# Szemerédi-Trotter bounds for tubes and applications

Ciprian Demeter  Department of Mathematics, Indiana University, 831 East 3rd
St., Bloomington IN 47405<EMAIL_ADDRESS>and  Hong Wang  Courant institute
of mathematical sciences, New York University<EMAIL_ADDRESS>
###### Abstract.

We prove sharp estimates for incidences involving planar tubes that satisfy
packing conditions. We apply them to improve the estimates for the Fourier
transform of fractal measures supported on planar curves.

###### Key words and phrases:

tube incidences, packing conditions, Fourier transform of measures

The first author is partially supported by the NSF grants DMS-2055156 and
DMS-2349828. The second author is supported by NSF CAREER DMS-2238818 and NSF
DMS-2055544

##  1\. Introduction

Given a collection of $\delta\times 1$ tubes in $[0,1]^{2}$ we will denote by
$\mathcal{P}_{r}(\mathcal{T})$ ($\mathcal{P}_{\gtrsim r}(\mathcal{T})$) the
collection of all $\delta$-squares in
$\mathcal{D}_{\delta}=\\{[n\delta,(n+1)\delta]\times[m\delta,(m+1)\delta]:\;0\leq
n,m\leq\delta^{-1}-1\\}$ that intersect $\sim r$ ($\gtrsim r$) of the tubes.

Here is our main result. We refer the unfamiliar reader to Section 2 for
terminology.

######  Theorem 1.1.

Let $s\in(0,1/2]$. Suppose $\Lambda\subset\mathbb{S}^{1}$ is a
$(\delta,s)$-set consisting of $\delta$-intervals (arcs) with cardinality
$|\Lambda|\sim\delta^{-s}$. Suppose that for each $\delta$-arc
$\theta\in\Lambda$ there is a $(\delta,1-s)$-set $\mathcal{T}_{\theta}$ of
$\delta$-tubes in direction (normal to) $\theta$, with
$|\mathcal{T}_{\theta}|\sim\delta^{-1+s}$. Let
$\mathcal{T}=\cup_{\theta\in\Lambda}\mathcal{T}_{\theta}$.

Then for each $\upsilon>0$ there is $C_{\upsilon}>0$ such that for each $1\leq
r\lesssim\delta^{-s}$ we have

(1.1) $|\mathcal{P}_{r}(\mathcal{T})|\leq
C_{\upsilon}\delta^{-\upsilon}\frac{|\mathcal{T}|^{2}}{r^{3}}.$

Theorem 1.1 can be seen as a tube analogue of the celebrated Szemerédi–Trotter
Theorem [17], which asserts that given a family $\mathcal{L}$ of lines in the
plane, the number $|\mathcal{P}_{r}(\mathcal{L})|$ of points that intersect
$r$ of these lines is
$O(\frac{|\mathcal{L}|^{2}}{r^{3}}+\frac{|\mathcal{L}|}{r})$. Note that in our
case $\frac{|\mathcal{T}|}{r}\leq\frac{|\mathcal{T}|^{2}}{r^{3}}$, since
$s\leq 1/2$ and $|\mathcal{T}|\sim\delta^{-1}$.

Let us draw a few comparisons with results in the literature. Inequality (1.1)
was proved in [7] for well spaced collections consisting of $\sim\delta^{-1}$
tubes. In particular, Theorem 1.1 in [7] with $W=\delta^{-1/2}$ proves our
Theorem 1.1 when $s=\frac{1}{2}$, in the particular case when both $\Lambda$
and each $\mathcal{T}_{\theta}$ are well spaced.

We remark that Theorem 1.1 implies (in fact it is equivalent to) the incidence
bound

(1.2)
$\mathcal{I}(\mathcal{P},\mathcal{T})=|\\{(p,T)\in\mathcal{P}\times\mathcal{T}:\;p\cap
T\not=\emptyset\\}|\lessapprox(|\mathcal{P}||\mathcal{T}|)^{2/3},$

for arbitrary collections $\mathcal{P}\subset\mathcal{D}_{\delta}$. Without
any requirement on $\mathcal{T}$, the best possible upper bound is
$\mathcal{I}(\mathcal{P},\mathcal{T})\lesssim\delta^{-1/3}(|\mathcal{P}||\mathcal{T}|)^{2/3}$.
See [5], Theorem 4.3 in [6], or Exercise 7.5 in [4] for slightly different
arguments.

Inequality (1.2) is known to hold when $\mathcal{T}$ is a
$(\delta,\alpha)$-set with cardinality $\sim\delta^{-\alpha}$ and
$\mathcal{P}$ is a $(\delta,\beta)$-set with cardinality
$\sim\delta^{-\beta}$, with $\alpha+\beta=3$. This is a particular case of
Theorem 4.3 in [6]. When $\alpha=1$, this would require $\mathcal{P}$ to have
essentially maximal cardinality $\delta^{-2}$.

It is also worth comparing our result with Theorem 5.2 in [6]. A particular
case of it asserts that if $\mathcal{T}$ is a $(\delta,1)$-set with
cardinality $\sim\delta^{-1}$ and if $\mathcal{P}_{T}$ is a
$(\delta,1-s,K)$-Katz-Tao set of $\delta$-squares intersecting $T$, then,
writing $\mathcal{P}=\cup_{T\in\mathcal{T}}\mathcal{P}_{T}$,

(1.3)
$\sum_{T\in\mathcal{T}}|\mathcal{P}_{T}|\lessapprox(\delta^{-1}K|\mathcal{P}||\mathcal{T}|)^{1/2}.$

It is easy to see that if $\mathcal{T}$ is as in Theorem 1.1, then
$\mathcal{P}_{T}:=\mathcal{P}_{r}(\mathcal{T})\cap T$ is a
$(\delta,1-s,\frac{\log\delta^{-1}}{r\delta^{s}})$-Katz-Tao set. Indeed, fix a
ball $B_{\rho}$ of radius $\rho$, and estimate $|\mathcal{P}_{T}\cap
B_{\rho}|$ by double counting incidences between the squares in
$\mathcal{P}_{T}$ and their associated bushes

$r|\mathcal{P}_{T}\cap B_{\rho}|\lesssim\sum_{\delta\lesssim\eta\lesssim
1:\;dyadic}(\eta/\delta)^{s}\frac{1}{\eta}(\rho\eta/\delta)^{1-s}.$

This is because

1\. there are $\lesssim(\eta/\delta)^{s}$ directions of tubes
$T^{\prime}\in\mathcal{T}$ intersecting $T$ at angle $\sim\eta$

2\. each such $T^{\prime}$ intersects $T$ along $\sim\eta^{-1}$ many
$\delta$-squares

3\. for each direction there are $\lesssim(\rho\eta/\delta)^{1-s}$ tubes in
this direction that intersect $B_{\rho}$.

Thus (1.3) gives

$r|\mathcal{P}_{r}(\mathcal{T})|\lessapprox(\delta^{-1}(r\delta^{s})^{-1}|\mathcal{P}_{r}(\mathcal{T})||\mathcal{T}|)^{1/2},$

which may be written as

$|\mathcal{P}_{r}(\mathcal{T})|\lessapprox\delta^{-s}\frac{|\mathcal{T}|^{2}}{r^{3}}.$

This however falls short of proving (1.1) by the factor $\delta^{-s}$.

Theorem 1.1 is sharp in a number of ways.

First, inequality (1.1) is false when $s>\frac{1}{2}$, even when
$\frac{|\mathcal{T}|^{2}}{r^{3}}$ is replaced with
$\frac{|\mathcal{T}|^{2}}{r^{3}}+\frac{|\mathcal{T}|}{r}$. Indeed, start with
a $(\delta,1-s)$-set $\mathcal{P}\subset\mathcal{D}_{\delta}$ of squares
intersecting $[0,1]\times\\{0\\}$, with cardinality $\delta^{s-1}$. Consider
also an AD-regular set $\Lambda$ as in Theorem 1.1. By this we mean that each
$\rho$-arc centered at some $\theta\in\Lambda$ intersects
$\sim(\rho/\delta)^{s}$ arcs in $\Lambda$. Finally, for each
$p\in\mathcal{P}$, let $\mathcal{T}_{p}$ be the collection of $\delta$-tubes
passing through $p$, with one tube for each direction in $\Lambda$. The
collection $\mathcal{T}=\cup_{p\in\mathcal{P}}\mathcal{T}_{p}$ is as in
Theorem 1.1. An easy computation shows that for each $p\in\mathcal{P}$ and
each $1\leq r\lesssim\delta^{-s}$ we have
$|\mathcal{P}_{r}(\mathcal{T}_{p})|\sim\delta^{-1-s}r^{-\frac{s+1}{s}}$. Thus
$|\mathcal{P}_{r}(\mathcal{T})|\gtrsim\delta^{-2}r^{-\frac{s+1}{s}}$, and this
is much larger than $\delta^{-2}r^{-3}+\delta^{-1}r^{-1}$ when $s>1/2$ and
$r\ll\delta^{-s}$. Note also that $\frac{s+1}{s}=3$ when $s=\frac{1}{2}$.

Second, given (1.1), the index $1-s$ for the dimension of
$|\mathcal{T}_{\theta}|$ cannot be replaced with any other $t$. Indeed, if
$t>1-s$ then we have the area estimate
$\sum_{T\in\mathcal{T}}|T|=\delta^{-1-s-t}\gg\delta^{-2}$. A random choice of
$\mathcal{T}$ subject to the constraints in our theorem would lead to the
(essentially uniform) scenario where a large fraction of
$\mathcal{D}_{\delta}$ is in $\mathcal{P}_{r}(\mathcal{T})$, with
$r\approx\delta^{-s}$. It is easy to see that (1.1) fails for such a
configuration. On the other hand, (1.1) shows that the number of incidences
between tubes in $\mathcal{T}$ and squares in $\mathcal{D}_{\delta}$ is
dominated by

$\sum_{r\geq 1}\frac{|\mathcal{T}|^{2}}{r^{2}}\lessapprox\delta^{-2(t+s)}.$

Since there are $\sim\delta^{-1}|\mathcal{T}|$ incidences, we find that $t$
must be $\geq 1-s$.

Third, it was pointed out to us by Joshua Zahl that Theorem 1.1 is false if
$\mathcal{T}$ is only assumed to be a $(\delta,1)$-set with cardinality
$\sim\delta^{-1}$, without our additional constraints on $\Lambda$ and
$\mathcal{T}_{\theta}$. Let us consider the following train track-type of
example. The set $\mathcal{T}$ will consist of $\delta^{-1/2}$ many bushes.
Each bush contains $\sim\delta^{-1/2}$ tubes that intersect a horizontal
segment (rectangle) $U$ of height $\delta$ and length $\delta^{1/2}$.
Consecutive segments $U$ have the vertical separation $\delta^{1/2}$ between
them. It is easy to see that this $\mathcal{T}$ is a $(\delta,1)$-set. Each of
the $\delta$-squares lying inside one of the segments $U$ is in
$\mathcal{P}_{r}(\mathcal{T})$, with $r\sim\delta^{-1/2}$. Since there are
$\delta^{-1}$ such squares, this violates (1.1). However, $\Lambda$ fails to
be a $(\delta,\frac{1}{2})$-set, as it consists of $\delta^{-1/2}$ consecutive
$\delta$-intervals.

As an application of our theorem we prove the following result.

######  Theorem 1.2.

Let $\Gamma$ be the graph of a $C^{3}$ function $\gamma:[-1,1]\to\mathbb{R}$
satisfying the nonzero curvature condition $\min_{-1\leq x\leq
1}|\gamma^{\prime\prime}(x)|>0$. Let $0<s\leq\frac{1}{2}$. Let $\mu$ be a
Borel measure supported on $\Gamma$ satisfying the Frostman condition

(1.4) $\mu(B(y,r))\lesssim r^{s}$

for each $y\in\mathbb{R}^{2}$ and each $r>0$. Then for each ball
$B_{R}\subset\mathbb{R}^{2}$ of radius $R$ and each $\epsilon>0$ we have

(1.5)
$\|\widehat{\mu}\|^{6}_{L^{6}(B_{R})}\lesssim_{\epsilon}R^{2-2s-\frac{s}{4}+\epsilon}.$

It was conjectured in [10] that the exponent of $R$ should be $2-3s$. In the
same paper, the exponent $2-2s$ is proved as a consequence of a sharp $L^{4}$
estimate. Also, an epsilon improvement over $2-2s$ was achieved when $\Gamma$
is the parabola. A similar improvement was recovered by a different argument
in [3], one that works for all $\Gamma$.

Basic constructions discussed in [10] show that the exponent $2-3s$ is best
possible in (1.5) for each $s\leq\frac{1}{2}$, while the exponent $1-s$ is
best possible in the range $\frac{1}{2}\leq s\leq 1$. The recent paper [13]
proves the expected result with essentially sharp exponent $1-s+\epsilon$ in
the range $s\geq\frac{2}{3}$, when $\Gamma$ is the parabola. The case
$s\leq\frac{1}{2}$ is expected to be more difficult, as it entails full square
root cancellation.

For context, Theorem 1.2 is of similar strength with following statement about
the $\delta$-energy of a finite $\delta$-separated set of points
$S\subset\Gamma$. This is also a consequence of our argument in the last
section.

######  Theorem 1.3.

Let $s\leq\frac{1}{2}$. Assume that for each $y\in\mathbb{R}^{2}$ and each
$\delta\leq r\lesssim 1$ we have

$|S\cap B(y,r)|\lesssim(r/\delta)^{s}.$

Let

$\mathbb{E}_{3,\delta}(S)=|\\{(s_{1},\ldots,s_{6})\in
S^{6}:\;|s_{1}+s_{2}+s_{3}-s_{4}-s_{5}-s_{6}|\lesssim\delta\\}|.$

Then

$\mathbb{E}_{3,\delta}(S)\lesssim_{\epsilon}(\delta^{-s})^{\frac{7}{2}+\frac{1}{4}+\epsilon}.$

The exponent $\frac{7}{2}+\frac{1}{4}$ is slightly worse than the one from the
estimate

$\mathbb{E}_{3}(S)=|\\{(s_{1},\ldots,s_{6})\in
S^{6}:\;s_{1}+s_{2}+s_{3}=s_{4}+s_{5}+s_{6}\\}|\lesssim|S|^{\frac{7}{2}},$

proved in [1] for arbitrary finite subsets of the circle, and then in [2] for
arbitrary finite subsets of the parabola. In both cases, the sharp exponent is
conjectured to be $3$.

It is possible that a refinement of our argument would lead to the better
exponents $\frac{7}{2}+\epsilon$ in Theorem 1.3 and
$2-2s-\frac{s}{2}+\epsilon$ in (1.5). This is certainly true in the AD-regular
case, see Remark 9.2.

One of the main tools we will use in the proof of Theorem 1.1 is the following
recent result of Kevin Ren and the second author, that solves the Furstenberg
set Conjecture.

######  Theorem 1.4 ([14]).

Let $0<s\leq 1$ and $0\leq t\leq 2$. Then for each $\eta,\eta_{1}>0$ there
exists $\delta(\eta,\eta_{1},s,t)>0$ and $\epsilon(\eta,\eta_{1},s,t)>0$ with
$\lim_{\eta\to 0\atop{\eta_{1}\to 0}}\epsilon(\eta,\eta_{1},s,t)=0$, such that
the following holds.

Consider a pair $(\mathcal{P},\mathcal{T})$ consisting of $\delta$-squares and
$\delta$-tubes in $[0,1]^{2}$ with $\delta<\delta(\eta,\eta_{1},s,t)$ such
that

(a) $\mathcal{P}$ is a $(\delta,t,\delta^{-\eta})$-set

(b) for each $p\in\mathcal{P}$ there is a $(\delta,s,\delta^{-\eta_{1}})$-set
$\mathcal{T}(p)\subset\mathcal{T}$ of tubes intersecting $p$, with cardinality
$\sim M$.

Then

(1.6)
$|\mathcal{T}|\gtrsim\delta^{-\min\\{t,\frac{s+t}{2},1\\}+\epsilon(\eta,\eta_{1},s,t)}M.$

The equivalent dual formulation (when $\eta_{1}=\eta$) of this result assumes
$\mathcal{T}$ is a $(\delta,t,\delta^{-\eta})$-set and each $T\in\mathcal{T}$
intersects all squares in some $(\delta,s,\delta^{-\eta})$-set
$\mathcal{P}_{T}\subset\mathcal{P}$ with cardinality $M$. Note that this
forces $M\gtrsim\delta^{-s+\eta}$. Then

$|\mathcal{P}|\gtrsim\delta^{-\min\\{t,\frac{s+t}{2},1\\}+\epsilon(\eta,s,t)}M.$

Our Theorem 1.1 easily implies (a more general form of) this estimate in the
case $t=1$ and $\frac{1}{2}\leq s\leq 1$, when $\mathcal{T}$ has the special
structure in Theorem 1.1 (with $s$ replaced by $1-s$).

######  Corollary 1.5.

Let $\eta>0$ and let $\frac{1}{2}\leq s\leq 1$. The following is true for
$\delta<\delta(\eta)$.

Assume $\mathcal{T}=\cup_{\theta\in\Lambda}\mathcal{T}_{\theta}$, where
$\Lambda$ is a $(\delta,1-s)$-set with size $\sim\delta^{s-1}$, and each
$\mathcal{T}_{\theta}$ is a $(\delta,s)$-set of tubes in the direction
$\theta$, with size $\sim\delta^{-s}$. Consider also a set $\mathcal{P}$ such
that each $T\in\mathcal{T}$ intersects at least $M=\delta^{-s+\eta}$ squares
$\mathcal{P}_{T}\subset\mathcal{P}$. Note that we do not make any non-
concentration assumption on $\mathcal{P}_{T}$.

Then

$|\mathcal{P}|\gtrsim\delta^{-\frac{s+1}{2}+\eta}M.$

###### Proof.

We double count the incidences. We have, with
$L\sim\delta^{\frac{s-1-2\eta}{2}}$

$\displaystyle\delta^{-1}M\leq\mathcal{I}(\mathcal{P},\mathcal{T})$
$\displaystyle\lesssim\sum_{r\text{:
dyadic}}r|\mathcal{P}_{r}(\mathcal{T})\cap\mathcal{P}|$
$\displaystyle=\sum_{r\text{: dyadic }\geq
L}r|\mathcal{P}_{r}(\mathcal{T})|+\sum_{r\text{: dyadic }<L}r|\mathcal{P}|$
$\displaystyle\lesssim\delta^{-\eta/2}\sum_{r\text{: dyadic }\geq
L}r\frac{\delta^{-2}}{r^{3}}+L|\mathcal{P}|$
$\displaystyle\lesssim\delta^{-\eta/2-2}\log(\delta^{-1})L^{-2}+L|\mathcal{P}|.$

We have used Theorem 1.1 for the first term. Our choice of $L$ (and small
$\delta$) implies that the second term must dominate, leading to

$|\mathcal{P}|\gtrsim\delta^{-1}M/L\gtrsim\delta^{-\frac{s+1}{2}+\eta}M.$

∎

Strategy: In Section 2 we recall the essential concepts and basic tools we
will employ throughout the paper. Most of them were known in the form we
present them. One example is the multi-scale decomposition in Lemma 2.10, that
produces both good and bad intervals. However, we also prove a new version of
it, Lemma 2.11, that delivers only good intervals. This will be used in the
form of Theorem 2.13.

Section 3 presents a few probabilistic constructions that are used in Section
4 to prove certain variants of Theorem 1.1. They will in turn be used in the
main argument in Section 8.

In Section 5 a version of the main theorem is analyzed in the simpler context
when the tubes intersecting the $r$-rich squares have certain regularity. It
is this part only that makes use of the estimate (1.6) on the size of
Furstenberg sets. Even in this special setting, Theorem 5.3 only delivers a
dichotomy. In one case, it gives (a superficially stronger form of) the
desired Szemerédi–Trotter-type bound. The second possible scenario is the high
concentration of the rich squares at a certain scale $\Delta$. In Section 6 we
use Fourier analysis to show how this scenario leads to an abundance of tubes
intersecting squares of a certain scale larger than $\delta$.

To make Theorem 5.3 applicable, in Section 7 we use Lemma 2.11 and a
combinatorial argument to create rich subcollections of two-ends tubes.

All these ingredients will be fed into the induction on scales argument
presented in Section 8. Instead of using the full multi-scale decomposition,
this latter argument relies on a two-scale increment provided by Theorem 2.13.
It is in this part of the argument that the novel decomposition in Theorem
2.13 is proving to be crucial. Our main assumption $s\leq\frac{1}{2}$ will
only be used in Step 3 of the argument in Section 8.

In Section 9 we combine the incidence bound in Theorem 1.1 with decoupling to
prove Theorem 1.2.

Most of the arguments are extremely delicate, as we need to keep track of
various epsilons and scales, and to sharply quantify losses in a way that
makes induction on scales viable.

Notation: For positive quantities $A,B$, typically depending on the scale
$\delta$, we write either $A\lesssim B$ or $A=O(B)$ if there is a universal
constant $C$ independent of the scale $\delta$ such that $A\leq CB$. The
notation $A\sim B$ means $A\lesssim B\lesssim A$.

We write $o_{\epsilon}(1)$ for a not necessarily positive quantity (e.g.
$\epsilon^{1/2}$, $-100\epsilon$) that goes to zero, as $\epsilon$ goes to
zero. In particular $O(\epsilon)$ implies $o_{\epsilon}(1)$, but not vice-
versa. Whenever we use $o_{\epsilon}(1)$, it will mean that the sign of this
quantity is irrelevant.

We write $A\sim B\delta^{o_{\epsilon}(1)}$ if

$A/B,B/A\lesssim\delta^{o_{\epsilon}(1)},$

with the implicit constants depending on $\epsilon$, but not on $\delta$.

The symbol $\lessapprox$ will be reserved only to denote arbitrarily small
$\delta^{-\upsilon}$ losses, such as $\log(1/\delta)^{O(1)}$ losses.

######  Acknowledgment.

We are grateful to Tuomas Orponen and Joshua Zahl for helpful conversations.

##  2\. Preliminaries

The reader may assume that all scales $\delta$, $\rho$, etc are dyadic,
meaning that they are in $2^{-\mathbb{N}}$. Smaller $\delta$-squares are
either inside or disjoint from a given larger $\rho$-square. We start by
reviewing standard concepts and terminology.

For a set $A$ in a metric space, we denote by $|A|_{\rho}$ the smallest number
of balls of radius $\rho$ needed to cover $A$. Typically for us, $A$ will be a
disjoint union of either $\delta$-intervals (in $\mathbb{R}$ or
$\mathbb{S}^{1}$), $\delta$-squares, or $\delta$-tubes, for some
$\delta\leq\rho$. There is a map sending points to lines

$(a,b)\mapsto\\{(x,y):\;y=ax+b\\}$

that generates a metric on the space of lines. $\delta$-tubes are (length one
segments of) images of $\delta$-squares under this map. We denote by
$\mathcal{D}^{dual}_{\delta}$ the collection of $\delta$-tubes $T$.

