lemmatheoremlemma definitiontheoremdefinition corollarytheoremcorollary
propositiontheoremproposition facttheoremfact

# On the stability of the area law for the entanglement entropy of the Landau
Hamiltonian

Paul Pfeiffer  Fakultät für Mathematik und Informatik, FernUniversität in
Hagen, Universitätsstraße 1, 58097, Hagen, Germany  paul.pfeiffer@fernuni-
hagen.de

(Date: 10th May 2020)

###### Abstract.

We consider the two-dimensional ideal Fermi gas subject to a magnetic field
which is perpendicular to the Euclidean plane $\mathbb{R}^{2}$ and whose
strength $B(x)$ at $x\in\mathbb{R}^{2}$ converges to some $B_{0}>0$ as
$\|x\|\to\infty$. Furthermore, we allow for an electric potential
$V_{\varepsilon}$ which vanishes at infinity. They define the single-particle
Landau Hamiltonian of our Fermi gas (up to gauge fixing). Starting from the
ground state of this Fermi gas with chemical potential $\mu\geq B_{0}$ we
study the asymptotic growth of its bipartite entanglement entropy associated
to $L\Lambda$ as $L\to\infty$ for some fixed bounded region
$\Lambda\subset\mathbb{R}^{2}$. We show that its leading order in $L$ does not
depend on the perturbations $B_{\varepsilon}:=B_{0}-B$ and $V_{\varepsilon}$
if they satisfy some mild decay assumptions. Our result holds for all
$\alpha$-Rényi entropies $\alpha>1/3$; for $\alpha\leq 1/3$, we have to assume
in addition some differentiability of the perturbations $B_{\varepsilon}$ and
$V_{\varepsilon}$. The case of a constant magnetic field $B_{\varepsilon}=0$
and with $V_{\varepsilon}=0$ was treated recently for general $\mu$ by
Leschke, Sobolev and Spitzer. Our result thus proves the stability of that
area law under the same regularity assumptions on the boundary
$\partial\Lambda$.

###### Contents

  1. 1 Introduction
  2. 2 Notations and preliminaries
  3. 3 Setting and main result
  4. 4 The Ansatz for the proof of Theorem 3.1
  5. 5 Kernel estimates
  6. 6 Proof of Theorem 4.1 and Theorem 4.3
  7. A
  8. B Proof of Section 3

##  1\. Introduction

Bipartite entanglement entropy is an important quantity that measures
correlations of particles inside a given region with the particles outside
that region. These non-trivial correlations are solely due to the Fermi–Dirac
statistics of the particles involved. In recent years there has been
considerable interest and progress in quantifying these correlations.
Mathematicians and physicists alike realized fascinating connections between
the large scale asymptotics of entanglement entropy and certain semi-classical
asymptotic formulas of traces of certain operators, mostly Toeplitz operators
in the discrete case and Wiener–Hopf operators in the continuous case.

In the discrete setting, Jin and Korepin related the Fisher–Hartwig conjecture
of Toeplitz matrices to the scaling of the entanglement entropy in the
$XY$-chain in a transverse magnetic field in [8]. More relevant to our
continuous setting here is the discovery of Gioev and Klich [5] that a
conjecture by Harold Widom (proved by Alexander V. Sobolev [22]) gives the
precise leading asymptotic growth of the bipartite entanglement entropy in
ground states of the free Fermi gas. It displays a logarithmically enhanced
area law of the order $L^{d-1}\ln(L)$, where $L$ is a scaling parameter, see
below. In [9], this was finally proved by Leschke, Sobolev and Spitzer. In
[14], [16], Müller and Schulte proved that this law is stable under a
perturbation by a compactly supported potential. The line of proof in their
first paper is also important for our model here.

