10000\tilde{T}}\psi_{T}*\eta^{\vee}_{\mathrm{low}}}_{\infty}$
$\displaystyle\leq\sum_{\tilde{T}\in\tilde{\mathbb{T}}_{\mathrm{light}}}\sum_{T\in\mathbb{T}:T\subseteq
10000\tilde{T}}\norm*{\psi_{T}*\eta^{\vee}_{\mathrm{low}}}_{\infty}$
$\displaystyle\lesssim(\delta^{-1}K^{-1})(c^{\prime}K^{3-\sqrt{\varepsilon}})K^{-2}$
$\displaystyle=c^{\prime}\delta^{-1+\varepsilon}.$

By the rapid decay of $\psi_{T}*\eta^{\vee}_{\mathrm{low}}$ and Observations 3
and 4 above, we have

$\displaystyle\left|\sum_{\tilde{T}\in\tilde{\mathbb{T}}}\sum_{T\in\mathbb{T}:T\subseteq
10000\tilde{T}}\psi_{T}*\eta^{\vee}_{\mathrm{low}}(x)\right|$
$\displaystyle\sim\sum_{\tilde{T}\in\tilde{\mathbb{T}}}\sum_{T\in\mathbb{T}:T\subseteq
10000\tilde{T}}\left|\psi_{T}*\eta^{\vee}_{\mathrm{low}(x)}\right|$
$\displaystyle\sim\sum_{T\in\mathbb{T}}\left|\psi_{T}*\eta^{\vee}_{\mathrm{low}}(x)\right|$
$\displaystyle\sim\left|\sum_{T\in\mathbb{T}}\psi_{T}*\eta^{\vee}_{\mathrm{low}}(x)\right|$
$\displaystyle=\left|f_{\mathrm{low}}(x)\right|\gtrsim
c\delta^{\varepsilon-1},$

for all $x\in F$, using (6.25). Thus the triangle inequality gives

$\left|\sum_{\tilde{T}\in\tilde{\mathbb{T}}_{\mathrm{heavy}}}\sum_{T\in\mathbb{T}:T\subseteq
10000\tilde{T}}\psi_{T}*\eta^{\vee}_{\mathrm{low}}(x)\right|\gtrsim
c\delta^{\varepsilon-1},\quad\forall x\in F,$ (6.33)

if $c^{\prime}$ is chosen to be small enough. Again by rapid decay of
$\psi_{T}*\eta^{\vee}_{\mathrm{low}}$, we have

$\sum_{\tilde{T}\in\tilde{\mathbb{T}}_{\mathrm{heavy}}}\sum_{T\in\mathbb{T}:T\subseteq
10000\tilde{T}}1_{T}(x)\gtrsim cK^{2}\delta^{\varepsilon-1},\quad\forall x\in
F.$ (6.34)

Using Observation 5 above, this implies that

$\sum_{\tilde{T}\in\tilde{\mathbb{T}}_{\mathrm{heavy}}}1_{10000\tilde{T}}(x)\gtrsim
cK^{-1}\delta^{\varepsilon-1},\quad\forall x\in F.$ (6.35)

####  6.5.3 Analysis at new scale

Our goal in this step is to let $\tilde{\mathbb{T}}$ satisfy the hypothesis of
Theorem 6.1 at a larger scale $10000K\delta=\delta^{1-\sqrt{\varepsilon}}$.

Let $\Theta^{\prime}$ denote the lattice $K\delta\mathbb{Z}\cap[0,1]$. Recall
that by our choice, $\tilde{\mathbb{T}}_{\mathrm{heavy}}$ can be written as
$\cup_{\theta^{\prime}\in\Theta^{\prime}}\tilde{\mathbb{T}}_{\theta^{\prime}}$,
where each $\tilde{\mathbb{T}}_{\theta^{\prime}}$ is a collection of disjoint
$K\delta\times K\delta\times 1$ tubes pointing in the direction
$\gamma(\theta^{\prime})$. By a trivial tiling argument and losing an absolute
constant, we can further assume
$\Theta^{\prime}=10^{4}K\delta\mathbb{Z}\cap[0,1]$.

Write

$\delta^{-1+\varepsilon}(10^{4}K)^{-1}=(10^{4}K\delta)^{-1+\tilde{\varepsilon}},\quad\tilde{\varepsilon}:=\frac{\varepsilon}{1-\sqrt{\varepsilon}}.$
(6.36)

Then we have, for some absolute constant $c_{0}$,

$\sum_{\tilde{T}\in\tilde{\mathbb{T}}_{\mathrm{heavy}}}1_{10^{4}\tilde{T}}(x)\geq
c_{0}c(10^{4}K\delta)^{-1+\tilde{\varepsilon}},\quad\forall
x\in\mathrm{supp}(\mu).$ (6.37)

Now we can apply the definition of
$M(10^{4}K\delta,\tilde{\varepsilon},c_{0}c)$ in (6.3) to get that

$\\#\tilde{\mathbb{T}}_{\mathrm{heavy}}\geq
M(10^{4}K\delta,\tilde{\varepsilon},c_{0}c)^{-1}\mu(\mathbb{R}^{3})\rho_{a}(\mu)^{-1}(10^{4}K\delta)^{-1-a}.$
(6.38)

By the definition of $\tilde{\mathbb{T}}_{\mathrm{heavy}}$, we have

$\displaystyle\\#\mathbb{T}$ $\displaystyle\geq
c(10^{4}K)^{3-\sqrt{\varepsilon}}M(10^{4}K\delta,\tilde{\varepsilon},c_{0}c)^{-1}\mu(\mathbb{R}^{3})\rho_{a}(\mu)^{-1}(10^{4}K\delta)^{-1-a}$
$\displaystyle\geq
c(10^{4}K)^{2-a-\sqrt{\varepsilon}}M(10^{4}K\delta,\tilde{\varepsilon},c_{0}c)^{-1}\mu(\mathbb{R}^{3})\rho_{a}(\mu)^{-1}\delta^{-1-a}.$

Thus in the low case, we have

$\displaystyle M(\delta,\varepsilon,c)$ $\displaystyle\leq
M(10^{4}K\delta,\tilde{\varepsilon},c_{0}c)c^{-1}(10^{4}K)^{a-2+\sqrt{\varepsilon}}$
$\displaystyle\leq
M(10^{4}K\delta,\tilde{\varepsilon},c_{0}c)c^{-1}\delta^{-\varepsilon}$
$\displaystyle\leq
M(\delta^{1-\sqrt{\varepsilon}},\tilde{\varepsilon},c_{0}c)\delta^{-2\varepsilon},$

where we have used $a-2\leq 0$ and used $\delta<\delta_{0}$ taking
$\delta_{0}$ to be small enough. Thus the result follows.

Combining the low case and the high and mixed case, we are done.

###  6.6 Corollary 2

We now restate and prove Corollary 2 in [GGG+22].

######  Corollary 6.5.

Let $a\in(0,3]$ and $a_{1}\in[0,\min\\{2,a\\})$. Let $\mu$ be a finite non-
zero Borel measure supported on the unit ball in $\mathbb{R}^{3}$ with
$\rho_{a}(\mu)\leq 1$. Let $\delta\in(0,1)$ and denote by $\Theta$ the
$\delta$-lattice in $[0,1]$. For each $\theta\in\Theta$, let
$\mathbb{S}_{\theta}$ be a disjoint collection of at most
$c\mu(\mathbb{R}^{3})\delta^{-a_{1}}$ squares of side length $\delta$ in
$\pi_{\theta}(\mathbb{R}^{3})$. Then there exists $\varepsilon\in(0,1)$,
depending only on $\gamma,c,a,a_{1}$, such that

$\delta\sum_{\theta\in\Theta}\pi_{\theta\\#}\mu\left(\bigcup\mathbb{S}_{\theta}\right)\leq
C(a,a_{1})\mu(\mathbb{R}^{3})\delta^{\varepsilon}.$ (6.39)

###### Proof.

We argue by contradiction. Suppose that there exist $a,a_{1}$ such that for
every $\varepsilon\in(0,1)$ and every large constant $C>10$, there exist
$\delta>0$ and a disjoint collection $\mathbb{S}_{\theta}$ of at most
$\mu(\mathbb{R}^{3})\delta^{-a_{1}}$ squares of radius $\delta$ in
$\pi_{\theta}(\mathbb{R}^{3})$ for each $\theta\in\Theta$, such that

$\delta\sum_{\theta\in\Theta}\pi_{\theta\\#}\mu\left(\bigcup\mathbb{S}_{\theta}\right)>C\mu(\mathbb{R}^{3})\delta^{\varepsilon}.$
(6.40)

Note that for each $S_{\theta}\in\mathbb{S}_{\theta}$,
$\pi_{\theta}^{-1}(S_{\theta})\cap B^{3}(0,1)$ is a $\delta$-tube. We denote
these tubes by

$\mathbb{T}_{\theta}:=\\{\pi_{\theta}^{-1}(S_{\theta})\cap
B^{3}(0,1):S_{\theta}\in\mathbb{S}_{\theta}\\},$ (6.41)

and write $\mathbb{T}:=\cup_{\theta\in\Theta}\mathbb{T}_{\theta}$. Thus (6.40)
becomes

$\delta\int\sum_{T\in\mathbb{T}}1_{T}(x)d\mu(x)=\delta\sum_{T\in\mathbb{T}}\mu(T)=\delta\sum_{\theta\in\Theta}\mu\left(\bigcup_{T_{\theta}\in\mathbb{T}_{\theta}}T_{\theta}\right)>C\mu(\mathbb{R}^{3})\delta^{\varepsilon}.$
(6.42)

We perform a pigeonholing argument here. Denote
$S(x):=\delta\sum_{T\in\mathbb{T}}1_{T}(x)$, which is bounded above by
$\delta\\#\Theta=1$. Define

$F:=\left\\{x\in\mathrm{supp}(\mu):\delta\sum_{T\in\mathbb{T}}1_{T}(x)\geq\frac{C\delta^{\varepsilon}}{2}\right\\}.$
(6.43)

We thus have

$\displaystyle\int S(x)d\mu(x)$
$\displaystyle=\int_{F}S(x)d\mu(x)+\int_{F^{c}}S(x)d\mu(x)$
$\displaystyle\leq\int_{F}1d\mu(x)+\int_{F^{c}}\frac{C\delta^{\varepsilon}}{2}d\mu(x)$
$\displaystyle\leq\mu(F)+\frac{C\delta^{\varepsilon}}{2}\mu(F^{c}).$

Using (6.42), this implies that

$\mu(F)\geq\frac{C\delta^{\varepsilon}}{2}\mu(\mathbb{R}^{3}).$ (6.44)

Now we define a new measure $\nu$ by

$\nu(A):=\mu(A\cap F),$ (6.45)

supported on $F\subseteq B^{3}(0,1)$, with
$\rho_{\min\\{2,a\\}}(\nu)\leq\rho_{a}(\nu)\leq\rho_{a}(\mu)\leq 1$. Also, the
definition of $F$ means that

$\sum_{T\in\mathbb{T}}1_{T}(x)\geq\frac{C}{2}\delta^{\varepsilon-1},\quad\forall
x\in\mathrm{supp}(\nu).$ (6.46)

Now we may apply Theorem 6.1 with $\nu$ in place of $\mu$ and $\min\\{2,a\\}$
in place of $a$ to obtain

$\\#\mathbb{T}\gtrsim_{\varepsilon,a}\nu(\mathbb{R}^{3})\delta^{-1-\min\\{2,a\\}+100\sqrt{\varepsilon}}\geq\frac{C\delta^{\varepsilon}}{2}\mu(\mathbb{R}^{3})\delta^{-1-\min\\{2,a\\}+100\sqrt{\varepsilon}}.$
(6.47)

However, since each $\\#\mathbb{S}_{\theta}\leq
c\mu(\mathbb{R}^{3})\delta^{-a_{1}}$ for each $\theta\in\Theta$, we have on
the other hand

$\\#\mathbb{T}\leq c\mu(\mathbb{R}^{3})\delta^{-1-a_{1}}.$ (6.48)

Compared with (6.47), we have there is a constant $C_{\varepsilon,c,a}$ such
that

$C\leq
C_{\varepsilon,c,a}\delta^{\min\\{2,a\\}-a_{1}-\varepsilon-100\sqrt{\varepsilon}}.$
(6.49)

Choosing $\varepsilon>0$ small enough such that
$\min\\{2,a\\}-a_{1}-\varepsilon-100\sqrt{\varepsilon}>0$ and then
$C>C_{\varepsilon,c,a}$, we arrive at a contradiction. ∎

###  6.7 Corollary 3

We now restate and prove Corollary 3 in [GGG+22].

######  Corollary 6.6.

Let $a\in[2,3]$ and $a_{1}\in[0,2)$. Let $\mu$ be a finite non-zero Borel
measure supported on the unit ball in $\mathbb{R}^{3}$ with
$\rho_{a}(\mu)<\infty$. Let $\delta\in(0,1)$ and denote by $\Delta$ the
$\delta^{1/2}$-lattice in $[0,1]$. For each $\theta\in\Delta$, let
$\mathbb{P}_{\theta}$ be a disjoint collection of at most
$\mu(\mathbb{R}^{3})\delta^{-\frac{a_{1}+1}{2}}$ rectangles of dimensions
$\delta\times\delta^{1/2}$ in $\pi_{\theta}(\mathbb{R}^{3})$ whose longer
sides point in the $\gamma^{\prime}(\theta)=\mathbf{e}_{2}(\theta)$ direction.
Then there exists $\varepsilon\in(0,1)$, depending only on $\gamma,a,a_{1}$,
such that

$\delta^{1/2}\sum_{\theta\in\Delta}\pi_{\theta\\#}\mu\left(\bigcup\mathbb{P}_{\theta}\right)\leq
C(a,a_{1})\rho_{a}(\mu)\mu(\mathbb{R}^{3})\delta^{\varepsilon}.$ (6.50)

###### Proof.

We may assume $\rho_{a}(\mu)=1$ without loss of generality. To apply Corollary
6.5, we need to construct for each
$\theta^{\prime}\in\Theta=\delta\mathbb{Z}\cap[0,1]$ a collection of squares
$\mathbb{S}_{\theta^{\prime}}$ of radius $\delta$ in
$\pi_{\theta^{\prime}}(\mathbb{R}^{3})$. We proceed as follows. For each
$\theta^{\prime}\in\Theta$, we choose $\theta\in\Delta$ to be the closest to
$\theta^{\prime}$, so that $|\theta^{\prime}-\theta|<\delta^{1/2}$. We denote
this relation by $\theta^{\prime}\prec\theta$. We then consider the set of
planks

$\mathbb{U}_{\theta}:=\\{U_{\theta}=\pi_{\theta}^{-1}(P_{\theta})\cap
B^{3}(0,1):P_{\theta}\in\mathbb{P}_{\theta}\\}.$ (6.51)

We partition each $100U_{\theta}$ into $(\delta/2)\times(\delta/2)\times 1$
tubes $T_{\theta}$ with direction $\gamma(\theta)$ and denote the collection
of all these tubes by $\mathbb{T}_{\theta}$. Then we define

$\mathbb{S}_{\theta^{\prime}}:=\\{\pi_{\theta^{\prime}}(T_{\theta}):T_{\theta}\in\mathbb{T}_{\theta}\\}.$
(6.52)

Strictly speaking, each element of $\mathbb{S}_{\theta^{\prime}}$ is a
possibly slightly distorted parallelogram, but we can slightly enlarge it to
become a square of side length $\delta$. The slightly enlarged collection of
squares will be at most $100$-overlapping since
$|\gamma(\theta^{\prime})-\gamma(\theta)|\leq|\theta^{\prime}-\theta|<\delta^{1/2}$;
by abuse of notation we will still denote it by
$\mathbb{S}_{\theta^{\prime}}$. We also have

$\\#\mathbb{S}_{\theta^{\prime}}=\\#\mathbb{T}_{\theta}\sim\delta^{-1/2}\\#\mathbb{P}_{\theta}\lesssim\mu(\mathbb{R}^{3})\delta^{-\frac{a_{1}+2}{2}}.$
(6.53)

Since $\frac{a_{1}+2}{2}<2=\min\\{2,a\\}$, we may apply Corollary 6.5 to
obtain

$\delta\sum_{\theta^{\prime}\in\Theta}\pi_{\theta^{\prime}\\#}\mu\left(\bigcup\mathbb{S}_{\theta^{\prime}}\right)\lesssim\mu(\mathbb{R}^{3})\delta^{\varepsilon},$
(6.54)

and so

$\displaystyle\mu(\mathbb{R}^{3})\delta^{\varepsilon}$
$\displaystyle\gtrsim\delta\sum_{\theta\in\Delta}\sum_{\theta^{\prime}\prec\theta}\mu\left(\pi_{\theta^{\prime}}^{-1}(\bigcup_{T_{\theta}\in\mathbb{T}_{\theta}}\pi_{\theta^{\prime}}(T_{\theta}))\right)$
$\displaystyle=\delta\sum_{\theta\in\Delta}\sum_{\theta^{\prime}\prec\theta}\mu\left(\pi_{\theta^{\prime}}^{-1}(\pi_{\theta^{\prime}}(\cup_{U_{\theta}\in\mathbb{U}_{\theta}}100U_{\theta}))\right)$
$\displaystyle\geq\delta\sum_{\theta\in\Delta}\sum_{\theta^{\prime}\prec\theta}\mu\left(\pi_{\theta}^{-1}(\pi_{\theta}(\cup_{U_{\theta}\in\mathbb{U}_{\theta}}U_{\theta}))\right)$
$\displaystyle\sim\delta^{1/2}\sum_{\theta\in\Delta}\mu\left(\pi_{\theta}^{-1}(\cup_{P_{\theta}\in\mathbb{P}_{\theta}}P_{\theta})\right)$
$\displaystyle=\delta^{1/2}\sum_{\theta\in\Delta}\pi_{\theta\\#}\mu\left(\bigcup\mathbb{P}_{\theta}\right),$

where the third line follows from a simple geometric observation. ∎

##  7 Proof of Theorem 8

We now prove Theorem 1.5.