Due to the loose distinction between relations such as $p\cap T\not=\emptyset$
and $p\subset CT$ (with $C=O(1)$), we will think of $\delta$-tubes as
$O(\delta)$-neighborhoods (in the Euclidean metric) of unit line segments.

If $\mathcal{T}$ are $\delta$-tubes, the quantity $|\mathcal{T}|_{\rho}$ is
comparable with the cardinality of the smallest family of $\rho$-tubes that
covers all $T\in\mathcal{T}$.

In all these instances, if $\rho=\delta$, then $|A|_{\delta}$ coincides with
the cardinality $|A|$ of $A$, understood as a collection of intervals,
squares, tubes. We will interchange the use of $|A|_{\delta}$ and $|A|$
throughout the arguments.

######  Definition 2.1 ($(\delta,s,C)$-set ).

For $s\in(0,d]$, a non-empty set $A\subset\mathbb{R}^{d}$ with diameter
$\lesssim 1$ is called a $(\delta,s)$-set if

$|A\cap B(x,r)|_{\delta}\leq Cr^{s}|A|_{\delta},\quad\forall
x\in\mathbb{R}^{d},r\in[\delta,1].$

Testing the inequality with $r\sim 1$ shows that $C\gtrsim 1$. When $C\sim 1$,
we simply call it a $(\delta,s)$-set.

######  Definition 2.2 ($(\delta,s,K)$-Katz-Tao set).

For $s\in(0,d]$, a non-empty set $A\subset\mathbb{R}^{d}$ with diameter
$\lesssim 1$ is called a $(\delta,s,K)$-Katz-Tao set if

$|A\cap B(x,r)|_{\delta}\leq K(r/\delta)^{s},\quad\forall
x\in\mathbb{R}^{d},r\in[\delta,1].$

Testing the inequality with $r\sim\delta$ shows that $K\gtrsim 1$. It will
always be implicitly assumed that $K=O(\delta^{s-d})$, as each set is
trivially Katz-Tao with respect to any larger $K$.

Let $\delta\leq\rho^{\prime}\leq\rho$. Given
$\mathcal{P}\subset\mathcal{D}_{\delta}$ we write
$\mathcal{P}_{\rho}=\\{\mathbf{p}\in\mathcal{D}_{\rho}:\;\mathbf{p}\text{
contains some }p\in\mathcal{P}\\}$. Note that
$\mathcal{P}_{\delta}=\mathcal{P}$. Given $\mathbf{p}\in\mathcal{D}_{\rho}$ we
write

$\mathcal{P}_{\rho^{\prime}}[\mathbf{p}]=\\{\mathbf{p}^{\prime}\in\mathcal{P}_{\rho^{\prime}}:\;\mathbf{p}^{\prime}\subset\mathbf{p}\\}$

and simply

$\mathcal{P}[\mathbf{p}]=\\{p\in\mathcal{P}:\;p\subset\mathbf{p}\\}.$

Given $\mathcal{T}\subset\mathcal{D}_{\delta}^{dual}$ we write
$\mathcal{T}^{\rho}=\\{\mathbf{T}\in\mathcal{D}_{\rho}^{dual}:\;\mathbf{T}\text{
contains some }T\in\mathcal{T}\\}$. Given
$\mathbf{T}\in\mathcal{D}_{\rho}^{dual}$ we write

$\mathcal{T}^{\rho^{\prime}}[\mathbf{T}]=\\{\mathbf{T}^{\prime}\in\mathcal{T}^{\rho^{\prime}}:\;\mathbf{T}^{\prime}\subset\mathbf{T}\\}$

and simply

$\mathcal{T}[\mathbf{T}]=\\{T\in\mathcal{T}:\;T\subset\mathbf{T}\\}.$

One of the typical defects of $\delta$-sets is that they may have different
concentrations inside disjoint balls of a given radius. Another one is that,
while it is not hard for a set $S$ to be a $(\delta,s,O(1))$-set for $s$ close
enough to zero, the size of $S$ may be much larger than $\delta^{-s}$. The
next definition and Lemma 2.7 repair the first defect, and take a step towards
fixing the second one.

######  Definition 2.3 (Uniform sets).

Given $\epsilon>0$, let $T_{\epsilon}$ satisfy
$T_{\epsilon}^{-1}\log(2T_{\epsilon})=\epsilon$. Given $0<\delta\leq
2^{-T_{\epsilon}}$, pick the largest natural number $m$ such that
$\frac{1}{m}\log_{2}\delta^{-1}\geq T_{\epsilon}$ and let
$T=T(\delta,\epsilon)=\frac{1}{m}\log_{2}\delta^{-1}$. Note that
$T_{\epsilon}\leq T\leq 2T_{\epsilon}$, and $\delta=2^{-mT}$.

A set $\mathcal{P}\subset\mathcal{D}_{\delta}$ is called $\epsilon$-uniform if
for each $\rho=\delta^{-jT}$, $0\leq j\leq m$ and each
$\mathbf{p}_{1},\mathbf{p}_{2}\in\mathcal{P}_{\rho}$ we have that

$|\mathcal{P}[\mathbf{p}_{1}]|=|\mathcal{P}[\mathbf{p}_{2}]|.$

A set $\mathcal{T}\subset\mathcal{D}_{\delta}^{dual}$ is called
$\epsilon$-uniform if for each $\rho=\delta^{-jT}$, $0\leq j\leq m$ and each
$\mathbf{T}_{1},\mathbf{T}_{2}\in\mathcal{T}^{\rho}$ we have that

$|\mathcal{T}[\mathbf{T}_{1}]|=|\mathcal{T}[\mathbf{T}_{2}]|.$

Given the correspondence between squares and tubes, the two definitions are in
sync with each other. We will typically formulate auxiliary results for
squares, but they hold equally well for tubes. A similar definition also holds
for intervals on the real line, or on $\mathbb{S}^{1}$.

From now on, it will always be implicitly assume that $\delta$ is small
enough, depending on the context.

######  Lemma 2.4.

Given $\epsilon>0$, there is $C(\epsilon)>0$ such that the following holds.
For each $\epsilon$-uniform set $\mathcal{P}\subset\mathcal{D}_{\delta}$, each
$\delta\leq\rho\leq 1$, and for each
$\mathbf{p}_{1},\mathbf{p}_{2}\in\mathcal{P}_{\rho}$ we have that

$C(\epsilon)^{-1}|\mathcal{P}[\mathbf{p}_{1}]|\leq|\mathcal{P}[\mathbf{p}_{2}]|\leq
C(\epsilon)|\mathcal{P}[\mathbf{p}_{1}]|.$

###### Proof.

For each $2^{-jT}<\rho<2^{-(j-1)T}$, we use the fact that each $\rho$-square
contains at most $2^{2T}$ many $2^{-jT}$-squares. Thus the inequalities are
satisfied with the constant $2^{2T}$. We take $C(\epsilon)=2^{4T_{\epsilon}}$,
since this only depends on $\epsilon$.

∎

######  Remark 2.5.

Lemma (2.4) shows that each $\epsilon$-uniform set is also an approximate
$\epsilon$-uniform set. By this we mean that the equality in Definition 2.3 is
replaced with a double inequality involving a constant $C(\epsilon)$. We will
apply this observation with $\upsilon=\upsilon(\epsilon)$. What matters is
that $C(\epsilon)=C(\epsilon^{-1}(\upsilon))=O_{\upsilon}(1)$, meaning that
the constant only depends on $\upsilon$. All results and arguments continue to
hold for such an approximate $\upsilon(\epsilon)$-uniform set, and the
implicit constants will carry further dependence on $\upsilon(\epsilon)$.

######  Lemma 2.6.

If $\mathcal{P}$ is an $\epsilon$-uniform $(\delta,s,C_{1})$-set and
$\delta\leq\rho$, then $\mathcal{P}_{\rho}$ is a
$(\rho,s,C(\epsilon)C_{1})$-set. Moreover, its cardinality satisfies
$|\mathcal{P}_{\rho}|\gtrsim C_{1}^{-1}\rho^{-s}$.

###### Proof.

Let $\rho\leq\rho^{\prime}\leq 1$ and let
$\mathbf{p}^{\prime}\in\mathcal{P}_{\rho^{\prime}}$. Let $n^{\prime}$ be the
average value of $|\mathcal{P}[\mathbf{p}]|$ for
$\mathbf{p}\in\mathcal{P}_{\rho}[\mathbf{p}^{\prime}]$. Then
$n^{\prime}|\mathcal{P}_{\rho}[\mathbf{p}^{\prime}]|\leq
C_{1}|\mathcal{P}|(\rho^{\prime})^{s}$. Let $n$ be the average value of
$|\mathcal{P}[\mathbf{p}]|$ for $\mathbf{p}\in\mathcal{P}_{\rho}$. Then
$n|\mathcal{P}_{\rho}|=|\mathcal{P}|$. Uniformity implies that $n\leq
C(\epsilon)n^{\prime}$. The combination of these shows that
$|\mathcal{P}_{\rho}[\mathbf{p}^{\prime}]|\leq
C(\epsilon)C_{1}|\mathcal{P}_{\rho}|(\rho^{\prime})^{s}$.

The inequality $|\mathcal{P}_{\rho}|\gtrsim C_{1}^{-1}\rho^{-s}$ follows since
$n\leq C_{1}|\mathcal{P}|\rho^{s}.$

∎

We also recall Lemma 2.15 in [12].

######  Lemma 2.7.

Let $\epsilon>0$ and $\delta<\delta(\epsilon)$. Then each set
$\mathcal{P}\subset\mathcal{D}_{\delta}$ has an $\epsilon$-uniform subset
containing at least a $\sim\delta^{\epsilon}$-fraction of the original set.

######  Lemma 2.8.

Each $\epsilon$-uniform $(\delta,s,C)$-set $S\subset\mathcal{D}_{\delta}$ with
cardinality $|S|\gtrsim\delta^{-s}$ can be partitioned into
$\sim_{\epsilon}|S|\delta^{s}$ sets $S_{i}$ which are
$(\delta,s,O_{\epsilon}(C))$-sets, each with cardinality
$\sim_{\epsilon}\delta^{-s}$.

###### Proof.

Recall that $\delta=2^{-Tm}$, with $T$ depending only on $\epsilon$.

Start testing from the bottom of the dyadic tree of scales. Let $\rho_{1}$ be
the smallest scale of the form $2^{-lm}$ such that the number $M_{1}$ of
squares in $S$ inside a nonempty $\rho_{1}$-square $Q_{1}$ (this number is the
same for all such $Q_{1}$) satisfies $M_{1}\geq C(\rho_{1}/\delta)^{s}$.
Create $M_{1}/(C(\rho_{1}/\delta)^{s})$ collections, each with
$C(\rho_{1}/\delta)^{s}$ squares inside each $Q_{1}$.

Let $\rho_{2}>\rho_{1}$ be the smallest scale of the form $2^{-lm}$ such that
the number $M_{2}$ of squares $Q_{1}$ inside a nonempty $\rho_{2}$-square
$Q_{2}$ satisfies $M_{2}\geq C(\rho_{2}/\rho_{1})^{s}$. Create
$M_{2}/(C(\rho_{2}/\rho_{1})^{s})$ collections of squares $Q_{1}$, each with
$C(\rho_{2}/\rho_{1})^{s}$ such squares inside each $Q_{2}$.

When this process stops, it produces scales $\rho_{1},\ldots,\rho_{L}$,
numbers $M_{1},\ldots,M_{L}$ and sets $S_{1},\ldots,S_{I}$. Each set $S_{i}$
satisfies for each $r\geq\delta$

$|S_{i}\cap B(x,r)|\lesssim_{\epsilon}C(r/\delta)^{s}.$

There are $L\leq T$ steps, and the size discrepancy at each step is at most a
factor of 2. So all sets $S_{i}$ have cardinality within a multiplicative
factor of size $O_{\epsilon}(1)$.

It remains to prove that $I\lesssim_{\epsilon}|S|\delta^{s}$, since this will
force each set to have cardinality $\gtrsim\delta^{-s}$ (thus in fact
$\sim\delta^{-s}$).

Note that (since $C\geq 1$)

$I\leq 2^{T}\frac{M_{1}\cdot\ldots\cdot M_{L}}{(\rho_{L}/\delta)^{s}}.$

The numerator is the number of $\delta$-squares inside a $\rho_{L}$-square. By
hypothesis, this is $\lesssim|S|\rho_{L}^{s}$.

∎

######  Lemma 2.9.

Let $\epsilon>0$. Each $(\delta,s,K)$-Katz-Tao set $S$ can be partitioned into
$O(\delta^{-\epsilon}\log(1/\delta))$ many $\epsilon$-uniform
$(\delta,s,K)$-Katz-Tao sets.

###### Proof.

Each subset of a $(\delta,s,K)$-Katz-Tao set inherits this property. Consider
an $\epsilon$-uniform subset $S_{1}$ of $S$, with size
$\gtrsim\delta^{\epsilon}|S|$, cf. Lemma 2.7. Then consider an
$\epsilon$-uniform subset $S_{2}$ of $S\setminus S_{1}$, with size
$\gtrsim\delta^{\epsilon}|S\setminus S_{1}|$. Repeat the process. Note that it
takes at most $n=O(\delta^{-\epsilon})$ steps to achieve half size
$|S\setminus(S_{1}\cup\ldots\cup S_{n})|<|S|/2$. We repeat this until the
leftover has size $O(\delta^{-\epsilon})$. Each of the remaining $s\in S$ is a
one-element uniform set.

∎

We recall the following multi-scale decomposition, whose formulation has
evolved in recent years, see e.g. [15], [16] and [12].

######  Lemma 2.10.

For every $\epsilon>0$ there is $\tau=\tau(\epsilon)>0$ such that the
following holds: for each non-decreasing $C$-Lipschitz function
$f:[0,1]\to[0,\infty)$ there is a partition

$0=a_{1}<a_{2}<\ldots<a_{J+1}=1$

and a sequence

$\sigma_{1}<\sigma_{2}<\ldots<\sigma_{J}\leq C$

such that

(i) for each $j$ we have $\tau\leq a_{j+1}-a_{j}$

(ii) for each $1\leq j_{1}<j_{0}\leq J$ we have

$\sum_{j=j_{1}}^{j_{0}}\sigma_{j}(a_{j+1}-a_{j})\geq
f(a_{j_{0}+1})-f(a_{j_{1}})-\epsilon$

(iii) for each $j$ we have

$f(x)\geq f(a_{j})+\sigma_{j}(x-a_{j}),\;a_{j}\leq x\leq a_{j+1}.$

###### Proof.

Items (i) and (iii) as well as (ii) for $j_{1}=1$, $j_{0}=J$ follow by
combining Lemma 5.20 and 5.21 in [16]. See also Lemma 2.10 in [12]. The fact
that (ii) holds for all $j_{0},j_{1}$ follows from this and summation of (iii)
(with $x=a_{j+1}$) over $j$.

∎

In previous literature Lemma 2.10 was typically used to classify the intervals
$[a_{j},a_{j+1}]$ as either “good” or “bad”. We will prove a more robust
version of the previous lemma, which guarantees all intervals are good. The
key difference is that between (ii) in Lemma 2.10 and the corresponding (2.10)
in Lemma 2.11, the latter being what defines an interval as being good. This
difference is very subtle, and can only be fully appreciated in the argument
from Section 8. More precisely, the $(\rho_{0}/\delta)^{o_{\epsilon}(1)}$-loss
in (8.58) is finely tuned into the induction on scales. On the other hand,
(ii) in Lemma 2.10 would only produce the very harmful
$(1/\delta)^{o_{\epsilon}(1)}$-loss. We mention that this refined version
proved in the next lemma is only needed once in our paper, in the afore-
mentioned argument from Section 8. That proof requires the first interval to
be good. For the other two applications in Sections 5 and 7, Lemma 2.10 would
suffice.

######  Lemma 2.11.

For each small enough $\epsilon>0$ we let

(2.7) $\epsilon_{0}=\epsilon^{2\epsilon^{-1}}.$

Then for each non-decreasing $1$-Lipschitz function $f:[0,1]\to[0,1]$ there
exists a partition

$0=A_{1}<A_{2}<\ldots<A_{L+1}=1$

and a sequence

$0\leq t_{1}<t_{2}<\ldots<t_{L}\leq 1$

such that and for each $1\leq l\leq L$

(2.8) $A_{l+1}-A_{l}\geq\epsilon_{0}\epsilon^{-1},$

(2.9) $f(x)\geq f(A_{l})+t_{l}(x-A_{l})-\epsilon(A_{l+1}-A_{l}),\;A_{l}\leq
x\leq A_{l+1},$ (2.10) $f(A_{l+1})\leq
f(A_{l})+(t_{l}+3\epsilon)(A_{l+1}-A_{l}),$

(2.11) $t_{1}\leq f(1)-f(0)+\epsilon.$

###### Proof.

Apply Lemma 2.10 to $f$ with $\epsilon$ replaced by $\epsilon_{0}$. The lemma
produces the numbers $0\leq a_{j}\leq 1$ and $0\leq\sigma_{j}\leq 1$. Only
(ii) and (iii) in the lemma will be used.

Write $s_{k}=\epsilon k$, $0\leq k\leq\epsilon^{-1}-1$. Let $I_{k}$ be the
union of the intervals $[a_{j},a_{j+1}]$ with $\sigma_{j}\in[s_{k},s_{k+1})$.
When $k=\epsilon^{-1}-1$, we consider the closed interval
$\sigma_{j}\in[s_{k},s_{k+1}]$. Since $\sigma_{j}<\sigma_{j+1}$ for all $1\leq
j\leq J-1$, $I_{k}$ is a union of consecutive intervals, and so it is either a
nonempty interval, or the empty set. Also,
$\\{I_{k}\\}_{k=0}^{\epsilon^{-1}-1}$ form a partition of $[0,1]$, and
$I_{{k+1}}$ follows right after $I_{k}$.

Let $k^{(1)}$ be the index that corresponds to the largest value $|I_{k}|$.
Write $I_{k^{(1)}}=[a,b]$. If there are multiple choices, we choose an
arbitrary one. We have $|I_{k^{(1)}}|\geq\epsilon$. In particular, due to
(2.7) we have

(2.12) $|I_{k^{(1)}}|\geq\epsilon_{0}\epsilon^{-1}.$

Let $I_{k_{l}}=[a_{ll},a_{lr}]$ be an interval on the left hand side of
$I_{k^{(1)}}$ such that $|I_{k_{l}}|\geq\epsilon^{2}|I_{k^{(1)}}|$ and
$|I_{k}|<\epsilon^{2}|I_{k^{(1)}}|$ for each $k_{l}<k<k^{(1)}$. Such an
interval may not exist, in which case we let $a_{ll}=a_{lr}=a$. But if it
does, it is unique.

Similarly, we let $I_{k_{r}}=[a_{rl},a_{rr}]$ be an interval on the right hand
side of $I_{k^{(1)}}$ such that $|I_{k_{r}}|\geq\epsilon^{2}|I_{k^{(1)}}|$ and
$|I_{k}|<\epsilon^{2}|I_{k^{(1)}}|$ for each $k^{(1)}<k<k_{r}$. If such an
interval does not exist, we let $a_{rl}=a_{rr}=b$.

We choose $[a_{lr},a_{rl}]$ to be the first interval in our partition of
$[0,1]$, call it $[A_{l_{1}},A_{l_{1}+1}]$. We assign it the value
$t_{l_{1}}=s_{k^{(1)}}$. Note that (2.8) is satisfied due to (2.12).

We claim that

(2.13) $f(x)\geq
f(a_{lr})+s_{k^{(1)}}(x-a_{lr})-\epsilon(a_{rl}-a_{lr}),\;\;a_{lr}\leq x\leq
a_{rl}$

and

(2.14) $f(a_{rl})\leq f(a_{lr})+(s_{k^{(1)}}+3\epsilon)(a_{rl}-a_{lr}).$

We first prove (2.13). Recall that $[a_{lr},a]$ is the union of (at most
$\epsilon^{-1}$) intervals $I_{k}$, each with length
$<\epsilon^{2}|I_{k^{(1)}}|$. Thus

(2.15) $a-a_{lr}\leq\epsilon|I_{k^{(1)}}|\leq\epsilon(a_{rl}-a_{lr}).$

Similarly,

(2.16) $a_{rl}-b\leq\epsilon(a_{rl}-a_{lr}).$

Recall that $s_{k^{(1)}}\leq 1$. Thus, when $x\in(a_{lr},a]$, by monotonicity
and (2.15)

$f(x)\geq f(a_{lr})\geq
f(a_{lr})+s_{k^{(1)}}(x-a_{lr})-\epsilon(a_{rl}-a_{lr}).$

When $x\in[a,a_{rl}]$, then due to Lemma 2.10 (iii) and (2.15)

$\displaystyle f(x)$ $\displaystyle\geq f(a)+s_{k^{(1)}}(x-a)$
$\displaystyle\geq
f(a_{lr})+s_{k^{(1)}}(x-a)+s_{k^{(1)}}(a-a_{lr})-\epsilon(a_{rl}-a_{lr})$
$\displaystyle=f(a_{lr})+s_{k^{(1)}}(x-a_{lr})-\epsilon(a_{rl}-a_{lr}).$

Next we verify (2.14). By first using the 1-Lipschitz property, then Lemma
2.10 (ii) we find

$f(a_{rl})\leq
f(b)+(a_{rl}-b)\leq\epsilon_{0}+f(a_{lr})+(s_{k^{(1)}}+\epsilon)(b-a_{lr})+(a_{rl}-b).$

Since (recall (2.16)) $a_{rl}-b\leq\epsilon(a_{rl}-a_{lr})$ and (recall
(2.12)) $\epsilon_{0}\leq\epsilon(b-a)\leq\epsilon(a_{rl}-a_{lr})$, we have

$f(a_{rl})\leq f(a_{lr})+(s_{k^{(1)}}+3\epsilon)(a_{rl}-a_{lr}).$

If $a_{ll}>0$, we iterate the same process to merge $I_{k_{l}}:=I_{k^{(2)}}$
with intervals $I_{k}$ on the left hand side of $I_{k_{l}}$ that satisfy
$|I_{k}|<\epsilon^{2}|I_{k_{l}}|$. Similarly, if $a_{rr}<1$, we merge
$I_{k_{r}}:=I_{k^{(3)}}$ with intervals on the right hand side of $I_{k_{r}}$
that satisfy $|I_{k}|<\epsilon^{2}|I_{k_{r}}|$. This will produce the next two
intervals $[A_{l_{2}},A_{l_{2}+1}],\;[A_{l_{3}},A_{l_{3}+1}]$ in our
partition. Recall that we have control over the drop in size

$|I_{k^{(2)}}|,\;|I_{k^{(3)}}|\geq\epsilon^{2}|I_{k^{(1)}}|\geq\epsilon^{3}.$

We iterate this process until we exhaust all intervals. We create the last
interval of the partition by using $I_{k^{(L)}}$. Note that
$|I_{k^{(L)}}|\geq|I_{k^{(1)}}|\epsilon^{2(L-1)}\geq\epsilon^{2L-1}$. Since
there are only $\epsilon^{-1}$ intervals $I_{k}$ to start with, we have that
$L\leq\epsilon^{-1}$. When combined with (2.7) this gives

$|I_{k^{(L)}}|\geq\epsilon_{0}\epsilon^{-1}.$

As we have seen earlier, it is this inequality that allows for the
verification of (2.14) for the last selected interval
$[A_{l_{L}},A_{l_{L+1}}]$. Our choice for small $\epsilon_{0}$ guarantees that
this inequality is preserved throughout the whole selection process. The key
is that we knew a priori the upper bound $\epsilon^{-1}$ on the number of
steps. Note also that (2.8) is immediate and (2.9) follows as in the first
step of the iteration.