A ground state of a non-interacting fermions on $\mathbb{R}^{2}$ with single-
particle Hamiltonian $H$ as in our model is given by the (Fermi) spectral
projection $1_{\leq\mu}(H)$, where $\mu\in\mathbb{R}$. The function
$1_{\leq\mu}$ is the indicator function of the set
$(-\infty,\mu]\subset\mathbb{R}$ and the number $\mu$ is called the Fermi
energy. Let $\alpha>0$ and let $h_{\alpha}$ be the Rényi entropy function, see
(3.3). For a given bounded region $\Lambda\subset\mathbb{R}^{2}$ we denote by
$1_{\Lambda}$ the (multiplication operator associated to the) indicator
function on $\Lambda$. Then we define the local entropy (or entanglement
entropy) $S_{\alpha}(\Lambda)$ to be the (usual Hilbert space) trace of
$h_{\alpha}$ applied to the spatially to $\Lambda$ reduced Fermi projection,
that is,

$S_{\alpha}(\Lambda):=\operatorname{tr}h_{\alpha}(1_{\Lambda}1_{\leq\mu}(H)1_{\Lambda})\,.$
(1.1)

At positive temperature a definition of entanglement entropy or mutual
information needs to be amended, see [10].

For a fixed region $\Lambda$, it is generally hard or impossible to calculate
the entropy. However, if we introduce a scaling parameter $L>0$ and consider
the leading order asymptotic expansion of (1.1) with $\Lambda$ replaced by
$L\Lambda$ for $L\to\infty$, there are interesting results. They all assume
some kind of regularity of the boundary $\partial\Lambda$, assume the
Hamiltonian $H$ to be of a certain form, and may restrict to the case
$\alpha=1$. For $H=-\nabla^{2}+V$, with some assumptions on $V$, there are
results presented in [4, 9, 10, 16, 15, 18, 19].

In this paper, we consider the Hamiltonian
$H=(-i\nabla-A)^{2}+V_{\varepsilon}$, which is a slight perturbation of the
Landau Hamiltonian $H_{0}$ for a constant magnetic field and no electric
field, see (3.2) and (3.10). Entanglement entropy of the ground state of the
latter Landau Hamiltonian (for the ground state with chemical potential
$\mu=B_{0}$) has been studied in [12, 21, 20] with some additional assumptions
on the region $\Lambda$. The case of $\mu=B_{0}$ has been solved by Charles
and Estienne in [3], and then the case of an arbitrary $\mu\geq B_{0}$ by
Leschke, Sobolev and Spitzer in [11], both under some regularity assumptions
on the boundary $\partial\Lambda$. Our main result is Section 3. It shows that
the leading order asymptotic growth of the entanglement entropy for arbitrary
$\alpha>0$ does not change, if we add such a slight perturbation in both the
magnetic field and the electric potential, assuming some differentiability of
these perturbations in the case $\alpha\leq\frac{1}{3}$, depending on
$\alpha$. Hence, we will not need to recalculate the value of the leading
term, as we only estimate that this perturbation leads to an error term of
smaller order in the scaling parameter $L$.

Our proof is based on a statement by Aleksandrov and Peller in [1], which is
Section 3 in this paper. With the help of this and approximations of the Rényi
entropy functions $h_{\alpha}$ (see (3.3)), we can reduce our result to some
$p$-Schatten (quasi-)norm estimates, as we prove in Section 3.

Proving these $p$-Schatten (quasi-)norm estimates relies on the fact that some
Sobolev embeddings on bounded subset of $\mathbb{R}^{d}$ are in some
$p$-Schatten classes, which we specify and prove in Appendix A. It is based on
a result by Gramsch in [6]. This allows us to estimate the $p$-Schatten
(quasi-)norms of operators with sufficiently differentiable kernels. To get a
representation of the kernel of the spectral projection of the perturbed
Hamiltonian, we use the contour integral representation and the resolvent
expansion. This has recently been done for perturbations of the free case
($H=-\nabla^{2}+V$) by Müller and Schulte in [16], which inspired me to try
this approach. In our case ($B_{0}>0$), we use an expanded resolvent
expansion. The discrete spectrum allows us to explicitly resolve the contour
integral for most terms. The general idea is explained in Section 4, while the
required kernel estimates are proven in the remaining sections.

The magnetic case (with an asymptotically constant magnetic field) appears
simpler and more stable than the free case with the (negative) Laplacian as
its single-particle Hamiltonian. From a technical point of view this is due to
the gaps in the purely essential spectrum and the exponential decay of
eigenfunctions of the Landau Hamiltonian. This is also the reason for an area
law growth (without any logarithmic enhancement as in the free case), see also
[17].