To avoid technical issues and to make the presentation slightly cleaner, we
only restrict ourselves to the case of a standard light cone. More precisely,
we consider $I_{0}:=[0,\sqrt{2}\pi]$ and
$\mathbf{e}_{1}=\gamma:I_{0}\to\mathbb{S}^{2}$ given by

$\gamma(\theta)=\frac{1}{\sqrt{2}}\left(\cos\sqrt{2}\theta,\sin\sqrt{2}\theta,1\right).$
(7.1)

In this case, it is easy to see that

$\begin{split}\mathbf{e}_{2}(\theta)&=\left(-\sin\sqrt{2}\theta,\cos\sqrt{2}\theta,0\right),\\\
\mathbf{e}_{3}(\theta)&=\frac{1}{\sqrt{2}}\left(-\cos\sqrt{2}\theta,-\sin\sqrt{2}\theta,1\right),\\\
\uptau(\theta)&=1.\end{split}$ (7.2)

The main idea is to decompose the frequency space into “good” and “bad” parts
according to the cone
$\Gamma=\\{r\mathbf{e}_{3}(\theta):r\in\mathbb{R},\theta\in I_{0}\\}$, and
estimate them separately.

We record some important geometric facts about the light cone.

######  Proposition 7.1.

Let $\xi=(x_{1},x_{2},x_{3})\in\mathbb{R}^{3}$. Then there exists a
$\xi^{\prime}:=r\mathbf{e}_{3}(\omega)\in\Gamma$ such that
$|\xi-\xi^{\prime}|=\mathrm{dist}(\xi,\Gamma)$, and also $\xi-\xi^{\prime}$ is
parallel to $\mathbf{e}_{1}(\omega)$. As a result,
$\mathrm{dist}(\xi,\Gamma)=|\xi\cdot\mathbf{e}_{1}(\omega)|$.

Meanwhile, we also have
$\mathrm{dist}(\xi,\Gamma)=2^{-1/2}|x_{3}-\sqrt{x_{1}^{2}+x_{2}^{2}}|$.

###### Proof.

This is easily seen using the cylindrical coordinates in $\mathbb{R}^{3}$. ∎

Maybe add a picture later

###  7.1 Construction of covering planks

Our goal is to construct planks in three steps. We give the main idea first.
Given $j\geq k\geq 0$, Step 1 is to construct a canonical family
$\Lambda_{j,k}$ of planks $P_{\theta,j,k}$ that cover the set

$\\{\xi\in\mathbb{R}^{3}:|\xi|\sim 2^{j},\mathrm{dist}(\xi,\Gamma)\sim
2^{j-k}\\},$ (7.3)

except for $k=j$, in which case we want to cover

$\\{\xi\in\mathbb{R}^{3}:|\xi|\sim 2^{j},\mathrm{dist}(\xi,\Gamma)\lesssim
1\\}.$

Each plank $P_{\theta,j,k}$ essentially has normal, aperture and radial
lengths $2^{j-k}$, $2^{j-k/2}$, $2^{j}$, respectively.

In Step 2, we define $\Lambda_{j}=\cup_{k\leq j}\Lambda_{j,k}$, so that
$\Lambda_{j}$ forms a canonical family of planks that cover the $2^{j}$
neighbourhood of $\\{\xi\in\mathbb{R}^{3}:|\xi|\sim 2^{j}\\}$.

In Step 3, we define $\Lambda=\cup_{j\in\mathbb{N}}\Lambda_{j}$, so that
$\Lambda$ forms a covering of the entire $\mathbb{R}^{3}$.

We will then define wave packets adapted to the planks above, and perform
subsequent Fourier analysis.

Now we come to the details and define the planks $P_{\theta,j,k}$. Fix
$\theta\in I_{0}$. For $0\leq k<j$, we define

$\displaystyle P_{\theta,j,k}$
$\displaystyle:=\Bigg{\\{}\sum_{i=1}^{3}\xi_{i}\mathbf{e}_{i}(\theta):2^{j-k-10}\leq|\xi_{1}|\leq
2^{j-k+10},$ $\displaystyle|\xi_{2}|\leq
2^{j-\frac{k}{2}-100},2^{j-10}\leq|\xi_{3}|\leq 2^{j+10}\Bigg{\\}}.$ (7.4)

For $0\leq k=j$, we define

$\displaystyle P_{\theta,j,j}$
$\displaystyle=\Bigg{\\{}\sum_{i=1}^{3}\xi_{i}\mathbf{e}_{i}(\theta):|\xi_{1}|\leq
2^{10},|\xi_{2}|\leq 2^{\frac{j}{2}-100},2^{j-10}\leq|\xi_{3}|\leq
2^{j+10}\Bigg{\\}}.$ (7.5)

Strictly speaking, each $P_{\theta,j,k}$ for $k<j$ is a union of four planks.
However, by symmetry we may assume $\xi_{1}>0$ and $\xi_{3}>0$, and abusing
notation we still denote this by $P_{\theta,j,k}$. Similar for
$P_{\theta,j,j}$.

Heuristically, $\xi_{1}\sim 2^{j-k}$ (or $|\xi_{1}|\lesssim 1$ when $j=k$) is
the distance between the plank $P_{\theta,j,k}$ and $\Gamma$,
$|\xi_{2}|\lesssim 2^{j-k/2}$ is the angular aperture, and $\xi_{3}\sim 2^{j}$
is the distance from the origin. The precise statement is as follows.

######  Lemma 7.2.

Fix $\theta\in I_{0}$. Let $0\leq k\leq j$ and $\xi\in P_{\theta,j,k}$. Then
the distance between $\xi$ and the cone $\Gamma$ is at most $2^{j-k+10}$.
Furthermore, if $j>k$, then the distance between $\xi$ and the cone $\Gamma$
is at least $2^{j-k-40}$.

###### Proof.

Fix $\xi\in P_{\theta,j,k}$ and write
$\xi=\sum_{i=1}^{3}\xi_{i}\mathbf{e}_{i}(\theta)=(x_{1},x_{2},x_{3})$ where
$x_{i}$ can be found using (7.1) and (7.2). Then Proposition 7.1 allow us to
compute directly

$\displaystyle\text{dist}(\xi,\Gamma)=$
$\displaystyle\frac{1}{2}\bigg{|}((\xi_{1}+\xi_{3})-\bigg{(}(\xi_{1}\cos\sqrt{2}\theta-\sqrt{2}\xi_{2}\sin\sqrt{2}\theta-\xi_{3}\cos\sqrt{2}\theta)^{2}$
$\displaystyle+((\xi_{1}\sin\sqrt{2}\theta+\sqrt{2}\xi_{2}\cos\sqrt{2}\theta-\xi_{3}\sin\sqrt{2}\theta)^{2}\bigg{)}^{1/2}\bigg{|}$
$\displaystyle=\frac{1}{2}\left|\xi_{1}+\xi_{3}-\sqrt{2\xi_{2}^{2}+(\xi_{3}-\xi_{1})^{2}}\right|$
$\displaystyle=|2\xi_{1}\xi_{3}-\xi_{2}^{2}|\left|\xi_{1}+\xi_{3}+\sqrt{2\xi_{2}^{2}+(\xi_{3}-\xi_{1})^{2}}\right|^{-1}.$

In particular, the term inside the absolute value is positive for $\xi\in
P_{\theta,j,k}$, and is a increasing function in $|\xi_{2}|$. For $j>k$, using
the upper and lower bounds of $|\xi_{i}|$, we thus have

$\mathrm{dist}(\xi,\Gamma)\in[2^{j-k-40},2^{j-k+10}],$ (7.6)

where the upper bound was obtained by considering $\xi_{2}=0$. The case $j=k$
is similar.

∎

Now we are ready to define the covering planks. Let $c=100^{-100}$. For each
$k\geq 0$, define the lattice

$\Theta_{k}:=c2^{-k/2}\mathbb{N}\cap I_{0}.$ (7.7)

Define

$\Lambda_{j,k}:=\\{P_{\theta,j,k}:\theta\in\Theta_{k}\\},$ (7.8)

and then

$\Lambda_{j}:=\bigcup_{k=0}^{j}\Lambda_{j,k},\quad\Lambda:=\bigcup_{j=0}^{\infty}\Lambda_{j}.$
(7.9)

Note that we always have

$\mathrm{dist}(\xi,\Gamma)\leq 2^{-1/2}|\xi|.$ (7.10)

Indeed, write $\xi=(x_{1},x_{2},x_{3})$. By radial symmetry we may assume
$x_{2}=0$, and by dilation symmetry we can assume $x_{1}^{2}+x_{3}^{2}=1$.
Then the distance is maximized exactly when $x_{1}=0$ or $x_{3}=1$, and the
distance is $2^{-1/2}$.

For $C\geq 1$, we also define for $j>k\geq 0$

$\displaystyle C\Gamma_{j,k}$
$\displaystyle:=\\{\xi\in\mathbb{R}^{3}:C^{-1}2^{j-1}\leq|\xi|\leq C2^{j},$
$\displaystyle\mathrm{dist}(\xi,\Gamma)\in[C^{-1}2^{-k-3/2}|\xi|,C2^{-k-1/2}|\xi|]\\},$
(7.11)

and for $j=k$

$C\Gamma_{j,j}:=\\{\xi\in\mathbb{R}^{3}:C^{-1}2^{j-1}\leq|\xi|\leq
C2^{j},\mathrm{dist}(\xi,\Gamma)\in[0,C2^{-j-1/2}|\xi|]\\}.$ (7.12)

When $C=1$, we simply denote $\Gamma_{j,k}$. We also denote

$C\Gamma_{j}=\\{\xi\in\mathbb{R}^{3}\colon C^{-1}2^{j-1}\leq|\xi|\leq
C2^{j}\\}.$

######  Lemma 7.3.

For all $j\geq k\geq 0$, we have

$\Gamma_{j,k}\subseteq\bigcup_{\theta\in\Theta_{k}}P_{\theta,j,k}\subseteq
A\Gamma_{j,k},$ (7.13)

for some absolute constant $A$. Moreover, the family $\Lambda_{j,k}$ has
$O(1)$-overlap.

###### Proof.

The second inclusion follows directly from Lemma 7.2 if $A$ is large enough.
The $O(1)$-overlap follows from essentially the same proof of that of Part 2
of Lemma 5.5 in Section 5.11.

To prove the first inclusion, let $\xi\in\Gamma_{j,k}$. Take
$\xi^{\prime}:=r\mathrm{e}_{3}(\omega)\in\Gamma$ that attains
$\mathrm{dist}(\xi,\Gamma)$. Take $\theta\in\Theta_{k}$ be closest to
$\omega$, so $|\theta-\omega|\leq c2^{-k/2}$. It suffices to prove that
$\xi\in P_{\theta,j,k}$.

We first consider the normal direction. For $j>k$, by Proposition 7.1, we have

$|\xi\cdot\mathbf{e}_{1}(\omega)|=|(\xi-\xi^{\prime})\cdot\mathbf{e}_{1}(\omega)|=\mathrm{dist}(\xi,\Gamma)\geq
2^{-k-3/2}|\xi|\geq 2^{j-k-5/2}.$ (7.14)

Also,

$|\xi\cdot\mathbf{e}_{2}(\omega)|=|(\xi-\xi^{\prime})\cdot\mathbf{e}_{2}(\omega)|\leq|\xi-\xi^{\prime}|\leq
2^{-k-1/2}|\xi|.$ (7.15)

Thus by Taylor expansion,

$\displaystyle|\xi\cdot\mathbf{e}_{1}(\theta)|$
$\displaystyle\geq|\xi\cdot\mathbf{e}_{1}(\omega)|-|\xi\cdot\mathbf{e}_{2}(\omega)||\omega-\theta|-2|\xi||\omega-\theta|^{2}$
$\displaystyle\geq 2^{j-k-5/2}-c2^{j-3k/2-1/2}-2c^{2}2^{j-k}$
$\displaystyle\geq 2^{j-k-10}.$

Similarly, for $j\geq k$, we also get an upper bound
$|\xi\cdot\mathbf{e}_{1}(\theta)|\leq 2^{j-k+10}$.

Now we come to the tangential direction. By Proposition 7.1, we have

$\xi\cdot\mathbf{e}_{2}(\omega)=(\xi-\xi^{\prime})\cdot\mathbf{e}_{2}(\omega)+\xi^{\prime}\cdot\mathbf{e}_{2}(\omega)=0.$
(7.16)

By Taylor expansion,

$|\xi\cdot\mathbf{e}_{2}(\theta)|\leq 2|\xi||\theta-\omega|\leq
2c2^{j-k/2}\leq 2^{-100}2^{j-k/2}.$ (7.17)

Lastly, we consider the radial direction. By Proposition 7.1, we have

$|\xi\cdot\mathbf{e}_{3}(\omega)|=|\xi^{\prime}\cdot\mathbf{e}_{3}(\omega)|=|\xi^{\prime}|.$
(7.18)

Also, recalling that $\mathrm{dist}(\xi,\Gamma)\leq 2^{-1/2}|\xi|$, we have

$|\xi^{\prime}|\geq|\xi|-|\xi^{\prime}-\xi|\geq(1-2^{-1/2})|\xi|\geq 2^{j-5}.$
(7.19)

Thus by Taylor expansion,

$|\xi\cdot\mathbf{e}_{3}(\theta)|\geq|\xi\cdot\mathbf{e}_{3}(\omega)|-|\xi||\theta-\omega|\geq
2^{j-5}-c2^{j-k/2}\geq 2^{j-10}.$ (7.20)

Similarly, we can get an upper bound $|\xi\cdot\mathbf{e}_{3}(\theta)|\leq
2^{j+10}$. ∎

######  Lemma 7.4.

We have

$\Gamma_{j}\subseteq\bigcup\Lambda_{j}\subseteq A\Gamma_{j}.$ (7.21)

Moreover, the family $\Lambda_{j}$ has $O(1)$-overlap.

###### Proof.

The inclusion follows by Lemma 7.3 and taking union $\cup_{k=0}^{j}$. The
bounded overlap of $\Lambda_{j}$ follows readily from the bounded overlap of
each $\Lambda_{j,k}$ established in Lemma 7.3 and the bounded overlap of
$\\{A\Gamma_{j,k}:0\leq k\leq j\\}$. ∎

######  Lemma 7.5.

The family of planks $\Lambda$ covers $\mathbb{R}^{3}\backslash B^{3}(0,1/2)$
and has $O(1)$-overlap.

###### Proof.

The bounded overlap of $\Lambda$ follows from the bounded overlap of each
$A\Lambda_{j}$ and the bounded overlap of $\\{\Gamma_{j}:j\geq 0\\}$.

To prove the covering, by Lemma 7.4, it suffices to prove that

$\bigcup_{j=0}^{\infty}\bigcup_{k=0}^{j}\Gamma_{j,k}\supseteq\mathbb{R}^{3}\backslash
B^{3}(0,1/2).$ (7.22)

To see this, given any $\xi\in\mathbb{R}^{3}\backslash B^{3}(0,1/2)$. Take
$j\geq 0$ such that $|\xi|\in[2^{j-1},2^{j}]$. Also, since
$\mathrm{dist}(\xi,\Gamma)\leq 2^{-1/2}|\xi|$, take $k^{\prime}\geq 0$ such
that

$\mathrm{dist}(\xi,\Gamma)\in[2^{-k^{\prime}-3/2}|\xi|,2^{-k^{\prime}-1/2}|\xi|].$
(7.23)

If $j\geq k^{\prime}$, then we just take $k=k^{\prime}$. If $j<k^{\prime}$,
then we take $k=j$. ∎

###  7.2 Construction of wave packets

Consider the collection $\Lambda$ of (infinitely many) planks we constructed
in (7.9).

We first define a smooth partition of unity adapted to the family $\Lambda$:
for each $P\in\Lambda$, let $\psi_{P}$ be a bump function that $\sim 1$ on $P$
and is supported in $2P$. Also, let $\psi_{0}$ be a bump function adapted to
the ball $B^{3}(0,1)$. Moreover, they can be chosen in a way that

$\psi_{0}+\sum_{P\in\Lambda}\psi_{P}=1.$ (7.24)

For technical issues, we need to introduce a tiny exponent $\beta>0$ to be
determined.

For each $P\in\Lambda$, let $\mathbb{T}_{P}$ be a tiling of $\mathbb{R}^{3}$
by planks $T$ nearly dual to $P=P_{\theta,j,k}$, that is, having dimensions

$2^{k-j+k\beta},\quad 2^{k/2-j+k\beta},\quad 2^{-j+k\beta}$ (7.25)

in the normal, tangential and radial directions, respectively. Strictly
speaking, they are not duals due to the tiny factor $\beta$. This is intended
so that we can use repeated integration by parts argument to prove Lemma 7.7
below.

For each $T\in\mathbb{T}_{P}$, we choose a Schwartz function $\eta_{T}$ which
$\sim 1$ on $T$ and is Fourier supported on the plank centred at $0$, with
sides parallel to $P$ but with dimensions $2^{j-k-k\beta}$, $2^{j-k/2-k\beta}$
and $2^{j-k\beta}$, respectively, such that we also have

$\sum_{T\in\mathbb{T}_{P}}\eta_{T}=1.$ (7.26)

Now for each $P\in\Lambda$ and each $T\in\mathbb{T}_{P}$, we define the wave
packet

$M_{T}\mu(x):=\eta_{T}(x)(\mu*\psi_{P}^{\vee})(x)=\eta_{T}(x)\int\psi_{P}^{\vee}(x-y)d\mu(y).$
(7.27)

We record some estimates on the wave packets.

######  Lemma 7.6.

For $P\in\Lambda_{j,k}$, $T\in\mathbb{T}_{P}$ and any $\theta\in I_{0}$, we
have

$\begin{split}\norm*{\pi_{\theta\\#}M_{T}\mu}_{L^{1}(\mathcal{H}^{2})}&\lesssim_{N,\beta}\int
w_{T,N}(y)d\mu(y)\\\
&\lesssim_{N,\beta}2^{-kN}\mu(\mathbb{R}^{3})+\mu(T),\quad\forall N\geq
1,\end{split}$ (7.28)

where $w_{T,N}(x)$ is any weight function adapted to $T$ (refer to (3.17)).

###### Proof.

Given a continuous function $f:\mathbb{R}^{3}\to\mathbb{C}$ of compact support
and $\theta\in I_{0}$, we may compute, using the definition of push-forward
measure via Riesz representation theorem

$\pi_{\theta\\#}f(x)=\int_{\mathbb{R}}f(x+t\gamma(\theta))dt,\quad\forall
x\in\pi_{\theta\\#}(\mathbb{R}^{3}).$ (7.29)

Thus

$\displaystyle\pi_{\theta\\#}M_{T}\mu(x)$
$\displaystyle=\int_{\mathbb{R}}M_{T}\mu(x+t\gamma(\theta))dt$
$\displaystyle=\int_{\mathbb{R}}\int\eta_{T}(x+t\gamma(\theta))\psi_{P}^{\vee}(x+t\gamma(\theta)-y)d\mu(y)dt.$

By Fubini’s theorem and writing $z=x+t\gamma(\theta)$, we have

$\int|\pi_{\theta\\#}M_{T}\mu(x)|d\mathcal{H}^{2}(x)=\int\int_{\mathbb{R}^{3}}\eta_{T}(z)\psi_{P}^{\vee}(z-y)dzd\mu(y).$
(7.30)

Since $\psi_{P}^{\vee}$ is an $L^{1}$-normalized Schwartz function essentially
supported on a slightly smaller rectangle than $T$ (due to the $\beta$
factor), an application of Proposition 3.6 (generalized to $\mathbb{R}^{3}$ in
an apparent way) gives $|\psi_{P}^{\vee}*\eta_{T}(y)|\lesssim_{N}w_{T,N}(y)$
for any choice of $N$. Thus the result follows.

