $p>4$, using the lower bound $\\#\Theta\gtrapprox\delta^{-t}$, we see that
for small enough $\delta$,

$(\\#\Theta)^{\frac{p-1}{p}}K^{\frac{2}{p}}\delta^{-\frac{s+1}{p}}\geq\delta^{-\varepsilon-\frac{s+1}{p}-t(\frac{1}{2}-\frac{1}{p})}(\\#\Theta)^{\frac{1}{2}}\lambda^{t(\frac{1}{2}-\frac{2}{p})},$
(5.64)

which means that the upper bound for the $\lambda$-mixed part is dominated by
the upper bound of the high part.

###  5.9 Conclusion

Since Theorem 5.1 holds at the sharp exponent $p=6$, by the discussion at the
end of Sections 5.7 and 5.8, we know that the high part dominates over both
the low part and the mixed parts. Since the number of dyadic
$\lambda\in[\delta^{1/2},K^{-1}]$ is $O(|\log\delta|)$, by (5.26), we have

$\int_{H}|f|^{p}\lesssim|\log\delta|(\\#\Theta)^{p-1}K^{2}\delta^{-1-s}.$
(5.65)

On the other hand, since $f(x)\gtrsim\\#\Theta$ on $H$, we trivially have the
lower bound

$\int_{H}|f|^{p}\gtrsim(\\#\Theta)^{p}|H|\sim(\\#\Theta)^{p}\delta^{-a}.$
(5.66)

Recalling (5.32) and $\\#\Theta\gtrapprox\delta^{-t}$, we have
$\delta^{-a-t}\lesssim_{\varepsilon}\delta^{-1-s-\varepsilon}$, as required.

###  5.10 Proof of dyadic pigeonholing

We now prove the statement leading to (5.39). Recall that by $\Omega$ we
denote the $\delta^{1/2}$-lattice contained in $[0,1]$. For each
$\omega\in\Omega$, by the $\delta$-separation condition of $\Theta$, we have
the trivial bound $\\#\\{\theta\prec\omega\\}\leq\delta^{-1/2}$.

Let $J=\log_{2}(\delta^{-1/2})$. For each integer $0\leq j\leq J$, we denote
the set

$\Omega_{j}:=\\{\omega\in\Omega:\\#\\{\theta\in\Theta:\theta\prec\omega\\}\in[2^{j},2^{j+1})\\}.$
(5.67)

Then it is easy to see that

$\sum_{j=1}^{J}2^{j}\\#\Omega_{j}\leq\\#\Theta\leq\sum_{j=1}^{J}2^{j+1}\\#\Omega_{j}$
(5.68)

since we only consider those $\omega$ such that
$\\{\theta\in\Theta:\theta\prec\omega\\}\neq\varnothing$.

Now by the pigeonholing principle, there exists some $i\in[1,J]$ such that

$\sum_{j=1}^{J}2^{j}\\#\Omega_{j}\leq J2^{i}\\#\Omega_{i}.$ (5.69)

This gives

$2^{i}\\#\Omega_{i}\leq\\#\Theta\leq J2^{i+1}\\#\Omega_{i},$ (5.70)

which is as required once we take $N=2^{i}$, since $J\sim|\log\delta|$.

###  5.11 Proof of covering lemmas

We now give proofs of the covering Lemmas 5.5 and 5.7.

###### Proof of Lemma 5.5.

We first prove Part 1. Let $\xi=\sum_{i=1}^{3}\xi_{i}\mathbf{e}_{i}(\theta)\in
P_{\theta,\delta^{1/2}}$. Then using Proposition 1.10,

$\displaystyle|\xi_{1}^{\prime}|:$
$\displaystyle=|\xi\cdot\mathbf{e}_{1}(\theta^{\prime})|\lesssim\delta\cdot
1+\delta^{1/2}\cdot\delta^{1/2}+1\cdot\delta\sim\delta,$
$\displaystyle|\xi_{2}^{\prime}|:$
$\displaystyle=|\xi\cdot\mathbf{e}_{2}(\theta^{\prime})|\lesssim\delta\cdot\delta^{1/2}+\delta^{1/2}\cdot
1+1\cdot\delta^{1/2}\sim\delta^{1/2},$ $\displaystyle\xi_{3}^{\prime}:$
$\displaystyle=\xi\cdot\mathbf{e}_{3}(\theta^{\prime})=O(\delta^{2})+O(\delta^{1/2}\cdot\delta^{1/2})+\xi_{3}(1+O(\delta))$
$\displaystyle\in[K^{-1}/2,2].$

