# Logarithmically enhanced area-laws for fermions in vanishing magnetic fields
in dimension two

Paul Pfeiffer and Wolfgang Spitzer  Fakultät für Mathematik und Informatik,
FernUniversität in Hagen, Universitätsstraße 1, 58097, Hagen, Germany
<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

We consider fermionic ground states of the Landau Hamiltonian, $H_{B}$, in a
constant magnetic field of strength $B>0$ in $\mathbb{R}^{2}$ at some fixed
Fermi energy $\mu>0$, described by the Fermi projection $P_{B}\coloneqq
1(H_{B}\leq\mu)$. For some fixed bounded domain $\Lambda\subset\mathbb{R}^{2}$
with boundary set $\partial\Lambda$ and an $L>0$ we restrict these ground
states spatially to the scaled domain $L\Lambda$ and denote the corresponding
localised Fermi projection by $P_{B}(L\Lambda)$. Then we study the scaling of
the Hilbert-space trace, $\mathrm{tr}f(P_{B}(L\Lambda))$, for polynomials $f$
with $f(0)=f(1)=0$ of these localised ground states in the joint limit
$L\to\infty$ and $B\to 0$. We obtain to leading order logarithmically enhanced
area-laws depending on the size of $LB$. Roughly speaking, if $1/B$ tends to
infinity faster than $L$, then we obtain the known enhanced area-law (by the
Widom–Sobolev formula) of the form $L\ln(L)a(f,\mu)|\partial\Lambda|$ as
$L\to\infty$ for the (two-dimensional) Laplacian with Fermi projection
$1(H_{0}\leq\mu)$. On the other hand, if $L$ tends to infinity faster than
$1/B$, then we get an area law with an $L\ln(\mu/B)a(f,\mu)|\partial\Lambda|$
asymptotic expansion as $B\to 0$. The numerical coefficient $a(f,\mu)$ in both
cases is the same and depends solely on the function $f$ and on $\mu$. The
asymptotic result in the latter case is based upon the recent joint work of
Leschke, Sobolev and the second named author [8] for fixed $B$, a proof of the
sine-kernel asymptotics on a global scale, and on the enhanced area-law in
dimension one by Landau and Widom. In the special but important case of a
quadratic function $f$ we are able to cover the full range of parameters $B$
and $L$. In general, we have a smaller region of parameters $(B,L)$ where we
can prove the two-scale asymptotic expansion $\mathrm{tr}f(P_{B}(L\Lambda))$
as $L\to\infty$ and $B\to 0$.

###### Key words and phrases:

Area laws, Landau Hamiltonian, asymptotic analysis

###### 2010 Mathematics Subject Classification:

Primary 47G30, 35S05; Secondary 45M05, 47B10, 47B35

###### Contents

  1. 1 Introduction
    1. 1.1 Some notations and preliminary definitions
    2. 1.2 Reduction to the case $\mu=2$, introduction of $K$ and $L$
  2. 2 Preliminary asymptotic results on the integral kernel of the Fermi projection
    1. 2.1 The case $x\leq\frac{1}{2}$
    2. 2.2 The case $x\geq\frac{1}{2}$
    3. 2.3 An integral bound and an identity on the kernel $G_{K}$
  3. 3 On the particle number fluctuations
  4. 4 The case $K\gg L$
  5. 5 The case $K\ll L$: Overview
  6. 6 $K\ll L$: Norm estimates
  7. 7 $K\ll L$: Proof of Theorem 5.3
    1. 7.1 Proof of Theorem 5.3 for polygons $\Lambda$
    2. 7.2 Proof of Theorem 5.3 for $\mathsf{C}^{2}$-smooth domains $\Lambda$
  8. 8 $K\ll L$: Proof of Theorem 5.4
  9. A Two simple trace(-norm) inequalities
  10. B Sine-kernel asymptotics and the leading asymptotic coefficient
    1. B.1 Sine-kernel asymptotics on a global scale
    2. B.2 Evaluation of the leading asymptotic coefficient $\mathsf{M}_{<K}(f)$
  11. C Proof of the expansion (3.6)
  12. D Concluding remarks

##  1\. Introduction

In recent years, there has been a lot of efforts devoted to entanglement
entropy (EE). The motivation of the present work is to understand the
transition between a (strict) area law and an enhanced area law for the EE of
fermionic ground states.

Even in the simplest situation when there are no particle interactions
present, the entanglement (or rather local) entropy of ground states is a
complicated and interesting function of the defining parameters. Also, the
asymptotic behaviour (for large domains) of this entropy is related to the
asymptotic behaviour of Szegő-type asymptotics of Toeplitz or Wiener–Hopf
operators which has been studied since more than a century. It was Harold
Widom who conjectured in 1990 (see [23]) a formula in the higher-dimensional
setting and proved a special case. That conjecture led D. Gioev and I. Klich
[3] to conjecture in 2006 the asymptotic expansion of the EE of ground states
of the ideal Fermi gas. In 2013, A.V. Sobolev [17] proved Widom’s conjecture
which in turn paved the way to prove the conjecture on the EE by Leschke,
Sobolev and one of the present authors in [6].

In 2021, the same authors proved the area law for the EE of ground states of
the ideal Fermi gas in a constant magnetic field in the two-dimensional case,
see [8]. In the three-dimensional case the situation is different from the
start since the spectrum of the Landau Laplacian is now purely absolutely
continuous. We proved a logarithmically enhanced area-law recently in [15].

Some connections between the appearance of a strict area-law versus a
logarithmically enhanced area-law are obvious. For example, if the off-
diagonal integral kernel of the Fermi projection (characterizing the ground
state) is decaying fast (exponentially, say) then an area law holds. On the
other hand, purely absolutely continuous spectrum does not guarantee a
logarithmically enhanced area-law.

Let us recall some more mathematical results that add to the understanding of
EE of non-interacting Fermi gases. In [16], Pfirsch and Sobolev treat a
periodic (electric) potential $V$ in dimension one and prove a logarithmically
enhanced area-law. What is particularly interesting is that the second-order
term (or “surface” term of the order $\ln(L)$) is the same as for the
Laplacian, that is, with $V=0$. The higher dimensional case remains an open
problem.

Stability of the enhanced area-law by a local (compactly supported)
perturbation $V$ was proved by Müller and Schulte in [12, 10]. Motivated by
these papers, the first named author of the present paper proved the stability
of the area law for the two-dimensional Landau Hamiltonian by allowing a
perturbation on the magnetic potential and a perturbation by an electric
potential, see [14].

There are also results on the EE of random systems described by an Anderson-
type Hamiltonian. They concern the surprising logarithmic enhancement of EE in
the one-dimensional dimer model at a certain Fermi energy proved by Müller,
Pastur and Schulte [11]. In a more general case, Pastur and Slavin [13] and
Elgart, Pastur and Shcherbina [2] proved an area law for the EE at a Fermi
energy in the localisation regime for an Anderson model on the lattice
$\mathbb{Z}^{d}$. However, their formula for the leading coefficient is not
very explicit and it is not known how it depends on the disorder parameter. No
rigorous result is available when the Fermi energy lies in the delocalisation
regime of a random Hamiltonian, but this question touches on the notoriously
difficult problem of the existence of such a regime in the first place.

We continue here the study of the local or entanglement entropy of ground
states of the ideal Fermi gas in a constant magnetic field in dimension two.
To this end, we fix some Fermi energy $\mu>0$ and denote by $P_{B}\coloneqq
1(H_{B}\leq\mu)$ the spectral (or Fermi) projection of the Landau Hamiltonian

$H_{B}\coloneqq(-i\nabla-a)^{2}\,,$ (1.1)

where $i$ is the imaginary unit and
$\nabla=(\partial_{x_{1}},\partial_{x_{2}})$ is the gradient. We choose the
symmetric gauge $a(x)=(a_{1}(x),a_{2}(x))\coloneqq(x_{2},-x_{1})B/2$ for the
vector potential $a:\mathbb{R}^{2}\to\mathbb{R}^{2}$ generating the constant
magnetic field (vector) perpendicular to the plane with Cartesian coordinates
$x=(x_{1},x_{2})$. The strength of magnetic field is given by the real number
$B>0$. On (a suitable domain of) $\mathsf{L}^{2}(\mathbb{R}^{2})$, $H_{B}$
acts as a (positive) self-adjoint operator.

For some (bounded) Borel set $\Lambda\subset\mathbb{R}^{2}$ with Lebesgue
volume $|\Lambda|$, we consider the localised Fermi projection

$P_{B}(\Lambda)\coloneqq 1_{\Lambda}\,P_{B}\,1_{\Lambda}\,,$ (1.2)

where $1_{\Lambda}$ is the multiplication operator with the indicator function
of $\Lambda$. For our asymptotic results we assume in addition that $\Lambda$
is an open domain (that is, $\Lambda$ has only finitely many connected
components) with some “smoothness” properties of the boundary,
$\partial\Lambda$. The latter may be piecewise $\mathsf{C}^{2}$-smooth in our
first main result, piecewise $\mathsf{C}^{1}$-smooth in our second main result
and a polygon or $\mathsf{C}^{2}$-smooth in our third main result.

For some (suitable) “test” function $f$ we are then interested in the Hilbert-
space trace $\mathrm{tr}f(P_{B}(\Lambda))$. The most relevant cases are the
quadratic polynomial $f(t)=t(1-t)$ related to particle number fluctuations and
$f(t)=-t\ln(t)-(1-t)\ln(1-t)$ related to the von Neumann EE. For fixed
$\Lambda$, this trace is too complicated but it is of interest to study the
behaviour for large domains. To this end, we introduce a scaling parameter
$L>0$ and consider, for fixed $\Lambda$ and fixed $\mu$, the function
$(L,B)\mapsto\mathrm{tr}f(P_{B}(L\Lambda))$ as we let $L\to\infty$. This has
been completely analysed in [8] for fixed $B$, namely, the following area law
has been proved (under the condition that $\partial\Lambda$ is
$\mathsf{C}^{3}$-smooth),

$\mathrm{tr}f(P_{B}(L\Lambda))=L^{2}B\frac{|\Lambda|}{2\pi}(n+1)f(1)+L\sqrt{B}\,|\partial\Lambda|\,\mathsf{M}_{\leq
n}(f)+o(L)\,,$ (1.3)

see [8, Theorem 2]. Here, $n\coloneqq\lfloor(\mu/B-1)/2\rfloor$ is the number
of Landau levels below $\mu$ and the coefficient $\mathsf{M}_{\leq n}(f)$ is
defined in (8.3). Hence, if $f(1)=0$, then the leading contribution as
$L\to\infty$ is of the order $L|\partial\Lambda|$, which is the reason why it
is called an area law. We speak of an enhanced area-law if the leading term is
larger. The most prominent example is when there is an extra factor of
$\ln(L)$. Such a logarithmically enhanced area-law is present for the (free)
Laplacian (set $B=0$ in the above Hamiltonian $H_{B}$), as was conjectured by
Gioev and Klich in [3] and proved in [6].

The concrete purpose of the present paper is to study the transition from an
area law to a logarithmically enhanced area-law as $B$ vanishes and the number
of Landau levels $n$ tends to infinity. The results may also be interpreted as
a high energy limit where the magnetic strength $B$ is kept fixed and the
Fermi energy (or the number of Landau levels) and the scaling parameter $L$
tend to infinity.

We cannot use the above result (1.3) from the constant $B$ case directly as we
have no control over lower order error terms, which depend, in general, on $n$
and might, a priori, blow up as $n\to\infty$. The joint limit $B\to 0$ and
$L\to\infty$ (for fixed $\mu$) depends crucially on the product $BL$. We
venture to state the following conjecture, which we will prove in certain
relevant circumstances, in particular only for polynomials $f$.

######  Conjecture 1.1.

For any Hölder-continuous function $f$ with Hölder exponent strictly bigger
than $0$, which satisfies $f(0)=f(1)=0$, any bounded Lipschitz domain
$\Lambda$, and any $\mu>0$ we have the asymptotic expansion

$\displaystyle\operatorname{tr}f(1_{L\Lambda}1(H_{B}\leq\mu)1_{L\Lambda})=\begin{cases}\lvert\partial\Lambda\rvert\,\frac{2\sqrt{\mu}}{\pi}\,\mathsf{I}(f)L\ln(\sqrt{\mu}L)+o(\sqrt{\mu}L\ln(\sqrt{\mu}L))\,&\text{
if }BL<\sqrt{\mu}\,,\\\
\lvert\partial\Lambda\rvert\,\frac{2\sqrt{\mu}}{\pi}\,\mathsf{I}(f)L\ln(\mu/B)+o(\sqrt{\mu}L\ln(\mu/B))\,&\text{
if }BL\geq\sqrt{\mu}\,,\end{cases}$ (1.4)

as $L\to\infty$ and $B\to 0$, where we defined the functional

$\displaystyle
f\mapsto\mathsf{I}(f)\coloneqq\frac{1}{4\pi^{2}}\int_{0}^{1}\frac{f(t)}{t(1-t)}\,\mathrm{d}t\,.$
(1.5)

In order to explain the numerical factors in this formula we recall the
asymptotic formula for the Laplacian in two dimensions. As mentioned, it was
proved that for a Hölder-continuous function $f$ with $f(0)=f(1)=0$ (using the
notation of [6, (7)])

$\displaystyle\operatorname{tr}f(1_{L\Lambda}1(-\Delta\leq\mu)1_{L\Lambda})=J(\partial\Gamma,\partial\Lambda)\,\mathsf{I}(f)\sqrt{\mu}L\ln(\sqrt{\mu}L)+o(\sqrt{\mu}L\ln(\sqrt{\mu}L))\,,$
(1.6)

as $L\to\infty$, where $\Gamma\coloneqq\\{p\in\mathbb{R}^{2}:p^{2}\leq\mu\\}$,
$\hbar=1$, and

$\displaystyle
J(\partial\Gamma,\partial\Lambda)=\frac{2}{(\frac{1}{2})!}\left(\frac{\mu}{4\pi}\right)^{\frac{1}{2}}|\partial\Lambda|=\frac{2\sqrt{\mu}}{\pi}\,|\partial\Lambda|\,,$
(1.7)

with $(1/2)!\coloneqq\sqrt{\pi}/2$ being defined by Euler’s gamma function.

This transition of area laws is similar in spirit to the transition that
happens for the free Laplacian as one lets the temperature $T$ go to zero in
the study of the EE of equilibrium states. As proved by A.V. Sobolev [20] in
arbitrary spatial dimension and prior by Leschke, Sobolev and one of the
present authors in [7] in dimension one, there are two regions for $(T,L)$: if
$TL\to 0$, then there is an enhanced area-law of the order $L^{d-1}\ln(L)$. On
the other hand, if $TL\to\infty$, then we have an enhanced area-law of the
order $L^{d-1}\ln(T_{0}/T)$, for some temperature $T_{0}>0$. To draw the
connection between the two scenario one could identify $T$ with $B$.

However, from a technical point of view, the two cases differ in the sense
that there is a full-fledged pseudo-differential calculus for the study of the
Laplacian, or more general, of translation-invariant operators. This is not so
much the case for the magnetic Laplacian $H_{B}$ but we can rely on well-
established (asymptotic) properties of Hermite and Laguerre polynomials.

At the end of this section, we will argue by scaling that it suffices to
consider $\mu=2$, only. Moreover, we find it more convenient to switch from
$B$ to $\mu/(2B)\approx K\in\mathbb{N}$ and hence let $K\to\infty$. Our main
results in this paper are therefore formulated for $\mu=2$ and in the joint
limit $L\to\infty,K\to\infty$.

Our first main result is Theorem 3.1 which deals with particle number
fluctuations (that is, the function $f(t)=t(1-t)$) and the full parameter set
of $K$ and $L$. We can handle the quadratic test function $f$ because the
phase $\exp(ix\wedge y/(2K))$, which appears in the integral kernel of
$P_{B}$, cancels in the computation of the trace and is out of our way.
Nevertheless, it is an important case and we believe that is yields the right
picture in the general case. Therefore, we venture to state 1.1. We discuss
the quadratic case in Section 3.

Our second main result Theorem 4.1 is the logarithmically enhanced area-law of
the order $L\ln(L)$ if $K$ is much larger than $L$ and the Landau Hamiltonian
$H_{B}$ is “close” to the (free) two-dimensional Laplacian. Of course, for any
finite $K$ (or strictly positive $B$), the spectrum of $H_{B}$ is never
anything like that of the Laplacian and the off-diagonal integral kernel
$P_{B}(x,y)$ decays exponentially to 0 as the distance $\|x-y\|$ tends to
infinity. But the rate is given by $1/K$ and that goes to 0 in the end and we
do get the convergence to the integral kernel of $1(-\Delta\leq\mu)$. More
precisely, we are able to prove the enhanced area-law under the condition
$L\leq CK^{2/5}$ but, as just said, we believe that this holds true up to the
transition line $K=CL$. We should note that we use (but do not reprove) the
known result for the Laplacian ($1/K=0$ in a way) and consider the case with
small $1/K$ as a perturbation.

Our third main result is Theorem 5.2 and deals with the region $K\ll L$. Here,
due to the slow vanishing of the magnetic field, we are more in the regime of
a constant magnetic field and the area law is of the order $L\ln(K)$.
Interestingly, the $\ln(K)$ is a result of an enhanced area-law for the one-
dimensional Laplacian where $K$ is the effective scaling parameter. Distilling
the one-dimensional Laplacian is the result of the so-called sine-kernel
asymptotics for Hermite and Laguerre functions. The difficulty here is that we
need this asymptotics on a global scale, which takes up some space to prove.
We succeed to prove 1.1 (almost) over the full range of parameters $K\leq CL$
when the domain $\Lambda$ is a polygon; in fact, we have to assume $K\leq
CL/\ln(L)$. When $\partial\Lambda$ is $\mathsf{C}^{2}$-smooth we lose control
over some error terms and end up with the restriction $K^{2}\leq CL$.

We return to some open questions in Appendix D.

###  1.1. Some notations and preliminary definitions

A _domain_ $\Lambda$ is a (non-empty) bounded, open set in the two-dimensional
Euclidean space $\mathbb{R}^{2}$ with finitely many connected components. It
is called $\mathsf{C}^{r}$-smooth or piecewise $\mathsf{C}^{r}$-smooth if the
boundary $\partial\Lambda=\bar{\Lambda}\setminus\Lambda$ is a
$\mathsf{C}^{r}$-smooth curve, respectively a piecewise
$\mathsf{C}^{r}$-smooth curve, for some $r\in\mathbb{N}$. $\Lambda$ is called
Lipschitz if the boundary is Lipschitz continuous. The surface area
$|\partial\Lambda|$ is the one-dimensional Hausdorff measure of
$\partial\Lambda$.

By $D_{R}(x)$ we denote the open disk of radius $R>0$ at the centre
$x\in\mathbb{R}^{2}$ and by $D_{R}(S)\coloneqq\bigcup_{x\in S}D_{R}(x)$ the
$R$-neighbourhood of a set $S\subset\mathbb{R}^{2}$.

We denote the set of natural numbers by $\mathbb{N}\coloneqq\\{1,2,\ldots\\}$
and $\mathbb{N}_{0}\coloneqq\mathbb{N}\cup\\{0\\}$ the set including $0$.

The parameter $L$ is a positive real number which scales the domain $\Lambda$
and goes to infinity in our asymptotic results. The parameter $K$ is another
positive (in most cases natural) number and the inverse of the magnetic field
strength, which also tends to infinity (in most statements). It determines the
Fermi projection $P_{K}$, see (1.9).

The indicator function of a set $I\subset\mathbb{R}^{m}$ is denoted by $1_{I}$
and our notation does not distinguish between this function and the
multiplication operator by this function. The identity operator is denoted by
$\mathds{1}$.

We use the standard big-$\mathcal{O}$ and little-$o$ notation. That is, for
two functions $f$ and $g>0$, $f=\mathcal{O}(g)$ if $|f(L)|\leq Cg(L)$ for some
(finite) constant $C$ and sufficiently large $L$ and $f=o(g)$ if
$\limsup|f|/g(L)=0$. In the latter case, we also write $f\ll g$ or $g\gg f$.
In a series of estimates the specific value of a constant may change from line
to line without changing its name. Constants are always finite real numbers
and usually strictly positive.

###  1.2. Reduction to the case $\mu=2$, introduction of $K$ and $L$

As is well-known, the spectrum of the Landau Hamiltonian $H_{B}$ of (1.1)
equals $B(2\mathbb{N}_{0}+1)$. Each eigenvalue is infinitely degenerate. Let
$\Pi_{\ell,B}$ be the projection onto the eigenspace with eigenvalue
$B(2\ell+1)$ for $\ell\in\mathbb{N}_{0}$. For some fixed (Fermi energy)
$\mu>0$ we work with the spectral projection
$1(H_{B}\leq\mu)=\sum_{\ell=0}^{\nu}\Pi_{\ell,B}$ with
$\nu\coloneqq\lfloor(\mu/B-1)/2\rfloor$, also called Fermi projection.

The expressions we are interested in are only dependent on the eigenvalues of
the operator $1_{\Lambda}1(H_{B}\leq\mu)1_{\Lambda}$ for some domain
$\Lambda\subset\mathbb{R}^{2}$. For any $\lambda\in\mathbb{R}^{+}$, this
operator is unitarily equivalent to
$1_{\lambda^{-1}\Lambda}1(H_{\lambda^{2}B}\leq\lambda^{2}\mu)1_{\lambda^{-1}\Lambda}$.
We define $K\coloneqq\lfloor(\mu/B-1)/2\rfloor+1$. We can then assume without
loss of generality that $B=1/K$, as both sides of (1.4) are invariant under
this scaling. For the rescaled $\mu^{\prime}$ we observe
$K-1=\lfloor(\mu^{\prime}K-1)/2\rfloor=\lfloor(2K-1)/2\rfloor$ and this
implies $1(H_{1/K}\leq\mu^{\prime})=1(H_{1/K}\leq 2)$. This shows
$\mu^{\prime}=2+\mathcal{O}(1/K)$ and thus, replacing $\mu^{\prime}$ by $2$ on
the right-hand side of (1.4) only changes the leading term of the right-hand
side by an additive error term of order $L\ln(\min(L,K))/K=\mathcal{O}(L)$.
This is why, in the following, we always assume

$\displaystyle\mu=2\quad\mbox{ and }\quad B=1/K\;\mbox{ for
}\;K\in\mathbb{N}\,.$ (1.8)

Finally, we redefine the projection

$P_{K}\coloneqq 1(H_{B}\leq 2)=\sum_{\ell=0}^{K-1}\Pi_{\ell,1/K}\,.$ (1.9)

##  2\. Preliminary asymptotic results on the integral kernel of the Fermi
projection

This section starts with the integral kernel of the Fermi projector $P_{K}$
and collects various estimates on this projector, which are needed throughout
the paper. Most of them are probably well-known and we list them here for
completeness and the convenience of the reader. We do not claim any novelty.