∎

The next is an almost orthogonality lemma.

######  Lemma 7.7.

For each $P\in\Lambda_{j,k}$, denote by $\theta_{P}$ the angle in $I_{0}$ such
that $P=P_{\theta_{P},j,k}$. Then for any $\theta\in I_{0}$ such that
$|\theta-\theta_{P}|\geq C2^{-k/2}$ (here $C$ is a large enough absolute
constant) and any $x\in\pi_{\theta}(B^{3}(0,1))$, we have

$\sum_{T\in\mathbb{T}_{P}}|\pi_{\theta\\#}M_{T}\mu(x)|\lesssim_{N}2^{-kN}|P|\mu(\mathbb{R}^{3}),\quad\forall
N\geq 1.$ (7.31)

###### Proof.

It suffices to prove that for any rectangle $T\in\mathbb{T}_{P}$ centred at
$c_{T}$ and any $x\in\pi_{\theta}(B^{3}(0,1))$, we have

$|\pi_{\theta\\#}M_{T}\mu(x)|\lesssim_{\beta,N}2^{-kN}(1+|x-c_{T}|2^{j-k-k\beta})^{-N}|P|\mu(\mathbb{R}^{3}),\quad\forall
N\geq 1.$ (7.32)

Using the inverse Fourier transform formula, we compute directly that

$\displaystyle\pi_{\theta\\#}M_{T}\mu(x)$
$\displaystyle=\int\int\eta_{T}(x+t\gamma(\theta))\widehat{\mu}(\xi)\psi_{P}(\xi)e^{i(x+t\gamma(\theta))\cdot\xi}dtd\xi$
$\displaystyle\leq\int|\widehat{\mu}(\xi)\psi_{P}(\xi)|\left|\int\eta_{T}(x+t\gamma(\theta)e^{it\gamma(\theta))\cdot\xi}dt\right|d\xi.$

Using $\|\hat{\mu}\|_{\infty}\leq\mu(\mathbb{R}^{3})$, it suffices to prove
that

$\left|\int\eta_{T}(x+t\gamma(\theta))e^{it\gamma(\theta)\cdot\xi}dt\right|\lesssim_{\beta,N}2^{-kN}(1+|x-c_{T}|2^{j-k-k\beta})^{-N},\quad\forall
N\geq 1.$ (7.33)

To prove this, we first prove the following claim.

######  Claim 7.8.

Let $\theta\in I_{0}$ and $|\theta-\theta_{P}|\geq C2^{-k/2}$. If $C$ is large
enough, then we have

$|\xi\cdot\gamma(\theta)|\gtrsim(\theta-\theta_{P})^{2}2^{j},\quad\forall\xi\in
P.$ (7.34)

###### Proof of claim.

Write $\xi=\sum_{i=1}^{3}\xi_{i}\mathbf{e}_{i}(\theta_{P})\in P$. For $i=3$,
by Taylor expansion at the centre $\theta_{P}$,

$\displaystyle\mathbf{e}_{3}(\theta_{P})\cdot\gamma(\theta)$
$\displaystyle=\mathbf{e}_{3}(\theta_{P})\cdot\bigg{(}\mathbf{e}_{1}(\theta_{P})+\mathbf{e}_{2}(\theta_{P})(\theta-\theta_{P})$
$\displaystyle+(-\mathbf{e}_{1}(\theta_{P})+\uptau(\theta_{P})\mathbf{e}_{3}(\theta_{P}))(\theta-\theta_{P})^{2}\bigg{)}+o((\theta-\theta_{P})^{2})$
$\displaystyle\sim(\theta-\theta_{P})^{2}.$

Since $\xi_{3}\sim 2^{j}$, we have

$|\xi_{3}\mathbf{e}_{3}(\theta_{P})\cdot\gamma(\theta)|\gtrsim(\theta-\theta_{P})^{2}2^{j}.$
(7.35)

We also have

$|\xi_{1}\mathbf{e}_{1}(\theta_{P})\cdot\gamma(\theta)|\lesssim 2^{j-k},$
(7.36)

and

$|\xi_{2}\mathbf{e}_{2}(\theta_{P})\cdot\gamma(\theta)|\lesssim
2^{j-k/2}|\theta-\theta_{P}|.$ (7.37)

If $C$ is large enough, then

$|\xi\cdot\gamma(\theta)|\sim|\xi_{3}\mathbf{e}_{3}(\theta_{P})\cdot\gamma(\theta)|\gtrsim(\theta-\theta_{P})^{2}2^{j}.$
(7.38)

∎

We also need to estimate the derivative of $\eta_{T}(x+t\gamma(\theta))$ with
respect to $t$. Without loss of generality we may write

$\displaystyle\eta_{T}(x)$
$\displaystyle=a_{1}(2^{j-k-k\beta}(x-c_{T})\cdot\mathbf{e}_{1}(\theta_{P}))a_{2}(2^{j-k/2-k\beta}(x-c_{T})\cdot\mathbf{e}_{2}(\theta_{P}))$
$\displaystyle\cdot
a_{3}(2^{j-k\beta}(x-c_{T})\cdot\mathbf{e}_{3}(\theta_{P})),$

for some suitable bump functions $a_{1},a_{2},a_{3}$ supported on $(-1,1)$.
Thus for $x\in B^{3}(c_{T},2^{k-j+k\beta})$, we have

$\displaystyle\left|\frac{d}{dt}\eta_{T}(x+t\gamma(\theta))\right|$
$\displaystyle\lesssim
2^{j-k-k\beta}|\mathbf{e}_{1}(\theta_{P})\cdot\mathbf{e}_{1}(\theta)|+2^{j-k/2-k\beta}|\mathbf{e}_{2}(\theta_{P})\cdot\mathbf{e}_{1}(\theta)|$
$\displaystyle+2^{j-k\beta}|\mathbf{e}_{3}(\theta_{P})\cdot\mathbf{e}_{1}(\theta)|$
$\displaystyle\lesssim
2^{j-k-k\beta}+2^{j-k/2-k\beta}|\theta-\theta_{P}|+2^{j-k\beta}|\theta-\theta_{P}|^{2}$
$\displaystyle\lesssim 2^{j-k\beta}|\theta-\theta_{P}|^{2},$

and for $x\notin B^{3}(c_{T},2^{k-j+k\beta})$ we have rapid decay in $x$ since
$\eta$ is Schwartz.

With Claim 7.8, we can apply integration by parts repeatedly to estimate the
left hand side of (7.33). Each time we integrate by parts, we gain a factor
$\lesssim 2^{-k\beta}$, from which the result follows.

∎

###  7.3 Good part and bad part

The main idea is to divide the wave packets into two parts, called the good
part and the bad part. We will prove an $L^{1}$ estimate for the bad part and
an $L^{2}$ estimate for the good part.

Let $a>2$ be as in Theorem 1.5. Let $\varepsilon,a_{1}$ be small numbers to be
determined later.

For each $P\in\Lambda$, define the set of heavy planks in $\mathbb{T}_{P}$

$\mathbb{T}_{P,b}:=\left\\{T\in\mathbb{T}_{P}:\mu(T)\geq
2^{-k\frac{a_{1}+1}{2}-a(j-k)}\right\\},\quad\mathbb{T}_{P,g}:=\mathbb{T}_{P}\backslash\mathbb{T}_{P,b}.$
(7.39)

Decompose $\mu=\mu_{b}+\mu_{g}:=\mu_{b}+\mu_{0}+\mu_{g1}+\mu_{g_{2}}$, where

$\displaystyle\mu_{0}$ $\displaystyle:=\mu*\psi_{0}^{\vee},$ (7.40)
$\displaystyle\mu_{b}$ $\displaystyle:=\sum_{j\geq
0}\sum_{k\in[j\varepsilon,j]}\sum_{P\in\Lambda_{j,k}}\sum_{T\in\mathbb{T}_{P,b}}M_{T}\mu,$
(7.41) $\displaystyle\mu_{g_{1}}$ $\displaystyle:=\sum_{j\geq
0}\sum_{k\in[j\varepsilon,j]}\sum_{P\in\Lambda_{j,k}}\sum_{T\in\mathbb{T}_{P,g}}M_{T}\mu,$
(7.42) $\displaystyle\mu_{g_{2}}$ $\displaystyle:=\sum_{j\geq
0}\sum_{k\in[0,j\varepsilon)}\sum_{P\in\Lambda_{j,k}}\sum_{T\in\mathbb{T}_{P}}M_{T}\mu.$
(7.43)

We remark that the wave packets of $\mu_{b}$ are those that have heavy mass
and are not too far away from the cone $\Gamma$ (as $k$ is not too small).

Note that each $M_{T}\mu$ is a function but $\mu_{g},\mu_{b}$ are in general
measures. We also remark that $\mu_{g}=\mu_{g,\beta,a,a_{1},\varepsilon}$ and
$\mu_{b}=\mu_{b,a,\beta,a,a_{1},\varepsilon}$ depend on parameters
$\beta,a,a_{1},\varepsilon$, but for simplicity we just omit them.

We now state the main estimates we want to prove.

######  Lemma 7.9 (Lemma 6 of [GGG+22]).

Let $a>2$, $\varepsilon\ll 1$ and $a_{1}<2$. For all Borel measures $\mu$
supported on the unit ball in $\mathbb{R}^{3}$ with $\rho_{a}(\mu)\leq 1$, it
holds that

$\sum_{j\geq
0}\sum_{k\in[j\varepsilon,j]}\sum_{P\in\Lambda_{j,k}}\sum_{T\in\mathbb{T}_{P,b}}\int\int|\pi_{\theta\\#}M_{T}\mu|d\mathcal{H}^{2}d\theta\lesssim
1.$ (7.44)

Let $\\{\phi_{n}:n\geq 1\\}$ be an approximation to the identity in
$\mathbb{R}^{3}$, and then define $\mu^{(n)}_{g}:=\mu_{g}*\phi_{n}$ to be a
mollified version of $\mu_{g}$, such that each $\mu^{(n)}_{g}$ is a function.

######  Lemma 7.10 (Lemma 7 of [GGG+22]).

Let $a>2$. Then for $\varepsilon>0$ and $2-a_{1}>0$ small enough depending on
$a$, all Borel measures $\mu$ supported on the unit ball in $\mathbb{R}^{3}$
with $\rho_{a}(\mu)\leq 1$, it holds that

$\int\int|\pi_{\theta\\#}\mu^{(n)}_{g}|^{2}d\mathcal{H}^{2}d\theta\lesssim 1,$
(7.45)

where the constant is independent of $n$.

We now prove Theorem 1.5 assuming Lemmas 7.9 and 7.10.

###### Proof of Theorem 1.5.

Take $\varepsilon,2-a_{1}$ sufficiently small such that both Lemmas 7.9 and
7.10 hold.

By Lemma 7.9, in particular, for a.e. $\theta$ we have

$\sum_{j\geq
0}\sum_{k\in[j\varepsilon,j]}\sum_{P\in\Lambda_{j,k}}\sum_{T\in\mathbb{T}_{P,b}}\int|\pi_{\theta\\#}M_{T}\mu|d\mathcal{H}^{2}<\infty,$
(7.46)

which implies that $\pi_{\theta\\#}\mu_{b}$ is absolutely continuous with
respect to $\mathcal{H}^{2}$.

By Lemma 7.10, the sequence of functions $\\{\pi_{\theta\\#}\mu^{(n)}_{g}\\}$
is bounded in $L^{2}(\mathcal{H}^{2}\times I_{0})$, and so by weak
compactness, there exists an $L^{2}(\mathcal{H}^{2}\times I_{0})$ function $h$
such that $\pi_{\theta\\#}\mu^{(n)}_{g}\to h$ weakly in
$L^{2}(\mathcal{H}^{2}\times I_{0})$ passing to a subsequence. However, by
properties of approximate identity we know that $\mu^{(n)}_{g}\to\mu_{g}$
weakly in $L^{2}(\mathcal{H}^{2})$, whence
$\pi_{\theta\\#}\mu^{(n)}_{g}\to\pi_{\theta\\#}\mu_{g}$ weakly in
$L^{2}(\mathcal{H}^{2}\times I_{0})$. By uniqueness of limits, we have
$h=\pi_{\theta\\#}\mu_{g}$, and so $\pi_{\theta\\#}\mu_{g}$ is actually a
function in $L^{2}(\mathcal{H}^{2}\times I_{0})$. Since
$\mathcal{H}^{2}|_{B^{2}(0,1)}\times I_{0}$ is a finite measure space,
Hölder’s inequality implies that

$\int\int|\pi_{\theta\\#}\mu_{g}|d\mathcal{H}^{2}d\theta\lesssim 1.$ (7.47)

In particular, for a.e. $\theta\in I_{0}$ we have

$\int|\pi_{\theta\\#}\mu|d\mathcal{H}^{2}<\infty,$ (7.48)

and so $\pi_{\theta\\#}\mu_{g}$ is absolutely continuous with respect to
$\mathcal{H}^{2}$. Hence $\pi_{\theta\\#}\mu$ is absolutely continuous with
respect to $\mathcal{H}^{2}$. ∎

###  7.4 Controlling bad part

We now prove Lemma 7.9. The main ingredient is Corollary 6.6. We have

$\displaystyle\int\sum_{j\geq
0}\sum_{k\in[j\varepsilon,j]}\sum_{P\in\Lambda_{j,k}}\sum_{T\in\mathbb{T}_{P,b}}\int|\pi_{\theta\\#}M_{T}\mu|d\mathcal{H}^{2}d\theta$
$\displaystyle=\sum_{j\geq
0}\sum_{k\in[j\varepsilon,j]}\int\int\sum_{\stackrel{{\scriptstyle
P\in\Lambda_{j,k}}}{{|\theta_{P}-\theta|<C2^{-k/2}}}}\sum_{T\in\mathbb{T}_{P,b}}|\pi_{\theta\\#}M_{T}\mu|d\mathcal{H}^{2}d\theta$
(7.49) $\displaystyle+\sum_{j\geq
0}\sum_{k\in[j\varepsilon,j]}\int\int\sum_{\stackrel{{\scriptstyle
P\in\Lambda_{j,k}}}{{|\theta_{P}-\theta|\geq
C2^{-k/2}}}}\sum_{T\in\mathbb{T}_{P,b}}|\pi_{\theta\\#}M_{T}\mu|d\mathcal{H}^{2}d\theta.$
(7.50)

####  7.4.1 Far part

To bound (7.50), we use Lemma 7.7 to get

$\displaystyle\sum_{\stackrel{{\scriptstyle
P\in\Lambda_{j,k}}}{{|\theta_{P}-\theta|\geq
C2^{-k/2}}}}\sum_{T\in\mathbb{T}_{P}}|\pi_{\theta\\#}M_{T}\mu(x)|$
$\displaystyle\lesssim_{N,\beta}(\\#\Lambda_{j,k})2^{-kN+3j}\mu(\mathbb{R}^{3})$
$\displaystyle\lesssim_{N,\beta}2^{-kN+3j},$

where we have used $\\#\Lambda_{j,k}\lesssim 2^{k/2}$ and
$\mu(\mathbb{R}^{3})\lesssim 1$ since $\rho_{a}(\mu)\leq 1$.

Thus we have

$\displaystyle\eqref{eqn_proof_lem6_02}$
$\displaystyle\lesssim_{N,\beta}\sum_{j\geq
0}\sum_{k\in[j\varepsilon,j]}\int\int 2^{-kN+3j}d\mathcal{H}^{2}d\theta$
$\displaystyle\lesssim\sum_{j\geq 0}2^{3j-3j\varepsilon N}$
$\displaystyle\lesssim 1,$

if we choose $N>10\varepsilon^{-1}$.

####  7.4.2 Near part

To estimate (7.49), we apply Lemma 7.6 and get

$\sum_{\stackrel{{\scriptstyle P\in\Lambda_{j,k}}}{{|\theta_{P}-\theta|\leq
C2^{-k/2}}}}\sum_{T\in\mathbb{T}_{P,b}}\int|\pi_{\theta\\#}M_{T}\mu|d\mathcal{H}^{2}\lesssim_{\beta,N}\sum_{\stackrel{{\scriptstyle
P\in\Lambda_{j,k}}}{{|\theta_{P}-\theta|\leq
C2^{-k/2}}}}\sum_{T\in\mathbb{T}_{P,b}}2^{-kN}+\mu(T).$ (7.51)

The rapidly decaying term $2^{-kN}$ can be dealt with easily, since we also
have

$\\#\mathbb{T}_{P,b}\leq 2^{k\frac{a_{1}+1}{2}+a(j-k)}$ (7.52)

by the heavy mass in our definition of $\mathbb{T}_{P,b}$. The remaining
argument is divided into the following steps.

Step 1: Discretization.

To estimate the other term, we define

$F(\theta,C):=\sum_{\stackrel{{\scriptstyle
P\in\Lambda_{j,k}}}{{|\theta_{P}-\theta|\leq
C2^{-k/2}}}}\sum_{T\in\mathbb{T}_{P,b}}\mu(T)=\sum_{\stackrel{{\scriptstyle
P\in\Lambda_{j,k}}}{{|\theta_{P}-\theta|\leq
C2^{-k/2}}}}\mu(\cup\mathbb{T}_{P,b}).$ (7.53)

It is easy to see that for $|\theta-\theta^{\prime}|\leq C2^{-k/2}$, we have

$F(\theta,C)\leq F(\theta^{\prime},2C).$ (7.54)

Using this and denoting $\Theta^{\prime}_{k}$ to be the $C2^{-k}$ lattice on
$[0,1]$, we have

$\displaystyle\int F(\theta,C)d\theta$
$\displaystyle=\sum_{\theta^{\prime}\in\Theta^{\prime}_{k}}\int_{\theta^{\prime}-C2^{-k-1}}^{\theta^{\prime}+C2^{-k-1}}F(\theta,C)d\theta$
$\displaystyle\leq\sum_{\theta^{\prime}\in\Theta^{\prime}_{k}}\int_{\theta^{\prime}-C2^{-k-1}}^{\theta^{\prime}+C2^{-k-1}}F(\theta^{\prime},2C)d\theta$
$\displaystyle\sim\sum_{\theta^{\prime}\in\Theta^{\prime}_{k}}2^{-k}F(\theta^{\prime},2C)$
$\displaystyle\sim 2^{-k}\sum_{P\in\Lambda_{j,k}}\mu(\cup\mathbb{T}_{P,b}),$
(7.55)

where in the last line we have used that $\Theta_{k}$ is a $c2^{-k/2}$
lattice.

