$\mathcal{O}\left(r^{-3}|p|^{2}\right)\times\int_{\mathbb{R}^{2}}(|u-x|+|p|)\mathsf{1}_{u\in\mathrm{B}_{2r},\mathrm{dist}(u,\partial\mathrm{B}_{2r})\geq|p|}|K_{\ell}(x-u)|\mathtt{d}u=\mathcal{O}\left(\frac{|p|^{2}}{r^{3}\ell}\right)+\mathcal{O}\left(\frac{|p|^{2}d}{r^{3}}\right),$

and since $d\geq\bar{d}\geq\ell^{-1}$ the second error term is the largest
one.

Inserting those controls into (A.15), taking the expectation, returning to
(A.14), and keeping only the dominant error term (using that
$d\geq\bar{d}=\sqrt{r\ell^{-1}}$, we have: $\frac{1}{r\ell
d^{2}}\geq\frac{d}{r^{3}}\geq\frac{1}{r^{2}\ell d}$), we obtain as desired:

$\mathbb{E}\left[\left(\mathcal{A}(x+p)-\mathcal{A}(x)\right)\mathsf{1}_{|p|\leq\frac{d}{2}}\right]=\mathcal{O}\left(\frac{1}{r\ell
d^{2}}\right)+\mathcal{O}(r^{-1})\times\mathbb{E}\left[|\mathrm{p}_{0}|\mathsf{1}_{|\mathrm{p}_{0}|\geq\frac{d}{4}}\right].$

∎

#### Analysis outside $\mathrm{B}_{2r}$.

######  Claim A.7.

If $x\notin\mathrm{B}_{2r}$, with $d:=|x|-2r\geq\bar{d}$, we have:

$\mathbb{E}\left[\left(\mathcal{A}(x+p)-\mathcal{A}(x)\right)\mathsf{1}_{|\mathrm{p}_{x}-\mathrm{p}_{0}|\leq{\frac{d}{2}}}\right]=\mathcal{O}\left(\frac{1}{r\ell
d^{2}}\right)+\mathcal{O}(r^{-1})\times\mathbb{E}\left[{|\mathrm{p}_{0}|\mathsf{1}_{|\mathrm{p}_{0}|\geq\frac{d}{4}}}\right].$

###### Proof of Claim A.7.

The analysis here is similar to, and somewhat simpler than the previous case.
Since $|p|\leq{\frac{d}{2}}$ the points $x$ and $x+p$ are both outside
$\mathrm{B}_{2r}$ hence the difference
$\widehat{j_{r}}(x+p)-\widehat{j_{r}}(x)$ is exactly $0$ because both terms
are $0$. On the other hand, we still have:

$\widehat{j_{r}}\ast K_{\ell}(x+p)-\widehat{j_{r}}\ast
K_{\ell}(x)=\int_{\mathbb{R}^{2}}\left(\widehat{j_{r}}(u+p)-\widehat{j_{r}}(u)\right)K_{\ell}(x-u)\mathtt{d}u,$

so in order to control the difference
$\widehat{j_{r}}(u+p)-\widehat{j_{r}}(u)$ for $u\in\mathbb{R}^{2}$ we will
again distinguish three cases:

  * •

If $u$ is outside $\mathrm{B}_{2r}$ and
$\mathrm{dist}(u,\partial\mathrm{B}_{2r})\geq|p|$ then both terms are $0$ and
so is their difference, which we can certainly write as
$\widehat{j_{r}}(u+p)-\widehat{j_{r}}(u)=0=\nabla\widehat{j_{r}}(u)\cdot p+0$
since $\nabla\widehat{j_{r}}(u)=0$.

  * •

If $\mathrm{dist}(u,\partial\mathrm{B}_{2r})\leq|p|$ we can always use the
Lipschitz bound on $\widehat{j_{r}}$, which controls the difference by
$\mathcal{O}(r^{-1})\times|p|$. For convenience, we write this as:

$\widehat{j_{r}}(u+p)-\widehat{j_{r}}(u)=\nabla\widehat{j_{r}}(u)\cdot
p+\mathcal{O}(r^{-1})\times|p|,$

where in fact both terms in the right-hand side are
$\mathcal{O}(r^{-1})\times|p|$.

  * •

Finally, if $u$ is outside $\mathrm{B}_{2r}$ and
$\mathrm{dist}(u,\partial\mathrm{B}_{2r})\geq|p|$, we can use a second-order
Taylor’s expansion of $\widehat{j_{r}}$ and write:

$\widehat{j_{r}}(u+p)-\widehat{j_{r}}(u)=\nabla\widehat{j_{r}}(u)\cdot
p+\mathcal{O}(r^{-2})\times|p|^{2}.$

Using these informations we get:

$\int_{\mathbb{R}^{2}}\left(\widehat{j_{r}}(u+p)-\widehat{j_{r}}(u)\right)K_{\ell}(x-u)\mathtt{d}u=\int\nabla\widehat{j_{r}}(u)K_{\ell}(x-u)\mathtt{d}u\cdot
p\\\
+\mathcal{O}(r^{-1})\times|p|\times\int_{\mathbb{R}^{2}}|K_{\ell}(x-u)|\mathsf{1}_{\mathrm{dist}(u,\partial\mathrm{B}_{2r})\leq|p|}\mathtt{d}u+\mathcal{O}(r^{-2}|p|^{2})\times\int_{\mathbb{R}^{2}}|K_{\ell}(x-u)|\mathsf{1}_{u\in\mathrm{B}_{2r}}\mathtt{d}u.$

As in (A.13), (A.14), when taking the expectation, the term of first order in
$p$ vanishes - up to an error controlled as above by
$\mathcal{O}(r^{-1})\times\mathbb{E}\left[|\mathrm{p}_{0}|\mathsf{1}_{|\mathrm{p}_{0}|\geq\frac{d}{4}}\right]$).
The other error terms are handled as in the previous step: on the one hand,
using Claim A.3, we have:

$\mathcal{O}(r^{-1})\times|p|\times\int_{\mathbb{R}^{2}}|K_{\ell}(x-u)|\mathsf{1}_{\mathrm{dist}(u,\partial\mathrm{B}_{2r})\leq|p|}\mathtt{d}u=\mathcal{O}(r^{-1})\times|p|\times\mathcal{O}\left(\frac{|p|}{\ell
d^{2}}\right)=\mathcal{O}\left(\frac{|p|^{2}}{r\ell d^{2}}\right),$

whose expectation is simply $\mathcal{O}\left(\frac{1}{r\ell d^{2}}\right)$,
and on the other hand, using Claim A.4:

$\mathcal{O}(r^{-2}|p|^{2})\times\int_{\mathbb{R}^{2}}|K_{\ell}(x-u)|\mathsf{1}_{u\in\mathrm{B}_{2r}}\mathtt{d}u=\mathcal{O}(r^{-2}|p|^{2})\times\mathcal{O}\left(\frac{1}{\ell
d}\right)=\mathcal{O}\left(\frac{|p|^{2}}{\ell dr^{2}}\right),$

whose expectation is smaller than the previous error term since $r\geq d$. ∎

#### Application: proof of Lemma A.1

Combining Claim A.2 with Claim A.6 (for $x\in\mathrm{B}_{2r}$) or Claim A.7
(for $x\notin\mathrm{B}_{2r}$), we obtain that:

$\mathcal{B}_{3}(x)\preceq\frac{1}{r\ell
d^{2}}+{r^{-1}\times\mathbb{E}\left[|\mathrm{p}_{0}|\mathsf{1}_{|\mathrm{p}_{0}|\geq\frac{d}{4}}\right]}.$

It now remains to compute the sum over
$\\{x\in\mathcal{L}\setminus\\{0\\},d\geq\bar{d},|x|\leq 100r\\}$.

The first sum is comparable to an integral:

$\sum_{x\in\mathcal{L}\setminus\\{0\\},d\geq\bar{d},|x|\leq
100r}\frac{1}{r\ell d^{2}}\preceq\frac{1}{\ell
r}\times\int_{u=\bar{d}}^{r}\frac{1}{u^{2}}\times
r\mathtt{d}u=\mathcal{O}\left(\frac{1}{\ell\bar{d}}\right).$

The second sum can be compared to:

$\sum_{x\in\mathcal{L}\setminus\\{0\\},d\geq\bar{d},|x|\leq
100r}r^{-1}\times\mathbb{E}\left[|\mathrm{p}_{0}|\mathsf{1}_{|\mathrm{p}_{0}|\geq\frac{d}{4}}\right]\preceq\int_{s\geq\frac{\bar{d}}{4}}r^{-1}\times\mathbb{E}\left[{|\mathrm{p}_{0}|\mathsf{1}_{|\mathrm{p}_{0}|\geq
s}}\right]\times
r\mathtt{d}s\preceq\mathbb{E}\left[|\mathrm{p}_{0}|^{2}\mathsf{1}_{|\mathrm{p}_{0}|\geq\frac{\bar{d}}{4}}\right].$

This concludes the proof of Lemma A.1. ∎

###  A.2 Close, but not too close to the boundary - Preparation

From now on, we will assume that $0<d\leq\bar{d}$ (recall that $d=d(x)$
denotes the distance from $x$ to the boundary $\partial\mathrm{B}_{2r}$ and
that $\bar{d}$ is a lengthscale chosen in (A.1)). We will often use that this
implies that $r\leq|x|\leq 3r$.

#### Preliminaries.

We separate the characteristic function of $\mathrm{p}_{x}-\mathrm{p}_{0}$
into real and imaginary parts and write:

$\varphi_{\mathrm{p}_{x}-\mathrm{p}_{0}}(t)-1=\mathbb{E}\left[e^{-it\cdot(\mathrm{p}_{x}-\mathrm{p}_{0}})-1\right]=:\Psi^{r}_{x}(t)-i\Psi^{i}_{x}(t),$

where $\Psi^{r}_{x},\Psi^{i}_{x}$ are the following functions (when the
lattice point $x$ that we are considering is unambiguous, we will simply write
$p$ instead of $\mathrm{p}_{x}-\mathrm{p}_{0}$):

$\Psi^{r}_{x}(t):=\mathbb{E}\left[\cos(t\cdot
p)-1\right],\quad\Psi^{i}_{x}(t):=\mathbb{E}\left[\sin(t\cdot p)\right].$

Clearly $\Psi^{r}_{x}$ is even, $\Psi^{i}_{x}$ is odd. We now record some
properties of $\Psi^{r}_{x},\Psi^{i}_{x}$ which will be needed later.

######  Lemma A.8.

With implicit multiplicative constants depending only on the distribution of a
single perturbation, we have:

  * •

For $|t|\leq 1$:

$|\Psi^{r}_{x}(t)|+|\Psi^{i}_{x}(t)|=\mathcal{O}(|t|^{2}),\quad|\nabla\Psi^{r}_{x}(t)|+|\nabla\Psi^{i}_{x}(t)|=\mathcal{O}(|t|),\quad|\mathrm{Hess}\Psi^{r}_{x}(t)|+|\mathrm{Hess}\Psi^{i}_{x}(t)|=\mathcal{O}(1),$
(A.16)

  * •

For $|t|\geq 1$:

$|\Psi^{r}_{x}(t)|+|\Psi^{i}_{x}(t)|=\mathcal{O}(1),\quad|\nabla\Psi^{r}_{x}(t)|+|\nabla\Psi^{i}_{x}(t)|=\mathcal{O}(1),\quad\quad|\mathrm{Hess}\Psi^{r}_{x}(t)|+|\mathrm{Hess}\Psi^{i}_{x}(t)|=\mathcal{O}(1).$
(A.17)

  * •

For $\Psi^{r}_{x}$, working in polar coordinates $(\rho,\theta)$:

$\left|\frac{\partial}{\partial\theta}\Psi^{r}_{x}(\rho,\theta)\right|=\mathcal{O}(\rho^{2}).$
(A.18)

###### Proof of Lemma A.8.

For $t$ near $0$, we can expand $e^{-it\cdot(\mathrm{p}_{x}-\mathrm{p}_{0})}$
as
$1-it\cdot(\mathrm{p}_{x}-\mathrm{p}_{0})+\mathcal{O}\left(\frac{|t|^{2}}{2}(\mathrm{p}_{x}-\mathrm{p}_{0})^{2}\right)$,
which gives
$|\varphi_{\mathrm{p}_{x}-\mathrm{p}_{0}}(t)-1|=\mathcal{O}(|t|^{2})$ because
$(\mathrm{p}_{x}-\mathrm{p}_{0})$ is centered and has finite second moment,
and we proceed similarly for the first and second derivatives (using
expansions to lower order each time). This yields the bounds (A.16) on
$\Psi^{r}_{x},\Psi^{i}_{x}$.

For $|t|$ large, we can use the fact that $\mathrm{p}_{x}-\mathrm{p}_{0}$ has
a second moment to see that $\Psi^{r}_{x},\Psi^{i}_{x}$ as well as their first
and second derivatives are bounded, which yields (A.17).

It remains to prove (A.18). Working in polar coordinates
$t=(\rho\cos\theta,\rho\sin\theta)$, and letting $P$ be the distribution of
$\mathrm{p}_{x}-\mathrm{p}_{0}$, we can write:

$\Psi^{r}_{x}(\rho,\theta)=\int_{s>0}\int_{u\in[0,2\pi]}\cos\left(\rho\cos(\theta-u)s\right)s\mathtt{d}P(s,u),$

and taking a derivative with respect to $\theta$ yields (using the inequality
$|\sin x|\leq|x|$ in the last step):

$\frac{\partial}{\partial\theta}\Psi^{r}_{x}(\rho,\theta)=-\int_{s>0}\int_{u}\rho
s\sin(\theta-u)\sin\left(\rho\cos(\theta-u)s\right)s\mathtt{d}P(s,u)\preceq\rho^{2}\int_{s>0}\int_{u}s^{3}\mathtt{d}P(s,u)=\mathcal{O}(\rho^{2}),$

because $|\mathrm{p}_{x}-\mathrm{p}_{0}|$ has a finite second moment hence
$\iint s^{3}\mathtt{d}P(s,u)$ is finite. Note that in this last computation,
if $\rho\geq 1$, it would be beneficial to not use $|\sin x|\leq|x|$ and to
simply write:

$\frac{\partial}{\partial\theta}\Psi^{r}_{x}(\rho,\theta)\preceq\rho\iint
s^{2}\mathtt{d}P(s,u)=\mathcal{O}(\rho),$

however, using the $\mathcal{O}(\rho^{2})$ bound everywhere will be enough for
us and this way we avoid distinguishing more cases. ∎

Denote by $\Im$ the imaginary part of a complex number. We will use later the
following slight improvement:

######  Lemma A.9 (A more accurate bound for $\Psi^{i}_{x}$).

For $t\in\mathbb{R}^{2}$ let $\mathcal{K}^{i}_{x}(t):=\Im\left[e^{-it\cdot
p}+it\cdot p\right]$. We have:

$|\mathcal{K}^{i}_{x}(t)|\preceq\min\left(|t||p|,|t|^{3}|p|^{3}\right),\quad|\nabla\mathcal{K}^{i}_{x}(t)|\preceq\min\left(|p|,|t|^{2}|p|^{3}\right),\quad|\mathrm{Hess}\mathcal{K}^{i}_{x}(t)|\preceq\min({|p|^{2}},|t||p|^{3}).$
(A.19)

###### Proof of Lemma A.9.

This follows from Taylor’s expansions, and by separating cases $|t||p|\leq 1$,
$|t||p|\leavevmode\nobreak\ \geq\leavevmode\nobreak\ 1$. ∎

Since $p=\mathrm{p}_{x}-\mathrm{p}_{0}$ is centered, we have
$\Psi^{i}_{x}(t)=-\mathbb{E}[\mathcal{K}^{i}_{x}(t)]$ for all $t$ and thus the
bounds of (A.19) translate into bounds for $\Psi^{i}_{x}$ (as well as its
first and second derivatives), which slightly improve on (A.16) because by the
dominated convergence theorem, as $t\to 0$:

${\mathbb{E}\left[\min\left(|t||p|,|t|^{3}|p|^{3}\right)\right]=o(|t|^{2}),\quad\mathbb{E}\left[\min\left(|p|,|t|^{2}|p|^{3}\right)\right]=o(|t|)},\quad\mathbb{E}\left[\min(1,|t||p|^{3})\right]=o(1).$

#### Decomposition on Fourier side.

Recall that:

$\mathcal{B}_{3}(x)=\int_{t\in\mathbb{R}^{2}}{e^{-it\cdot
x}}j_{r}(t)\left(\varphi_{\mathrm{p}_{x}-\mathrm{p}_{0}}(t)-1\right)(1-\chi_{\ell}(t))\mathtt{d}t\\\
=\int_{t\in\mathbb{R}^{2}}\left(\cos(t\cdot x)-i\sin(t\cdot
x)\right)j_{r}(t)\left(\Psi^{r}_{x}(t)-i\Psi^{i}_{x}(t)\right)(1-\chi_{\ell}(t))\mathtt{d}t.$

Since the quantity we ultimately want to bound, namely the rescaled number
variance, is real, we can focus on controlling the real part of
$\mathcal{B}_{3}(x)$. We have:

$\Re(\mathcal{B}_{3}(x))=\int_{t\in\mathbb{R}^{2}}\cos(t\cdot
x)j_{r}(t)\Psi^{r}_{x}(t)(1-\chi_{\ell}(t))\mathtt{d}t-\int_{t\in\mathbb{R}^{2}}\sin(t\cdot
x)j_{r}(t)\Psi^{i}_{x}(t)(1-\chi_{\ell}(t))\mathtt{d}t.$

We denote the two integrals in the right-hand side by
$\mathcal{B}^{r},-\mathcal{B}^{i}$ (they are both real but one contains
$\Psi^{r}_{x}$, the other contains $\Psi^{i}_{x}$). Since $1-\chi_{\ell}$
vanishes on $\\{|t|\leq r^{-1}\\}$, we may insert the second order asymptotics
for the Bessel function $\mathrm{J}_{1}^{2}$ (see (2.9)) and write:

$\mathcal{B}^{r}(x)=\int_{t\in\mathbb{R}^{2}}\cos(t\cdot
x)\frac{\mathrm{J}_{1}^{2}(rt)}{|t|^{2}}\Psi^{r}_{x}(t)(1-\chi_{\ell}(t))\mathtt{d}t=I^{r}(x)+II^{r}(x)+III^{r}(x),$
(A.20)

where we set:

$I^{r}(x):=a\int_{t\in\mathbb{R}^{2}}\cos(t\cdot
x)\frac{1}{r|t|^{3}}\sin^{2}\left(r|t|-\frac{\pi}{4}\right)\Psi^{r}_{x}(t)(1-\chi_{\ell}(t))\mathtt{d}t,\\\
II^{r}(x):=b\int_{t\in\mathbb{R}^{2}}\cos(t\cdot
x)\frac{1}{r^{2}|t|^{4}}\cos(2r|t|)\Psi^{r}_{x}(t)(1-\chi_{\ell}(t))\mathtt{d}t,\\\
III^{r}(x):=\int_{t\in\mathbb{R}^{2}}\mathcal{O}\left(\frac{1}{r^{3}|t|^{5}}\right)\Psi^{r}_{x}(t)(1-\chi_{\ell}(t))\mathtt{d}t.$
(A.21)

The term $\mathcal{B}^{i}(x)$ has a similar expression, with $\sin(t\cdot x)$
instead of $\cos(t\cdot x)$ and $\Psi^{i}_{x}$ instead of $\Psi^{r}_{x}$, and
we decompose it accordingly into a sum of three terms denoted by
$I^{i}(x),II^{i}(x),III^{i}(x)$. For convenience, we will refer to the terms
$I^{r}(x),I^{i}(x)$ as “$I(x)$” and similarly for $II(x)$, $III(x)$.