Next, we prove Part 2 by contradiction. Let
$\xi=\sum_{i=1}^{3}\xi_{i}\mathbf{e}_{i}(\theta)\in
C_{1}P_{\theta,\delta^{1/2}}$ and also
$\xi=\sum_{i=1}^{3}\xi^{\prime}_{i}\mathbf{e}_{i}(\theta^{\prime})\in
C_{1}P_{\theta^{\prime},\delta^{1/2}}$. It turns out that we only need to
study $|\xi^{\prime}_{2}|=|\xi\cdot\mathbf{e}_{2}(\theta^{\prime})|\leq
C_{1}\delta^{1/2}$. First, since $\xi\in C_{1}P_{\theta,\delta^{1/2}}$, we
have $\xi_{1}=O(\delta)$ and $\xi_{2}=O(\delta^{1/2})$, and so

$\xi_{2}^{\prime}=\xi\cdot\mathbf{e}_{2}(\theta^{\prime})=\xi_{3}\mathbf{e}_{3}(\theta)\cdot\mathbf{e}_{2}(\theta^{\prime})+O(\delta^{1/2}),$
(5.71)

and so by $\xi_{3}\gtrsim K^{-1}$ we have

$|\mathbf{e}_{3}(\theta)\cdot\mathbf{e}_{2}(\theta^{\prime})|\lesssim
K\delta^{1/2}.$ (5.72)

By Taylor expansion for $\mathbf{e}_{3}$ around $\theta^{\prime}$, we get

$\mathbf{e}_{3}(\theta)=\mathbf{e}_{3}(\theta^{\prime})+\mathbf{e}_{3}^{\prime}(\theta^{\prime})(\theta-\theta^{\prime})+o(\theta-\theta^{\prime})$
(5.73)

where $\frac{|o(\theta-\theta^{\prime})|}{|\theta-\theta^{\prime}|}\rightarrow
0$ as $\theta\rightarrow\theta^{\prime}$. Then

$\begin{split}K\delta^{1/2}\gtrsim&|\mathbf{e}_{3}(\theta)\cdot\mathbf{e}_{2}(\theta^{\prime})|\\\
=&|\\{\mathbf{e}_{3}(\theta^{\prime})+\mathbf{e}_{3}^{\prime}(\theta^{\prime})(\theta-\theta^{\prime})+o(\theta-\theta^{\prime})\\}\cdot\mathbf{e}_{2}(\theta^{\prime})|\\\
=&|\\{-\uptau(\theta^{\prime})\mathbf{e}_{2}(\theta^{\prime})(\theta-\theta^{\prime})+o(\theta-\theta^{\prime})\\}\cdot\mathbf{e}_{2}(\theta^{\prime})|\\\
\gtrsim&|\uptau(\theta^{\prime})||\theta-\theta^{\prime}|-|o(\theta-\theta^{\prime})|\end{split}$

By the condition
$\frac{|o(\theta-\theta^{\prime})|}{|\theta-\theta^{\prime}|}\rightarrow 0$,
combined with $\uptau\sim 1$, we can get

$\begin{split}K\delta^{1/2}\gtrsim|\theta-\theta^{\prime}|\end{split}$ (5.74)

which leads to the contradiction by the assumption on
$|\theta-\theta^{\prime}|$.

∎

###### Proof of Lemma 5.7.

For the first part, let $\xi=\sum_{i=1}^{3}\xi_{i}\mathbf{e}_{i}(\theta)\in
Q_{\theta,\lambda}$. Then using Proposition 1.10 with $\lambda^{-2}\delta^{2}$
in place of $\delta$,

$\displaystyle|\xi_{1}^{\prime}|:$
$\displaystyle=|\xi\cdot\mathbf{e}_{1}(\theta^{\prime})|\lesssim\delta\cdot
1+\lambda\cdot\lambda^{-1}\delta+1\cdot\lambda^{-2}\delta^{2}\sim\delta,$
$\displaystyle\xi_{2}^{\prime}:$
$\displaystyle=\xi\cdot\mathbf{e}_{2}(\theta^{\prime})=O(\delta)+\xi_{2}(1+O(\lambda^{-2}\delta^{2}))+O(\lambda^{-1}\delta)\in[\lambda/4,4\lambda],$
if $C_{0}$ chosen at the beginning of Section 5.8 is large enough,
$\displaystyle\xi_{3}^{\prime}:$
$\displaystyle=\xi\cdot\mathbf{e}_{3}(\theta^{\prime})=O(\lambda^{-2}\delta^{3})+O(\delta)+\xi_{3}(1+O(\lambda^{-2}\delta^{2}))$
$\displaystyle\in[K^{-1}/2,2].$

Next, we prove the disjointedness by contradiction. Let
$\xi=\sum_{i=1}^{3}\xi_{i}\mathbf{e}_{i}(\theta)\in C_{1}Q_{\theta,\lambda}$
and also $\xi=\sum_{i=1}^{3}\xi^{\prime}_{i}\mathbf{e}_{i}(\theta^{\prime})\in
C_{1}Q_{\theta^{\prime},\lambda}$.