Step 2: Rescaling.

Now we seek to apply Corollary 6.6 with $\delta^{1/2}\sim 2^{-k/2}$, where we
take $\Delta=\Theta_{k}$.

Let $B_{l}$ be finitely overlapping balls of radius $2^{k-j+k\beta}$ that
cover $B^{3}(0,1)$. For each $\theta$ and $l$ we have

$\mu(\cup\mathbb{T}_{P,b})\leq\sum_{B_{l}}\mu(\cup\mathbb{T}_{P,l,b}),$ (7.56)

where

$\mathbb{T}_{P,l,b}:=\\{T\in\mathbb{T}_{P,b}:T\cap B_{l}\neq\varnothing\\}.$
(7.57)

It is easy to see that $\cup\mathbb{T}_{P,l,b}\subseteq 2^{1000}B_{l}$ where
$2^{1000}B_{l}$ denotes the concentric dilation of $B_{l}$; also
$2^{1000}B_{l}$’s have finite overlap.

Now we perform a rescaling. Let $\lambda(x):=2^{-j+k-k\beta}x$ and $\mu_{j,k}$
be the measure on $\mathbb{R}^{3}$ defined by

$\mu_{j,k}(A):=\mu(\lambda(A)).$ (7.58)

Also, for each $l$, define $\tilde{B}_{l}=\lambda(2^{1000}B_{l})$ and

$\mu_{j,k,l}:=2^{a(j-k)}\mu_{j,k}1_{\tilde{B}_{l}},\quad\tilde{\mathbb{T}}_{P,l,b}:=\\{\lambda(T):T\in\mathbb{T}_{P,l,b}\\}.$
(7.59)

We can check that the set $\cup\tilde{\mathbb{T}}_{P,l,b}$ is a disjoint union
of planks of dimensions $2^{-k}\times 2^{-k/2}\times 1$ pointing in the
directions $\mathbf{e}_{i}(\theta_{P})$, $i=1,2,3$ respectively, and it is
contained in $\tilde{B}_{l}$. If we project every plank along the direction
$\mathbf{e}_{3}(\theta_{P})$, then we obtain a disjoint family of planar
rectangles of dimensions $2^{-k}\times 2^{-k/2}$ pointing in the directions
$\mathbf{e}_{i}(\theta_{P})$, $i=1,2$ respectively.

Meanwhile, in view of the heavy mass as in the definition $\mathbb{T}_{P,b}$,
we have $\mu_{j,k,l}(\lambda(T))\gtrsim 2^{-k\frac{a_{1}+1}{2}}$, and so we
have

$\\#(\tilde{\mathbb{T}}_{P,l,b})\lesssim
2^{k\frac{a+1}{2}}\mu_{j,k,l}(\mathbb{R}^{3}).$ (7.60)

It is also direct to see that $\rho_{a}(\mu_{j,k,l})\lesssim 2^{ka\beta}$.

Step 3: Applying Corollary 6.6.

Now we use Corollary 6.6 with $\delta^{1/2}=c2^{-k/2}$ to get some small
exponent $\varepsilon^{\prime}>0$, depending only on $a,a_{1}$, such that

$\sum_{P\in\Lambda_{j,k}}\pi_{\theta_{P}\\#}\mu_{j,k,l}\left(\bigcup\pi_{\theta_{P}\\#}\tilde{\mathbb{T}}_{P,l,b}\right)\lesssim
2^{-\frac{k}{2}-k\varepsilon^{\prime}+ka\beta}\mu_{j,k,l}(\mathbb{R}^{3}).$
(7.61)

But by direct computation,

$\pi_{\theta_{P}\\#}\mu_{j,k,l}\left(\bigcup\pi_{\theta_{P}\\#}\tilde{\mathbb{T}}_{P,l,b}\right)=\mu_{j,k,l}\left(\bigcup\tilde{\mathbb{T}}_{P,l,b}\right)\gtrsim
2^{a(j-k)}\mu(\cup\mathbb{T}_{P,l,b}).$ (7.62)

Summing over $B_{l}$ and using (7.56), we thus have

$\sum_{P\in\Lambda_{j,k}}\mu(\cup\mathbb{T}_{P,b})\leq\sum_{B_{l}}\mu(\cup\mathbb{T}_{P,l,b})\lesssim
2^{-\frac{k}{2}-k\varepsilon^{\prime}}\mu(\mathbb{R}^{3})\leq
2^{-\frac{k}{2}-k\varepsilon^{\prime}+ak\beta}.$ (7.63)

Continuing estimating from (7.51) and using (7.53) and (7.55), we have

$\sum_{\stackrel{{\scriptstyle P\in\Lambda_{j,k}}}{{|\theta_{P}-\theta|\leq
C2^{-k/2}}}}\sum_{T\in\mathbb{T}_{P,b}}\mu(T)\lesssim
2^{ak\beta-k\varepsilon^{\prime}}.$ (7.64)

Now it suffices to choose $\beta=\varepsilon^{\prime}/100$ and sum over
$k\in[j\varepsilon,j]$ and then $j\in\mathbb{N}$. This concludes the $L^{1}$
estimate of the bad part.

###  7.5 Controlling good part

We now prove Lemma 7.10. By the functional analytic proof of Theorem 1.5, we
may abuse notation and treat $\mu_{g}$ as a function as well.

Fix $\theta$ and $x\in\pi_{\theta\\#}(\mathbb{R}^{3})$. We have

$\pi_{\theta\\#}\mu_{g}(x)=\int\mu_{g}(x+t\gamma(\theta))dt.$ (7.65)

With the coordinate system $\\{\mathbf{e}_{i}(\theta):i=1,2,3\\}$ and slightly
abusing notation, any $x\in\pi_{\theta\\#}(\mathbb{R}^{3})$ can be written as
$(0,x_{2},x_{3})$. Thus the above equality can be rewritten as

$\pi_{\theta\\#}\mu_{g}(x_{2},x_{3})=\int\mu_{g}(t,x_{2},x_{3})dt.$ (7.66)

Doing the Fourier transform in the $(\mathbf{e}_{2},\mathbf{e}_{3})$-plane, we
have

$(\pi_{\theta\\#}\mu_{g})^{\wedge}(\eta_{2},\eta_{3})=\widehat{\mu_{g}}(0,\eta_{2},\eta_{3})=\widehat{\mu_{g}}(\eta_{2}\mathbf{e}_{2}(\theta)+\eta_{3}\mathbf{e}_{3}(\theta)).$
(7.67)

By Plancherel’s theorem,

$\int\int_{\mathbb{R}^{2}}|\pi_{\theta\\#}\mu_{g}|^{2}d\mathcal{H}^{2}d\theta=\int\iint|\widehat{\mu_{g}}(\eta_{2}\mathbf{e}_{2}(\theta)+\eta_{3}\mathbf{e}_{3}(\theta))|^{2}d\eta
d\theta.$ (7.68)

By the triangle inequality, we will estimate $\mu_{0}$, $\mu_{g1}$ and
$\mu_{g2}$ respectively.

We first settle the easiest case, namely, for $\mu_{0}$. We have
$\widehat{\mu_{0}}=\widehat{\mu}\psi_{0}$. Since
$\sup|\widehat{\mu}|\leq\mu(\mathbb{R}^{3})<\infty$, the integral over
$|\eta|\lesssim 1$ and $\theta\in I_{0}$ is clearly finite.

####  7.5.1 Preparation

To estimate the other two terms, we first perform a change of variables

$\xi=(\xi_{1},\xi_{2},\xi_{3})=\eta_{2}\mathbf{e}_{2}(\theta)+\eta_{3}\mathbf{e}_{3}(\theta),$
(7.69)

which has Jacobian

$|\det(\mathbf{e}_{2},\mathbf{e}_{3},\eta_{2}(-\mathbf{e}_{1}+\uptau\mathbf{e}_{3})-\eta_{3}\uptau\mathbf{e}_{2})|=|\eta_{2}\det(\mathbf{e}_{1},\mathbf{e}_{2},\mathbf{e}_{3})|=|\eta_{2}|.$
(7.70)

Thus we have

$\int\iint|\widehat{\mu_{g}}(\eta_{2}\mathbf{e}_{2}(\theta)+\eta_{3}\mathbf{e}_{3}(\theta))|^{2}d\eta
d\theta=\int_{\mathbb{R}^{3}}|\widehat{\mu_{g}}(\xi)|^{2}|\eta_{2}|^{-1}d\xi.$
(7.71)

We also need the following lemma.

######  Lemma 7.11.

Let $P\in\Lambda_{j,k}$. If there exist $\theta\in I_{0}$ and
$(\eta_{2},\eta_{3})$ satisfying

$\eta_{2}\mathbf{e}_{2}(\theta)+\eta_{3}\mathbf{e}_{3}(\theta)\in\mathrm{supp}(\psi_{P}),$
(7.72)

as well as $|\theta-\theta_{P}|\ll 1$, then it holds that

$|\theta-\theta_{P}|\lesssim 2^{-\frac{k}{2}},\quad|\eta_{2}|\lesssim
2^{j-\frac{k}{2}},\quad|\eta_{3}|\sim 2^{j}.$ (7.73)

Furthermore, if $j>k$, then it holds that

$|\theta-\theta_{P}|\sim 2^{-\frac{k}{2}},\quad|\eta_{2}|\sim
2^{j-\frac{k}{2}},\quad|\eta_{3}|\sim 2^{j}.$ (7.74)

###### Proof.

Write $\xi=\eta_{2}\mathbf{e}_{2}(\theta)+\eta_{3}\mathbf{e}_{3}(\theta)$.
Since $\xi\in\mathrm{supp}(\psi_{P})=2P$, we have (without loss of generality)

$\displaystyle|\xi\cdot\mathbf{e}_{1}(\theta_{P})|\lesssim 2^{j-k},$ (7.75)
$\displaystyle|\xi\cdot\mathbf{e}_{2}(\theta_{P})|\ll 2^{j-\frac{k}{2}},$
(7.76) $\displaystyle\xi\cdot\mathbf{e}_{3}(\theta_{P})\sim 2^{j}.$ (7.77)

In particular, $|\xi|\sim|\eta_{2}|+|\eta_{3}|\sim 2^{j}$. Assume towards
contradiction that $|\theta-\theta_{P}|\gg 2^{-k/2}$. Using (7.76), by Taylor
expansion for $\mathbf{e}_{2}(\theta_{P})$ at $\theta$ we have

$|\eta_{2}+\eta_{3}(\theta_{P}-\theta)+O(\theta_{P}-\theta)^{2}|\ll
2^{j-\frac{k}{2}}.$ (7.78)

Using (7.77), by Taylor expansion for $\mathbf{e}_{1}(\theta_{P})$ at $\theta$
we have

$\displaystyle 2^{j-k}$
$\displaystyle\gtrsim|\xi\cdot[\mathbf{e}_{1}(\theta)+\mathbf{e}_{2}(\theta)(\theta_{P}-\theta)+(-\mathbf{e}_{1}(\theta)+\uptau(\theta)\mathbf{e}_{3}(\theta))(\theta_{P}-\theta)^{2}$
$\displaystyle+|\xi|o((\theta_{P}-\theta)^{2})]|$
$\displaystyle=\eta_{2}(\theta_{P}-\theta)+\eta_{3}(\theta_{P}-\theta)^{2}(1/2+o(1)).$

Using $|\theta-\theta_{P}|\gg 2^{-k/2}$, we have

$|\eta_{2}+\eta_{3}(\theta_{P}-\theta)(1/2+o(1))|\ll 2^{j-\frac{k}{2}}.$
(7.79)

Combining this with (7.78) and using the triangle inequality, we have

$|\eta_{2}|\ll 2^{j-k/2},$ (7.80)

and plugging this into (7.78) again, we get

$\eta_{3}\lesssim|\theta_{P}-\theta|^{-1}2^{j-k/2}\ll 2^{j},$ (7.81)

using $|\theta-\theta_{P}|\ll 1$. This leads to a contradiction since
$|\eta_{2}|+|\eta_{3}|\sim 2^{j}$. Thus we must have
$|\theta-\theta_{P}|\lesssim 2^{-k/2}$.

By Taylor approximation, it is easy to get $|\eta_{2}|\lesssim 2^{j-k/2}$ and
$\eta_{3}\sim 2^{j}$.

Lastly, we come to the case $j>k$. Assume towards contradiction that
$|\theta-\theta_{P}|\ll 2^{-k/2}$. Then by Taylor expansion,

$|\xi\cdot\mathbf{e}_{1}(\theta)-\xi\cdot\mathbf{e}_{1}(\theta_{P})|\leq|\xi\cdot\mathbf{e}_{2}(\theta_{P})||\theta-\theta_{P}|\ll
2^{j-k},$ (7.82)

and so $|\xi\cdot\mathbf{e}_{1}(\theta)|\sim 2^{j-k}$, contradicting the
assumption that $\xi\cdot\mathbf{e}_{1}(\theta)=0$. Thus
$|\theta-\theta_{P}|\gtrsim 2^{-k/2}$. By Taylor approximation, it is easy to
get $|\eta_{2}|\sim 2^{j-k/2}$ and $\eta_{3}\sim 2^{j}$.

∎

####  7.5.2 Estimate of $\mu_{g2}$

For $\mu_{g2}$ we have $k\leq j\varepsilon$, namely, the planks $P$ are
extremely far away from the cone $\Gamma$. We estimate (7.68) with $\mu_{g2}$
in place of $\mu_{g}$. Since $k\geq j\varepsilon$, Lemma 7.11 implies that

$|\eta_{2}|\gtrsim|\xi|^{1-\varepsilon/2}.$ (7.83)

Thus, by (7.71), we have

$\int_{\mathbb{R}^{3}}|\widehat{\mu_{g}}(\xi)|^{2}|\eta_{2}|^{-1}d\xi\lesssim\int_{\mathbb{R}^{3}}|\widehat{\mu}(\xi)|^{2}|\xi|^{\varepsilon/2-1}d\xi,$
(7.84)

which is finite if we choose $\varepsilon$ small enough that
$a-\varepsilon/2>2$, since $\rho_{a}(\mu)\lesssim 1$ and $a>2$ (see, for
instance, Theorems 2.7 and 3.10 of [Mat15]).

####  7.5.3 Estimate of $\mu_{g1}$

This part estimates the contribution from the planks that have light mass and
are not too far away from the cone. By definition, we have

$\widehat{\mu_{g}}(\xi)=\sum_{j\geq
0}\sum_{k\in[j\varepsilon,k]}\sum_{P\in\Lambda_{j,k}}\sum_{T\in\mathbb{T}_{P,g}}\widehat{M_{T}\mu}$
(7.85)

where we have

$\widehat{M_{T}\mu}(\xi)=\widehat{\eta_{T}}*(\widehat{\mu}(\xi)\psi_{P}(\xi)),$
(7.86)

which is supported on $3P$ where $P\in\Lambda_{j,k}$. By Plancherel’s
identity, we have

$\int_{\mathbb{R}^{3}}|\widehat{\mu_{g}}(\xi)|^{2}|\eta_{2}|^{-1}d\xi=\sum_{j\geq
0}\sum_{k\in[j\varepsilon,k]}\sum_{P\in\Lambda_{j,k}}\int_{\mathbb{R}^{3}}\bigg{|}\sum_{T\in\mathbb{T}_{P,g}}\widehat{M_{T}\mu}(\xi)\bigg{|}^{2}|\eta_{2}|^{-1}d\xi.$
(7.87)

We further divide the argument into the following steps.

Step 1: Using orthorgonality.