Finally, the fact that $t_{l}$ is increasing follows from the fact that
$\sigma_{j}$ is increasing, while (2.11) follows by summing (2.9) with
$x=A_{l+1}$ and using the fact that $t_{1}$ is the smallest among all $t_{l}$.

∎

######  Definition 2.12.

We define the branching function $\beta:[0,m]\to[0,m]$ of an
$\epsilon$-uniform set $\mathcal{P}$ of $\delta$-intervals with parameters
$T,m$ to be

$\beta(j)=\frac{\log_{2}|\mathcal{P}|_{2^{-jT}}}{T},\;0\leq j\leq m$

for integers, and then we interpolate linearly. A similar definition holds for
tubes and squares.

This function is $1$-Lipschitz.

The next theorem is one of the main novelties in our paper. It is a direct
consequence of Lemma 2.11 applied to the branching function. Predecessors of
this result have established themselves as chief tools in fractal analysis.
They essentially produce a sequence of scales
$\delta=\delta^{A_{L+1}}>\delta^{A_{L}}>\ldots>\delta^{A_{1}}=1$ such that,
apart from negligible $\epsilon$-losses, the dyadic tree of each uniform set
$S$ consists of $O(1)$ many layers, each of which is a (rescaled)
$(\delta^{A_{l+1}-A_{l}},t_{l},O(1))$-set with cardinality
$\sim\delta^{(A_{l}-A_{l+1})t_{l}}$. This is almost as good as $S$ being a
$(\delta,t,O(1))$-set with size $\sim\delta^{-t}$. Such a set would only have
one layer, $L=1$, and $t_{1}=t$.

######  Theorem 2.13 (Multi-scale decomposition with good intervals).

For each small enough $\epsilon>0$ we let

(2.17) $\epsilon_{0}=\epsilon^{2\epsilon^{-1}}.$

Then for each $\epsilon_{0}$-uniform set $S$ of $\delta$-intervals (or
equivalently, $\delta$-tubes intersecting a fixed $\delta$-square) there
exists a partition

$0=A_{1}<A_{2}<\ldots<A_{L+1}=1$

and a sequence

$0\leq t_{1}<t_{2}<\ldots<t_{L}\leq 1$

such that

(i) $A_{l+1}-A_{l}\geq\psi(\epsilon):=\epsilon_{0}\epsilon^{-1}$,

(ii)
$\log_{1/\delta}(\frac{|S_{\delta^{A_{l+1}}}|}{|S_{\delta^{A_{l}}}|})\leq(t_{l}+3\epsilon)(A_{l+1}-A_{l})$,

(iii) for each $I\in S_{\delta^{A_{l}}}$ the $\times\delta^{-A_{l}}$ rescaled
copy of $S_{\delta^{A_{l+1}}}[I]$ is a
$(\delta^{A_{l+1}-A_{l}},t_{l},O(\delta^{(A_{l}-A_{l+1})\varphi(\epsilon)}))$-set
with size in the interval

$[O(\delta^{(A_{l}-A_{l+1})(t_{l}-\varphi(\epsilon))}),O(\delta^{(A_{l}-A_{l+1})(t_{l}+\varphi(\epsilon))})]$

(iv) $t_{1}\leq\log_{1/\delta}|S|+\epsilon$.

The non negative function $\varphi(\epsilon)$ satisfies $\lim_{\epsilon\to
0}\varphi(\epsilon)=0$.

While we may take $\varphi(\epsilon)=3\epsilon$, the exact value of
$\varphi(\epsilon)$ will be irrelevant. The same applies to $\psi$, whose only
requirement is to be $>0$. Calling these quantities $\varphi(\epsilon)$ and
$\psi(\epsilon)$ will help us distinguish their role in our later arguments.
See for example (8.60) and (8.61).

There are similar versions of Lemma 2.11 and of Theorem 2.13 for the branching
functions of higher dimensional collections, such as squares and tubes.

##  3\. Probabilistic arguments

Katz-Tao sets may sometimes have too small of a size, that is not reflective
of their dimension. In this section we use random rigid motions to correct
this deficit.

######  Lemma 3.1.

Let $S$ be an $\epsilon$-uniform $(\delta,s,K)$-Katz-Tao set of intervals in
$\mathcal{D}_{\delta}$. Let $\upsilon>0$. Then, for each
$\delta<\delta(\epsilon,\upsilon)$ there is
$T\subset[0,1]\cap\delta\mathbb{Z}$ with size
$N\sim\frac{\delta^{-s}K}{|S|_{\delta}}$ such that the set

$A:=\cup_{t\in T}(t+S)$

(a) has size
$K\delta^{-s}\gtrsim|A|_{\delta}\gtrsim\delta^{\upsilon}K\delta^{-s}$

(b) is a $(\delta,s,O_{\epsilon}(K\delta^{-\upsilon}))$-Katz-Tao set

(c) each $\delta$-interval is in at most $\delta^{-\upsilon}$ of the
translates $t+S$, $t\in T$.

###### Proof.

As a result of Stirling’s formula, for each $1\leq M\leq N$

${N\choose M}\lesssim(\frac{Ne}{M})^{M+\frac{1}{2}}.$

We select a random set $T$ of $N\sim\frac{\delta^{-s}K}{|S|_{\delta}}$
distinct translations. First, let us gauge the size of the resulting set $A$.
Fix a $\delta$-interval $I_{\delta}=[n\delta,(n+1)\delta]$. The probability
that a random translate $t+S$ will contain $I_{\delta}$ is at most
$\delta|S|_{\delta}$. The probability that at least $\delta^{-\upsilon}$
translates will contain $I_{\delta}$ is at most

$\sum_{M\geq\delta^{-\upsilon}}{N\choose M}(\delta|S|_{\delta})^{M}\lesssim
N^{3/2}\max_{M\geq\delta^{-\upsilon}}(\frac{Ne\delta|S|_{\delta}}{M})^{M}.$

Since $K\lesssim\delta^{s-1}$ and $N\leq\delta^{-1}$, this number is
$O(\delta^{-\frac{3}{2}+\upsilon\delta^{-\upsilon}})$, and thus
$O(\delta^{100})$, assuming $\delta$ is small enough.

It follows that the probability that at least $\delta^{-\upsilon}$ of the $N$
chosen translates will contain some $I_{\delta}$ is $O(\delta^{99})$.
Equivalently, with probability $1-O(\delta^{99})$, the set $T$ will satisfy

(3.18) $\|\sum_{t\in T}1_{t+S}\|_{\infty}\leq\delta^{-\upsilon}.$

Then (a) follows by combining this with

$\delta N|S|_{\delta}=\|\sum_{t\in
T}1_{t+S}\|_{1}\leq\delta|A|_{\delta}\|\sum_{t\in T}1_{t+S}\|_{\infty}.$

Note also that (c) follows from (3.18). For (b), it suffices to consider
$I_{\rho}$ with length $\rho=2^{-jT}\geq\delta$. The set $S$ may be covered
with $|S|_{\rho}$ $\rho$-intervals, each containing $|S|_{\delta}/|S|_{\rho}$
intervals $I_{\delta}$. We have two possibilities.

1\. If $|S|_{\rho}\rho^{s}\delta^{-\upsilon}\geq 1$, then any choice of $N$
translates will contribute at most

$\leq N\frac{|S|_{\delta}}{|S|_{\rho}}\lesssim
K\delta^{-\upsilon}(\frac{\rho}{\delta})^{s}$

$\delta$-intervals to $I_{\rho}$. Thus, the non-concentration condition is
verified in this case, with probability 1.

2\. Assume $|S|_{\rho}\rho^{s}\delta^{-\upsilon}\leq 1$. The probability that
$t+S$ intersects $I_{\rho}$ is at most $\rho|S|_{\rho}$. If it does intersect
$I_{\rho}$, it contributes at most $|S|_{\delta}/|S|_{\rho}$
$\delta$-intervals. We have

${\mathbb{P}}(I_{\rho}\text{ contains
}\geq\delta^{-\upsilon}K(\rho/\delta)^{s}\text{ intervals from }A)\leq$
${\mathbb{P}}(I_{\rho}\cap(t+S)\not=\emptyset\text{ for
}M\geq\delta^{-\upsilon}K(\rho/\delta)^{s}|S|_{\rho}/|S|_{\delta}\text{
translates }t).$

Our non-concentration hypothesis together with the main assumption in this
case implies that

$N\geq\delta^{-\upsilon}K(\rho/\delta)^{s}|S|_{\rho}/|S|_{\delta}\geq\delta^{-\upsilon}.$

The same computation shows $M\geq\delta^{-\upsilon}$. The fact that $N$ is at
least as large as the smallest admissible $N$ shows that the above probability
is nonzero, so it demands an estimate. As before, this probability is

$\displaystyle\lesssim
N^{3/2}\max_{M\geq\delta^{-\upsilon}K(\rho/\delta)^{s}|S|_{\rho}/|S|_{\delta}}(\frac{N}{M}\rho
e|S|_{\rho})^{M}$ $\displaystyle\lesssim
N^{3/2}\max_{M\geq\delta^{-\upsilon}}(\frac{N}{M}\rho e|S|_{\rho})^{M}$
$\displaystyle\lesssim\delta^{-3/2}(e\rho^{1-s}\delta^{\upsilon})^{\delta^{-\upsilon}}\leq\delta^{100},$

when $\delta$ is small enough.

We only need to test (b) for (at most $O(\delta^{-1})$) intervals $I_{\rho}$,
with $\rho$ of the form $2^{-jT}$. For all other scales, the non-concentration
inequality will be forced to hold with $O_{\epsilon}(1)$ losses.

We conclude that (a) and (b) can be simultaneously achieved, with probability
very close to 1.

∎

######  Remark 3.2.

Since $(\log\delta^{-1})^{\log\delta^{-1}}\gtrsim\delta^{-100}$, the factor
$\delta^{-\upsilon}$ may be replaced with $\log\delta^{-1}$ in the previous
lemma. This remains true for the applications of this result in this section
and the following one.

######  Corollary 3.3.

Suppose $\Lambda\subset\mathbb{S}^{1}$ is a $(\delta,s,K_{1})$-Katz-Tao
$\epsilon$-uniform set such that for each $\delta$-arc $\theta\in\Lambda$,
there is a $(\delta,1-s,K_{2})$-Katz-Tao $\epsilon$-uniform set
$\mathcal{T}_{\theta}$ of $\delta$-tubes in the direction $\theta$. Assume all
$|\mathcal{T}_{\theta}|_{\delta}$ are comparable. Let
$\mathcal{T}=\cup_{\theta\in\Lambda}\mathcal{T}_{\theta}$.

Let $\upsilon>0$. Then there is a collection of
$N\sim\frac{\delta^{-1}K_{1}K_{2}}{|\mathcal{T}|_{\delta}}$ rigid motions
$\mathcal{A}_{i}$ such that

$\mathcal{T}_{new}=\cup_{i=1}^{N}\mathcal{A}_{i}(\mathcal{T})$

is a collection of tubes $($in the slightly larger set $[-2,2]^{2})$ having
the following structure: there is a
$(\delta,s,O(K_{1}\delta^{-\upsilon}))$-Katz-Tao set
$\Lambda_{new}\subset\mathbb{S}^{1}$ with cardinality
$\delta^{\upsilon}K_{1}\delta^{-s}\lesssim|\Lambda_{new}|\lesssim
K_{1}\delta^{-s}$ and for each $\delta$-arc $\theta\in\Lambda_{new}$ there is
a $(\delta,1-s,O(K_{2}\delta^{-\upsilon}))$-Katz-Tao set
$\mathcal{T}_{\theta,new}$ of tubes with cardinality
$\delta^{\upsilon}K_{2}\delta^{s-1}\lesssim|\mathcal{T}_{\theta,new}|\lesssim
K_{2}\delta^{s-1}$ such that

$\mathcal{T}_{new}=\bigcup_{\theta\in\Lambda_{new}}\mathcal{T}_{\theta,new}.$

Moreover, each $T\in\mathcal{T}_{new}$ appears in at most
$O(\delta^{-\upsilon})$ many collections $\mathcal{A}_{i}(\mathcal{T})$. All
implicit constants are allowed to depend on $\upsilon,\epsilon$.

###### Proof.

This is a two-step procedure. First, we use (a close variant of) Lemma 3.1 to
find a collection $\mathcal{R}$ of
$N_{1}\sim\frac{\delta^{-s}K_{1}}{|\Lambda|_{\delta}}$ rotations, such that

$\Lambda_{new}=\bigcup_{r\in{\mathcal{R}}}r(\Lambda)$

satisfies the desired properties. For $\theta^{\prime}\in\Lambda_{new}$,
$\theta^{\prime}=r(\theta)$ with $\theta\in\Lambda$ and $r\in\mathcal{R}$, we
define $\mathcal{T}_{\theta^{\prime}}$ to be $r(\mathcal{T}_{\theta})$.

Second, we use again Lemma 3.1 to find
$N_{2}\sim\frac{\delta^{s-1}K_{2}}{|\mathcal{T}_{\theta}|_{\delta}}$
translates $T$ such that for each $\theta^{\prime}\in\Lambda_{new}$, the set

$\mathcal{T}_{\theta^{\prime},new}=\cup_{t\in
T}(t+\mathcal{T}_{\theta^{\prime}})$

satisfies the desired properties. The fact that we can realize this with the
same $T$ for each $\theta^{\prime}\in\Lambda_{new}$ is due the high
probabilities we have computed in the proof of the lemma.

The rigid motions $\mathcal{A}_{i}$ will be all possible combinations of
rotations $r\in\mathcal{R}$ followed by translations $t\in T$. Note that
$N\sim N_{1}N_{2}$.

∎

We are now able to formulate a sharper version of Theorem 1.4, when it comes
to the dependence of $\epsilon$ on $\eta$. While this will not be needed in
this paper, it may be useful in future applications.

######  Theorem 3.4.

Let $0<s\leq 1$ and $0\leq t\leq 2$. Then for each $\upsilon_{1},\eta_{1}>0$
there is $\epsilon_{1}(\upsilon_{1},\eta_{1},s,t)$ with
$\lim_{\upsilon_{1},\eta_{1}\to 0}\epsilon_{1}(\upsilon_{1},\eta_{1},s,t)=0$,
such that the following holds for each $\eta>0$ and each (small enough)
$\delta$.

Consider a pair $(\mathcal{P},\mathcal{T})$ consisting of $\delta$-squares and
$\delta$-tubes in $[0,1]^{2}$ such that

(a) $\mathcal{P}$ is a $(\delta,t,\delta^{-\eta})$-set

(b) for each $p\in\mathcal{P}$ there is a $(\delta,s,\delta^{-\eta_{1}})$-set
$\mathcal{T}(p)\subset\mathcal{T}$ of tubes intersecting $p$, with cardinality
$\sim M$.

Then for each $\upsilon_{1}>0$

$|\mathcal{T}|\gtrsim\delta^{-\min\\{t,\frac{s+t}{2},1\\}+\eta+\epsilon_{1}(\upsilon_{1},\eta_{1},s,t)}M.$

###### Proof.

Let first $\mathcal{P}_{unif}$ be a $\upsilon_{1}$-uniform subset of
$\mathcal{P}$ with
$|\mathcal{P}_{unif}|\gtrsim\delta^{-\upsilon_{1}}|\mathcal{P}|$. We apply
(the two dimensional version of) Lemma 3.1 to $\mathcal{P}_{unif}$ with
$K=|\mathcal{P}_{unif}|\delta^{t}\delta^{-\eta-\upsilon_{1}}$. Using a
collection $T$ of $N\sim\delta^{-\eta-\upsilon_{1}}$ translates, we get a
$(\delta,t,O(K\delta^{-\upsilon_{1}}))$-Katz-Tao set
$\mathcal{P}_{new}=\cup_{t\in T}(t+\mathcal{P}_{unif})$ with

$K\delta^{-t+\upsilon_{1}}\lesssim|\mathcal{P}_{new}|\lesssim K\delta^{-t}.$

Note that these turn $\mathcal{P}_{new}$ into a
$(\delta,t,\delta^{-2\upsilon_{1}})$-set. Note also that
$|\mathcal{P}_{new}|\lesssim|\mathcal{P}|\delta^{-\eta-\upsilon_{1}}$.

For each $p\in\mathcal{P}_{new}$ lying in some translate
$t^{\prime}+\mathcal{P}_{unif}$, $p=t^{\prime}+p^{\prime}$, define
$\mathcal{T}(p)=t^{\prime}+\mathcal{T}(p^{\prime})$. If there is more than one
such pair $(p^{\prime},t^{\prime})$, just pick any of them.

We apply Theorem 1.4 to this configuration to get

$|\bigcup_{p\in\mathcal{P}_{new}}\mathcal{T}(p)|\gtrsim\delta^{-\min\\{t,\frac{s+t}{2},1\\}+\epsilon(2\upsilon_{1},\eta_{1},s,t)}M.$

Note also that

$\displaystyle|\bigcup_{p\in\mathcal{P}_{new}}\mathcal{T}(p)|$
$\displaystyle\leq|\bigcup_{p^{\prime}\in\mathcal{P}_{unif}}\bigcup_{t^{\prime}\in
T}(t^{\prime}+\mathcal{T}(p^{\prime}))|$
$\displaystyle=|\bigcup_{t^{\prime}\in
T}(t^{\prime}+\bigcup_{p^{\prime}\in\mathcal{P}_{unif}}\mathcal{T}(p^{\prime}))|$
$\displaystyle\leq|T||\bigcup_{p^{\prime}\in\mathcal{P}_{unif}}\mathcal{T}(p^{\prime})|.$

Combining these we conclude that

$|\mathcal{T}|\geq|\bigcup_{p^{\prime}\in\mathcal{P}_{unif}}\mathcal{T}(p^{\prime})|\gtrsim\delta^{-\min\\{t,\frac{s+t}{2},1\\}+\eta+\upsilon_{1}+\epsilon(2\upsilon_{1},\eta_{1},s,t)}M.$

Our result is thus verified with
$\epsilon_{1}(\upsilon_{1},\eta_{1},s,t)=\epsilon(2\upsilon_{1},\eta_{1},s,t)+\upsilon_{1}$.
∎

##  4\. Consequences of Theorem 1.1

Let us write $\mathcal{S}\mathcal{T}(\delta)$ for the smallest constant that
satisfies the inequality

(4.19)
$|\mathcal{P}_{r}(\mathcal{T})|_{\delta}\leq\mathcal{S}\mathcal{T}(\delta)\frac{|\mathcal{T}|^{2}}{r^{3}}$

for all $\Lambda,\mathcal{T},r$ as in Theorem 1.1. Recall that we need to
prove that $\mathcal{S}\mathcal{T}(\delta)\lessapprox 1$.

We will derive similar inequalities for collections of tubes that are slightly
less structured than those in Theorem 1.1. These inequalities will be used for
larger scales, in the final argument. In this section we will not impose the
restrictive assumption $s\leq 1/2$, as we only compare, rather than explicitly
estimate various upper bounds.

######  Proposition 4.1.

Let $s\in(0,1)$. Suppose $\Lambda\subset\mathbb{S}^{1}$ is a
$(\delta,s,1)$-set such that for each $\delta$-arc $\theta\in\Lambda$, there
is a $(\delta,1-s,1)$-set $\mathcal{T}_{\theta}$ of $\delta$-tubes in
direction $\theta$. We assume $|\mathcal{T}_{\theta}|$ are comparable. Let
$\mathcal{T}=\cup_{\theta\in\Lambda}\mathcal{T}_{\theta}$. Then for each
$r\geq 1$

$|\mathcal{P}_{r}(\mathcal{T})|\lessapprox\mathcal{S}\mathcal{T}(\delta)\delta\frac{|\mathcal{T}|^{3}}{r^{3}}.$

Note that our bound $\delta|\mathcal{T}|^{3}/r^{3}$ is larger than
$|\mathcal{T}|^{2}/r^{3}$, since the hypothesis forces
$|\mathcal{T}|\gtrsim\delta^{-1}$.

###### Proof.

Step 1. We assume both $\Lambda$ and each $\mathcal{T}_{\theta}$ are
$\eta$-uniform for some fixed $\eta$. Apply Lemma 2.8 first to the direction
set, then to each $\mathcal{T}_{\theta}$ to decompose
$\mathcal{T}=\cup_{i=1}^{M}\mathcal{T}_{i}$ such that each $\mathcal{T}_{i}$
is a set of tubes satisfying the assumption of Theorem 1.1 and
$M\sim\delta|\mathcal{T}|$.

By dyadic pigeonholing, there exists a number $N\leq M$ and a subset
$\mathcal{P}\subset\mathcal{P}_{r}(\mathcal{T})$ such that
$|\mathcal{P}|\gtrsim(\log(1/\delta)^{-1}|\mathcal{P}_{r}(\mathcal{T})|$ and
for each $p\in\mathcal{P}$ there exist $\gtrsim N/\log(1/\delta)$ sets
$\mathcal{T}_{i}$ such that $p\in\mathcal{P}_{r/N}(\mathcal{T}_{i})$. Apply
double counting and Theorem 1.1 as follows

$\displaystyle|\mathcal{P}_{r}(\mathcal{T})|\lesssim|\mathcal{P}|\log(1/\delta)$
$\displaystyle\lesssim\log(1/\delta)^{2}N^{-1}\sum_{i=1}^{M}|\mathcal{P}_{r/N}(\mathcal{T}_{i})|$
$\displaystyle\lessapprox\mathcal{S}\mathcal{T}(\delta)\frac{M}{N}\frac{\delta^{-2}}{(r/N)^{3}}\leq\mathcal{S}\mathcal{T}(\delta)\frac{\delta^{-2}M^{3}}{r^{3}}$
$\displaystyle\sim\mathcal{S}\mathcal{T}(\delta)\delta\frac{|\mathcal{T}|^{3}}{r^{3}}.$

Step 2. We prove the general case. For $K_{1},K_{2}\geq 1$, let
$C(K_{1},K_{2})$ be the smallest constant such that

$|\mathcal{P}_{r}(\mathcal{T})|\leq
C(K_{1},K_{2})\delta\frac{|\mathcal{T}|^{3}}{r^{3}}$

holds for each $\mathcal{T}$ as in our hypothesis, where
$K_{1}=|\Lambda|_{\delta}\delta^{s}$,
$K_{2}\sim|\mathcal{T}_{\theta}|_{\delta}$. Let us call such $\mathcal{T}$ a
$(K_{1},K_{2})$-set.