## Acknowledgment

I would like to thank Wolfgang Spitzer for introducing me to this topic, proof
reading multiple previous versions, and generally providing advice.

##  2\. Notations and preliminaries

Let $\mathbb{N}=\\{0,1,2,\dots\\}$ be the natural numbers and $\mathbb{Z}^{+}$
be the positive integers.

Let $n,d$ be positive integers and $k$ be a natural number. For
$x\in\mathbb{R}^{n}$ or $x\in\mathbb{C}^{n}$, let $\lVert x\rVert$ be its
$2$-norm. The space of $p$-integrable (respectively essentially bounded if
$p=\infty$), complex valued functions on $\mathbb{R}^{n}$ is called
$\mathrm{L}^{p}(\mathbb{R}^{n})$. The Sobolev space $W^{k,p}(\mathbb{R}^{n})$
is the subspace of $\mathrm{L}^{p}(\mathbb{R}^{n})$, such that their first $k$
distributional derivatives in any combination of directions are in
$\mathrm{L}^{p}(\mathbb{R}^{n})$. We define
$C^{k}_{b}(\mathbb{R}^{n},\mathbb{C}^{d})$ as the subspace of
$C^{k}(\mathbb{R}^{n},\mathbb{C}^{d})$, such that all derivatives of order
$0\leq j\leq n$ are bounded.

For any non-empty set $\Lambda\subset\mathbb{R}^{n}$ and any point
$x\in\mathbb{R}^{n}$, we define the distance as

$\displaystyle\operatorname{dist}(x,\Lambda)\coloneqq\inf_{y\in\Lambda}\lVert
x-y\rVert,$ (2.1)

and for any $r>0$ we define the $r$-neighbourhood of $\Lambda$ as

$\displaystyle
D_{r}(\Lambda)\coloneqq\\{y\in\mathbb{R}^{n}|\operatorname{dist}(y,\Lambda)<r\\}.$
(2.2)

Furthermore, $1_{\Lambda}\colon\mathbb{R}^{n}\to\\{0,1\\}\subset\mathbb{R}$ is
the indicator function of $\Lambda$,
$\Lambda^{\complement}\coloneqq\mathbb{R}^{n}\setminus\Lambda$ is the
complement of $\Lambda$, and if $\Lambda$ is measurable, let
$\lvert\Lambda\rvert$ be its $n$-dimensional Lebesgue measure. If $\Lambda$
has Lipschitz-boundary $\partial\Lambda$, let $\lvert\partial\Lambda\rvert$ be
the $(n-1)$-dimensional Hausdorff measure of $\partial\Lambda$.

For any $x\in\mathbb{R}^{n}$, we define the disk $D_{r}(x)=D_{r}(\\{x\\})$.
For any $x\in\mathbb{C}^{n},j\in\mathbb{N}$, we inductively define $x^{\otimes
j}\in\left(\mathbb{C}^{n}\right)^{\otimes j}\cong\mathbb{C}^{n^{j}}$ by
setting $x^{\otimes 0}\coloneqq
1\in\mathbb{C}\eqqcolon\left(\mathbb{C}^{n}\right)^{\otimes 0}$ and
$x^{\otimes(j+1)}\coloneqq x^{\otimes j}\otimes
x\in\left(\mathbb{C}^{n}\right)^{\otimes
j}\otimes\mathbb{C}^{n}\eqqcolon\left(\mathbb{C}^{n}\right)^{\otimes(j+1)}\cong\mathbb{C}^{n^{j+1}}$.
Every appearance of ${\makebox[4.30554pt]{{$\cdot$}}}^{\otimes j}$ refers to
this tensor product.

By $J$ we denote the matrix

$\displaystyle J\coloneqq\begin{pmatrix}0&1\\\ -1&0\end{pmatrix}.$ (2.3)

For a complex number $\zeta$, let $\Re\zeta$ be its real part.

For a multiplication operator with a function
$G\colon\mathbb{R}^{2}\to\mathbb{C}^{n}$, we use a slight abuse of notation
and call it $G$ as well. This is relevant to decide, whether we are applying
an operator to the underlying function or taking the composition of a
multiplication operator and any other operator. Whenever there are both
multiplication operators and other operators present in an expression, we
regard $G$ as the multiplication operator, unless we write $G({\cdot})$.