Fix $k\in[j\varepsilon,k]$. We consider the cases $k<j$ and $k=j$ separately.
In the former case, using Lemma 7.11, we have

$\displaystyle\int_{\mathbb{R}^{3}}\bigg{|}\sum_{T\in\mathbb{T}_{P,g}}\widehat{M_{T}\mu}(\xi)\bigg{|}^{2}|\eta_{2}|^{-1}d\xi$
$\displaystyle\sim
2^{-j+k/2}\int_{\mathbb{R}^{3}}\bigg{|}\sum_{T\in\mathbb{T}_{P,g}}\widehat{M_{T}\mu}(\xi)\bigg{|}^{2}d\xi$
$\displaystyle=2^{-j+k/2}\int_{\mathbb{R}^{3}}\bigg{|}\sum_{T\in\mathbb{T}_{P,g}}M_{T}\mu(x)\bigg{|}^{2}dx$
$\displaystyle\sim
2^{-j+k/2}\sum_{T\in\mathbb{T}_{P,g}}\int_{\mathbb{R}^{3}}|M_{T}\mu(x)|^{2}dx.$

Summing over $P\in\Lambda_{j,k}$, we thus have

$\sum_{P\in\Lambda_{j,k}}\int_{\mathbb{R}^{3}}\bigg{|}\sum_{T\in\mathbb{T}_{P,g}}\widehat{M_{T}\mu}(\xi)\bigg{|}^{2}|\eta_{2}|^{-1}d\xi\lesssim
2^{-j+\frac{k}{2}}\sum_{P\in\Lambda_{j,k}}\sum_{T\in\mathbb{T}_{P,g}}\int_{\mathbb{R}^{3}}|M_{T}\mu(x)|^{2}dx.$
(7.88)

For $j=k$ we have (7.88) holds as well. To see this, fix $P$ and we go back to
the form

$\int\iint\bigg{|}\sum_{T\in\mathbb{T}_{P,g}}\widehat{M_{T}\mu}(\eta_{2}\mathbf{e}_{2}(\theta)+\eta_{3}\mathbf{e}_{3}(\theta))\bigg{|}^{2}d\eta
d\theta.$ (7.89)

By Lemma 7.11, we must have $|\theta-\theta_{P}|\lesssim 2^{-j/2}$. For every
such $\theta$, we apply Plancherel’s identity in the $(\eta_{2},\eta_{3})$
plane to get

$\iint\bigg{|}\sum_{T\in\mathbb{T}_{P,g}}\widehat{M_{T}\mu}(\eta_{2}\mathbf{e}_{2}(\theta)+\eta_{3}\mathbf{e}_{3}(\theta))\bigg{|}^{2}d\eta=\iint|F(x_{2},x_{3})|^{2}dx_{2}dx_{3},$
(7.90)

where

$\displaystyle F(x_{2},x_{3})$
$\displaystyle:=\sum_{T\in\mathbb{T}_{P,g}}\iint\widehat{M_{T}\mu}(\eta_{2}\mathbf{e}_{2}(\theta)+\eta_{3}\mathbf{e}_{3}(\theta))e(x_{2}\eta_{2}+x_{3}\eta_{3})d\eta_{2}d\eta_{3}$
$\displaystyle=\sum_{T\in\mathbb{T}_{P,g}}\int_{\mathbb{R}^{3}}M_{T}\mu(y)\iint
e(-y\cdot(\eta_{2}\mathbf{e}_{2}(\theta)+\eta_{3}\mathbf{e}_{3}(\theta)))dy$
$\displaystyle\cdot e(x_{2}\eta_{2}+x_{3}\eta_{3})d\eta_{2}d\eta_{3}$
$\displaystyle=\sum_{T\in\mathbb{T}_{P,g}}\int_{\mathbb{R}^{3}}M_{T}\mu(y)\delta_{0}(x_{2}-y\cdot\mathbf{e}_{2}(\theta))\delta_{0}(x_{3}-y\cdot\mathbf{e}_{3}(\theta))dy$
$\displaystyle=\sum_{T\in\mathbb{T}_{P,g}}\int_{\mathbb{R}}M_{T}\mu(t\mathbf{e}_{1}(\theta)+x_{2}\mathbf{e}_{2}(\theta)+x_{3}\mathbf{e}_{2}(\theta))dt.$

Plugging back into (7.90), we get

$\displaystyle\iint|F(x_{2},x_{3})|^{2}dx_{2}dx_{3}$ $\displaystyle=$
$\displaystyle\iint F(x_{2},x_{3})\overline{F(x_{2},x_{3})}dx_{2}dx_{3}$
$\displaystyle=$
$\displaystyle\sum_{T\in\mathbb{T}_{P,g}}\sum_{T^{\prime}\in\mathbb{T}_{P,g}}\iint\int_{\mathbb{R}}\int_{\mathbb{R}}M_{T}\mu(t\mathbf{e}_{1}(\theta)+x_{2}\mathbf{e}_{2}(\theta)+x_{3}\mathbf{e}_{3}(\theta))$
$\displaystyle\cdot\overline{M_{T^{\prime}}\mu(t^{\prime}\mathbf{e}_{1}(\theta)+x_{2}\mathbf{e}_{2}(\theta)+x_{3}\mathbf{e}_{3}(\theta))}dtdt^{\prime}dx_{2}dx_{3}.$

For a fixed $T$, recall that $M_{T}\mu$ is essentially supported on $T$. Since
$|\theta-\theta_{P}|\lesssim 2^{-j/2}$, we can show that

$|M_{T}\mu(t\mathbf{e}_{1}(\theta)+x_{2}\mathbf{e}_{2}(\theta)+x_{3}\mathbf{e}_{3}(\theta))|\lesssim|\tilde{M}_{T}\mu(t\mathbf{e}_{1}(\theta_{P})+x_{2}\mathbf{e}_{2}(\theta_{P})+x_{3}\mathbf{e}_{3}(\theta_{P}))|$
(7.91)

where $\tilde{M}_{T}:=w_{T}(\mu*\psi_{P}^{\vee})$ and $w_{T}$ is a suitable
weight function adapted to $T$. For fixed $x_{2},x_{3}$, we thus have
$T,T^{\prime}$ must only differ at most in the $\mathbf{e}_{1}(\theta_{P})$
position (otherwise we have rapid decay which can be treated as $0$; we leave
the rigorous argument to the reader; cf. Section 3.8). However, since we are
in the case $k=j$ and we may assume $T\subseteq B^{3}(0,1)$ (otherwise we have
rapid decay), we must have $T=T^{\prime}$. Thus we can continue estimating

$\displaystyle\iint|F(x_{2},x_{3})|^{2}dx_{2}dx_{3}$
$\displaystyle\lesssim\sum_{T\in\mathbb{T}_{P,g}}\int_{\mathbb{R}}\int_{\mathbb{R}}\left|\int_{\mathbb{R}}M_{T}\mu(t\mathbf{e}_{1}(\theta_{P})+x_{2}\mathbf{e}_{2}(\theta_{P})+x_{3}\mathbf{e}_{3}(\theta_{P}))dt\right|^{2}dx_{2}dx_{3}$
$\displaystyle\lesssim\sum_{T\in\mathbb{T}_{P,g}}\int_{\mathbb{R}}\int_{\mathbb{R}}\int_{\mathbb{R}}\left|M_{T}\mu(t\mathbf{e}_{1}(\theta_{P})+x_{2}\mathbf{e}_{2}(\theta_{P})+x_{3}\mathbf{e}_{3}(\theta_{P}))\right|^{2}dtdx_{2}dx_{3},$

where in the last line we have used the fact that $t$ is essentially supported
on the unit interval. Integrating in $\theta$, using the discretization and
summing over $P\in\Lambda_{j,k}$, we have (7.88).

We continue to estimate (7.88) and do not distinguish $j>k$ and $j=k$ anymore.
Thus it suffices to estimate the term

$\sum_{P\in\Lambda_{j,k}}\sum_{T\in\mathbb{T}_{P,g}}\int_{\mathbb{R}^{3}}|M_{T}\mu(x)|^{2}dx.$
(7.92)

Step 2: Preparation for refined Strichartz inequality.

We hope to apply Theorem 8.2 stated in Section 8 below. Before that we need to
do some preparation work.

Write

$f_{T}:=(\eta_{T}M_{T}\mu)*\psi_{P}^{\vee},$ (7.93)

so that

$\sum_{P\in\Lambda_{j,k}}\sum_{T\in\mathbb{T}_{P,g}}\int_{\mathbb{R}^{3}}|M_{T}\mu(x)|^{2}dx=\int\sum_{P\in\Lambda_{j,k}}\sum_{T\in\mathbb{T}_{P,g}}f_{T}d\mu.$
(7.94)

Similar to the proof of Step 2 of Section 7.4.2, we cover the unit ball by
$O(1)$-overlapping small balls $B_{l}$ of radius $2^{k-j+k\beta}$ and let
$\nu_{l}$ be the restriction of $\mu$ to $B_{l}$. By Cauchy-Schwarz, we have

$\int\sum_{P\in\Lambda_{j,k}}\sum_{T\in\mathbb{T}_{P,g}}f_{T}d\mu\leq\sum_{l}\mu(B_{l})^{1/2}\left(\int\left|\sum_{P\in\Lambda_{j,k}}\sum_{T\in\mathbb{T}_{P,g}}f_{T}\right|^{2}d\nu_{l}\right)^{1/2}.$
(7.95)

Let $\zeta_{j}$ be a non-negative bump function such that
$\widehat{\zeta_{j}}(\xi)=1$ for $|\xi|\leq 2^{j+k\beta+10}$. By the Fourier
support condition of $f_{T}$, we have

$\int\left|\sum_{P\in\Lambda_{j,k}}\sum_{T\in\mathbb{T}_{P,g}}f_{T}\right|^{2}d\nu_{l}=\int\left|\sum_{P\in\Lambda_{j,k}}\sum_{T\in\mathbb{T}_{P,g}}f_{T}\right|^{2}d(\nu_{l}*\zeta_{j}).$
(7.96)

By dyadic pigeonholing (see Section 5.10 for the technique), we can find a
subset

$\mathbb{W}_{l}\subseteq\bigcup_{P\in\Lambda_{j,k}}\\{T\in\mathbb{T}_{P,g}:T\cap
B_{l}\neq\varnothing\\}$ (7.97)

such that $\norm*{f_{T}}_{2}$ is a constant $N$ up to a factor of $2$ as $T$
varies over $\mathbb{W}_{l}$, and

$\int\left|\sum_{P\in\Lambda_{j,k}}\sum_{T\in\mathbb{T}_{P,g}}f_{T}\right|^{2}d(\nu_{l}*\zeta_{j})\lesssim
j\int\left|\sum_{T\in\mathbb{W}_{l}}f_{T}\right|^{2}d(\nu_{l}*\zeta_{j}).$
(7.98)

Now we take some $p\geq 2$ to be determined. By Hölder’s inequality, we have

$\int\left|\sum_{T\in\mathbb{W}_{l}}f_{T}\right|^{2}d(\nu_{l}*\zeta_{j})\leq\norm*{\sum_{T\in\mathbb{W}_{l}}f_{T}}_{p}^{2}\left(\int(\nu_{l}*\zeta_{j})^{\frac{p}{p-2}}\right)^{1-\frac{2}{p}}.$
(7.99)

We can view $\sum_{T\in\mathbb{W}_{l}}f_{T}$ as essentially supported on
$B^{3}(0,1)$ by the rapid decay outside. Thus, by dyadic pigeonholing again,
there is some dyadic $M\geq 1$ and a disjoint union $Y$ of balls $Q$ of radius
$2^{-j}$, such that each $Q\subseteq Y$ intersects $\sim M$ planks $T$ as they
vary over $\mathbb{W}_{l}$, and that

$\norm*{\sum_{T\in\mathbb{W}_{l}}f_{T}}_{p}^{2}\left(\int(\nu_{l}*\zeta_{j})^{\frac{p}{p-2}}\right)^{1-\frac{2}{p}}\lesssim
j\norm*{\sum_{T\in\mathbb{W}_{l}}f_{T}}_{L^{p}(Y)}^{2}\left(\int_{Y}(\nu_{l}*\zeta_{j})^{\frac{p}{p-2}}\right)^{1-\frac{2}{p}}.$
(7.100)

Combining all the above estimates, we have in particular

$\displaystyle\sum_{P\in\Lambda_{j,k}}\sum_{T\in\mathbb{T}_{P,g}}\int_{\mathbb{R}^{3}}|M_{T}\mu(x)|^{2}dx$
$\displaystyle\lesssim_{\beta}2^{j\beta}\sum_{l}\mu(B_{l})^{1/2}\norm*{\sum_{T\in\mathbb{W}_{l}}f_{T}}_{L^{p}(Y)}\left(\int_{Y}(\nu_{l}*\zeta_{j})^{\frac{p}{p-2}}\right)^{\frac{1}{2}-\frac{1}{p}}.$
(7.101)

Step 3: Applying refined Strichartz inequality.

Now we apply Theorem 8.2 (to be proved later) to the rescaled functions
$f_{T}(2^{-j}\cdot)$ to get

$\displaystyle\norm*{\sum_{T\in\mathbb{W}_{l}}f_{T}}_{L^{p}(Y)}$
$\displaystyle\lesssim_{\beta}2^{j\beta+(3j-\frac{3k}{2})(\frac{1}{2}-\frac{1}{p})}\left(\frac{M}{\\#\mathbb{W}_{l}}\right)^{\frac{1}{2}-\frac{1}{p}}\left(\sum_{T\in\mathbb{W}_{l}}\norm*{f_{T}}_{2}^{2}\right)^{1/2}$
$\displaystyle\sim
2^{j\beta+(3j-\frac{3k}{2})(\frac{1}{2}-\frac{1}{p})}\left(\frac{M}{\\#\mathbb{W}_{l}}\right)^{\frac{1}{2}-\frac{1}{p}}(\\#\mathbb{W}_{l})^{\frac{1}{2}}N$
$\displaystyle=2^{j\beta+(3j-\frac{3k}{2})(\frac{1}{2}-\frac{1}{p})}M^{\frac{1}{2}-\frac{1}{p}}(\\#\mathbb{W}_{l})^{\frac{1}{2}}N.$
(7.102)

We also need to estimate the term involving $\nu*\zeta_{j}$. Using the
assumption that $\rho_{a}(\mu)\leq 1$, we have $\rho_{a}(\nu)\leq 1$, and so
by the rapid decay of $\zeta_{j}$ outside $B^{3}(0,2^{-j-k\beta})$,

$\displaystyle\left|\int\zeta_{j}(x-y)d\nu_{l}(y)\right|$
$\displaystyle\lesssim\nu_{l}(B^{3}(x,2^{-j-k\beta}))2^{3(j+k\beta)}\lesssim
2^{3(j+k\beta)-a(j-k\beta)}.$

Hence by the light mass in the definition of $\mathbb{T}_{P,g}$, we have

$\displaystyle\int_{Y}(\nu_{l}*\zeta_{j})^{\frac{p}{p-2}}$
$\displaystyle\lesssim\int_{Y}\nu_{l}*\zeta_{j}\cdot
2^{\frac{6(j+k\beta)-2a(j-k\beta)}{p-2}}$ $\displaystyle\sim
2^{\frac{6(j+k\beta)-2a(j-k\beta)}{p-2}}M^{-1}\sum_{T\in\mathbb{W}_{l}}\int_{T}\nu_{l}*\zeta_{j}$
$\displaystyle\lesssim
2^{\frac{6(j+k\beta)-2a(j-k\beta)}{p-2}}M^{-1}(\\#\mathbb{W}_{l})\cdot
2^{-k\frac{a_{1}+1}{2}-a(j-k)}.$

Using (7.101) and (7.102), we have

$\displaystyle\sum_{P\in\Lambda_{j,k}}\sum_{T\in\mathbb{T}_{P,g}}\int_{\mathbb{R}^{3}}|M_{T}\mu(x)|^{2}dx$
(7.103)
$\displaystyle\lesssim_{\beta}2^{j\beta}\sum_{l}\mu(B_{l})^{1/2}2^{j\beta+(3j-\frac{3k}{2})(\frac{1}{2}-\frac{1}{p})}M^{\frac{1}{2}-\frac{1}{p}}(\\#\mathbb{W}_{l})^{\frac{1}{2}}N$
$\displaystyle\cdot\left(2^{\frac{6(j+k\beta)-2a(j-k\beta)}{p-2}}M^{-1}(\\#\mathbb{W}_{l})\cdot
2^{-k\frac{a_{1}+1}{2}-a(j-k)}\right)^{\frac{1}{2}-\frac{1}{p}}$
$\displaystyle\lesssim
2^{O(j\beta)}2^{j\frac{3-a}{2}}2^{k(\frac{1}{2}-\frac{1}{p})(a-2-\frac{a_{1}}{2})}\sum_{l}\mu(B_{l})^{1/2}(\\#\mathbb{W}_{l})^{1/2}N$
$\displaystyle\leq
2^{O(j\beta)}2^{j\frac{3-a}{2}}2^{k(\frac{1}{2}-\frac{1}{p})(a-2-\frac{a_{1}}{2})}\mu(\mathbb{R}^{3})^{\frac{1}{2}}\left(\sum_{l}(\\#\mathbb{W}_{l})N^{2}\right)^{\frac{1}{2}},$
(7.104)

where we have used Cauchy-Schwarz in the last inequality. Recalling that
$N\sim\norm*{f_{T}}_{2}$, we have

$\displaystyle\sum_{l}(\\#\mathbb{W}_{l})N^{2}$
$\displaystyle\sim\sum_{l}\sum_{T\in\mathbb{W}_{l}}\norm*{f_{T}}^{2}_{2}$
$\displaystyle\lesssim\sum_{l}\sum_{T\in\mathbb{W}_{l}}\norm*{M_{T}\mu}^{2}_{2}$
$\displaystyle\lesssim\sum_{P\in\Lambda_{j,k}}\sum_{T\in\mathbb{T}_{P,g}}\int_{\mathbb{R}^{3}}|M_{T}\mu|^{2}.$

Plugging this back into (7.104) and noting this coincides with the term
(7.104) at the beginning, we obtain

$\sum_{P\in\Lambda_{j,k}}\sum_{T\in\mathbb{T}_{P,g}}\int_{\mathbb{R}^{3}}|M_{T}\mu(x)|^{2}dx\lesssim_{\beta}2^{O(j\beta)}2^{j(3-a)}2^{k(1-\frac{2}{p})(a-2-\frac{a_{1}}{2})}\mu(\mathbb{R}^{3}).$
(7.105)

Step 4: Conclusion.

Using $\mu(\mathbb{R}^{3})\lesssim 1$, (7.88), (7.71) and (7.68), we have

$\int\int_{\mathbb{R}^{2}}|\pi_{\theta\\#}\mu_{g}|^{2}d\mathcal{H}^{2}d\theta\lesssim_{\beta}\sum_{j=0}^{\infty}\sum_{k\in[j\varepsilon,j]}2^{O(j\beta)}2^{j(3-a)}2^{k(1-\frac{2}{p})(a-2-\frac{a_{1}}{2})}2^{-j+\frac{k}{2}}.$
(7.106)

We may assume $a$ and $a_{1}$ are very close to $2$. Thus we have

$(a-2-\frac{a_{1}}{2})(1-\frac{2}{p})+\frac{1}{2}=-\frac{1}{2}+\frac{2}{p}+\beta^{\prime},$
(7.107)

where $\beta^{\prime}>0$ can be chosen to be arbitrarily close to $0$. If we
choose any $p>4$ and $\beta^{\prime}\ll p-4$, then we have

$\displaystyle\sum_{k\in[j\varepsilon,j]}\int\int_{\mathbb{R}^{2}}|\pi_{\theta\\#}\mu_{g}|^{2}d\mathcal{H}^{2}d\theta$
$\displaystyle\lesssim_{\beta}2^{O(j\beta)}2^{j(3-a)-j}2^{j\varepsilon(1-\frac{2}{p})(-\frac{1}{2}+\frac{2}{p}+\beta^{\prime}))+\frac{j\varepsilon}{2}}$
$\displaystyle\lesssim 2^{O(j\beta+j\varepsilon)}2^{j(2-a)}.$

Choosing $\beta$ and $\varepsilon$ small enough (depending on $a,a_{1}$ only),
we have

$O(\beta)+O(\varepsilon)+2-a<0,$ (7.108)

and so summing over $j$ gives the claim. This finishes the estimate of
$\mu_{g1}$ and concludes the proof of Lemma 7.10 and hence Theorem 1.5.

##  8 Refined Strichartz inequality

We begin our study of the refined Strichartz inequality. This type of estimate
was first introduced in the paper [GIOW18] and the subsequent argument follows
section 4 of [GIOW18] closely. The reader is encouraged to compare the two
papers to understand the general strategy behind proving this theorem. For
this exposition we omit some of the computations and focus on the general
proof strategy. The reader is referred to the original paper [GGG+22] for
details.