It turns out that we only need to study
$|\xi^{\prime}_{1}|=|\xi\cdot\mathbf{e}_{1}(\theta^{\prime})|\leq
C_{1}\delta$. Using Taylor expansion for $\mathbf{e}_{1}(\theta^{\prime})$
around $\theta$, Lemma 1.8 and the assumption that $\xi\in
C_{1}Q_{\theta,\lambda}$, this implies

$|\xi_{2}(\theta^{\prime}-\theta)+\xi_{2}O(|\theta^{\prime}-\theta|^{2})+\xi_{3}O(\theta^{\prime}-\theta)^{2}|=O(\delta),$
(5.75)

and so using $|\xi_{3}|\lesssim 1$, $|\theta-\theta^{\prime}|\leq
C_{2}^{-1}\ll 1$ and $\xi_{2}\sim\lambda$, we get

$\lambda|\theta^{\prime}-\theta|=O(\delta+|\theta^{\prime}-\theta|^{2}).$
(5.76)

Thus we have either $|\theta^{\prime}-\theta|\lesssim\lambda^{-1}\delta$, or
$|\theta^{\prime}-\theta|\gtrsim\lambda$, but both contradict the assumption
if $C_{2}$ is large enough.

∎

##  6 Proof of Theorem 7

The goal of this section is to prove Theorem 7 in [GGG+22] and its corollaries
2 and 3. Corollary 3 was a necessary ingredient for the proof of Theorem 8.
One can keep in mind the following implications where the theorem numbers
correspond to the ones in [GGG+22]:

Theorem 7 $\Rightarrow$ Corollary 2 $\Rightarrow$ Corollary 3 $\Rightarrow$
Theorem 8 $\Rightarrow$ Theorem 2.

######  Theorem 6.1 (Theorem 7 in [GGG+22]).

Fix $0\leq a\leq 2$. For each $\varepsilon>0$ and each $0<c\leq 1$, there
exists a constant $b(\varepsilon,a,c)>0$ so that the following holds. Let
$\delta>0$. Let $\mu$ be a finite nonzero Borel measure supported in the unit
ball in $\mathbb{R}^{3}$ with $\rho_{a}(\mu)<\infty$ (recall (1.14)).

Let $\Theta$ be a maximally $\delta$-separated subset of $[0,1]$. For each
$\theta\in\Theta$, let $\mathbb{T}_{\theta}$ be a collection of disjoint
$\delta\times\delta\times 1$ tubes pointing in the direction $\gamma(\theta)$,
and let $\mathbb{T}=\cup_{\theta\in\Theta}\mathbb{T}_{\theta}$. Suppose that

$\sum_{T\in\mathbb{T}}1_{T}(x)\geq c\delta^{\varepsilon-1},\quad\forall
x\in\mathrm{supp}(\mu).$ (6.1)

Then

$\\#\mathbb{T}\geq
b(\varepsilon,a,c)\mu(\mathbb{R}^{3})\rho_{a}(\mu)^{-1}\delta^{-1-a+100\sqrt{\varepsilon}}.$
(6.2)

The main argument is a high-low decomposition, combined with an induction on
scales argument. Decoupling is also used in the analysis of high part. Note
that we have rescaled the physical space back to the unit ball, and so the
frequency scale will be in $B^{3}(0,\delta^{-1})$.

###  6.1 A bootstrap inequality

To facilitate the induction, we first prove the following bootstrap
inequality.

######  Lemma 6.2.

With the statement of Theorem 6.1, let
$M(\delta,\varepsilon,c)=M_{a}(\delta,\varepsilon,c)$ be the smallest constant
for which

$M(\delta,\varepsilon,c)\\#\mathbb{T}\geq\mu(\mathbb{R}^{3})\rho_{a}(\mu)^{-1}\delta^{-1-a}.$
(6.3)

Then there exists a small $\delta_{0}=\delta_{0}(\varepsilon,a,c)\in(0,1)$
such that for all $\delta\in(0,\delta_{0})$ we have

$M(\delta,\varepsilon,c)\leq\delta^{-2\varepsilon}M\left(\delta^{1-\sqrt{\varepsilon}},\frac{\varepsilon}{1-\sqrt{\varepsilon}},c_{0}c\right)+c^{-4}A(\varepsilon)\delta^{-10\sqrt{\varepsilon}},$
(6.4)

where $c_{0}\in(0,1)$ is an absolute constant and $A(\varepsilon)$ is
decreasing in $\varepsilon$.

Remark. By (6.1) we see that $c\delta^{\varepsilon-1}\leq\\#\Theta$ so

$M(\delta,\varepsilon,c)\leq
c^{-1}\mu(\mathbb{R}^{3})\rho_{a}(\mu)^{-1}\delta^{-a-\varepsilon},$ (6.5)

which is in particular less than $\delta^{-3}$ for $\delta$ small enough.