We introduce the function $F_{K}^{(\alpha)}$ on $\mathbb{R}^{+}$, which is
related to the (generalised) Laguerre polynomial of degree $K-1$. Then we
study its asymptotic properties as $K$ becomes large. This is split into two
subsections, one is devoted to small arguments $x$, that is, to $x\leq 1/2$
and the other one to large arguments, that is, to $x\geq 1/2$. The main
results on the asymptotic expansion of the translation invariant part,
$G_{K}$, of the integral kernel of $P_{K}$ are collected in Theorem 2.4 and
2.11. The last subsection contains an integral bound on $G_{K}$, which is of
immediate use in the next section on particle number fluctuations.

In this section, we study the integral kernel of the Fermi projection and in
particular, how it behaves asymptotically for small magnetic fields. We will
see in which specific way it converges to the free projector. This kernel is
given for both $x=(x_{1},x_{2})$ and $y=(y_{1},y_{2})$ in $\mathbb{R}^{2}$ by

$\displaystyle P_{K}(x,y)$ $\displaystyle=1(H_{1/K}\leq 2)(x,y)$ (2.1)
$\displaystyle=\frac{1}{2\pi K}\exp\left(-\frac{1}{4K}\lVert
x-y\rVert^{2}+i\frac{1}{2K}x\wedge
y\right)\mathcal{L}_{K-1}^{(1)}\left(\frac{1}{2K}\lVert x-y\rVert^{2}\right)$
(2.2) $\displaystyle=\exp\left(i\frac{1}{2K}x\wedge y\right)G_{K}(\lVert
x-y\rVert/\sqrt{8})\,,$ (2.3)

where $x\wedge y\coloneqq x_{1}y_{2}-x_{2}y_{1}$,

$\displaystyle\mathcal{L}_{K-1}^{(\alpha)}(t)\coloneqq\sum_{j=0}^{K-1}\frac{(-1)^{j}}{j!}{K-1+\alpha\choose
K-1-j}t^{j}\,,\quad t\geq 0$

is the (generalised) Laguerre polynomial of degree $K-1$ (for any
$\alpha\in\mathbb{R}$), see [21, (5.1.6)] and

$\displaystyle G_{K}(t)\coloneqq\frac{1}{2\pi
K}\exp\left(-4K\frac{t^{2}}{2K^{2}}\right)\mathcal{L}_{K-1}^{(1)}\left(4K\frac{t^{2}}{K^{2}}\right),\quad
t\geq 0\,.$ (2.4)

We need to be a bit more general and define for $\alpha\in\mathbb{R}$ and
$\nu\coloneqq 4K+2(\alpha-1)$ the function

$\displaystyle F_{K}^{(\alpha)}(x)\coloneqq
2^{\alpha}\sqrt{\nu}x^{\frac{1}{2}\alpha+\frac{1}{4}}\lvert
1-x\rvert^{\frac{1}{4}}\exp(-\nu x/2)\mathcal{L}_{K-1}^{(\alpha)}(\nu
x)\,,\quad x>0\,.$ (2.5)

The definition of this function is based on the asymptotic analysis of
$G_{K}$, which can be understood using $\alpha=1$. It can be found in [1,
(18.15.(iv))]. We will consider the equations (18.15.17) to (18.15.23) in [1].
In particular, the equations (18.15.19) and (18.15.22) provide us with the
asymptotics for $G_{K}(t)$. We start by identifying the variables and
parameters. According to (18.15.17) and trivialities, we see that

$\displaystyle\alpha=1\,,\quad n=K-1\,,\quad\nu=4K\,,\quad
x=\frac{t^{2}}{K^{2}}\,.$ (2.6)

Later on, we will also use the cases $\alpha=\pm\frac{1}{2}$ to find the
asymptotic expansion of the Hermite polynomials.

This allows us to rewrite $G_{K}$ as

$\displaystyle G_{K}(t)=\frac{1}{8\pi t^{3/2}\left\lvert
1-\frac{t^{2}}{K^{2}}\right\rvert^{1/4}}F_{K}^{(1)}\left(\frac{t^{2}}{K^{2}}\right).$
(2.7)

We intend to show

$\displaystyle
F_{K}^{(\alpha)}(x)=\begin{cases}\sqrt{\nu\xi(x)}J_{\alpha}(\nu\xi(x))+\mathcal{O}(1/K)&\text{
if }0\leq x\leq 1/2\,,\\\
\sqrt{2/\pi}\cos(\nu\xi(x)+g(\alpha))+\mathcal{O}(1/(1+K\sqrt{x}))&\text{ if
}0\leq x\leq 1/2\,,\\\
\sqrt{2/\pi}\cos(\nu\xi(x)+g(\alpha))+\mathcal{O}(1/(1+K(1-x)^{3/2}))&\text{
if }1/2\leq x\leq 1\,,\\\ 0+\mathcal{O}(\exp(-K(x-1)^{3/2}/10^{4})&\text{ if
}1\leq x\leq 3/2\,,\\\ 0+\mathcal{O}(\exp(-Kx/10^{4})&\text{ if }3/2\leq
x\,.\end{cases}$ (2.8)

The factors $10^{4}$ are obviously placeholders and can depend on $\alpha$,
but not on $K$. The function $\xi$ is given in (2.9) below (or [1,
(18.15.18)]) and $g(\alpha)=-\pi\alpha/2-\pi/4$ is affine linear.

$\displaystyle\xi(x)\coloneqq\frac{1}{2}\left(\sqrt{x-x^{2}}+\operatorname{arcsin}(\sqrt{x})\right),\quad
x\in[0,1]\,.$ (2.9)

Let us also define $\eta\colon[-1,\infty)\to\mathbb{C}$ by $\eta(0)=0$ and for
any $t\in(-1,\infty)$

$\displaystyle\eta^{\prime}(t)\coloneqq\begin{cases}\sqrt{1-t^{2}}\,,\quad&\text{
if }-1\leq t\leq 1\,,\\\ i\sqrt{t^{2}-1}\,,\quad&\text{ if }t\geq
1\,.\end{cases}$ (2.10)

An explicit formula for $\eta$ is given by

$\displaystyle\eta(t)=\begin{cases}1/2\left(t\sqrt{1-t^{2}}+\arcsin(t)\right),\quad&\text{
if }-1\leq t\leq 1\,,\\\
i/2\left(t\sqrt{t^{2}-1}-\operatorname{arccosh}(t)\right)+\pi/4\,,\quad&\text{
if }t\geq 1\,.\end{cases}$ (2.11)

We see that this function satisfies

$\displaystyle\eta(\sqrt{t})=\xi(t)\quad\mbox{for any }t\in[0,1]\mbox{ and
}\eta(1)=\pi/4\,.$ (2.12)

At this point, let us also define the function $\zeta$

$\displaystyle\zeta(s)\coloneqq-\left[3/2(\pi/4-\eta(\sqrt{s}))\right]^{2/3}\,,\quad
s\geq 0\,.$ (2.13)

In the last equation, the terms inside the brackets is either in
$\mathbb{R}^{+}$ or in $(-i)\mathbb{R}^{+}$ and thus the two thirds power has
a unique real value with the branch chosen by $(-i)^{2/3}=-1$. Thus, for
$s\geq 1$, we get

$\displaystyle\zeta(s)=-\left[3/2(-i)\,\mathfrak{Im}(\eta(\sqrt{s}))\right]^{2/3}=\left[3/2\,\mathfrak{Im}(\eta(\sqrt{s}))\right]^{2/3}\,,$
(2.14)

where $\mathfrak{Im}$ refers to the imaginary part.

######  Lemma 2.1.

For $h\in(-1,1)$, the function $\eta$ satisfies the following properties:

$\displaystyle\eta(h)$ $\displaystyle=h+\mathcal{O}(h^{3})\,,$ (2.15)
$\displaystyle|\eta(h)|$ $\displaystyle\geq(\pi/4)|h|\,,$ (2.16)
$\displaystyle 3/2\big{(}\pi/4-\eta(1-h)\big{)}$
$\displaystyle=\sqrt{2}\Big{(}\sqrt{h}^{3}-\frac{3}{20}\sqrt{h}^{5}+\mathcal{O}(\sqrt{\lvert
h\rvert}^{7})\Big{)}\,,$ (2.17)

where $\sqrt{h}=i\sqrt{-h}$ for $h\leq 0$. Furthermore, $\eta^{-1}$ is
$2/3$-Hölder continuous on $\eta([-1,2])$.

###### Proof.

The expansion for $\eta(h)$ is trivial. The lower bound for $\eta(h)/h$ holds
for positive $h$, as $\eta$ is concave on $[0,1]$ and thus, the line from
$(0,0)$ to $(1,\pi/4)$ lies below the graph of $\eta$. For negative $h$, the
claim follows as $\eta(-h)=-\eta(h)$.

The expansion for $\eta(1-h)$ follows from $\eta(1)=\pi/4$ and

$\displaystyle\eta^{\prime}(1-h)=\sqrt{1-(1-h)^{2}}=\sqrt{h(2-h)}=\sqrt{2}\sqrt{h}-\frac{\sqrt{2}}{4}\sqrt{h}^{3}+\mathcal{O}(|h|^{5/2})\,,$
(2.18)

where we choose the branch of the square root, which satisfies
$\sqrt{-t}=i\sqrt{t}$ for any $t\geq 0$. Integrating this equation leads to

$\displaystyle
3/2\big{(}\pi/4-\eta(1-h)\big{)}=\sqrt{2}\Big{(}\sqrt{h}^{3}-\frac{3}{20}\sqrt{h}^{5}+\mathcal{O}(|h|^{7/2})\Big{)}\,.$
(2.19)

The inverse $\eta^{-1}$ is locally Lipschitz continuous away from
$\eta(1)=\pi/4$, as $\eta^{\prime}$ only vanishes at $1$. From the expansion
of $\eta(1-h)$, we can thus conclude that $\eta^{-1}$ is $2/3$-Hölder
continuous on $\eta([-1,2])$. ∎

With these preparations we discuss in the following two subsections pointwise
estimates on $G_{K}$. In the last subsection we present an estimate on the
integral of $|G_{K}(t)|^{2}$ and a simple integral identity. All these
estimates will be useful in later sections.

###  2.1. The case $x\leq\frac{1}{2}$

In this subsection, we will establish the necessary understanding of [1,
(18.15.19)]. We would like to establish an asymptotic expansion up to a
sufficiently small error term.

As [1, (18.15.19)] reduces the Laguerre polynomial asymptotic to the Bessel
functions $J_{\alpha}$, we take a look at their asymptotics.

######  Proposition 2.2.

For $s\geq 0$, we have

$\displaystyle\sqrt{s}J_{\alpha}(s)=\begin{cases}\sqrt{2/\pi}\cos(s-\alpha\pi/2-\pi/4)&\text{
if }\alpha=\pm 1/2\,,\\\
\sqrt{2/\pi}\cos(s-\alpha\pi/2-\pi/4)+\mathcal{O}(1/(1+s))&\text{ if
}\alpha=1\,.\end{cases}$ (2.20)

###### Proof.

The first part is just [1, (10.16.1)] and the second part is [1, (10.17.3)]. ∎

######  Lemma 2.3.

For $0\leq x\leq 1/2$ and $\alpha\in\\{\pm 1/2,1\\}$, we have

$\displaystyle
F_{K}^{(\alpha)}(x)=\sqrt{\nu\xi(x)}J_{\alpha}(\nu\xi(x))+\mathcal{O}\left(1/K\right).$
(2.21)

In particular, for $\alpha=\pm 1/2$, we get

$\displaystyle
F_{K}^{(\alpha)}(x)=\sqrt{2/\pi}\cos(\nu\xi(x)-\alpha\pi/2-\pi/4)+\mathcal{O}\left(1/K\right).$
(2.22)

Furthermore, if $\alpha=1$, as $x\to 0$, the error term is at most
$\mathcal{O}(\sqrt{K}x^{3/4})$.

###### Proof.

The asymptotic expansion [1, (18.15.19)] with $\nu=4K+2(\alpha-1)$ and $M=1$
can be solved for $F_{K}^{(\alpha)}$ and provides us with

$\displaystyle
F_{K}^{(\alpha)}(x)=\sqrt{\nu}\left(\sqrt{\xi(x)}J_{\alpha}(\nu\xi(x))+{\xi(x)^{-\frac{1}{2}}}J_{\alpha+1}(\nu\xi(x))\frac{B_{0}(\xi(x))}{\nu}+\sqrt{\xi(x)}\,\mathrm{env}J_{\alpha}(\nu\xi(x))\mathcal{O}\left(\frac{1}{\nu}\right)\right).$
(2.23)

We are left to establish upper bounds for the last two summands. We need to
bound the coefficient $B_{0}$. For that, we recall (2.12) and (2.15) to
observe

$\displaystyle\xi(x)$
$\displaystyle=\eta(\sqrt{x})=\sqrt{x}+\mathcal{O}\left(x^{\frac{3}{2}}\right).$
(2.24)

Hence, the coefficient described in [1, (18.15.20)],

$\displaystyle
B_{0}(\xi(x))\coloneqq-\frac{1}{2}\left(\frac{1-4\alpha^{2}}{8}+\xi(x)\left(\frac{1-x}{x}\right)^{\frac{1}{2}}\left(\frac{4\alpha^{2}-1}{8}+\frac{1}{4}\frac{x}{1-x}+\frac{5}{24}\left(\frac{x}{1-x}\right)^{2}\right)\right),$
(2.25)

satisfies $\lvert B_{0}(\xi(x))\rvert\leq Cx\leq C\xi(x)^{2}$ for $0\leq
x\leq\frac{1}{2}$ with a constant $C$ depending on $\alpha$.

Using [1, (2.8.32–34)] and [1, (10.17.3–4)], we see that for $s\geq 1$, we
have

$\displaystyle\max\\{\lvert J_{\alpha}(s)\rvert,\lvert
J_{\alpha+1}(s)\rvert,\lvert\text{env}J_{\alpha}(s)\rvert\\}$
$\displaystyle\leq C/\sqrt{s}$ (2.26)

and for $0\leq s\leq 1$, [1, (10.2.2)] tells us that

$\displaystyle\max\\{\lvert J_{\alpha}(s)\rvert,\lvert
J_{\alpha+1}(s)\rvert,\lvert\text{env}J_{\alpha}(s)\rvert\\}$
$\displaystyle\leq Cs^{\alpha}\,.$ (2.27)

Thus, as $\alpha\geq-1/2$, (2.26) holds for any $s\geq 0$. Hence, for any
$0\leq x\leq 1/2$, we observe

$\displaystyle\sqrt{\nu}$
$\displaystyle\left({\xi(x)^{-\frac{1}{2}}}J_{\alpha+1}(\nu\xi(x))\frac{B_{0}(\xi(x))}{\nu}+\sqrt{\xi(x)}\,\text{env}J_{\alpha}(\nu\xi(x))\,\mathcal{O}\left(\frac{1}{\nu}\right)\right)$
(2.28)
$\displaystyle=\mathcal{O}\left({\xi(x)^{-\frac{1}{2}}}\frac{\sqrt{\nu}}{\sqrt{\nu\xi(x)}}\frac{\xi(x)^{2}}{\nu}+\sqrt{\nu\xi(x)}\frac{1}{\sqrt{\nu\xi(x)}}\frac{1}{\nu}\right)$
(2.29) $\displaystyle=\mathcal{O}((1+\xi(x))/\nu)=\mathcal{O}(1/K)\,.$ (2.30)

This proves the first claim. The second one follows by (2.20). Let $\alpha=1$.
If $\nu\xi(x)\geq 1$, we have $\sqrt{x}\geq C/K$, which implies
$\sqrt{K}x^{3/4}>C/K$ and thus we may assume $\nu\xi(x)\leq 1$. Then, we get

$\displaystyle\sqrt{\nu}$
$\displaystyle\left({\xi(x)^{-\frac{1}{2}}}J_{\alpha+1}(\nu\xi(x))\frac{B_{0}(\xi(x))}{\nu}+\sqrt{\xi(x)}\,\text{env}J_{\alpha}(\nu\xi(x))\,\mathcal{O}\left(\frac{1}{\nu}\right)\right)$
(2.31)
$\displaystyle=\mathcal{O}\left({\xi(x)^{-\frac{1}{2}}}\sqrt{\nu}(\nu\xi(x))^{1}\frac{\xi(x)^{2}}{\nu}+\sqrt{\nu\xi(x)}(\nu\xi(x))^{1}\frac{1}{\nu}\right)$
(2.32)
$\displaystyle=\mathcal{O}(\sqrt{\nu}\xi(x)^{3/2})=\mathcal{O}(\sqrt{K}x^{3/4})\,.$
(2.33)

This was the third claim. ∎

This brings us to the first main result on the translation-invariant part of
the integral kernel of the Fermi projection in case the argument $t$ is small
compared to $K$.

######  Theorem 2.4.

We have for $K\in\mathbb{N}$, we have

$\displaystyle
G_{K}(t)=\begin{cases}\frac{\cos(\omega_{K}(t)-3\pi/4)}{4\sqrt{2}\pi^{\frac{3}{2}}t^{\frac{3}{2}}\left(1-\frac{t^{2}}{K^{2}}\right)^{\frac{1}{4}}}+\mathcal{O}\left(\frac{1}{t^{\frac{5}{2}}}\right)\quad&\text{
if }1\leq t\leq\sqrt{\frac{1}{2}}K\,,\\\ \frac{J_{1}(4t)}{4\pi
t}+\mathcal{O}\left(\frac{(1+t)^{\frac{3}{2}}}{K^{2}}+\frac{1}{K(1+t)^{\frac{3}{2}}}\right)&\text{
if }0\leq t\leq K^{2/3}\,,\end{cases}$ (2.34)

where

$\displaystyle\omega_{K}(t)\coloneqq
4K\xi\left(\frac{t^{2}}{K^{2}}\right)=4K\eta(t/K)=2\left(t\sqrt{1-\frac{t^{2}}{K^{2}}}+K\operatorname{arcsin}\left(\frac{t}{K}\right)\right).$
(2.35)

The regions mentioned in this theorem overlap for $K>1$. The second case
provides the way, in which the kernel converges to the free kernel, that is,
to the integral kernel of the two-dimensional Laplacian.

###### Proof.

As $\nu=4K$, we have

$\displaystyle\omega_{K}(t)=\nu\xi(t^{2}/K^{2})\,.$ (2.36)

Since $t\leq K/\sqrt{2}$, we have $1/2\leq 1-t^{2}/K^{2}\leq 1$. We use
$\nu=4K$, (2.7) and 2.3 and see

$\displaystyle G_{K}(t)$ $\displaystyle=\frac{1}{8\pi
t^{3/2}\left(1-\frac{t^{2}}{K^{2}}\right)^{1/4}}\left(\sqrt{\omega_{K}(t)}J_{1}(\omega_{K}(t))+\mathcal{O}\left(\min\\{1/K,\sqrt{K}(t^{2}/K^{2})^{3/4}\\}\right)\right)$
(2.37) $\displaystyle=\frac{\sqrt{\omega_{K}(t)}J_{1}(\omega_{K}(t))}{8\pi
t^{3/2}\left(1-\frac{t^{2}}{K^{2}}\right)^{1/4}}+\mathcal{O}\left(\min\\{1/(Kt^{3/2}),1/K\\}\right)$
(2.38) $\displaystyle=\frac{\sqrt{\omega_{K}(t)}J_{1}(\omega_{K}(t))}{8\pi
t^{3/2}\left(1-\frac{t^{2}}{K^{2}}\right)^{1/4}}+\mathcal{O}\left(1/(K(1+t^{3/2}))\right).$
(2.39)

As $K>t$, we have $1/(K(1+t^{3/2}))\leq 1/t^{5/2}$. Thus, the error term is
currently good enough for both results. Due to (2.36) and (2.24), we have

$\displaystyle\omega_{K}(t)=4t+\mathcal{O}(t^{3}/K^{2}).$ (2.40)

We also have $\omega_{K}(t)\geq Ct$, see (2.16), which implies
$1/(1+\omega_{K}(t))\leq C/(1+t)$.

We consider the case $t\geq 1$. According to 2.2, we observe

$\displaystyle G_{K}(t)=$
$\displaystyle\frac{\sqrt{2/\pi}\cos(\omega_{K}(t)-3\pi/4)+\mathcal{O}(1/(1+\omega_{K}(t)))}{8\pi
t^{3/2}\left(1-\frac{t^{2}}{K^{2}}\right)^{1/4}}+\mathcal{O}\left(1/(K(1+t^{3/2}))\right)$
(2.41) $\displaystyle=$
$\displaystyle\frac{\cos(\omega_{K}(t)-3\pi/4)}{4\sqrt{2}\pi^{\frac{3}{2}}t^{\frac{3}{2}}\left(1-\frac{t^{2}}{K^{2}}\right)^{\frac{1}{4}}}+\mathcal{O}\left(\frac{1}{t^{\frac{5}{2}}}\right).$
(2.42)

This finishes the proof of the first claim.

We are left with the case $0\leq t\leq K^{\frac{2}{3}}$. In this case, we want
to fully eliminate the dependency of the leading term on $K$. We consider the
function $s\mapsto\sqrt{s}J_{1}(s)$ and want to study its derivative. For
that, we note that $J_{1}^{\prime}(s)=-J_{2}(s)+J_{1}(s)/s$ (see [1,
(10.6.2)]). We get

$\displaystyle\left(\sqrt{s}J_{1}(s)\right)^{\prime}=\frac{J_{1}(s)}{2\sqrt{s}}+\sqrt{s}J_{1}^{\prime}(s)=\frac{J_{1}(s)}{2\sqrt{s}}-\sqrt{s}J_{2}(s)+\frac{J_{1}(s)}{\sqrt{s}}=\frac{3J_{1}(s)-2sJ_{2}(s)}{2\sqrt{s}}\,.$
(2.43)

With (2.26) and (2.27), we see that $\left(\sqrt{s}J_{1}(s)\right)^{\prime}$
is bounded independently of $s$. Thus, using (2.40), we arrive at

$\displaystyle\sqrt{\omega_{K}(t)}J_{1}(\omega_{K}(t))=\sqrt{4t}J_{1}(4t)+\mathcal{O}(t^{3}/K^{2})\,.$
(2.44)

Using (2.26) and (2.27) again, we also see that

$\displaystyle\frac{\lvert J_{1}(4t)\rvert}{t}\leq\frac{C}{(1+t)^{3/2}}\,.$
(2.45)

Let us now deal with the denominator. We observe

$\displaystyle\frac{1}{\left(1-t^{2}/K^{2}\right)^{1/4}}=$ $\displaystyle
1+\mathcal{O}(t^{2}/K^{2})\,.$ (2.46)

Combining these results, we arrive at

$\displaystyle G_{K}(t)$
$\displaystyle=\frac{\sqrt{\omega_{K}(t)}J_{1}(\omega_{K}(t))}{8\pi
t^{3/2}\left(1-\frac{t^{2}}{K^{2}}\right)^{1/4}}+\mathcal{O}\left(1/(K(1+t^{3/2}))\right)$
(2.47) $\displaystyle=\frac{\sqrt{4t}J_{1}(4t)+\mathcal{O}(t^{3}/K^{2})}{8\pi
t^{3/2}}\left(1+\mathcal{O}(t^{2}/K^{2})\right)+\mathcal{O}\left(1/(K(1+t^{3/2}))\right)$
(2.48) $\displaystyle=\frac{J_{1}(4t)}{4\pi
t}\left(1+\mathcal{O}(t^{2}/K^{2})\right)+\mathcal{O}\left(\frac{t^{3/2}}{K^{2}}+\frac{1}{K(1+t)^{3/2}}\right)$
(2.49) $\displaystyle=\frac{J_{1}(4t)}{4\pi
t}+\mathcal{O}\left(\frac{t^{2}}{K^{2}(1+t)^{3/2}}+\frac{t^{3/2}}{K^{2}}+\frac{1}{K(1+t)^{3/2}}\right)$
(2.50) $\displaystyle=\frac{J_{1}(4t)}{4\pi
t}+\mathcal{O}\left(\frac{t^{3/2}}{K^{2}}+\frac{1}{K(1+t)^{3/2}}\right).$
(2.51)

The first part of the supposed error term is only smaller than the order of
the main term $\mathcal{O}(1/t^{\frac{3}{2}})$, if $t<K^{\frac{2}{3}}$. ∎

###  2.2. The case $x\geq\frac{1}{2}$

In this section, we intend to understand [1, (18.15.22)] sufficiently well.