Recall that we had that for each $\theta$,
$e_{1}(\theta)=\gamma(\theta),e_{2}(\theta)=\gamma^{\prime}(\theta),e_{3}(\theta)=\gamma(\theta)\times\gamma^{\prime}(\theta)$
forms an orthonormal basis for $\mathbb{R}^{3}$.

For each $R\geq 1$, let $\Pi_{R}=\\{jR^{-1/2}:j\in\mathbb{Z}\\}\cap[0,a]$. For
each $\theta\in\Pi_{R}$ we let

$\tau_{R}(\theta)=\\{x_{1}e_{1}(\theta)+x_{2}e_{2}(\theta)+x_{3}e_{3}(\theta):1\leq
x_{1}\leq 2,|x_{2}|\leq R^{-1/2},|x_{3}|\leq R\\}.$

For the rest of this section we fix $R$ and thus we may abbreviate
$\tau_{R}(\theta)$ by $\tau(\theta)$. Let
$\mathcal{P}_{R^{-1}}=\\{\tau(\theta):\theta\in\Pi_{R}\\}$. If
$\tau=\tau(\theta)\in\mathcal{P}_{R^{-1}}$ for some fixed $\tau$ then we
define $\theta_{\tau}=\theta$ for this choice of $\tau$. Let

$T^{\circ}_{\tau,0}=\\{x_{1}e_{3}(\theta_{\tau})+x_{2}e_{2}(\theta_{\tau})+x_{3}e_{1}(\theta_{\tau}):|x_{1}|\leq
R,|x_{2}|\leq R^{1/2},|x_{3}|\leq 1\\}.$

Notice that $T^{\circ}_{\tau,0}$ are planks that are dual to
$\tau(\theta_{\tau})$. We denote by $\mathbb{T}^{\circ}_{\tau}$ the collection
of translates of $T^{\circ}_{\tau,0}$ that cover $B(0,R)$. For a fixed small
constant $\delta>0$, we let $T_{\tau,0}=R^{\delta}T^{\circ}_{\tau,0}$, and
$\mathbb{T}_{\tau}:=\\{R^{\delta}T:T\in\mathbb{T}^{\circ}_{\tau}\\}$. For a
given plank in the covering $T\in\mathbb{T}_{\tau}$, we set $\tau(T):=\tau$.

######  Definition 8.1.

Fix $T\in\mathbb{T}_{\tau}$. We say that a function
$f_{T}:\mathbb{R}^{3}\rightarrow\mathbb{C}$ is a $T-$ function if
$\hat{f_{T}}$ is supported on $\tau(T)$ and

$\norm*{f_{T}}_{L^{\infty}(B(0,R)\setminus
T)}\lesssim_{\delta}R^{-1000}\norm*{f_{T}}_{2}.$

We are now ready to state our refined Strichartz inequality.

######  Theorem 8.2.

Let $\gamma:[a,b]\rightarrow S^{2}$ be a non-degenerate $C^{2}$ curve (i.e.
with $\det(\gamma,\gamma^{\prime},\gamma^{\prime\prime})$ non vanishing. As a
consequence let $B$ be such that

$|\det(\gamma,\gamma^{\prime},\gamma^{\prime\prime})|\geq B^{-1}$ (8.1)

and

$\norm*{\gamma}_{C^{2}([a,b])}\leq B.$ (8.2)

Let $R\geq 1$ and suppose that $f=\sum_{T\in\mathbb{W}}f_{T}$, where each
$f_{T}$ is a $T-$ function and

$\mathbb{W}\subseteq\bigcup_{\tau\in\mathcal{P}_{R^{-1}}}\mathbb{T}_{\tau}$

Also assume that for all $T,T^{\prime}\in\mathbb{W}$,

$\norm*{f_{T}}_{2}\sim\norm*{f_{T^{\prime}}}_{2}.$

Let $Y$ be a disjoint union of unit balls in $B(0,R)$ each of which intersects
$M$ sets of the form $2T$ where $T\in\mathbb{W}$. Then for $2\leq p\leq 6$, we
have

$\norm*{f}_{L^{p}(Y)}\lesssim_{\epsilon,\delta}R^{\epsilon}\left(\frac{M}{|\mathbb{W}|}\right)^{1/2-1/p}\left(\sum_{T\in\mathbb{W}}\norm*{f}_{p}^{2}\right)^{1/2}.$

###### Proof.

Assume that $[a,b]=[-1,1]$. We fix
$\epsilon\in(0,1/2),\delta_{0}=\epsilon^{100}$ and $\delta\in(0,\delta_{0})$.
We choose $R$ to be sufficiently large (the specific choice does not matter
for this presentation).

We assume that the following theorem holds for $K^{2}$ cubes instead of unit
cubes where $K=R^{\delta^{2}}$ and for all scales
$\tilde{R}=\frac{R}{K^{2}}=R^{1-2\delta^{2}}$ and for all curves $\gamma$
satisfying the two conditions (8.1) and (8.2), and for all $B\geq 1$. We shall
show that this assumption implies the theorem. As a consequence of inducting
on scales, this implies the full theorem, for an introduction to the concept
of inducting on scales we refer the reader to [Tao19].

For each $\tau\in\mathcal{P}_{R^{-1/2}}$, let
$\kappa=\kappa(\tau)\in\mathcal{P}_{K}^{-1}$ which minimizes
$|\theta_{\tau}-\theta_{\kappa}|$. For each such $\kappa$, let

$\displaystyle\square_{\kappa,0}$
$\displaystyle=\bigg{\\{}x_{1}\frac{(\gamma\times\gamma^{\prime})(\theta_{\kappa})}{|(\gamma\times\gamma^{\prime})(\theta_{\kappa})|}+x_{2}\frac{\gamma^{\prime}(\theta_{\kappa})}{|\gamma^{\prime}(\theta_{\kappa})|}+x_{3}\gamma(\theta_{\kappa}):$
$\displaystyle|x_{1}|\leq R^{1+\delta},|x_{2}|\leq R^{1+\delta}/K,|x_{3}|\leq
R^{1+\delta}/K^{2}\bigg{\\}}.$

and

$\displaystyle\mathcal{P}_{\kappa}$
$\displaystyle=\bigg{\\{}\square=a\gamma(\theta_{\kappa})+b\frac{\gamma^{\prime}(\theta_{\kappa})}{|\gamma^{\prime}(\theta_{\kappa})|}+\square_{\kappa,0}:$
$\displaystyle
a\in((1/10)R^{1+\delta}K^{-2})\mathbb{Z},b\in((1/10)R^{1+\delta}K^{-1})\mathbb{Z}\bigg{\\}}.$

We let
$\mathbb{P}=\bigcup_{\kappa\in\mathcal{P}_{K^{-1}}}\mathbb{P}_{\kappa}$. Given
any $\tau$ and corresponding $\kappa=\kappa(\tau)$, a routine computation
shows that

$|\langle\gamma\times\gamma^{\prime}(\theta_{\tau}),\gamma^{\prime}(\theta_{\kappa})\rangle|\leq
B^{-7}K^{-1}$ (8.3)

and

$|\langle(\gamma\times\gamma^{\prime})(\theta_{\tau}),\gamma(\theta_{\kappa})\rangle|\leq
B^{-7}K^{-2}.$ (8.4)

From here we can see that for each $T\in\mathbb{T}_{\tau}$ there are $\sim 1$
sets $\square\in\mathbb{P}_{\kappa(\tau)}$ with $T\cap 10\square\neq\phi$ and
moreover $T\subseteq 100\square$ whenever $T\cap 10\square\neq\phi$ and
moreover $T\subseteq 100\square$ whenever $T\cap 10\square\neq\phi$. For each
such $T$, let $\square=\square(T)\in\mathbb{P}_{\kappa}$ be some choice such
that $T\cap 10\square\neq\phi$, and let $\mathbb{W}_{\square}$ be the set of
$T$’s that associated to the set $\square$.

For each $\kappa$ and each $\square\in\mathbb{P}_{\kappa}$, let
$\\{\mathcal{Q}_{\square}\\}$ be a finitely overlapping cover of $100\square$
by translates of the ellipsoid given by:

$\left\\{x_{1}\gamma(\theta_{\kappa})+x_{2}\frac{\gamma^{\prime}(\theta_{\kappa})}{\absolutevalue{\gamma^{\prime}(\theta_{\kappa})}}+x_{3}\frac{\gamma\times\gamma^{\prime}(\theta_{\kappa})}{\absolutevalue{\gamma\times\gamma^{\prime}(\theta_{\kappa})}}:\absolutevalue{x_{1}}^{2}+(\absolutevalue{x_{2}}K^{-1})^{2}+(\absolutevalue{x_{3}}^{1/2}K^{-1})^{2}\leq\tilde{K}^{2}\right\\}.$

Now let
$\\{\eta_{\mathcal{Q}_{\square}}\\}_{{Q_{\square}\in\mathcal{Q}_{\square}}}$
be a smooth partition of unity such that on $\square$, we have that

$\sum_{{\eta}_{Q_{\square}\in\mathcal{Q}_{\square}}}\eta_{{Q}_{\square}}=1$

with the property that each $\eta_{\mathcal{Q}_{\square}}$ satisfies the
following two bounds:

$\norm*{\eta_{Q_{\square}}}\lesssim
1,\quad\norm*{\eta_{\mathcal{Q}_{\square}}}_{L^{\infty}(\mathbb{R}^{3}\setminus
T)}\lesssim R^{-1000}$

and

$\absolutevalue{\eta_{\mathcal{Q}_{\square}}(x)}\lesssim\mathrm{dist}(x,\mathcal{Q}_{\square})^{-1000},\quad\forall
x\in\mathbb{R}^{3}.$

we may also ensure that $\widehat{\eta_{Q_{\square}}}$ is supported in the
following set:

$\left\\{x_{1}\gamma(\theta_{\kappa})+x_{2}\frac{\gamma^{\prime}(\theta_{\kappa})}{\absolutevalue{\gamma^{\prime}(\theta_{\kappa})}}+x_{3}\frac{\gamma\times\gamma^{\prime}(\theta_{\kappa})}{\absolutevalue{\gamma\times\gamma^{\prime}(\theta_{\kappa})}}:\absolutevalue{x_{1}}\leq\tilde{K},\absolutevalue{x_{2}}\leq\tilde{K}K,\absolutevalue{x_{3}}\leq\tilde{K}K^{2}\right\\}.$

By dyadic pigeonholing we have

$\norm*{f}_{L^{p}(Y)}\lesssim\log
R\,\left\|\sum_{\square}\sum_{T\in\mathbb{W}_{\square}}\eta_{Y_{\square}}f_{T}\right\|_{L^{p}(Y)}+R^{-1000}\left(\sum_{T\in\mathbb{W}}\norm*{f_{T}}_{p}^{2}\right)^{1/2},$

where for each $\square$ we have that $Y_{\square}$ is a union of over a
smaller subset $Q_{\square}$ and each $Q_{\square}\subseteq Y_{\square}$
intersects a number of sets of the form $T$ (with $T\in\mathbb{W}_{\square}$
where the number of such intersections is contained in the set
$[M^{\prime}(\square),2M^{\prime}(\square))$ are constant.
$\square\in\mathbb{B}$ (upto a possible factor of 2).

We dyadic pigeonhole again to yield that

$\left\|\sum_{\square}\sum_{T\in\mathbb{W}_{\square}}\eta_{Y_{\square}}f_{T}\right\|_{L^{p}(Y)}\lesssim(\log
R)^{2}\left\|\sum_{\square\in\mathbb{B}}\sum_{T\in\mathbb{W}_{\square}}\eta_{Y_{\square}}f_{T}\right\|_{L^{p}(Y)},$

where now $\absolutevalue{\mathbb{W}_{\square}}$ and $M^{\prime}(\square)$ are
constant over $\square\in\mathbb{B}$ (once again, upto a factor of 2). We do
one more pigeonholing step

$\left\|\sum_{\square}\sum_{T\in\mathbb{W}_{\square}}\eta_{Y_{\square}}f_{T}\right\|_{L^{p}(Y)}\lesssim(\log
R)\left\|\sum_{\square\in\mathbb{B}}\sum_{T\in\mathbb{W}_{\square}}\eta_{Y_{\square}}f_{T}\right\|_{L^{p}(Y^{\prime})},$

where $Y^{\prime}$ is now a union over $K^{2}$ balls $Q\subseteq Y$ such that
each ball $2Q$ intersects a number of sets of the form $Y_{\square}$, where
the number of intersections is contained in the interval
$[M^{\prime\prime},2M^{\prime\prime})$. Let us fix such a $Q\subseteq
Y^{\prime}$. Now using the decoupling theorem for $C^{2}$ cones (discussion in
section 9.3) and applying Hölder’s inequality, we have

$\displaystyle\left\|\sum_{\square}\sum_{T\in\mathbb{W}_{\square}}\eta_{Y_{\square}}f_{T}\right\|_{L^{p}(Q)}$
$\displaystyle\lesssim_{\epsilon}(M^{\prime\prime})^{1/2-1/p}\left(\sum_{\square\in\mathbb{B}}\left\|\sum_{T\in\mathbb{W}_{\square}}\eta_{{Y}_{\square}}f_{T}\right\|_{L^{p}(2Q)}^{p}\right)^{1/p}$
$\displaystyle+R^{-900}\left(\sum_{T\in\mathbb{W}}\norm*{f_{T}}_{p}^{2}\right)^{1/2}.$

Now summing over all $Q$ gives us

$\displaystyle\norm*{f}_{L^{p}(Y)}$ $\displaystyle\lesssim(\log
R)^{O(1)}B^{100}K^{\epsilon/100}(M^{\prime\prime})^{1/2-1/p}$
$\displaystyle\times\left(\left\|\sum_{T\in\mathbb{W}_{\square}}\eta_{{Y}_{\square}}f_{T}\right\|_{L^{p}(Y_{\square})}^{p}\right)^{1/p}+R^{-800}\left(\sum_{T\in\mathbb{W}}\norm*{f_{T}}_{p}^{2}\right)^{1/2}.$

We will bound this by using the inductive assumption following the use of a
Lorentz rescaling map. For each $\theta\in[-1,1]$, define the Lorentz
rescaling map $L_{\theta}$ at $\theta$ by

$\displaystyle
L_{\theta}\left(x_{1}\gamma(\theta_{\kappa})+x_{2}\frac{\gamma^{\prime}(\theta_{\kappa})}{\absolutevalue{\gamma^{\prime}(\theta_{\kappa})}}+x_{3}\frac{\gamma\times\gamma^{\prime}(\theta_{\kappa})}{\absolutevalue{\gamma\times\gamma^{\prime}(\theta_{\kappa})}}\right)$
$\displaystyle=$ $\displaystyle
x_{1}\gamma(\theta_{\kappa})+Kx_{2}\frac{\gamma^{\prime}(\theta_{\kappa})}{\absolutevalue{\gamma^{\prime}(\theta_{\kappa})}}+K^{2}x_{3}\frac{\gamma\times\gamma^{\prime}(\theta_{\kappa})}{\absolutevalue{\gamma\times\gamma^{\prime}(\theta_{\kappa})}}$

For each $\square\in\mathbb{B}$ and $T\in\mathbb{W}_{\square}$, let
$g_{T}=f_{T}\circ L$, where $L=L_{\theta_{k}(\square)}$. Then it can be shown
that (as a consequence of a calculation in [GGG+22])

$\left\|\sum_{T\in\mathbb{W}_{\square}}f_{T}\right\|_{L^{p}(Y_{\square})}\leq
K^{3/p}\left\|\sum_{T\in\mathbb{W}_{\square}}g_{T}\right\|_{L^{p}(L^{-1}Y_{\square})}$

. A consequence of the inequalities $(\ref{1})$ and $(\ref{2})$ one can show
that for each $T\in\mathbb{W}_{\square}$, the set $L^{-1}(T)$ is equivalent to
a plank of length of
$\tilde{R}^{1+\delta}\times\tilde{R}^{1/2-\delta}\times\tilde{R}^{\delta}$,
where the longest direction parallel to
$L^{-1}(\gamma\times\gamma^{\prime})(\theta_{\tau(T)})$. This allows us to
reduce to our initial setup so that we can use our inductive assumption. In
[GGG+22] it is shown that

$\displaystyle L(\tau)$
$\displaystyle\subseteq\bigg{\\{}x_{1}\gamma(\theta_{\kappa})+x_{2}\frac{\gamma^{\prime}(\theta_{\kappa})}{\absolutevalue{\gamma^{\prime}(\theta_{\kappa})}}+x_{3}\frac{\gamma\times\gamma^{\prime}(\theta_{\kappa})}{\absolutevalue{\gamma\times\gamma^{\prime}(\theta_{\kappa})}}:$
$\displaystyle 1\leq\absolutevalue{x_{1}}\leq
2.01,\absolutevalue{x_{2}}\leq(1.01)\tilde{R}^{-1/2},\absolutevalue{x_{3}}\leq\tilde{R}^{-1}\bigg{\\}}.$

This neatly sets up the induction step. Inductively applying the theorem at
scale $\tilde{R}$ gives that

$\left\|\sum_{T\in\mathbb{W}_{\square}}f_{T}\right\|_{L^{p}(Y_{\square})}\lesssim_{\epsilon,\delta}R^{\epsilon}K^{-2\epsilon}\left(\frac{M^{\prime}}{\absolutevalue{\mathbb{W_{\square}}}}\right)^{1/2-1/p}\left(\sum_{T\in\mathbb{W}_{\square}}\norm*{f_{T}}_{p}^{2}\right)^{1/2}.$

Thus summing over $\square\in\mathbb{B}$ we have that

$\norm*{f}_{L^{p}(Y)}\lesssim_{\epsilon,\delta}B^{10^{10}/\epsilon}K^{-\epsilon}\left(\frac{M^{\prime}M^{\prime\prime}}{\absolutevalue{\mathbb{W}_{\square}}}\right)^{1/2-1/p}\left(\sum_{\square\in\mathbb{B}}\left(\sum_{T\in\mathbb{T\in\mathbb{W}_{\square}}}\norm*{f_{T}}_{p}^{2}\right)^{p/2}\right)^{1/p}$

By the dyadically constant property of $\norm*{f_{T}}_{p}$ (this is a
consequence of the final pigeonholing step), we note that this is

$\lesssim_{\epsilon,\delta}\left(\frac{M^{\prime}M^{\prime\prime}}{\absolutevalue{\mathbb{W}}}\right)^{1/2-1/p}\left(\frac{\absolutevalue{\mathbb{B}\mathbb{W_{\square}}}}{\absolutevalue{\mathbb{W}}}\right)^{1/p}\left(\sum_{T\in\mathbb{T\in\mathbb{W}_{\square}}}\norm*{f_{T}}_{p}^{2}\right)^{p/2}.$