#### Rewriting $I(x),II(x)$ and bounding $III(x)$.

###### Rewriting $I(x)$.

By elementary trigonometry, we have:

$\sin^{2}\left(r|t|-\frac{\pi}{4}\right)=\frac{1}{2}\left(1-\cos\left(2r|t|-\frac{\pi}{2}\right)\right)=\frac{1}{2}-\frac{1}{2}\sin\left(2r|t|\right),$

and moreover $\cos(t\cdot
x)\sin\left(2r|t|\right)=\frac{1}{2}\left(\sin\left(2r|t|+t\cdot
x\right)+\sin\left(2r|t|-t\cdot x\right)\right)$, thus we may re-write
$I^{r}(x)$ as:

$I^{r}(x)=\frac{a}{2}\int_{t\in\mathbb{R}^{2}}\cos(t\cdot
x)\frac{1}{r|t|^{3}}\Psi^{r}_{x}(t)(1-\chi_{\ell}(t))\mathtt{d}t-\frac{a}{2}\int_{t\in\mathbb{R}^{2}}\sin\left(2r|t|-t\cdot
x\right)\frac{1}{r|t|^{3}}\Psi^{r}_{x}(t)(1-\chi_{\ell}(t))\mathtt{d}t,$
(A.22)

where we used the parity of $\Psi^{r}_{x}$ and $\chi_{\ell}$ in order to
combine $\sin\left(2r|t|\pm t\cdot x\right)$ into a single contribution. In
the sequel, we will study both integrals in (A.22) separately and denote:

$I^{r}_{1}(x):=\int_{t\in\mathbb{R}^{2}}\cos(t\cdot
x)\frac{1}{r|t|^{3}}\Psi^{r}_{x}(t)(1-\chi_{\ell}(t))\mathtt{d}t,\\\
I^{r}_{2}(x):=\int_{t\in\mathbb{R}^{2}}\sin\left(2r|t|-t\cdot
x\right)\frac{1}{r|t|^{3}}\Psi^{r}_{x}(t)(1-\chi_{\ell}(t))\mathtt{d}t.$
(A.23)

For $I^{i}(x)$, we obtain instead (using this time that $\Psi^{i}_{x}$ is
odd):

$I^{i}_{1}(x):=\int_{t\in\mathbb{R}^{2}}\sin(t\cdot
x)\frac{1}{r|t|^{3}}\Psi^{i}_{x}(t)(1-\chi_{\ell}(t))\mathtt{d}t,\quad
I^{i}_{2}(x):=\int_{t\in\mathbb{R}^{2}}\cos\left(2r|t|-t\cdot
x\right)\frac{1}{r|t|^{3}}\Psi^{i}_{x}(t)(1-\chi_{\ell}(t))\mathtt{d}t.$
(A.24)

###### Rewriting $II(x)$.

Still by elementary trigonometry, we have:

$\cos(t\cdot x)\cos(2r|t|)=\frac{1}{2}\left(\cos(2r|t|+t\cdot
x)+\cos(2r|t|-t\cdot x)\right).$

Using the parity of $\Psi^{r}_{x}$ and $\chi_{\ell}$ in order to combine both
contributions, we may re-write $II^{r}(x)$ as:

$II^{r}(x)=b\int_{t\in\mathbb{R}^{2}}\cos(2r|t|-t\cdot
x)\frac{1}{r^{2}|t|^{4}}\Psi^{r}_{x}(t)(1-\chi_{\ell}(t))\mathtt{d}t.$ (A.25)

Similarly, for $II^{i}(x)$ we obtain (using this time that $\Psi^{i}_{x}$ is
odd):

$II^{i}(x)=b\int_{t\in\mathbb{R}^{2}}\sin(2r|t|-t\cdot
x)\frac{1}{r^{2}|t|^{4}}\Psi^{i}_{x}(t)(1-\chi_{\ell}(t))\mathtt{d}t.$ (A.26)

######  Remark A.10.

At this point, applying a rough bound is not quite good enough, as one would
get, using Lemma A.8:

$|II^{r}(x)|+|II^{i}(x)|\preceq\int_{t\in\mathbb{R}^{2},\ell\leq|t|\leq
1}\frac{1}{r^{2}|t|^{4}}t^{2}\mathtt{d}t+\int_{t\in\mathbb{R}^{2},|t|\geq
1}\frac{1}{r^{2}|t|^{4}}\mathtt{d}t=\mathcal{O}\left(\frac{1}{r^{2}}\log\ell\right),$
(A.27)

and summing over the $\mathcal{O}(r\bar{d})$ lattice points which are at
distance $\leq\bar{d}$ from the boundary, we would obtain a control of size
$\mathcal{O}\left(\frac{\bar{d}\log\ell^{-1}}{r}\right)$. Althhough we do know
that $\ell^{-1}\ll\bar{d}\ll r$, it is not clear that one can take $\ell^{-1}$
and $\bar{d}$ small enough compared to $r$ to ensure that
$\bar{d}\log\ell^{-1}\ll r$ (e.g. if $\ell^{-1}$ has to be at least
$\frac{r}{\log r}$ then that comparison is false), cf. the discussion after
Lemma 3.6, which ultimately governs the choice of $\ell$.

It is worth noticing that if the perturbations were assumed to have a moment
of order $2+\delta$ (one could even ask for less), then we could take $\ell$
as some power $r^{-1+\delta^{\prime}}$, and a rough bound like (A.27) would be
enough.

###### Bounding $III(x)$.

We may immediately bound the last term in (A.20), independently of $x$, by:

$III^{r}(x)=\int_{\ell\leq|t|\leq
1}\mathcal{O}\left(\frac{1}{r^{3}|t|^{5}}\right)\times\mathcal{O}(|t|^{2})\mathtt{d}t+\int_{\ell\leq|t|\leq
1}\mathcal{O}\left(\frac{1}{r^{3}|t|^{5}}\right)\times\mathcal{O}(1)\mathtt{d}t=\mathcal{O}(r^{-3}\ell^{-1}),$
(A.28)

and the exact same bound holds for $III^{i}(x)$, using Lemma A.8. In contrast
with the previous paragraph, here a rough bound is enough - one can sum
$\mathcal{O}(r^{2})$ such terms and still get $\mathcal{O}((r\ell)^{-1})$,
which tends to $0$ as $r\to\infty$.

#### Bounding $I_{1}(x)$.

Integrating by parts twice in $I^{r}_{1}(x)$ (see (A.23)) gives:

$I^{r}_{1}(x)=\frac{-1}{r|x|^{2}}\int_{t\in\mathbb{R}^{2}}\cos(t\cdot
x)\Delta\left(\frac{1}{|t|^{3}}\Psi^{r}_{x}(t)(1-\chi_{\ell}(t))\right)\mathtt{d}t.$

We now want to evaluate the size of the Laplacian in the integrand. Since
$1-\chi_{\ell}$ is supported on $\\{|t|\geq\ell\\}$, so is that Laplacian, and
we make several observations:

###### For $\frac{1}{2}\ell\leq|t|\leq 1$.

  * •

The $k$-th derivatives of $1-\chi_{\ell}(t)$ are $0$ for
$|t|\leq\frac{1}{2}\ell$ ($\chi_{\ell}\equiv 1$) and $|t|\geq\ell$
($\chi_{\ell}\equiv 0$) and of order $\ell^{-k}$ for
$|t|\in[\frac{1}{2}\ell,\ell]$ and $k\geq 0$. Note that for
$|t|\in[\frac{1}{2}\ell,\ell]$, the quantities $\ell^{-k}$ and $|t|^{-k}$ are
of the same order.

  * •

The $k$-th derivatives of $\Psi^{r}_{x}$ are of order $|t|^{2-k}$ for
$k=0,1,2$, see Lemma A.8.

  * •

Of course, the $k$-th derivatives of $\frac{1}{|t|^{3}}$ are of order
$|t|^{-3-k}$ for $k\geq 0$.

When taking a second derivative of
$\frac{1}{|t|^{3}}\Psi^{r}_{x}(t)(1-\chi_{\ell}(t))$ for
$\frac{1}{2}\ell\leq|t|\leq 1$, we may thus lose up to two powers of $|t|$,
and the pointwise bound is thus
$\mathcal{O}(|t|^{-3+2-2})=\mathcal{O}(|t|^{-3})$.

###### For $|t|\geq 1$.

  * •

$1-\chi_{\ell}$ is identically $1$ and its derivatives are thus $0$.

  * •

$\Psi^{r}_{x}$ is bounded and so are its first and second derivatives, see
Lemma A.8.

  * •

The $k$-th derivatives of $\frac{1}{|t|^{3}}$ are still of order $|t|^{-3-k}$
for $k\geq 0$.

When taking a second derivative of
$\frac{1}{|t|^{3}}\Psi^{r}_{x}(t)(1-\chi_{\ell}(t))$ for $|t|\geq 1$, the
worst case is thus when both derivatives hit $\Psi^{r}_{x}$, in which case the
pointwise bound in $\mathcal{O}(|t|^{-3})$.

###### Application to bounding $I^{r}_{1}(x)$.

We can thus bound
$\Delta\left(\frac{1}{|t|^{3}}\Psi^{r}_{x}(t)(1-\chi_{\ell}(t))\right)$
pointwise by $\mathcal{O}\left(|t|^{-3}\right)$, uniformly with respect to
$x\in\mathcal{L}$. We obtain (keeping in mind that $|x|\geq r$):

$I^{r}_{1}(x)\preceq\frac{1}{r|x|^{2}}\int_{t\in\mathbb{R}^{2},|t|\geq\ell}|t|^{-3}\mathtt{d}t=\mathcal{O}\left(\frac{1}{\ell
r^{3}}\right).$ (A.29)

The same integration by parts works with $\sin(t\cdot x)$ instead of
$\cos(t\cdot x)$, and the estimates of Lemma A.8 also apply to $\Psi^{i}_{x}$,
thus $I^{i}_{1}(x)$ can be bounded the same way. Note that the right-hand side
of (A.29) is comparable to the error term we found in (A.28) for $III^{r}(x)$.
We move on to studying of $I^{r}_{2}(x),I^{i}_{2}(x)$.

###  A.3 Preliminaries to the study of $I_{2}(x)$

Recall that:

$I^{r}_{2}(x):=\int_{t\in\mathbb{R}^{2}}\sin\left(2r|t|-t\cdot
x\right)\frac{1}{r|t|^{3}}\Psi^{r}_{x}(t)(1-\chi_{\ell}(t))\mathtt{d}t,$

and $I^{i}_{2}(x)$ is similar, with $\cos$ instead of $\sin$ and
$\Psi^{i}_{x}$ instead of $\Psi^{r}_{x}$.

#### Using polar coordinates.

We introduce the angle $\theta_{x}$ as an argument of $x$, and re-write
$I^{r}_{2}(x)$ (see (A.23)) in polar coordinates as:

$I^{r}_{2}(x)=\frac{1}{r}\int_{\theta\in[0,2\pi]}\mathtt{d}\theta\int_{\rho=0}^{+\infty}\sin\left(2r\rho-\rho|x|\cos(\theta-\theta_{x})\right)\frac{1}{\rho^{3}}\left(\varphi_{\mathrm{p}_{x}-\mathrm{p}_{0}}(\rho,\theta)-1\right)(1-\chi_{\ell}(\rho))\rho\mathtt{d}\rho.$

The $\frac{1}{\rho^{3}}$ term coming from the Bessel function and the $\rho$
term coming from (the Jacobian of) polar coordinates can be combined. We also
apply a translation $\theta\mapsto\theta+\theta_{x}$ on the angle coordinate,
and re-write $I^{r}_{2}(x)$ as:

$I^{r}_{2}(x)=\frac{1}{r}\int_{-\pi}^{\pi}\mathtt{d}\theta\int_{\rho=0}^{+\infty}\sin\left(2r\rho-\rho|x|\cos\theta\right)\frac{1}{\rho^{2}}\Psi^{r}_{x}(\rho,\theta+\theta_{x})(1-\chi_{\ell}(\rho))\mathtt{d}\rho.$
(A.30)

The expression for $I^{i}_{2}(x)$ is similar, with
$\cos\left(2r\rho-\rho|x|\cos\theta\right)$ instead of
$\sin\left(2r\rho-\rho|x|\cos\theta\right)$ and $\Psi^{i}_{x}$ instead of
$\Psi^{r}_{x}$.

#### Auxiliary lengthscales and associated cut-off.

Let $\rho_{\star}$ be some positive length scale and let
$\chi_{\rho_{\star}}=\chi\left(\frac{\cdot}{\rho_{\star}}\right)$ be a cut-off
at scale $\rho_{\star}$ (for convenience we use the same reference cutoff
$\chi$ as in Section 3.1). We will choose $\rho_{\star}$ later, possibly
depending on $x,d$ and on the angle coordinate $\theta$. We can then always
split $I^{r}_{2}(x),I^{i}_{2}(x)$ in two parts: the “small-$\rho$” one
corresponding to $\chi_{\rho_{\star}}$ and the “large-$\rho$” one
corresponding to $1-\chi_{\rho_{\star}}$:

$I^{r}_{2}(x)=\frac{1}{r}\int_{-\pi}^{\pi}\mathtt{d}\theta\Bigg{(}\int_{\rho=0}^{+\infty}\sin\left(2r\rho-\rho|x|\cos\theta\right)\frac{1}{\rho^{2}}\Psi^{r}_{x}(\rho,\theta+\theta_{x})(1-\chi_{\ell}(\rho))\chi_{\rho_{\star}}(\rho)\mathtt{d}\rho\\\
+\int_{\rho=0}^{+\infty}\sin\left(2r\rho-\rho|x|\cos\theta\right)\frac{1}{\rho^{2}}\Psi^{r}_{x}(\rho,\theta+\theta_{x})(1-\chi_{\ell}(\rho))(1-\chi_{\rho_{\star}}(\rho))\mathtt{d}\rho\Bigg{)},$
(A.31)

and similarly for $I^{i}_{2}(x)$.

After choosing $\rho_{\star}$ depending on $\theta$ we will treat those two
parts differently.

  * •

The “small-$\rho$” term gets evaluated through a rough bound (see (A.32)),
except in one case where we will need to look more closely.

  * •

The “large-$\rho$” term gets integrated by parts (usually twice), and (most of
the time) we then apply a rough bound (see (A.34)) on the resulting
expression.

Finally, it will remain to integrate those bounds over $\theta$.

#### Rough bound on the “small-$\rho$” contribution.

######  Claim A.11.

Let $\rho_{\star}$ be a lengthscale and $\chi_{\rho_{\star}}$ the associated
truncation as introduced above. We have:

$\int_{\rho=0}^{+\infty}\sin\left(2r\rho-\rho|x|\cos\theta\right)\frac{1}{\rho^{2}}\Psi^{r}_{x}(\rho,\theta+\theta_{x})(1-\chi_{\ell}(\rho))\chi_{\rho_{\star}}(\rho)\mathtt{d}\rho=\mathcal{O}\left(\min(\rho_{\star},1)\right).$
(A.32)

###### Proof.

It follows from a direct computation. If $\rho_{\star}\leq 1$, we have (using
Lemma A.8):

$\int_{\rho=0}^{+\infty}\sin\left(2r\rho-\rho|x|\cos\theta\right)\frac{1}{\rho^{2}}\Psi^{r}_{x}(\rho,\theta+\theta_{x})(1-\chi_{\ell}(\rho))\chi_{\rho_{\star}}(\rho)\mathtt{d}\rho\leq\int_{0}^{\rho_{\star}}\mathcal{O}\left(\frac{1}{\rho^{2}}\right)\times\mathcal{O}(\rho^{2})\mathtt{d}\rho=\mathcal{O}(\rho_{\star}),$

uniformly with respect to $\theta$ and $x$. If $\rho_{\star}$ is larger than
$1$, we get $\mathcal{O}(1)$ instead of $\mathcal{O}(\rho_{\star})$ in (A.32),
because then we can use that $\Psi^{r}_{x}(\rho,\theta)=\mathcal{O}(1)$ (and
not $\mathcal{O}(\rho^{2})$) on $\\{|t|\in(1,\rho_{\star})\\}$. ∎

This rough bound works the same with $\cos$ instead of $\sin$ and replacing
$\Psi^{r}_{x}$ by $\Psi^{i}_{x}$, which is also covered by Lemma A.8.