###### Proof of Theorem 6.1 assuming Lemma 6.2.

The base cases when $\delta\geq\delta_{0}$ or $\varepsilon>1/2$ are trivial,
so it suffices to assume $\delta<\delta_{0}$ and $\varepsilon<1/2$ and iterate
the bootstrap inequality (6.4) in Lemma 6.2.

Let $\delta<\delta_{0}$. We define three sequences
$\delta_{n},\varepsilon_{n},c_{n}$, $n\in\mathbb{N}$ as follows:

$\displaystyle\delta_{1}:=\delta,\quad\varepsilon_{1}:=\varepsilon,\,c_{1}=c$
(6.6)
$\displaystyle\varepsilon_{n+1}:=\frac{\varepsilon_{n}}{1-\sqrt{\varepsilon_{n}}},$
(6.7) $\displaystyle\delta_{n+1}:=\delta_{n}^{1-\sqrt{\varepsilon_{n}}},$
(6.8) $\displaystyle c_{n+1}:=c_{0}^{n}c.$ (6.9)

Then for each $N\in\mathbb{N}$ such that $\delta_{N}<\delta_{0}$ and
$\varepsilon_{N}<1/2$, we have, by direct computation

$M(\delta,\varepsilon,c)\leq\delta^{-2N\varepsilon}M(\delta_{N+1},\varepsilon_{N+1},c_{N+1})+\sum_{n=1}^{N}c_{n}^{-4}A(\varepsilon_{n})\delta^{-2(n-1)\varepsilon}\delta_{n}^{-10{\sqrt{\varepsilon_{n}}}}.$
(6.10)

It is easy to see that $\varepsilon_{n}$ is increasing and bounded from below
by $\varepsilon$. Also, it diverges to $\infty$. By the definition of the
$\delta_{n}$, we have

$\delta_{n+1}=\delta_{n}^{\varepsilon_{n}/\varepsilon_{n+1}}=(\delta_{n-1})^{\frac{\varepsilon_{n-1}}{\varepsilon_{n}}\frac{\varepsilon_{n}}{\varepsilon_{n+1}}}=\delta_{n-1}^{\frac{\varepsilon_{n-1}}{\varepsilon_{n+1}}},$

and by iterating this a few times
$\delta_{n}=\delta^{\frac{\varepsilon}{\varepsilon_{n}}}$. Therefore

$\delta_{n}^{-10\sqrt{\varepsilon_{n}}}=\delta^{-10\frac{\varepsilon}{\sqrt{\varepsilon_{n}}}}\leq\delta^{-10\sqrt{\varepsilon}}.$
(6.11)

If we let $N$ be the smallest integer such that $\varepsilon_{N+1}\geq
1/2>\varepsilon_{N}$, then we have the following claim:

$N\leq 5\varepsilon^{-1/2}.$ (6.12)

Assuming this, using $M(\delta_{N+1},\varepsilon_{N+1},c_{N+1})\lesssim 1$ and
inequality (6.11), we can estimate from (6.10) that

$\displaystyle M(\delta,\varepsilon,c)$
$\displaystyle\lesssim\delta^{-10\sqrt{\varepsilon}}\left(1+\sum_{n=1}^{N}(c_{0}^{n-1}c)^{-4}A(\varepsilon_{n})\delta^{-10\sqrt{\varepsilon}}\right)$
$\displaystyle\leq\delta^{-10\sqrt{\varepsilon}}\left(1+N(c_{0}^{N-1}c)^{-4}A(\varepsilon)\delta^{-10\sqrt{\varepsilon}}\right)$
$\displaystyle\lesssim_{\varepsilon,c}\delta^{-100\sqrt{\varepsilon}},$

which proves Theorem 6.1. It remains to prove (6.12). Write
$x_{n}=\varepsilon_{n}^{-1}$, so we have the iterative formula

$x_{n+1}=x_{n}-\sqrt{x_{n}}.$ (6.13)

Since
$\varepsilon_{N+1}=\frac{\varepsilon_{N}}{1-\sqrt{\varepsilon_{N}}}<\frac{\sqrt{2}}{2(\sqrt{2}-1)}$,
one has $x_{N+1}=\varepsilon_{N+1}^{-1}>2-\sqrt{2}>1$.

Thus $2<x_{n}\leq\varepsilon^{-1}$ for every $1\leq n\leq N$,
$x_{N+1}\in(2-\sqrt{2},2]$, and $(x_{n})$ is decreasing from $1$ to $N+1$. The
function $f(t)=t-\sqrt{t}$ is strictly increasing for $t>1$, and so (6.13)
implies

$x_{n}=x_{n+1}+\frac{1}{2}+\frac{\sqrt{4x_{n+1}+1}}{2}:=f^{-1}(x_{n+1}).$
(6.14)

In reverse order, we write $y_{1}=x_{N+1}$ and $y_{n+1}=f^{-1}(y_{n})$, so
$y_{N+1}=\varepsilon^{-1}$.