We begin by showing that
$\frac{\absolutevalue{\mathbb{B}}\absolutevalue{\mathbb{W_{\square}}}}{\absolutevalue{\mathbb{W}}}\lesssim
1$, This is because

$\absolutevalue{W}=\sum_{T\in\mathbb{W}}1\geq\sum_{\square\in\mathbb{B}}\sum_{\begin{subarray}{c}T\in\mathbb{W}\\\
\square=\square(T)\end{subarray}}\gtrsim\absolutevalue{\mathbb{B}}\absolutevalue{\mathbb{W}}.$

Thus in order to complete the inductive argument we must show that
$M^{\prime}M^{\prime\prime}\lesssim M$. Let $Q\subseteq Y^{\prime}$ where $Q$
is a ball of radius $R^{1/2}$. We have, by the definition of $M$ that

$\displaystyle M$
$\displaystyle\gtrsim\sum_{\begin{subarray}{c}T\in\mathbb{W}\\\ 2T\cap
Q\neq\emptyset\end{subarray}}\sum_{\begin{subarray}{c}\square\in\mathbb{B}\\\
\square=\square(T)\end{subarray}}1$
$\displaystyle=\sum_{\square\in\mathbb{B}}\sum_{\begin{subarray}{c}\square\in\mathbb{B}\\\
\square=\square(T)\\\ 2T\cap Q\neq\emptyset\end{subarray}}1$
$\displaystyle\geq\sum_{\square\in\mathbb{B}}\sum_{\begin{subarray}{c}\square\in\mathbb{B}\\\
2T\cap Q\neq\emptyset\end{subarray}}1.$

Using this we perform the following computation (by beginning with recalling
the definition of $M^{\prime}$ and $M^{\prime\prime}$),

$\displaystyle M^{\prime}M^{\prime\prime}$
$\displaystyle\sim\sum_{\begin{subarray}{c}\square\in\mathbb{B}\\\
m(Y_{\square}\cap Q)>0\end{subarray}}M^{\prime}$
$\displaystyle\leq\sum_{\begin{subarray}{c}\square\in\mathbb{B}\\\
m(Y_{\square}\cap Q)>0\end{subarray}}\sum_{Q_{\square}\subseteq
Y_{\square}}M^{\prime}\frac{m(Q_{\square}\cap 2Q)}{m(Y_{\square}\cap 2Q)}$
$\displaystyle\sim\sum_{\begin{subarray}{c}\square\in\mathbb{B}\\\
m(Y_{\square}\cap Q)>0\end{subarray}}\sum_{Q_{\square}\subseteq
Y_{\square}}\sum_{\begin{subarray}{c}T\in\mathbb{W}_{\square}\\\
Q_{\square}\cap(1.5)T\neq\emptyset\end{subarray}}\frac{m(Q_{\square}\cap
2Q)}{m(Y_{\square}\cap 2Q)}$
$\displaystyle=\sum_{\begin{subarray}{c}\square\in\mathbb{B}\\\
m(Y_{\square}\cap
Q)>0\end{subarray}}\sum_{T\in\mathbb{W}_{\square}}\sum_{\begin{subarray}{c}Q_{\square}\subseteq
Y_{\square}\\\
Q_{\square}\cap(1.5)T\neq\emptyset\end{subarray}}\frac{m(Q_{\square}\cap
2Q)}{m(Y_{\square}\cap 2Q)}$
$\displaystyle\leq\sum_{\begin{subarray}{c}\square\in\mathbb{B}\\\
m(Y_{\square}\cap
Q)>0\end{subarray}}\sum_{\begin{subarray}{c}T\in\mathbb{W}_{\square}\\\
2T\cap\neq\emptyset\end{subarray}}\sum_{Q_{\square}\subseteq
Y_{\square}}\frac{m(Q_{\square}\cap 2Q)}{m(Y_{\square}\cap 2Q)}$
$\displaystyle\lesssim\sum_{\square\in\mathbb{B}}\sum_{\begin{subarray}{c}T\in\mathbb{W}_{\square}\\\
2T\cap Q\neq\emptyset\end{subarray}}1$ $\displaystyle\lesssim M,$

where in the second to last inequality we have used the observation that if
$Q_{\square}\cap 2Q\neq\emptyset$ and if $T\in\mathbb{W}_{\square}$ is such
that $Q_{\square}\cap(1.5)T\neq\emptyset$, then we must have that $2T\cap
Q\neq\emptyset$.

This closes the induction and completes the proof. ∎

##  9 Appendix

###  9.1 Proof of flat decoupling

We give a proof of Theorem 5.2. First, by a simple tiling argument and the
triangle and Hölder’s inequalities, we may assume $K=1$.

The idea is to use interpolation. The case $p=2$ follows from Plancherel’s
identity, and the case $p=\infty$ follows from the triangle and Hölder’s
inequalities. However, there are a few technical obstacles we need to
consider. First, it is not of the form of a general sublinear operator since
we have extra Fourier support conditions on each $f_{T}$. Also, on the right
hand side of the decoupling inequality we have a mixed norm. The full detail
is as follows.

###### Rigorous proof of flat decoupling.

For each $T\in\mathbb{T}$, let $\psi_{T}$ be a Schwartz function such that
$\widehat{\psi_{T}}$ is supported on $CT$ and equals $1$ on $T$. Define the
vector-input linear operator $L$ by

$L((f_{T})_{T\in\mathbb{T}})=\sum_{T\in\mathbb{T}}f_{T}*\psi_{T},$

where $f_{T}$ are arbitrary functions (with no Fourier support assumption) in
$L^{p}(\mathbb{R}^{3})$. Then by Plancherel’s identity, we have

$\norm*{L((f_{T})_{T\in\mathbb{T}})}_{L^{2}(\mathbb{R}^{3})}=\norm*{\norm*{f_{T}*\psi_{T}}_{L^{2}(\mathbb{R}^{3})}}_{\ell^{2}(T\in\mathbb{T})}\lesssim_{C}\norm*{\norm*{f_{T}}_{L^{2}(\mathbb{R}^{3})}}_{\ell^{2}(T\in\mathbb{T})},$
(9.1)

using Young’s inequality, since $\norm*{\psi_{T}}_{L^{1}(\mathbb{R}^{3})}\sim
1$. On the other hand, by the triangle and Hölder’s inequalities, we have

$\displaystyle\norm*{L((f_{T})_{T\in\mathbb{T}})}_{L^{\infty}(\mathbb{R}^{3})}$
$\displaystyle\leq(\\#\mathbb{T})^{1/2}\norm*{\norm*{f_{T}*\psi_{T}}_{L^{\infty}(\mathbb{R}^{3})}}_{\ell^{2}(T\in\mathbb{T})}$
$\displaystyle\lesssim_{C}(\\#\mathbb{T})^{1/2}\norm*{\norm*{f_{T}}_{L^{\infty}(\mathbb{R}^{3})}}_{\ell^{2}(T\in\mathbb{T})}.$

Thus, using the interpolation theorem for mixed normed spaces (see, for
instance, Theorem 2 of Section 7 of [BP61]), we have for every $2\leq
p\leq\infty$ that

$\norm*{L((f_{T})_{T\in\mathbb{T}})}_{L^{p}(\mathbb{R}^{3})}\lesssim_{C}(\\#\mathbb{T})^{\frac{1}{2}-\frac{1}{p}}\norm*{\norm*{f_{T}}_{L^{p}(\mathbb{R}^{3})}}_{\ell^{2}(T\in\mathbb{T})}.$
(9.2)

Lastly, given any $f_{T}$ Fourier supported on $T$, we have
$f_{T}=f_{T}*\psi_{T}$. Thus

$\norm*{\sum_{T\in\mathbb{T}}f_{T}}_{L^{p}(\mathbb{R}^{3})}=\norm*{\sum_{T\in\mathbb{T}}f_{T}*\psi_{T}}_{L^{p}(\mathbb{R}^{3})}\lesssim_{C}(\\#\mathbb{T})^{\frac{1}{2}-\frac{1}{p}}\norm*{\norm*{f_{T}}_{L^{p}(\mathbb{R}^{3})}}_{\ell^{2}(T\in\mathbb{T})}.$
(9.3)

∎

###  9.2 Proof of decoupling for $C^{2}$ curves

In this section, we roughly sketch the proof of Theorem 10 using the language
of [Dem20], and point out the changes we need to make when we only have
$C^{2}$ regularity. For simplicity of notation, we take $p=6$.

Remark. Unless otherwise stated, an uncoloured reference without a hyperlink
will mean a label in [Dem20].

Below and henceforth, for each Schwartz function
$f:\mathbb{R}^{2}\to\mathbb{C}$ and each interval $I\subseteq\mathbb{R}$, we
denote by $f_{I}$ the Fourier restriction of $f$ onto $I\times\mathbb{R}$,
namely, $\widehat{f_{I}}:=\widehat{f}1_{I\times\mathbb{R}}$.

######  Theorem 9.1 (Bourgain-Demeter decoupling for planar $C^{2}$ curves).

Let $\gamma:[0,1]\to\mathbb{R}$ be a $C^{2}$-curve such that
$|\gamma^{\prime\prime}(t)|\sim 1$. Let $\delta\in 4^{-\mathbb{N}}$ and let
$N_{\gamma}(\delta)$ be the (vertical) $\delta$-neighbourhood of the graph of
$\gamma$ over $[0,1]$, namely,

$N_{\gamma}(\delta):=\\{(t,\gamma(t)+s):t\in[0,1],|s|\leq\delta\\}.$ (9.4)

Let $\mathcal{I}_{\delta^{1/2}}$ be a tiling of $[0,1]$ into intervals of
length $\delta^{1/2}$. Let $C>0$ and denote by $D_{\gamma}(\delta)$ the
smallest constant such that the following inequality holds for all $f$ Fourier
supported on $N_{\gamma}(C\delta)$:

$\norm*{f}_{L^{6}(\mathbb{R}^{2})}\leq
D_{\gamma}(\delta)\norm*{\norm*{f_{I}}_{L^{6}(\mathbb{R}^{2})}}_{\ell^{2}(I\in\mathcal{I}_{\delta^{1/2}})}.$
(9.5)

Then the following holds: for every $\varepsilon>0$ there exists some
$C_{\varepsilon,\gamma,C}$ such that

$D_{\gamma}(\delta)\leq
C_{\varepsilon,\gamma,C}\delta^{-\varepsilon},\quad\forall\delta\in
4^{-\mathbb{N}}.$ (9.6)

The following corollary will be useful when the curve is given by a parametric
equation.

######  Corollary 9.2.

Let $\phi:[0,1]\to\mathbb{R}^{2}$ be a $C^{2}$-curve with nonzero velocity and
nonzero curvature. Let $\delta\in 4^{-\mathbb{N}}$ and let
$N^{\prime}_{\phi}(\delta)$ be the $\delta$-neighbourhood of the image of
$\phi$ in $\mathbb{R}^{2}$, namely,

$N^{\prime}_{\phi}(\delta):=\\{\phi(t)+s:t\in[0,1],|s|\leq\delta\\}.$ (9.7)

Let $\mathcal{T}_{\delta^{1/2}}$ be a covering of $N^{\prime}_{\phi}(C\delta)$
by $O(1)$-overlapping rectangles $T$ of dimensions $O(\delta^{1/2})\times
O(\delta)$, such that $\cup\mathcal{T}_{\delta^{1/2}}\subseteq
N^{\prime}_{\phi}(O(\delta))$. Then

$\norm*{f}_{L^{6}(\mathbb{R}^{2})}\lesssim_{\varepsilon}\delta^{-\varepsilon}\norm*{\norm*{\mathcal{R}_{T}f}_{L^{6}(\mathbb{R}^{2})}}_{\ell^{2}(T\in\mathcal{T}_{\delta^{1/2}})},\quad\forall\varepsilon>0,$
(9.8)

where $\mathcal{R}_{T}f:=(\hat{f}1_{T})^{\vee}$ and the implicit constant also
depends on $C$ and all implicit constants in the big $O$ notations above.

###### Proof.

Since $|\phi^{\prime}|=1$, by the implicit function theorem and a compactness
argument, we can cut the image of $\phi$ into $O(1)$ many pieces, such that
each piece after a suitable rotation becomes the graph of some $C^{2}$
function $\gamma$ with non-vanishing curvature. By the triangle and Cauchy-
Schwarz inequalities, it suffices to consider each such piece. However, after
rotation, we can view the tiling $\mathcal{T}_{\delta^{1/2}}$ as a $O(1)$
overlapping cover of $N_{\gamma}(\delta)$ by rectangles of dimensions
$\delta^{1/2}\times\delta$ that are all contained in $N_{\gamma}(O(\delta))$.
By a simple tiling argument, we may apply Theorem 9.1 to get our desired
conclusion, since decoupling is invariant under rotations. ∎

Now we come to the proof of Theorem 9.1. One of the most important
modifications is for Proposition 10.2.

Let $A>1$ be a constant, and denote by $\mathcal{F}(A)$ the family of all
$C^{2}$ functions $\gamma:[0,1]\to\mathbb{R}$ satisfying

$\sup_{t\in[0,1]}|\gamma^{\prime\prime}(t)|+\left(\inf_{t\in[0,1]}|\gamma^{\prime\prime}(t)|\right)^{-1}\leq
A.$ (9.9)

######  Proposition 9.3 (Parabolic rescaling for $\mathcal{F}(A)$).

Let $\sigma\in[\delta,1]\cap 4^{-\mathbb{N}}$. For each
$J\in\mathcal{I}_{\sigma^{1/2}}$, denote by $\mathcal{I}_{\delta^{1/2}}(J)$
the collection of all intervals $I\in\mathcal{I}_{\delta^{1/2}}$ that are also
contained in $J$.

Then there exists some $\tilde{\gamma}\in\mathcal{F}(A)$ such that for every
Schwartz function $f$ Fourier supported on $N_{\gamma}(\delta)$, we have

$\norm*{f_{J}}_{L^{6}(\mathbb{R}^{2})}\leq
D_{\tilde{\gamma}}(\sigma^{-1}\delta)\norm*{\norm*{f_{I}}_{L^{6}(\mathbb{R}^{2})}}_{\ell^{2}(I\in\mathcal{I}_{\delta^{1/2}}(J))}.$
(9.10)

###### Proof.

Write $J=[c,c+\sigma^{1/2}]$ and let

$\widehat{g}(t,s):=\widehat{f}(\sigma^{1/2}t+c,\sigma s),$ (9.11)

which is supported on $N_{\tilde{\gamma}}(\sigma^{-1}\delta)$ where

$\tilde{\gamma}(t):=\sigma^{-1}\gamma(\sigma^{1/2}t+c).$ (9.12)

If $\gamma\in\mathcal{F}(A)$, we can check that
$\tilde{\gamma}\in\mathcal{F}(A)$ as well. Applying the definition of
$D_{\tilde{\gamma}}(\sigma^{-1}\delta)$ to $g$, we have

$\norm*{g}_{L^{6}(\mathbb{R}^{2})}\leq
D_{\tilde{\gamma}}(\sigma^{-1}\delta)\norm*{\norm*{g_{\tilde{I}}}_{L^{6}(\mathbb{R}^{2})}}_{\ell^{2}(\tilde{I}\in\mathcal{I}_{\sigma^{-1/2}\delta^{1/2}})}.$
(9.13)

Reversing the change of variables and cancelling the same Jacobian factor on
both sides, we obtain the desired result. ∎

The next modification will be for Proposition 10.14.

Fix $I_{1}=[0,1/4]$ and $I_{2}=[1/2,1]$. Define $BD_{\gamma}(\delta)$ to be
the smallest constant such that the inequality

$\norm*{|f_{1}f_{2}|^{1/2}}_{L^{6}(\mathbb{R}^{2})}\leq
BD_{\gamma}(\delta)\prod_{i=1}^{2}\norm*{\norm*{f_{I}}_{L^{6}(\mathbb{R}^{2})}}^{1/2}_{\ell^{2}(I\in\mathcal{I}_{\delta^{1/2}}(I_{i}))}$
(9.14)

holds for all $f_{i}$ with Fourier support on
$N_{\gamma}(\delta)\cap(I_{i}\times\mathbb{R})$, $i=1,2$ respectively. For our
convenience, we also define

$D(\delta):=\sup_{\gamma\in\mathcal{F}(A)}D_{\gamma}(\delta),\quad
BD(\delta):=\sup_{\gamma\in\mathcal{F}(A)}BD_{\gamma}(\delta).$ (9.15)

We also have the following bilinear analogue of Proposition 9.3.

######  Proposition 9.4 (Bilinear parabolic rescaling for $\mathcal{F}(A)$).

Let $\sigma\in[\delta,1/16]\cap 4^{-\mathbb{N}}$. For $i=1,2$, intervals
$J_{i}\in\mathcal{I}_{\sigma^{1/2}}(I_{i})$ and every Schwartz function $f$
Fourier supported on $N_{\gamma}(\delta)$, we have

$\norm*{|f_{J_{1}}f_{J_{2}}|^{1/2}}_{L^{6}(\mathbb{R}^{2})}\leq
BD(\sigma^{-1}\delta)\prod_{i=1}^{2}\norm*{\norm*{f_{I}}_{L^{6}(\mathbb{R}^{2})}}^{1/2}_{\ell^{2}(I\in\mathcal{I}_{\delta^{1/2}}(J_{i}))}.$
(9.16)

###### Proof.

The proof is similar to that of Proposition 9.3, so we leave it as an
exercise. ∎

######  Lemma 9.5 (Sub-multiplicativity of decoupling constants).

For $\delta_{1},\delta_{2}\in 4^{-\mathbb{N}}$ we have

$D(\delta_{1}\delta_{2})\leq D(\delta_{1})D(\delta_{2}),\quad
BD(\delta_{1}\delta_{2})\leq BD(\delta_{1})BD(\delta_{2}).$ (9.17)

###### Proof.

The first inequality follows from Minkowski’s inequality and Proposition 9.3.
The second inequality follows from Hölder’s inequality, Minkowski’s inequality
and Proposition 9.4. ∎

######  Proposition 9.6 (Reduction to bilinear decoupling).

For each $\varepsilon>0$, there exists $C_{\varepsilon}$ such that for all
$\delta\in 4^{-\mathbb{N}}$ we have

$D(\delta)\leq
C_{\varepsilon}\delta^{-\varepsilon}\max_{\sigma\in[\delta,1]\cap
4^{-\mathbb{N}}}BD(\sigma).$ (9.18)

###### Proof.

Everything in the proof of Proposition 10.14 should remain unchanged (modulo
difference in notation), except that we now use Proposition 9.4 for the family
$\mathcal{F}(A)$. ∎

The content of Corollary 10.18 is unchanged.