We need to prove that
$C(K_{1},K_{2})\lessapprox\mathcal{S}\mathcal{T}(\delta)$, with implicit
constant independent of $K_{1},K_{2}$. We achieve this by proving a
bootstrapping inequality.

Apply Lemma 2.9 to write each $\mathcal{T}_{\theta}$ as a union of
$O(\delta^{-2\eta})$ $\eta$-uniform $(\delta,1-s,K_{2})$-Katz-Tao sets
$\mathcal{T}_{\theta,i}$. By allowing some of these sets to be empty, we split
$\mathcal{T}$ into $O(\delta^{-2\eta}\log(1/\delta))$ sets $\mathcal{T}_{j}$
such that each $\mathcal{T}_{j}$ contains at most one $\mathcal{T}_{\theta,i}$
for each $\theta$ (by that we mean, it either contains all tubes from some
$\mathcal{T}_{\theta,i}$ and no tubes from any other
$\mathcal{T}_{\theta,i^{\prime}}$, or it contains no tubes in the direction
$\theta$). Moreover, we require that $|\mathcal{T}_{\theta,i}|$ have similar
size for all $\theta$ contributing to $\mathcal{T}_{j}$. This is achieved by
selecting sets $\mathcal{T}_{\theta,i}$ with size inside a fixed dyadic level.

For each $\mathcal{T}_{j}$ we call $\Lambda_{j}\subset\Lambda$ the
corresponding set of contributing directions. We apply Lemma 2.9 again to
partition each $\Lambda_{j}$ into $O(\delta^{-2\eta})$ many $\eta$-uniform
$(\delta,s,K_{1})$-Katz-Tao sets $\Lambda_{j,k}$. Overall, we have a partition
of $\mathcal{T}$ into $O(\delta^{-5\eta})$ such sets, and we call them
$\mathcal{T}^{l}$. We split indices $l$ into two categories:
$\mathcal{L}_{large}$ will be those $l$ for which
$|\mathcal{T}^{l}|\geq\delta^{C_{1}\eta}|\mathcal{T}|$, where $C_{1}$ is a
large enough absolute constant to be chosen later. Also, we will call
$\mathcal{L}_{small}$ those $l$ for which
$|\mathcal{T}^{l}|<\delta^{C_{1}\eta}|\mathcal{T}|$. The starting point is the
following inequality

(4.20)
$|\mathcal{P}_{r}(\mathcal{T})|\lesssim_{\eta}\delta^{-5\eta}\max_{l}|\mathcal{P}_{r\delta^{5\eta}}(\mathcal{T}^{l})|.$

We first evaluate $|\mathcal{P}_{r\delta^{5\eta}}(\mathcal{T}^{l})|$ when
$l\in\mathcal{L}_{large}$. We note that each such $\mathcal{T}^{l}$ falls into
the category analyzed in Step 1, modulo $\delta^{-C_{1}\eta}$ losses in
constants. More precisely, we must have
$|\Lambda^{l}|\gtrsim\delta^{C_{1}\eta}|\Lambda|$ and
$|\mathcal{T}^{l}_{\theta}|\gtrsim\delta^{C_{1}\eta}|\mathcal{T}_{\theta}|$
for each $\theta\in\Lambda^{l}$. This shows that $\Lambda^{l}$ is a
$(\delta,s,\delta^{-C_{1}\eta})$-set, and each $\mathcal{T}^{l}_{\theta}$ is a
$(\delta,1-s,\delta^{-C_{1}\eta})$-set. A small modification of the argument
in Step 1 shows that

(4.21)
$\max_{l\in\mathcal{L}_{large}}|\mathcal{P}_{r\delta^{5\eta}}(\mathcal{T}^{l})|\lesssim\mathcal{S}\mathcal{T}(\delta)\delta^{1-C_{2}\eta}\frac{|\mathcal{T}^{l}|^{3}}{(r\delta^{5\eta)^{3}}}\leq\mathcal{S}\mathcal{T}(\delta)\delta^{1-C_{3}\eta}\frac{|\mathcal{T}|^{3}}{r^{3}}.$

Here $C_{2}$ is an absolute constant, and $C_{3}=C_{2}+15$.

Next we evaluate $|\mathcal{P}_{r\delta^{5\eta}}(\mathcal{T}^{l})|$ when
$l\in\mathcal{L}_{small}$. The previous argument is no longer efficient, since
the sets $\mathcal{T}^{l}$ are now much smaller than the original
$\mathcal{T}$. To make them large enough, we apply Corollary 3.3 to each
$\mathcal{T}^{l}$. This result provides
$N\sim\frac{\delta^{-1}K_{1}K_{2}}{|\mathcal{T}^{l}|}$ rigid motions
$\\{\mathcal{A}_{i}\\}_{i=1}^{M}$ and a set $\mathcal{T}^{l}_{new}$ with size
$\sim\delta^{-1}K_{1}K_{2}$ that satisfies the hypothesis of our corollary
with $\delta^{-\eta}$-losses. In other words, $\mathcal{T}^{l}_{new}$ is a
$(K_{1}\delta^{-O(\eta)},K_{2}\delta^{-O(\eta}))$-set. It is not difficult to
prove that

$C(K_{1}\delta^{-O(\eta)},K_{2}\delta^{-O(\eta}))\lesssim\delta^{-C_{4}\eta}C(K_{1},K_{2}),$

for some absolute constant $C_{4}$.

Note that the sets
$\mathcal{P}_{i}=\mathcal{P}_{r\delta^{5\eta}}(\mathcal{A}_{i}(\mathcal{T}^{l}))$
have the same cardinality as $\mathcal{P}_{r\delta^{5\eta}}(\mathcal{T}^{l})$,
as they are rigid motions of the latter. We use the fact that

$N|\mathcal{P}_{r\delta^{5\eta}}(\mathcal{T}^{l})|=\sum_{i=1}^{N}|\mathcal{P}_{i}|=\sum_{1\leq
M\leq N}M\\#\\{q:\;q\text{ is in }M\text{ sets }\mathcal{P}_{i}\\}.$

Since Corollary 3.3 guarantees that each tube in $\mathcal{T}^{l}_{new}$
appears in at most $O(\delta^{-\eta})$ of the sets
$\mathcal{A}_{i}(\mathcal{T}^{l})$, it follows that each $q$ that is in $M$
sets $\mathcal{P}_{i}$ is in fact in $\mathcal{P}_{L}(\mathcal{T}^{l}_{new})$,
for some $Mr\delta^{6\eta}\lesssim L\lesssim Mr\delta^{5\eta}$. It follows
that

$\sum_{1\leq M\leq N}M\\#\\{q:\;q\text{ is in }M\text{ sets
}\mathcal{P}_{i}\\}\lesssim\log(1/\delta)\delta^{-6\eta}\max_{L\gtrsim
r\delta^{6\eta}}\frac{L}{r}|\mathcal{P}_{L}(\mathcal{T}^{l}_{new})|.$

Then

$\frac{L}{r}|\mathcal{P}_{L}(\mathcal{T}_{new}^{l})|\lesssim
C(K_{1},K_{2})\delta^{1-C_{4}\eta}\frac{|\mathcal{T}_{new}^{l}|^{3}}{rL^{2}}$
$\lesssim
C(K_{1},K_{2})\delta^{1-(C_{4}+12)\eta}\frac{(K_{1}K_{2}\delta^{-1})^{3}}{r^{3}}.$

We conclude that

$\max_{l\in\mathcal{L}_{small}}|\mathcal{P}_{r\delta^{5\eta}}(\mathcal{T}^{l})|\lesssim
C(K_{1},K_{2})\frac{\log(1/\delta)\delta^{1-(C_{4}+18)\eta}}{N}\frac{|\mathcal{T}|^{3}}{r^{3}}$
$\lesssim
C(K_{1},K_{2})\log(1/\delta)\delta^{1+(C_{1}-C_{4}-18)\eta}\frac{|\mathcal{T}|^{3}}{r^{3}}.$

We have used the fact that $N\gtrsim\delta^{-C_{1}\eta}$ for each
$l\in\mathcal{L}_{small}$. At this point, we choose $C_{1}=C_{4}+100$.

Finally, we combine this inequality with (4.20), (4.21) and the definition of
$C(K_{1},K_{2})$ to get

$C(K_{1},K_{2})\leq
C_{\eta}\delta^{-O(\eta)}\mathcal{S}\mathcal{T}(\delta)+\frac{C(K_{1},K_{2})}{2}.$

Since the argument works for all $\eta>0$, we conclude that
$C(K_{1},K_{2})\lessapprox\mathcal{S}\mathcal{T}(\delta)$, as desired.

∎

We next use the previous proposition to prove the following more general
version. See also Remark 4.3.

######  Proposition 4.2.

Let $s\in(0,1)$. Suppose $\Lambda\subset\mathbb{S}^{1}$ is a
$(\delta,s,K_{1})$-Katz-Tao set such that for each $\delta$-arc
$\theta\in\Lambda$, there is a $(\delta,1-s,K_{2})$-Katz-Tao set
$\mathcal{T}_{\theta}$ of $\delta$-tubes in direction $\theta$. Let
$\mathcal{T}=\cup_{\theta\in\Lambda}\mathcal{T}_{\theta}$. Then for each
$r\geq 1$

$|\mathcal{P}_{r}(\mathcal{T})|\lessapprox\mathcal{S}\mathcal{T}(\delta)(K_{1}K_{2})^{2}\delta^{-1}\frac{|\mathcal{T}|}{r^{3}}.$

Note that when $K_{1},K_{2}\sim 1$, this estimate is superficially weaker than
$|\mathcal{T}|^{2}/r^{3}$.

###### Proof.

To get a first glimpse into the numerology, we test the inequality with $r\sim
1$. In this case the trivial bound
$|\mathcal{P}_{r}(\mathcal{T})|\lesssim\delta^{-1}|\mathcal{T}|$ implies the
desired estimate, since $K_{1},K_{2}\gtrsim 1$.

The forthcoming argument recycles the elements of the proof of the previous
Proposition.

Step 1. We first prove the result under the assumptions that $\Lambda$ and
each $\mathcal{T}_{\theta}$ are $\eta$-uniform, and moreover, all
$|\mathcal{T}_{\theta}|$ are comparable. Here $\eta$ is arbitrarily small.

We apply Corollary 3.3. It provides
$N\sim\frac{\delta^{-1}K_{1}K_{2}}{|\mathcal{T}|}$ rigid motions
$\\{\mathcal{A}_{i}\\}_{i=1}^{M}$ and a set $\mathcal{T}_{new}$ satisfying the
assumptions of Proposition 4.1, with $\delta^{-\eta}$-losses.

Note that the sets
$\mathcal{P}_{i}=\mathcal{P}_{r}(\mathcal{A}_{i}(\mathcal{T}))$ have the same
cardinality as $\mathcal{P}_{r}(\mathcal{T})$, as they are rigid motions of
the latter. We use the fact that

$N|\mathcal{P}_{r}(\mathcal{T})|=\sum_{i=1}^{N}|\mathcal{P}_{i}|=\sum_{1\leq
M\leq N}M\\#\\{q:\;q\text{ is in }M\text{ sets }\mathcal{P}_{i}\\}.$

Since Corollary 3.3 guarantees that each tube in $\mathcal{T}_{new}$ appears
in at most $O(\delta^{-\eta})$ of the sets $\mathcal{A}_{i}(\mathcal{T})$, it
follows that each $q$ that is in $M$ sets $\mathcal{P}_{i}$ is in fact in
$\mathcal{P}_{L}(\mathcal{T}_{new})$, for some $Mr\delta^{\eta}\lesssim
L\lesssim Mr$. It follows that

$\sum_{1\leq M\leq N}M\\#\\{q:\;q\text{ is in }M\text{ sets
}\mathcal{P}_{i}\\}\lesssim\log(1/\delta)\delta^{-\eta}\max_{L\gtrsim
r\delta^{\eta}}\frac{L}{r}|\mathcal{P}_{L}(\mathcal{T}_{new})|.$

Then by Proposition 4.1,

$\frac{L}{r}|\mathcal{P}_{L}(\mathcal{T}_{new})|\lessapprox\mathcal{S}\mathcal{T}(\delta)\delta\frac{|\mathcal{T}_{new}|^{3}}{rL^{2}}\lesssim\mathcal{S}\mathcal{T}(\delta)\delta^{1-3\eta}\frac{|\mathcal{T}_{new}|^{3}}{r^{3}}$
$\lesssim\mathcal{S}\mathcal{T}(\delta)\delta^{1-3\eta}\frac{(K_{1}K_{2}\delta^{-1})^{3}}{r^{3}}.$

We conclude that

$|\mathcal{P}_{r}(\mathcal{T})|\lesssim\mathcal{S}\mathcal{T}(\delta)\log(1/\delta)\delta^{1-4\eta}\frac{(K_{1}K_{2}\delta^{-1})^{3}}{Nr^{3}}$
$\sim\mathcal{S}\mathcal{T}(\delta)\log(1/\delta)\delta^{-1-4\eta}(K_{1}K_{2})^{2}\frac{|\mathcal{T}|}{r^{3}}.$

Step 2. (the general case) Fix $\eta>0$. Apply Lemma 2.9 like in the proof of
Corollary 4.1, to partition $\mathcal{T}$ into $O(\delta^{-5\eta})$ sets
$\mathcal{T}^{l}$ that fall into the category covered by Step 1. Using Step 1,
we conclude that

$|\mathcal{P}_{r}(\mathcal{T})|\lesssim_{\eta}\delta^{-5\eta}\max_{l}|\mathcal{P}_{r\delta^{5\eta}}(\mathcal{T}^{l})|\lesssim_{\eta}\mathcal{S}\mathcal{T}(\delta)(K_{1}K_{2})^{2}\delta^{-1-O(\eta)}\frac{|\mathcal{T}|}{r^{3}}.$

The conclusion follows since $\eta>0$ is arbitrary.

∎

######  Remark 4.3.

The following general version of Proposition 4.1 is an immediate consequence
of Proposition 4.2. Suppose $\Lambda\subset\mathbb{S}^{1}$ is a
$(\delta,s,C_{1})$-set such that for each $\delta$-arc $\theta\in\Lambda$,
there is a $(\delta,1-s,C_{2})$-set $\mathcal{T}_{\theta}$ of $\delta$-tubes
in direction $\theta$, with all $|\mathcal{T}_{\theta}|$ of similar size. Then
for any $r\geq 1$

$|\mathcal{P}_{r}(\mathcal{T})|\lessapprox\mathcal{S}\mathcal{T}(\delta)(C_{1}C_{2})^{2}\delta\frac{|\mathcal{T}|^{3}}{r^{3}}.$

Indeed, apply Proposition 4.2 to $\mathcal{T}$ with
$K_{1}=C_{1}\delta^{s}|\Lambda|$,
$K_{2}=C_{2}\delta^{1-s}|\mathcal{T}_{\theta}|$.

We close this section with one last result of the same type.

######  Proposition 4.4.

Suppose $\Lambda\subset\mathbb{S}^{1}$ is an $\eta$-uniform $(\delta,s,C)$-set
with $|\Lambda|\sim\delta^{-s}$ and suppose that for each $\delta$-arc
$\theta\in\Lambda$ there is a $(\delta,1-s,C)$-set $\mathcal{T}_{\theta}$ of
$\delta$-tubes in direction $\theta$ with
$|\mathcal{T}_{\theta}|\sim\delta^{-1+s}$. Let
$\mathcal{T}=\cup_{\theta\in\Lambda}\mathcal{T}_{\theta}$, so
$|\mathcal{T}|\sim\delta^{-1}$. Assume $\mathcal{T}$ is $\eta$-uniform.

Let $\rho_{0}\geq\delta$, and consider the collection $\mathcal{T}^{\rho_{0}}$
of $\rho_{0}$-thickenings of $\mathcal{T}$.

Then for any $\mathbf{r}\geq 1$,

$|\mathcal{P}_{\mathbf{r}}(\mathcal{T}^{\rho_{0}})|\lessapprox\mathcal{S}\mathcal{T}(\rho_{0})\rho_{0}\frac{|\mathcal{T}^{\rho_{0}}|^{3}}{\mathbf{r}^{3}}.$

The implicit constants will depend on $C$ and $\eta$.

###### Proof.

It suffices to prove that the collection $\mathcal{T}^{\rho_{0}}$ satisfies
the assumptions of Proposition 4.1 with $\delta$ replaced with $\rho_{0}$.

First, by Lemma 2.6 the set $\Lambda_{\rho_{0}}$ is a $(\rho_{0},s,O(C))$-set.

Second, we claim that for any $\rho_{0}$-arc $\tau\in\Lambda_{\rho_{0}}$,
$\mathcal{T}^{\rho_{0}}_{\tau}:=\\{T_{\rho_{0}}\in\mathcal{T}^{\rho_{0}}:\text{dir}(T_{\rho_{0}})=\tau\\}$
is a $(\rho_{0},1-s,O(C))$-set. Indeed, since $\mathcal{T}$ is uniform, the
number $N=\\{T\in\mathcal{T}:\;T\subset T_{\rho_{0}}\\}$ is roughly the same
for each $T_{\rho_{0}}\in\mathcal{T}^{\rho_{0}}$. There are
$\sim\\#\\{\theta\subset\tau\\}\delta^{s-1}$ tubes $T\in\mathcal{T}$ with
direction $\theta\in\tau$. Each of these $T$ lies inside some
$T_{\rho_{0}}\in\mathcal{T}_{\tau}^{\rho_{0}}$. Thus, there are
$\sim\frac{1}{N}\\#\\{\theta\subset\tau\\}\delta^{s-1}$ fat tubes in
$\mathcal{T}_{\tau}^{\rho_{0}}$.

Similarly, for each $\delta\leq\Delta\leq 1$ there are
$\lesssim\frac{1}{N}\\#\\{\theta\subset\tau\\}(\delta/\Delta)^{s-1}$ fat tubes
$T_{\rho_{0}}\in\mathcal{T}_{\tau}^{\rho_{0}}$ inside each $\Delta\times
1$-box in the direction $\tau$. Combining these it follows that
$\mathcal{T}_{\tau}^{\rho_{0}}$ is a $(\rho_{0},1-s,O(C))$-set.

Third, since $\Lambda$ is uniform, the number
$\frac{1}{N}\\#\\{\theta\subset\tau\\}\delta^{s-1}$ is roughly the same for
each $\tau\in\Lambda_{\rho_{0}}$.

∎

##  5\. Incidence results via Furstenberg set estimates

In many of the forthcoming arguments we will consider collections of tubes
$\mathcal{T}$, squares $\mathcal{P}$, and subsets
$\mathcal{T}_{p}\subset\mathcal{T}$ with the property that $p\cap
T\not=\emptyset$ for each $T\in\mathcal{T}_{p}$. Typically, $\mathcal{T}_{p}$
may not contain all $T\in\mathcal{T}$ intersecting $p$. The incidences between
$\mathcal{P}$ and $\mathcal{T}$ (relative to the family of sets
$\mathcal{T}_{p}$) are defined as

$\mathcal{I}(\mathcal{P},\mathcal{T})=\\{(p,T)\in\mathcal{P}\times\mathcal{T}:\;T\in\mathcal{T}_{p}\\}.$

We will often use double counting

$|\mathcal{I}(\mathcal{P},\mathcal{T})|=\sum_{p\in\mathcal{P}}|\mathcal{T}_{p}|=\sum_{T\in\mathcal{T}}|\\{p:\;T\in\mathcal{T}_{p}\\}|.$

The key concept used in this section is that of two-ends. This is a rather
weak form of non-concentration. Here is how it plays into our argument. The
main result in this section, Theorem 5.3, will use the estimate (1.6). The
minimum there contains three terms. Working with two-ends tubes will narrow it
down to two terms. Thus, Theorem 5.3 will deliver a dichotomy, rather than a
trichotomy.

######  Definition 5.1 (Two-ends).

Let $T\in\mathcal{D}_{\delta}^{dual}$ and let
$\mathcal{P}_{T}\subset\mathcal{D}_{\delta}$ be a nonempty set of
$\delta$-squares intersecting $T$. We say that $T$ is $\epsilon$-two-ends
(with respect to $\mathcal{P}_{T}$) if for each $\rho\in(\delta,1)$ and each
ball $B_{\rho}$,

$|\mathcal{P}_{T}\cap
B_{\rho}|_{\delta}\lesssim\rho^{\epsilon}\delta^{-5\epsilon^{3}}|\mathcal{P}_{T}|_{\delta}.$

Note that being $\epsilon$-two-ends implies that no
$\delta^{\epsilon}$-segment of $T$ contains the whole $\mathcal{P}_{T}$. It
also implies that $|\mathcal{P}_{T}|\gtrsim\delta^{-\epsilon}$.

The two-ends condition will be used in our proof that Theorem
5.4$\implies$Theorem 5.3, via the following simple observation.

######  Lemma 5.2.

Let $\mathcal{P}$ be a collection of $\delta$-squares. Assume
$\mathcal{T}_{p}$ is a $(\delta,s)$-set of tubes intersecting a fixed
$\delta$-square $p$. Let $\mathcal{P}_{T}=T\cap\mathcal{P}$, the squares
intersecting $\mathcal{T}$. Assume there is
$\mathcal{T}_{p}^{\prime}\subset\mathcal{T}_{p}$ with
$|\mathcal{T}_{p}^{\prime}|\gtrsim\delta^{O(\epsilon)}|\mathcal{T}_{p}|$ such
that each $T\in\mathcal{T}_{p}^{\prime}$ is $\epsilon$-two-ends.

Then for each $\rho\gtrsim\delta$ we have

$|\mathcal{P}|_{\rho}\gtrsim\rho^{-s}\delta^{O(\epsilon)}.$

###### Proof.

Note that $|\mathcal{T}_{p}^{\prime}|\gtrsim\delta^{-s+O(\epsilon)}$. Consider
a collection of $\gtrsim\rho^{-s}\delta^{O(\epsilon)}$ directions that are
$\delta^{-\epsilon}\rho$-separated, such that there is
$T\in\mathcal{T}^{\prime}_{p}$ pointing in each of these directions. For each
such tube, consider its segment that is $\gtrsim\delta^{\epsilon}$-away from
$p$. Our hypothesis forces each segment to contain at least one square in
$\mathcal{P}$. The $\rho$-neighborhoods of these segments are pairwise
disjoint. ∎

Here is our main result in this section.

######  Theorem 5.3.

Let $\epsilon,\eta>0$. Let $\mathcal{P}_{\delta,s}$ be a set of
$\delta$-squares $p$. For each $p\in\mathcal{P}_{\delta,s}$, let
$\mathcal{T}_{p}$ be a $(\delta,s,\delta^{o_{\epsilon}(1)})$-set of tubes
intersecting $p$, with $|\mathcal{T}_{p}|\sim r$.