In accordance with [1, (18.15.21)] (see (2.11), (2.13), and (2.56)) we have

$\displaystyle\zeta(x)=\begin{cases}-\left(\frac{3}{4}\left(\arccos(\sqrt{x})-\sqrt{x-x^{2}}\right)\right)^{\frac{2}{3}}\quad&\text{
if }0\leq x\leq 1\,,\\\
\left(\frac{3}{4}\left(\sqrt{x^{2}-x}-\operatorname{arccosh}(\sqrt{x})\right)\right)^{\frac{2}{3}}\quad&\text{
if }x\geq 1\,.\end{cases}$ (2.52)

Therefore, $\zeta$ is negative on $(0,1)$ and positive on $(1,\infty)$ with a
(unique) zero at $1$.

We first want to put this in relation to $\xi$ and present the following
simple

######  Lemma 2.5.

For $0\leq x\leq 1$ and $\nu=4K+2(\alpha-1)$ with $K\in\mathbb{N}$ and
$\alpha\in\mathbb{R}$, we have

$\displaystyle\cos(\nu\xi(x)-\alpha\pi/2-\pi/4)=(-1)^{K-1}\cos(2\nu/3(-\zeta(x))^{3/2}-\pi/4)\,.$
(2.53)

###### Proof.

We first consider

$\displaystyle\xi(x)+2/3(-\zeta(x))^{3/2}$
$\displaystyle=\frac{1}{2}\left(\sqrt{x-x^{2}}+\arcsin(\sqrt{x})\right)+\frac{1}{2}\left(\arccos(\sqrt{x})-\sqrt{x-x^{2}}\right)$
(2.54)
$\displaystyle=\frac{1}{2}\left(\arcsin(\sqrt{x})+\arccos(\sqrt{x})\right)$
(2.55) $\displaystyle=\frac{\pi}{4}\,.$ (2.56)

We proceed to add the two arguments inside the cosines. Thus, we observe

$\displaystyle\nu\xi(x)-\alpha\pi/2-\pi/4+2\nu/3(-\zeta(x))^{3/2}-\pi/4$
$\displaystyle=(4K+2(\alpha-1))\pi/4-\alpha\pi/2-\pi/2$ (2.57)
$\displaystyle=(K-1)\pi\,.$ (2.58)

Let $s,t$ be the arguments of the cosines. We just showed $s+t=(K-1)\pi$.
Thus, $t=(K-1)\pi-s$, which implies
$\cos(t)=(-1)^{K-1}\cos(-s)=(-1)^{K-1}\cos(s)$, which is the claim. ∎

The following expansion to second order is only needed to prove 2.8. Beyond
that, the rougher estimates in 2.7 are sufficient.

######  Lemma 2.6.

The function $\zeta$ satisfies the expansion

$\displaystyle\zeta(1+s)=\frac{s}{2^{\frac{2}{3}}}\left(1-s/5\right)+\mathcal{O}(s^{3})$
(2.59)

for $\lvert s\rvert\leq\frac{1}{2}$.

###### Proof.

We recall (2.17), that is,

$\displaystyle
3/2(\pi/4-\eta(1-h))=\sqrt{2}\left(\sqrt{h}^{3}-\frac{3}{20}\sqrt{h}^{5}+\mathcal{O}(|h|^{7/2})\right),$
(2.60)

which implies

$\displaystyle\zeta((1-h)^{2})=-2^{1/3}h\left(1-\frac{h}{10}+\mathcal{O}(h^{2})\right).$
(2.61)

If we insert $h=1-\sqrt{1+s}=-s/2+s^{2}/8+\mathcal{O}(s^{3})$, we arrive at
the claimed expansion. ∎

We now need to study $\zeta$ for $x$ further away from $1$.

######  Lemma 2.7.

On $\mathbb{R}^{+}$, the function $\zeta$ is strictly increasing. Furthermore,
there are positive constants $C_{1},C_{2}$, such that for any $x>\frac{3}{2}$,
we have

$\displaystyle C_{1}\leq\frac{\zeta(x)^{\frac{3}{2}}}{x}\leq C_{2}\,,$ (2.62)

and for $x\in\left(\frac{1}{2},\frac{3}{2}\right)$, we have

$\displaystyle C_{1}\leq\frac{\zeta(x)}{x-1}\leq C_{2}\,.$ (2.63)

###### Proof.

We know that $\zeta(1)=0$ and $\zeta^{\prime}(1)=2^{-\frac{2}{3}}>0$ (see
2.6). Then, we would like to estimate the derivative of
$x\mapsto\lvert\zeta(x)\rvert^{\frac{3}{2}}$ for $x\not=1$ by using the
derivative of $\eta$, see (2.10). For $0<x<1$, it is given by

$\displaystyle\left((-\zeta(x))^{\frac{3}{2}}\right)^{\prime}=\frac{3}{2}\left(\frac{\pi}{4}-\eta(\sqrt{x})\right)^{\prime}=-\frac{3}{4}\eta^{\prime}(\sqrt{x})\frac{1}{\sqrt{x}}=-\frac{3}{4}\sqrt{1-x}\frac{1}{\sqrt{x}}<0\,.$
(2.64)

Similarly, for $x>1$, due to (2.14), it is given by

$\displaystyle\left(\zeta(x)^{\frac{3}{2}}\right)^{\prime}=\frac{3}{2}\,\left(\mathfrak{Im}\left(\eta(\sqrt{x})\right)\right)^{\prime}=\frac{3}{2}\,\mathfrak{Im}(\eta^{\prime}(\sqrt{x}))\frac{1}{2\sqrt{x}}=\frac{3}{4}\,\sqrt{x-1}\,\frac{1}{\sqrt{x}}>0\,.$
(2.65)

Altogether this implies that $\zeta$ is strictly increasing and that (2.63)
holds.

The last equation also shows that for $x\geq 3/2$,

$\frac{\sqrt{3}}{4}\leq\left(\zeta(x)^{\frac{3}{2}}\right)^{\prime}=\frac{3}{4}\sqrt{1-x^{-1}}\leq\frac{3}{4}\,.$
(2.66)

As $\zeta$ is strictly increasing and thus $\zeta(\frac{3}{2})>0$, we have
proved (2.62). This is the end of the proof. ∎

Next, we consider the coefficient described in [1, (18.15.23)], and define the
function $F_{0}$,

$\displaystyle
F_{0}(x)\coloneqq-\frac{5}{48\zeta(x)^{2}}+\sqrt{\frac{x-1}{x\zeta(x)}}\left(\frac{1}{2}-\frac{1}{8}-\frac{1}{4}\frac{x}{x-1}+\frac{5}{24}\left(\frac{x}{x-1}\right)^{2}\right),\quad
x\in(0,\infty)\setminus\\{1\\}\,,$ (2.67)

and $F_{0}(1)\coloneqq\lim_{x\to 1}F_{0}(x)$, see the next lemma. This
function has, of course, nothing to do with the function $F_{K}^{(\alpha)}$.

######  Lemma 2.8.

There is a constant $C>0$ such that $F_{0}(x)$ for any $x>\frac{1}{2}$
satisfies

$\displaystyle\lvert F_{0}(x)\rvert\leq C\,.$ (2.68)

###### Proof.

Due to 2.7 and $\zeta(1)=0$, we know that $\zeta$ has a unique zero at $x=1$.
Let $A\subset[\frac{1}{2},\infty)$ be a closed subset with $1\not\in A$. We
see that $F_{0}$ is bounded on $A$, as the functions
$x\mapsto\frac{1}{\zeta(x)},x\mapsto\frac{x}{x-1},x\mapsto\frac{x-1}{x}$ are
all bounded on $A$. We employ 2.6 and see that as $s\to 0$, we have

$\displaystyle\frac{1}{\zeta(1+s)^{2}}=$
$\displaystyle\frac{1}{\left(2^{-\frac{2}{3}}s\left(1-\frac{s}{5}\right)+\mathcal{O}(s^{3})\right)^{2}}=\frac{2^{\frac{4}{3}}}{s^{2}}\frac{1}{\left(1-\frac{s}{5}+\mathcal{O}(s^{2})\right)^{2}}=\frac{2^{\frac{4}{3}}}{s^{2}}\left(1+\frac{2s}{5}\right)+\mathcal{O}(1)\,.$
(2.69)

Similarly, we observe

$\displaystyle\sqrt{\frac{s}{(1+s)\zeta(1+s)}}$
$\displaystyle=\sqrt{\frac{s}{(1+s)\left(2^{-\frac{2}{3}}s\left(1-\frac{s}{5}\right)+\mathcal{O}(s^{3})\right)}}$
(2.70)
$\displaystyle=\frac{1}{\sqrt{(1+s)\left(2^{-\frac{2}{3}}\left(1-\frac{s}{5}\right)+\mathcal{O}(s^{2})\right)}}=2^{\frac{1}{3}}\left(1-\frac{2s}{5}\right)+\mathcal{O}(s^{2})\,.$
(2.71)

We conclude

$\displaystyle 2^{-\frac{1}{3}}$ $\displaystyle F_{0}(1+s)$ (2.72)
$\displaystyle=2^{-\frac{1}{3}}\left(-\frac{5}{48\zeta(1+s)^{2}}+\sqrt{\frac{s}{(1+s)\zeta(1+s)}}\left(\frac{1}{2}-\frac{1}{8}-\frac{1}{4}\frac{1+s}{s}+\frac{5}{24}\left(\frac{1+s}{s}\right)^{2}\right)\right)$
(2.73)
$\displaystyle=-\frac{5}{24s^{2}}-\frac{1}{12s}+\left(1-\frac{2s}{5}\right)\left(-\frac{1}{4s}+\frac{5(1+2s)}{24s^{2}}\right)+\mathcal{O}(1)$
(2.74)
$\displaystyle=-\frac{5}{24s^{2}}-\frac{1}{12s}-\frac{1}{4s}+\frac{5}{24s^{2}}+\frac{5}{12s}-\frac{1}{12s}+\mathcal{O}(1)=\mathcal{O}(1)\,\,.$
(2.75)

Thus, $F_{0}$ can be defined at $1$ as a limit. Moreover, we have shown that
$F_{0}$ is bounded on some neighbourhood of $1$ and on any closed subset of
$[\frac{1}{2},\infty)$ not containing $1$, which implies that $F_{0}$ is
bounded on $[\frac{1}{2},\infty)$, which was the claim. ∎

As [1, (18.15.22)] reduces our $F_{K}^{(\alpha)}$ to the Airy function, its
derivative and its envelope, we should now take a look at the asymptotics of
these functions.

######  Proposition 2.9.

Let $s>0$ and let $\mathcal{A}(s)\coloneqq\frac{2}{3}s^{3/2}$ (see [1,
(9.7.1)]). Then, there are (positive) constants $C$ (which may vary from line
to line) such that

$\displaystyle\operatorname{Ai}(-s)s^{1/4}$
$\displaystyle=1/\sqrt{\pi}\cos(\mathcal{A}(s)-\pi/4)+\mathcal{O}(1/(1+\mathcal{A}(s)))\,,$
(2.76) $\displaystyle\lvert\operatorname{Ai}^{\prime}(-s)s^{-1/4}\rvert$
$\displaystyle\leq C(1+s^{-1/4})\,,$ (2.77) $\displaystyle
0\leq\operatorname{env}\operatorname{Ai}(-s)s^{1/4}$ $\displaystyle\leq C\,,$
(2.78) $\displaystyle\lvert\operatorname{Ai}(s)\rvert s^{1/4}$
$\displaystyle\leq C\exp(-\mathcal{A}(s))\,,$ (2.79)
$\displaystyle\lvert\operatorname{Ai}^{\prime}(s)s^{-1/4}\rvert$
$\displaystyle\leq C\exp(-\mathcal{A}(s))(1+s^{-1/4})\,,$ (2.80)
$\displaystyle 0\leq\operatorname{env}\operatorname{Ai}(s)s^{1/4}$
$\displaystyle\leq C\exp(-\mathcal{A}(s))\,.$ (2.81)

###### Proof.

All of these follow from [1, (9.7.5–11)] for the asymptotics of the Airy
functions $\operatorname{Ai},\operatorname{Bi}$ and [1, (2.8.19–21)] for the
definition of the envelope $\text{env}\operatorname{Ai}$, which can be
expressed in $\operatorname{Ai}$ and $\operatorname{Bi}$. ∎

We are now prepared to take on [1, (18.15.22)].

######  Lemma 2.10.

Let $\alpha\in\\{\pm 1/2,1\\}$ and $\nu=4K+2(\alpha-1)$. For $1/2\leq x\leq
1$, we have

$\displaystyle F_{K}^{(\alpha)}(x)$
$\displaystyle=\sqrt{2/\pi}\cos(\nu\xi(x)-\alpha\pi/2-\pi/4)+\mathcal{O}(1/(1+K(1-x)^{3/2})\,,$
(2.82)

for $1\leq x\leq 3/2$, it holds

$\displaystyle\left\lvert F_{K}^{(\alpha)}(x)\right\rvert\leq C\exp(-\beta
K(x-1)^{3/2})\,,$ (2.83)

and in the case $x>3/2$, we get

$\displaystyle\left\lvert F_{K}^{(\alpha)}(x)\right\rvert\leq C\exp(-\beta
Kx)$ (2.84)

for some constants $\beta>0$ and $C<\infty$.

###### Proof.

We note that $\zeta(x)/(x-1)=\lvert\zeta(x)\rvert/\lvert x-1\rvert$ due to
2.7. Solving [1, (18.15.22)] for $F_{K}^{(\alpha)}(x)$ provides us with

$\displaystyle F_{K}^{(\alpha)}$ $\displaystyle(x)(-1)^{K-1}\sqrt{2}^{-1}$
(2.85)
$\displaystyle=\sqrt{\nu}\lvert\zeta(x)\rvert^{\frac{1}{4}}\left(\frac{\operatorname{Ai}(\nu^{2/3}\zeta(x))}{\nu^{1/3}}+\frac{\operatorname{Ai}^{\prime}(\nu^{2/3}\zeta(x))}{\nu^{5/3}}F_{0}(x)+\text{env}\operatorname{Ai}(\nu^{2/3}\zeta(x))\mathcal{O}(\nu^{-4/3})\right)$
(2.86)
$\displaystyle=\left(\nu^{2/3}\lvert\zeta(x)\rvert\right)^{1/4}\operatorname{Ai}(\nu^{2/3}\zeta(x))$
(2.87)
$\displaystyle+\mathcal{O}\left(\frac{\operatorname{Ai}^{\prime}(\nu^{2/3}\zeta(x))}{\left(\nu^{2/3}\lvert\zeta(x)\rvert\right)^{1/4}}\frac{\sqrt{\lvert\zeta(x)\rvert}}{\nu}+\left(\nu^{2/3}\lvert\zeta(x)\rvert\right)^{1/4}\text{env}\operatorname{Ai}(\nu^{2/3}\zeta(x))\nu^{-1}\right).$
(2.88)

We have regrouped the terms such that we get the left-hand sides of the
formulas in 2.9.

For $1/2\leq x\leq 1$, due to 2.7, we get $-\infty<\zeta(1/2)\leq\zeta(x)\leq
0$ and thus we can bound the contribution of the $\operatorname{Ai}^{\prime}$
and $\text{env}\operatorname{Ai}$ term by $C/K$ using (2.77) and (2.78). For
the $\operatorname{Ai}$ term, we use (2.76) with $s=-\nu^{2/3}\zeta(x)>0$ and
2.5, which yield

$\displaystyle F_{K}^{(\alpha)}(x)$
$\displaystyle=\sqrt{2}(-1)^{K-1}\left(\nu^{2/3}\lvert\zeta(x)\rvert\right)^{1/4}\operatorname{Ai}(\nu^{2/3}\zeta(x))+\mathcal{O}(1/K)$
(2.89)
$\displaystyle=\sqrt{2/\pi}(-1)^{K-1}\cos(2\nu/3(-\zeta(x))^{3/2}-\pi/4)+\mathcal{O}(1/(1+K\lvert\zeta(x)\rvert^{3/2})+1/K)$
(2.90)
$\displaystyle=\sqrt{2/\pi}\cos(\nu\xi(x)-\alpha\pi/2-\pi/4)+\mathcal{O}(1/(1+K(1-x)^{3/2}))\,.$
(2.91)

For $x\geq 1$, we use (2.79), (2.80) and (2.81) to see

$\displaystyle\left\lvert F_{K}^{(\alpha)}(x)\right\rvert$ $\displaystyle\leq
C\exp(-\mathcal{A}(\nu^{2/3}\zeta(x)))\left(1+\frac{\sqrt{\lvert\zeta(x)\rvert}}{\nu}+\frac{1}{\nu}\right)$
(2.92) $\displaystyle\leq C\exp(-2\nu/3\,\zeta(x)^{3/2})(1+\sqrt{\zeta(x)})$
(2.93) $\displaystyle\leq
C\exp(-\nu/2\,\zeta(x)^{3/2})\sup_{s>0}\left(\exp(-\nu
s^{3}/6)\,(1+s)\right).$ (2.94)

As $\nu=4K+2(\alpha-1)\geq 4-3=1$, the supremum at the end is bounded
independently of $\nu$. To get a better understanding of the exponential
decay, we need 2.7. For $1\leq x\leq 3/2$, we get

$\displaystyle\left\lvert F_{K}^{(\alpha)}(x)\right\rvert$ $\displaystyle\leq
C\exp(-\nu/2\,\zeta(x)^{3/2})$ (2.95) $\displaystyle\leq
C\exp(-\nu/2\,C_{1}^{3/2}(x-1)^{3/2})$ (2.96) $\displaystyle\leq C\exp(-\beta
K(x-1)^{3/2})\,,$ (2.97)

where $\beta>0$ is some constant and where we used $\nu\geq 4K-3\geq K$.

For $3/2\leq x<\infty$, we use the other estimate in 2.7 and see

$\displaystyle\left\lvert F_{K}^{(\alpha)}(x)\right\rvert$ $\displaystyle\leq
C\exp(-\nu/2\,\zeta(x)^{3/2})$ (2.98) $\displaystyle\leq
C\exp(-\nu/2\,C_{1}x)$ (2.99) $\displaystyle\leq C\exp(-\beta Kx)\,,$ (2.100)

with a possibly different $\beta>0$.

∎

This leads to the following second main result on $G_{K}(t)$ in case the
argument $t$ is large compared to $K$,

######  Corollary 2.11.

There exists a constant $\beta>0$ and a constant $C<\infty$ independent of $K$
such that we have the following estimates for the function $G_{K}$ from (2.7),

$\displaystyle\lvert G_{K}(t)\rvert\leq\begin{cases}C\exp(-\beta
t)\quad&\text{ if }t>\sqrt{\frac{3}{2}}K\,,\\\
\frac{C}{K^{\frac{3}{2}}\left\lvert
1-\frac{t}{K}\right\rvert^{\frac{1}{4}}}\quad&\text{ if
}\sqrt{\frac{1}{2}}K\leq t\leq\sqrt{\frac{3}{2}}K\,.\end{cases}$ (2.101)

###### Proof.

We recall $x=t^{2}/K^{2}$ and (2.7), which states that

$\displaystyle G_{K}(t)=\frac{1}{8\pi t^{3/2}\left\lvert
1-\frac{t^{2}}{K^{2}}\right\rvert^{1/4}}F_{K}^{(1)}\left(\frac{t^{2}}{K^{2}}\right).$
(2.102)

If $t>\sqrt{3/2}K$, we have $x>3/2$. Thus, the fraction in front is bounded by
$C/K^{3/2}\leq C$ and 2.10 tells us that

$\displaystyle F_{K}^{(1)}(t^{2}/K^{2})\leq C\exp(-\beta Kx)\leq C\exp(-\beta
K\sqrt{x})=C\exp(-\beta t)\,.$ (2.103)

This finishes the case $t>\sqrt{3/2}K$. In the case $\sqrt{1/2}K\leq
t\leq\sqrt{3/2}K$, 2.10 implies that $\lvert
F_{K}^{(1)}(t^{2}/K^{2})\rvert\leq C$ and thus, we can bound

$\displaystyle\lvert G_{K}(t)\rvert\leq\frac{C}{t^{3/2}\left\lvert
1-\frac{t^{2}}{K^{2}}\right\rvert^{1/4}}\leq\frac{C}{K^{3/2}\left\lvert
1-\frac{t}{K}\right\rvert^{1/4}}.$ (2.104)

∎

###  2.3. An integral bound and an identity on the kernel $G_{K}$

We start with an integral estimate, which will help to calculate some
Hilbert–Schmidt norms.