######  Proposition 9.7 (Corollary 10.18).

Assume $\delta=4^{-2^{u}}$ for some $u\in\mathbb{N}$. Then for each $1\leq
s\leq u$,

$BD(\delta)\lesssim_{s,\varepsilon}\delta^{-\varepsilon-2^{-s-2}}\prod_{l=1}^{s}D(\delta^{1-2^{-l}})^{2^{l-s-1}}.$
(9.19)

###### Proof of Proposition 9.7.

All arguments in [Dem20] leading to Corollary 10.18 are the same, except the
paragraph right after (10.16). In this case, the longer side of each
$T\in\mathcal{F}_{I}$ is now pointing in the direction
$(\gamma^{\prime}(c_{I}),-1)$. Since $|\gamma^{\prime\prime}|\sim 1$, when
$|c_{I}-c_{\tilde{I}}|\sim 1$, the directions are also $\sim 1$ separated.
This ensures the bilinear Kakeya inequality in Theorem 10.21 is applicable. We
also remark here that the analogue of the rigorous proof of Theorem 11.10 also
remains unchanged. ∎

Now we come to the proof of Theorem 9.1.

###### Proof.

Iterating Proposition 9.7 in the same way as in [Dem20], we obtain
$BD(\delta)\lesssim_{\varepsilon}\delta^{-\varepsilon}$ when $\delta\in
4^{-2^{\mathbb{N}}}$.

We still need to prove $BD(\delta)\lesssim_{\varepsilon}\delta^{-\varepsilon}$
for a general $\delta\in 4^{-\mathbb{N}}$. This part was not given in [Dem20],
so we give it here for completeness. The idea is a simple induction on scales
argument.

Let $m\geq 1$ and assume we have proved
$BD(\delta)\lesssim_{\varepsilon}\delta^{-\varepsilon}$ for every
$\delta=4^{-n}$ where $0\leq n<2^{m}$. We need to show
$BD(\delta)\lesssim_{\varepsilon}\delta^{-\varepsilon}$ for every
$\delta=4^{-n}$ where $2^{m}<n<2^{m+1}$. Then $n-2^{m}\in(0,2^{m})$, and so by
Lemma 9.5, we have

$BD(\delta)=BD(4^{-2^{m}}4^{2^{m}-n})\leq
BD(4^{-2^{m}})BD(4^{2^{m}-n})\lesssim_{\varepsilon}4^{2^{m}\varepsilon}4^{(n-2^{m})\varepsilon}=\delta^{-\varepsilon},$
(9.20)

using both the result already established for $\delta\in 4^{-2^{\mathbb{N}}}$
and the induction hypothesis. ∎

###  9.3 Proof of decoupling for $C^{2}$ cones

In this section, we give a proof of Theorem 5.1. Although the most important
idea of the proof is already given in the appendix of [GGG+22], we would like
to provide more details. We also refer the reader to Section 12.2 of [Dem20]
as well as Section 8 of [BD15].

We first prove Part 1. For the covering part, let $\theta\in[0,1]$ be
arbitrary, we choose $\omega\in\delta^{1/2}\mathbb{Z}\cap[0,1]$ that is
closest to $\theta$. Then the result follows from Proposition 1.10. The
bounded overlap follows exactly from Part 2 of Lemma 5.5. (Note the planks
$A_{\omega,\delta^{1/2}}$ are denoted $C_{1}P_{\omega,\delta^{1/2}}$ there.)

####  9.3.1 Preliminary reductions

To prove Part 2, we use an iteration argument known as Pramanik-Seeger type
iteration [PS07]. For simplicity of notation we only consider $p=6$. By
splitting $\Gamma$ into $\lesssim|\log K|$ many pieces and rescaling each
piece back to $1\leq\xi_{3}\leq 2$, we may assume $K=1$, which leads to a loss
of $O(K^{2}|\log K|)$ by the triangle and Cauchy-Schwarz inequalities (so
$K^{10}$ is safe). Lastly, also by a simple tiling argument and triangle and
Cauchy-Schwarz inequalities, we may assume the collection
$\\{A_{\omega,\delta^{1/2}}\\}$ is disjoint.

####  9.3.2 Introducing intermediate scale

For $\delta\in 4^{-\mathbb{N}}$, we denote by $D(\delta)$ the smallest
constant such that for every family of Schwartz functions
$\\{f_{\omega}:\omega\in\Omega\\}$ where each $f_{\omega}$ has Fourier support
in $A_{\omega,\delta^{1/2}}$, we have

$\left\|\sum_{\omega\in\Omega}f_{\omega}\right\|_{L^{6}(\mathbb{R}^{3})}\leq
D(\delta)\norm*{\norm*{f_{\omega}}_{L^{6}(\mathbb{R}^{3})}}_{\ell^{2}(\omega\in\Omega)}.$
(9.21)

Our goal is to show that
$D(\delta)\lesssim_{\varepsilon}\delta^{-\varepsilon}$ for every
$\varepsilon>0$.

To this end, we define an intermediate scale

$\lambda:=\delta^{(1-\varepsilon)/2},$ (9.22)

and consider the lattice $\Sigma=\lambda\mathbb{Z}\cap[0,1]$. The choice of
$\lambda$ will be clear soon. We then recall (5.50) (with $K=1$):

$CP_{\sigma,\lambda}=\left\\{\sum_{i=1}^{3}\xi_{i}\mathbf{e}_{i}(\sigma):|\xi_{1}|\leq
C\lambda^{2},|\xi_{2}|\leq C\lambda,C^{-1}\leq\xi_{3}\leq C\right\\}.$ (9.23)

Since $A_{\omega,\delta^{1/2}}\subseteq C_{1}P_{\omega,\lambda}$, by Lemma
5.8, we know that the collection
$\\{C_{3}P_{\sigma,\lambda}:\sigma\in\Sigma\\}$ forms a $O(1)$-overlapping
cover of $\cup_{\omega\in\Omega}A_{\omega,\delta^{1/2}}$.

Recall the notation $\omega\prec\sigma$, which means that
$\omega-\sigma\in(\lambda/2,\lambda/2]$. Define for each $\sigma\in\Sigma$
that

$f_{\sigma}=\sum_{\omega\prec\sigma}f_{\omega},$ (9.24)

which has Fourier support in $C_{3}P_{\sigma,\lambda}$ by Part 2 of Lemma 5.8.
Also, we have

$\sum_{\omega\in\Omega}f_{\omega}=\sum_{\sigma\in\Sigma}f_{\sigma}.$ (9.25)

Thus, we may first apply the definition of $D(\lambda^{2})$ to get

$\left\|\sum_{\omega\in\Omega}f_{\omega}\right\|_{L^{6}(\mathbb{R}^{3})}\leq
D(\lambda^{2})\norm*{\norm*{f_{\sigma}}_{L^{6}(\mathbb{R}^{3})}}_{\ell^{2}(\sigma\in\Sigma)}.$
(9.26)

It remains to estimate each $\norm*{f_{\sigma}}_{L^{6}(\mathbb{R}^{3})}$.

####  9.3.3 Approximation by cylinder

We also encounter some technical difficulty here, namely, the curve
$\mathbf{e}_{3}$ is merely $C^{1}$ by Lemma 1.8 since the torsion function
$\uptau$ may not be differentiable. This will hinder applications of Theorem
9.1 as well as multiple Taylor expansions.

To deal with this issue, we will reparametrize $\mathbf{e}_{3}$; note that the
planks remain unchanged. Let us be more precise. Recall by Lemma 1.8 that
$\mathbf{e}^{\prime}_{3}=-\uptau e_{2}$. Put

$s=s(\theta)=\int_{0}^{\theta}-\frac{1}{\uptau(t)}dt,$ (9.27)

and redefine

$\tilde{\mathbf{e}}_{i}(\theta)=\mathbf{e}_{i}(s(\theta)),\quad i=1,2,3.$
(9.28)

In this way, we obtain the new Frenet frame formula:

$\tilde{\mathbf{e}}^{\prime}_{1}=-\uptau^{-1}\tilde{\mathbf{e}}^{\prime}_{2},\quad\tilde{\mathbf{e}}^{\prime}_{2}=\uptau^{-1}\tilde{\mathbf{e}}_{1}-\tilde{\mathbf{e}}_{3},\quad\tilde{\mathbf{e}}^{\prime}_{3}=\tilde{\mathbf{e}}_{2}.$
(9.29)

Thus $\tilde{\mathbf{e}}_{3}$ becomes $C^{2}$, at the cost that
$\tilde{\mathbf{e}}_{1}$ may not necessarily be $C^{2}$.

For simplicity, we abuse notation and still use $\mathbf{e}_{i}$ to mean
$\tilde{\mathbf{e}}_{i}$.

Fix $\sigma\in\Sigma$. For simplicity of notation we will assume without loss
of generality that

$\mathbf{e}_{1}(\sigma)=(1,0,0),\quad\mathbf{e}_{2}(\sigma)=(0,1,0),\quad\mathbf{e}_{3}(\sigma)=(0,0,1).$
(9.30)

Also write

$\mathbf{e}_{i}=(e_{i1},e_{i2},e_{i3}).$ (9.31)

Let $\varepsilon>0$ be arbitrary. Since we allow a loss of the form
$\delta^{-\varepsilon}$, we may cut $[0,1]$ into smaller intervals of length
$\delta^{\varepsilon}$. By the triangle and Cauchy-Schwarz inequalities, it
suffices to assume all angles
$\theta,\omega,\sigma\in[0,\delta^{\varepsilon}]$. For the same reason, we may
partition each $A_{\omega,\delta^{1/2}}$ and assume
$|\xi_{3}-1|\leq\delta^{\varepsilon}$ on $A_{\omega,\delta^{1/2}}$.

The key is the following geometric observation. It is more complicated than
the argument outlined in the proof of Theorem 11 in [GGG+22], since we have
chosen to perform the approximation directly, instead of first reducing to the
form $\mathbf{e}_{3}(t)=(1,t,\phi(t))$ for some $\phi\in C^{2}$.

######  Lemma 9.8.

For each $\omega\prec\sigma$, the plank $A_{\omega,\delta^{1/2}}$ is contained
in the $O(\delta)$ neighbourhood of the part of the cylinder

$\displaystyle R_{\sigma,\omega,\lambda}$
$\displaystyle:=\left\\{(e_{31}(\theta),e_{32}(\theta),\xi_{3}):|\theta-\omega|\lesssim\delta^{1/2},|\xi_{3}-1|\lesssim\delta^{\varepsilon}\right\\}.$
(9.32)

###### Proof of Lemma 9.8.

Let $\xi=\sum_{i=1}^{3}\xi_{i}\mathbf{e}_{i}(\omega)\in
A_{\omega,\delta^{1/2}}$. It suffices to find $\theta,\xi^{\prime}_{3}$ such
that
$|\xi-(e_{31}(\theta),e_{32}(\theta),\xi^{\prime}_{3})|\lesssim\lambda^{2}\delta^{\varepsilon}$.

We Taylor expand $\mathbf{e}_{3}$ with respect to the centre $\omega$. Using
(9.29), $|\xi_{1}|\lesssim\delta$ and $|\omega-\theta|\leq\delta^{1/2}$, we
get

$\displaystyle\sum_{i=1}^{3}\xi_{1}\mathbf{e}_{i}(\omega)-(e_{31}(\theta),e_{32}(\theta),\xi_{3}^{\prime})$
$\displaystyle=$
$\displaystyle\Big{(}\xi_{2}e_{21}(\omega)+(\xi_{3}-1)e_{31}(\omega)+e_{21}(\theta-\omega)+O(\delta),$
$\displaystyle\xi_{2}e_{22}(\omega)+(\xi_{3}-1)e_{32}(\omega)+e_{22}(\omega)(\theta-\omega)+O(\delta)$
$\displaystyle\xi_{2}e_{23}(\omega)+\xi_{3}e_{33}(\omega)-\xi_{3}^{\prime}+O(\delta)\Big{)}.$
(9.33)

We thus choose $\xi^{\prime}_{3}=\xi_{2}e_{23}(\omega)+\xi_{3}e_{33}(\omega)$,
which satisfies $|\xi^{\prime}_{3}-1|\lesssim\delta^{\varepsilon}$ since
$|e_{33}(\omega)-1|=|e_{33}(\omega)-e_{33}(\sigma)|\lesssim\delta^{\varepsilon}$.
This makes the third coordinate of (9.33) of the order $O(\delta)$.

Next, we put

$\theta=\omega+\frac{\xi_{2}e_{22}(\omega)-(\xi_{3}-1)e_{32}(\omega)}{e_{22}(\omega)},$
(9.34)

which satisfies $|\theta-\omega|\lesssim\delta^{1/2}$. Indeed, the denominator
is $\sim 1$ since $e_{22}(\omega)\sim 1$. For the numerator, we have
$|\xi_{2}|\lesssim\delta^{1/2}$ and $|\xi_{3}-1|\leq\delta^{\varepsilon}$, and
using Lemma 9.30 and (9.29), we get

$|e_{32}(\omega)|=|e_{32}(\omega)-e_{32}(\sigma)|\lesssim|\omega-\sigma|\leq\lambda,$
(9.35)

and so $|\theta-\omega|\lesssim\delta^{\varepsilon}\lambda\leq\delta^{1/2}$.
This makes the second coordinate of (9.33) of the order $O(\delta)$.

It remains to estimate the first coordinate. Using (9.34), the first
coordinate of (9.33) is within $O(\delta)$ of

$e_{22}(\omega)^{-1}(\xi_{3}-1)(e_{31}(\omega)e_{22}(\omega)-e_{21}(\omega)e_{32}(\omega)).$
(9.36)

Now we recall $|\omega-\sigma|\leq\lambda$, and use (9.29), Taylor expansion
and (9.30) to get

$e_{31}(\omega)=e_{31}(\sigma)+e_{21}(\sigma)(\omega-\sigma)+O(\lambda^{2})=O(\lambda^{2}).$
(9.37)

Similarly, we also have $e_{32}(\omega)=O(\lambda^{2})$. Thus (9.36) is
bounded above by $O(\lambda^{2})$, and so the first coordinate of (9.33) is of
the order $O(\lambda^{2}\delta^{\varepsilon})=O(\delta)$. ∎

####  9.3.4 Applying decoupling for $C^{2}$ curves

The most important consequence of Lemma 9.8 is that if we project every plank
$R_{\sigma,\omega,\lambda}$ onto $\mathbf{e}_{3}(\sigma)^{\perp}$, then we are
essentially in the case of decoupling for $C^{2}$ curves as in Corollary 9.2.
This is a general philosophy of decoupling called the projection principle,
namely, the decoupling constant for a family of subsets is dominated by the
decoupling constant for the same family of subsets under any fixed orthogonal
projection. Using Corollary 9.2 and noting the $O(1)$ overlap, the result is
that

$\left\|f_{\sigma}\right\|_{L^{6}(\mathbb{R}^{3})}\lesssim_{\varepsilon}\delta^{-\varepsilon^{2}}\norm*{\norm*{f_{\omega}}_{L^{6}(\mathbb{R}^{3})}}_{\ell^{2}(\omega\prec\sigma)}.$
(9.38)

The choice of $\delta^{-\varepsilon^{2}}$ but not $\delta^{-\varepsilon}$ is
necessary for the bootstrap inequality in the next subsection to give the
desired conclusion.

We give more details of the proof of (9.38) here. Fix $z\in\mathbb{R}$. For
each $\omega\prec\sigma$, define $g_{z,\omega}:\mathbb{R}^{2}\to\mathbb{C}$ by

$\widehat{g_{z,\omega}}(\xi_{1},\xi_{2}):=\int_{\mathbb{R}}\widehat{f_{\omega}}(\xi_{1},\xi_{2},\xi_{3})e(z\xi_{3})d\xi_{3}.$
(9.39)

Then by Lemma 9.8, $\widehat{g_{z,\omega}}$ is supported on the $O(\delta)$
neighbourhood of

$\\{(e_{31}(\theta),e_{32}(\theta)):|\theta-\omega|\lesssim\delta^{1/2}\\},$
(9.40)

which has $O(1)$ overlap, by the same proof of Lemma 5.5 as in Section 5.11.
The planar curve $(e_{31}(\theta),e_{32}(\theta))$ is $C^{2}$ and has
curvature comparable to $1$. Thus we may apply Corollary 9.2 to the family
$\\{g_{z,\omega}:\omega\prec\sigma\\}$ to obtain

$\left\|\sum_{\omega\prec\sigma}g_{z,\omega}\right\|_{L^{6}(\mathbb{R}^{2})}\lesssim_{\varepsilon}\delta^{-\varepsilon^{2}}\norm*{\norm*{g_{z,\omega}}_{L^{6}(\mathbb{R}^{2})}}_{\ell^{2}(\omega\prec\sigma)}.$
(9.41)

The desired inequality (9.38) then follows by Fubini’s theorem and Minkowski’s
inequality.

####  9.3.5 Iteration

Combining (9.26) and (9.38), we thus get

$\left\|\sum_{\omega\in\Omega}f_{\omega}\right\|_{L^{6}(\mathbb{R}^{3})}\leq
D(\lambda^{2})C_{\varepsilon}\delta^{-\varepsilon^{2}}\norm*{\norm*{f_{\omega}}_{L^{6}(\mathbb{R}^{3})}}_{\ell^{2}(\omega\in\Omega)},\quad\forall\varepsilon>0,$
(9.42)

which, by definition of $D(\delta)$, implies the following bootstrap
inequality:

$D(\delta)\leq
D(\delta^{1-\varepsilon})C_{\varepsilon}\delta^{-\varepsilon^{2}},\quad\forall\varepsilon>0.$
(9.43)

It suffices to iterate this bootstrap inequality. For $\delta\geq 1/2$ we
trivially have $D(\delta)\sim 1$. If $\delta<1/2$, we let $N$ be the largest
integer such that $\delta^{(1-\varepsilon)^{N}}<1/2$, whence
$N\lesssim\varepsilon^{-1}\log\log\delta^{-1}$, and

$D(\delta)\leq
D(\delta^{(1-\varepsilon)^{N+1}})C_{\varepsilon}^{N}\delta^{-\varepsilon^{2}(1+(1-\varepsilon)+\cdots+(1-\varepsilon)^{N})}\lesssim_{\varepsilon}\delta^{-2\varepsilon}.$
(9.44)

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