The heuristic idea is as follows. Since $f^{-1}(y)>y+\sqrt{y}$, if we view
$y_{n}$ as a continuous function in $n$, then the iterative formula roughly
says $\frac{dy}{dn}>y^{1/2}$, which gives $y_{n}>n^{2}/4$.

The detail is as follows. We actually prove that
$y_{n}>\frac{n^{2}y_{1}}{14}$. The base case $n=1$ is trivial and the case
$n=2$ follows from $y_{2}=x_{N}>2$ and $y_{1}=x_{N+1}\leq 2$. Assume it holds
for some $n\geq 2$, then we have

$\displaystyle y_{n+1}-\frac{(n+1)^{2}y_{1}}{14}$
$\displaystyle=f^{-1}(y_{n})-\frac{(n+1)^{2}y_{1}}{14}$
$\displaystyle>y_{n}+\sqrt{y_{n}}-\frac{(n+1)^{2}y_{1}}{14}$
$\displaystyle\geq\frac{n^{2}y_{1}}{14}+\sqrt{\frac{n^{2}y_{1}}{14}}-\frac{(n+1)^{2}y_{1}}{14}$
$\displaystyle=\frac{\sqrt{y_{1}}}{14}((\sqrt{14}-2\sqrt{y_{1}})n-\sqrt{y_{1}}),$

which is positive since $y_{1}=x_{N+1}\leq 2$ and $n\geq 2$. Using
$y_{N}>N^{2}y_{1}/14$, we thus have

$N<\sqrt{\frac{14y_{N}}{y_{1}}}\leq\sqrt{\frac{14y_{N}}{2-\sqrt{2}}}\leq
5\sqrt{y_{N}}\leq 5\varepsilon^{-1/2}.$ (6.15)

∎

The remaining of this section is devoted to the proof of Lemma 6.2.

###  6.2 Construction of wave packets

This step is the same as Section 5.2 in the proof of Theorem 5.3 except
rescaling.

Fix $\theta\in\Theta$. We define a frequency slab $P_{\theta}$ centred at the
origin that has dimensions $\delta^{-1}\times\delta^{-1}\times 1$, and its
shortest direction is parallel to $\mathbf{e}_{1}(\theta)$. That is,
$P_{\theta}$ is dual to every $T_{\theta}\in\mathbb{T}_{\theta}$. We may also
assume without loss of generality that $\mathbb{T}_{\theta}$ is a
subcollection of the lattice that tiles $[-10,10]^{3}$.

For each $T_{\theta}\in\mathbb{T}_{\theta}$, define a Schwartz function
$\psi_{T_{\theta}}:\mathbb{R}^{3}\to\mathbb{R}$ Fourier supported on
$P_{\theta}$ such that $\psi_{T_{\theta}}\gtrsim 1$ on $T_{\theta}$ and decays
rapidly outside $T_{\theta}$. With this, we define

$f_{\theta}=\sum_{T_{\theta}\in\mathbb{T}_{\theta}}\psi_{T_{\theta}},\quad
f=\sum_{\theta\in\Theta}f_{\theta}.$ (6.16)

Thus $f_{\theta}$ is also Fourier supported on $P_{\theta}$, but physically
essentially supported on $\cup\mathbb{T}_{\theta}$. The function $f$ is
Fourier supported on $\cup_{\theta\in\Theta}P_{\theta}$, which is a union of
$\delta^{-1}\times\delta^{-1}\times 1$ slabs of the same size centered at $0$,
but with normal orientations prescribed by $\mathbf{e}_{1}(\theta)$.

Thus by definition, for every $x\in\mathrm{supp}(\mu)$ we have

$c\delta^{\varepsilon-1}\leq f(x).$ (6.17)

In particular, for every $p<\infty$,

$\mu(\mathbb{R}^{3})c^{p}\delta^{p(\varepsilon-1)}\leq\int|f|^{p}d\mu.$ (6.18)

###  6.3 High-low decomposition

For each $\delta>0$, $\varepsilon>0$, let

$K=K(\delta,\varepsilon)=10^{-4}\delta^{-\sqrt{\varepsilon}}.$ (6.19)

With this, decompose $P_{\theta}$ into:

  1. 1.

The high part:

$P_{\theta,\mathrm{high}}:=\left\\{\sum_{i=1}^{3}\xi_{i}\mathbf{e}_{i}(\theta):|\xi_{1}|\leq
1,K^{-1}\delta^{-1}\leq|\xi_{2}|\leq\delta^{-1},|\xi_{3}|\leq\delta^{-1}\right\\}.$
(6.20)

  2. 2.