######  Lemma 2.12.

There are constants $C<\infty$ and $\beta>0$, independent of $K$ such that for
any $R\geq 0$

$\displaystyle\int_{R}^{\infty}\lvert
G_{K}(t)\rvert^{2}t\,\mathrm{d}t\leq\begin{cases}\frac{C}{1+R}\quad&\text{ if
}R<2K\,,\\\ C\exp(-\beta R)&\text{ if }R\geq 2K\,.\end{cases}$ (2.105)

###### Proof.

Combining 2.11 and Theorem 2.4, we obtain the universal upper bound

$\displaystyle\lvert
G_{K}(t)\rvert\leq\begin{cases}\frac{C}{(1+t)^{3/2}\left\lvert
1-\frac{t}{K}\right\rvert^{1/4}}\quad&\text{ if }t<2K\,,\\\ C\exp(-\beta
t)&\text{ if }t\geq 2K\,.\end{cases}$ (2.106)

For $R\geq 2K$, we simply observe

$\displaystyle\int_{R}^{\infty}\lvert G_{K}(t)\rvert^{2}t\,\mathrm{d}t\leq
C\int_{R}^{\infty}\exp(-2\beta t)t\,\mathrm{d}t\leq C\exp(-\beta R)\,.$
(2.107)

First, we assume $1<R<2K$ and estimate

$\displaystyle\int_{R}^{2K}\lvert G_{K}(t)\rvert^{2}t\,\mathrm{d}t$
$\displaystyle\leq\int_{R}^{2K}\frac{Ct}{t^{3}\left\lvert
1-\frac{t}{K}\right\rvert^{1/2}}\,\mathrm{d}t$ (2.108)
$\displaystyle=\int_{R/K}^{2}\frac{C}{(Ks)^{2}\left\lvert
1-s\right\rvert^{1/2}}K\,\mathrm{d}s$ (2.109)
$\displaystyle\leq\frac{C}{K}\int_{R/K}^{2}\frac{1}{s^{2}\lvert
1-s^{2}\rvert^{1/2}}\,\mathrm{d}s$ (2.110)
$\displaystyle=\frac{C}{K}\left(\frac{\sqrt{3}}{2}+\frac{\sqrt{1-R^{2}/K^{2}}}{R/K}\right)\leq\frac{C}{K}+\frac{C}{R}\leq\frac{C}{1+R}\,.$
(2.111)

Finally, for $0\leq R\leq 1$, it suffices to see that

$\displaystyle\int_{0}^{1}\lvert
G_{K}(t)\rvert^{2}t\,\mathrm{d}t\leq\int_{0}^{1}Ct\,\mathrm{d}t\leq C\,.$
(2.112)

This concludes the proof, as we can split the integral on $(R,\infty)$ into
the at most three parts $(R,1),(\max\\{1,R\\},2K)$ and
$(\max\\{2K,R\\},\infty)$ and the upper bound is decreasing in $R$. ∎

We finish this subsection with a simple and yet useful identity of the
localised Fermi projection.

######  Lemma 2.13.

For every $E\subset\mathbb{R}^{2}$ measurable and bounded, we have

$\displaystyle\left\lVert
1_{E}P_{K}\right\rVert_{2}^{2}=\operatorname{tr}1_{E}P_{K}1_{E}=\frac{\lvert
E\rvert}{2\pi}\,.$ (2.113)

###### Proof.

This follows from

$\displaystyle\operatorname{tr}1_{E}P_{K}1_{E}$
$\displaystyle=\int_{E}P_{K}(x,x)\,\mathrm{d}x=\int_{E}G_{K}(0)\,\mathrm{d}x=\lvert
E\rvert\frac{1}{2\pi K}\mathcal{L}_{K-1}^{(1)}(0)=\frac{\lvert
E\rvert}{2\pi}\,,$ (2.114)

where we used $\mathcal{L}_{K-1}^{(1)}(0)=(K-1)+1=K$, see [1, (18.6.1)] with
$(2)_{K-1}=((K-1)+1)!$, see [1, (5.2.iii)]. ∎

##  3\. On the particle number fluctuations

This section is devoted to the study of the asymptotic expansion of the trace
of $f(P_{K}(L\Lambda))$ when $f$ is a quadratic function. Since we assume
$f(0)=f(1)=0$, we may restrict to $f(t)=t(1-t)$. It is related to the
fluctuations of the local particle number in the ground state and hence of
physical interest.

######  Theorem 3.1.

Let $\Lambda\subset\mathbb{R}^{2}$ be a piecewise $\mathsf{C}^{2}$-smooth
domain. Then, with the above test function $f(t)=t(1-t)$,

$\displaystyle\operatorname{tr}f(P_{K}(L\Lambda))=\operatorname{tr}\left(1_{L\Lambda}P_{K}1_{L\Lambda^{\complement}}P_{K}1_{L\Lambda}\right)=\frac{1}{\sqrt{2}\pi^{3}}\begin{cases}L\lvert\partial\Lambda\rvert\ln(K)+\mathcal{O}(L)\quad&\text{
if }K<L\,,\\\ L\lvert\partial\Lambda\rvert\ln(L)+\mathcal{O}(L)\quad&\text{ if
}K>L\,.\end{cases}$ (3.1)

######  Remark.

In line with our 1.1 the value $\frac{1}{\sqrt{2}\pi^{3}}$ equals
$\frac{2\sqrt{2}}{\pi}\mathsf{I}(t\mapsto t(1-t))$.

###### Proof.

The first trace identity is rather obvious, see the short proof following
(5.2). As we are then just calculating the Hilbert–Schmidt norm of
$1_{L\Lambda}P_{K}1_{L\Lambda^{\complement}}$, we have (see (2.3))

$\displaystyle\operatorname{tr}\left(1_{L\Lambda}P_{K}1_{L\Lambda^{\complement}}P_{K}1_{L\Lambda}\right)$
$\displaystyle=\int_{L\Lambda}\mathrm{d}x\int_{L\Lambda^{\complement}}\mathrm{d}y\,\lvert
G_{K}(\|x-y\|/\sqrt{8})\rvert^{2}$ (3.2)
$\displaystyle=\int_{0}^{\infty}\mathrm{d}s\,\lvert
G_{K}(s/\sqrt{8})\rvert^{2}F(s)\,.$ (3.3)

The last step relies on changing to polar coordinates in $y$ and Fubini, where

$\displaystyle F(s)\coloneqq
s\int_{L\Lambda}\mathrm{d}x\int_{0}^{2\pi}\mathrm{d}\theta\,1_{L\Lambda^{\complement}}(x+s(\cos(\theta),\sin(\theta)))\,.$
(3.4)

Trivially, we have

$\displaystyle F(s)\leq s\int_{L\Lambda}\mathrm{d}x\,2\pi=2\pi
sL^{2}\lvert\Lambda\rvert=\mathcal{O}(sL^{2})\,.$ (3.5)

As $\partial\Lambda$ is piecewise $\mathsf{C}^{2}$-smooth, we have the
expansion

$\displaystyle F(s)=2s^{2}L\lvert\partial\Lambda\rvert+\mathcal{O}(s^{3})\,.$
(3.6)

See Appendix C for a proof. Thus, for any $s>0$, we observe

$\displaystyle F(s)\leq
C\min\big{\\{}2sL^{2},s^{2}(L+s)\big{\\}}=\begin{cases}2CsL^{2}&\text{ if
}s>L\,,\\\ Cs^{2}(L+s)\leq 2Cs^{2}L&\text{ if }s\leq L\,,\end{cases}$ (3.7)

or equivalently,

$\displaystyle F(s)\leq CsL\min\\{s,L\\}\,.$ (3.8)

We want to replace the integral in (3.3) over $\mathbb{R}^{+}$ by an integral
over $(\sqrt{8},\min\\{K,L\\})$. Let us consider the resulting error terms.
The first one is trivial, the second and third one rely upon 2.12.

$\displaystyle\int_{0}^{\sqrt{8}}\mathrm{d}s\,\lvert
G_{K}(s/\sqrt{8})\rvert^{2}F(s)$ $\displaystyle\leq
C\int_{0}^{\sqrt{8}}\mathrm{d}s\,s^{2}L\leq CL\,,$ (3.9)
$\displaystyle\int_{L}^{\infty}\mathrm{d}s\lvert
G_{K}(s/\sqrt{8})\rvert^{2}F(s)$
$\displaystyle=C\int_{L/\sqrt{8}}^{\infty}\mathrm{d}t\,\lvert
G_{K}(t)\rvert^{2}L^{2}t\leq CL\,,$ (3.10)
$\displaystyle\int_{K}^{10K}\mathrm{d}s\,\lvert
G_{K}(s/\sqrt{8})\rvert^{2}F(s)$ $\displaystyle\leq
CK\int_{K/\sqrt{8}}^{10K/\sqrt{8}}\mathrm{d}t\,\lvert G_{K}(t)\rvert^{2}Lt\leq
CL\,,$ (3.11) $\displaystyle\int_{10K}^{L}\mathrm{d}s\,\lvert
G_{K}(s/\sqrt{8})\rvert^{2}F(s)$ $\displaystyle\leq
C\int_{10K/\sqrt{8}}^{\infty}\mathrm{d}t\,\exp(-\beta t)t^{2}L\leq
CL\exp(-\beta K)\,.$ (3.12)

Thus, as the integrand is always positive, we get

$\displaystyle\int_{0}^{\infty}\mathrm{d}s\,\lvert
G_{K}(s/\sqrt{8})\rvert^{2}F(s)$
$\displaystyle=\int_{\sqrt{8}}^{\min\\{K,L\\}}\mathrm{d}s\,\lvert
G_{K}(s/\sqrt{8})\rvert^{2}\left(2s^{2}L\lvert\partial\Lambda\rvert+\mathcal{O}(s^{3})\right)+\mathcal{O}(L)$
(3.13)
$\displaystyle=\sqrt{8}\int_{1}^{\min\\{K,L\\}/\sqrt{8}}\mathrm{d}t\,\lvert
G_{K}(t)\rvert^{2}\left(16t^{2}L\lvert\partial\Lambda\rvert+\mathcal{O}(t^{3})\right)+\mathcal{O}(L)\,.$
(3.14)

Next, we utilize Theorem 2.4. As $1<t<K/\sqrt{8}$ the singularities of the
denominator are outside the integration domain. As the singularity around
$t=K$ is substantially away from the domain, we can easily bound that factor.
Thus, we observe

$\displaystyle\lvert G_{K}(t)\rvert^{2}$
$\displaystyle=\left\lvert\frac{\cos(\omega_{K}(t)-3\pi/4)}{4\sqrt{2}\pi^{\frac{3}{2}}t^{\frac{3}{2}}\left(1-\frac{t^{2}}{K^{2}}\right)^{\frac{1}{4}}}+\mathcal{O}\left(\frac{1}{t^{\frac{5}{2}}}\right)\right\rvert^{2}$
(3.15)
$\displaystyle=\frac{1}{32\pi^{3}t^{3}}\left(\cos(\omega_{K}(t)-3\pi/4)\left(1+\mathcal{O}\left(\frac{t^{2}}{K^{2}}\right)\right)+\mathcal{O}\left(\frac{1}{t}\right)\right)^{2}$
(3.16)
$\displaystyle=\frac{\cos(\omega_{K}(t)-3\pi/4)^{2}}{32\pi^{3}t^{3}}+\mathcal{O}\left(\frac{1}{tK^{2}}+\frac{1}{t^{4}}\right).$
(3.17)

We estimate the next batch of error terms of (3.14) and (3.17), respectively,

$\displaystyle\int_{1}^{\min\\{K,L\\}/\sqrt{8}}\mathrm{d}t\,\lvert
G_{K}(t)\rvert^{2}t^{3}$ $\displaystyle\leq C\int_{1}^{L}\mathrm{d}t\,1=CL\,,$
(3.18)
$\displaystyle\int_{1}^{\min\\{K,L\\}/\sqrt{8}}\mathrm{d}t\,\left(\frac{1}{tK^{2}}+\frac{1}{t^{4}}\right)Lt^{2}$
$\displaystyle\leq CL\left(\frac{K^{2}}{K^{2}}+1\right)=2CL\,.$ (3.19)

Thus, we have shown

$\displaystyle\int_{0}^{\infty}\mathrm{d}s\,\lvert
G_{K}(s/\sqrt{8})\rvert^{2}F(s)$
$\displaystyle=\sqrt{8}\int_{1}^{\min\\{K,L\\}/\sqrt{8}}\mathrm{d}t\,\frac{\cos(\omega_{K}(t)-3\pi/4)^{2}}{32\pi^{3}t^{3}}16t^{2}L\lvert\partial\Lambda\rvert+\mathcal{O}(L)$
(3.20)
$\displaystyle=\frac{\sqrt{2}L\lvert\partial\Lambda\rvert}{\pi^{3}}\int_{1}^{\min\\{K,L\\}/\sqrt{8}}\mathrm{d}t\,\frac{\cos(\omega_{K}(t)-3\pi/4)^{2}}{t}+\mathcal{O}(L)$
(3.21)
$\displaystyle=\frac{\sqrt{2}L\lvert\partial\Lambda\rvert}{\pi^{3}}\int_{\omega_{K}(1)}^{\omega_{K}(\min\\{K,L\\}/\sqrt{8})}\mathrm{d}s\,\frac{\cos(s-3\pi/4)^{2}}{\omega_{K}^{-1}(s)\omega_{K}^{\prime}(\omega_{K}^{-1}(s))}+\mathcal{O}(L)\,.$
(3.22)

For the last step, we need to show that $\omega_{K}$ is invertible on the
range $(1,K/\sqrt{8})$ and need to estimate the inverse function
$\omega_{K}^{-1}$ and the differential $\omega_{K}^{\prime}$. As
$\omega_{K}(t)=4K\eta(t/K)$ we observe that

$\displaystyle\omega_{K}^{\prime}(t)=4\eta^{\prime}(t/K)=4\sqrt{1-\frac{t^{2}}{K^{2}}}=4+\mathcal{O}\left(\frac{t^{2}}{K^{2}}\right).$
(3.23)

Since $\omega_{K}^{\prime}$ and $\omega_{K}^{-1}$ appear in the denominator,
we need to establish lower bounds for both of them. Thus, for $0\leq t\leq
K/\sqrt{8}$, we see that $4\geq\omega_{K}^{\prime}(t)\geq 4\sqrt{7/8}$. This
ensures that $\omega_{K}$ is invertible. Since $\omega_{K}(0)=0$, this also
implies

$\displaystyle 4t\geq\omega_{K}(t)\geq 4\sqrt{7/8}\,t\,.$ (3.24)

Thus, for $0\leq s\leq\omega_{K}(K/\sqrt{8})$, we have

$\displaystyle 4\geq\omega_{K}^{\prime}(\omega_{K}^{-1}(s))\geq
4\sqrt{7/8}\,,\quad\frac{1}{4}s\leq\omega_{K}^{-1}(s)\leq\frac{1}{4}\sqrt{\frac{8}{7}}s\,.$
(3.25)

Using the last two estimates, (3.23) and (2.40), we also obtain

$\displaystyle\frac{1}{\omega_{K}^{\prime}(\omega_{K}^{-1}(s))}=\frac{1}{4}+\mathcal{O}\left(\frac{s^{2}}{K^{2}}\right),\quad\frac{1}{\omega_{K}^{-1}(s)}=\frac{4}{s}+\mathcal{O}\left(\frac{s}{K^{2}}\right).$
(3.26)

Thus, we conclude

$\displaystyle\int_{0}^{\infty}\mathrm{d}s\,$ $\displaystyle\lvert
G_{K}(s/\sqrt{8})\rvert^{2}F(s)$ (3.27)
$\displaystyle=\frac{\sqrt{2}L\lvert\partial\Lambda\rvert}{\pi^{3}}\int_{\omega_{K}(1)}^{\omega_{K}(\min\\{K,L\\}/\sqrt{8})}\mathrm{d}s\,\frac{\cos(s-3\pi/4)^{2}}{\omega_{K}^{-1}(s)\omega_{K}^{\prime}(\omega_{K}^{-1}(s))}+\mathcal{O}(L)$
(3.28)
$\displaystyle=\frac{\sqrt{2}L\lvert\partial\Lambda\rvert}{\pi^{3}}\int_{\omega_{K}(1)}^{\omega_{K}(\min\\{K,L\\}/\sqrt{8})}\mathrm{d}s\,\Big{[}\frac{\cos(s-3\pi/4)^{2}}{s}+\mathcal{O}\left(\frac{s}{K^{2}}\right)\Big{]}+\mathcal{O}(L)$
(3.29)
$\displaystyle=\frac{\sqrt{2}L\lvert\partial\Lambda\rvert}{\pi^{3}}\int_{\omega_{K}(1)}^{\omega_{K}(\min\\{K,L\\}/\sqrt{8})}\mathrm{d}s\,\frac{1-\sin(2s)}{2s}+\mathcal{O}(L)$
(3.30)
$\displaystyle=\frac{L\lvert\partial\Lambda\rvert}{\sqrt{2}\pi^{3}}\left(\ln(\min\\{K,L\\})+\mathcal{O}(1)\right)+\mathcal{O}(L)$
(3.31)
$\displaystyle=\frac{L\lvert\partial\Lambda\rvert}{\sqrt{2}\pi^{3}}\ln(\min\\{K,L\\})+\mathcal{O}(L)\,.$
(3.32)

We used that $2\cos(a-3\pi/4)^{2}=1+\cos(2a-3\pi/2)=1-\sin(2a)$ and (3.24).

∎

##  4\. The case $K\gg L$

######  Theorem 4.1.

Suppose that $\Lambda$ has a piecewise $\mathsf{C}^{1}$-smooth boundary
$\partial\Lambda$. Let $L\leq CK^{2/5}$. Then, for any polynomial $f$ with
$f(0)=0$, we have the asymptotic expansion

$\displaystyle\operatorname{tr}f(1_{L\Lambda}P_{K}1_{L\Lambda})=L^{2}\frac{\lvert\Lambda\rvert}{2\pi}f(1)+L\ln(L)\lvert\partial\Lambda\rvert\,\frac{2\sqrt{2}}{\pi}\,\mathsf{I}(t\mapsto(f(t)-tf(1)))+\mathcal{O}(L)\,,$
(4.1)

as $L\to\infty$ (and hence $K\to\infty$) with
$\mathsf{I}({\makebox[4.30554pt]{{$\cdot$}}})$ defined in (1.5).

######  Remarks.

  1. (1)

Our proof relies on the same result for $B=0$, see [19, Theorem 2.2], where
the set $\Lambda$ plays the same role as here and $\Omega$ is the Fermi sea at
Fermi energy $2$, which is smooth. We will show that in the case $L\leq
CK^{2/5}$, the magnetic field only yields a small perturbation relative to the
free case $H=-\Delta$.

  2. (2)

With some efforts we could improve the result and relax the condition to
$L=\mathcal{O}(\sqrt{K})$. As for the particle number fluctuations with the
quadratic function $f$, we believe that the optimal condition is $L\leq CK$,
but we do not know how to prove this.

###### Proof.

For the integral kernel of the Fermi projection of the Laplacian we have the
explicit expression in terms of the Bessel function $J_{1}$,

$\displaystyle P_{\infty}(x,y)\coloneqq 1(-\Delta\leq
2)(x,y)=\begin{cases}\frac{J_{1}(\sqrt{2}\|x-y\|)}{\sqrt{2}\pi\|x-y\|}\,,\quad&x,y\in\mathbb{R}^{2}\,\text{
with }x\neq y\,,\\\ \frac{1}{2\pi}\,,\quad&x=y\in\mathbb{R}^{2}\,.\end{cases}$
(4.2)

The latter is also the pointwise limit of the integral kernel of $P_{K}$ as
$K\to\infty$, see Theorem 2.4. Formula (4.2) can be derived by a simple
Fourier transformation, the use of polar coordinates and the very definition
of the Bessel function $J_{1}$.

We consider the polynomials $f(t)=t^{m}$ for $m\in\mathbb{N}$ with $m\geq 2$;
the case $m=1$ is covered by 2.13 without mentioning $P_{\infty}$.

According to the above remark it suffices to show

$\displaystyle\operatorname{tr}(1_{L\Lambda}P_{K}1_{L\Lambda})^{m}=\operatorname{tr}(1_{L\Lambda}P_{\infty}1_{L\Lambda})^{m}+\mathcal{O}(L)\,.$
(4.3)

To this end, we use A.2 with $A_{1}=1_{L\Lambda}P_{K}1_{L\Lambda}$ and
$A_{2}=1_{L\Lambda}P_{\infty}1_{L\Lambda}$. Both operator norms are bounded by
$1$. Let $R_{0}>0$ satisfy $\Lambda\subset D_{R_{0}}(0)$. We begin by
estimating

$\displaystyle\lVert
1_{L\Lambda}(P_{K}-P_{\infty})1_{L\Lambda}\rVert_{2}^{2}=\int_{L\Lambda}\mathrm{d}x\int_{L\Lambda}\mathrm{d}y\,\lvert
P_{K}(x,y)-P_{\infty}(x,y)\rvert^{2}\,.$ (4.4)

For this, we need the assumption $L\leq CK^{0.4}$. For $x,y\in L\Lambda$,
according to (2.3) and Theorem 2.4, we have

$\displaystyle P_{K}(x,y)$ $\displaystyle=\exp\left(i\frac{1}{2K}x\wedge
y\right)G_{K}(\lVert x-y\rVert/\sqrt{8})$ (4.5)
$\displaystyle=\left(1+\mathcal{O}\left(\frac{\|x-y\|\|x\|}{K}\right)\right)\left(P_{\infty}(x,y)+\mathcal{O}\left(L^{1.5}K^{-2}+K^{-1}L^{-1.5}\right)\right)$
(4.6)
$\displaystyle=P_{\infty}(x,y)+\mathcal{O}\left(\frac{\|x-y\|\|x\|}{K}\frac{1}{(1+\|x-y\|)^{1.5}}+L^{1.5}K^{-2}+K^{-1}L^{-1.5}\right)$
(4.7)
$\displaystyle=P_{\infty}(x,y)+\mathcal{O}\left(\frac{L}{K(1+\|x-y\|)^{0.5}}+L^{1.5}K^{-2}+K^{-1}L^{-1.5}\right).$
(4.8)

Thus, we observe

$\displaystyle\lVert 1_{L\Lambda}$
$\displaystyle(P_{K}-P_{\infty})1_{L\Lambda}\rVert_{2}^{2}$ (4.9)
$\displaystyle=\int_{L\Lambda}\mathrm{d}x\int_{L\Lambda}\mathrm{d}y\,\lvert
P_{K}(x,y)-P_{\infty}(x,y)\rvert^{2}$ (4.10) $\displaystyle\leq
C\int_{D_{R_{0}L}(0)}\mathrm{d}x\int_{D_{R_{0}L}(0)}\mathrm{d}y\,\left(\frac{L}{K(1+\|x-y\|)^{0.5}}+L^{1.5}K^{-2}+K^{-1}L^{-1.5}\right)^{2}$
(4.11) $\displaystyle\leq
C\int_{D_{R_{0}L}(0)}\mathrm{d}x\int_{D_{R_{0}L}(0)}\mathrm{d}y\,\left(\frac{L^{2}}{K^{2}(1+\|x-y\|)}+L^{3}K^{-4}+K^{-2}L^{-3}\right)$
(4.12) $\displaystyle\leq
C\left(L^{5}/K^{2}+L^{7}/K^{4}+L^{1}/K^{2}\right)\leq C\,.$ (4.13)

Using the last estimate, (4.2) and (2.45), we can conclude

$\displaystyle\max\left(\lVert 1_{L\Lambda}P_{K}1_{L\Lambda}\rVert_{2},\lVert
1_{L\Lambda}P_{\infty}1_{L\Lambda}\rVert_{2}\right)^{2}$
$\displaystyle\leq\left(\lVert
1_{L\Lambda}P_{\infty}1_{L\Lambda}\rVert_{2}+C\right)^{2}$ (4.14)
$\displaystyle\leq
2C^{2}+2\int_{L\Lambda}\mathrm{d}x\int_{L\Lambda}\mathrm{d}y\,\lvert
P_{\infty}(x,y)\rvert^{2}$ (4.15) $\displaystyle\leq
C+\int_{L\Lambda}\mathrm{d}x\int_{L\Lambda}\mathrm{d}y\,\frac{C}{(1+\|x-y\|)^{3}}$
(4.16) $\displaystyle\leq
C+\int_{D_{LR_{0}}(0)}\mathrm{d}x\int_{\mathbb{R}^{2}}\mathrm{d}y\,\frac{C}{(1+\|x-y\|)^{3}}$
(4.17) $\displaystyle\leq CL^{2}\,.$ (4.18)

The error term in (A.7) is therefore of the order
$\mathcal{O}(\sqrt{CL^{2}}C)=\mathcal{O}(L)$ and we arrive at

$\displaystyle\operatorname{tr}(1_{L\Lambda}P_{K}1_{L\Lambda})^{m}=\operatorname{tr}(1_{L\Lambda}P_{\infty}1_{L\Lambda})^{m}+\mathcal{O}(L)\,,$
(4.19)

which, as we explained before, finishes the proof. ∎

##  5\. The case $K\ll L$: Overview

In this section we discuss the asymptotic expansion when $K$ grows
significantly slower than $L$. We first need to list some technical
conditions.