#### Integration by parts for “large-$\rho$”.

For any fixed $\theta\in[-\pi,\pi]$, we can integrate by parts twice with
respect to $\rho$ within the second line of (A.31), which corresponds to
“large” values of $\rho$ ($\rho\geq\rho_{\star}$) and write:

$\int_{\rho=0}^{+\infty}\sin\left(2r\rho-\rho|x|\cos\theta\right)\frac{1}{\rho^{2}}\Psi^{r}_{x}(\rho,\theta+\theta_{x})(1-\chi_{\ell}(\rho))(1-\chi_{\rho_{\star}}(\rho))\mathtt{d}\rho\\\
=\frac{-1}{\left(2r-|x|\cos\theta\right)^{2}}\int_{\rho=0}^{+\infty}\sin\left(2r\rho-\rho|x|\cos\theta\right)\\\
\times\frac{\partial^{2}}{\partial^{2}\rho}\left(\frac{1}{\rho^{2}}\Psi^{r}_{x}(\rho,\theta+\theta_{x})(1-\chi_{\ell}(\rho))(1-\chi_{\rho_{\star}}(\rho))\right)\mathtt{d}\rho.$
(A.33)

For $I^{i}_{2}(x)$, the integration by parts works similarly with
$\cos\left(2r\rho-\rho|x|\cos\theta\right)$ instead of
$\sin\left(2r\rho-\rho|x|\cos\theta\right)$.

We stress that in (A.33), the lengthscale $\rho_{\star}$ appearing in
$\chi_{\rho_{\star}}$ is allowed to depend on $\theta$.

#### Rough bound on the “large-$\rho$” contribution.

######  Claim A.12.

Let $\rho_{\star}$ be some lengthscale and $\chi_{\rho_{\star}}$ the
associated truncation as introduced above. We have:

$\int_{\rho=0}^{+\infty}\sin\left(2r\rho-\rho|x|\cos\theta\right)\frac{\partial^{2}}{\partial^{2}\rho}\left(\frac{1}{\rho^{2}}\Psi^{r}_{x}(\rho,\theta+\theta_{x})(1-\chi_{\ell}(\rho))(1-\chi_{\rho_{\star}}(\rho))\right)\mathtt{d}\rho\\\
=\mathcal{O}\left(\min\left(\frac{1}{\rho_{\star}},\frac{1}{\ell}\right)\right).$
(A.34)

###### Proof.

We first study the size of the second derivative with respect to $\rho$ (this
is similar to the argument leading to (A.29)). By definition of the cut-off
function, the $k$-th derivatives of $1-\chi_{\rho_{\star}}(\rho)$ are $0$ for
$|\rho|\leq\frac{1}{2}\rho_{\star}$ and $|\rho|\geq\rho_{\star}$ and of order
$\rho_{\star}^{-k}$ for $|\rho|\in[\frac{1}{2}\rho_{\star},\rho_{\star}]$ and
$k\geq 0$. Note that for $|\rho|\in[\frac{1}{2}\rho_{\star},\rho_{\star}]$,
the quantities $\rho_{\star}^{-k}$ and $|\rho|^{-k}$ are of the same order.
The same is true for $1-\chi_{\ell}(\rho)$ (with $\ell$ instead of
$\rho_{\star}$). Moreover:

###### For $|\rho|\leq 1$.

  * •

The $k$-th derivatives of $\Psi^{r}_{x}(\rho,\theta)$ with respect to $\rho$
are of order $|\rho|^{2-k}$ for $k=0,1,2$ (see Lemma A.8).

  * •

The $k$-th derivatives of $\frac{1}{\rho^{2}}$ are of course of order
$\rho^{-2-k}$ for $k\geq 0$.

When taking the second derivative of
$\rho\mapsto\frac{1}{\rho^{2}}\Psi^{r}_{x}(\rho,\theta)(1-\chi_{\ell}(\rho))(1-\chi_{\rho_{\star}}(\rho))$
with respect to $\rho$ for $\rho\leq 1$, we may thus lose up to two powers of
$\rho$ in which case the bound is $\mathcal{O}(\rho^{-2})$.

###### For $|\rho|\geq 1$.

  * •

The $k$-th derivatives of $\Psi^{r}_{x}$ are bounded for $k=0,1,2$, see Lemma
A.8.

  * •

The $k$-th derivatives of $\frac{1}{\rho^{2}}$ are still of order
$\rho^{-2-k}$ for $k\geq 0$.

When taking the second derivative of
$\rho\mapsto\frac{1}{\rho^{2}}\Psi^{r}_{x}(\rho,\theta)(1-\chi_{\ell}(\rho))(1-\chi_{\rho_{\star}}(\rho))$
with respect to $\rho$ for $\rho\geq 1$, the worst case is when both
derivatives hit $\Psi^{r}_{x}$, in which case the bound is
$\mathcal{O}(\rho^{-2})$.

###### Application.

Notice that $(1-\chi_{\ell}(\rho))(1-\chi_{\rho_{\star}}(\rho))$ is supported
on $\\{|\rho|\geq\frac{1}{2}\max(\rho_{\star},\ell)\\}$. We just saw that we
can bound the second derivative within the integral by
$\mathcal{O}(\rho^{-2})$ hence we obtain, uniformly in $\theta$:

$\int_{\rho=0}^{+\infty}\sin\left(2r\rho-\rho|x|\cos\theta\right)\frac{\partial^{2}}{\partial^{2}\rho}\left(\frac{1}{\rho^{2}}\Psi^{r}_{x}(\rho,\theta+\theta_{x})(1-\chi_{\ell}(\rho))(1-\chi_{\rho_{\star}}(\rho))\right)\mathtt{d}\rho\\\
\preceq\int_{\rho=\frac{1}{2}\max(\rho_{\star},\ell)}^{+\infty}\frac{1}{\rho^{2}}\mathtt{d}\rho=\mathcal{O}\left(\min\left(\frac{1}{\rho_{\star}},\frac{1}{\ell}\right)\right).$

∎ The same computation holds for the corresponding part of $I^{i}_{2}(x)$. We
now turn to bounding $I^{r}_{2}(x),I^{i}_{2}(x)$.

###  A.4 Studying $I_{2}(x)$ inside the disk.

######  Lemma A.13 (Control of $I_{2}(x)$ for $x$ in $\mathrm{B}_{2r}$).

Assume that $x\in B_{2r}$ and that $d:=2r-|x|\leq\bar{d}$. We have:

$|I^{r}_{2}(x)|+|I^{i}_{2}(x)|=\mathcal{O}\left(\frac{1}{r}\times\frac{1}{\sqrt{dr}}\right).$
(A.35)

###### Proof of Lemma A.13.

Recall that $d\leq\bar{d}\ll r$ implies $|x|\geq r$. We focus on
$I^{r}_{2}(x)$ (see (A.30)) - then $I^{i}_{2}(x)$ can be treated exactly the
same way.

#### Preliminary: the $\theta$-dependent term

Recall (see (A.30)) that in the course of integration by parts, a term
$\frac{1}{(2r-|x|\cos\theta)^{2}}$ appears. We start by bounding it, depending
on the values of $\theta$.

Since $\cos\theta=1-\frac{\theta^{2}}{2}+o(\theta^{2})$ as $\theta\to 0$,
there exists $\delta>0$ such that for $|\theta|\leq\delta$, we have
$\cos\theta\leq 1-\frac{\theta^{2}}{3}$ and thus we have the following lower
bound:

$2r-|x|\cos\theta\geq
2r-|x|\left(1-\frac{\theta^{2}}{3}\right)=d+\frac{|x|}{3}\theta^{2}.$ (A.36)

In particular, we can write:

$\frac{1}{(2r-|x|\cos\theta)^{2}}\preceq\frac{1}{d^{2}}\ \text{ for
$|\theta|\leq\sqrt{\frac{d}{r}}$},\quad\frac{1}{(2r-|x|\cos\theta)^{2}}\preceq\frac{1}{r^{2}\theta^{4}}\
\text{ for $|\theta|\geq\sqrt{\frac{d}{r}}$}.$ (A.37)

Moreover, there exists $\varepsilon>0$ such that $\cos\theta\leq
1-\varepsilon$ if $|\theta|\geq\delta$, and thus if $|\theta|\geq\delta$:

$\frac{1}{(2r-|x|\cos\theta)^{2}}\leq\frac{1}{(2r-(1-\varepsilon)|x|)^{2}}\leq\frac{1}{(d+\varepsilon|x|)^{2}}\leq\frac{1}{\varepsilon^{2}r^{2}}.$
(A.38)

We can choose here $\delta,\varepsilon$ globally, not depending on $r$ or any
other parameter. Hence, we do not highlight the dependence of the constants on
$\delta$ and $\varepsilon.$

We will distinguish three regimes for $\theta$: large, moderate and small.

#### Large values of $\theta$.

We start by the “large $\theta$” region where $|\theta|\geq\delta$. On this
region, we integrate by parts twice the entire $\rho$-integral (which amounts
to choosing $\rho_{\star}=\ell$ in (A.31)), we use the rough bound (A.34)
together with our estimate (A.38), and write:

$\frac{1}{r}\int_{\theta\in[-\pi,\pi],|\theta|\geq\delta}\int_{\rho=0}^{+\infty}\sin\left(2r\rho-\rho|x|\cos\theta\right)\times\left(\frac{1}{\rho^{2}}\Psi^{r}_{x}(\rho,\theta+\theta_{x})(1-\chi_{\ell}(\rho))\right)\mathtt{d}\rho\\\
=\frac{1}{r}\int_{\theta\in[-\pi,\pi],|\theta|\geq\delta}\frac{-1}{\left(2r-|x|\cos\theta\right)^{2}}\int_{\rho=0}^{+\infty}\sin\left(2r\rho-\rho|x|\cos\theta\right)\times\frac{\partial^{2}}{\partial^{2}\rho}\left(\frac{1}{\rho^{2}}\Psi^{r}_{x}(\rho,\theta+\theta_{x})(1-\chi_{\ell}(\rho))\right)\mathtt{d}\rho\\\
=\frac{1}{r}\times\mathcal{O}\left(\frac{1}{r^{2}}\right)\times\mathcal{O}\left(\frac{1}{\ell}\right)=\mathcal{O}\left(\frac{1}{r^{3}\ell}\right).$
(A.39)

Observe that this has the same order of magnitude the bounds on (A.28),
(A.29).

#### Moderate values of $\theta$.

We now consider a “moderate range” $\sqrt{\frac{d}{r}}\leq|\theta|\leq\delta$.
We then choose a $\theta$-dependent truncation at scale
$\rho_{\star}=\frac{1}{r\theta^{2}}$.

On the one hand, we get, using the rough bound (A.32) on the “small-$\rho$”
part:

$\int_{\rho=0}^{+\infty}\sin\left(2r\rho-\rho|x|\cos\theta\right)\frac{1}{\rho^{2}}\Psi^{r}_{x}(\rho,\theta+\theta_{x})(1-\chi_{\ell}(\rho))\chi_{\rho_{\star}}(\rho)\mathtt{d}\rho=\mathcal{O}\left(\frac{1}{r\theta^{2}}\right).$

On the other hand, integrating by parts (see (A.33)), using the rough bound
(A.34) and inserting (A.37):

$\frac{1}{\left(2r-|x|\cos\theta\right)^{2}}\int_{\rho=0}^{+\infty}\sin\left(2r\rho-\rho|x|\cos\theta\right)\times\frac{\partial^{2}}{\partial^{2}\rho}\left(\frac{1}{\rho^{2}}\Psi^{r}_{x}(\rho,\theta+\theta_{x})(1-\chi_{\ell}(\rho))(1-\chi_{\rho_{\star}}(\rho))\right)\mathtt{d}\rho\\\
\preceq\frac{1}{r^{2}\theta^{4}}\times
r\theta^{2}=\mathcal{O}\left(\frac{1}{r\theta^{2}}\right).$

We are left to compute an integral over $\theta$:

$\frac{1}{r}\int_{\sqrt{\frac{d}{r}}\leq|\theta|\leq\delta}\frac{1}{r\theta^{2}}\mathtt{d}\theta=\mathcal{O}\left(\frac{1}{r}\times\frac{1}{\sqrt{dr}}\right).$
(A.40)

#### Small values of $\theta$.

We are left with the range of “small” values $|\theta|\leq\sqrt{\frac{d}{r}}$.
We choose a truncation at scale $\rho_{\star}=\frac{1}{d}$. On the one hand,
we get, using the rough bound (A.32) on the “small-$\rho$” contribution:

$\int_{\rho=0}^{+\infty}\sin\left(2r\rho-\rho|x|\cos\theta\right)\frac{1}{\rho^{2}}\Psi^{r}_{x}(\rho,\theta+\theta_{x})(1-\chi_{\ell}(\rho))\chi_{\rho_{\star}}(\rho)\mathtt{d}\rho=\mathcal{O}\left(\frac{1}{d}\right),$

and on the other hand, integrating by parts, using the rough bound (A.34) and
inserting (A.37):

$\frac{1}{\left(2r-|x|\cos\theta\right)^{2}}\int_{\rho=0}^{+\infty}\sin\left(2r\rho-\rho|x|\cos\theta\right)\times\frac{\partial^{2}}{\partial^{2}\rho}\left(\frac{1}{\rho^{2}}\Psi^{r}_{x}(\rho,\theta+\theta_{x})(1-\chi_{\ell}(\rho))(1-\chi_{\rho_{\star}}(\rho))\right)\mathtt{d}\rho\\\
\preceq\frac{1}{d^{2}}\times d=\mathcal{O}\left(\frac{1}{d}\right).$

We are left to compute an integral over $\theta$:

$\frac{1}{r}\int_{|\theta|\leq\sqrt{\frac{d}{r}}}\frac{1}{d}\mathtt{d}\theta=\mathcal{O}\left(\frac{1}{r}\times\frac{1}{\sqrt{dr}}\right).$
(A.41)

#### Conclusion

Combining (A.39), (A.40) and (A.41) (and keeping only the dominant term)
yields (A.35) for $I^{r}_{2}(x)$. The term $I^{i}_{2}(x)$ can be treated
exactly the same way, as we only relied on the rough bounds of (A.32), (A.34).
This proves Lemma A.13. ∎

###  A.5 Studying $I_{2}(x)$ outside the disk.

######  Lemma A.14 (Control of $I_{2}(x)$ for $x$ outside $\mathrm{B}_{2r}$).

Assume that $x\in B_{2r}$ and that $d:=|x|-2r\leq\bar{d}$. We have:

$|I^{r}_{2}(x)|+|I^{i}_{2}(x)|=\mathcal{O}\left(\frac{1}{r}\times\frac{1}{\sqrt{dr}}\right).$
(A.42)

###### Proof of Lemma A.14.

We proceed in a similar fashion as for the previous lemma. The additional
difficulty when studying points _outside_ $\partial\mathrm{B}_{2r}$ is that
there is always an angle $\bar{\theta}$, defined by:

$\bar{\theta}:=\arccos\left(\frac{2r}{|x|}\right)=\arccos\left(1-\frac{d}{2r+d}\right),$
(A.43)

such that the denominator of $\frac{1}{\left(2r-|x|\cos\theta\right)^{2}}$,
which appears when integrating by parts, vanishes for $\theta=\bar{\theta}$.
In fact there are _two_ such angles (one in $[0,\pi]$ and one in $[-\pi,0]$),
so it is convenient to restrict ourselves to an interval where $\cos$ is one-
to-one. We decompose $I^{r}_{2}(x)$, using the parity of $\cos$ and the parity
of $\Psi^{r}_{x}(\rho,\cdot)$, as:

$I^{r}_{2}(x)=\frac{1}{r}\int_{0}^{\pi}\mathtt{d}\theta\int_{\rho=0}^{+\infty}\sin\left(2r\rho-\rho|x|\cos\theta\right)\frac{1}{\rho^{2}}\Psi^{r}_{x}(\rho,\theta+\theta_{x})(1-\chi_{\ell}(\rho))\mathtt{d}\rho\\\
+\frac{1}{r}\int_{0}^{\pi}\mathtt{d}\theta\int_{\rho=0}^{+\infty}\sin\left(2r\rho-\rho|x|\cos\theta\right)\frac{1}{\rho^{2}}\Psi^{r}_{x}(\rho,\theta-\theta_{x})(1-\chi_{\ell}(\rho))\mathtt{d}\rho.$

and we will focus on bounding the first term - one would proceed identically
for the second one. Similarly, we can write (this time $\Psi^{i}_{x}$ is odd):

$I^{i}_{2}(x)=\frac{1}{r}\int_{0}^{\pi}\mathtt{d}\theta\int_{\rho=0}^{+\infty}\cos\left(2r\rho-\rho|x|\cos\theta\right)\frac{1}{\rho^{2}}\Psi^{i}_{x}(\rho,\theta+\theta_{x})(1-\chi_{\ell}(\rho))\mathtt{d}\rho\\\
-\frac{1}{r}\int_{0}^{\pi}\mathtt{d}\theta\int_{\rho=0}^{+\infty}\cos\left(2r\rho-\rho|x|\cos\theta\right)\frac{1}{\rho^{2}}\Psi^{i}_{x}(\rho,\theta-\theta_{x})(1-\chi_{\ell}(\rho))\mathtt{d}\rho.$

Our goal is now to bound:

$\frac{1}{r}\int_{0}^{\pi}\mathtt{d}\theta\int_{\rho=0}^{+\infty}\sin\left(2r\rho-\rho|x|\cos\theta\right)\frac{1}{\rho^{2}}\Psi^{r}_{x}(\rho,\theta+\theta_{x})(1-\chi_{\ell}(\rho))\mathtt{d}\rho,\\\
\frac{1}{r}\int_{0}^{\pi}\mathtt{d}\theta\int_{\rho=0}^{+\infty}\cos\left(2r\rho-\rho|x|\cos\theta\right)\frac{1}{\rho^{2}}\Psi^{i}_{x}(\rho,\theta+\theta_{x})(1-\chi_{\ell}(\rho))\mathtt{d}\rho.$
(A.44)

As we will see below, both terms can be treated the same way _to some extent_
, but there is one region of the parameters $(\rho,\theta)$ where we need to
study those integrands more carefully, and separately.

#### Working with $\widetilde{\theta}$.

To emphasize the role of $\bar{\theta}$ as in (A.43), we set
$\theta=:\widetilde{\theta}+\bar{\theta}$, so $\widetilde{\theta}$ is our new
angle variable, ranging from $-\bar{\theta}$ to $\pi-\bar{\theta}$. Our goal
is thus now to study the following quantity (as well as its counterpart):

$\frac{1}{r}\int_{-\bar{\theta}}^{\pi-\bar{\theta}}\mathtt{d}\widetilde{\theta}\int_{\rho=0}^{+\infty}\sin\left(2r\rho-\rho|x|\cos(\bar{\theta}+\widetilde{\theta})\right)\frac{1}{\rho^{2}}\Psi^{r}_{x}(\rho,\bar{\theta}+\theta_{x}+\widetilde{\theta})(1-\chi_{\ell}(\rho))\mathtt{d}\rho,$
(A.45)

and to bound it as claimed in Lemma A.14.