Let $\bar{\mathcal{T}}_{p}\subset\mathcal{T}_{p}$ and write
$\mathcal{P}_{T}=\\{p:\;T\in\bar{\mathcal{T}}_{p}\\}$,
$\bar{\mathcal{T}}=\cup_{p}\bar{\mathcal{T}}_{p}$. Assume

(5.22) $T\in\bar{\mathcal{T}}\implies T\text{ is }\epsilon\text{-two-ends with
respect to }\mathcal{P}_{T},$ (5.23)
$\sum_{p}|\bar{\mathcal{T}}_{p}|\gtrsim\delta^{\eta}\sum_{p}|\mathcal{T}_{p}|.$

Then at least one of the following happens (these will be referred later as
Item (1) and Item (2)):

Item 1.

$|\mathcal{P}_{\delta,s}|\lesssim\delta^{s+o_{\epsilon}(1)+o_{\eta}(1)}\frac{|\bar{\mathcal{T}}|^{2}}{r^{2}}.$

Item 2. There exists a scale $\Delta\geq\delta^{1-\sqrt{\epsilon}}$ and a
family of $\Delta$-squares
$B_{\Delta}\subset(\mathcal{P}_{\delta,s})_{\Delta}$ such that these squares
cover at least a $\delta^{O(\epsilon+\eta)}$-fraction of
$\mathcal{P}_{\delta,s}$ and

$|\mathcal{P}_{\delta,s}\cap
B_{\Delta}|\gtrsim(\frac{\Delta}{\delta})^{2-s+\epsilon^{1/4}}.$

Let us say a few words about the relevance of this result in the grand scheme
of proving Theorem 1.1. Note that our hypothesis forces
$r\lessapprox\delta^{-s}$. So if Item (1) holds then we get the upper bound
$|\mathcal{T}|^{2}/r^{3}$ for $|\mathcal{P}_{\delta,s}|$. This matches the one
in Theorem 1.1. However, for the typical square
$p\in\mathcal{P}_{r}(\mathcal{T})$ in Theorem 1.1, the collection
$\mathcal{T}_{p}$ of tubes intersecting $p$ will not be a $(\delta,s,C)$-set
with small $C$, unless $r\approx\delta^{-s}$. We will have to carefully
manipulate the collection $\mathcal{T}$ to make Theorem 5.3 applicable.

The application of Theorem 5.3 will generate a dichotomy in our final argument
based on induction on scales. If Item (1) holds, it will lead to a more direct
closing of the induction. If Item (2) holds, we will use an elegant
application of the high-low method.

We deduce Theorem 5.3 from the following related result.

######  Theorem 5.4.

Let $\mathcal{T}$ be a set of $\delta$-tubes in the plane. Let $\mathcal{P}$
denote an $\epsilon$-uniform set of $\delta$-squares $p$ such that for each
$\rho\gtrsim\delta$ we have

(5.24) $|\mathcal{P}|_{\rho}\gtrsim\rho^{-s}\delta^{o_{\epsilon}(1)}.$

Let $\mathcal{T}_{p}\subset\mathcal{T}$ be a
$(\delta,s,\delta^{o_{\epsilon}(1)})$-set of tubes passing through $p$ with
$|\mathcal{T}_{p}|\sim r\delta^{o_{\epsilon}(1)}$. We assume each
$\mathcal{T}_{p}$ is $\epsilon$-uniform, with the same branching function
(i.e. for each $\rho=2^{-jT}$ and each $\mathbf{T}\in\mathcal{T}_{p}^{\rho}$,
$|\mathcal{T}_{p}[\mathbf{T}]|$ has a fixed value $N_{\rho}$, independent of
$p\in\mathcal{P}$).

Then at least one of the following happens:

  1. (1)

$|\mathcal{P}|\lesssim\delta^{s+o_{\epsilon}(1)}\frac{|\mathcal{T}|^{2}}{r^{2}}.$

  2. (2)

there exists $\Delta\gtrsim\delta^{1-\sqrt{\epsilon}}$ such that for each
$B_{\Delta}\in\mathcal{P}_{\Delta}$

$|\mathcal{P}\cap
B_{\Delta}|\gtrsim(\frac{\Delta}{\delta})^{2-s+\epsilon^{1/4}}.$

For later use in the final induction on scales argument, we quantify
explicitly the lower bounds for $\Delta$ and $|\mathcal{P}\cap B_{\Delta}|$.
The other $o_{\epsilon}(1)$ terms are irrelevant.

###### Proof.

( Theorem 5.4$\implies$Theorem 5.3 )

There are two steps.

1\. Uniformization: We reduce the theorem to proving it for $\epsilon$-uniform
collections of tubes and squares.

Write $P=|\mathcal{P}_{\delta,s}|$. We call a square $p$ incident to a tube
$T\in\bar{\mathcal{T}}$ if $T\in\bar{\mathcal{T}}_{p}$, or equivalently, if
$p\in\mathcal{P}_{T}$. Note that

$|\mathcal{I}(\mathcal{P}_{\delta,s},{\mathcal{T}})|\gtrsim\delta^{\eta}Pr.$

In the following we refine first $\bar{\mathcal{T}}$, then
$\mathcal{P}_{\delta,s}$, by essentially preserving the incidences. It will be
important that the refinement of $\mathcal{P}_{\delta,s}$ remains comparable
in size.

Select a number $t$ and a subset
$\mathcal{T}^{\prime}\subset\bar{\mathcal{T}}$ such that each
$T\in\mathcal{T}^{\prime}$ is incident to $\sim t$ squares in
$\mathcal{P}_{\delta,s}$ (thus $|\mathcal{P}_{T}|\sim t$) and moreover, the
number of incidences is essentially preserved

$|\mathcal{T}^{\prime}|t\sim\mathcal{I}(\mathcal{P}_{\delta,s},\mathcal{T}^{\prime})\gtrsim\delta^{\eta}(\log(1/\delta))^{-1}Pr.$

Note that the size of $\mathcal{T}^{\prime}$ might be much smaller than the
size of $\bar{\mathcal{T}}$, but this will only help us with our final
estimate.

Each $p$ was initially incident to $\sim r$ tubes. But since we trimmed the
collection of tubes, this is no longer the case. However, the following
satisfactory substitute holds.

Note that the subset $\mathcal{P}^{\prime}\subset\mathcal{P}_{\delta,s}$ of
those $p$ that are incident to $\gtrsim\delta^{\eta}(\log(1/\delta))^{-1}r$
tubes in $\mathcal{T}^{\prime}$ satisfies $|\mathcal{P}^{\prime}|\gtrsim
P\delta^{\eta}(\log(1/\delta))^{-1}$. This happens since

$\delta^{\eta}(\log(1/\delta))^{-1}Pr\lesssim\mathcal{I}(\mathcal{P}_{\delta,s},\mathcal{T}^{\prime})=\mathcal{I}(\mathcal{P}^{\prime},\mathcal{T}^{\prime})+\mathcal{I}(\mathcal{P}_{\delta,s}\setminus\mathcal{P}^{\prime},\mathcal{T}^{\prime}),$
$\mathcal{I}(\mathcal{P}_{\delta,s}\setminus\mathcal{P}^{\prime},\mathcal{T}^{\prime})\ll\delta^{\eta}(\log(1/\delta))^{-1}Pr,$

and thus

$|\mathcal{P}^{\prime}|r\gtrsim\mathcal{I}(\mathcal{P}^{\prime},\mathcal{T}^{\prime})\gtrsim
Pr\delta^{\eta}(\log(1/\delta))^{-1}.$

For each $p\in\mathcal{P}^{\prime}$ we write $\mathcal{T}_{p}^{\prime}$ for
the collection of tubes in $\mathcal{T}^{\prime}$ that $p$ is incident to.
Then $|\mathcal{T}_{p}^{\prime}|\gtrsim\delta^{\eta}(\log(1/\delta))^{-1}r$.

Find a subset $\mathcal{P}^{\prime\prime}\subset\mathcal{P}^{\prime}$ such
that $|\mathcal{P}^{\prime\prime}|\gtrsim P\delta^{O(\epsilon^{3})+\eta}$ and
for each $p\in\mathcal{P}^{\prime\prime}$, there is (cf. Lemma 2.7)
an-$\epsilon^{3}$-uniform set
$\mathcal{T}_{p,unif}\subset\mathcal{T}_{p}^{\prime}$ with
$|\mathcal{T}_{p,unif}|\gtrsim\delta^{\epsilon^{3}}|\mathcal{T}_{p}^{\prime}|$
and with branching function independent of $p$. Here is why this is possible.
Each $\mathcal{T}_{p,unif}$ is associated with a string
$(N_{1},\ldots,N_{m})$, with $N_{j}:=|\mathcal{T}_{p,unif}|_{2^{-jT}}$
satisfying $N_{j+1}/N_{j}=2^{c_{j}}$ for some nonnegative integer
$c_{j}\lesssim T$. There are $(O(T))^{m}\lesssim\delta^{-\epsilon^{3}}$ such
strings. Pigeonhole to pick one that represents many $p$. This gives a large
subset $\mathcal{P}^{\prime\prime}$.

Finally, find an $\epsilon^{3}$-uniform subset
$\mathcal{P}_{unif}\subset\mathcal{P}^{\prime\prime}$ with size $\gtrsim
P\delta^{O(\epsilon^{3})+\eta}$. Note that

$\mathcal{I}(\mathcal{P}_{unif},\mathcal{T}^{\prime})\gtrsim\delta^{O(\epsilon^{3})+\eta}|\mathcal{P}_{unif}|r\gtrsim\delta^{O(\epsilon^{3})+\eta}Pr.$

Note that for each $p\in\mathcal{P}_{unif}$ the collection
$\mathcal{T}_{p,unif}$ is a
$(\delta,s,\delta^{o_{\epsilon}(1)+o_{\eta}(1)})$-set with cardinality $\sim
r\delta^{o_{\epsilon}(1)+o_{\eta}(1)}$.

2\. Covering number estimates. We prove that for each $\rho\gtrsim\delta$ we
have

(5.25)
$|\mathcal{P}_{unif}|_{\rho}\gtrsim\rho^{-s}\delta^{o_{\epsilon}(1)+\eta}.$

Recall that each $T\in\mathcal{T}^{\prime}$ is incident to $\sim t$ squares in
$cP_{\delta,r}$, so

$Pr\gtrsim\mathcal{I}(\mathcal{P}_{\delta,r},\mathcal{T}^{\prime})\sim|\mathcal{T}^{\prime}|t.$

Recall also that

$\mathcal{I}(\mathcal{P}_{unif},\mathcal{T}^{\prime})\gtrsim\delta^{O(\epsilon^{3})+\eta}Pr.$

Thus,

$\mathcal{I}(\mathcal{P}_{unif},\mathcal{T}^{\prime})\gtrsim|\mathcal{T}^{\prime}|t\delta^{O(\epsilon^{3})+\eta}.$

It follows that, for $C_{1}$ large enough, the number of tubes in the
collection $\mathcal{T}^{\prime\prime}\subset\mathcal{T}^{\prime}$ that are
incident to $\gtrsim t\delta^{C_{1}\epsilon^{3}}$ squares in
$\mathcal{P}_{unif}$ is
$\gtrsim|\mathcal{T}^{\prime}|\delta^{O(\epsilon^{3})+\eta}$. For later use we
record that

(5.26)
$\mathcal{I}(\mathcal{P}_{unif},\mathcal{T}^{\prime\prime})\gtrsim|\mathcal{T}^{\prime}|t\delta^{O(\epsilon^{3})+\eta}Pr\gtrsim\delta^{O(\epsilon^{3})+2\eta}Pr.$

Note that for each $T\in\mathcal{T}^{\prime\prime}$, the squares in
$\mathcal{P}_{unif}$ that are incident to $T$ cannot all be contained in a
$\delta^{\epsilon}$-segment of $T$. Indeed, $T$ being two-ends with respect to
$\mathcal{P}_{T}$ guarantees that such a segment only contains
$\lesssim\delta^{\epsilon^{2}-5\epsilon^{3}}t$ squares even from
$\mathcal{P}_{T}$. And if $\epsilon$ is small enough we have
$\epsilon^{2}-5\epsilon^{3}>C_{1}\epsilon^{3}$.

Using (5.26) we get

$\sum_{p\in\mathcal{P}_{unif}}|\\{T\in\mathcal{T}^{\prime\prime}:\;p\cap
T\not=\emptyset\\}|\geq\mathcal{I}(\mathcal{P}_{unif},\mathcal{T}^{\prime\prime})\gtrsim\delta^{O(\epsilon^{3})+2\eta}Pr.$

It follows that there is $p$ such that

$|\\{T\in\mathcal{T}^{\prime\prime}:\;p\cap
T\not=\emptyset\\}|\gtrsim\delta^{O(\epsilon^{3})+2\eta}r.$

The desired (5.25) now follows from Lemma 5.2.

3\. Conclusion: We write $\mathcal{P}=\mathcal{P}_{unif}$. Remark 2.5 implies
that $\mathcal{P}$ is approximate $\epsilon$-uniform, satisfies (5.25) and

(5.27)
$|\mathcal{P}_{\delta,s}|\lesssim\delta^{-o_{\epsilon}(1)-\eta}|\mathcal{P}|.$

Each $\mathcal{T}_{unif,p}$ is approximate $\epsilon$-uniform and a
$(\delta,s,\delta^{o_{\epsilon}(1)+o_{\eta}(1)})$-set. The value of $r$ has
decreased by at most a factor of $\delta^{o_{\epsilon}(1)+\eta}$. We now apply
Theorem 5.4 to the collections $\mathcal{P}$ and $\mathcal{T}^{\prime}$. The
$\delta^{\eta}$-losses in constants will only incur $\delta^{O(\eta)}$-losses
in Item (1).

If Item (1) in that theorem holds, then when combined with (5.27) it
immediately implies (1) in Theorem 5.3.

Also, since $\mathcal{P}\subset\mathcal{P}_{\delta,p}$, if Item (2) in Theorem
5.4 holds, then Item (2) in Theorem 5.3 must also hold.

∎

###### Proof.

(of Theorem 5.4 ) Recall the function $\psi$ from Theorem 2.13. We let
$\upsilon=\upsilon(\epsilon)\geq\epsilon$ be such that
$\psi(\upsilon)\geq\sqrt{\epsilon}$ and $\lim_{\epsilon\to
0}\upsilon(\epsilon)=0.$ Each $\epsilon$-uniform set will also be
(approximate) $\upsilon$-uniform, see Remark 2.5. It will be useful to keep in
mind that being $o_{\upsilon}(1)$ is equivalent with being $o_{\epsilon}(1)$.

Apply (the two-dimensional version of) Theorem 2.13 to $\mathcal{P}$ and
$\upsilon$ (in place of $\epsilon$). Then, using (5.24), rearranging Theorem
2.13 (ii) and since $A_{1}=0$ we get

$A_{2}(t_{1}-t_{2})+A_{3}(t_{2}-t_{3})+\ldots+A_{l}(t_{l-1}-t_{l})+t_{l}A_{l+1}\geq
sA_{l+1}+o_{\upsilon}(1),\;1\leq l\leq L.$

Since $t_{l}<t_{l+1}$ this gives

$t_{l}A_{l+1}\geq sA_{l+1}+c_{\upsilon},$

with $c_{\upsilon}=o_{\upsilon}(1)$.

A closer inspection of the proof of Theorem 2.13 reveals that $c_{\upsilon}$
is negative, and this requires some attention. We pick $l_{0}$ such that on
the one hand $A_{l_{0}}=o_{\upsilon}(1)$, and on the other hand
$c_{\upsilon}/A_{l+1}=o_{\upsilon}(1)$ for each $l\geq l_{0}$. All intervals
$[A_{l},A_{l+1}]$ with $l<l_{0}$ will be moved into the “bad” category. Thus,
the combined total length of all the bad intervals is still $o_{\upsilon}(1)$.
All intervals $[A_{l},A_{l+1}]$ with $l\geq l_{0}$ will be called “good”.

Our choice forces that for each $l\geq l_{0}$,

(5.28) $t_{l}\geq s+o_{\upsilon}(1).$

For each good interval we will get essentially sharp estimates. For the bad
ones trivial estimates will suffice, since their lengths add up to a
negligible amount $o_{\upsilon}(1)$.

Case 1. When $t_{L}\geq 2-s+\epsilon^{1/4}+3\epsilon$, this gives Item $(2)$
with $\Delta=\delta^{A_{L}}$. Indeed,

$\Delta/\delta=(1/\delta)^{A_{L+1}-A_{L}}\geq(1/\delta)^{\psi(\upsilon)}\geq\delta^{-\sqrt{\epsilon}},$

and, by Theorem 2.13 (iii) with $I$ replaced by $B_{\Delta}$,

$|B_{\Delta}\cap\mathcal{P}|\gtrsim(\frac{1}{\delta})^{(A_{L+1}-A_{L})(t_{L}-3\epsilon)}=(\Delta/\delta)^{t_{L}-3\epsilon}\geq(\Delta/\delta)^{2-s+\epsilon^{1/4}}.$

If we are in Case 1, this is the end of the whole argument.

Case 2. When

(5.29) $t_{L}\leq 2-s+\epsilon^{1/4}+3\epsilon,$

the only implication we use is that

(5.30) $t_{l}\leq 2-s+o_{\upsilon}(1)$

for each $1\leq l\leq L$.

We write $\mathbf{p}_{l}$ for a generic square in
$\mathcal{P}_{\delta^{A_{l}}}$. In particular, $\mathbf{p}_{L+1}$ is an
alternative notation for an arbitrary square $p\in\mathcal{P}$.

The following argument is applicable to each $l$. Let us explain the main step
for $[A_{L},A_{L+1}]$. Consider
$\mathbf{p}_{L}\in\mathcal{P}_{\delta^{A_{L}}}$. For
$p\in\mathcal{P}\cap\mathbf{p}_{L}$, call $\mathcal{U}_{p}$ the collection of
$\delta\times\delta^{A_{L}}$-tube segments $U=T\cap\mathbf{p}_{L}$ with
$T\in\mathcal{T}_{p}$. Let also $\mathcal{U}[\mathbf{p}_{L}]$ denote the
collection of all these distinct segments for
$p\in\mathbf{p}_{L}\cap\mathcal{P}$.

By Lemma 2.6, $\mathcal{U}_{p}$ is a (rescaled)
$(\delta^{A_{L+1}-A_{L}},s,\delta^{-(A_{L+1}-A_{L})\eta_{1}})$-set, with
$\eta_{1}=o_{\epsilon}(1)=o_{\upsilon}(1).$ The set
$\mathbf{p}_{L}\cap\mathcal{P}$ is a rescaled
$(\delta^{A_{L+1}-A_{L}},t_{L},\delta^{-(A_{L+1}-A_{L})\eta})$-set, with
$\eta=o_{\upsilon}(1)$.

Using (5.28) and (5.29) we find

$\min(\frac{3s+t_{L}}{2},s+t_{L},s+1)=\frac{3s+t_{L}}{2}+o_{\upsilon}(1).$

To uniformize existing notation with the forthcoming one, we denote
$\mathcal{U}[\mathbf{p}_{L}]$ by $\mathcal{T}_{\mathbf{p}_{L}}$ and
$|\mathcal{T}_{p}|_{\delta^{A_{L+1}-A_{L}}}$ by $M_{\mathbf{p}_{L}}$.

By the Furstenberg set estimate Theorem 1.4 applied to (the rescaled version
of) $(\mathbf{p}_{L}\cap\mathcal{P},\mathcal{T}_{\mathbf{p}_{L}})$ we get

$|\mathcal{T}_{\mathbf{p}_{L}}|\gtrsim\delta^{-(A_{L+1}-A_{L})(\frac{t_{L}+s+o_{\upsilon}(1))}{2}+\epsilon(o_{\upsilon}(1),o_{\upsilon}(1),s,t_{L}))}M_{\mathbf{p}_{L}}.$

Let $\bar{\epsilon}(\eta,\eta_{1},s)=\max_{s\leq t\leq
2-s}\epsilon(\eta,\eta_{1},s,t)$. We have $\lim_{\eta,\eta_{1}\to
0}\bar{\epsilon}(\eta,\eta_{1},s)=0$, so in particular,
$\bar{\epsilon}(o_{\upsilon}(1),o_{\upsilon}(1),s)=o_{\upsilon}(1)$. Given
(5.28) and (5.30), we have
$\epsilon(\eta,\eta_{1},s,t_{l})\leq\bar{\epsilon}(\eta,\eta_{1},s)$ for each
$l\geq l_{0}$. Recalling also that by (iii) of Theorem 2.13 we have
$|\mathbf{p}_{L}\cap\mathcal{P}|_{\delta}\sim\delta^{-(A_{L+1}-A_{L})(t_{L}+o_{\upsilon}(1))}$,
we conclude that

$\\#(\mathbf{p}_{L+1}\subset\mathbf{p}_{L})=|\mathbf{p}_{L}\cap\mathcal{P}|\lesssim\delta^{(A_{L+1}-A_{L})(s+o_{\upsilon}(1))}(\frac{|\mathcal{T}_{\mathbf{p}_{L}}|}{M_{\mathbf{p}_{L}}})^{2}.$

We repeat this argument for the next scale. We use [11, Proposition 4.1], to
cover the $\delta$-tubes through each $\mathbf{p}_{L}\subset\mathbf{p}_{L-1}$
with $M_{\mathbf{p}_{L-1}}$ many $\delta^{A_{L}}\times\delta^{A_{L-1}}$-tube
segments lying inside $\mathbf{p}_{L-1}$. We write
$\mathcal{T}_{\mathbf{p}_{L-1}}$ for the collection of all these tube
segments. Applying the argument from the first step to
$\mathbf{p}_{L-1}\in\mathcal{P}_{\delta^{A_{L-1}}}$ and
$\mathcal{T}_{\mathbf{p}_{L-1}}$, we are led to

$\\#(\mathbf{p}_{L}\subset\mathbf{p}_{L-1})\lesssim\delta^{(A_{L}-A_{L-1})(s+o_{\upsilon}(1))}(\frac{|\mathcal{T}_{\mathbf{p}_{L-1}}|}{M_{\mathbf{p}_{L-1}}})^{2}.$

We repeat this step, with the last inequality in the iteration being (recall
that $\mathbf{p}_{1}=[0,1]^{2}$)

$\\#(\mathbf{p}_{2}\subset\mathbf{p}_{1})\lesssim\delta^{(A_{2}-A_{1})(s+o_{\upsilon}(1))}(\frac{|\mathcal{T}_{\mathbf{p}_{1}}|}{M_{\mathbf{p}_{1}}})^{2}.$

Proposition 5.2 [11] explains how to efficiently use Proposition 4.1 from the
same paper, in a way that the product of these estimates is kept under
control. The following is recorded in Proposition 3.8 of [14]

$\prod_{l=1}^{L}\max_{\mathbf{p}_{l}}\frac{|\mathcal{T}_{\mathbf{p}_{l}}|}{M_{\mathbf{p}_{l}}}\lessapprox\frac{|\mathcal{T}|}{r},$

with $\lessapprox$ hiding multiple logarithmic losses.