######  Condition 5.1.

We say a triple $(\Lambda,K,L)$, consisting of a domain $\Lambda$, an integer
$K\geq 3$ and a real number $L\geq 100$ satisfies

Condition A:

$\Lambda$ is a polygon (see 7.1 for a formal definition) and $K<CL/\ln(L)$ for
some (finite) constant $C$, or

Condition B:

$\Lambda$ is a $\mathsf{C}^{2}$-smooth domain and $K^{2}\leq CL$ for some
(finite) constant $C$.

Our third main result of this paper is the following theorem for polynomials
$f$. It verifies our 1.1 under the above Condition.

######  Theorem 5.2.

Let the triple $(\Lambda,K,L)$ satisfy either one of the conditions in 5.1.
Then, for any polynomial $f$ with $f(0)=f(1)=0$, we have the asymptotic
expansion

$\displaystyle\operatorname{tr}f(1_{L\Lambda}P_{K}1_{L\Lambda})=L\ln(K)\lvert\partial\Lambda\rvert\,\frac{2\sqrt{2}}{\pi}\,\mathsf{I}(f)+\mathcal{O}(L\ln\ln(K))\,,$
(5.1)

as $K\to\infty$ (and hence $L\to\infty$). The coefficient $\mathsf{I}(f)$ is
defined in (1.5).

The proof of this theorem spans the remaining chapters of this paper. Our
approach essentially approximates the boundary curve $\partial\Lambda$ by a
straight line and this leads to an $\mathcal{O}(K^{2})$ error term for
$\mathsf{C}^{2}$-smooth domains. When we already have straight boundary lines
as in polygons then our error term is much smaller, or put differently, we can
allow for a larger $K$. The difference in results may only be due to our
methods of proof.

The proof of these theorems is rather long, so we begin with a short summary.
We will consider the polynomials $f(t)=(1-t)^{m}t$ with $m\in\mathbb{N}$. For
them, we get

$\displaystyle\operatorname{tr}f(1_{L\Lambda}P_{K}1_{L\Lambda})=$
$\displaystyle\operatorname{tr}1_{L\Lambda}(P_{K}1_{L\Lambda^{\complement}})^{m}P_{K}1_{L\Lambda}\,.$
(5.2)

To see this identity, we note that the eigenvalues of
$1_{L\Lambda}P_{K}1_{L\Lambda}$ agree with the eigenvalues of
$P_{K}1_{L\Lambda}P_{K}$, as both are jut the squares of the principal values
of $P_{K}1_{L\Lambda}$. Thus, we get

$\displaystyle\operatorname{tr}f(1_{L\Lambda}P_{K}1_{L\Lambda})$
$\displaystyle=\operatorname{tr}f(P_{K}1_{L\Lambda}P_{K})$
$\displaystyle=\operatorname{tr}(\mathds{1}-P_{K}1_{L\Lambda}P_{K})^{m}P_{K}1_{L\Lambda}P_{K}$
$\displaystyle=\operatorname{tr}1_{L\Lambda}P_{K}(\mathds{1}-P_{K}1_{L\Lambda}P_{K})^{m}P_{K}1_{L\Lambda}$
$\displaystyle=\operatorname{tr}1_{L\Lambda}P_{K}(P_{K}-P_{K}1_{L\Lambda}P_{K})^{m}P_{K}1_{L\Lambda}$
$\displaystyle=\operatorname{tr}1_{L\Lambda}P_{K}(P_{K}(\mathds{1}-1_{L\Lambda})P_{K})^{m}P_{K}1_{L\Lambda}$
$\displaystyle=\operatorname{tr}1_{L\Lambda}(P_{K}1_{L\Lambda^{\complement}})^{m}P_{K}1_{L\Lambda}\,.$

We have used that $P_{K}1_{L\Lambda}P_{K}$ commutes with $P_{K}$.

For fixed $K$ and $L\to\infty$, the asymptotics of the trace of this
operator111Apart from the different base of polynomials $f(t)=t^{m}$ was
reduced to an integral depending only on $K$ and $m$ using Roccaforte’s
formula in [8]. They calculated the said integral. The strong (exponential)
decay of the integral kernel $P_{K}$ for fixed $K$ was used to deal with the
error terms originating from Roccaforte’s formula. As we consider the limit
$K\to\infty$, this is not directly possible. We can, however, get to the same
integral using some geometric manipulations and operator estimates before
switching to the integral. This will allow us to show the same leading term
with an error bound, that depends on $K$ in a good manner. After that, we
still have to study the asymptotics of the formula in [8, Theorem 2] (for
fixed $K$) as $K\to\infty$.

Let us define for any measurable $E,E^{\prime}\subset\mathbb{R}^{2}$ and
$m\in\mathbb{N}$,

$\displaystyle\mathcal{J}_{m}(E,E^{\prime};K)\coloneqq\operatorname{tr}1_{E}(P_{K}1_{E^{\prime}})^{m}P_{K}1_{E}\,,$
(5.3)

which takes values in $[0,\infty]$, as we shall soon see.

Then, with $f_{m}(t)\coloneqq t(1-t)^{m}$,

$\displaystyle\operatorname{tr}f_{m}(1_{L\Lambda}P_{K}1_{L\Lambda})=\operatorname{tr}1_{L\Lambda}(P_{K}1_{L\Lambda^{\complement}})^{m}P_{K}1_{L\Lambda}=\mathcal{J}_{m}(L\Lambda,L\Lambda^{\complement};K)\,.$
(5.4)

The proof of Theorem 5.2 relies on two main steps, which will be proved in the
remainder of this paper. The first one is

######  Theorem 5.3.

Let the triple $(\Lambda,K,L)$ satisfy either one of the conditions in 5.1.
Let $m\in\mathbb{N}$ and $f_{m}(t)=t(1-t)^{m}$. Then, we have the asymptotic
expansion

$\displaystyle\operatorname{tr}f_{m}(1_{L\Lambda}P_{K}1_{L\Lambda})=\mathcal{J}_{m}(L\Lambda,L\Lambda^{\complement};K)=L\lvert\partial\Lambda\rvert\,\mathcal{J}_{m}\left([0,1)\times\mathbb{R}^{-},\mathbb{R}\times\mathbb{R}^{+};K\right)+\mathcal{O}(L)\,.$
(5.5)

This result achieves that the leading term of the asymptotic expansion of
$\mathrm{tr}f_{m}(1_{L\Lambda}P_{K}1_{L\Lambda})$ is a product of the surface
area of $L\Lambda$ and a term $\mathcal{J}_{m}(E,E^{\prime};K)$ with fixed
sets $E$ and $E^{\prime}$ which depends on $K$ but is independent of $L$ and
$\Lambda$.

This is proved in Sections 6 and 7. The second main step is

######  Theorem 5.4.

With $\mathcal{J}_{m}$ defined in (5.3) and $\mathsf{I}$ defined in (1.5), we
have as $K\to\infty$

$\displaystyle\mathcal{J}_{m}([0,1)\times\mathbb{R}^{-},\mathbb{R}\times\mathbb{R}^{+};K)=\frac{2\sqrt{2}}{\pi}\,\mathsf{I}(t\mapsto
t(1-t)^{m})\,\ln(K)+\mathcal{O}(\ln\ln(K))\,.$ (5.6)

This is proved in Section 8 and Appendix B. With these theorems, we can easily
conclude the

###### Proof of Theorem 5.2.

Note that any polynomial $f$ with $f(0)=f(1)=0$ can be written as a finite
linear combination of the basis polynomials $f_{m}(t)=t(1-t)^{m}$ for
$m\in\mathbb{N}$. Since both sides of (5.2) are linear in $f$, it suffices to
show the identity for $f=f_{m}$, which Theorem 5.3 and Theorem 5.4 do. ∎

Now we collect some properties of $\mathcal{J}_{m}$.

######  Lemma 5.5.

Let
$E,E_{1},E_{2},E^{\prime},E_{1}^{\prime},E_{2}^{\prime}\subset\mathbb{R}^{2}$
be measurable. Then,

  1. (1)

$\mathcal{J}_{m}(E,E^{\prime};K)\in[0,\infty]$ is well-defined.

  2. (2)

$\mathcal{J}_{m}({\makebox[4.30554pt]{{$\cdot$}}},E^{\prime};K)$ is additive,
that is, if $E_{1},E_{2}$ are disjoint, we have

$\displaystyle\mathcal{J}_{m}(E_{1}\cup
E_{2},E^{\prime};K)=\mathcal{J}_{m}(E_{1},E^{\prime};K)+\mathcal{J}_{m}(E_{2},E^{\prime};K)\,.$
(5.7)

  3. (3)

$\mathcal{J}_{m}$ satisfies the a-priori Hilbert–Schmidt norm estimate,

$\displaystyle\mathcal{J}_{m}(E,E^{\prime};K)\leq\mathcal{J}_{1}(E,E^{\prime};K)=\lVert
1_{E}P_{K}1_{E^{\prime}}\rVert_{2}^{2}\,.$ (5.8)

  4. (4)

$\mathcal{J}_{m}$ satisfies a Lipschitz-type estimate, in the sense of

$\displaystyle\lvert\mathcal{J}_{m}(E_{1},E_{1}^{\prime};K)-\mathcal{J}_{m}(E_{2},E_{2}^{\prime};K)\rvert\leq\lvert
E_{1}\Delta E_{2}\rvert/(2\pi)+m\lvert E_{1}^{\prime}\Delta
E_{2}^{\prime}\rvert/(2\pi)\,,$ (5.9)

where $E_{1}\Delta E_{2}$ is the symmetric difference. Consequently, we can
always change the sets $E$ and $E^{\prime}$ in the two arguments by sets of
(Lebesgue) measure zero without changing $\mathcal{J}_{m}$.

  5. (5)

For any unitary $\mathcal{A}\colon\mathbb{R}^{2}\to\mathbb{R}^{2}$ which is
affine-linear, that is, $\lVert\mathcal{A}(x)-\mathcal{A}(y)\rVert=\lVert
x-y\rVert$, we have

$\displaystyle\mathcal{J}_{m}(E,E^{\prime};K)=\mathcal{J}_{m}(\mathcal{A}(E),\mathcal{A}(E^{\prime});K)\,.$
(5.10)

###### Proof.

We have

$\displaystyle\mathcal{J}_{m}(E,E^{\prime};K)=\operatorname{tr}1_{E}(P_{K}1_{E^{\prime}})^{m}P_{K}1_{E}$
(5.11)
$\displaystyle=\begin{cases}\operatorname{tr}1_{E}(P_{K}1_{E^{\prime}})^{m/2}P_{K}P_{K}(1_{E^{\prime}}P_{K})^{m/2}1_{E}\,\,=\lVert
1_{E}(P_{K}1_{E^{\prime}})^{m/2}P_{K}\rVert_{2}^{2}\quad&\text{ if $m$ is
even\,,}\\\
\operatorname{tr}1_{E}(P_{K}1_{E^{\prime}})^{(m+1)/2}(1_{E^{\prime}}P_{K})^{(m+1)/2}1_{E}=\lVert
1_{E}(P_{K}1_{E^{\prime}})^{(m+1)/2}\rVert_{2}^{2}\quad&\text{ if $m$ is
odd\,.}\end{cases}$ (5.12)

Thus, $\mathcal{J}_{m}(E,E^{\prime};K)$ is the trace of a positive operator,
which has a well-defined value in $[0,\infty]$.

As $\operatorname{tr}AB=\operatorname{tr}BA$, we have

$\displaystyle\mathcal{J}_{m}(E,E^{\prime};K)=\operatorname{tr}1_{E^{\prime}}P_{K}1_{E}(P_{K}1_{E^{\prime}})^{m}\,.$
(5.13)

For any disjoint $E_{1},E_{2}$ we have $1_{E_{1}\cup
E_{2}}=1_{E_{1}}+1_{E_{2}}$ and thus, as the trace is linear, we get

$\displaystyle\mathcal{J}_{m}(E_{1}\cup
E_{2},E^{\prime};K)=\mathcal{J}_{m}(E_{1},E^{\prime};K)+\mathcal{J}_{m}(E_{2},E^{\prime};K)\,.$
(5.14)

For the Hilbert–Schmidt norm estimate, we just use that
$1_{E},1_{E^{\prime}}$, and $P_{K}$ are projections (and thus have norm $1$).
Then, we get

$\displaystyle\mathcal{J}_{m}(E,E^{\prime};K)=\lVert
1_{E}(P_{K}1_{E^{\prime}})^{m}P_{K}1_{E}\rVert_{1}\leq\lVert
1_{E}P_{K}1_{E^{\prime}}\rVert_{2}^{2}\,,$ (5.15)

where the first equality holds by positivity.

For the Lipschitz-type estimate, we first consider

$\displaystyle\left\lvert\mathcal{J}_{m}(E_{1},E_{1}^{\prime};K)-\mathcal{J}_{m}(E_{2},E_{1}^{\prime};K)\right\rvert$
$\displaystyle=\lvert\mathcal{J}_{m}(E_{1}\setminus
E_{2},E_{1}^{\prime};K)-\mathcal{J}_{m}(E_{2}\setminus
E_{1},E_{1}^{\prime};K)\rvert$ (5.16) $\displaystyle\leq\lVert
1_{E_{1}\setminus E_{2}}P_{K}1_{E_{1}^{\prime}}\rVert_{2}^{2}+\lVert
1_{E_{2}\setminus E_{1}}P_{K}1_{E_{1}^{\prime}}\rVert_{2}^{2}$ (5.17)
$\displaystyle\leq\frac{1}{2\pi}\left(\lvert E_{1}\setminus E_{2}\rvert+\lvert
E_{2}\setminus E_{1}\rvert\right)=\frac{1}{2\pi}\lvert E_{1}\Delta
E_{2}\rvert\,.$ (5.18)

This relies upon 2.13 and the additivity above. For the estimate in the second
component, we see222The start of this calculation is copied from the proof of
6.2, with some renaming of the sets $E$.

$\displaystyle\big{\lvert}\mathcal{J}_{m}$
$\displaystyle(E_{2},E_{1}^{\prime};K)-\mathcal{J}_{m}(E_{2},E_{2}^{\prime};K)\big{\rvert}$
(5.19)
$\displaystyle=\left\lvert\sum_{k=1}^{m}\operatorname{tr}1_{E_{2}}(P_{K}1_{E_{1}^{\prime}})^{k-1}P_{K}(1_{E_{1}^{\prime}}-1_{E_{2}^{\prime}})(P_{K}1_{E_{2}^{\prime}})^{m-k}P_{K}1_{E_{2}}\right\rvert$
(5.20) $\displaystyle\leq\sum_{k=1}^{m}\left\lVert
1_{E_{2}}(P_{K}1_{E_{1}^{\prime}})^{k-1}P_{K}(1_{E_{1}^{\prime}}-1_{E_{2}^{\prime}})(P_{K}1_{E_{2}^{\prime}})^{m-k}P_{K}1_{E_{2}}\right\rVert_{1}$
(5.21) $\displaystyle\leq m\lVert
P_{K}(1_{E_{1}^{\prime}}-1_{E_{2}^{\prime}})P_{K}\rVert_{1}$ (5.22)
$\displaystyle\leq m\left(\lVert P_{K}1_{E_{1}^{\prime}\setminus
E_{2}^{\prime}}\rVert_{2}^{2}+\lVert P_{K}1_{E_{2}^{\prime}\setminus
E_{1}^{\prime}}\rVert_{2}^{2}\right)=\frac{m}{2\pi}\lvert E_{1}^{\prime}\Delta
E_{2}^{\prime}\rvert\,.$ (5.23)

Combining the two estimates with the triangle inequality finishes the
Lipschitz-type estimate.

For the last point, let $\mathcal{S}_{\mathcal{A}}$ be the unitary operator on
$\mathrm{L}^{2}(\mathbb{R}^{2})$ that maps $f\mapsto(x\mapsto
f(\mathcal{A}(x)))$. We immediately see that for any measurable $E$, we have

$\displaystyle(\mathcal{S}_{\mathcal{A}})^{-1}1_{E}\,\mathcal{S}_{\mathcal{A}}=1_{\mathcal{A}(E)}\,.$
(5.24)

As the set of all affine linear maps, under which the claim holds, is closed
under composition, we may assume that $\mathcal{A}$ is either a translation or
a reflection333Any rotation is the composition of two reflections. Let
$\mathcal{A}$ be a reflection, that is, a $2\times 2$ orthogonal matrix with
determinant $-1$. For any $x,y\in\mathbb{R}^{2}$, we observe

$\displaystyle\left(\mathcal{S}_{\mathcal{A}}P_{K}\mathcal{S}_{\mathcal{A}}^{-1}\right)(x,y)=P_{K}(\mathcal{A}(x),\mathcal{A}(y))=\overline{P_{K}(x,y)}\,,$
(5.25)

as $\lVert x-y\rVert=\lVert\mathcal{A}x-\mathcal{A}y\rVert$ and $x\wedge
y=-(\mathcal{A}x)\wedge(\mathcal{A}y)$. By Mercer’s theorem, we have

$\displaystyle\mathcal{J}_{m}(E,E^{\prime};K)$
$\displaystyle=\operatorname{tr}1_{E}(P_{K}1_{E^{\prime}})^{m}P_{K}1_{E}$
(5.26)
$\displaystyle=\int_{E}\mathrm{d}x\int_{E^{\prime}}\mathrm{d}x_{1}\int_{E^{\prime}}\mathrm{d}x_{2}\cdots\int_{E^{\prime}}\mathrm{d}x_{m}\,P_{K}(x,x_{1})P_{K}(x_{1},x_{2})\cdots
P_{K}(x_{m},x)\,.$ (5.27)

Thus, we have
$\mathcal{J}_{m}(E,E^{\prime};K)=\overline{\mathcal{J}_{m}(\mathcal{A}(E),\mathcal{A}(E^{\prime});K)}$.
As $\mathcal{J}_{m}(E,E^{\prime};K)$ is always real, this implies
$\mathcal{J}_{m}(E,E^{\prime};K)=\mathcal{J}_{m}(\mathcal{A}(E),\mathcal{A}(E^{\prime});K)$.

We now consider the case that $\mathcal{A}$ is a translation, that is,
$\mathcal{A}(x)=x-x_{0}$ for some $x_{0}\in\mathbb{R}^{2}$. Here, we need to
use the magnetic translation operator $\mathcal{S}_{x_{0},K}$, which is given
by

$\displaystyle(\mathcal{S}_{x_{0},K}\varphi)(x)\coloneqq\exp(ix\wedge
x_{0}/(2K))\,\varphi(x-x_{0})\,,\quad x\in\mathbb{R}^{2}\,,$ (5.28)

for $\varphi\in\mathrm{L}^{2}(\mathbb{R}^{2})$. For any measurable set $E$, it
satisfies $\mathcal{S}_{x_{0},K}1_{E}\mathcal{S}_{x_{0},K}^{-1}=1_{E+x_{0}}$
and commutes with $P_{K}$. Thus, we observe

$\displaystyle\mathcal{J}_{m}(E,E^{\prime};K)$
$\displaystyle=\operatorname{tr}1_{E^{\prime}}P_{K}1_{E}(P_{K}1_{E^{\prime}})^{m}$
(5.29)
$\displaystyle=\operatorname{tr}\mathcal{S}_{x_{0},K}1_{E^{\prime}}P_{K}1_{E}(P_{K}1_{E^{\prime}})^{m}\mathcal{S}_{x_{0},K}^{-1}$
(5.30)
$\displaystyle=\operatorname{tr}1_{E^{\prime}+x_{0}}P_{K}1_{E+x_{0}}(P_{K}1_{E^{\prime}+x_{0}})^{m}$
(5.31) $\displaystyle=\mathcal{J}_{m}(E+x_{0},E^{\prime}+x_{0};K)\,.$ (5.32)

∎

##  6\. $K\ll L$: Norm estimates

In this section we prove norm estimates of various combinations and powers of
$1_{E}$ (for certain sets $E$) and $P_{K}$, that will be useful later.

######  Lemma 6.1.

Let $K,m\in\mathbb{N}$. If $m>1$, then we have arbitrary measurable sets
$A_{i}\subset\mathbb{R}^{2}$ for $i=2,\dots,m$. Let $R>0$ and
$x_{0}\in\mathbb{R}^{2}$. Then, there are constants $C<\infty$ and $\beta>0$
such that

$\displaystyle\left\lVert
1_{D_{R}(x_{0})}P_{K}\left(\prod_{i=2}^{m}1_{A_{i}}P_{K}\right)1_{D_{(m+1)R}^{\complement}(x_{0})}\right\rVert_{2}^{2}\leq
Cm^{4}\begin{cases}R\quad&\text{ if }R<10K\,,\\\ \exp(-\beta R)&\text{ if
}R\geq 10K\,.\end{cases}$ (6.1)

If $m=1$, then the product $\prod_{i=2}^{m}1_{A_{i}}P_{K}$ is interpreted as
the identity.