Recall the expansion $\arccos(1-t)=\sqrt{2t}+\mathcal{O}(t^{3/2})$. Since
$d\leq\bar{d}\ll r$, we have $\bar{\theta}\sim\sqrt{\frac{d}{r}}$ as
$r\to\infty$, and $\sin\bar{\theta}\sim\bar{\theta}\sim\sqrt{\frac{d}{r}}$.

#### Preliminary: the main $\theta$-dependent term.

We study here not only the term
$\frac{1}{\left(2r-|x|\cos(\bar{\theta}+\widetilde{\theta})\right)^{2}}$ that
appears in the integration by parts, but also the argument
$2r\rho-\rho|x|\cos(\bar{\theta}+\widetilde{\theta})$ appearing in (A.45).

We start with an elementary Taylor’s expansion:

$\cos(\bar{\theta}+\widetilde{\theta})=\cos\bar{\theta}-\widetilde{\theta}\sin\bar{\theta}+\mathcal{O}(\widetilde{\theta}^{2})=\cos\bar{\theta}-\widetilde{\theta}\sin\bar{\theta}-\frac{1}{2}\widetilde{\theta}^{2}\cos\bar{\theta}+\mathcal{O}(\widetilde{\theta}^{3}),$
(A.46)

and keep in mind that $\sin\bar{\theta}\sim\sqrt{\frac{d}{r}}$ while
$\cos\bar{\theta}\sim 1$ as $r\to\infty$.

We choose $\delta>0$ such that for all $r$ large enough and for all
$|\widetilde{\theta}|\leq\delta$, it holds:

$\cos(\bar{\theta}+\widetilde{\theta})\leq\cos\bar{\theta}-\widetilde{\theta}\sin\bar{\theta}-\frac{1}{4}\widetilde{\theta}^{2}.$
(A.47)

We also observe that for $r$ large enough, if
$|\widetilde{\theta}|\leq\bar{\theta}\sim\sqrt{\frac{d}{r}}\to 0$ then:

$\cos(\bar{\theta}+\widetilde{\theta})\leq\cos\bar{\theta}-\frac{1}{2}\widetilde{\theta}\sin\bar{\theta}\quad\text{(for
$\widetilde{\theta}\geq
0$)},\quad\cos(\bar{\theta}+\widetilde{\theta})\geq\cos\bar{\theta}-\frac{1}{2}\widetilde{\theta}\sin\bar{\theta}\quad\text{(for
$\widetilde{\theta}\leq 0$)}.$ (A.48)

###### Very large values.

If $\delta\leq\widetilde{\theta}\leq\pi-\bar{\theta}$, then
$\cos(\bar{\theta}+\widetilde{\theta})\leq\cos(\bar{\theta}+\delta)\leq(1-\varepsilon)\cos\bar{\theta}$
for some $\varepsilon>0$ (depending only on $\delta$). This does not follow
from Taylor’s expansions mentioned above, but simply from the identity
$\cos(\bar{\theta}+\widetilde{\theta})=\cos(\bar{\theta})\cos(\widetilde{\theta})-\sin(\bar{\theta})\sin(\widetilde{\theta})$,
with $\sin(\bar{\theta}),\sin(\widetilde{\theta})\geq 0$ and
$\cos(\widetilde{\theta})\leq(1-\varepsilon)$. As a consequence, we have:

$\text{For
}\delta\leq\widetilde{\theta}\leq\pi-\bar{\theta},\quad\frac{1}{\left(2r-|x|\cos(\bar{\theta}+\widetilde{\theta})\right)^{2}}\leq\frac{1}{\left(2r-|x|\times\frac{2r}{|x|}\times(1-\varepsilon)\right)^{2}}=\mathcal{O}\left(\frac{1}{r^{2}}\right).$
(A.49)

###### Large values.

If $\bar{\theta}\leq\widetilde{\theta}\leq\delta$, we use (A.47) and write:

$2r-|x|\cos(\bar{\theta}+\widetilde{\theta})\geq|x|\left(\widetilde{\theta}\sin\bar{\theta}+\frac{1}{4}\widetilde{\theta}^{2}\right).$

We used that $2r-|x|\cos(\bar{\theta})=0$ by definition of $\bar{\theta}$ in
(A.43). The quadratic term dominates (because
$\sin\bar{\theta}\sim\bar{\theta}\sim\sqrt{\frac{d}{r}}$ and
$\widetilde{\theta}\geq\bar{\theta}$), and we get (since $|x|\geq r$):

$\text{For
}\bar{\theta}\leq\widetilde{\theta}\leq\delta,\quad\frac{1}{\left(2r-|x|\cos(\bar{\theta}+\widetilde{\theta})\right)^{2}}=\mathcal{O}\left(\frac{1}{r^{2}\widetilde{\theta}^{2}}\right).$
(A.50)

For simplicity, we combine (A.49) and (A.50) and use (A.50) as a bound on the
whole range $\widetilde{\theta}\in[\bar{\theta},\pi-\bar{\theta}]$.

###### Moderate values.

If $|\widetilde{\theta}|\leq\bar{\theta}$, we use (A.48), distinguish two
cases depending on the sign of $\widetilde{\theta}$, and get that:

$\left|2r-|x|\cos(\bar{\theta}+\widetilde{\theta})\right|\geq|x|\times\frac{1}{2}\times|\widetilde{\theta}|\sin\bar{\theta}.$

Using that $|x|\geq r$ and that $\sin\bar{\theta}\sim\sqrt{\frac{d}{r}}$, we
obtain:

$\text{For
}|\widetilde{\theta}|\leq\bar{\theta},\quad\frac{1}{\left(2r-|x|\cos(\bar{\theta}+\widetilde{\theta})\right)^{2}}=\mathcal{O}\left(\frac{1}{rd\widetilde{\theta}^{2}}\right).$
(A.51)

We can now start evaluating the integrals in (A.44).

#### Large & very large values
($\bar{\theta}\leq\widetilde{\theta}\leq\pi-\bar{\theta}$).

In this regime, we integrate by parts twice the entire $\rho$-integral in
(A.44) (which amounts to choosing $\rho_{\star}=\ell$ in (A.31)) and use the
rough bound from (A.34), which gives:

$\int_{\rho=0}^{+\infty}\sin\left(2r\rho-\rho|x|\cos(\bar{\theta}+\widetilde{\theta})\right)\frac{\partial^{2}}{\partial^{2}\rho}\left(\frac{1}{\rho^{2}}\Psi^{r}_{x}(\rho,\bar{\theta}+\widetilde{\theta}+\theta_{x})(1-\chi_{\ell}(\rho))\right)\mathtt{d}\rho=\mathcal{O}\left(\frac{1}{\ell}\right),$

then we integrate over $\widetilde{\theta}$ using the control (A.50) on
$\frac{1}{\left(2r-|x|\cos(\bar{\theta}+\widetilde{\theta})\right)^{2}}$ (see
(A.33)):

$\frac{1}{r}\int_{\bar{\theta}}^{\pi-\bar{\theta}}\frac{1}{\left(2r-|x|\cos(\bar{\theta}+\widetilde{\theta})\right)^{2}}\times\mathcal{O}\left(\frac{1}{\ell}\right)\mathtt{d}\widetilde{\theta}=\frac{1}{r}\times\int_{\bar{\theta}}^{\pi-\bar{\theta}}\mathcal{O}\left(\frac{1}{r^{2}\widetilde{\theta}^{2}}\right)\times\mathcal{O}\left(\frac{1}{\ell}\right)\mathtt{d}\widetilde{\theta}=\mathcal{O}\left(\frac{1}{r}\times\frac{1}{r^{3/2}\sqrt{d}\ell}\right),$

using the fact that $\bar{\theta}\sim\sqrt{\frac{d}{r}}$.

Since $\ell r\geq 1$, we can simplify this bound into
$\mathcal{O}\left(\frac{1}{r}\times\frac{1}{\sqrt{dr}}\right)$, which will be
enough for us. In conclusion:

$\frac{1}{r}\int_{\bar{\theta}}^{\pi-\bar{\theta}}\mathtt{d}\widetilde{\theta}\int_{\rho=0}^{+\infty}\sin\left(2r\rho-\rho|x|\cos(\bar{\theta}+\widetilde{\theta})\right)\frac{1}{\rho^{2}}\Psi^{r}_{x}(\rho,\bar{\theta}+\widetilde{\theta}+\theta_{x})(1-\chi_{\ell}(\rho))\mathtt{d}\rho=\mathcal{O}\left(\frac{1}{r}\times\frac{1}{\sqrt{dr}}\right).$
(A.52)

This computation works identically for the corresponding part of
$I^{i}_{2}(x)$, because we are only relying on the rough bounds (A.32),
(A.34).

#### Moderate values: $|\widetilde{\theta}|\leq\bar{\theta}$ (${r}$-part).

This is the most subtle part of the analysis, and the only moment where there
is a significant difference of treatment between the “${r}$-part” (term with
$\Psi^{r}_{x}$) and the “${i}$-part” (term with $\Psi^{i}_{x}$).

An important observation is that this range of values for $\widetilde{\theta}$
is symmetric around $0$. We choose the following truncation:

$\rho_{\star}:=\frac{1}{|\widetilde{\theta}|^{\frac{4}{3}}r^{\frac{2}{3}}d^{\frac{1}{3}}}.$
(A.53)

It is not hard to check that with this choice of $\rho_{\star}$, and for
$\widetilde{\theta}\leq\bar{\theta}\sim\sqrt{\frac{d}{r}}$ we have:

$\rho_{\star}r\widetilde{\theta}^{2}\preceq 1,\quad|\widetilde{\theta}|\preceq
1,\quad\rho_{\star}|\widetilde{\theta}|\preceq
r\widetilde{\theta}^{2}\rho_{\star}^{2}.$ (A.54)

###### Small values of $\rho$.

We first consider the part corresponding to $\chi_{\rho_{\star}}$ in (A.44),
namely (for fixed $\widetilde{\theta}$):

$\int_{\rho=0}^{+\infty}\sin\left(2r\rho-\rho|x|\cos(\bar{\theta}+\widetilde{\theta})\right)\frac{1}{\rho^{2}}\Psi^{r}_{x}(\rho,\widetilde{\theta}+\bar{\theta}+\theta_{x})(1-\chi_{\ell}(\rho))\chi_{\rho_{\star}}(\rho)\mathtt{d}\rho.$

In the rest of this paragraph, we may assume that
$\rho_{\star}\geq{\frac{1}{2}}\ell$, otherwise this integral vanishes.

Here we do not use the rough bound of (A.32). Instead, we rely on expanding
the integrand and on a parity argument to cancel the first order term. First,
we expand the oscillating term, using (A.46), as:

$\sin\left(2r\rho-\rho|x|\cos(\bar{\theta}+\widetilde{\theta})\right)=\sin\left(\rho{|x|}(\widetilde{\theta}\sin\bar{\theta}+\mathcal{O}(\widetilde{\theta}^{2}))\right)=\sin\left(\rho{|x|}\sin\bar{\theta}\cdot\widetilde{\theta}\right)+\mathcal{O}(\rho
r\widetilde{\theta}^{2}).$ (A.55)

We also use Lemma A.8 to expand $\Psi^{r}_{x}$ to first order around
$\widetilde{\theta}=0$, which yields:

$\Psi^{r}_{x}(\rho,\bar{\theta}+\theta_{x}+\widetilde{\theta})=\Psi^{r}_{x}(\rho,\bar{\theta}+\theta_{x})+\widetilde{\theta}\times\mathcal{O}\left(\rho^{2}\right).$
(A.56)

We know a priori that $\Psi^{r}_{x}$ is $\mathcal{O}(\rho^{2})$ (see Lemma
A.8). Inserting both expansions (A.55) and (A.56) (the first two inequalities
in (A.54) indicate that these are meaningful expansions) into the integral we
find:

$\int_{0}^{+\infty}\sin\left(2r\rho-\rho|x|\cos(\bar{\theta}+\widetilde{\theta})\right)\frac{1}{\rho^{2}}\Psi^{r}_{x}(\rho,\widetilde{\theta}+\bar{\theta}+\theta_{x})(1-\chi_{\ell}(\rho))\chi_{\rho_{\star}}(\rho)\mathtt{d}\rho\\\
=\int_{0}^{+\infty}\left(\sin\left(\rho{|x|}\sin\bar{\theta}\cdot\widetilde{\theta}\right)+\mathcal{O}(\rho
r\widetilde{\theta}^{2})\right)\times\left(\Psi^{r}_{x}(\rho,\bar{\theta}+\theta_{x})+\widetilde{\theta}\times\mathcal{O}\left(\rho^{2}\right)\right)\times\frac{1}{\rho^{2}}(1-\chi_{\ell}(\rho))\chi_{\rho_{\star}}(\rho)\mathtt{d}\rho\\\
=\int_{\ell}^{\rho_{\star}}\sin\left(\rho{|x|}\sin\bar{\theta}\cdot\widetilde{\theta}\right)\Psi^{r}_{x}(\rho,\bar{\theta}+\theta_{x})\frac{1}{\rho^{2}}(1-\chi_{\ell}(\rho))\chi_{\rho_{\star}}(\rho)\mathtt{d}\rho\\\
+\int_{\ell}^{\rho_{\star}}\mathcal{O}(\widetilde{\theta})\mathtt{d}\rho+\int_{\ell}^{\rho_{\star}}\mathcal{O}(\rho
r\widetilde{\theta}^{2})\mathtt{d}\rho.$

Remark: it might happen that $\rho_{\star}\geq 1$ \- this is the case if
$|\widetilde{\theta}|$ is smaller than
$\frac{1}{r^{\frac{1}{2}}d^{\frac{1}{4}}}$ \- in which case for the part of
the integral corresponding to $1\leq\rho\leq\rho_{\star}$, writing that
$\Psi^{r}_{x}=\mathcal{O}(\rho^{2})$ and
$\partial_{\theta}\Psi^{r}_{x}=\mathcal{O}(\rho^{2})$ is sub-optimal: in view
of Lemma A.8 we could in fact use $\Psi^{r}_{x}=\mathcal{O}(1)$ and
$\partial_{\theta}\Psi^{r}_{x}=\mathcal{O}(\rho)$. However, the bounds we use
here yield a result that is enough for us and we avoid distinguishing more
cases.

The function
$\widetilde{\theta}\mapsto\int_{\ell}^{\rho_{\star}}\sin\left(\rho{|x|}\sin\bar{\theta}\cdot\widetilde{\theta}\right)\Psi^{r}_{x}(\rho,\bar{\theta}+\theta_{x})\frac{1}{\rho^{2}}(1-\chi_{\ell}(\rho))\chi_{\rho_{\star}}(\rho)\mathtt{d}\rho$
_is odd_ (because $\sin$ is, and the dependency of $\rho_{\star}$ is even) and
thus (using that the domain of integration is symmetric around $0$):

$\frac{1}{r}\int_{|\widetilde{\theta}|\leq\bar{\theta}}\mathtt{d}\widetilde{\theta}\int_{\ell}^{\rho_{\star}}\sin\left(\rho{|x|}\sin\bar{\theta}\cdot\widetilde{\theta}\right)\Psi^{r}_{x}(\rho,\bar{\theta}+\theta_{x})\frac{1}{\rho^{2}}(1-\chi_{\ell}(\rho))\chi_{\rho_{\star}}(\rho)\mathtt{d}\rho=0.$

On the other hand, we can directly estimate the two other integrals:

$\int_{\ell}^{\rho_{\star}}\widetilde{\theta}\mathtt{d}\rho=\mathcal{O}(\widetilde{\theta}\rho_{\star}),\quad\int_{\ell}^{\rho_{\star}}\rho
r\widetilde{\theta}^{2}\mathtt{d}\rho=\mathcal{O}(r\widetilde{\theta}^{2}\rho_{\star}^{2}).$

In view of (A.54), the second term dominates. We are left to compute (using
$\bar{\theta}\sim\sqrt{\frac{d}{r}}$):

$\frac{1}{r}\int_{|\widetilde{\theta}|\leq\bar{\theta}}r\widetilde{\theta}^{2}\rho_{\star}^{2}\mathtt{d}\widetilde{\theta}=\frac{1}{r}\int_{|\widetilde{\theta}|\leq\bar{\theta}}r\widetilde{\theta}^{2}\frac{1}{|\widetilde{\theta}|^{\frac{8}{3}}r^{\frac{4}{3}}d^{\frac{2}{3}}}\mathtt{d}\widetilde{\theta}\preceq\frac{1}{r}\times\frac{1}{\sqrt{dr}}.$
(A.57)

In conclusion, we have bounded the contribution coming from “small values” of
$\rho$ as follows:

$\frac{1}{r}\int_{|\widetilde{\theta}|\leq\bar{\theta}}\mathtt{d}\widetilde{\theta}\int_{\rho=0}^{+\infty}\sin\left(2r\rho-\rho|x|\cos(\bar{\theta}+\widetilde{\theta})\right)\Psi^{r}_{x}(\rho,\widetilde{\theta}+\bar{\theta}+\theta_{x})\frac{1}{\rho^{2}}(1-\chi_{\ell}(\rho))\chi_{\rho_{\star}}(\rho)\mathtt{d}\rho\\\
=\mathcal{O}\left(\frac{1}{r}\frac{1}{\sqrt{dr}}\right).$ (A.58)

###### High values of $\rho$.

We now consider the part corresponding to $1-\chi_{\rho_{\star}}$, we
integrate it twice by parts as in (A.33) and we apply the rough bound from
(A.34) together with the estimate (A.51). We then integrate this bound over
$\widetilde{\theta}$. We get:

$\frac{1}{r}\int_{|\widetilde{\theta}|\leq\bar{\theta}}\mathtt{d}\widetilde{\theta}\int_{\rho=0}^{+\infty}\sin\left(2r\rho-\rho|x|\cos(\bar{\theta}+\widetilde{\theta})\right)\frac{1}{\rho^{2}}\Psi^{r}_{x}(\rho,\widetilde{\theta}+\bar{\theta}+\theta_{x})(1-\chi_{\ell}(\rho))(1-\chi_{\rho_{\star}}(\rho))\mathtt{d}\rho\\\
=\frac{1}{r}\int_{|\widetilde{\theta}|\leq\bar{\theta}}\mathtt{d}\widetilde{\theta}\frac{-1}{\left(2r-|x|\cos(\bar{\theta}+\widetilde{\theta})\right)^{2}}\times\mathcal{O}\left(\frac{1}{\rho_{\star}}\right)\preceq\frac{1}{r}\int_{|\widetilde{\theta}|\leq\bar{\theta}}\frac{1}{dr\widetilde{\theta}^{2}}\times\frac{1}{\rho_{\star}}\mathtt{d}\widetilde{\theta}.$

Remark: if $\rho_{\star}$ happens to be smaller than $\ell$ we could use
$\mathcal{O}\left(\frac{1}{\ell}\right)$ instead of
$\mathcal{O}\left(\frac{1}{\rho_{\star}}\right)$, but we won’t need it.