The low part:

$P_{\theta,\mathrm{low}}:=\left\\{\sum_{i=1}^{3}\xi_{i}\mathbf{e}_{i}(\theta):|\xi_{1}|\leq
1,|\xi_{2}|\leq K^{-1}\delta^{-1},|\xi_{3}|\leq K^{-1}\delta^{-1}\right\\}.$
(6.21)

  3. 3.

The $\delta^{1/2}$-mixed part:

$P_{\theta,\delta^{1/2}}:=\left\\{\sum_{i=1}^{3}\xi_{i}\mathbf{e}_{i}(\theta):|\xi_{1}|\leq
1,|\xi_{2}|\leq\delta^{-1/2},K^{-1}\delta^{-1}\leq|\xi_{3}|\leq\delta^{-1}\right\\}.$
(6.22)

  4. 4.

For dyadic numbers $\lambda\in(\delta^{1/2},K^{-1}]$, the $\lambda$-mixed
part:

$P_{\theta,\lambda}:=\left\\{\sum_{i=1}^{3}\xi_{i}\mathbf{e}_{i}(\theta):|\xi_{1}|\leq
1,\frac{\lambda}{2}\delta^{-1}\leq|\xi_{2}|\leq\lambda\delta^{-1},K^{-1}\delta^{-1}\leq|\xi_{3}|\leq\delta^{-1}\right\\}.$
(6.23)

We define decompositions as in (5.14) through (5.17). With this, we have the
following pigeonholing argument:

######  Claim 6.3.

Either one of the following is correct:

  * •

High case: there exists a Borel set $F$ with
$\mu(F)\geq\mu(\mathbb{R}^{3})/2$, such that

$\left|f_{\mathrm{high}}(x)+\sum_{\text{dyadic
}\lambda\in[\delta^{1/2},K^{-1}]}f_{\lambda}(x)\right|\geq\frac{|f(x)|}{2}\geq\frac{c\delta^{\varepsilon-1}}{2},\quad\forall
x\in F.$ (6.24)

  * •

Low case: there exists a Borel set $F$ with $\mu(F)\geq\mu(\mathbb{R}^{3})/2$,
such that

$\left|f_{\mathrm{low}}(x)\right|\geq\frac{|f(x)|}{2}\geq\frac{c\delta^{\varepsilon-1}}{2},\quad\forall
x\in F.$ (6.25)

###### Proof.

Define $F=\\{x\in\mathrm{supp}(\mu):|f_{\mathrm{low}}(x)|\geq|f(x)|/2\\}$. If
$\mu(F)\geq\mu(\mathbb{R}^{3})/2$, then we are in the latter case. Otherwise,
$\mu(\mathrm{supp}(\mu)\backslash F)\geq\mu(\mathbb{R}^{3})/2$, and for
$x\in\mathrm{supp}(\mu)\backslash F$, we are in the former case with
$\mathrm{supp}(\mu)\backslash F$ in place of $F$. ∎

###  6.4 Analysis of high and mixed parts

Assume that we are in the high case of Claim 6.3. Note that we have
incorporated the mixed parts into the high part which will make our argument
much simpler.

By the lower bound in (6.17) and the triangle inequality with the fact that we
only have $O(|\log\delta|)$ many dyadic scales $\lambda$, we have

$\int|f_{\mathrm{high}}|^{p}d\mu+\sum_{\text{dyadic
}\lambda\in[\delta^{1/2},K^{-1}]}|f_{\lambda}|^{p}d\mu\gtrapprox
c^{p}\mu(\mathbb{R}^{3})\delta^{p(\varepsilon-1)}.$ (6.26)

One good thing about the high case and the mixed cases is that we can use the
uncertainty principle. More precisely, we have the following lemma.

######  Lemma 6.4.

Let $f$ be a bounded integrable function and $\eta$ be an $L^{1}$-normalised
Schwartz function Fourier supported on a ball $B\subseteq\mathbb{R}^{d}$ of
radius $\delta^{-1}$. Let $a\in(0,d]$ and $\mu$ be a finite Borel measure such
that $\rho_{a}(\mu)<\infty$. Then for each $p\in[1,\infty)$ we have

$\int|f*\eta(x)|^{p}d\mu(x)\lesssim\rho_{a}(\mu)\delta^{a-d}\int|f*\tilde{\eta}(x)|^{p}dx,$
(6.27)

where $\tilde{\eta}$ is a Schwartz function whose Fourier transform is
supported on $2B$ and equals $1$ on $B$.