We have described what happens if all intervals are good. For the bad
intervals, a trivial inequality will hold, with the bound
$\delta^{(A_{l}-A_{l-1})s}$ replaced by the crude estimate
$\delta^{-O(A_{l}-A_{l-1})}$,

$\\#(\mathbf{p}_{l}\subset\mathbf{p}_{l-1})\lesssim\delta^{-O(A_{l}-A_{l-1})}(\frac{|\mathcal{T}_{\mathbf{p}_{l-1}}|}{M_{\mathbf{p}_{l-1}}})^{2}.$

The contribution from the bad intervals is thus $\delta^{o_{\upsilon}(1)}$.
Combining all these inequalities we get

$|\mathcal{P}|=\Pi_{l=2}^{L+1}\\#(\mathbf{p}_{l}\subset\mathbf{p}_{l-1})\lessapprox\delta^{s+o_{\upsilon}(1)}\frac{|\mathcal{T}|^{2}}{r^{2}}.$

In addition to factors of the form $\delta^{(A_{l+1}-A_{l})o_{\upsilon}(1)}$
incurred at each step $l$, that contribute a total
$\delta^{o_{\upsilon}(1)}$-loss, there are implicit constants hidden in the
notation $\lesssim$, that are $O(1)$. Recall that $L$ is independent of
$\delta$, and only depends on $\upsilon$, so this contribution is
$O(1)^{L}=O_{\upsilon}(1)=O_{\epsilon}(1)$. Finally, recall that
$o_{\upsilon}(1)=o_{\epsilon}(1)$.

∎

##  6\. The high-low method

The following result traces its origins to [7]. We will use it in the next
section to perform induction on scales, in the case when Item (2) in Theorem
5.3 holds.

######  Lemma 6.1 (the high-low method).

Let $\beta>0$. Let $\mathcal{T}$ be a collection of $\delta$-tubes and let
$r_{0}\geq 1$. Suppose
$|\mathcal{P}_{r_{0}}(\mathcal{T})|\geq\delta^{-2-\beta}/r_{0}$. Then there
exists a scale $1\geq\tilde{\delta}\geq\delta^{1-\frac{\beta}{2}}$ such that
for at least half of the $\delta$-squares $q$ (with center $c_{q}$) in
$\mathcal{P}_{r_{0}}(\mathcal{T})$, the number of tubes in $\mathcal{T}$ that
intersect the ball $B(c_{q},\tilde{\delta}\delta^{-\upsilon})$ is
$\gtrsim_{\upsilon}r_{0}\tilde{\delta}/\delta$, for each $\upsilon>0$.

###### Proof.

Our argument follows [7]. We fix a smooth, positive
$\phi:\mathbb{R}^{2}\to[0,\infty)$ with $\phi\geq 1_{B(0,1)}$ and with Fourier
transform supported inside $B(0,1)$. Let $f=\sum_{T\in\mathcal{T}}\phi_{T}$,
where $\phi_{T}$ is the affine rescaling of $\phi$ that satisfies
$\phi_{T}\geq 1_{T}$ and has Fourier support inside the $1\times\delta^{-1}$
rectangle centered at the origin, dual to $T$.

If $|\mathcal{T}|\gtrsim\delta^{-1}r_{0}$, we are done by taking
$\tilde{\delta}=1$. So we may assume that $|\mathcal{T}|\ll\delta^{-1}r_{0}$.
Assume $\psi$ is smooth and $1_{B(0,\delta^{-1+\beta/2})}\leq\psi\leq
1_{B(0,2\delta^{-1+\beta/2})}$. Decompose $f=f^{h}+f^{l}$ into the high
frequency part $f^{h}$ and low frequency part $f^{l}$ where
$f^{l}:=(\widehat{f}\psi)^{\vee}$. For each $x$, at least one of the
inequalities $|f(x)|\leq 2|f^{h}(x)|$ and $|f(x)|\leq 2|f^{l}(x)|$ will hold.

Assume the former holds for a subset of $\mathcal{P}_{r_{0}}(\mathcal{T})$
with at least half its measure. We show below that this forces that

(6.31)
$r_{0}^{2}\delta^{2}|\mathcal{P}_{r_{0}}(\mathcal{T})|\lesssim\int|f^{h}|^{2}\lesssim\delta^{1-\beta}|\mathcal{T}|.$

However, this cannot hold because
$|\mathcal{P}_{r_{0}}(\mathcal{T})|\geq\delta^{-2-\beta}/r_{0}$ and
$|\mathcal{T}|\ll\delta^{-1}/r_{0}$.

To see the second inequality in (6.31), partition $\mathbb{S}^{1}$ into
$\delta^{1-\frac{\beta}{2}}$-arcs. For each such arc, consider a tiling of the
plane with thin $\delta^{1-\frac{\beta}{2}}\times 1$ rectangles $R$, with the
long side pointing in the direction normal to the arc.

Group the tubes in $\mathcal{T}$ according to which $R$ they belong to. Write

$\phi_{T}^{l}=(\widehat{\phi_{T}}\psi)^{\vee}$

and $f_{R}^{h}=\sum_{T\subset R}(\phi_{T}-\phi_{T}^{l})=f_{R}-f_{R}^{l}$. It
is easy to see that the functions $f_{R}^{h}$ form an almost orthogonal
family, due to the combination of spatial and frequency localization.

Assume $R$ contains $M_{R}$ tubes. Note that $M_{R}\lesssim\delta^{-\beta}$.
Each $f_{R}^{l}$ is essentially supported in (a slight enlargement of) $R$ and

$\|f_{R}^{l}\|_{\infty}\lesssim M_{R}\delta^{\beta/2}.$

Thus

$\|f_{R}^{l}\|_{2}^{2}\lesssim M_{R}^{2}\delta^{\beta}|R|\sim
M_{R}^{2}\delta^{1+\frac{\beta}{2}},$

and

$\sum_{R}\|f_{R}^{l}\|_{2}^{2}\lesssim\delta^{1+\frac{\beta}{2}}\max_{R}M_{R}\sum_{R}M_{R}\lesssim\delta^{1-\frac{\beta}{2}}|\mathcal{T}|.$

Also,

$\sum_{R}\|\sum_{T\subset
R}\phi_{T}\|_{2}^{2}\lesssim\sum_{R}M_{R}\|\sum_{T\subset
R}\phi_{T}\|_{1}\lesssim\delta^{1-\beta}|\mathcal{T}|.$

Combining these two inequalities with almost orthogonality we find

$\|f^{h}\|_{2}^{2}\lesssim\sum_{R}\|f_{R}^{h}\|_{2}^{2}\lesssim\sum_{R}\|f_{R}^{l}\|_{2}^{2}+\sum_{R}\|f_{R}\|_{2}^{2}\lesssim\delta^{1-\beta}|\mathcal{T}|.$

We have learned that $f(x)\leq 2f^{l}(x)$ for $x$ in a subset of
$\mathcal{P}_{r_{0}}(\mathcal{T})$ with at least half its size. It follows
that for at least half of the squares $q$ we have

$r_{0}\lesssim\delta/\tilde{\delta}\cdot\\#\\{T\in\mathcal{T}:T\cap
B(c_{q},\tilde{\delta}\delta^{-\upsilon})\neq\emptyset\\}$

for $\tilde{\delta}=\delta^{1-\frac{\beta}{2}}$. This is because
$\|\phi_{T}^{l}\|_{\infty}\lesssim\delta/\tilde{\delta}$, and $\phi_{T}^{l}$
is negligible outside the $\delta^{-\frac{\beta}{2}-\upsilon}$-thickening of
$T$.

∎

##  7\. Two-ends reduction

We start with some notation. Consider a collection $\mathcal{T}_{p}$ of
$\delta$-tubes intersecting some $p\in\mathcal{D}_{\delta}$. Given a scale
$\delta\leq\rho\leq 1$, we let $\mathbf{p}_{\rho}$ be the dyadic square with
side-length $\rho$ containing $p$. We introduce the collection of
$\delta\times\rho$-segments

$\mathcal{U}_{\rho}(\mathcal{T}_{p})=\\{U_{\rho}=T\cap\mathbf{p}_{\rho},\text{
for some }T\in\mathcal{T}_{p}\\}.$

We will use that

(7.32)
$|\mathcal{U}_{\rho}(\mathcal{T}_{p})|=|\mathcal{T}_{p}^{\delta/\rho}|=|\mathcal{T}_{p}|_{\delta/\rho}.$

Assume $\mathcal{T}_{p}$ is uniform and assume
$\tilde{\mathcal{T}}_{p}\subset\mathcal{T}_{p}$ satisfies
$|\tilde{\mathcal{T}}_{p}|\geq C^{-1}|\mathcal{T}_{p}|$. Using (7.32) it is
immediate that

(7.33) $|\mathcal{U}_{\rho}(\tilde{\mathcal{T}}_{p})|\gtrsim
C^{-1}|\mathcal{U}_{\rho}(\mathcal{T}_{p})|.$

The implicit constant in the inequality $\gtrsim$ depends on the uniformity
constants.

The following theorem shows how to find subcollections of tube segments that
are two-ends, see (7.36) and preserve incidences, see (7.37). Moreover, (7.38)
guarantees that the average number of incidences for tube segments is at least
as large as the average for tubes. The proof combines graph theory methods
(Lemma 8.2 in [18] and double counting for various subgraphs) with the
decomposition in Lemma 2.11, that will help us identify the scale
$\tilde{\rho}$. The reader eager to get to the main argument in Section 8 may
skip this proof.

######  Theorem 7.1.

Fix $\epsilon>0$ small enough. Consider a collection
$\mathcal{P}\subset\mathcal{D}_{\delta}$. For each $p\in\mathcal{P}$ we let
$\mathcal{T}_{p}$ be an $\epsilon$-uniform set of $\delta$-tubes intersecting
$p$. We assume all $\mathcal{T}_{p}$ have the same branching function. Assume

(7.34)
$\sum_{p\in\mathcal{P}}|\mathcal{T}_{p}|\geq\delta^{-2\epsilon}|\bigcup_{p\in\mathcal{P}}\mathcal{T}_{p}|.$

Then there is a scale

(7.35) $\delta^{1-\epsilon}\leq{\tilde{\rho}}\leq 1$

such that the following holds. Call
$\mathcal{U}_{{\tilde{\rho}},p}=\mathcal{U}_{\tilde{\rho}}(\mathcal{T}_{p})$
and
$\mathcal{U}_{\tilde{\rho}}=\cup_{p\in\mathcal{P}}\mathcal{U}_{{\tilde{\rho}},p}$.
For each $p\in\mathcal{P}$ there is
$\bar{\mathcal{U}}_{\tilde{\rho},p}\subset\mathcal{U}_{{\tilde{\rho}},p}$ with
the following properties. For each
$U_{\tilde{\rho}}\in\bar{\mathcal{U}}_{\tilde{\rho}}:=\cup_{p\in\mathcal{P}}\bar{\mathcal{U}}_{{\tilde{\rho}},p}$,
letting

$\mathcal{P}_{U_{\tilde{\rho}}}=\\{p\in\mathcal{P}:\;U_{\tilde{\rho}}\in\bar{\mathcal{U}}_{{\tilde{\rho}},p}\\},$

we have

(7.36) $|\mathcal{P}_{U_{\tilde{\rho}}}\cap
B_{{\tilde{\rho}}^{\prime}}|\lesssim(\frac{{\tilde{\rho}}^{\prime}}{\tilde{\rho}})^{\epsilon}(\frac{\tilde{\rho}}{\delta})^{5\epsilon^{3}}|\mathcal{P}_{U_{\tilde{\rho}}}|,\;\;{\tilde{\rho}}^{\prime}\geq\delta,$
(7.37)
$\sum_{p\in\mathcal{P}}|\bar{\mathcal{U}}_{{\tilde{\rho}},p}|\gtrsim(\delta/{\tilde{\rho}})^{\epsilon^{3}}\sum_{p\in\mathcal{P}}|\mathcal{U}_{{\tilde{\rho}},p}|,$
(7.38)
$\delta^{o_{\epsilon}(1)}\frac{\sum_{p\in\mathcal{P}}|\mathcal{T}_{p}|}{|\bigcup_{p\in\mathcal{P}}\mathcal{T}_{p}|}\lesssim\frac{\sum_{p\in\mathcal{P}}|{\mathcal{U}}_{\tilde{\rho},p}|}{|\bar{\mathcal{U}}_{\tilde{\rho}}|}.$

###### Proof.

Step 1. Uniformization.

Call $\mathcal{T}=\cup_{p\in\mathcal{P}}\mathcal{T}_{p}$. For each
$T\in\mathcal{T}$, define $Y(T):=\\{p\in\mathcal{P}:T\in\mathcal{T}_{p}\\}$.
Find an $\epsilon^{4}$-uniform subset $Y^{\prime}(T)\subset Y(T)$ such that
$|Y^{\prime}(T)|\gtrsim\delta^{\epsilon^{4}}|Y(T)|$. Let
$\mathcal{T}^{\prime}\subset\mathcal{T}$ be such that

(1) for each $T\in\mathcal{T}^{\prime}$, $Y^{\prime}(T)$ has the same
branching function, in particular the same size, which in the future will be
denoted by $|Y^{\prime}(T)|$.

(2)
$|\mathcal{T}^{\prime}|\cdot|Y^{\prime}(T)|=\sum_{T\in\mathcal{T}^{\prime}}|Y^{\prime}(T)|\gtrsim\delta^{\epsilon^{4}}|\mathcal{P}|\cdot|\mathcal{T}_{p}|.$

Let $G$ be the incidence graph between $\mathcal{P}$ and
$\mathcal{T}^{\prime}$ defined as $(p,T)\in G$ iff $p\in Y^{\prime}(T)$. So
$|G|\sim|\mathcal{T}^{\prime}||Y^{\prime}(T)|$. We apply Lemma 8.2 in [18] to
$G$ and obtain a subgraph
$G^{\prime}\subset(\mathcal{P}_{1}\times\mathcal{T}_{1})\cap G$ such that

(3) $|G^{\prime}|\gtrsim|G|$

(4) for each $p\in\mathcal{P}_{1}$,
$\mathcal{T}_{p}^{\prime}=\\{T^{\prime}\in\mathcal{T}_{p}\cap\mathcal{T}_{1}:(p,T^{\prime})\in
G^{\prime}\\}$ satisfies
$|\mathcal{T}_{p}^{\prime}|\gtrsim\frac{|G|}{|\mathcal{P}|}\gtrsim\delta^{\epsilon^{4}}|\mathcal{T}_{p}|$

(5) for each $T\in\mathcal{T}_{1}$, $Y^{\prime\prime}(T)=\\{p\in
Y^{\prime}(T)\cap\mathcal{P}_{1}:(p,T)\in G^{\prime}\\}$ satisfies
$|Y^{\prime\prime}(T)|\gtrsim\frac{|G|}{|\mathcal{T}^{\prime}|}\sim|Y^{\prime}(T)|.$

It follows that

(7.39) $|\mathcal{P}_{1}|\gtrsim\delta^{\epsilon^{4}}|\mathcal{P}|$

and

$|\mathcal{T}_{1}|\gtrsim\delta^{\epsilon^{4}}|\mathcal{T}^{\prime}|.$

Also, for each $(p,T)\in\mathcal{P}_{1}\times\mathcal{T}_{1}$ we have
$T\in\mathcal{T}_{p}^{\prime}$ iff $p\in Y^{\prime\prime}(T)$.

Step 2. The definition of ${\tilde{\rho}}$. Proof that
${\tilde{\rho}}\geq\delta^{1-\epsilon}$.

Consider the branching function $f$ of $Y^{\prime}(T)$. Then using (2)
followed by (7.34) we find

(7.40)
$\delta^{f(0)-f(1)}=|Y^{\prime}(T)|\gtrsim\frac{\delta^{\epsilon^{4}}|\mathcal{P}||\mathcal{T}_{p}|}{|\mathcal{T}|}\geq\delta^{\epsilon^{4}-2\epsilon}>\delta^{-\epsilon-\epsilon^{4}}.$

Apply Lemma 2.11 to $f$ to find a decomposition $\\{[a_{j},a_{j+1}]\\}$ of
$[0,1]$ and a sequence of $\sigma_{j}<\sigma_{j+1}$ such that for each
$x\in[a_{j},a_{j+1}]$

(7.41) $f(x)\geq f(a_{j})+\sigma_{j}(x-a_{j})-\epsilon^{4}(a_{j+1}-a_{j})$

and

(7.42) $f(a_{j+1})\leq f(a_{j})+(\sigma_{j}+\epsilon^{4})(a_{j+1}-a_{j}).$

Let $b$ be the smallest $a_{j}$ such that $\sigma_{j}\geq\epsilon$. Such a $b$
exists because of(7.42) and (7.40). It is possible that $b=0$, but this is
harmless. We write ${\tilde{\rho}}=\delta^{b}$.

To prove that ${\tilde{\rho}}\geq\delta^{1-\epsilon}$ we use three things. On
the one hand, (7.42) and the definition of $b$ imply that $f(b)-f(0)\leq
b(\epsilon+\epsilon^{4}).$ On the other hand, since $f$ is 1-Lipschitz,
$f(1)-f(b)\leq 1-b.$ When combined with (7.40) we get

$2\epsilon-\epsilon^{4}\leq f(1)-f(0)\leq 1+b(\epsilon+\epsilon^{4}-1),$

which gives
$1-b\geq\frac{\epsilon-2\epsilon^{4}}{1-\epsilon-\epsilon^{4}}\geq\epsilon.$

Step 3. The construction of $\mathcal{T}_{1,p}$.

For each $U_{\tilde{\rho}}\in\mathcal{U}_{\tilde{\rho}}$, define
$\mathcal{T}(U_{\tilde{\rho}})$ as the collection of those
$T\in\mathcal{T}_{1}$ such that $U_{\tilde{\rho}}\subset T$ and
$U_{\tilde{\rho}}\cap Y^{\prime\prime}(T)\neq\emptyset$. Also, for
$p\in\mathcal{P}_{1}$ define a maximal subset
$\mathcal{T}_{1,p}\subset\mathcal{T}_{p}^{\prime}$ such that for each
$\mathbf{T}\in\mathcal{T}^{\delta/{\tilde{\rho}}}_{1,p}$ we have
$\mathbf{T}\cap\mathcal{T}_{1,p}=\mathbf{T}\cap\mathcal{T}_{p}^{\prime}$ and

(7.43)
$|\mathbf{T}\cap\mathcal{T}_{1,p}|\geq\delta^{2\epsilon^{4}}|\mathcal{T}_{p}|\cdot|\mathcal{T}_{p}|_{\delta/{\tilde{\rho}}}^{-1}.$

Note that, due to (4)

(7.44)
$|\mathcal{T}_{p}^{\prime}\setminus\mathcal{T}_{1,p}|=\sum_{\mathbf{T}\in{\mathcal{T}_{p}^{\prime}}^{\delta/{\tilde{\rho}}}\setminus\mathcal{T}^{\delta/{\tilde{\rho}}}_{1,p}}|\mathbf{T}\cap\mathcal{T}^{\prime}_{p}|\leq\sum_{\mathbf{T}\in\mathcal{T}_{p}^{\delta/{\tilde{\rho}}}}\delta^{2\epsilon^{4}}|\mathcal{T}_{p}|\cdot|\mathcal{T}_{p}|_{\delta/{\tilde{\rho}}}^{-1}=\delta^{2\epsilon^{4}}|\mathcal{T}_{p}|\lesssim\delta^{\epsilon^{4}}|\mathcal{T}_{p}^{\prime}|.$

Step 4. The construction of $Y_{1}(T)$

For each $T\in\mathcal{T}_{1}$ we prove the existence of $Y_{1}(T)\subset
Y^{\prime\prime}(T)$ such that for each
$U_{\tilde{\rho}}\in\mathcal{U}_{\tilde{\rho}}$ with $U_{\tilde{\rho}}\subset
T$ and $U_{\tilde{\rho}}\cap Y_{1}(T)\not=\emptyset$ we have

$U_{\tilde{\rho}}\cap Y_{1}(T)=\\{p\in
U_{\tilde{\rho}}:T\in\mathcal{T}_{1,p}\\},$ (7.45) $|U_{\tilde{\rho}}\cap
Y_{1}(T)|\geq\delta^{\epsilon^{4}}\frac{|Y^{\prime}(T)|_{\delta}}{|Y^{\prime}(T)|_{{\tilde{\rho}}}},$

and defining $\mathcal{T}_{1}(U_{\tilde{\rho}})$ as the set of those
$T\in\mathcal{T}_{1}$ such that $U_{\tilde{\rho}}\subset T$ and
$U_{\tilde{\rho}}\cap Y_{1}(T)\neq\emptyset$, then

(7.46)
$|\mathcal{T}_{1}(U_{\tilde{\rho}})|\geq\delta^{\epsilon^{4}}|\mathcal{T}(U_{\tilde{\rho}})|.$

More precisely, for each $U_{\tilde{\rho}}\subset T\in\mathcal{T}_{1}$, define
$Y_{0}(U_{\tilde{\rho}},T)=\\{p\in Y^{\prime\prime}(T)\cap
U_{\tilde{\rho}}:T\in\mathcal{T}_{1,p}\\}$. Define
$Y_{0}(T)=\cup_{U_{\tilde{\rho}}\subset T}Y_{0}(U_{\tilde{\rho}},T)$, where
the union is over all $U_{\tilde{\rho}}\subset T$ with
$|Y_{0}(U_{\tilde{\rho}},T)|\geq\delta^{\epsilon^{4}}\frac{|Y^{\prime}(T)|_{\delta}}{|Y^{\prime}(T)|_{{\tilde{\rho}}}}$.
For each $U_{\tilde{\rho}}$, define $\mathcal{T}_{1}(U_{\tilde{\rho}})$ as the
set of those $T\in\mathcal{T}_{1}$ such that $U_{\tilde{\rho}}\subset T$ and
$U_{\tilde{\rho}}\cap Y_{0}(T)\neq\emptyset$. Define
$Y_{1}(T)=\cup_{U_{\tilde{\rho}}\subset T}Y_{0}(U_{\tilde{\rho}},T)$, where
the union is over all $U_{\tilde{\rho}}\subset T$ with
$|\mathcal{T}_{1}(U_{\tilde{\rho}})|\geq\delta^{\epsilon^{4}}|\mathcal{T}(U_{\tilde{\rho}})|$.
Then $Y_{1}(T)$ and $\mathcal{T}_{1}(U_{\tilde{\rho}})$ satisfy all the
requirements.