###### Proof.

We will prove this by induction over $m$. We begin with a slight
generalisation of the case $m=1$. Let $l\in\mathbb{N}$. Then, we observe

$\displaystyle\left\lVert
1_{D_{lR}(x_{0})}P_{K}1_{D_{(l+1)R}^{\complement}(x_{0})}\right\rVert_{2}^{2}$
$\displaystyle=\int_{D_{lR}(x_{0})}\mathrm{d}x\int_{D_{(l+1)R}^{\complement}(x_{0})}\mathrm{d}y\,\lvert
P_{K}(x,y)\rvert^{2}$ (6.2)
$\displaystyle=\int_{D_{lR}(0)}\mathrm{d}x\int_{D_{(l+1)R}(0)^{\complement}}\mathrm{d}y\,\lvert
G_{K}(\lVert x-y\rVert/\sqrt{8})\rvert^{2}$ (6.3) $\displaystyle\leq\pi
l^{2}R^{2}\int_{D_{R}(0)^{\complement}}\mathrm{d}y\,\lvert G_{K}(\lVert
y\rVert/\sqrt{8})\rvert^{2}$ (6.4) $\displaystyle\leq
Cl^{2}R^{2}\int_{R/\sqrt{8}}^{\infty}\mathrm{d}t\,2\pi t\,\lvert
G_{K}(t)\rvert^{2}$ (6.5) $\displaystyle\leq Cl^{2}\begin{cases}R\quad&\text{
if }R<10K\,,\\\ \exp(-\beta R)&\text{ if }R\geq 10K\,.\end{cases}$ (6.6)

We used (2.3) and 2.12 noting that $R>10K$ implies $R/\sqrt{8}>2K$.

We continue with the induction step. We have

$\displaystyle\left\lVert
1_{D_{R}(x_{0})}P_{K}\left(\prod_{i=2}^{m+1}1_{A_{i}}P_{K}\right)1_{D_{(m+2)R}^{\complement}(x_{0})}\right\rVert_{2}$
(6.7) $\displaystyle\leq\left\lVert
1_{D_{R}(x_{0})}P_{K}\left(\prod_{i=2}^{m}1_{A_{i}}P_{K}\right)1_{D_{(m+1)R}^{\complement}(x_{0})}1_{A_{m+1}}P_{K}1_{D_{(m+2)R}^{\complement}(x_{0})}\right\rVert_{2}$
(6.8) $\displaystyle+\left\lVert
1_{D_{R}(x_{0})}P_{K}\left(\prod_{i=2}^{m}1_{A_{i}}P_{K}\right)1_{A_{m+1}}1_{D_{(m+1)R}(x_{0})}P_{K}1_{D_{(m+2)R}^{\complement}(x_{0})}\right\rVert_{2}$
(6.9) $\displaystyle\leq\left\lVert
1_{D_{R}(x_{0})}P_{K}\left(\prod_{i=2}^{m}1_{A_{i}}P_{K}\right)1_{D_{(m+1)R}^{\complement}(x_{0})}\right\rVert_{2}\left\lVert
1_{A_{m+1}}P_{K}1_{D_{(m+2)R}^{\complement}(x_{0})}\right\rVert$ (6.10)
$\displaystyle+\left\lVert
1_{D_{R}(x_{0})}P_{K}\left(\prod_{i=2}^{m}1_{A_{i}}P_{K}\right)1_{A_{m+1}}\right\rVert\left\lVert
1_{D_{(m+1)R}(x_{0})}P_{K}1_{D_{(m+2)R}^{\complement}(x_{0})}\right\rVert_{2}$
(6.11) $\displaystyle\leq\left\lVert
1_{D_{R}(x_{0})}P_{K}\left(\prod_{i=2}^{m}1_{A_{i}}P_{K}\right)1_{D_{(m+1)R}^{\complement}(x_{0})}\right\rVert_{2}+\left\lVert
1_{D_{(m+1)R}(x_{0})}P_{K}1_{D_{(m+2)R}^{\complement}(x_{0})}\right\rVert_{2}$
(6.12)
$\displaystyle\leq(\sqrt{C_{m}}+(m+1)\sqrt{C_{1}})\begin{cases}\sqrt{R}\quad&\text{
if }R<10K\,,\\\ C\exp(-\beta R/2)&\text{ if }R\geq 10K\,.\end{cases}$ (6.13)

The first step relies on the triangle inequality and the fact that
multiplication operators commute. The second step relies on the Hölder
inequality and the third step uses that all the operators $1_{A},P_{K}$ have
operator norm $1$, as they are projections.

In the last step, we used the induction hypothesis and (6.6) with $l=m+1$
(taking the square root of both in the process). Here, $C_{1},C_{m}$ are the
$m$-dependent constants in the claim. The recursion
$\sqrt{C_{m+1}}=\sqrt{C_{m}}+(m+1)\sqrt{C_{1}}$ implies
$\sqrt{C_{m}}=\frac{m^{2}+m}{2}\sqrt{C_{1}}$ and thus
$C_{m}=\frac{(m^{2}+m)^{2}}{4}C_{1}$. This explains the power $m^{4}$ in the
claim and thus finishes the proof. ∎

######  Lemma 6.2.

For $R>0,x_{0}\in\mathbb{R}^{2}$, let $E\subset D_{R}(x_{0})$ and let
$E^{\prime}\subset\mathbb{R}^{2}$ be measurable. Then, with $\mathcal{J}_{m}$
defined in (5.3) we have for some $\beta>0$,

$\displaystyle\mathcal{J}_{m}(E,E^{\prime};K)=\mathcal{J}_{m}(E,E^{\prime}\cap
D_{(m+1)R}(x_{0});K)+\begin{cases}\mathcal{O}(m^{5}R)\quad&\text{ if
}R<10K\,,\\\ \mathcal{O}(m^{5}\exp(-\beta R))&\text{ if }R\geq
10K\,.\end{cases}$ (6.14)

###### Proof.

Let $E^{\prime\prime}\coloneqq E^{\prime}\cap D_{(m+1)R}(x_{0})$. Using the
definition of $\mathcal{J}_{m}$, we can write the telescope sum

$\displaystyle\left\lvert\mathcal{J}_{m}(E,E^{\prime};K)-\mathcal{J}_{m}(E,E^{\prime\prime};K)\right\rvert$
(6.15)
$\displaystyle=\left\lvert\sum_{k=1}^{m}\operatorname{tr}1_{E}(P_{K}1_{E^{\prime}})^{k-1}P_{K}(1_{E^{\prime}}-1_{E^{\prime\prime}})(P_{K}1_{E^{\prime\prime}})^{m-k}P_{K}1_{E}\right\rvert$
(6.16) $\displaystyle\leq\sum_{k=1}^{m}\left\lVert
1_{E}(P_{K}1_{E^{\prime}})^{k-1}P_{K}(1_{E^{\prime}}-1_{E^{\prime\prime}})(P_{K}1_{E^{\prime\prime}})^{m-k}P_{K}1_{E}\right\rVert_{1}$
(6.17) $\displaystyle\leq\sum_{k=1}^{m}\left\lVert
1_{E}(P_{K}1_{E^{\prime}})^{k-1}P_{K}1_{E^{\prime}\setminus
E^{\prime\prime}}\right\rVert_{2}\left\lVert 1_{E^{\prime}\setminus
E^{\prime\prime}}(P_{K}1_{E^{\prime\prime}})^{m-k}P_{K}1_{E}\right\rVert_{2}$
(6.18) $\displaystyle\leq\sum_{k=1}^{m}\left\lVert
1_{D_{R}(x_{0})}(P_{K}1_{E^{\prime}})^{k-1}P_{K}1_{D_{(m+1)R}(x_{0})^{\complement}}\right\rVert_{2}\left\lVert
1_{D_{(m+1)R}(x_{0})^{\complement}}(P_{K}1_{E^{\prime\prime}})^{m-k}P_{K}1_{D_{R}(x_{0})}\right\rVert_{2}$
(6.19) $\displaystyle\leq\begin{cases}Cm^{5}R\quad&\text{ if }R<10K\,,\\\
Cm^{5}\exp(-\beta R)&\text{ if }R\geq 10K\,.\end{cases}$ (6.20)

We used that $E^{\prime}\subset E^{\prime\prime}$ and thus
$1_{E^{\prime}}-1_{E^{\prime\prime}}=1_{E^{\prime}\setminus E^{\prime\prime}}$
and that $E^{\prime}\setminus E^{\prime\prime}\subset
D_{(m+1)R}(x_{0})^{\complement}$ by construction of $E^{\prime\prime}$. We
also used that whenever $A\subset A^{\prime}$ and $B\subset B^{\prime}$, for
any operator $T$, we have $\lVert 1_{A}T1_{B}\rVert_{2}=\lVert
1_{A}1_{A^{\prime}}T1_{B^{\prime}}1_{B}\rVert_{2}\leq\lVert
1_{A^{\prime}}T1_{B^{\prime}}\rVert_{2}$. The last step is of course 6.1. ∎

We need some more properties of $\mathcal{J}_{m}$, where $E^{\prime}$ is a
half space.

######  Lemma 6.3.

Let $a,b\in[0,\infty)$ and $c\in(0,\infty]$ with $b<c$. Then, with
$\mathcal{J}_{m}$ defined in (5.3) we have

$\displaystyle\mathcal{J}_{m}\left([0,a)\times(-c,-b),\mathbb{R}\times\mathbb{R}^{+};K\right)=a\mathcal{J}_{m}\left([0,1)\times(-c,-b),\mathbb{R}\times\mathbb{R}^{+};K\right),$
(6.21)

and there is a constant $\beta>0$ such that

$\displaystyle\mathcal{J}_{m}\left([0,a)\times(-c,-b),\mathbb{R}\times\mathbb{R}^{+};K\right)\leq\begin{cases}Ca\ln\left(\frac{20K}{b+1}\right)&\text{
if }b<10K\,,\\\ Ca\exp(-\beta b)&\text{ if }b\geq 10K\,.\end{cases}$ (6.22)

###### Proof.

We prove the upper bound first. As $\mathcal{J}_{m}$ is additive in the first
component and non-negative, as shown in 5.5, it suffices to consider
$c=\infty$. We observe that by the Hilbert–Schmidt norm estimate

$\displaystyle\mathcal{J}_{m}\left([0,a)\times(-\infty,-b),\mathbb{R}\times\mathbb{R}^{+};K\right)$
$\displaystyle\leq\int_{0}^{a}\mathrm{d}x_{1}\int_{-\infty}^{-b}\mathrm{d}x_{2}\int_{\mathbb{R}\times\mathbb{R}^{+}}\mathrm{d}y\,\lvert
P_{K}((x_{1},x_{2}),y)\rvert^{2}$ (6.23) $\displaystyle\leq
a\int_{-\infty}^{-b}\mathrm{d}x_{2}\int_{D_{\lvert
x_{2}\rvert}(0)^{\complement}}\mathrm{d}y\,\lvert
G_{K}(\|y\|/\sqrt{8})\rvert^{2}$ (6.24) $\displaystyle=16\pi
a\int_{b}^{\infty}\mathrm{d}s\,\int_{s/\sqrt{8}}^{\infty}\mathrm{d}t\,t\lvert
G_{K}(t)\rvert^{2}$ (6.25) $\displaystyle\leq
Ca\begin{cases}\ln\left(\frac{20K}{b+1}\right)&\text{ if }b<10K\,,\\\
\exp(-\beta b)&\text{ if }b\geq 10K\,.\end{cases}$ (6.26)

The last step relies upon 2.12 and an easy estimate for the integral over $s$.
The first claim just says that for any $b,c$, the function

$\displaystyle a\mapsto
f_{b,c}(a)\coloneqq\mathcal{J}_{m}\left([0,a)\times(-c,-b),\mathbb{R}\times\mathbb{R}^{+};K\right)$
(6.27)

is linear. As we have already shown, it has a linear upper bound, and hence it
suffices to show that it satisfies the Cauchy functional equation, that is,
$f_{b,c}(a)+f_{b,c}(a^{\prime})=f_{b,c}(a+a^{\prime})$. For shortness, we
write $I\coloneqq(-c,-b)$. We use the additivity and translational invariance
of $\mathcal{J}_{m}$, as proved in 5.5. We observe

$\displaystyle f_{b,c}(a+a^{\prime})$
$\displaystyle=\mathcal{J}_{m}\left([0,a+a^{\prime})\times
I,\mathbb{R}\times\mathbb{R}^{+};K\right)$ (6.28)
$\displaystyle=\mathcal{J}_{m}\left([0,a)\times
I,\mathbb{R}\times\mathbb{R}^{+};K\right)+\mathcal{J}_{m}\left([a,a+a^{\prime})\times
I,\mathbb{R}\times\mathbb{R}^{+};K\right)$ (6.29)
$\displaystyle=f_{b,c}(a)+\mathcal{J}_{m}\left([0,a^{\prime})\times
I,(\mathbb{R}-a)\times\mathbb{R}^{+};K\right)$ (6.30)
$\displaystyle=f_{b,c}(a)+\mathcal{J}_{m}\left([0,a^{\prime})\times
I,\mathbb{R}\times\mathbb{R}^{+};K\right)$ (6.31)
$\displaystyle=f_{b,c}(a)+f_{b,c}(a^{\prime})\,.$ (6.32)

This completes the proof. ∎

##  7\. $K\ll L$: Proof of Theorem 5.3

The goal of this section is to prove Theorem 5.3, which stated See 5.3

Let us also recall 5.1: See 5.1 The proof of Theorem 5.3 splits into the two
cases according to this condition. Under Condition A, it follows from 7.2
below and under Condition B it is implied by 7.4 below.

###  7.1. Proof of Theorem 5.3 for polygons $\Lambda$

The time has come for the formal definition of polygons.

######  Definition 7.1.

A polygon $\Lambda$ is a bounded, connected Lipschitz domain in
$\mathbb{R}^{2}$ with boundary $\partial\Lambda$ such that there is a finite
set $\mathcal{V}=\\{x_{1},\ldots,x_{V}\\}$ of points in $\mathbb{R}^{2}$ with
$\partial\Lambda$ being the union of the closed line segments
$\gamma_{i}=[x_{i},x_{i+1}],i=1,\ldots,V$ between $x_{i}$ and $x_{i+1}$ with
the convention $x_{V+1}\coloneqq x_{0}$. The points in $\mathcal{V}$ (or the
boundary of the edges) of are called corners and the line segments
$\gamma_{i}$ are called edges. The number $V$ is chosen minimal so that none
of the interior angles, denoted by $\theta_{i}$, (at the corner $x_{i}$) are
$\pi$ or $2\pi$.

By definition, a polygon is open and simply-connected. By the Lipschitz
property, two edges can only intersect in a common corner.

All our claims below can be extended easily to polygons that are not simply-
connected but this would be more cumbersome from a notational point of view.

######  Lemma 7.2.

Let $\Lambda\subset\mathbb{R}^{2}$ be a polygon according to 7.1. Let
$K,m\in\mathbb{N},L\in\mathbb{R}$ with $L>2$. Then, as $L\to\infty$ and with
$K\leq CL/\ln(L))$ (for some finite constant $C$), we have the asymptotic
expansion

$\displaystyle\mathcal{J}_{m}(L\Lambda,L\Lambda^{\complement};K)$
$\displaystyle=L\lvert\partial\Lambda\rvert\,\mathcal{J}_{m}\left([0,1)\times\mathbb{R}^{-},\mathbb{R}\times\mathbb{R}^{+};K\right)+\mathcal{O}((\ln(L)^{3}+K\ln(K)))\,.$
(7.1)

######  Remark.

The condition $K\leq CL/\ln(L)$ fits the condition for polygons in the main
Theorem 5.2. We could extend 7.2 to $K,L$ such that $K/L\to 0$ as $L\to\infty$
with essentially the same proof at the expense of a larger error term.

###### Proof.

As we have additivity in the first argument of $\mathcal{J}_{m}$, we will
decompose $L\Lambda$ as a (disjoint) union of certain sets, called edge sets,
corner sets, the essential interior and a (remaining) null set. First, we
consider the collar neighbourhood of the boundary $L\Lambda$, namely
$D_{R}(\partial L\Lambda)\cap L\Lambda$, define $E_{\text{int}}\coloneqq
L\Lambda\setminus\overline{D_{R}(\partial L\Lambda)}$ and call this the
essential interior. We want to cover (up to null sets) the collar
neighbourhood by (lots of) open squares of length $R$, so that these squares
are all inside $L\Lambda$, one side lies on an edge of $L\Lambda$ and there is
a ball of radius $R(m+1)$ around the centre of the square, in which $L\Lambda$
looks like a half-space. To enable this condition and prevent squares of
different edges from intersecting we stop in a safe distance from the corners.
Around each corner of $L\Lambda$ we will cut out some measurable subset of a
disk of radius $3R(m+1)/\varepsilon$ centred at the corner and call the
intersection with $L\Lambda$ a corner set. This distance depends also on the
angle at a corner and therefore we introduce an $\varepsilon$. The variables
$R$ and $\varepsilon$ will be chosen accordingly in the proof. A null set is
included to really cover $L\Lambda$ by all these components.

Figure 1. Two examples of a corner set (dotted), some edge sets (hatched) and
the essential interior (covered by a hexagonal grid) near a corner with
different angles. The dotted circles show the balls of radius $R(m+1)$ around
the centre of edge sets and the circular sectors are the disks of radius
$R(m+1)/\varepsilon$ around the corners intersected with $\Lambda$.

As $\Lambda$ is a polygon, there is an $\varepsilon\in(0,1)$ with the
following _intersection property_ : For each $\delta<\varepsilon$,
$x\in\mathbb{R}^{2}$ such that the disk $D_{\varepsilon\delta}(x)$ intersects
multiple edges, it intersects exactly two edges and the corner between these
edges is at most $\delta$ away from $x$. One such $\varepsilon$ can be
constructed as follows. Let $d_{i}$ be the smallest (Euclidean) distance
between the corner $x_{i}$ and any edge of the polygon not containing $x_{i}$.
Then we set (recall, $\theta_{i}$ is the interior angle at $x_{i}$)

$\displaystyle\varepsilon\coloneqq\min\\{|\sin(\theta_{i})|,d_{i}:i=1,\ldots,V\\}\,.$
(7.2)

Since $K<L/\ln(L)$ and since we only care for the asymptotic behaviour as
$L\to\infty$, we can assume

$\displaystyle 30K(m+1)<\varepsilon^{2}L\,.$ (7.3)

Let $R\coloneqq\max(10K,\varepsilon^{2}\ln(L)^{2}/(3(m+1)))$. This ensures
that $R$ behaves nicely with the domain, as

$\displaystyle 3R(m+1)<\varepsilon^{2}L$ (7.4)

and that the error term in 6.2, which is mainly $\exp(-\beta R)$ decays faster
than $L^{-1}$ as $L\to\infty$.

We will now construct a covering of $L\Lambda$ with the properties we just
described. Let $\gamma\subset\partial\Lambda$ be an edge of length
$|\gamma|=2\lambda$. After an isometry, we may assume that
$\gamma=(-\lambda,\lambda)\times\\{0\\}$ and that the outward unit normal
vector of $\Lambda$ at $0$ is $+e_{2}$. Scaling up with $L$, the edge
$L\gamma$ is now $(-L\lambda,L\lambda)\times\\{0\\}$. The candidates for our
edge sets are now the sets
$E_{\gamma,j}\coloneqq(R(j-1/2),R(j+1/2))\times(-R,0)$ for some
$j\in\mathbb{Z}$.

Let $\lvert j\rvert R\leq L\lambda-2R(m+1)/\varepsilon$. In order to show that
$E_{\gamma,j}$ is an edge set, according to the above definition, we assume,
by contradiction, that $D_{R(m+1)}((Rj,-R/2))\cap\partial(L\Lambda)\setminus
L\gamma$ is non-empty and contains the point $x$. Note that $+e_{2}$ is the
outward unit normal vector of $L\Lambda$ at $(Rj,0)$. Then, we have

$\displaystyle\delta\coloneqq\operatorname{dist}(x/L,(jR/L,0))/\varepsilon$
$\displaystyle\leq
1/L\left(\operatorname{dist}(x,(jR,-R/2))+R/2\right)/\varepsilon$ (7.5)
$\displaystyle<R(m+2)/(\varepsilon L)<3R(m+1)/(\varepsilon L)<\varepsilon\,.$
(7.6)

The last inequality follows by (7.4). As the points $x/L\in\partial\Lambda$
and $(jR/L,0)\in\partial\Lambda$ have distance $\varepsilon\delta$, we know
that the distance from $(jR/L,0)$ to the next corner, which is
$(\pm\lambda,0)$, is at most $\delta<2R(m+1)/L\varepsilon$, by the above
intersection property of $\Lambda$. This contradicts the assumption $\lvert
j\rvert R\leq L\lambda-2R(m+1)/\varepsilon$. Thus, under this assumption,
$E_{\gamma,j}$ is indeed an edge set.

For an edge $\gamma$, let
$N_{\gamma}\coloneqq\left\lfloor\left(L\lambda-2R(m+1)/\varepsilon\right)/R\right\rfloor$
(with $\lambda=\lvert\gamma\rvert/2$). Thus, we have $2N_{\gamma}+1$ edge sets
along $\gamma$ and we observe that

$\displaystyle(2N_{\gamma}+1)R=L\lvert\gamma\rvert+\mathcal{O}(R)\,.$ (7.7)

Furthermore, together with a Lebesgue null set $E_{\gamma,\text{null}}$, they
cover

$\displaystyle E_{\gamma}\coloneqq
E_{\gamma,\text{null}}\cup\bigcup_{j=-N_{\gamma}}^{N_{\gamma}}E_{\gamma,j}=[-(N_{\gamma}+1/2)R,(N_{\gamma}+1/2)R]\times[-R,0)\,.$
(7.8)

The next claim is that any two edge sets
$E_{\gamma,j},E_{\gamma^{\prime},j^{\prime}}$ are disjoint. For
$\gamma=\gamma^{\prime}$, this is obvious. Assume
$\overline{E_{\gamma,j}}\cap\overline{E_{\gamma^{\prime},j^{\prime}}}\neq\emptyset$
with $\gamma\neq\gamma^{\prime}$. Let $x$ be the centre of $E_{\gamma,j}$.
Then, the set $\overline{D_{(1+\sqrt{2}/2)R}(x)}\cap\partial L\Lambda$
contains points on the edge $L\gamma^{\prime}$ and thus, as
$(1+\sqrt{2}/2)R<2R\leq R(m+1)$, we can conclude that $E_{\gamma,j}$ is not an
edge set, which is a contradiction.

Let us now define the corner sets. For each $i=1,2,\dots,V$, let the corner
set $E_{x_{i}}$ be defined by

$\displaystyle E_{x_{i}}\coloneqq
D_{3R(m+1)/\varepsilon}(Lx_{i})\cap\left(L\Lambda\setminus\left(E_{\text{int}}\cup\bigcup_{i^{\prime}=1}^{V}E_{\gamma_{i^{\prime}}}\right)\right).$
(7.9)

We will now show that $L\Lambda$ is covered by the sets
$E_{\text{int}},E_{\gamma_{i}}$ and $E_{x_{i}}$ for $i\in\\{1,2,\dots,V\\}$.