Our specific choice (A.53) of $\rho_{\star}$ happens to be such that
$\frac{1}{dr\widetilde{\theta}^{2}}\times\frac{1}{\rho_{\star}}=r\widetilde{\theta}^{2}\rho_{\star}^{2}$
and we are thus facing (A.57) again. In conclusion, we also have:

$\frac{1}{r}\int_{|\widetilde{\theta}|\leq\bar{\theta}}\mathtt{d}\widetilde{\theta}\int_{\rho=0}^{+\infty}\sin\left(2r\rho-\rho|x|\cos(\bar{\theta}+\widetilde{\theta})\right)\frac{1}{\rho^{2}}\Psi^{r}_{x}(\rho,\widetilde{\theta}+\bar{\theta}+\theta_{x})(1-\chi_{\ell}(\rho))(1-\chi_{\rho_{\star}}(\rho))\mathtt{d}\rho\\\
=\mathcal{O}\left(\frac{1}{r}\frac{1}{\sqrt{dr}}\right).$ (A.59)

#### Moderate values: $|\widetilde{\theta}|\leq\bar{\theta}$ (“$i$-part”).

Let us now focus on the corresponding range of values for $\widetilde{\theta}$
in the $i$-part of (A.44), we want to prove:

$\frac{1}{r}\int_{|\widetilde{\theta}|\leq\bar{\theta}}\mathtt{d}\widetilde{\theta}\int_{\rho=0}^{+\infty}\cos\left(2r\rho-\rho|x|\cos(\bar{\theta}+\widetilde{\theta})\right)\frac{1}{\rho^{2}}\Psi^{i}_{x}(\rho,\widetilde{\theta}+\bar{\theta}+\theta_{x})(1-\chi_{\ell}(\rho))\mathtt{d}\rho=\mathcal{O}\left(\frac{1}{r}\frac{1}{\sqrt{dr}}\right).$
(A.60)

One could try to proceed similarly as above, but the parity argument fails to
apply in this case. In order to make the necessary gain we will rely instead
on Lemma A.9. We start by writing $\Psi^{i}_{x}$ as
$-\mathbb{E}[\mathcal{K}^{i}_{x}(\cdot)]$ with the random variable
$\mathcal{K}^{i}_{x}(t):=\Im\left[e^{-it\cdot p}+it\cdot p\right]$ as in Lemma
A.9, and we recall that (almost surely):

$|\mathcal{K}^{i}_{x}(t)|\preceq\min\left(\rho|p|,\rho^{3}|p|^{3}\right),\quad|\nabla\mathcal{K}^{i}_{x}(t)|\preceq\min\left(|p|,\rho^{2}|p|^{3}\right),\quad|\mathrm{Hess}\mathcal{K}^{i}_{x}(t)|\preceq\min(|p|^{2},\rho|p|^{3}).$
(A.61)

For any $\widetilde{\theta}$ the $\rho$-integral in (A.60) can be re-written
as:

$\int_{\rho=0}^{+\infty}\cos\left(2r\rho-\rho|x|\cos(\bar{\theta}+\widetilde{\theta})\right)\frac{1}{\rho^{2}}\Psi^{i}_{x}(\rho,\widetilde{\theta}+\bar{\theta}+\theta_{x})(1-\chi_{\ell}(\rho))\mathtt{d}\rho\\\
=-\mathbb{E}\left[\int_{\rho=0}^{+\infty}\cos\left(2r\rho-\rho|x|\cos(\bar{\theta}+\widetilde{\theta})\right)\frac{1}{\rho^{2}}\mathcal{K}^{i}_{x}(\rho,\widetilde{\theta}+\bar{\theta}+\theta_{x})(1-\chi_{\ell}(\rho))\mathtt{d}\rho\
\mathsf{1}_{|p|\geq\sqrt{dr}|\widetilde{\theta}|}\right]\\\
-\mathbb{E}\left[\int_{\rho=0}^{+\infty}\cos\left(2r\rho-\rho|x|\cos(\bar{\theta}+\widetilde{\theta})\right)\frac{1}{\rho^{2}}\mathcal{K}^{i}_{x}(\rho,\widetilde{\theta}+\bar{\theta}+\theta_{x})(1-\chi_{\ell}(\rho))\mathtt{d}\rho\
\mathsf{1}_{|p|\leq\sqrt{dr}|\widetilde{\theta}|}\right],$ (A.62)

where we simply introduced a distinction according to the respective sizes of
$|p|={|\mathrm{p}_{x}-\mathrm{p}_{0}|}$ and $\sqrt{dr}|\widetilde{\theta}|$.

##### When $|p|$ is large.

To treat the first expectation in (A.62), we work here on the event
$\left\\{|p|\geq\sqrt{dr}|\widetilde{\theta}|\right\\}$. We want to study

$\int_{\rho=0}^{+\infty}\cos\left(2r\rho-\rho|x|\cos(\bar{\theta}+\widetilde{\theta})\right)\frac{1}{\rho^{2}}\mathcal{K}^{i}_{x}(\rho,\widetilde{\theta}+\bar{\theta}+\theta_{x})(1-\chi_{\ell}(\rho))\mathtt{d}\rho\
\mathsf{1}_{|p|\geq\sqrt{dr}|\widetilde{\theta}|},$ (A.63)

and then take the expectation. We introduce a cut-off at scale

$\rho_{\star}:=\frac{1}{\sqrt{dr}|\widetilde{\theta}|}.$

This divides the $\rho$-integral of (A.63) into “small-$\rho$” and
“large-$\rho$” contributions.

###### Small-$\rho$ for large $|p|$.

In view of (A.61), it is natural to split the “small-$\rho$” contribution
further into:

$\int_{\rho=0}^{+\infty}\cos\left(2r\rho-\rho|x|\cos(\bar{\theta}+\widetilde{\theta})\right)\frac{1}{\rho^{2}}\mathcal{K}^{i}_{x}(\rho,\widetilde{\theta}+\bar{\theta}+\theta_{x})(1-\chi_{\ell}(\rho))\chi_{\rho_{\star}}(\rho)\mathtt{d}\rho\\\
\leq\int_{\rho=0}^{\frac{1}{|p|}}\frac{1}{\rho^{2}}|\mathcal{K}^{i}_{x}(\rho,\widetilde{\theta}+\bar{\theta}+\theta_{x})|\mathtt{d}\rho+\int_{\rho=\frac{1}{|p|}}^{\rho_{\star}}\frac{1}{\rho^{2}}|\mathcal{K}^{i}_{x}(\rho,\widetilde{\theta}+\bar{\theta}+\theta_{x})|\mathtt{d}\rho.$

We now use the estimates of Lemma A.9 recalled in (A.61) to bound
$\mathcal{K}^{i}_{x}$ and obtain (for almost every $p$):

$\int_{\rho=0}^{+\infty}\cos\left(2r\rho-\rho|x|\cos(\bar{\theta}+\widetilde{\theta})\right)\frac{1}{\rho^{2}}\mathcal{K}^{i}_{x}(\rho,\widetilde{\theta}+\bar{\theta}+\theta_{x})(1-\chi_{\ell}(\rho))\chi_{\rho_{\star}}(\rho)\mathtt{d}\rho\\\
\preceq\int_{\rho=0}^{\frac{1}{|p|}}\frac{1}{\rho^{2}}\rho^{3}|p|^{3}\mathtt{d}\rho+\int_{\rho=\frac{1}{|p|}}^{\rho_{\star}}\frac{1}{\rho^{2}}\rho|p|\mathtt{d}\rho\preceq|p|+|p|\log(\rho_{\star}|p|),$

the second term being the dominant one when $|p|\geq\frac{1}{\rho_{\star}}$.
We may now insert the value of
$\rho_{\star}:=\frac{1}{\sqrt{dr}|\widetilde{\theta}|}$, take the expectation
and integrate over $\widetilde{\theta}$, which leads us to compute:

$\frac{1}{r}\int_{|\widetilde{\theta}|\leq\bar{\theta}}\mathbb{E}\left[|p|\log\left(\frac{|p|}{|\widetilde{\theta}|\sqrt{dr}}\right)\mathsf{1}_{|p|\geq|\widetilde{\theta}|\sqrt{dr}}\right]\mathtt{d}\widetilde{\theta}=\frac{1}{r}\mathbb{E}\left[\int_{|\widetilde{\theta}|\leq\bar{\theta}}|p|\log\left(\frac{|p|}{|\widetilde{\theta}|\sqrt{dr}}\right)\mathsf{1}_{|p|\geq|\widetilde{\theta}|\sqrt{dr}}\mathtt{d}\widetilde{\theta}\right]\\\
\preceq\frac{1}{r}\mathbb{E}\left[\int_{\widetilde{\theta}=0}^{\frac{|p|}{\sqrt{dr}}}|p|\log\left(\frac{|p|}{\widetilde{\theta}\sqrt{dr}}\right)\mathtt{d}\widetilde{\theta}\right]=\frac{1}{r}\frac{1}{\sqrt{dr}}\mathbb{E}\left[|p|^{2}\right]\times\int_{v=0}^{1}-\log
v\mathtt{d}v,$

where we used the change of variables
$v=\frac{\sqrt{dr}\widetilde{\theta}}{|p|}$. This is indeed
$\mathcal{O}\left(\frac{1}{r}\frac{1}{\sqrt{dr}}\right)$.

###### Large-$\rho$ for large $|p|$.

We now look at the “large-$\rho$” contribution in (A.63), which we integrate
by parts _but this time only once_ i.e. we write:

$\int_{\rho=0}^{+\infty}\cos\left(2r\rho-\rho|x|\cos(\bar{\theta}+\widetilde{\theta})\right)\frac{1}{\rho^{2}}\mathcal{K}^{i}_{x}(\rho,\widetilde{\theta}+\bar{\theta}+\theta_{x})(1-\chi_{\ell}(\rho))(1-\chi_{\rho_{\star}}(\rho))\mathtt{d}\rho\\\
=\frac{1}{2r-|x|\cos(\bar{\theta}+\widetilde{\theta})}\int_{\rho=\frac{1}{2}\rho_{\star}}^{+\infty}\sin\left(2r\rho-\rho|x|\cos(\bar{\theta}+\widetilde{\theta})\right)\frac{\partial}{\partial\rho}\left(\frac{1}{\rho^{2}}\mathcal{K}^{i}_{x}(\rho,\widetilde{\theta}+\bar{\theta}+\theta_{x})(1-\chi_{\ell}(\rho))(1-\chi_{\rho_{\star}}(\rho))\right)\mathtt{d}\rho.$
(A.64)

We can use (A.51) to bound
$\frac{1}{2r-|x|\cos(\bar{\theta}+\widetilde{\theta})}$ by
$\mathcal{O}\left(\frac{1}{\sqrt{dr}|\widetilde{\theta}|}\right)$ and our next
task is to bound the first derivative appearing in the integrand. To do so, we
recall that according to Lemma A.9 (as recalled in (A.61)),
$\mathcal{K}^{i}_{x}$ is bounded by $\rho|p|$ and $\nabla\mathcal{K}^{i}_{x}$
by $|p|$ almost surely, so we have the pointwise, almost sure bound (compare
with the computation leading to (A.34)):

$\frac{\partial}{\partial\rho}\left(\frac{1}{\rho^{2}}\mathcal{K}^{i}_{x}(\rho,\widetilde{\theta}+\bar{\theta}+\theta_{x})(1-\chi_{\ell}(\rho))(1-\chi_{\rho_{\star}}(\rho))\right)\preceq\frac{1}{\rho^{2}}|p|.$

As a consequence, we get (using (A.64)) an almost sure bound:

$\int_{\rho=0}^{+\infty}\cos\left(2r\rho-\rho|x|\cos(\bar{\theta}+\widetilde{\theta})\right)\frac{1}{\rho^{2}}\mathcal{K}^{i}_{x}(\rho,\widetilde{\theta}+\bar{\theta}+\theta_{x})(1-\chi_{\ell}(\rho))(1-\chi_{\rho_{\star}}(\rho))\mathtt{d}\rho\\\
\preceq\frac{1}{\sqrt{dr}|\widetilde{\theta}|}\times\int_{\rho=\frac{1}{2}\rho_{\star}}^{+\infty}\frac{1}{\rho^{2}}|p|\mathtt{d}\rho=\frac{1}{\sqrt{dr}|\widetilde{\theta}|}\times\frac{2}{\rho_{\star}}\times|p|=\mathcal{O}(|p|)$

because we chose $\rho_{\star}=\frac{1}{\sqrt{dr}|\widetilde{\theta}|}$. We
can then easily integrate over $|\widetilde{\theta}|\leq\bar{\theta}$:

$\frac{1}{r}\int_{|\widetilde{\theta}|\leq\bar{\theta}}\mathbb{E}\left[\int_{\rho=0}^{+\infty}\cos\left(2r\rho-\rho|x|\cos(\bar{\theta}+\widetilde{\theta})\right)\frac{1}{\rho^{2}}\mathcal{K}^{i}_{x}(\rho,\widetilde{\theta}+\bar{\theta}+\theta_{x})(1-\chi_{\ell}(\rho))(1-\chi_{\rho_{\star}}(\rho))\mathtt{d}\rho\
\mathsf{1}_{|p|\geq|\widetilde{\theta}|\sqrt{dr}}\right]\mathtt{d}\widetilde{\theta}\\\
\preceq\frac{1}{r}\mathbb{E}\left[\int_{|\widetilde{\theta}|\leq\bar{\theta}}|p|\mathsf{1}_{|p|\geq|\widetilde{\theta}|\sqrt{dr}}\right]=\frac{1}{r}\times\frac{1}{\sqrt{dr}}\mathbb{E}\left[|p|\int_{|v|\leq|p|}1\mathtt{d}v\right]=\frac{1}{r}\times\frac{1}{\sqrt{dr}}\mathbb{E}\left[|p|^{2}\right],$

(where we have set $v=|\widetilde{\theta}|\sqrt{dr}$), which is again
$\mathcal{O}\left(\frac{1}{r}\frac{1}{\sqrt{dr}}\right)$.

##### When $|p|$ is small.

We return to (A.62) and study the second expectation, namely we work here on
the event $\left\\{|p|\leq\sqrt{dr}|\widetilde{\theta}|\right\\}$, we study

$\int_{\rho=0}^{+\infty}\cos\left(2r\rho-\rho|x|\cos(\bar{\theta}+\widetilde{\theta})\right)\frac{1}{\rho^{2}}\mathcal{K}^{i}_{x}(\rho,\widetilde{\theta}+\bar{\theta}+\theta_{x})(1-\chi_{\ell}(\rho))\mathtt{d}\rho\
\mathsf{1}_{|p|\leq\sqrt{dr}|\widetilde{\theta}|},$ (A.65)

and then take the expectation. We use the same cut-off
$\rho_{\star}:=\frac{1}{\sqrt{dr}|\widetilde{\theta}|}$ as before.

###### Small-$\rho$ for small $|p|$.

Observe that now, if $\rho\leq\rho_{\star}$, then $\rho\leq\frac{1}{|p|}$.
Using (A.61), we can thus bound the “small-$\rho$” contribution by:

$\int_{\rho=0}^{+\infty}\cos\left(2r\rho-\rho|x|\cos(\bar{\theta}+\widetilde{\theta})\right)\frac{1}{\rho^{2}}\mathcal{K}^{i}_{x}(\rho,\widetilde{\theta}+\bar{\theta}+\theta_{x})(1-\chi_{\ell}(\rho))\chi_{\rho_{\star}}(\rho)\mathtt{d}\rho\leq\int_{\rho=0}^{\rho_{\star}}\rho|p|^{3}\mathtt{d}\rho=\mathcal{O}\left(\frac{1}{rd\widetilde{\theta}^{2}}\right)\times|p|^{3},$

which we can then integrate over $\widetilde{\theta}$:

$\frac{1}{r}\int_{|\widetilde{\theta}|\leq\bar{\theta}}\mathbb{E}\left[\int_{\rho=0}^{+\infty}\cos\left(2r\rho-\rho|x|\cos(\bar{\theta}+\widetilde{\theta})\right)\frac{1}{\rho^{2}}\mathcal{K}^{i}_{x}(\rho,\widetilde{\theta}+\bar{\theta}+\theta_{x})(1-\chi_{\ell}(\rho))\chi_{\rho_{\star}}(\rho)\mathtt{d}\rho\
\mathsf{1}_{|p|\leq|\widetilde{\theta}|\sqrt{dr}}\right]\mathtt{d}\widetilde{\theta}\\\
\preceq\frac{1}{r}\mathbb{E}\left[\int_{|\widetilde{\theta}|\leq\bar{\theta}}\frac{1}{rd\widetilde{\theta}^{2}}|p|^{3}\
\mathsf{1}_{|p|\leq|\widetilde{\theta}|\sqrt{dr}}\mathtt{d}\widetilde{\theta}\right]\preceq\frac{1}{r}\times\frac{1}{\sqrt{dr}}\mathbb{E}\left[|p|^{3}\int_{v=|p|}^{+\infty}\frac{1}{v^{2}}\mathtt{d}v\right]=\frac{1}{r}\times\frac{1}{\sqrt{dr}}\mathbb{E}\left[|p|^{2}\right],$

(where we have set $v=|\widetilde{\theta}|\sqrt{dr}$), which is
$\mathcal{O}\left(\frac{1}{r}\times\frac{1}{\sqrt{dr}}\right)$.