###### Proof of lemma.

Since $f*\tilde{\eta}*\eta=f*\eta$, it suffices to prove (6.27) with the
integral on the right hand side replaced by $\int|f|^{p}dx$, and then replace
$f$ by $f*\tilde{\eta}$. Next, by Hölder’s inequality we have

$|f*\eta(x)|^{p}\lesssim|f|^{p}*|\eta|(x)\cdot\|\eta\|_{L^{1}}^{p-1}\lesssim|f|^{p}*|\eta|(x),$

and so we may assume $p=1$. By Minkowski’s inequality, it then suffices to
prove

$\int|\eta(y)|d\mu(y)\lesssim\rho_{a}(\mu)\delta^{a-d}.$ (6.28)

To check that, use that from the support assumption on $\eta$ one can write
$\eta(y)=\delta^{-d}\eta_{0}(\delta^{-1}y)$ where $\eta_{0}$ is Fourier
supported in a ball of radius $1$. Then

$\int|\eta(y)|d\mu(y)\lesssim\delta^{-d}\left\\{\int_{|y|\leq\delta}\|\eta_{0}\|_{\infty}d\mu(y)+\sum_{i\geq
0}\int_{|y|\sim\delta 2^{i}}|\eta(\delta^{-1}y)|d\mu(y)\right\\}$

so (6.28) follows by the rapid decay of $\eta_{0}$ and the definition of
$\rho_{a}(\mu)$. ∎

Using the lemma to bound the left hand side of (6.26) (strictly speaking,
$f_{\mathrm{high}}$ and $f_{\lambda}$ should be replaced by functions Fourier
supported on a slightly larger slab, but that does not affect the estimates),
we have

$\int|f_{\mathrm{high}}|^{p}dx+\sum_{\text{dyadic
}\lambda\in[\delta^{1/2},K^{-1}]}|f_{\lambda}|^{p}dx\gtrapprox
c^{p}\mu(\mathbb{R}^{3})\rho_{a}(\mu)^{-1}\delta^{p(\varepsilon-1)+3-a}.$
(6.29)

Suppose $p\geq 4$. We can just use (5.27), (5.44) and (5.62) with $t=1$ and
$\\#\Theta=\delta^{-1}$, noting there is a scaling difference, to get

$\displaystyle\int|f_{\mathrm{high}}|^{p}dx+\sum_{\text{dyadic
}\lambda\in[\delta^{1/2},K^{-1}]}|f_{\lambda}|^{p}dx$
$\displaystyle\lesssim_{\varepsilon}\delta^{3}\left(K^{2}\delta^{-p+1}\left(\sum_{\theta\in\Theta}\\#\mathbb{T}_{\theta}\right)+\sum_{\text{dyadic
}\lambda\in[\delta^{1/2},K^{-1}]}\delta^{-\varepsilon-p+1}\left(\sum_{\theta\in\Theta}\\#\mathbb{T}_{\theta}\right)\lambda^{\frac{p}{2}-2}\right)$
$\displaystyle\lesssim\delta^{-2\sqrt{\varepsilon}-p+4}\left(\sum_{\theta\in\Theta}\\#\mathbb{T}_{\theta}\right)$
$\displaystyle=\delta^{-2\sqrt{\varepsilon}-p+4}\\#\mathbb{T},$

where in the second to last line we used (6.19).

We may take, say, $p=4$. Using this and (6.29), we thus have, for some large
constant $A(\varepsilon)$ independent of $\delta,c$,

$\\#\mathbb{T}\geq
A(\varepsilon)^{-1}c^{4}\mu(\mathbb{R}^{3})\rho_{a}(\mu)^{-1}\delta^{10\sqrt{\varepsilon}-1-a}.$
(6.30)

We have just absorbed the term $\delta^{-4\varepsilon-2\sqrt{\varepsilon}}$
and the logarithmic loss into the term $\delta^{-10\sqrt{\varepsilon}}$, using
$\delta<\delta_{0}$ if $\delta_{0}$ is small enough. This gives in the high
case

$M(\delta,\varepsilon,c)\leq
A(\varepsilon)c^{-4}\delta^{-10\sqrt{\varepsilon}}.$ (6.31)

###  6.5 Analysis of low part

####  6.5.1 Maximally incomparable collection of thickened tubes

For each $T_{\theta}\in\mathbb{T}_{\theta}$, we can view the essential support
of $\psi_{T_{\theta}}*\eta_{\theta,\mathrm{low}}^{\vee}$ as a thickened tube
$\tilde{T}_{\theta}$ with dimensions $1\times K\delta\times K\delta$. Let
$\tilde{\mathbb{T}}_{\theta}$ and thus $\tilde{\mathbb{T}}$ denote the
collection of these thickened tubes.

With this, we say two tubes $\tilde{T}$, $\tilde{T}^{\prime}$ are
$C$-comparable if one is contained in the $C$-time dilation of the other. We
then choose from $\tilde{\mathbb{T}}$ a maximally $1000$-incomparable
subcollection; we still denote this by $\tilde{\mathbb{T}}$ slightly abusing
notation.