Let $x\in L\Lambda$ with $x\not\in E_{\text{int}}$ and for all
$i\in\\{1,2,\dots,V\\}$, $x\not\in E_{\gamma_{i}}$. As $x\in L\Lambda\setminus
D_{R}(\partial L\Lambda)$, we know that $\operatorname{dist}(x,\partial
L\Lambda)<R$. Let $y\in\partial L\Lambda$ be a closest point to $x$, that is,
$\|x-y\|=\operatorname{dist}(x,\partial L\Lambda)<R$. Let
$\gamma\subset\partial L$ be an edge such that $y\in L\gamma$ 444There is only
one such edge unless $y$ is a corner, in which case, there are two possible
$\gamma$.. Assume without loss of generality that
$\gamma=(-\lambda,\lambda)\times\\{0\\}$. As $x\not\in E_{\gamma}$, we can
conclude
$y\in[-L\lambda,L\lambda]\setminus[-(N_{\gamma}+1/2)R,(N_{\gamma}+1/2)R]$.
Thus, the distance of $y$ to the closest corner $x_{i_{0}}$ is at most
$2R(m+1)/\varepsilon+R$ and the distance of $x$ to $x_{i_{0}}$ is at most
$2R(m+1)/\varepsilon+2R\leq 3R(m+1)/\varepsilon$. This implies $x\in
E_{x_{i_{0}}}$ and thus

$\displaystyle
L\Lambda=E_{\text{int}}\cup\bigcup_{i=1}^{V}\left(E_{\gamma_{i}}\cup
E_{x_{i}}\right).$ (7.10)

Using the Hilbert–Schmidt norm estimate $\mathcal{J}_{m}\leq\mathcal{J}_{1}$
in 5.5 and 2.12, we now estimate

$\displaystyle\mathcal{J}_{m}(E_{\text{int}},L\Lambda^{\complement};K)$
$\displaystyle\leq\int_{E_{\text{int}}}\mathrm{d}x\int_{L\Lambda^{\complement}}\mathrm{d}y\,\lvert
P_{K}(x,y)\rvert^{2}$ (7.11)
$\displaystyle\leq\int_{E_{\text{int}}}\mathrm{d}x\int_{D_{R}(x)^{\complement}}\mathrm{d}y\,\lvert
G_{K}(\|y-x\|/\sqrt{8})\rvert^{2}$ (7.12) $\displaystyle=C\lvert
E_{\text{int}}\rvert\int_{R/\sqrt{8}}^{\infty}\mathrm{d}s\,s\,\lvert
G_{K}(s)\rvert^{2}$ (7.13) $\displaystyle\leq
CL^{2}\lvert\Lambda\rvert\exp(-\beta R)$ (7.14) $\displaystyle\leq C\,.$
(7.15)

In the final step, we used that $R\geq\varepsilon^{2}\ln(L)^{2}/(3(m+1))$ and
in the step before, we used $R\geq 10K$. The important part is the
$\ln(L)^{2}$, which leads to an annihilation of the polynomial growth.

For a corner set $E_{x_{i}}$, let $\mathcal{A}(\theta_{i},r)$ be the circular
sector centred at the corner with radius $r$ and opening angle $\theta_{i}$
between the two edges touching this corner. We observe

$\displaystyle
E_{x_{i}}\subset\mathcal{A}(\theta_{i},3R(m+1)/\varepsilon)\subset
L\Lambda\,.$ (7.16)

Due to translational and rotational invariance of $\mathcal{J}_{m}$, we can
assume that the corner is at $0$ and one edge goes in direction $e_{1}$. We
observe

$\displaystyle\mathcal{J}_{m}(E_{x_{i}},L\Lambda^{\complement};K)$
$\displaystyle\leq\mathcal{J}_{1}(E_{x_{i}},L\Lambda^{\complement};K)$ (7.17)
$\displaystyle\leq\mathcal{J}_{1}(\mathcal{A}(\theta_{i},3R(m+1)/\varepsilon),L\Lambda^{\complement};K)$
(7.18)
$\displaystyle\leq\mathcal{J}_{1}(\mathcal{A}(\theta_{i},3R(m+1)/\varepsilon),\mathcal{A}(\theta_{i},3R(m+1)/\varepsilon)^{\complement};K)$
(7.19) $\displaystyle\leq CR\ln(R)\,.$ (7.20)

We used the Hilbert–Schmidt norm estimate and could then enlarge the domains
in both arguments of $\mathcal{J}_{1}$, as
$\mathcal{J}_{1}(E,E^{\prime};K)=\lVert
1_{E}P_{K}1_{E^{\prime}}\rVert_{2}^{2}$. The final step is just an application
of Theorem 3.1 with scaling parameter $R$ instead of $L$. The constant $C$
depends only on the angle $\theta_{i}$, $m$ and $\varepsilon$.555The constant
is continuous in $\theta_{i}\in[0,2\pi]$.

We denote the total number of edge sets by $N$ and get immediately that
$N=\sum_{i=1}^{V}(2N_{\gamma_{i}}+1)$. Using (7.7), we conclude that

$\displaystyle
RN=\sum_{i=1}^{V}R(2N_{\gamma_{i}}+1)=L\lvert\partial\Lambda\rvert+\mathcal{O}(VR)\,,$
(7.21)

where we recall that $V$ is the number of edges (or corners) of $\Lambda$.

Finally, for an edge set $E_{\gamma_{i},j}$, we translate and rotate it such
that $E_{\gamma_{i},j}=[0,R]\times[-R,0]$ and $L\partial\Lambda\cap
E_{\text{edge}}=[0,R]\times\\{0\\}$. Let
$x_{0}\coloneqq(R/2,-R/2)\in\mathbb{R}^{2}$. Due to one of the defining
properties of edge sets, we have

$\displaystyle
D_{R(m+1)}(x_{0})\cap(L\Lambda)^{\complement}=D_{R(m+1)}(x_{0})\cap\left(\mathbb{R}\times\mathbb{R}^{+}\right)\,.$
(7.22)

Thus, using the translational and rotational invariance of $\mathcal{J}_{m}$,
as well as applying 6.2, we can conclude

$\displaystyle\mathcal{J}_{m}\left(E_{\gamma_{i},j},L\Lambda^{\complement};K\right)=\mathcal{J}_{m}\left([0,R)\times(-R,0),\mathbb{R}\times\mathbb{R}^{+};K\right)+\mathcal{O}(\exp(-\beta
R))\,,$ (7.23)

where the error term is uniform in $i$ and $j$.

We will now use the additivity of $\mathcal{J}_{m}$ and that it vanishes on
zero sets in combination with 6.3 to see that

$\displaystyle\mathcal{J}_{m}$
$\displaystyle\left([0,R)\times(-R,0),\mathbb{R}\times\mathbb{R}^{+};K\right)$
(7.24)
$\displaystyle=\mathcal{J}_{m}\left([0,R)\times(-\infty,0),\mathbb{R}\times\mathbb{R}^{+};K\right)-\mathcal{J}_{m}\left([0,R)\times(-\infty,-R),\mathbb{R}\times\mathbb{R}^{+};K\right)$
(7.25)
$\displaystyle=R\mathcal{J}_{m}\left([0,1)\times(-\infty,0),\mathbb{R}\times\mathbb{R}^{+};K\right)+\mathcal{O}(R\exp(-\beta
R))\,.$ (7.26)

Again, as $R\geq C\ln(L)^{2}$, we the error term decays at least as
$\mathcal{O}(L^{-1})$ as $L\to\infty$.

We recall that $V$ is the number of corners and
$N=\sum_{i=1}^{V}(2N_{\gamma_{i}}+1)$ is the number of edge sets. Combining
everything we have just shown, we arrive at

$\displaystyle\mathcal{J}_{m}(L\Lambda,L\Lambda^{\complement};K)$ (7.27)
$\displaystyle=\mathcal{J}_{m}(E_{\text{int}},L\Lambda^{\complement};K)+\sum_{i=1}^{V}\Big{(}\mathcal{J}_{m}(E_{x_{i}},L\Lambda^{\complement};K)+\sum_{j=-N_{\gamma_{i}}}^{N_{\gamma_{i}}}\mathcal{J}_{m}(E_{\gamma_{i},j},L\Lambda^{\complement};K)\Big{)}$
(7.28)
$\displaystyle=\mathcal{O}(1)+V\mathcal{O}(R\ln(R))+N\left(R\mathcal{J}_{m}\left([0,1)\times(-\infty,0),\mathbb{R}\times\mathbb{R}^{+};K\right)+\mathcal{O}(L^{-1})\right)$
(7.29)
$\displaystyle=NR\mathcal{J}_{m}\left([0,1)\times(-\infty,0),\mathbb{R}\times\mathbb{R}^{+};K\right)+\mathcal{O}(VR\ln(R)+1)$
(7.30)
$\displaystyle=(L\lvert\partial\Lambda\rvert+\mathcal{O}(VR))\mathcal{J}_{m}\left([0,1)\times(-\infty,0),\mathbb{R}\times\mathbb{R}^{+};K\right)+\mathcal{O}(VR\ln(R))$
(7.31)
$\displaystyle=L\lvert\partial\Lambda\rvert\mathcal{J}_{m}\left([0,1)\times(-\infty,0),\mathbb{R}\times\mathbb{R}^{+};K\right)+\mathcal{O}(VR(\ln(K)+\ln(R))\,.$
(7.32)

We used 6.3 to deal with the error term stemming from the expansion of $N$,
which comes from (7.21).

The main term already agrees with the claim, but the error term still contains
$R$. We now use that $R\geq K$ and $R\approx\max(K,\ln(L)^{2})$ to conclude

$\displaystyle VR(\ln(K)+\ln(R))\leq CR\ln(R)\leq
C(K\ln(K)+\ln(L)^{2}\ln(\ln(L)^{2})\leq C(K\ln(K)+\ln(L)^{3})\,.$ (7.33)

The $\ln(L)^{3}$ is obviously not optimal, but it is not a relevant error term
for our application. ∎

###  7.2. Proof of Theorem 5.3 for $\mathsf{C}^{2}$-smooth domains $\Lambda$

The proof is similar to the polygonal case. However, instead of estimating the
contribution of corners, we need to deal with flattening the
$\mathsf{C}^{2}$-smooth boundary curve. We use rather crude estimates for the
error terms, which will lead to the assumption $K^{2}<L$. The error estimates
for some contributions will actually be rather sharp, but we guess that these
error terms cancel each other and thus we could allow for a weaker assumption,
at best $K\leq CL$.

Let us begin with some technical results concerning $\mathsf{C}^{2}$-smooth
domains. They are mainly stated for the convenience of the reader and
introduce the notation that is used later on.

######  Lemma 7.3 ($\mathsf{C}^{2}$-smooth tubular neighbourhood theorem).

Let $\Lambda$ be a $\mathsf{C}^{2}$-smooth domain. Let $S_{1},\dots,S_{r}$ be
the connected components of $\partial\Lambda$. Then, there are
$\varepsilon>0,C_{\Lambda}<\infty$ and for each $i=1,\dots,r$ a
$\mathsf{C}^{1}$-smooth function $g_{i}\colon[0,\lvert
S_{i}\rvert)\times(-\varepsilon,\varepsilon)\to\mathbb{R}^{2}$, such that for
any $i=1,\dots,r$, any $t,t_{0}\in[0,\lvert S_{i}\rvert)$ we have

  1. (1)

$Dg_{i}(t,0)\in O(2)$, the orthogonal group,

  2. (2)

$g_{i}$ is injective, it and its inverse have an $\varepsilon$-local Lipschitz
constant of at most $2$ and for $i\neq i^{\prime}$ the images of $g_{i}$ and
$g_{i^{\prime}}$ do not intersect,

  3. (3)

$s=0$ if and only if $g_{i}(t,s)\in\partial\Lambda$,

  4. (4)

$s<0$ if and only if $g_{i}(t,s)\in\Lambda$,

  5. (5)

$\operatorname{dist}(g(t,s),\partial\Lambda)=\lvert s\rvert$ and the image of
$g_{i}$ is the $\varepsilon$-neighbourhood $D_{\varepsilon}(S_{i})$,

  6. (6)

$\lVert Dg_{i}(t_{0},0)-Dg_{i}(t,s)\rVert\leq C_{\Lambda}(\lvert
s\rvert+\lvert t-t_{0}\rvert)$,

  7. (7)

$\lVert Dg_{i}(0,0)-Dg_{i}(\lvert S_{r}\rvert-t,s)\rVert\leq
C_{\Lambda}(\lvert s\rvert+\lvert t\rvert)$.

###### Proof.

Each of the connected components $S_{1},S_{2},\dots,S_{r}$ is a closed loop in
$\mathbb{R}^{2}$. For each such loop $S_{i}$, as $\partial\Lambda$ is
$\mathsf{C}^{2}$-smooth, we can choose a parametrisation, that is, a periodic
$\mathsf{C}^{2}$-smooth function $f_{i}\colon\mathbb{R}\to S_{i}$ which
satisfies $\lVert f_{i}^{\prime}(t)\rVert=1$ for all $t\in\mathbb{R}$. Thus,
its period is $\lvert S_{i}\rvert$, the length of the loop and we can regard
$f$ as an injective $\mathsf{C}^{2}$-smooth function on
$T_{i}\coloneqq[0,\lvert S_{i}\rvert]/\\{0,\lvert S_{i}\rvert\\}$, the
interval with identified endpoints. We proceed to define a function
$g_{i}\colon T_{i}\times[-1,1]\to\mathbb{R}^{2}$ by setting

$\displaystyle g_{i}(t,s)\coloneqq f_{i}(t)+s\mathcal{R}f_{i}^{\prime}(t)\,,$
(7.34)

where $\mathcal{R}=\begin{pmatrix}0&1\\\ -1&0\end{pmatrix}$ is the matrix
associated to a $-\pi/2$ rotation. We know that $\mathcal{R}f_{i}^{\prime}(t)$
has norm $1$, is continuous in $t$ and is always orthogonal to (the tangent
line at) $S_{i}$. Thus, with respect to $\partial\Lambda$, it is either the
inward normal vector for all $t\in T_{i}$ or the outward normal vector for all
$t\in T_{i}$. We assume that it is the outward normal vector. We observe that
$g_{i}$ is $\mathsf{C}^{1}$-smooth and that

$\displaystyle
Dg_{i}(t,0)=\left(f_{i}^{\prime}(t),\mathcal{R}f_{i}^{\prime}(t)\right)$
(7.35)

is an orthogonal matrix, as $\lVert f_{i}^{\prime}(t)\rVert=1$ and thus the
two column vectors form an orthonormal basis. In particular, $Dg_{i}(t,0)$ is
always invertible. Thus, for each $t\in T_{i}$, there is an
$\varepsilon_{t}>0$ such that $g_{i}$ is injective on
$D_{\varepsilon_{t}}((t,0))\subset T_{i}\times[-1,1]$. This forms an open
cover of the compact set $T_{i}\times\\{0\\}$, which means that there is a
fixed $\varepsilon>0$, such that $g_{i}$ is injective on any disk of radius
$7\varepsilon$. Assume $g_{i}(t,s)=g_{i}(t^{\prime},s^{\prime})$ for
$(t,s),(t^{\prime},s^{\prime})\in T_{i}\times[-\varepsilon,\varepsilon]$. As
$\lVert g_{i}(t,s)-f_{i}(t)\rVert\leq\varepsilon$, we have $\lVert
f_{i}(t)-f_{i}(t^{\prime})\rVert\leq 2\varepsilon$. Since $f_{i}$ is
injective, $\mathsf{C}^{2}$-smooth and $\lVert f_{i}^{\prime}(t)\rVert=1$, for
sufficiently small $\varepsilon>0$, we can conclude that
$\operatorname{dist}_{T_{i}}(t,t^{\prime})=\min\left(\lvert
t-t^{\prime}\rvert,\lvert S_{i}\rvert-\lvert t-t^{\prime}\rvert\right)\leq
4\varepsilon$. Thus, the distance between $(t,s)$ and
$(t^{\prime},s^{\prime})$ is at most $6\varepsilon$, which implies
$(t,s)=(t^{\prime},s^{\prime})$, as $6<7$.

We have now proved that $g_{i}$ is injective on
$T_{i}\times[-\varepsilon,\varepsilon]$ for sufficiently small
$\varepsilon>0$. By choosing $\varepsilon_{0}>0$ even smaller, we can ensure
that for any $\varepsilon<\varepsilon_{0}$ this holds for all $i=1,\dots,r$
simultaneously, that $g_{i}$ and $g_{i}^{-1}$ have $2$ as an
$\varepsilon$-local Lipschitz constant666This is possible as $Dg_{i}(t,0)\in
O(2)$ preserves the Euclidean norm. and that
$g_{i}(T_{i}\times(-\varepsilon,\varepsilon))\cap
g_{i^{\prime}}(T_{i^{\prime}}\times(-\varepsilon,\varepsilon))=\emptyset$ for
$i\neq i^{\prime}$.

If $g_{i}(t,s)\in\partial\Lambda$, then, as there are $i^{\prime},t^{\prime}$
with $g_{i}(t,s)=g_{i^{\prime}}(t^{\prime},0)$, due to injectivity, we know
that $(i,t,s)=(i^{\prime},t^{\prime},0)$ and thus $s=0$ if and only if
$g_{i}(t,s)\in\partial\Lambda$. Because we choose
$\mathcal{R}f_{i}^{\prime}(t)$ to be the outward normal vector, we can
conclude that for each fixed $t$, (4) holds whenever $\lvert s\rvert$ is
sufficiently small and thus, by continuity, it holds for all $\lvert
s\rvert<\varepsilon$, as $g_{i}(t,s)\in\partial\Lambda$ implies $s=0$.

For any $x\in D_{\varepsilon}(\partial\Lambda)$, the closest boundary point
$y\in\partial\Lambda$ has to be in some $S_{i}$ and thus $y=g_{i}(t,0)$.
Furthermore, the line from $x$ to $y$ has to intersect $\partial\Lambda$
orthogonally at $y$. However, by definition, that is the line $s\mapsto
g_{i}(t,s)$. Thus, as $\lvert s\rvert<\varepsilon$, the injectivity tells us
that $g_{i}(t,\pm s)=x$, where the sign is negative if and only if
$x\in\Lambda$.

For the final two claims, we observe that

$\displaystyle\left\lVert Dg_{i}(t,s)-Dg_{i}(t_{0},0)\right\rVert$
$\displaystyle=\left\lVert\left(f_{i}^{\prime}(t)-f_{i}^{\prime}(t_{0})+s\mathcal{R}f_{i}^{\prime\prime}(t),\quad\mathcal{R}f_{i}^{\prime}(t)-\mathcal{R}f_{i}^{\prime}(t_{0})\right)\right\rVert$
(7.36) $\displaystyle\leq 2\lVert
f_{i}^{\prime}(t)-f_{i}^{\prime}(t_{0})\rVert+\lvert s\rvert\lVert
f^{\prime\prime}(t)\rVert\,.$ (7.37)

As $f^{\prime\prime}$ is uniformly bounded, due to the mean value theorem on
one of the intervals $(t,t_{0}),(t_{0},t)$ or $(t,\lvert S_{r}\rvert+t_{0})$,
the last one using periodicity of $f_{i}$, we can conclude the final two
claims. ∎

######  Lemma 7.4.

Let $\Lambda$ be a $\mathsf{C}^{2}$-smooth domain. Let $m,K\in\mathbb{N}$ and
$L\in\mathbb{R}^{+}$. Then, asymptotically as $L\to\infty$ and uniformly in
$K$ as long as $K^{2}=o(L)$, we have

$\displaystyle\mathcal{J}_{m}(L\Lambda,L\Lambda^{\complement};K)=L\lvert\partial\Lambda\rvert\mathcal{J}_{m}\left([0,1)\times\mathbb{R}^{-},\mathbb{R}\times\mathbb{R}^{+};K\right)+\mathcal{O}(K^{2}+\ln(L)^{4})\,.$
(7.38)

###### Proof.

As mentioned, this proof is quite similar to the polygonal case, 7.2. Now, let
$C_{\Lambda},\varepsilon,(S_{i},g_{i})_{i=1}^{r}$ be given by 7.3,
$R\coloneqq\max(10K,\ln(L)^{2}),\delta\coloneqq R/L$ and let
$Q_{R}\coloneqq[0,R)\times(-R,0)$.777This square shows up a lot in this proof
and we hope this notation improves readability.

The claimed error term is just $\mathcal{O}(R^{2})$. As in the polygonal case,
we can use Theorem 3.1, 5.5 and 6.3 to get the a-priori estimates

$\displaystyle\mathcal{J}_{m}(L\Lambda,L\Lambda^{\complement};K)$
$\displaystyle\leq CL\ln(L)\,,$ (7.39) $\displaystyle
L\lvert\partial\Lambda\rvert\,\mathcal{J}_{m}\left([0,1)\times\mathbb{R}^{-},\mathbb{R}\times\mathbb{R}^{+};K\right)$
$\displaystyle\leq CL\ln(K)\,.$ (7.40)

Thus, we can assume that $\delta<\varepsilon/(12(m+1))$ and
$288C_{\Lambda}\delta(m+1)^{2}<1/2$, as otherwise the error term is larger
than both the main term and the actual result.