###### Large-$\rho$ for small $|p|$.

Let us finally study the large-$\rho$” regime: there, we perform a double
integration by parts and write:

$\int_{\rho=0}^{+\infty}\cos\left(2r\rho-\rho|x|\cos(\bar{\theta}+\widetilde{\theta})\right)\frac{1}{\rho^{2}}\mathcal{K}^{i}_{x}(\rho,\widetilde{\theta}+\bar{\theta}+\theta_{x})(1-\chi_{\ell}(\rho))(1-\chi_{\rho_{\star}}(\rho))\mathtt{d}\rho\\\
=\frac{-1}{(2r-|x|\cos(\bar{\theta}+\widetilde{\theta}))^{2}}\int_{\rho=0}^{+\infty}\cos\left(2r\rho-\rho|x|\cos(\bar{\theta}+\widetilde{\theta})\right)\frac{\partial^{2}}{\partial^{2}\rho}\left(\frac{1}{\rho^{2}}\mathcal{K}^{i}_{x}(\rho,\widetilde{\theta}+\bar{\theta}+\theta_{x})(1-\chi_{\ell}(\rho))(1-\chi_{\rho_{\star}}(\rho))\right)\mathtt{d}\rho.$
(A.66)

In order to estimate the size of the second derivative, we use (A.61) and
distinguish two regions:

  1. 1.

If $\rho_{\star}\leq\rho\leq\frac{1}{|p|}$, the $k^{th}$ derivative of
$\mathcal{K}^{i}_{x}$ (with respect to $\rho$) is bounded by
$\rho^{3-k}|p|^{3}$ (for $k=0,1,2$) and we get a pointwise bound of the form:

$\frac{\partial^{2}}{\partial^{2}\rho}\left(\frac{1}{\rho^{2}}\mathcal{K}^{i}_{x}(\rho,\widetilde{\theta}+\bar{\theta}+\theta_{x})(1-\chi_{\ell}(\rho))(1-\chi_{\rho_{\star}}(\rho))\right)\preceq\frac{1}{\rho}|p|^{3},$

  2. 2.

If $\frac{1}{|p|}\leq\rho$, the $k^{th}$ derivative of $\mathcal{K}^{i}_{x}$
(with respect to $\rho$) is bounded by $\rho^{1-k}|p|$ (for $k=0,1$) and
$|p|^{2}$ ($k=2$) and we get instead (there are two different terms, depending
on where the derivatives hit):

$\frac{\partial^{2}}{\partial^{2}\rho}\left(\frac{1}{\rho^{2}}\mathcal{K}^{i}_{x}(\rho,\widetilde{\theta}+\bar{\theta}+\theta_{x})(1-\chi_{\ell}(\rho))(1-\chi_{\rho_{\star}}(\rho))\right)\preceq{\frac{|p|}{\rho^{3}}+\frac{|p|^{2}}{\rho^{2}}},$

the second term being the dominant one since $\rho\leq\frac{1}{|p|}$.

Inserting both informations into the integral, we obtain:

$\int_{\rho=0}^{+\infty}\cos\left(2r\rho-\rho|x|\cos(\bar{\theta}+\widetilde{\theta})\right)\frac{\partial^{2}}{\partial^{2}\rho}\left(\frac{1}{\rho^{2}}\mathcal{K}^{i}_{x}(\rho,\widetilde{\theta}+\bar{\theta}+\theta_{x})(1-\chi_{\ell}(\rho))(1-\chi_{\rho_{\star}}(\rho))\right)\mathtt{d}\rho\\\
{\preceq\int_{\rho=\rho_{\star}}^{\frac{1}{|p|}}\frac{1}{\rho}|p|^{3}\mathtt{d}\rho+\int_{\rho=\frac{1}{|p|}}^{+\infty}\frac{|p|^{2}}{\rho^{2}}\mathtt{d}\rho=\mathcal{O}\left(|p|^{3}\left(\log\left(\frac{1}{|p|\rho_{\star}}\right)+1\right)\right)},$

Inserting this into (A.66), taking the expectation and integrating over
$\widetilde{\theta}$, we obtain:

$\frac{1}{r}\int_{|\widetilde{\theta}|\leq\bar{\theta}}\mathbb{E}\left[\int_{\rho=0}^{+\infty}\cos\left(2r\rho-\rho|x|\cos(\bar{\theta}+\widetilde{\theta})\right)\frac{1}{\rho^{2}}\mathcal{K}^{i}_{x}(\rho,\widetilde{\theta}+\bar{\theta}+\theta_{x})(1-\chi_{\ell}(\rho))(1-\chi_{\rho_{\star}}(\rho))\mathtt{d}\rho\
\mathsf{1}_{|p|\leq|\widetilde{\theta}|\sqrt{dr}}\right]\mathtt{d}\widetilde{\theta}\\\
\preceq\frac{1}{r}\mathbb{E}\left[\int_{|\widetilde{\theta}|\leq\bar{\theta}}\frac{1}{dr\widetilde{\theta}^{2}}|p|^{3}\left(\log\left(\frac{\sqrt{dr}\widetilde{\theta}}{|p|}\right)+1\right)\
\mathsf{1}_{|p|\leq|\widetilde{\theta}|\sqrt{dr}}\mathtt{d}\widetilde{\theta}\right]\\\
=\frac{1}{r}\times\frac{1}{\sqrt{dr}}\left(\mathbb{E}\left[|p|^{2}\right]\int_{w=1}^{+\infty}\frac{1}{w^{2}}(\log
w+1)\mathtt{d}w\right),$

where we used the change of variable
$w=\frac{\sqrt{dr}\widetilde{\theta}}{|p|}$. This last expression is again
$\mathcal{O}\left(\frac{1}{r}\times\frac{1}{\sqrt{dr}}\right)$.

We have thus proven (A.60).

#### Conclusion.

Combining (A.52) (which also applies to the “$i$-part”) with either (A.58) and
(A.59) (for the “$r$-part”) or (A.60) (for the “$i$-part”), we obtain (A.42),
and prove Lemma A.14. ∎

###  A.6 Bounding $II(x)$

We now turn to studying $II^{r}(x)$ and $II^{i}(x)$. We will prove the
following estimate:

######  Lemma A.15.

Assume that $x$ is such that $d\leq\bar{d}$. We have:

$|II^{r}(x)|+|II^{i}(x)|=\mathcal{O}\left(\frac{1}{r}\times\frac{1}{\sqrt{dr}}\right)+\mathcal{O}\left(\frac{1}{r^{2}}\right)\times\max\left(\log\frac{1}{\ell
d},0\right).$ (A.67)

#### Preliminaries

We want to study the following integral defining $II^{r}(x)$ (see (A.25)):

$II^{r}(x):=\int_{t\in\mathbb{R}^{2}}\cos(2r|t|-t\cdot
x)\frac{1}{r^{2}|t|^{4}}\Psi^{r}_{x}(t)(1-\chi_{\ell}(t))\mathtt{d}t,$

and similarly for $II^{i}(x)$ (strictly speaking, there is a multiplicative
constant $b$ in front of the integral, which we can safely forget about). The
approach here is very close to the analysis of $I_{2}(x)$ done previously, but
much shorter.

We use again polar coordinates and write:

$II^{r}(x):=\frac{1}{r^{2}}\int_{0}^{2\pi}\mathtt{d}\theta\int_{0}^{+\infty}\cos(2r\rho-\rho|x|\cos(\theta))\frac{1}{\rho^{3}}\Psi^{r}_{x}(\rho,\theta+\theta_{x})(1-\chi_{\ell}(\rho))\mathtt{d}\rho.$
(A.68)

We will use truncations $\chi_{\rho_{\star}}$ at certain lengthscales
$\rho_{\star}$ as before. The analogue of (A.32) is now:

$\int_{0}^{+\infty}\cos(2r\rho-\rho|x|\cos(\theta))\frac{1}{\rho^{3}}\Psi^{r}_{x}(\rho,\theta+\theta_{x})(1-\chi_{\ell}(\rho))\chi_{\rho_{\star}}(\rho)\mathtt{d}\rho=\mathcal{O}\left(\max\left(\log\left(\frac{\rho_{\star}}{\ell}\right),0\right)\right),$
(A.69)

and the analogue of (A.34) becomes:

$\int_{0}^{+\infty}\cos(2r\rho-\rho|x|\cos(\theta))\frac{\partial^{2}}{\partial^{2}\rho}\left(\frac{1}{\rho^{3}}\Psi^{r}_{x}(\rho,\theta+\theta_{x})(1-\chi_{\ell}(\rho))(1-\chi_{\rho_{\star}})(\rho)\right)\mathtt{d}\rho=\mathcal{O}\left(\frac{1}{\rho_{\star}^{2}}\right).$
(A.70)

(Note that we are, in both cases, losing a power of $\rho$ compared to
$I^{r}_{2}(x)$).

#### Points inside $\mathrm{B}_{2r}$.

To simplify, we take the same truncation length $\rho_{\star}=\frac{1}{d}$ for
all $\theta$. On the one hand, using (A.69), we get:

$\frac{1}{r^{2}}\int_{0}^{2\pi}\mathtt{d}\theta\int_{0}^{+\infty}\cos(2r\rho-\rho|x|\cos(\theta))\frac{1}{\rho^{3}}\Psi^{r}_{x}(\rho,\theta+\theta_{x})(1-\chi_{\ell}(\rho))\chi_{\rho_{\star}}(\rho)\mathtt{d}\rho=\frac{1}{r^{2}}\max\left(\log\left(\frac{1}{\ell
d}\right),0\right).$

On the other hand, using (A.70) and the rough bound
$\frac{1}{(2r-|x|\cos\theta)^{2}}\leq\frac{1}{d^{2}}$:

$\frac{1}{r^{2}}\int_{0}^{2\pi}\mathtt{d}\theta\frac{1}{(2r-|x|\cos\theta)^{2}}\int_{0}^{+\infty}\cos(2r\rho-\rho|x|\cos(\theta))\frac{\partial^{2}}{\partial^{2}\rho}\left(\frac{1}{\rho^{3}}\Psi^{r}_{x}(\rho,\theta+\theta_{x})(1-\chi_{\ell}(\rho))(1-\chi_{\rho_{\star}})(\rho)\right)\mathtt{d}\rho\\\
\preceq\frac{1}{r^{2}}\times\frac{1}{d^{2}}\times d^{2}=\mathcal{O}(r^{-2}).$

In conclusion, for points insde $\mathrm{B}_{2r}$ we obtain:

$II^{r}(x)=\mathcal{O}\left(\frac{1}{r^{2}}\right)\times\left(\max\left(\log\left(\frac{1}{\ell
d}\right),0\right)+1\right).$ (A.71)

The same analysis holds for $II^{i}(x)$.

#### Points outside $\mathrm{B}_{2r}$.

The proof goes along the same lines as Section A.5.

##### Large values ($\widetilde{\theta}\geq\bar{\theta}$).

As before, on this region we integrate by parts twice the entire
$\rho$-integral (i.e. we take $\rho_{\star}=\ell$), then use (A.70) (instead
of (A.34)), which gives:

$\int_{0}^{+\infty}\cos(2r\rho-\rho|x|\cos(\bar{\theta}+\widetilde{\theta}))\frac{\partial^{2}}{\partial^{2}\rho}\left(\frac{1}{\rho^{3}}\Psi^{r}_{x}(\rho,\bar{\theta}+\widetilde{\theta}+\theta_{x})(1-\chi_{\ell}(\rho))\right)\mathtt{d}\rho=\mathcal{O}\left(\frac{1}{\ell^{2}}\right),$

and using the control (A.50) on
$\frac{1}{(2r-|x|\cos(\bar{\theta}+\widetilde{\theta}))^{2}}$ we compute the
integral over $\widetilde{\theta}$ as:

$\frac{1}{r^{2}}\int_{\bar{\theta}}^{\pi-\bar{\theta}}\mathtt{d}\widetilde{\theta}\frac{1}{(2r-|x|\cos(\bar{\theta}+\widetilde{\theta}))^{2}}\int_{0}^{+\infty}\cos(2r\rho-\rho|x|\cos(\bar{\theta}+\widetilde{\theta}))\frac{\partial^{2}}{\partial^{2}\rho}\left(\frac{1}{\rho^{3}}\Psi^{r}_{x}(\rho,\bar{\theta}+\widetilde{\theta}+\theta_{x})(1-\chi_{\ell}(\rho))\right)\mathtt{d}\rho\\\
\preceq\frac{1}{r^{2}}\int_{\bar{\theta}}^{\pi-\bar{\theta}}\frac{1}{r^{2}\widetilde{\theta}^{2}}\times\frac{1}{\ell^{2}}\mathtt{d}\widetilde{\theta}=\mathcal{O}\left(\frac{1}{r}\times\frac{1}{r^{\frac{5}{2}}d^{\frac{1}{2}}\ell^{2}}\right)=\mathcal{O}\left(\frac{1}{r}\times\frac{1}{\sqrt{dr}}\right),$
(A.72)

using that $r\ell\gg 1$. This should be compared with the computation leading
to (A.52).

##### Moderate values ($|\widetilde{\theta}|\leq\bar{\theta}$).

Compared to what we did for the same regime when studying $I_{2}(x)$, things
are simpler and in particular we do not need to expand the integrand or to
make any real distinction between “$r$-part” and “$i$-part”. We choose
$\rho_{\star}=\frac{1}{\sqrt{dr}|\widetilde{\theta}|}$, and using (A.69),
(A.70) and (A.51) we obtain:

$\frac{1}{r^{2}}\int_{0}^{\bar{\theta}}\mathtt{d}\widetilde{\theta}{\int_{0}^{+\infty}\cos(2r\rho-\rho|x|\cos(\bar{\theta}+\widetilde{\theta}))\frac{1}{\rho^{3}}\Psi^{r}_{x}(\rho,\bar{\theta}+\widetilde{\theta}+\theta_{x})(1-\chi_{\ell}(\rho))}\mathtt{d}\rho\\\
=\frac{1}{r^{2}}\int_{0}^{\bar{\theta}}\max\left(\log\frac{1}{\sqrt{dr}\widetilde{\theta}\ell},0\right)\mathtt{d}\widetilde{\theta}+\frac{1}{r^{2}}\int_{0}^{\bar{\theta}}\frac{1}{rd\widetilde{\theta}^{2}}\times(dr\widetilde{\theta}^{2})\mathtt{d}\widetilde{\theta}.$
(A.73)

(estimating the “small-$\rho$” and “large-$\rho$” contributions separately).

The last integral gives a
$\mathcal{O}\left(\frac{1}{r}\times\frac{d^{\frac{1}{2}}}{r^{\frac{3}{2}}}\right)$
contribution, which is negligible compared to
$\frac{1}{r}\times\frac{1}{\sqrt{dr}}$ because $d\leq\bar{d}\ll r$, and we
forget about it. The first integral in the right-hand side of (A.73) can be
evaluated directly:

  * •

If $d\leq\ell^{-1}$ then the integrand is always equal to
$\log\frac{1}{\sqrt{dr}\widetilde{\theta}\ell}$ and we get, after a natural
change of variables:

$\frac{1}{r^{2}}\int_{0}^{\bar{\theta}}\log\frac{1}{\sqrt{dr}\widetilde{\theta}\ell}\mathtt{d}\widetilde{\theta}\preceq\frac{1}{r}\frac{1}{\sqrt{dr}r\ell}\int_{u=0}^{d\ell}-\log
u\mathtt{d}u\leq\frac{1}{r}\frac{1}{\sqrt{dr}r\ell}\int_{u=0}^{1}-\log
u\mathtt{d}u=o\left(\frac{1}{r}\frac{1}{\sqrt{dr}}\right),$

where we used that $\bar{\theta}\sim\sqrt{\frac{d}{r}}$.

  * •

If $d\geq\ell^{-1}$, then we should restrict the integral to
$\widetilde{\theta}\leq\frac{1}{\sqrt{dr}\ell}$ which gives again:

$\frac{1}{r^{2}}\int_{0}^{\frac{1}{\sqrt{dr}\ell}}\log\frac{1}{\sqrt{dr}\widetilde{\theta}\ell}\mathtt{d}\widetilde{\theta}=\frac{1}{r}\frac{1}{\sqrt{dr}r\ell}\int_{u=0}^{1}-\log
u\mathtt{d}u=o\left(\frac{1}{r}\frac{1}{\sqrt{dr}}\right).$

So in conclusion, we have:

${\frac{1}{r^{2}}\int_{0}^{\bar{\theta}}\mathtt{d}\widetilde{\theta}{\int_{0}^{+\infty}\cos(2r\rho-\rho|x|\cos(\bar{\theta}+\widetilde{\theta}))\frac{1}{\rho^{3}}\Psi^{r}_{x}(\rho,\bar{\theta}+\widetilde{\theta}+\theta_{x})(1-\chi_{\ell}(\rho))}\mathtt{d}\rho}=o\left(\frac{1}{r}\frac{1}{\sqrt{dr}}\right).$
(A.74)

The same computation can be done for the “$i$-part”.

#### Conclusion

Using (A.71) (if $x\in\mathrm{B}_{2r}$) or (A.72) and (A.74) (if
$x\notin\mathrm{B}_{2r}$) we obtain (A.67) and prove Lemma A.15.

###  A.7 Conclusion: proof of Proposition 3.7

#### Points that are far from the boundary

By Lemma A.1, we have (using the value $\bar{d}=\sqrt{r\ell^{-1}}$):

$\sum_{x\in\mathcal{L}\setminus\\{0\\},d\geq\bar{d},|x|\leq
100r}\mathcal{B}_{3}(x)\preceq\frac{1}{\sqrt{r\ell}}+\mathbb{E}\left[|\mathrm{p}_{0}|^{2}\mathsf{1}_{|\mathrm{p}_{0}|\geq\frac{\bar{d}}{2}}\right].$
(A.75)

#### Points that are close, but not too close to the boundary

######  Claim A.16.

We have:

$\sum_{x\in\mathcal{L},\tau\leq
d\leq\bar{d}}\mathcal{B}_{3}(x)\preceq\frac{1}{(r\ell)^{\frac{1}{4}}}+\frac{1}{r^{0.45}}.$
(A.76)

###### Proof of Claim A.16.

By (A.20) it is enough to study the sums of $I(x),II(x),III(x)$ separately.
Let us start with the latter: using (A.28) and the fact that there are
$\mathcal{O}(r^{2})$ points at distance $\leq\bar{d}$ from the boundary, we
have:

$\sum_{x\in\mathcal{L},\tau\leq d\leq\bar{d}}III(x)\preceq r^{2}\times
r^{-3}\ell^{-1}=\frac{1}{r\ell}.$

We split $I(x)$ further into $I_{1}(x)+I_{2}(x)$. Using (A.29) we see already
that $\sum_{x\in\mathcal{L},\tau\leq
d\leq\bar{d}}I_{1}(x)\preceq\frac{1}{r\ell}$.