We also perform a preliminary reduction here: for each
$\tilde{T}=\tilde{T}_{\theta}\in\tilde{\mathbb{T}}$, we replace it by some
concentric $\tilde{T}_{\theta^{\prime}}$ where $\theta^{\prime}\in
K\delta\mathbb{Z}\cap[0,1]$ is the closest lattice point to $\theta$. Note
that $\tilde{T}_{\theta}$ and $\tilde{T}_{\theta^{\prime}}$ are
$10$-comparable since $|\gamma^{\prime}|=1$. Also, the collection of slightly
distorted tubes $\\{\tilde{T}_{\theta^{\prime}}\\}$ is $10$-incomparable,
since the original collection is $1000$-incomparable.

Abusing notation again, we may thus assume without loss of generality that
$\tilde{\mathbb{T}}=\cup_{\theta\in
K\delta\mathbb{Z}\cap[0,1]}\tilde{\mathbb{T}}_{\theta}$. Also, by losing an
absolute constant we may assume $\mathrm{diam}(\Theta)\ll_{\gamma}1$, so that
we have the relation
$|\gamma(\theta)-\gamma(\theta^{\prime})|\sim|\theta-\theta^{\prime}|$ for all
$\theta,\theta^{\prime}\in\Theta$.

We have the following observations.

  1. 1.

If two tubes $\tilde{T}_{\theta}$, $\tilde{T}^{\prime}_{\theta}$ are from one
single $\tilde{\mathbb{T}}_{\theta}$, then they are $\geq K\delta$ apart.

  2. 2.

If two tubes $\tilde{T}_{\theta}$, $\tilde{T}_{\theta^{\prime}}$ intersect,
then $|\theta-\theta^{\prime}|\geq|\gamma(\theta)-\gamma(\theta^{\prime})|\geq
K\delta$.

  3. 3.

Every $T\in\mathbb{T}$ is contained in the 10000-time dilation of some
$\tilde{T}\in\tilde{\mathbb{T}}$, with possibly different orientations (with
difference less than $O(K\delta)$).

  4. 4.

As a result of Observations 2 and 3, each $T\in\mathbb{T}$ is contained in at
most $O(1)$ thickened tubes of the form $10000\tilde{T}$.

  5. 5.

The 10000-dilation of every $\tilde{T}\in\tilde{\mathbb{T}}$ contains at most
$O(K^{3})$ tubes $T\in\mathbb{T}$. Indeed, each $\tilde{T}$ contains at most
$O(K^{2})$ tubes $T$ from each single direction. Meanwhile, since $\Theta$ is
$\delta$-separated, we have $\\{\gamma(\theta):\theta\in\Theta\\}$ is also
$\delta$-separated. Thus by Observation 3, each $\tilde{T}$ could only contain
tubes $T\in\mathbb{T}$ from at most $O(K)$ directions, from which the result
follows.

  6. 6.

For each $x\in\mathbb{R}^{3}$, it is contained in at most
$O(K^{-1}\delta^{-1})$ thickened tubes from $\tilde{\mathbb{T}}$. Indeed, for
each direction $\theta\in\Theta$, Observation 1 implies that $x$ is contained
in at most $1$ thickened tube from $\tilde{\mathbb{T}}_{\theta}$. Observation
2 implies that if $|\theta-\theta^{\prime}|<K\delta$, i.e.
$|\gamma(\theta)-\gamma(\theta^{\prime})|<K\delta$, then $x$ cannot be
contained in both $\tilde{T}_{\theta}$ and $\tilde{T}_{\theta^{\prime}}$. Thus
there are at most $O(K^{-1}\delta^{-1})$ directions $\theta$ for which there
is a thickened tube in that direction that contains $x$.

####  6.5.2 A pigeonholing argument

Consider Observation 3 above. A priori, each $10000\tilde{T}$ may contain
different numbers of $T\in\mathbb{T}$. We now use a pigeonholing argument to
reduce to the case that each $10000\tilde{T}$ contains at least a large
portion of tubes in $T\in\mathbb{T}$.

Write $\tilde{\mathbb{T}}$ as a disjoint union

$\tilde{\mathbb{T}}=\tilde{\mathbb{T}}_{\mathrm{heavy}}\cup\tilde{\mathbb{T}}_{\mathrm{light}},$
(6.32)

where $\tilde{\mathbb{T}}_{\mathrm{light}}$ is the collection of thickened
tubes $\tilde{\mathbb{T}}$ whose 10000-dilation contains $\leq
c^{\prime}K^{3-\sqrt{\varepsilon}}$ tubes from $\mathbb{T}$. Here $c^{\prime}$
is a small constant to be chosen right below. Note that, using Observation 6,

$\displaystyle\norm*{\sum_{\tilde{T}\in\tilde{\mathbb{T}}_{\mathrm{light}}}\sum_{T\in\mathbb{T}:T\subseteq