For $i=1,\dots,r$, we have $\delta<\varepsilon<\lvert S_{i}\rvert$ and we can
thus choose $\delta_{i}\in[\delta,2\delta)$, such that $N_{i}\coloneqq\lvert
S_{i}\rvert/\delta_{i}$ is an integer. For any $j\in\mathbb{N}$ with $0<j\leq
N_{i}$, we define

$\displaystyle E_{ij}\coloneqq
Lg_{i}\left([\delta_{i}(j-1),\delta_{i}j)\times(-\delta_{i},0)\right)\subset
L\Lambda\,.$ (7.41)

These correspond to the _edge sets_ in the polygonal case. Let $R_{i}\coloneqq
L\delta_{i}$. Then, we have

$\displaystyle\sum_{i=1}^{r}N_{i}R_{i}=L\lvert\partial\Lambda\rvert\,.$ (7.42)

We now define our _essential interior_. It is given by

$\displaystyle E_{\text{int}}\coloneqq$ $\displaystyle
L\left(\Lambda\setminus\bigcup_{i=1}^{r}D_{\delta_{i}}(S_{i})\right)\subset
L\Lambda\setminus D_{R}(L\partial\Lambda)\,,$ (7.43)

since $D_{R}(L\partial\Lambda)\subset\bigcup_{i=1}^{r}LD_{\delta_{i}}(S_{i})$
as $\delta_{i}\geq\delta=R/L$. Its contribution to $\mathcal{J}_{m}$ can be
estimated identically to the polygonal case, see (7.11)–(7.15). This yields

$\displaystyle\mathcal{J}_{m}(E_{\text{int}},L\Lambda^{\complement};K)\leq
C\,.$ (7.44)

As a simple consequence of the properties in 7.3, we have
$D_{R_{i}}(LS_{i})\cap(L\Lambda)=L(D_{\delta_{i}}(S_{i})\cap\Lambda)=\bigcup_{i=1}^{r}\bigcup_{j=1}^{N_{i}}E_{ij}$
and thus

$\displaystyle
L\Lambda=E_{\text{int}}\cup\bigcup_{i=1}^{r}\bigcup_{j=1}^{N_{i}}E_{ij}\,,$
(7.45)

being a disjoint union888In particular, we do not get any _corner sets_ , as
$\mathsf{C}^{2}$-smooth domains do not have corners.. Thus, as
$\mathcal{J}_{m}({\makebox[4.30554pt]{{$\cdot$}}},L\Lambda^{\complement};K)$
is additive, we have

$\displaystyle\mathcal{J}_{m}(L\Lambda,L\Lambda^{\complement};K)=\mathcal{J}_{m}(E_{\text{int}},L\Lambda^{\complement};K)+\sum_{i=1}^{r}\sum_{j=1}^{N_{i}}\mathcal{J}_{m}(E_{ij},L\Lambda^{\complement};K)\,.$
(7.46)

We have already seen that $E_{\text{int}}$ can be absorbed into the error
term. We will now show that

$\displaystyle\mathcal{J}_{m}(E_{ij},L\Lambda^{\complement};K)=R_{i}\mathcal{J}_{m}([0,1)\times\mathbb{R}^{-},\mathbb{R}\times\mathbb{R}^{+};K)+\mathcal{O}(\delta
R^{2})\,,$ (7.47)

where the upper bound for the error is independent of $j$. In combination with
(7.42), this leads to the claim. Unlike in the polygonal case, the boundary
curve is not straight. However, due to the assumption $K^{2}\ll L$, we can
approximate it sufficiently well by a smooth curve. Due to additivity and 6.3,
we know that

$\displaystyle\mathcal{J}_{m}(Q_{R_{i}},\mathbb{R}\times\mathbb{R}^{+};K)$
$\displaystyle=\mathcal{J}_{m}([0,R_{i})\times\mathbb{R}^{-},\mathbb{R}\times\mathbb{R}^{+};K)-\mathcal{J}_{m}([0,R_{i})\times(-\infty,R_{i}),\mathbb{R}\times\mathbb{R}^{+};K)$
(7.48)
$\displaystyle=R_{i}\mathcal{J}_{m}([0,1)\times\mathbb{R}^{-},\mathbb{R}\times\mathbb{R}^{+};K)+\mathcal{O}(1)\,.$
(7.49)

We now fix $i,j$ and take care that the error term bounds only depend on
$i,R,L$ and the constants $\varepsilon,C_{\Lambda}$ in 7.3.

By choosing an appropriate affine-linear unitary transformation
$\mathcal{A}_{ij}$, we may assume $g_{i}(\delta_{i}(j-1),0)=0$ and
$Dg_{i}(\delta_{i}(j-1),0)=\text{Id}$ (the $2\times 2$ identity matrix)
without changing the constants $\varepsilon,C$ in 7.3. As $g_{i}$ has a
Lipschitz constant of at most $2$, we know
$E_{ij}=Lg_{i}((\delta_{i}(j-1),\delta_{i}j),(-R_{i},0))\subset
LD_{3\delta_{i}}(0)=D_{3R_{i}}(0)$. We recall that
$3R_{i}(m+1)<6R(m+1)<\varepsilon L/2$. Let $(E,E^{\prime})$ be one of the set
of pairs $(E_{ij},L\Lambda^{\complement})$ and
$\left(Q_{R_{i}},\mathbb{R}\times\mathbb{R}^{+}\right)$. Thus $E\subset
D_{3R_{i}}(0)$. We utilize 6.2 to obtain

$\displaystyle\mathcal{J}_{m}(E,E^{\prime};K)=\mathcal{J}_{m}(E,E^{\prime}\cap
D_{3R_{i}(m+1)}(0);K)+\mathcal{O}(m^{5}\exp(-\beta R_{i}))\,.$ (7.50)

Using the Lipschitz-type property in 5.5, we see that

$\displaystyle\left\lvert\mathcal{J}_{m}(E_{ij},L\Lambda^{\complement};K)-\mathcal{J}_{m}\left(Q_{R_{i}},\mathbb{R}\times\mathbb{R}^{+};K\right)\right\rvert$
(7.51) $\displaystyle\leq\mathcal{O}(m^{5}\exp(-\beta
R_{i}))+\frac{1}{2\pi}\left\lvert E_{ij}\Delta
Q_{R_{i}}\right\rvert+\frac{m}{2\pi}\left\lvert\left(L\Lambda^{\complement}\cap
D_{3R_{i}(m+1)}(0)\right)\Delta\left(\mathbb{R}\times\mathbb{R}^{+}\cap
D_{3R_{i}(m+1)}(0)\right)\right\rvert.$ (7.52)

Thus, we have reduced our claim to a purely geometric estimate999This estimate
is probably rather rough and could be the reason we require $K^{2}\ll L$. It
only checks how close the boundary curve of $L\Lambda$ and
$L\Lambda^{\complement}$ approach the same line locally, without fully using
the fact that it is the same boundary curve. We would not be surprised if the
error terms in this step cancel out to some order, but we have not yet found a
better way to estimate them..

We define $\hat{g}_{ij}\colon D_{6R_{i}(m+1)}(0)\to\mathbb{R}^{2}$ by

$\displaystyle\hat{g}_{ij}(x_{1},x_{2})\coloneqq
Lg_{i}(x_{1}/L+\delta(j-1),x_{2}/L)\,.$ (7.53)

As $D_{6R_{i}(m+1)}(0)\subset LD_{\varepsilon}(\Lambda)$, we know that
$\hat{g}_{ij}^{-1}$ is well-defined on $D_{6R_{i}(m+1)}(0)$. As $g_{i}$ and
$g_{i}^{-1}$ have Lipschitz constants of at most $2$, so do the rescaled
functions $\hat{g}_{ij},\hat{g}_{ij}^{-1}$. As $g_{ij}(0)=0$, we conclude that
$D_{3R_{i}(m+1)}(0)\subset\hat{g}_{ij}(D_{6R_{i}(m+1)}(0))$ and
$\hat{g}_{ij}(D_{3R_{i}(m+1)}(0))\subset D_{6R_{i}(m+1)}(0)$. This allows us
to write

$\displaystyle E_{ij}$ $\displaystyle=\hat{g}_{ij}\left(Q_{R_{i}}\right),$
(7.54) $\displaystyle L\Lambda^{\complement}\cap D_{3R_{i}(m+1)}(0)$
$\displaystyle=\hat{g}_{ij}\left(\mathbb{R}\times\mathbb{R}^{+}\cap
D_{6R_{i}(m+1)}(0)\right)\cap D_{3R_{i}(m+1)}(0)\,.$ (7.55)

As $Dg_{i}(\delta(j-1),0)=\text{Id}$, we have for any $x\in
D_{6R_{i}(m+1)}(0)$,

$\displaystyle\lVert D\hat{g}_{ij}(x_{1},x_{2})-\text{Id}\rVert$
$\displaystyle=\lVert
Dg_{i}(x_{1}/L+\delta(j-1),x_{2}/L)-Dg_{i}(\delta(j-1),0)\rVert$ (7.56)
$\displaystyle\leq C_{\Lambda}(\lvert x_{1}/L\rvert+\lvert x_{2}/L\rvert)\leq
4C_{\Lambda}(3R_{i}(m+1))/L\leq 24C_{\Lambda}\delta(m+1)\,.$ (7.57)

Let $h_{ij}(x)\coloneqq\hat{g}_{ij}(x)-x$. Then, by the mean value theorem we
can conclude for any $x\in D_{6R_{i}(m+1)}(0)$,

$\displaystyle\lVert\hat{g}_{ij}(x)-x\rVert$ $\displaystyle=\lVert
h_{ij}(x)\rVert=\lVert
h_{ij}(x)-h_{ij}(0)\rVert\leq(24C_{\Lambda}\delta(m+1))(6R_{i}(m+1))$ (7.58)
$\displaystyle\leq 288C_{\Lambda}\delta^{2}(m+1)^{2}L<R/2\,.$ (7.59)

The last inequality is based on the assumption
$288C_{\Lambda}\delta(m+1)^{2}<1/2$, which is stated earlier in the proof.

Let $r\coloneqq 288C_{\Lambda}\delta^{2}(m+1)^{2}L$ and let
$\mathcal{E}\subset\mathbb{R}^{2}$. We want to estimate
$\left\lvert\left(\mathcal{E}\Delta\hat{g}_{ij}(\mathcal{E})\right)\cap
D_{3R_{i}(m+1)}(0)\right\rvert$ for $\mathcal{E}=Q_{R_{i}}$ and
$\mathcal{E}=\mathbb{R}\times\mathbb{R}^{+}$.

Let $y\in\mathcal{E}\setminus\hat{g}_{ij}(\mathcal{E})$ with $y\in
D_{3R_{i}(m+1)}(0)$. Then, there is an $x\in
D_{6R_{i}(m+1)}(0)\cap\mathcal{E}^{\complement}$ with $\hat{g}_{ij}(x)=y$. As
$\lVert x-y\rVert=\lVert x-\hat{g}_{ij}(x)\rVert\leq r$, we see that $y\in
D_{r}(\mathcal{E}^{\complement})$. Thus, as $y\in\mathcal{E}$, we see $y\in
D_{r}(\partial\mathcal{E})$. On the other hand, if
$y\in\hat{g}_{ij}(\mathcal{E})\setminus\mathcal{E}$ with $y\in
D_{3R_{i}(m+1)}(0)$, there is an $x\in D_{6R_{i}(m+1)}(0)\cap\mathcal{E}$ with
$\hat{g}_{ij}(x)=y$ and thus $\lVert x-y\rVert\leq r$, which implies $y\in
D_{r}(\partial\mathcal{E})$, again. Thus, we have shown that

$\displaystyle\left(\mathcal{E}\Delta\hat{g}_{ij}(\mathcal{E})\right)\cap
D_{3R_{i}(m+1)}(0)\subset D_{r}(\partial\mathcal{E})\cap
D_{3R_{i}(m+1)}(0)\,,$ (7.60)

which leads to

$\displaystyle\left\lvert\left(\mathcal{E}\Delta\hat{g}_{ij}(\mathcal{E})\right)\cap
D_{3R_{i}(m+1)}(0)\right\rvert\leq\left\lvert D_{r}(\partial\mathcal{E})\cap
D_{3R_{i}(m+1)}(0)\right\rvert\,.$ (7.61)

Using this for the half space, we easily get

$\displaystyle\left\lvert\left(L\Lambda^{\complement}\cap
D_{3R_{i}(m+1)}(0)\right)\Delta\left(\mathbb{R}\times\mathbb{R}^{+}\cap
D_{3R_{i}(m+1)}(0)\right)\right\rvert$ (7.62) $\displaystyle\leq\left\lvert
D_{r}(\mathbb{R}\times\\{0\\})\cap D_{3R_{i}(m+1)}(0)\right\rvert$ (7.63)
$\displaystyle\leq 4r(3R_{i}(m+1))\leq 4\cdot
288C_{\Lambda}\delta^{2}(m+1)^{2}L\cdot 6R(m+1)<Cm^{3}\delta R^{2}\,.$ (7.64)

In the last step we used that $\delta=R/L$.

For the square, we use that $r<R/2$, which means that $r$ is less than half of
the side length of the square. This allows us to estimate

$\displaystyle\left\lvert E_{ij}\Delta Q_{R_{i}}\right\rvert\leq\left\lvert
D_{r}\left(\partial Q_{R_{i}}\right)\right\rvert=8Rr-(4-\pi)r^{2}<8Rr=8R\cdot
288C_{\Lambda}\delta^{2}(m+1)^{2}L<Cm^{2}\delta R^{2}\,.$ (7.65)

Thus, we have completed the proof of the asymptotic expansion

$\displaystyle\mathcal{J}_{m}(E_{ij},L\Lambda^{\complement};K)=R_{i}\mathcal{J}_{m}([0,1)\times\mathbb{R}^{-},\mathbb{R}\times\mathbb{R}^{+};K)+\mathcal{O}(\delta
R^{2})\,,$ (7.66)

where the upper bound for the error term is independent of $i,j$. Thus, we can
sum this expression over $i,j$ and recalling (7.46), (7.44) and (7.42) to
observe

$\displaystyle\mathcal{J}_{m}(L\Lambda,L\Lambda^{\complement};K)$
$\displaystyle=\sum_{i=1}^{r}\sum_{j=1}^{N_{i}}\mathcal{J}_{m}(E_{ij},L\Lambda^{\complement};K)+\mathcal{J}_{m}(E_{\text{int}},L\Lambda^{\complement};K)$
(7.67)
$\displaystyle=\left(\sum_{i=1}^{r}\sum_{j=1}^{N_{i}}R_{i}\mathcal{J}_{m}([0,1)\times\mathbb{R}^{-},\mathbb{R}\times\mathbb{R}^{+};K)+\mathcal{O}(\delta
R^{2})\right)+\mathcal{O}(1)$ (7.68)
$\displaystyle=\left(\sum_{i=1}^{r}N_{i}R_{i}\right)\mathcal{J}_{m}([0,1)\times\mathbb{R}^{-},\mathbb{R}\times\mathbb{R}^{+};K)+\mathcal{O}(R^{2}\sum_{i=1}^{r}N_{i}\delta_{i})+\mathcal{O}(1)$
(7.69)
$\displaystyle=L\lvert\partial\Lambda\rvert\mathcal{J}_{m}([0,1)\times\mathbb{R}^{-},\mathbb{R}\times\mathbb{R}^{+};K)+\mathcal{O}(R^{2}L\lvert\partial\Lambda\rvert/L)+\mathcal{O}(1)$
(7.70)
$\displaystyle=L\lvert\partial\Lambda\rvert\mathcal{J}_{m}([0,1)\times\mathbb{R}^{-},\mathbb{R}\times\mathbb{R}^{+};K)+\mathcal{O}(R^{2})\,.$
(7.71)

In (7.69), we used $\delta<\delta_{i}$ in the error term to apply (7.42). As
$R^{2}\leq C(K^{2}+\ln(L)^{4})$, this finishes the proof of this lemma. ∎

Altogether, 7.2 and 7.4 finally prove Theorem 5.3.

##  8\. $K\ll L$: Proof of Theorem 5.4

In this section we will prove Theorem 5.4. Let us recall that it states

See 5.4

In accordance with [8, (2.11)], we write

$\displaystyle\mathcal{K}_{K}\coloneqq\sum_{\ell=0}^{K-1}|\psi_{\ell}\rangle\langle\psi_{\ell}|\,,$
(8.1)

where $\psi_{\ell}$ are the Hermite functions (as in [8, (2.9)]) given by

$\displaystyle\psi_{\ell}(s)\coloneqq\left(\sqrt{\pi}2^{\ell}\ell!\right)^{-1/2}H_{\ell}(s)\exp(-s^{2}/2)\,,\quad
s\in\mathbb{R}\,.$ (8.2)

For any polynomial $f$ with $f(0)=f(1)=0$, we define

$\displaystyle\mathsf{M}_{<K}(f)\coloneqq\frac{1}{2\pi}\int_{\mathbb{R}}\mathrm{d}p\,\operatorname{tr}f(1_{>p}\mathcal{K}_{K}1_{>p})\,,$
(8.3)

where $1_{>p}\coloneqq 1_{(p,\infty)}$. This $\mathsf{M}_{<K}(f)$ agrees with
$\mathsf{M}_{\leq K-1}(f)$ defined in [8, (2.12)].

While there are asymptotic formulas for the Hermite polynomials directly, we
did not find the exact statement we needed. However, as we already go into
detail on the asymptotics for the Laguerre polynomials, it is convenient to
reduce the Hermite polynomials to Laguerre polynomials and only go deep into
the asymptotic expansion of one of these polynomials.

######  Lemma 8.1.

The asymptotic scaling coefficient for fixed $K$ agrees with the one shown in
[8]. That is,

$\displaystyle\mathcal{J}_{m}([0,1)\times\mathbb{R}^{-},\mathbb{R}\times\mathbb{R}^{+};K)=\frac{1}{\sqrt{K}}\mathsf{M}_{<K}(t\mapsto
t(1-t)^{m})\,.$ (8.4)

######  Remark.

We will present a full proof of this statement, which is based on a simple
comparison of coefficients and a sketch of an alternative proof.

###### Proof.

Let $f_{m}(t)\coloneqq t(1-t)^{m}$ for any $t\in[0,1]$. Consider the domain
$\Lambda_{0}=D_{1}(0)$, the unit disk. This is a $\mathsf{C}^{\infty}$-smooth
domain and thus, we can apply both 7.4 as well as [8, Theorem 2] to it. We now
fix $K\in\mathbb{N}$ and consider

$\displaystyle\lim_{L\to\infty}\frac{1}{2\pi
L}\mathcal{J}_{m}(L\Lambda,L\Lambda^{\complement};K)\,.$ (8.5)

On the one hand, according to 7.4, we observe

$\displaystyle\lim_{L\to\infty}\frac{1}{2\pi
L}\mathcal{J}_{m}(L\Lambda,L\Lambda^{\complement};K)=\mathcal{J}_{m}([0,1)\times\mathbb{R}^{-},\mathbb{R}\times\mathbb{R}^{+};K)\,.$
(8.6)

On the other hand, according to [8, Theorem 2], we have

$\displaystyle\lim_{L\to\infty}\frac{1}{2\pi
L}\mathcal{J}_{m}(L\Lambda,L\Lambda^{\complement};K)=\lim_{L\to\infty}\frac{1}{2\pi
L}\operatorname{tr}f_{m}(1_{L\Lambda}P_{K}1_{L\Lambda})=\frac{1}{\sqrt{K}}\,\mathsf{M}_{<K}(f_{m})\,,$
(8.7)

where we used that $K=1/B$. This completes the coefficient comparison proof. ∎

###### Sketch of an alternative proof.

One can also directly transform the two integral representations into one
another. Most of this work has been executed in the proof of Lemmata 5 & 6 in
[8]. However, they use Roocaforte’s approximation and get an error term, that
they bound using the exponential decay of $P_{K}$ for fixed $K$, while our
error term needs to be bounded in $K$.

Let $E\coloneqq[0,1)\times\mathbb{R}^{-}$ and
$E^{\prime}\coloneqq\mathbb{R}\times\mathbb{R}^{+}$. According to Mercer’s
theorem, we get

$\displaystyle\mathcal{J}_{m}$
$\displaystyle\left([0,1)\times\mathbb{R}^{-},\mathbb{R}\times\mathbb{R}^{+};K\right)\,=\,\operatorname{tr}1_{E}(P_{K}1_{E^{\prime}})^{m}P_{K}1_{E}$
(8.8)
$\displaystyle=\int_{E}\mathrm{d}x\int_{E^{\prime}}\mathrm{d}x_{1}\int_{E^{\prime}}\mathrm{d}x_{2}\cdots\int_{E^{\prime}}\mathrm{d}x_{m}\,P_{K}(x,x_{1})P_{K}(x_{1},x_{2})\cdots
P_{K}(x_{m},x)\,.$ (8.9)

This integral now looks very similar to the one studied in [8, Proof of Lemma
5]. The key advantage is that due to our choice of $E,E^{\prime}$, the
Roccaforte approximation is exact, which means basically their (3.7) does not
carry an error term. However, as we are studying a different base polynomial
($f_{m}$ instead of $t\mapsto t^{m}$), it looks slightly different. The main
idea is that the intersection of the offset sets depends entirely linearly on
the maximum offset in $e_{2}$ direction.

They have performed all the remaining integral transformations to carry out
the remaining parts of the proof. ∎

The symmetry of the coefficient $\mathsf{M}_{<K}(f)$ in the next lemma is of
independent interest. It is inspired by the same and obvious symmetry relation
of the functional $\mathsf{I}$ of (1.5). But is is also useful from a
technical point of view as it simplifies the construction of suitable
intervals as in B.4.

######  Lemma 8.2.

Let $K\in\mathbb{N}$ and let $f$ be a function with $f(0)=f(1)=0$ and $\lvert
f(t)\rvert\leq Ct^{\alpha}(1-t)^{\alpha}$ for any $t\in[0,1]$ and some
$\alpha>0$. We define the function $g$ by $g(t)\coloneqq f(1-t)$ for any
$t\in[0,1]$. Then, the following equalities hold,

$\displaystyle\mathsf{M}_{<K}(f)=\mathsf{M}_{<K}(g)=\frac{1}{2\pi}\int_{0}^{\infty}\mathrm{d}p\,\big{[}\operatorname{tr}f(1_{>p}\mathcal{K}_{K}1_{>p})+\operatorname{tr}g(1_{>p}\mathcal{K}_{K}1_{>p})\big{]}\,.$
(8.10)

Due to the symmetry of $M_{\leq K-1}(f)$ in $f$, meaning
$\mathsf{M}_{<K}(f)=\mathsf{M}_{<K}(t\mapsto f(1-t))$, for any polynomial $f$
with $f(0)=f(1)=0$, it suffices to consider $\mathsf{M}_{<K}(f)$ for symmetric
polynomials (meaning $f(t)=f(1-t)$). Thus, we can restrict it to the case
$f(t)=(t(1-t))^{m}$ for some $m\in\mathbb{N}$. Using the same idea as in
(5.2), we can establish

$\displaystyle
2\pi\,\mathsf{M}_{<K}(t\mapsto[t(1-t)]^{m})=\int_{\mathbb{R}}\mathrm{d}p\,\operatorname{tr}\left(1_{<p}\mathcal{K}_{K}1_{>p}\mathcal{K}_{K}1_{<p}\right)^{m}\,.$
(8.11)

The advantage of the expression is that each occurrence of $\mathcal{K}_{K}$
is flanked by $1_{<\xi}$ and $1_{\geq\xi}$, which should help with estimating
the Hilbert–Schmidt norms of error terms in the asymptotic expansion of the
kernel $\mathcal{K}_{K}$ for large $K$.

###### Proof of 8.2.

Let us first point out that $g$ also satisfies $\lvert g(t)\rvert\leq
Ct^{\alpha}(1-t)^{\alpha}$ for any $t\in[0,1]$. Thus, [8, Lemma 3] tells us
that $\mathsf{M}_{<K}(f)$ and $\mathsf{M}_{<K}(g)$ are well-defined.

We recall that by definition

$\displaystyle
2\pi\,\mathsf{M}_{<K}(f)=\int_{\mathbb{R}}\mathrm{d}p\,\operatorname{tr}f(1_{>p}\mathcal{K}_{K}1_{>p})\,.$
(8.12)

For any $\ell\in\\{0,1,\dots,K-1\\}$ and any $x,y\in\mathbb{R}$, we have
$\psi_{\ell}(-x)=(-1)^{\ell}\psi_{\ell}(-x)$ and thus

$\displaystyle\mathcal{K}_{K}(-x,-y)=\sum_{\ell=0}^{K-1}(-1)^{2\ell}\psi_{\ell}(x)\psi_{\ell}(y)=\mathcal{K}_{K}(x,y)\,.$
(8.13)

Thus, $\mathcal{K}_{K}$ commutes with the reflection operator $\mathcal{R}$ on