##### Sum of $I_{2}(x)$.

Using Lemma A.13 and Lemma A.14 we have:

$\sum_{x\in\mathcal{L},\tau\leq
d\leq\bar{d}}|I^{r}_{2}(x)|+|I^{i}_{2}(x)|\preceq\sum_{x\in\mathcal{L},\tau\leq
d\leq\bar{d}}\frac{1}{r}\times\frac{1}{\sqrt{dr}}.$ (A.77)

First, we restrict ourselves to $d\geq 1$ and compare that part of the sum in
to an integral:

$\sum_{x\in\mathcal{L},1\leq
d\leq\bar{d}}\frac{1}{r}\times\frac{1}{\sqrt{dr}}\preceq\int_{x=1}^{\bar{d}}\frac{1}{\sqrt{xr}}\mathtt{d}x=\mathcal{O}\left(\sqrt{\frac{\bar{d}}{r}}\right)=\mathcal{O}\left(\frac{1}{(r\ell)^{\frac{1}{4}}}\right)\to
0.$

The part of the sum (A.77) corresponding to $\tau\leq d\leq 1$ can be
estimated by saying that there are $\mathcal{O}(r)$ lattice points in that
annulus and thus, with our choice of $\tau=r^{-0.1}\gg r^{-1}$:

$\sum_{x\in\mathcal{L},\tau\leq d\leq
1}\frac{1}{r}\times\frac{1}{\sqrt{dr}}\preceq\frac{1}{r^{0.45}}.$

This concludes the evaluation of the sum of $I_{2}(x)$.

##### Sum of $II(x)$.

We use Lemma A.15 and write:

$\sum_{x\in\mathcal{L},\tau\leq
d\leq\bar{d}}{|II^{r}(x)|+|II^{i}(x)|}\preceq\sum_{x\in\mathcal{L},\tau\leq
d\leq\bar{d}}\frac{1}{r}\times\frac{1}{\sqrt{dr}}+\frac{1}{r^{2}}\max\left(\log\frac{1}{\ell
d},0\right).$

The first part of the sum can be treated exactly as above, the only annoying
new term is the $\log$ one. This term only appears if $d\leq\ell^{-1}$ and by
comparison to an integral we have:

$\frac{1}{r^{2}}\sum_{x\in\mathcal{L},1\leq
d\leq\frac{1}{\ell}}\log\frac{1}{\ell
d}\preceq\frac{1}{r}\int_{x=1}^{\ell^{-1}}-\log(x\ell)\mathtt{d}x=\mathcal{O}\left(\frac{1}{r\ell}\right).$

In this last sum we only considered $d\geq 1$ so we are left with:

$\sum_{x\in\mathcal{L},\tau\leq d\leq 1}\frac{1}{r^{2}}\times\log\frac{1}{\ell
d}\preceq
r\times\frac{1}{r^{2}}\times\log\left(\frac{r}{\tau}\right)=\mathcal{O}\left(\frac{\log(r)}{r}\right),$

using our choice of $\tau\gg r^{-1}$.

Collecting all error terms and discarding negligible ones, we obtain (A.76),
which proves Claim A.16. ∎

#### Points that are very close to the boundary

To treat the contribution of the sum of $\mathcal{B}_{3}(x)$ for $d\leq\tau$,
we rely on the “number-theoretic” result stated in Lemma 2.2 to count those
lattice points and on the rough bound (3.24). We obtain, since
$\tau=r^{-0.1}\ll 1$:

$\sum_{x\in\mathcal{L},d\leq\tau}\mathcal{B}_{3}(x)=\left(\mathcal{O}(\tau
r)+\mathcal{O}(r^{1-\gamma})\right)\times\mathcal{O}(r^{-1})=\mathcal{O}(r^{-\delta}),$
(A.78)

for $\delta=\min(\gamma,0.1)>0$.

Combining (A.75), (A.76) and (A.78), and discarding negligible terms, we
obtain (for some $\delta>0$):

$\sum_{x\in\mathcal{L},x\neq 0,|x|\leq
100r}\mathcal{B}_{3}(x)\preceq\mathbb{E}\left[|\mathrm{p}_{0}|^{2}\mathsf{1}_{|\mathrm{p}_{0}|\geq\frac{\bar{d}}{2}}\right]+r^{-\delta}\preceq\mathbb{E}\left[|\mathrm{p}_{0}|^{2}\mathsf{1}_{|\mathrm{p}_{0}|\geq\ell^{-1}}\right]+r^{-\delta},$

using $\bar{d}\gg\ell^{-1}$, which gives (3.25) and proves Proposition 3.7.

##  Appendix B The case of dimension 1

######  Proposition B.1.

If $\mathbb{E}[|\mathrm{p}_{0}|]$ is finite, then $\mathbf{PL}$ is
hyperuniform. If $\mathbb{E}[|\mathrm{p}_{0}|^{2}]$ is finite, then
$\mathbf{PL}$ is class-I hyperuniform.

###### Proof.

We start as in the proof of Proposition 3.1 by writing:

$\frac{\mathrm{Var}\left[|\mathbf{PL}\cap\mathrm{B}_{r}|\right]}{|\mathrm{B}_{r}|}=\mathbb{E}\sum_{x\in\mathcal{L}\setminus\\{0\\}}\widehat{j_{r}}(x+\mathrm{p}_{x}-\mathrm{p}_{0})+1-|\mathrm{B}_{r}|=A+B,$

where $A$ is a deterministic contribution coming from the lattice $\mathbb{Z}$
and $B$ is the “error term” due to the perturbations:

$A:=\sum_{x\in\mathbb{Z}\setminus\\{0\\}}\widehat{j_{r}}(x)+1-|\mathrm{B}_{r}|,\quad
B:=\mathbb{E}\sum_{x\in\mathbb{Z}\setminus\\{0\\}}\widehat{j_{r}}(x+\mathrm{p}_{x}-\mathrm{p}_{0})-\widehat{j_{r}}(x).$

The term $A$ is $\mathcal{O}(r^{-1})$ by the same computation as in the
$\mathsf{d}=2$ case and we focus on $B$.

We first use Lemma 2.9 and write:

$B=\mathbb{E}\sum_{x\in\mathbb{Z}\setminus\\{0\\},|x|\leq
100r}\widehat{j_{r}}(x+\mathrm{p}_{x}-\mathrm{p}_{0})-\widehat{j_{r}}(x)+\mathcal{O}\left(\mathbb{E}\left[|\mathrm{p}_{0}|\mathsf{1}_{|\mathrm{p}_{0}|\geq
10r}\right]\right).$ (B.1)

Note that if $\mathbb{E}[|\mathrm{p}_{0}|]$ is finite then the error term
tends to $0$ as $r\to\infty$ and if $\mathbb{E}[|\mathrm{p}_{0}|^{2}]$ is
finite, then by Markov’s inequality it is actually $\mathcal{O}(r^{-1})$.

The function $\widehat{j_{r}}$ is a “tent” supported on $[-2r,2r]$ and whose
expression is given by $\widehat{j_{r}}(t)=1-\frac{|t|}{2r}$ on that interval.
In particular, it is continuous, $\mathcal{O}(r^{-1})$-Lipschitz and piecewise
affine, with a lack of differentiability at $0$ and $\pm 2r$. For
$x\in\mathbb{Z}$, we let $d:=\mathrm{dist}\left(x,\\{0,-2r,2r\\}\right)$. We
introduce some lengthscale $\bar{d}$ to be chosen later with $1\leq\bar{d}\ll
r$. First, using the Lipschitz bound on $\widehat{j_{r}}$ we have:

$\mathbb{E}\sum_{x\in\mathbb{Z},d\leq\bar{d}}\widehat{j_{r}}(x+\mathrm{p}_{x}-\mathrm{p}_{0})-\widehat{j_{r}}(x)\preceq\frac{\bar{d}}{r}\times\mathbb{E}[|\mathrm{p}_{0}|].$
(B.2)

Next, if $x$ is such that $d\geq\bar{d}$, we write:

$\mathbb{E}[\widehat{j_{r}}(x+\mathrm{p}_{x}-\mathrm{p}_{0})-\widehat{j_{r}}(x)]=\mathbb{E}\left[\left(\widehat{j_{r}}(x+\mathrm{p}_{x}-\mathrm{p}_{0})-\widehat{j_{r}}(x)\right)\mathsf{1}_{|\mathrm{p}_{x}-\mathrm{p}_{0}|\leq
d}\right]+\mathbb{E}\left[\left(\widehat{j_{r}}(x+\mathrm{p}_{x}-\mathrm{p}_{0})-\widehat{j_{r}}(x)\right)\mathsf{1}_{|\mathrm{p}_{x}-\mathrm{p}_{0}|\geq
d}\right].$

Since $\widehat{j_{r}}$ is affine on $(x-d,x+d)$ with slope
$\mathcal{O}(r^{-1})$, and $\mathrm{p}_{x}-\mathrm{p}_{0}$ is centered, we
have:

$\mathbb{E}\left[\left(\widehat{j_{r}}(x+\mathrm{p}_{x}-\mathrm{p}_{0})-\widehat{j_{r}}(x)\right)\mathsf{1}_{|\mathrm{p}_{x}-\mathrm{p}_{0}|\leq
d}\right]\leq\frac{1}{r}\mathbb{E}\left[|\mathrm{p}_{x}-\mathrm{p}_{0}|\mathsf{1}_{|\mathrm{p}_{x}-\mathrm{p}_{0}|\geq
d}\right]\preceq\frac{1}{r}\mathbb{E}\left[|\mathrm{p}_{0}|\mathsf{1}_{|\mathrm{p}_{0}|\geq\frac{d}{2}}\right].$

It remains to see that, since $\widehat{j_{r}}$ is
$\mathcal{O}(r^{-1})$-Lipschitz:

$\mathbb{E}\left[\left(\widehat{j_{r}}(x+\mathrm{p}_{x}-\mathrm{p}_{0})-\widehat{j_{r}}(x)\right)\mathsf{1}_{|\mathrm{p}_{x}-\mathrm{p}_{0}|\geq
d}\right]\preceq\frac{1}{r}\mathbb{E}\left[|\mathrm{p}_{x}-\mathrm{p}_{0}|\mathsf{1}_{|\mathrm{p}_{x}-\mathrm{p}_{0}|\geq
d}\right]\preceq\frac{1}{r}\mathbb{E}\left[|\mathrm{p}_{0}|\mathsf{1}_{|\mathrm{p}_{0}|\geq\frac{d}{2}}\right],$
(B.3)

from which we can derive the following simple bound:

$\mathbb{E}\sum_{x\in\mathbb{Z},d\geq\bar{d},|x|\leq
100r}\widehat{j_{r}}(x+\mathrm{p}_{x}-\mathrm{p}_{0})-\widehat{j_{r}}(x)\preceq\mathbb{E}\left[|\mathrm{p}_{0}|\mathsf{1}_{|\mathrm{p}_{0}|\geq\frac{\bar{d}}{2}}\right].$
(B.4)

If $\mathbb{E}[|\mathrm{p}_{0}|]$ is finite, the right-hand side goes to $0$
as soon as $\bar{d}\to\infty$. On the other hand, if
$\mathbb{E}[|\mathrm{p}_{0}|^{2}]$ is finite, we might prefer to return to
(B.3) and compute instead the sum:

$\frac{1}{r}\sum_{x\in\mathbb{Z},|x|\leq
100r,d\geq\bar{d}}\mathbb{E}\left[|\mathrm{p}_{0}|\mathsf{1}_{|\mathrm{p}_{0}|\geq\frac{d}{2}}\right]\preceq\frac{1}{r}\mathbb{E}\left[|\mathrm{p}_{0}|\sum_{d=\bar{d}}^{100r}\mathsf{1}_{|\mathrm{p}_{0}|\geq\frac{d}{2}}\right]\preceq\frac{1}{r}\mathbb{E}\left[|\mathrm{p}_{0}|^{2}\right]=\mathcal{O}(r^{-1}).$
(B.5)

In summary, if $\mathbb{E}[|\mathrm{p}_{0}|]$ is finite, we can choose
$1\ll\bar{d}\ll r$ in (B.2), (B.4) and see that the term $B$ goes to $0$ as
$r\to\infty$, so $\mathbf{PL}$ is hyperuniform. If moreover
$\mathbb{E}[|\mathrm{p}_{0}|^{2}]$ is finite then we can choose $\bar{d}=1$ in
(B.2), (B.5) and we find that $B=\mathcal{O}(r^{-1})$ thus $\mathbf{PL}$ is
class-I hyperuniform. ∎

### Acknowledgement

_DD and DF were supported in part by the Labex CEMPI (ANR-11-LABX-0007-01),
the ANR project RANDOM (ANR-19-CE24-0014) and by the CNRS GdR 3477 GeoSto._

_DF is supported by the Czech Science Foundation, project no. 22-15763S and
thanks the University of Lille for its hospitality._

_MH is supported by the Deutsche Forschungsgemeinschaft (DFG, German Research
Foundation) through the SPP 2265 Random Geometric Systems as well as under
Germany’s Excellence Strategy EXC 2044 -390685587, Mathematics Münster:
Dynamics–Geometry–Structure._

_TL acknowledges the support of JCJC grant ANR-21-CE40-0009 from Agence
Nationale de la Recherche, as well as the hospitality of Mathematics Münster,
where this project was initiated._

## References

  * [AGL21] Kartick Adhikari, Subhroshekhar Ghosh, and Joel L Lebowitz.  Fluctuation and entropy in spectrally constrained random fields.  Communications in Mathematical Physics, 386:749–780, 2021.
  * [AS68] Milton Abramowitz and Irene A Stegun.  Handbook of mathematical functions with formulas, graphs, and mathematical tables, volume 55.  US Government printing office, 1968.
  * [BDGZ24] Raphael Butez, Sandrine Dallaporta, and David García-Zelada.  On the Wasserstein distance between a hyperuniform point process and its mean.  https://arxiv.org/pdf/2404.09549.pdf, 04 2024.
  * [Cos21] Simon Coste.  Order, fluctuations, rigidities, 2021.
  * [Dur19] Rick Durrett.  Probability: Theory and Examples.  Cambridge University Press, April 2019.
  * [EHJM23] Matthias Erbar, Martin Huesmann, Jonas Jalowy, and Bastian Müller.  Optimal transport of stationary point processes: Metric structure, gradient flow and convexity of the specific entropy.  https://arxiv.org/pdf/2304.11145.pdf, 04 2023.
  * [Fol16] Gerald B Folland.  A course in abstract harmonic analysis, volume 29.  CRC press, 2016.
  * [Gab04] Andrea Gabrielli.  Point processes and stochastic displacement fields.  Physical Review E, 70(6):066131, 2004.
  * [GJL02] Andrea Gabrielli, Michael Joyce, and Francesco Sylos Labini.  Glass-like universe: Real-space correlation properties of standard cosmological models.  Physical Review D, 65(8):083523, 2002.
  * [Gne99] Tilmann Gneiting.  Radial positive definite functions generated by Euclid’s hat.  Journal of Multivariate Analysis, 69(1):88–119, 1999.
  * [GS75] Peter Gacs and Domokos Szász.  On a problem of Cox concerning point processes in $\mathbb{R}^{k}$ of “controlled variability”.  The Annals of Probability, 3(4):597–607, 1975.
  * [HL24] Martin Huesmann and Thomas Leblé.  The link between hyperuniformity, Coulomb energy, and Wasserstein distance to lebesgue for two-dimensional point processes.  https://arxiv.org/pdf/2404.18588.pdf, 04 2024.
  * [Hux96] Martin Neil Huxley.  Area, lattice points, and exponential sums, volume 13.  Clarendon Press, 1996.
  * [KT18] Jaeuk Kim and Salvatore Torquato.  Effect of imperfections on the hyperuniformity of many-body systems.  Physical Review B, 97(5):054105, 2018.
  * [Lan15] Edmund Landau.  Zur analytischen Zahlentheorie der definiten quadratischen Formen (Über die Gitterpunkte in einem mehrdimensionalen Ellipsoid).  Akad. Wiss., 31:458–476, 1915.
  * [Leb83] Joel Lebowitz.  Charge fluctuations in Coulomb systems.  Journal of Statistical Physics, (27):1491–1494, 1983.
  * [LP17] Günter Last and Mathew Penrose.  Lectures on the Poisson process, volume 7.  Cambridge University Press, 2017.
  * [LRY24] Raphaël Lachièze-Rey and D. Yogeshwaran.  Hyperuniformity and optimal transport of point processes.  https://arxiv.org/pdf/2402.13705.pdf, 02 2024.
  * [MY80] Philippe André Martin and Talat Yalcin.  The charge fluctuations in classical Coulomb systems.  Phys. Rev. A, (22):435–463, 1980.
  * [Sos00] Alexander Soshnikov.  Determinantal random point fields.  Russian Mathematical Surveys, 55(5):923, 2000.
  * [SW71] Elias M Stein and Guido Weiss.  Introduction to Fourier analysis on Euclidean spaces, volume 1.  Princeton university press, 1971.
  * [Tor18] Salvatore Torquato.  Hyperuniform states of matter.  Physics Reports, 745:1–95, 2018.
  * [TS03] Salvatore Torquato and Frank H Stillinger.  Local density fluctuations, hyperuniformity, and order metrics.  Physical Review E, 68(4):041113, 2003.


# A remark on a proof of $[G,L]=0$ for a Lie group $G$

Haruo Minami  H. Minami: Professor Emeritus, Nara University of Education
<EMAIL_ADDRESS>
###### Abstract.

In this note we give an improvement of our proof of $[G,L]=0$ for a compact
framed Lie group $(G,L)$, which depends heavily on the choice of a circle
subgroup $S\subset G$. We attempt here to make a more suitable choice of this
circle subgroup.

###### 2010 Mathematics Subject Classification:

57R15, 22E46, 19L20

##  1\. Introduction

Let $G$ be a compact connected Lie group and $L$ be its left invariant
framing. We denote by $[G,{\it L}]$ the framed bordism class of $(G,L)$. In
[3] we gave a proof of the following result, based on a proof technique
proposed in [4].