We prove that

$\displaystyle\sum_{T\in\mathcal{T}_{1}}|Y^{\prime\prime}(T)\setminus
Y_{1}(T)|$
$\displaystyle\leq|\mathcal{P}_{1}|\cdot\delta^{\epsilon^{4}}{|\mathcal{T}_{p}^{\prime}|}+|\mathcal{T}_{1}|\cdot\delta^{\epsilon^{4}}\frac{|Y^{\prime}(T)|_{\delta}}{|Y^{\prime}(T)|_{{\tilde{\rho}}}}\cdot|Y^{\prime}(T)|_{{\tilde{\rho}}}$
$\displaystyle+\delta^{\epsilon^{4}}\sum_{U_{\tilde{\rho}}}|\mathcal{T}(U_{\tilde{\rho}})|\cdot\frac{|Y^{\prime}(T)|_{\delta}}{|Y^{\prime}(T)|_{{\tilde{\rho}}}}.$

The three terms record the three sources of edge losses in the graph
$G^{\prime}$, whose size we recall to be
$|G^{\prime}|=\sum_{T\in\mathcal{T}_{1}}|Y^{\prime\prime}(T)|$. The first term
comes from (7.44), the second from (7.45), while the third from (7.46).

We next observe that each of the three terms is (see (5) for the $\sim$)

$\lessapprox_{\delta}\delta^{\epsilon^{4}}\sum_{T\in\mathcal{T}_{1}}|Y^{\prime}(T)|\sim\delta^{\epsilon^{4}}\sum_{T\in\mathcal{T}_{1}}|Y^{\prime\prime}(T)|.$

It suffices to check this for the last term. Indeed, note that

$\sum_{U_{\tilde{\rho}}}|\mathcal{T}(U_{\tilde{\rho}})|=\sum_{U_{\tilde{\rho}}}\sum_{T\in\mathcal{T}_{1}:\atop{U_{\tilde{\rho}}\subset
T\atop{U_{\tilde{\rho}}\cap
Y^{\prime\prime}(T)\not=\emptyset}}}1=\sum_{T\in\mathcal{T}_{1}}\sum_{U_{\tilde{\rho}}\subset
T:\atop{U_{\tilde{\rho}}\cap
Y^{\prime\prime}(T)\not=\emptyset}}1=\sum_{T\in\mathcal{T}_{1}}|Y^{\prime\prime}(T)|_{\tilde{\rho}}\leq\sum_{T\in\mathcal{T}_{1}}|Y^{\prime}(T)|_{\tilde{\rho}}.$

Step 5. Definition of $\bar{\mathcal{U}}_{{\tilde{\rho}},p}$. Verification of
(7.37).

Let $\mathcal{T}_{1,p}^{\prime}=\\{T\in\mathcal{T}_{1}:\;p\in Y_{1}(T)\\}$.
Then Step 4 shows that

$\sum_{p\in\mathcal{P}_{1}}|\mathcal{T}_{1,p}^{\prime}|=\sum_{T\in\mathcal{T}_{1}}|Y_{1}(T)|\sim\sum_{T\in\mathcal{T}_{1}}|Y^{\prime\prime}(T)|.$

Since the last term equals
$|G^{\prime}|\overset{(3)}{\sim}|G|\geq\sum_{p\in\mathcal{P}_{1}}|\mathcal{T}_{p}|$,
we find that

$\sum_{p\in\mathcal{P}_{1}}|\mathcal{T}_{1,p}^{\prime}|\gtrsim\sum_{p\in\mathcal{P}_{1}}|\mathcal{T}_{p}|.$

It follows that there is $\mathcal{P}_{1}^{\prime}\subset\mathcal{P}_{1}$ with
$|\mathcal{P}_{1}^{\prime}|\gtrsim|\mathcal{P}_{1}|$ such that
$|\mathcal{T}_{1,p}^{\prime}|\gtrsim|\mathcal{T}_{p}|$ for each
$p\in\mathcal{P}_{1}^{\prime}$. When combined with (7.33), this shows that for
each $p\in\mathcal{P}_{1}^{\prime}$ we have
$|\mathcal{U}_{\tilde{\rho}}(\mathcal{T}^{\prime}_{1,p})|\gtrsim|\mathcal{U}_{{\tilde{\rho}},p}|$
(recall our notation
$\mathcal{U}_{{\tilde{\rho}},p}=\mathcal{U}_{\tilde{\rho}}(\mathcal{T}_{\tilde{\rho}})$).

We let $\bar{\mathcal{U}}_{{\tilde{\rho}},p}$ equal
$\mathcal{U}_{{\tilde{\rho}}}(\mathcal{T}_{1,p}^{\prime})$ if
$p\in\mathcal{P}_{1}$ and $\bar{\mathcal{U}}_{{\tilde{\rho}},p}=\emptyset$ if
$p\in\mathcal{P}\setminus\mathcal{P}_{1}$. Using (7.35) and (7.39), the
inequality (7.37) is now immediate, as $|\mathcal{U}_{{\tilde{\rho}},p}|$ is
constant for $p\in\mathcal{P}$.

Step 6. Verification of (7.36).

Fix a segment $U_{\tilde{\rho}}\in\bar{\mathcal{U}}_{\tilde{\rho}}$, so
$\mathcal{P}_{U_{\tilde{\rho}}}\not=\emptyset$. That means that there is
$T\in\mathcal{T}_{1}$ such that $U_{\tilde{\rho}}\subset T$ and
$U_{\tilde{\rho}}\cap Y_{1}(T)\not=\emptyset$. Define
$\mathcal{I}(U_{\tilde{\rho}})=\cup_{T\in\mathcal{T}_{1}(U_{\tilde{\rho}})}(Y_{1}(T)\cap
U_{\tilde{\rho}})$. Note that
$\mathcal{I}(U_{\tilde{\rho}})=\mathcal{P}_{U_{\tilde{\rho}}}$. Let

$X(U_{\tilde{\rho}})=\\{(p,T)\in\mathcal{I}(U_{\tilde{\rho}})\times\mathcal{T}_{1}(U_{\tilde{\rho}}):p\in
Y_{1}(T)\\}.$

Then, by (7.45)

$|X(U_{\tilde{\rho}})|\geq\sum_{T\in\mathcal{T}_{1}(U_{\tilde{\rho}})}|Y_{1}(T)\cap
U_{\tilde{\rho}}|\gtrsim\delta^{\epsilon^{4}}\delta^{-f(1)+f(b)}|\mathcal{T}_{1}(U_{\tilde{\rho}})|.$

Since for each $p\in\mathcal{I}(U_{\tilde{\rho}})$ the number of $T$ such that
$(p,T)\in X(U_{\tilde{\rho}})$ is
$\leq|\mathcal{T}|_{p}\cdot|\mathcal{T}_{p}|_{\delta/{\tilde{\rho}}}^{-1}$, we
have

$|X(U_{\tilde{\rho}})|\leq|\mathcal{I}(U_{\tilde{\rho}})||\mathcal{T}|_{p}\cdot|\mathcal{T}_{p}|_{\delta/{\tilde{\rho}}}^{-1}.$

Combining the upper and lower bounds we find

(7.47)
$|\mathcal{I}(U_{\tilde{\rho}})|\gtrsim\frac{\delta^{-f(1)+f(b)+\epsilon^{4}}|\mathcal{T}_{1}(U_{\tilde{\rho}})|}{|\mathcal{T}_{p}|\cdot|\mathcal{T}_{p}|_{\delta/{\tilde{\rho}}}^{-1}}.$

For any ${\tilde{\rho}}^{\prime}=\delta^{c}<\delta^{b}$ and each square
$B_{{\tilde{\rho}}^{\prime}}$, let

$X^{\prime}(U_{\tilde{\rho}})=\\{(p,T)\in(\mathcal{I}(U_{\tilde{\rho}})\cap
B_{{\tilde{\rho}}^{\prime}})\times\mathcal{T}(U_{\tilde{\rho}}):p\in
Y^{\prime\prime}(T)\\}.$

Then

$|X^{\prime}(U_{\tilde{\rho}})|\leq\sum_{T\in\mathcal{T}(U_{\tilde{\rho}})}|Y^{\prime\prime}(T)\cap
B_{{\tilde{\rho}}^{\prime}}|\leq\sum_{T\in\mathcal{T}(U_{\tilde{\rho}})}|Y^{\prime}(T)\cap
B_{{\tilde{\rho}}^{\prime}}|\lesssim\delta^{-f(1)+f(c)}\cdot|\mathcal{T}(U_{\tilde{\rho}})|.$

Moreover, for each $p\in\mathcal{I}(U_{\tilde{\rho}})\cap
B_{{\tilde{\rho}}^{\prime}}$, by (7.43), the number of
$T\in\mathcal{T}(U_{\tilde{\rho}})$ such that $p\in Y^{\prime\prime}(T)$ is
$\gtrsim\delta^{2\epsilon^{4}}|\mathcal{T}_{p}|\cdot|\mathcal{T}_{p}|_{\delta/{\tilde{\rho}}}^{-1}$.
Therefore,

(7.48) $|\mathcal{I}(U_{\tilde{\rho}})\cap
B_{{\tilde{\rho}}^{\prime}}|\lesssim\frac{\delta^{-f(1)+f(c)-2\epsilon^{4}}\cdot|\mathcal{T}(U_{\tilde{\rho}})|}{|\mathcal{T}_{p}|\cdot|\mathcal{T}_{p}|_{\delta/{\tilde{\rho}}}^{-1}}.$

By (7.46),(7.47) and (7.48) we conclude that

$|\mathcal{I}(U_{\tilde{\rho}})\cap
B_{{\tilde{\rho}}^{\prime}}|\lesssim\delta^{f(c)-f(b)-4\epsilon^{4}}|\mathcal{I}(U_{\tilde{\rho}})|.$

Recall that $c>b$ and that $\sigma_{j}\geq\epsilon$ on $[b,1]$. Combining this
with (7.41) it follows that $f(c)-f(b)\geq\epsilon(c-b)-\epsilon^{4}$. We thus
have

(7.49) $|\mathcal{I}(U_{\tilde{\rho}})\cap
B_{{\tilde{\rho}}^{\prime}}|\lesssim\delta^{\epsilon(c-b)-5\epsilon^{4}}|\mathcal{I}(U_{\tilde{\rho}})|=\delta^{-5\epsilon^{4}}({\tilde{\rho}}^{\prime}/{\tilde{\rho}})^{\epsilon}|\mathcal{I}(U_{\tilde{\rho}})|.$

(7.36) is now a consequence of this and (7.35).

Step 7. Verification of (7.38).

By (7.42) and our choice of $b$ we infer that $f(b)\leq
b(\epsilon+\epsilon^{4})$. Thus, for each $T\in\mathcal{T}_{1}$,

(7.50)
$|Y^{\prime}(T)|_{{\tilde{\rho}}}=\delta^{-f(b)}\leq\delta^{-b(\epsilon+\epsilon^{4})}=(\frac{1}{{\tilde{\rho}}})^{\epsilon+\epsilon^{4}}.$

Fix any $U_{\tilde{\rho}}\in\bar{\mathcal{U}}_{\tilde{\rho}}$. Then, as in
Step 6, taking ${\tilde{\rho}}^{\prime}={\tilde{\rho}}$

$|\mathcal{I}(U_{\tilde{\rho}})|\delta^{2\epsilon^{4}}|\mathcal{T}_{p}|\cdot|\mathcal{T}_{p}|_{\delta/{\tilde{\rho}}}^{-1}\lesssim|X^{\prime}(U_{\tilde{\rho}})|\lesssim|\mathcal{T}(U_{\tilde{\rho}})|\frac{|Y^{\prime}(T)|}{|Y^{\prime}(T)|_{\tilde{\rho}}}.$

Since by (7.45)

$|\mathcal{I}(U_{\tilde{\rho}})|\geq\delta^{\epsilon^{4}}\frac{|Y^{\prime}(T)|}{|Y^{\prime}(T)|_{{\tilde{\rho}}}},$

we find that

(7.51)
$|\mathcal{T}(U_{\tilde{\rho}})|\gtrsim\delta^{3\epsilon^{4}}|\mathcal{T}_{p}|\cdot|\mathcal{T}_{p}|_{\delta/{\tilde{\rho}}}^{-1}.$

Another double counting shows that

$|\bar{\mathcal{U}}_{\tilde{\rho}}|\min_{U_{\tilde{\rho}}\in\bar{\mathcal{U}}_{\tilde{\rho}}}|\mathcal{T}(U_{\tilde{\rho}})|\leq|\\{(T,U_{\tilde{\rho}})\in\mathcal{T}_{1}\times\bar{\mathcal{U}}_{\tilde{\rho}}:\;U_{\tilde{\rho}}\subset
T\text{ and }U_{\tilde{\rho}}\cap
Y_{1}^{\prime\prime}(T)\not=\emptyset\\}|\leq|\mathcal{T}_{1}||Y^{\prime}(T)|_{\tilde{\rho}},$

which when combined with (7.50) and (7.51) shows that

$|\bar{\mathcal{U}}_{\tilde{\rho}}||\mathcal{T}_{p}|\lesssim|\mathcal{T}_{1}||\mathcal{T}_{p}|_{\delta/{\tilde{\rho}}}\delta^{-4\epsilon^{4}-\epsilon}.$

This gives (7.38), as
$|\mathcal{U}_{{\tilde{\rho}},p}|=|\mathcal{T}_{p}|_{\delta/{\tilde{\rho}}}.$
∎

##  8\. Proof of Theorem 1.1

The argument in this section will recycle ideas and estimates established
earlier in the paper. Parts of the presentation will be kept slightly less
formal.

In order to simplify the iteration in Step 3 below, we will use the fact that
$\mathcal{S}\mathcal{T}(\rho)\lesssim\mathcal{S}\mathcal{T}(\delta)$ if
$\delta<\rho$. We sketch a heuristic argument for this. Fix $r_{\rho}$ and a
configuration $\mathcal{T}(\rho)$ of $\rho$-tubes $T_{\rho}$ as in Theorem
1.1. We call $\Lambda(\rho)\subset\mathbb{S}^{1}$ the associated collection of
$\rho$-arcs. For each $I\in\Lambda(\rho)$ we consider a (rescaled)
$(\delta/\rho,s)$-set $\Lambda_{I}$ of $\delta$-intervals inside $I$, with
cardinality $(\rho/\delta)^{s}$. The set
$\Lambda(\delta)=\cup_{I\in\Lambda(\rho)}\Lambda_{I}$ is a $(\delta,s)$-set
with cardinality $\sim\delta^{-s}$. For each $T_{\rho}\in\mathcal{T}(\rho)$ in
the direction $I$ and for each $J\in\Lambda_{I}$ we consider a (rescaled)
$(\delta/\rho,1-s)$-set $\mathcal{T}(T_{\rho},J)$ of $\delta$-tubes $T$ in the
direction $J$, lying inside $T_{\rho}$ and with cardinality
$(\rho/\delta)^{1-s}$. The union $\mathcal{T}$ of these tubes satisfies the
requirements of Theorem 1.1 at scale $\delta$. Fix a $\rho$-square
$\mathbf{p}_{\rho}$ intersecting some $T_{\rho}$. Since $T_{\rho}$ contains
$\sim\rho/\delta$ tubes $T\in\mathcal{T}$, the corresponding
$\delta\times\rho$ segments $T\cap\mathbf{p}_{\rho}$ have a total volume of
$\sim\rho^{2}$, comparable to the area of $\mathbf{p}_{\rho}$. A random choice
of the tubes $T$ inside $T_{\rho}$ will allow us to assume that a large
fraction of (the $\delta$-squares in) $\mathbf{p}_{\rho}$ intersects at least
one such $T$. Thus, for generic
$\mathbf{p}_{\rho}\in\mathcal{P}_{r_{\rho}}(\mathcal{T}(\rho))$, a large
fraction of its $\delta$-squares will intersect $\sim r_{\rho}$ tubes
$T\in\mathcal{T}$. This leads to

$|\mathcal{P}_{r_{\rho}}(\mathcal{T}(\rho))|(\rho/\delta)^{2}\lesssim|\mathcal{P}_{\sim
r_{\rho}}(\mathcal{T})|\leq\mathcal{S}\mathcal{T}(\delta)\frac{|\mathcal{T}|^{2}}{(\sim
r_{\rho})^{3}}\sim\mathcal{S}\mathcal{T}(\delta)\frac{\delta^{-2}}{r_{\rho}^{3}},$

showing that
$|\mathcal{P}_{r_{\rho}}(\mathcal{T}(\rho))|\lesssim\mathcal{S}\mathcal{T}(\delta)\frac{|\mathcal{T}(\rho)|^{2}}{r_{\rho}^{3}}$,
as desired.

We need to verify that for each $\upsilon>0$

(8.52) $\mathcal{S}\mathcal{T}(\delta)\lesssim_{\upsilon}\delta^{-\upsilon}.$

To achieve this, we fix $r_{0}$, $\delta$ and an arbitrary collection
$\mathcal{T}$ as in Theorem 1.1. We aim to get an upper bound for
$|\mathcal{P}_{r_{0}}(\mathcal{T})|$. There are several steps.

Step 1. Uniformization:

Fix $\epsilon>0$. Let $\epsilon_{0}$ satisfy (2.17).

We will assume $\mathcal{T}$ and $\Lambda$ are $\epsilon_{0}$-uniform. This
will incur negligible $\delta^{-O(\epsilon_{0})}$-losses. We will shortly see
that these are consistent with the bound (8.57) we aim to prove.

By dropping to a $\delta^{\epsilon_{0}}$ smaller collection of squares
$\mathcal{P}\subset\mathcal{P}_{r_{0}}(\mathcal{T})$, we may assume as before
(see the proof of Theorem 5.4$\implies$Theorem 5.3) that the set of tubes
through each $p\in\mathcal{P}$ admit an $\epsilon_{0}$-uniform set
$\mathcal{T}_{p}$ with a universal branch function. By an additional
refinement we may also assume $\mathcal{P}$ is $\epsilon_{0}$-uniform. In
particular, call $r$ the common values of $|\mathcal{T}_{p}|$. This produces
another affordable $\delta^{-\epsilon_{0}}$-loss, this time in the value of
$r_{0}$. The best we can say is that

(8.53) $r\gtrsim r_{0}\delta^{\epsilon_{0}}.$

Write $P=|\mathcal{P}|$. Since $\epsilon_{0}$ is a function of $\epsilon$, all
$\epsilon_{0}$-uniform sets will also be approximate $\epsilon$-uniform, see
Remark 2.5. All implicit constants will depend only on $\epsilon$.

Step 2: Scale inflation.

We apply Theorem 2.13 to $\mathcal{T}_{p}$. In this application of the
theorem, we only need to work with the interval $[A_{1},A_{2}]$. The numbers
$A_{2}$ and $t_{1}$ are independent of $p$. Recall that

$A_{2}\geq\epsilon^{-1}\epsilon_{0},\;\;t_{1}\leq\log_{1/\delta}r+\epsilon.$

We write $\rho_{0}=\delta^{1-A_{2}}$ and
$\psi(\epsilon)=\epsilon^{-1}\epsilon_{0}=\epsilon^{2\epsilon^{-1}-1}$. We
have that

(8.54)
$\delta^{-\epsilon_{0}}\leq(\frac{\rho_{0}}{\delta})^{\epsilon\psi(\epsilon)}.$

Step 3: Induction on scales.

We will prove that

(8.55)
$\begin{split}P&\leq\frac{|\mathcal{T}|^{2}}{r^{3}}C_{\epsilon}(\log\delta^{-1})^{O(1)}\big{\\{}(\rho_{0}/\delta)^{a_{\epsilon}}\max_{\delta\leq\rho\leq\rho_{0}}(\rho_{0}/\rho)^{t_{1}-\frac{1}{2}}[\mathcal{S}\mathcal{T}(\rho_{0})\mathcal{S}\mathcal{T}(\rho)]^{1/2}\\\
&+\max_{\delta(\rho_{0}/\delta)^{c_{\epsilon}}\lesssim\tilde{\delta}\leq\rho_{0}}({\tilde{\delta}}/{\delta})^{b_{\epsilon}}\mathcal{S}\mathcal{T}(\tilde{\delta})\big{\\}}.\end{split}$

The two terms on the right-hand side trace their origins to the two items in
Theorem 5.3. The first comes from Item (1), the second from Item (2).

The absolute constants $a_{\epsilon},b_{\epsilon},c_{\epsilon}$ satisfy
$\lim_{\epsilon\to 0}a_{\epsilon},b_{\epsilon}=0$ and $c_{\epsilon}>0$.
Moreover,

$\lim_{\epsilon\to 0}\frac{\epsilon\psi(\epsilon)}{c_{\epsilon}}=0.$

See (8.61) and (8.74). When combined with (8.54), it shows that the
$\delta^{-O(\epsilon_{0})}$ losses are negligible. More precisely,

$\delta^{-O(\epsilon_{0})}\lesssim(\frac{\tilde{\delta}}{\delta})^{o_{\epsilon}(1)},\text{
whenever }\delta(\rho_{0}/\delta)^{c_{\epsilon}}\lesssim\tilde{\delta}.$

This together with the estimate
$|\mathcal{P}_{r_{0}}(\mathcal{T})|\lesssim\delta^{-O(\epsilon_{0})}P$, (8.53)
and (8.55) leads to

(8.56)
$\begin{split}|\mathcal{P}_{r_{0}}(\mathcal{T})|&\leq\frac{|\mathcal{T}|^{2}}{r_{0}^{3}}C_{\epsilon}(\log\delta^{-1})^{O(1)}\big{\\{}(\rho_{0}/\delta)^{a_{\epsilon}}\max_{\delta\leq\rho\leq\rho_{0}}(\rho_{0}/\rho)^{t_{1}-\frac{1}{2}}[\mathcal{S}\mathcal{T}(\rho_{0})\mathcal{S}\mathcal{T}(\rho)]^{1/2}\\\
&+\max_{\delta(\rho_{0}/\delta)^{c_{\epsilon}}\lesssim\tilde{\delta}\leq\rho_{0}}({\tilde{\delta}}/{\delta})^{b_{\epsilon}}\mathcal{S}\mathcal{T}(\tilde{\delta})\big{\\}}.\end{split}$

The constant $C_{\epsilon}$, $a_{\epsilon},b_{\epsilon}$ have changed slightly
to incorporate the $\delta^{-O(\epsilon_{0})}$ loss, but their properties
remain unchanged. The parameters $\rho_{0},t_{1}$ depend on $r$, thus on
$r_{0}$, but in a uniform way. More precisely, using our chief assumption that
$s\leq\frac{1}{2}$, we find that $r\lesssim\delta^{-1/2}$, and thus
$t_{1}\leq\frac{1}{2}+o_{\epsilon}(1)$ (cf. (iv) in Theorem 2.13). Since also
$\rho_{0}/\delta=\delta^{-A_{2}}\geq\delta^{-\psi(\epsilon)}$ with
$\psi(\epsilon)$ independent of $r$, inequality (8.56) in fact gives the
($r_{0}$-independent) inequality (with a new $a_{\epsilon}$ that still goes to
zero)

$\mathcal{S}\mathcal{T}(\delta)\leq
C_{\epsilon}(\log\delta^{-1})^{O(1)}\max_{\rho_{0}\geq\delta^{1-\psi(\epsilon)}}\big{\\{}(\rho_{0}/\delta)^{a_{\epsilon}}[\mathcal{S}\mathcal{T}(\rho_{0})\mathcal{S}\mathcal{T}(\delta)]^{1/2}$
$+\max_{\delta(\rho_{0}/\delta)^{c_{\epsilon}}\lesssim\tilde{\delta}\leq\rho_{0}}({\tilde{\delta}}/{\delta})^{b_{\epsilon}}\mathcal{S}\mathcal{T}(\tilde{\delta})\big{\\}}.$

To streamline a bit the inequality, in the first line we have used that
$\mathcal{S}\mathcal{T}(\rho)\lesssim\mathcal{S}\mathcal{T}(\delta)$.