Theorem. For $G=SU(n)$, $SO(n)$, $Spin(n)$ $(n\geq 8)$; $Sp(n)$ $(n\geq 4)$;
$F_{4},$, $E_{6}$, $E_{7}$, $E_{8}$, we have $[G,L]=0$.

Here we give a direct proof by making a more suitable choice of the circle
subgroup $S\subset G$ which is a key ingredient of the formula used in [4].

From [4, 2] we know that $[G,L]_{(p)}=0$ for all primes $p\geq 5$, so we
restrict ourselves to the cases $p=2,3$ where $x_{(p)}$ is the image of $x$ by
the localizing homomorphism at $p$.

Suppose we are given a circle subgroup $S\subset G$ and a one-dimensional
representation $\gamma:S\to U(1)$. Let $\xi$ be the complex line bundle
associated with the principal $S$-bundle $G\to G/S$ by $\gamma$. Assuming that
its sphere bundle $S(\xi)\to G/S$ is isomorphic to $G\to G/S$, we consider the
Kronecker product of $[G/S]\in\pi_{d-1}^{S}(G/S^{+})$ and
$J(\beta\xi)\in\pi_{S}^{d}(S^{1}(G/S^{+}))$ ($d=\dim G$). Then by [4] we have

$[G,L]=-\langle J(\beta\xi)),[G/S]\rangle\quad\text{in}\ \pi_{d}^{S}.$

Here $G/S$ is a framed manifold with the framing inherited from $G$ in the
natural way and let $\beta$ be the Bott element and $J$ be the complex $J$
homomorphism.

In view of this formula, in order to prove that $[G,L]=0$ we show that
$J(\beta\xi)=0$ holds for $S\subset G$ specified depending on each $G$. But in
fact for the reasons mentioned above we show that $J(\beta\xi)_{(p)}=0$ holds
only for $p=2,3$.

For $j\in\mathbf{Z}$ we set $t^{j}=J(\beta\xi^{j})$ (where $\xi^{0}$ is the
trivial line bundle $G/S\times\mathbf{C}$). Then by the solution of the Adams
conjecture we have

$t_{(p)}^{j}=kt_{(p)}^{kj}\quad(k,j\neq 0)\quad\text{if}\ (p,k)=1.$

Also, since $J(\beta)$ becomes a generator of $\pi^{S}_{1}=\mathbf{Z}_{2}$, we
have

$2\\!\cdot\\!1=0$

where $1=t^{0}$. Applying these two relations to (1) below enables us to
calculate that $t_{(p)}=0$. But in the calculation below we drop for brevity
the subscript of $t_{(p)}$ except in some exceptional cases.

For any given $k_{1},\cdots,k_{\ell}\in\mathbf{Z}$ we write
$S(k_{1},\cdots,k_{\ell})$ for the circle subgroup of $G$ generated by ${\rm
diag}(z^{k_{1}},\cdots,z^{k_{\ell}})$, $z\in U(1)$, where ${\rm
diag}(c_{1},\cdots,c_{\ell})$ denote the diagonal matrix whose $ii$-entry is
$c_{i}$. We take $S=S(k_{1},\cdots,k_{\ell})$ for
$k_{1},\cdots,k_{\ell}\in\mathbf{Z}$ with $k_{j}=1$ for some $j$ and then
define $\gamma:S\to U(1)$ by ${\rm diag}(z^{k_{1}},\cdots,z^{k_{\ell}})\mapsto
z$. Suppose here that there exits a complex representation $\rho:G\to U(n)$
satisfying

$\rho|S=\gamma^{k_{1}}+\cdots+\gamma^{k_{\ell}}+(n-s)\quad(s\geq 0).$

Then it holds that
$\xi^{k_{1}}\oplus\cdots\oplus\xi^{k_{\ell}}\oplus(n-s)\xi^{0}\cong n\xi^{0}$
and therefore we have

(1) $t^{k_{1}+i}+\cdots+t^{k_{\ell}+i}=\ell t^{i}\quad(i\geq 0)$

Below we write $(1)_{i}$ for this equality to clarify that it belongs to the
formula of $t^{i}$. Besides in some cases, this equality is used in
combination with the one obtained by using $\lambda^{j}\rho$ instead of
$\rho$. .

##  2\. Proof of Theorem for classical Lie groups

###### Proof for the case $G=Sp(n)$.

Let $Sp(n)$ be embedded in $SU(2n)$ in the standard way. Let $\rho:Sp(n)\to
U(2n)$ be the restriction of the inclusion homomorphism $SU(2n)\to U(2n)$ to
$Sp(n)$. Take $S=S(1,2,3,-6,0,\cdots,0)\subset Sp(n)$. Then this circle
subgroup corresponds to $S(1,-1,2,-2,3,-3,-6,6,0,\cdots,0)$ in $SU(2n)$ via
the embedding above, so by (1) we have

(2)
$t^{1+i}+t^{2+i}+t^{3+i}+t^{6+i}+t^{-1+i}+t^{-2+i}+t^{-3+i}+t^{-6+i}=8t^{i}\quad(i\geq
0).$

Case $p=2$. From ${(2)}_{1}$, ${(2)}_{2}$, ${(2)}_{4}$ we have

$105t^{4}=916t+1,\ 45t^{4}+60t^{2}=76t+1,\ 840t^{4}-56t^{2}=176t$

(where the subscript of $t_{(2)}$ is omitted as noted above). By eliminating
$t^{2}$ and $t^{4}$ from these equations we have

$16t=0$

and therefore $t^{4}=4t+1$, $16t^{2}=0$. Substituting these equalities into
${(2)}_{5}$, ${(2)}_{2}$ we have $4t^{2}+8t=0$, $t^{8}=8t+1$. Finally,
substituting all these equalities into $(2)_{6}$ we obtain $t=0$, i.e.,
$t_{(2)}=0$.

Case $p=3$. From $(2)_{1}$, $(2)_{2}$, $(2)_{3}$ (with $t$ replaced by
$t_{(3)}$) we have similarly

$3t=0,\ t^{3}=0,\ t^{9}=0.$

In the above, replacing $\rho$ by $\lambda^{2}\rho$, we have a similar
equality to (2):

$\begin{array}[]{l}2t^{1+i}+t^{2+i}+2t^{3+i}+2t^{4+i}+2t^{5+i}+t^{7+i}+t^{8+i}+t^{9+i}+2t^{-1+i}\\\
+t^{-2+i}+2t^{-3+i}+2t^{-4+i}+2t^{-5+i}+t^{-7+i}+t^{-8+i}+t^{-9+i}=24t^{i}\quad(i\geq
0).\end{array}$

By substituting the three equalities obtained above into this equality for
$i=1$ we obtain $t=0$, i.e., $t_{(3)}=0$. ∎

###### Proof for the case $G=SU(n)$.

Let $\rho$ be the inclusion homomorphism $SU(n)\to U(n)$. We take
$S=S(1,-1,2,-2,3,-3,-6,6,0,\cdots,0)\subset SU(n)$ which has the same form as
that in the case $G=Sp(n)$. Then this choice of $S$ allows us to apply the
same argument and consequently leads us to the result. ∎

###### Proof for the case $G=SO(n)$.

Let $\rho:SO(n)\to U(n)$ be the complexification of the real inclusion
homomorphism $SO(n)\to O(n)$. Let $S\subset SO(n)$ be the circle subgroup
corresponding to $S(1,-1,2,-2,3,-3,-6,6,0,\cdots,0)$ in $SU(n)$ via the
canonical embedding $\iota:SO(n)\to SU(n)$. For the same reason as for the
above, applying the argument as in the above case we can obtain the result
similarly. ∎

###### Proof for the case $G=Spin(n)$.

Let $\rho=i\circ\pi:Spin(n)\to U(n)$ where $\pi$ denotes the natural covering
morphism $Spin(n)\to SO(n)$ and $i:SO(n)\to U(n)$ the complexification of the
inclusion homomorphism $SO(n)\to O(n)$. Let $S\subset Spin(n)$ be the circle
subgroup such that its image by $\pi$ corresponds to
$S(2,-2,2,-2,4,-4,6,-6,0,\cdots,0)$ in $SU(n)$ via $\iota$ above. It is clear
that by definition $S$ contains $-1\in Spin(n)$. Therefore by applying (1) we
have

(3) $2t^{2+i}+t^{4+i}+t^{6+i}+2t^{-2+i}+t^{-4+i}+t^{-6+i}=8t^{i}\quad(i\geq
0).$

Case $p=2$. From $(3)_{1}$ we have

$8t=0.$

In the above, replacing $\rho$ by $\lambda^{4}\rho$ we have

$\begin{array}[]{l}t^{2+i}+2t^{4+i}+5t^{6+i}+4t^{8+i}+4t^{10+i}+t^{14+i}+t^{-2+i}\\\
+2t^{-4+i}+5t^{-6+i}+4t^{-8+i}+4t^{-10+i}+t^{-14+i}=34t^{i}\quad(i\geq
0).\end{array}$

Substituting $8t=0$ above into this equality for $i=1$ we have $t=0$, i.e.,
$t_{(2)}=0$.

Case $p=3$. From $(3)_{1}$ we have $t_{(3)}=0$ by a simple calculation. ∎

##  3\. Proof of Theorem for exceptional Lie groups

###### Proof for the case $G=F_{4}$.

We know [4] that $F_{4}$ has $Spin(9)$ as a subgroup and a representation
$U:F_{4}\to U(26)$ such that its restriction to this $Spin(9)$ is
$1+\lambda^{1}+\Delta$, where $\lambda^{1}=i\circ\pi:Spin(9)\to SO(9)\to U(9)$
(with the notation above) and $\Delta$ is the spin representation. Take
$\rho=U$. Then, if we choose $S\subset F_{4}$ so that its image by $\pi$
corresponds to $S(2.-2,2.-2,2.-2,4.-4,0)$ in $SU(9)$ via the canonical
embedding $\iota$, then we have

$\rho\\!\mid\\!Spin(9)=(\gamma+\gamma^{-1})^{3}(\gamma^{2}+\gamma^{-2})+3\gamma^{2}+3\gamma^{-2}+\gamma^{4}+\gamma^{-4},$

so by (1) we have

(4)
$\begin{array}[]{l}4t^{1+i}+3t^{2+i}+3t^{3+i}+t^{4+i}+t^{5+i}+4t^{-1+i}+3t^{-2+i}\\\
+3t^{-3+i}+t^{-4+i}+t^{-5+i}=24t^{i}\quad(i\geq 0).\end{array}$

By replacing $\rho$ by $\lambda^{2}\rho$ we also have

(4’)
$\begin{array}[]{l}25t^{1+i}+24t^{2+i}+19t^{3+i}+19t^{4+i}+13t^{5+i}+10t^{6+i}+6t^{7+i}+3t^{8+i}\\\
+t^{9+i}+25t^{-1+i}+24t^{-2+i}+19t^{-3+i}+19t^{-4+i}+13t^{-5+i}+10t^{-6+i}\\\
+6t^{-7+i}+3t^{-8+i}+t^{-9+i}=240t^{i}\quad(i\geq 0).\end{array}$

Case $p=2$. From $(4)_{1}$, $(4)_{2}$ we have $20t^{2}+30t^{4}=392t$,
$2590t^{2}-315t^{4}=288t$, respectively. Further, calculating both $(4)_{3}$
and $(4)_{4}$ we have $7840t^{4}=3128t$. By eliminating $t^{2}$, $t^{4}$ from
these equations we have $8t=0$, so $32t^{2}=0$, $2t^{2}=5t^{4}$. Calculating
$(4)_{3}$ and $(4)_{4}$ again by use of these equalities, we have

$8t=0,\ 8t^{2}=0,\ t^{4}=2t^{2},\ t^{8}=4t^{2}+1.$

Using $(4^{\prime})_{1}$ the second equality is refined into $4t^{2}=0$, so it
follows that $t^{8}=1$. Substituting these equalities consequently we have
$t=0$, i.e., $t_{(2)}=0$.

Case $p=3$. From $(4)_{1}$, $(4)_{2}$ we have $3t=0$, $t^{3}=0$. Using these
equalities, from $(4)_{4}$ we have $t^{9}=0$. Substituting these equalities
into $(4^{\prime})_{1}$ yields $t=0$, i.e., $t_{(3)}=0$. ∎

###### Proof for the case $G=E_{6}$.

From [4] we know that $E_{6}$ has a subgroup $F_{4}$ and a representation
$W:E_{6}\to U(27)$ such that its restriction to $F_{4}\subset E_{6}$ is $1+U$.
This means that it enables us to apply the proof of the case $G=F_{4}$ to the
case here. Consequently we obtain the result. ∎

###### Proof for the case $G=E_{7}$.

By [4, Theorem. 11.1], $E_{7}$ contains $SU(8)/\\{\pm I\\}$ as a sybgroup
where $I\in SU(8)$ is the identity and a representation $\rho:E_{7}\to U(56)$
such that its restriction to this subgroup group is $\lambda^{2}+\lambda^{4}$.
Here $\lambda^{j}$ denotes the $j$-th exterior power of the standard
representation of $SU(8)$ on $\mathbf{C}^{8}$. Take $S=S(1,\cdots 1,-7)/\\{\pm
I\\}\subset E_{7}$. Then by (1) we have

(5) $21t^{1+i}+7t^{3+i}+21t^{-1+i}+7t^{-3+i}=52t^{i}\quad(i\geq 0).$

Here $t$ is the replacement of the square of $t$ defined for $S(1,\cdots,1-7)$
in (1). But by definition of (1) we find that in order to obtain the required
result it suffices to prove that $t_{(2)}=0$ and $t_{(3)}=0$ for this $t$.

In addition, replacing $\rho$ by $\lambda^{3}\rho$ in the above, we also have

(5’)
$\begin{array}[]{l}3t^{1+i}+t^{3+i}+6t^{5+i}+t^{7+i}+3t^{9+i}+6t^{-1+i}+3t^{-1+i}\\\
+t^{-3+i}+6t^{-5+i}+t^{-7+i}+3t^{-9+i}+6t^{1+i}=0\ \mod 8\qquad(i\geq
0).\end{array}$

Case $p=2$. From $(5)_{1}$, $(5)_{2}$ and $(5)_{3}$ we have $16t=0$, so
$8t^{2}=0$ and $t^{4}+2t^{2}=8t+1$. Using $(5)_{4}$ the first equality is
refined into $4t=0$, so the last one becomes $t^{4}+2t^{2}=1$. From $(5)_{5}$
we have $t^{8}=4t^{2}+1$. Substituting these equalities, from
$(5^{\prime})_{3}$, $(5^{\prime})_{4}$ we have $1=0$ and $t=1$, respectively.
Combining these two results we have $t=0$, i.e,. $t_{(2)}=0$.

Case $p=3$. From the calculation of $(5)_{1}$, $(5)_{2}$, $(5)_{3}$ we have
$t=0$, i.e., $t_{(3)}=0$. ∎

###### Proof for the case $G=E_{8}$.

We know [1] that $E_{8}$ contains $Spin(16)$ as a subgroup and the restriction
of the adjoint representation of $E_{8}$ to this $Spin(16)$ is
$\lambda^{2}+\Delta^{+}$ where $\lambda^{2}$ is the adjoint representation of
$Spin(16)$ and $\Delta^{+}$ the positive spinor representation. We choose here
a different $S\subset E_{8}$ in each case.

Case $p=2$. Let $S\subset E_{8}$ be the circle subgroup of $Spin(16)$ such
that its image by $\iota\circ\pi:Spin(16)\to SO(16)\to SU(16)$ is
$S(2,-2,\cdots,2,-2,6,-6,0,0)\subset SU(16)$. Then by (1) we have

(6)
$\begin{array}[]{l}21t^{1+i}+12t^{2+i}+21t^{3+i}+21t^{4+i}+15t^{5+i}+2t^{6+i}+6t^{7+i}+6t^{8+i}+t^{9+i}\\\
+21t^{-1+i}+12t^{-2+i}+21t^{-3+i}+21t^{-4+i}+15t^{-5+i}+2t^{-6+i}+6t^{-7+i}\\\
+6t^{-8+i}+t^{-9+i}=210t^{i}\quad(i\geq 0).\end{array}$

Calculating $(6)_{1}$, $(6)_{2}$, $(6)_{3}$, $(6)_{4}$ we have

$2^{5}\cdot 10223201t^{2}=2\cdot 1331664213t,\ \ 2^{2}\cdot
248349949873t^{2}=2^{2}\cdot 104608953537t.$

From these equalities it follows that $2t=0$ and thereby $4t^{2}=0$,
$2t^{4}=0$, $t^{8}=0$. Further, by using these equalities, from $(6)_{7}$ we
have $t^{16}=2t^{2}+1$. Finally, substituting these equalities into $(6)_{8}$
we obtain $t=0$, i.e., $t_{(2)}=0$.

Case $p=3$. We choose the circle subgroup of $Spin(16)$ as $S\subset E_{8}$
such that its image by $\iota\circ\pi:Spin(16)\to SO(16)\to SU(16)$ is
$S(2,-2,\cdots,2,-2,-2,2)\subset SU(16)$. Then by (1) we have

(7)
$56t^{2+2i}+28t^{8+2i}+8t^{6+2i}+56t^{-2+2i}+28t^{-8+2i}+8t^{-6+2i}=184t^{2i}\quad(i\geq
0).$

Here, thinking of $t^{2}$ as $t$ we prove that $t_{(3)}^{2}=0$, which means
that $t_{(3)}=0$, because of $2t_{(3)}^{2}=t_{(3)}$,

Combining the resuts of calculations of ${(7)}_{2i}$ for $i=1,2,3,4$, we have
$3t^{2}=0$, $t^{6}=0$. Calculating ${(7)}_{10}$ by use of these equalities we
have $t^{18}=0$. Substituting all these equalities into ${(7)}_{14}$, we
obtain $t^{2}=0$, i.e., $t^{2}_{(3)}=0$, so as stated above we conclude that
$t_{(3)}=0$. This completes the proof of the theorem. ∎

## References

  * [1] J. F. Adams, Lectures on Exceptional Lie Groups, Chicago Lectures in Math., University of Chicago Press, Chicago, 1996.
  * [2] K. Knapp, On odd-primary components of Lie groups Proc. Amer. Math. Soc. 79 (1980) 147–152.
  * [3] H. Minami, On framed simple Lie groups, J. Math. Soc. Japan 65 (2016) 1219–1229.
  * [4] E. Ossa, Lie groups as framed manifolds, Topology 21 (1982) 315–323.