If the first line dominates, then we have
$\mathcal{S}\mathcal{T}(\delta)\lesssim(\log\delta^{-1})^{O(1)}(\rho_{0}/\delta)^{o_{\epsilon}(1)}\mathcal{S}\mathcal{T}(\rho_{0})$.
Thus, the above inequality can be rewritten as

$\displaystyle\mathcal{S}\mathcal{T}(\delta)$ $\displaystyle\leq
C_{\epsilon}(\log\delta^{-1})^{O(1)}\max_{\rho_{0}\geq\delta^{1-\psi(\epsilon)}}\big{\\{}(\rho_{0}/\delta)^{a_{\epsilon}}\mathcal{S}\mathcal{T}(\rho_{0})$
$\displaystyle+\max_{\delta(\rho_{0}/\delta)^{c_{\epsilon}}\lesssim\tilde{\delta}\leq\rho_{0}}({\tilde{\delta}}/{\delta})^{b_{\epsilon}}\mathcal{S}\mathcal{T}(\tilde{\delta})\big{\\}}.$

Call $d_{\epsilon}=\psi(\epsilon)c_{\epsilon}>0$. Note that
$\rho_{0}/\delta,\tilde{\delta}/\delta\gtrsim\delta^{-d_{\epsilon}}$. The
above inequality implies that for each $\epsilon$ and $\delta$

(8.57) $\mathcal{S}\mathcal{T}(\delta)\leq
C_{\epsilon}\max_{\tilde{\delta}/\delta\gtrsim\delta^{-d_{\epsilon}}}({\tilde{\delta}}/{\delta})^{e_{\epsilon}}\mathcal{S}\mathcal{T}(\tilde{\delta}),$

for some $e_{\epsilon}\to 0$. The log terms have been absorbed into
$({\tilde{\delta}}/{\delta})^{e_{\epsilon}}$.

Note the delicate nature of this inequality. It contains no generic
$\delta^{o_{\epsilon}(1)}$-losses, only the carefully calibrated term
$(\tilde{\delta}/\delta)^{e_{\epsilon}}$. This by itself is enough to conclude
(8.52) via a bootstrapping argument, as follows.

Let $\upsilon_{0}$ be the infimum of the following nonempty set (this is an
interval)

$\\{\upsilon>0:\;\exists\;C_{\upsilon}<\infty\text{ such that
}\mathcal{S}\mathcal{T}(\delta)\leq C_{\upsilon}\delta^{-\upsilon}\text{ for
each }\delta\leq 1\\}.$

Assume for contradiction that $\upsilon_{0}>0$. This allows us to pick
$\epsilon$ sufficiently small so that $e_{\epsilon}<\upsilon_{0}/2$. Then we
pick $\upsilon$ larger than, but sufficiently close to $\upsilon_{0}$, so that
$\upsilon(1-\frac{d_{\epsilon}}{2})<\upsilon_{0}$. We use
$\mathcal{S}\mathcal{T}(\tilde{\delta})\leq
C_{\upsilon}\tilde{\delta}^{-\upsilon}$ in (8.57) with the $\epsilon$ picked
above and get

$\mathcal{S}\mathcal{T}(\delta)\leq
C_{\epsilon}C_{\upsilon}(\delta^{-1})^{\upsilon-(\upsilon-
e_{\epsilon})d_{\epsilon}}.$

Since $\upsilon-(\upsilon-
e_{\epsilon})d_{\epsilon}<\upsilon(1-\frac{d_{\epsilon}}{2})<\upsilon_{0}$,
this leads to a contradiction.

Step 4: Proof of (8.55). This occupies the remainder of this section.

There are $\sim\delta^{-A_{2}(t_{1}+o_{\epsilon}(1))}$ tubes
$\mathbf{T}\in\mathcal{T}_{p}^{\delta/\rho_{0}}\subset\mathcal{T}^{\delta/\rho_{0}}$.
We write $\mathcal{T}_{p}\cap\mathbf{T}$ for the collection
$\mathcal{T}_{p}[\mathbf{T}]$ of tubes $T\in\mathcal{T}_{p}$ lying inside
$\mathbf{T}$. Due to uniformity, $|\mathcal{T}_{p}\cap\mathbf{T}|$ is either
zero or $\sim\tilde{r}$. This $\tilde{r}$ is independent of $p$, but whether
$|\mathcal{T}_{p}\cap\mathbf{T}|$ is zero or $\sim\tilde{r}$ depends on $p$
(for each fixed $\mathbf{T}$). We have that

(8.58) $\tilde{r}=|\mathcal{T}_{p}^{\rho_{0}/\delta}|\sim
r(\delta/\rho_{0})^{t_{1}+o_{\epsilon}(1)}$

and $|\mathcal{T}_{p}\cap\mathbf{T}|\sim\tilde{r}$ for
$\sim\frac{r}{\tilde{r}}$ of the tubes
$\mathbf{T}\in\mathcal{T}^{\delta/\rho_{0}}$.

Throughout the argument we may afford $(\log\delta^{-1})^{O(1)}$ losses,
denoted by the symbol $\lessapprox$. These will arise in a few places, due to
pigeonholing and the application of the auxiliary results from Section 4 (see
also Remark 3.2).

We tile each $\mathbf{T}\in\mathcal{T}^{\delta/\rho_{0}}$ with
$\delta\times\rho_{0}$-segments $U$ having the same orientation as
$\mathbf{T}$. The length $\rho_{0}$ of these segments is informed by the fact
that for each maximal cluster of tubes $T\subset\mathbf{T}$ intersecting a
fixed square in the core $\frac{1}{2}\mathbf{T}$ of $\mathbf{T}$, the
intersection of the tubes in the cluster is a
$\sim\delta\times\rho_{0}$-segment.

For each $p\in\mathcal{P}$ call $\mathcal{U}_{p}=\mathcal{U}_{\rho_{0},p}$ the
collection of all such segments containing $p$, one for each
$\mathbf{T}\in\mathcal{T}_{p}^{\delta/\rho_{0}}$. Call
$\mathcal{U}=\cup_{p\in\mathcal{P}}\mathcal{U}_{p}$. Recall that

(8.59) $|\mathcal{U}_{p}|\sim r/\tilde{r}.$

We partition $[0,1]^{2}$ into squares $\mathbf{p}_{\rho_{0}}$ with side length
$\rho_{0}$, and call $\mathcal{U}[\mathbf{p}_{\rho_{0}}]$
($\mathcal{P}[\mathbf{p}_{\rho_{0}}]$) those segments from $\mathcal{U}$
(squares in $\mathcal{P}$) that lie inside $\mathbf{p}_{\rho_{0}}$. Write

$V_{\mathbf{p}_{\rho_{0}}}=\frac{\sum_{p\in\mathcal{P}[\mathbf{p}_{\rho_{0}}]}|\mathcal{U}_{p}|}{|\mathcal{U}[\mathbf{p}_{\rho_{0}}]|}.$

Recall the function $\varphi$ from Theorem 2.13 and the function $\psi$ from
(8.54). We let $h(\epsilon)$ be a positive quantity converging to zero slowly
enough as $\epsilon\to 0$, such that

(8.60) $\lim_{\epsilon\to 0}\frac{\varphi(\epsilon)}{h(\epsilon)^{7/4}}=0,$

and

(8.61) $\lim_{\epsilon\to
0}\frac{\epsilon\psi(\epsilon)}{h(\epsilon)^{7/4}}=0.$

We let $\mathcal{P}_{1}$ be those $p\in\mathcal{P}$ that lie inside a
$\mathbf{p}_{\rho_{0}}$ with
$V_{\mathbf{p}_{\rho_{0}}}\lesssim(\rho_{0}/\delta)^{2h(\epsilon)}$. Then
$\mathcal{P}_{2}=\mathcal{P}\setminus\mathcal{P}_{1}$.

Case 1. Estimates for $|\mathcal{P}_{1}|$.

Each $U\in\mathcal{U}$ is contained inside at least $\tilde{r}$ tubes $T$,
coming from exactly one $\mathbf{T}\in\mathcal{T}^{\delta/\rho_{0}}$.

Observe that $\mathcal{T}[\mathbf{T}]$, after parabolic rescaling by
$\rho_{0}/\delta$, satisfies the assumption of Proposition 4.2, with scale
$\delta$ replaced by $\rho_{0}$ and $K_{1},K_{2}\lesssim 1$. This rescaling
maps $U$ to a $\rho_{0}$-square and the collection $\mathcal{T}[\mathbf{T}]$
to the collection of $\rho_{0}\times 1$ tubes
$\mathcal{T}_{resc}(\mathbf{T})$. We find that

(8.62)
$|\mathcal{P}_{\geq\tilde{r}}(\mathcal{T}_{resc}(\mathbf{T}))|\lessapprox\mathcal{S}\mathcal{T}(\rho_{0})\rho_{0}^{-1}\frac{|\mathcal{T}[\mathbf{T}]|_{\delta}}{\tilde{r}^{3}}.$

On the other hand, using (8.59) we find

(8.63)
$\frac{r}{\tilde{r}}|\mathcal{P}_{1}|\sim\sum_{V_{\mathbf{p}_{\rho_{0}}}\lesssim(\rho_{0}/\delta)^{2h(\epsilon)}}V_{\mathbf{p}_{\rho_{0}}}\cdot|\mathcal{U}[\mathbf{p}_{\rho_{0}}]|\leq(\rho_{0}/\delta)^{o_{\epsilon}(1)}\sum_{\mathbf{T}\in\mathcal{T}^{\delta/\rho_{0}}}|\mathcal{P}_{\geq\tilde{r}}(\mathcal{T}_{resc}(\mathbf{T}))|.$

Using (8.58), (8.62) and (8.63) we find after summation in $\mathbf{T}$

$\displaystyle|\mathcal{P}_{1}|$
$\displaystyle\lessapprox\mathcal{S}\mathcal{T}(\rho_{0})\rho_{0}^{-1}(\rho_{0}/\delta)^{o_{\epsilon}(1)}\frac{|\mathcal{T}|}{r\tilde{r}^{2}}$
$\displaystyle\sim\mathcal{S}\mathcal{T}(\rho_{0})(\frac{\rho_{0}}{\delta})^{-1+o_{\epsilon}(1)}\frac{|\mathcal{T}|^{2}}{r\tilde{r}^{2}}\text{,
since }|\mathcal{T}|\sim\delta^{-1}$
$\displaystyle\sim\mathcal{S}\mathcal{T}(\rho_{0})(\frac{\rho_{0}}{\delta})^{2t_{1}-1+o_{\epsilon}(1)}\frac{|\mathcal{T}|^{2}}{r^{3}}.$

This verifies (8.55) in this case, as $|\mathcal{P}_{r}(\mathcal{T})|$ is
dominated by the first term in (8.55), with the maximum evaluated at
$\rho=\delta$. Indeed, this follows by taking the geometric average of the
estimate from above, with the trivial estimate
$P\leq\mathcal{S}\mathcal{T}(\delta)\frac{|\mathcal{T}|^{2}}{r^{3}}.$

Case 2. Estimates for $|\mathcal{P}_{2}|$ .

Here is a brief overview of the argument in this case. We first pursue the
two-ends reduction from Section 7. This will create the first intermediate
scale, called $\rho$. At this scale Theorem 5.3 becomes applicable. It leads
to a dichotomy that we analyze in Case 2(a) and Case 2(b). In the first case
we get another intermediate scale $\Delta$. The application of Lemma 6.1 at
scale $\Delta$ creates the third intermediate scale $\tilde{\delta}$. It will
be important to verify that all these scales are quantitatively larger than
the initial scale $\delta$.

Here are the details. We apply the following procedure to each
$\mathbf{p}_{\rho_{0}}$ with
$V_{\mathbf{p}_{\rho_{0}}}\gtrsim(\rho_{0}/\delta)^{2h(\epsilon)}$. We rescale
$\mathbf{p}_{\rho_{0}}$ by $1/\rho_{0}$ and map it to $[0,1]^{2}$. The
segments $U\in\mathcal{U}[\mathbf{p}_{\rho_{0}}]$ will become
$\delta\rho_{0}^{-1}\times 1$ tubes $T$. We apply Theorem 7.1 to these tubes,
and the collection of rescaled squares in
$\mathcal{P}[\mathbf{p}_{\rho_{0}}]$. The scale $\delta$ in Theorem 7.1 is
replaced with $\delta\rho_{0}^{-1}$. The hypothesis (7.34) is satisfied with
$\epsilon$ replaced by $h(\epsilon)$. Then Theorem 7.1 delivers a scale
$\tilde{\rho}$ such that, according to (7.35),

$\tilde{\rho}\geq(\frac{\delta}{\rho_{0}})^{1-h(\epsilon)}.$

Subject to a logarithmic loss we may and will assume that this scale is the
same for each $\mathbf{p}_{\rho_{0}}$. When we rescale back, this scale
becomes $\rho:=\tilde{\rho}\rho_{0}$. Note that

(8.64) $\frac{\rho}{\delta}\geq(\frac{\rho_{0}}{\delta})^{h(\epsilon)}.$

For each $p\in\mathcal{P}$, let $\mathcal{U}_{1,p}$ be the collection of all
$\delta\times\rho$ mini-segments $U_{1}$ containing $p$, one for each
$\mathbf{T}_{1}\in\mathcal{T}_{p}^{\delta/\rho}$.

Partition $\mathbf{p}_{\rho_{0}}$ into $\rho$-squares $\mathbf{p}_{\rho}$.
Since $\mathcal{P}$ is uniform, each $\mathbf{p}_{\rho}$ intersecting
$\mathcal{P}$ will contain roughly the same number of squares
$\mathcal{P}[\mathbf{p}_{\rho}]$ from $\mathcal{P}$. For each such
$\mathbf{p}_{\rho}$, we call
$\mathcal{U}_{1}[\mathbf{p}_{\rho}]=\cup_{p\in\mathbf{p}_{\rho}}\mathcal{U}_{1,p}$.
We apply Theorem 5.3 to the $\times\rho^{-1}$ rescaled version of
$\mathbf{p}_{\rho}$. The scale $\delta$ there is replaced with $\delta/\rho$.
The collections $\mathcal{T}_{p}$ in Theorem 5.3 are the rescaled copies of
the mini-segments $\mathcal{U}_{1,p}[\mathbf{p}_{\rho}]$. The $\epsilon$ in
that theorem is replaced with $\epsilon^{\prime}:=h(\epsilon)$. Let us note
the following.

$\bullet$ The two-ends hypothesis (5.22) is satisfied due to (7.36).

$\bullet$ Due to (7.37), the hypothesis (5.23) is satisfied with
$\eta=(\epsilon^{\prime})^{3}$, for at least a $(\delta/\rho)^{\eta}$ fraction
of the squares $\mathbf{p}_{\rho}$. This is good enough for us, as each
$\mathbf{p}_{\rho}$ contains the same number of $p\in\mathcal{P}_{2}$, and we
will see that $(\delta/\rho)^{\eta}$ is a negligible loss.

$\bullet$ $\mathcal{U}_{1,p}$ can be identified with
$\mathcal{T}_{p}^{\rho/\delta}$. Theorem 2.13 (iii) together with Lemma 2.6
imply that its $1/\rho$-rescaled copy is a
$(\delta/\rho,t_{1},(\rho_{0}/\delta)^{\varphi(\epsilon)})$-set. Combining
(8.60) and (8.64) we conclude that this is a
$(\delta/\rho,t_{1},(\rho/\delta)^{o_{\epsilon}(1)})$-set. Thus, it is also a
$(\delta/\rho,t_{1},(\rho/\delta)^{o_{\epsilon^{\prime}}(1)})$-set.

$\bullet$ We let $\bar{\mathcal{U}}_{1}[\mathbf{p}_{\rho}]$ be the subset of
$\mathcal{U}_{1}[\mathbf{p}_{\rho}]$ obtained by rescaling back to
$\mathbf{p}_{\rho}$ the collection $\bar{\mathcal{U}}_{\tilde{\rho}}$ provided
by Theorem 7.1. Then (7.38) implies that for a $\gtrsim 1$ fraction of the
contributing $\mathbf{p}_{\rho}\subset\mathbf{p}_{\rho_{0}}$ we have

(8.65)
$(\delta/\rho)^{o_{\epsilon}(1)}V_{\mathbf{p}_{\rho_{0}}}\lesssim\frac{\sum_{p\in\mathcal{P}[\mathbf{p}_{\rho}]}|\mathcal{U}_{1,p}|}{|\bar{\mathcal{U}}_{1}[\mathbf{p}_{\rho}]|}.$

This is because the numerator on the right-hand side has the same value for
all contributing $\mathbf{p}_{\rho}$ (since $\mathcal{P}$ is uniform).

Call $\mathcal{P}_{\rho,gen}$ the collection of all $\mathbf{p}_{\rho}$ with
$\mathbf{p}_{\rho}\cap\mathcal{P}_{2}\not=\emptyset$, that satisfy (5.23)
(after rescaling) and (8.65). With another logarithmic loss, we may also
assume that

(8.66)
$\frac{\sum_{p\in\mathcal{P}[\mathbf{p}_{\rho}]}|\mathcal{U}_{1,p}|}{|\bar{\mathcal{U}}_{1}[\mathbf{p}_{\rho}]|}\sim
W$

for each $\mathbf{p}_{\rho}\in\mathcal{P}_{\rho,gen}$.

We have

(8.67)
$|\mathcal{P}_{2}|\lessapprox(\rho/\delta)^{h(\epsilon)^{3}}\sum_{\mathbf{p}_{\rho}\in\mathcal{P}_{\rho,gen}}|\mathbf{p}_{\rho}\cap\mathcal{P}_{2}|.$

Theorem 5.3 provides a dichotomy for each
$\mathbf{p}_{\rho}\in\mathcal{P}_{\rho,gen}$, and we split the analysis in two
subcases.

Case 2(a): Assume Item $(2)$ of Theorem 5.3 happens for at least half of
$\mathbf{p}_{\rho}$. This means that there is a scale

(8.68) $\delta(\frac{\rho}{\delta})^{\sqrt{h(\epsilon)}}\leq\Delta\leq\rho$

such that at least a $(\delta/\rho)^{O(h(\epsilon))}$-fraction
$\mathcal{P}_{\Delta,\mathbf{p}_{\rho}}$ of
$\mathcal{P}_{2}\cap\mathbf{p}_{\rho}$ is covered by squares $B_{\Delta}$
satisfying

(8.69) $|\mathcal{P}_{2}\cap
B_{\Delta}|\gtrsim(\frac{\Delta}{\delta})^{2-t_{1}+h(\epsilon)^{1/4}}.$

We have, using (8.68)

(8.70)
$|\mathcal{P}_{\Delta,\mathbf{p}_{\rho}}|\gtrsim(\delta/\Delta)^{\sqrt{h(\epsilon)}}|\mathcal{P}_{2}\cap\mathbf{p}_{\rho}|.$

We now focus our analysis inside a fixed $B_{\Delta}$, in preparation for the
application of Lemma 6.1. There are
$r_{3}\sim|\mathcal{T}_{p}^{\delta/\Delta}|$ distinct
$\delta\times\Delta$-segments that intersect each
$p\in\mathcal{P}^{\prime}\cap B_{\Delta}$. Recall that
$\mathcal{T}_{p}^{\delta/\rho_{0}}$ is a
$(\delta/\rho_{0},t_{1},(\rho_{0}/\delta)^{\varphi(\epsilon)})$-set. By Lemma
2.6
$r_{3}\gtrsim(\delta/\rho_{0})^{\varphi(\epsilon)}(\frac{\Delta}{\delta})^{t_{1}}.$
Using (8.64), (8.68), then (8.60) we find

$({\rho_{0}}/\delta)^{\varphi(\epsilon)}\leq(\Delta/\delta)^{\frac{\varphi(\epsilon)}{h(\epsilon)^{3/2}}}\leq(\Delta/\delta)^{\frac{1}{2}h(\epsilon)^{1/4}},$

and thus $r_{3}\gtrsim(\Delta/\delta)^{t_{1}-\frac{1}{2}h(\epsilon)^{1/4}}$.
When combined with (8.69) it shows that

$|\mathcal{P}_{2}\cap
B_{\Delta}|\gtrsim\frac{1}{r_{3}}(\frac{\Delta}{\delta})^{2+\frac{1}{2}h(\epsilon)^{1/4}}.$

We may apply Lemma 6.1 to the $1/\Delta$-rescaled copy of $B_{\Delta}$, with
$\beta=\frac{1}{2}h(\epsilon)^{1/4}$. We find a scale

(8.71) $\delta(\Delta/\delta)^{\beta/2}\lesssim\tilde{\delta}\leq\Delta$

such that for at least a half of $p\in\mathcal{P}_{2}\cap B_{\Delta}$, the
heavy ball $B(c_{p},\tilde{\delta})$ (we ignore the extra
$(\Delta/\delta)^{\upsilon}$ in the definition of the radius, as it is
negligible) intersects $\gtrsim r_{3}\tilde{\delta}/\delta$ of the
$\delta\times\Delta$-segments. Each of these segments intersects $\sim
r/r_{3}$ tubes $T\in\mathcal{T}$, since each $p$ intersects $\sim r$ tubes in
$T\in\mathcal{T}$, and the tubes corresponding to distinct segments are
themselves distinct. We thus have

$\gtrsim r/r_{3}\times r_{3}\tilde{\delta}/\delta=r\tilde{\delta}/\delta$

tubes $T\in\mathcal{T}$ intersecting $B(c_{p},\tilde{\delta})$.

We summarize (8.64), (8.68) and (8.71)

(8.72)
$\rho_{0}/\delta\lesssim(\rho/\delta)^{\frac{1}{h(\epsilon)}}\lesssim(\Delta/\delta)^{\frac{1}{h(\epsilon)^{3/2}}}\lesssim(\tilde{\delta}/\delta)^{\frac{4}{h(\epsilon)^{7/4}}}.$

Recall $\mathcal{T}$ is (approximate) $\epsilon$-uniform. Let
$N=|\mathcal{T}[\mathbf{T}_{\tilde{\delta}}]|$ be the number of tubes
$\mathcal{T}$ inside a nonempty
$\mathbf{T}_{\tilde{\delta}}\in\mathcal{T}^{\tilde{\delta}}$ (this number is
roughly the same, within $O_{\epsilon}(1)$-losses). Then the number
$\mathbf{r}$ of $\mathbf{T}_{\tilde{\delta}}\in\mathcal{T}^{\tilde{\delta}}$
that intersect such a heavy $B(c_{p},\tilde{\delta})$ is