$C$ will always refer to a generic constant, that may depend on some, but
never on all variables. $F$ will be used similarly, but the dependency on one
complex variable will be important, which is why we write $F$ as a function of
that variable. Both may change from line to line.

For any compact operator $S$ and any $p\in\mathbb{R}^{+}$, we define the
$p$-Schatten von Neumann (quasi-)norm by the expression

$\displaystyle\lVert
S\rVert_{p}^{p}\coloneqq\sum_{n\in\mathbb{Z}^{+}}s_{n}(S)^{p},$ (2.4)

where $(s_{n}(S))_{n\in\mathbb{Z}^{+}}$ is the decreasing sequence of singular
values of $S$ counted with multiplicity. The operator norm of $S$ is written
as $\lVert S\rVert_{\infty}$. We say an operator is in the $p$-Schatten class,
if its $p$-Schatten norm is finite. For any pair of Hilbert spaces
$\mathcal{H}_{1},\mathcal{H}_{2}$, let
$S_{p}(\mathcal{H}_{1},\mathcal{H}_{2})$ be the (quasi-)normed space of all
$p$-Schatten class operators from $\mathcal{H}_{1}$ to $\mathcal{H}_{2}$.

We recall some properties of the $p$-Schatten von Neumann (quasi-) norms. In
the following, we will refer to them as $p$-Schatten norms.

######  Proposition .

Let $0<p\leq q\leq\infty$ and let $S,T$ be operators on a Hilbert space. The
$p$-Schatten norm satisfies the properties

Monotonicity I:

$\lVert S\rVert_{p}\geq\lVert S\rVert_{q}$,

Monotonicity II:

If $S\geq T\geq 0$, then $\lVert S\rVert_{p}\geq\lVert T\rVert_{p}$,

Triangle inequality:

If $p\geq 1$, then $\lVert S+T\rVert_{p}\leq\lVert S\rVert_{p}+\lVert
T\rVert_{p}$,

$p$-triangle inequality:

If $p\leq 1$, then $\lVert S+T\rVert_{p}^{p}\leq\lVert S\rVert_{p}^{p}+\lVert
T\rVert_{p}^{p}$,

Powers:

If $S\geq 0$, then $\lVert S^{p}\rVert_{q}^{q}=\lVert S^{q}\rVert_{p}^{p}$,

Square:

$\lVert S\rVert_{p}^{2}=\lVert S^{*}S\rVert_{p/2}$, where $S^{*}$ denotes the
adjoint of $S$,

Adjoint:

$\lVert S^{*}\rVert_{p}=\lVert S\rVert_{p}$.

Hölder I:

Let $\frac{1}{r}=\frac{1}{p}+\frac{1}{q}$. Then $\lVert ST\rVert_{r}\leq\lVert
S\rVert_{p}\lVert T\rVert_{q}$.

Hölder II:

Let $\frac{1}{r}=\frac{\alpha}{p}+\frac{1-\alpha}{q}$ with $0<\alpha<1$. Then
$\lVert S\rVert_{r}\leq\lVert S\rVert_{p}^{\alpha}\lVert
S\rVert_{q}^{1-\alpha}$.

Hilbert–Schmidt kernel:

If
$T\colon\mathrm{L}^{2}(\mathbb{R}^{d_{1}})\to\mathrm{L}^{2}(\mathbb{R}^{d_{2}})$
has an integral kernel $t$, which is square integrable, then $\lVert
T\rVert_{2}=\lVert t\rVert_{\mathrm{L}^{2}(\mathbb{R}^{d_{1}+d_{2}})}$.

Orthogonality:

If $ST^{*}=0$ or $S^{*}T=0$, then $\lVert S\rVert_{p}\leq\lVert
S+T\rVert_{p}$.

Most of these have for example been proven by McCarthy in [13]. We will now
briefly prove the remaining ones.

###### Proof.

“Monotonicity II” follows, as the inequality holds for the ordered sequence of
singular values. “Hölder II” is an application of “Hölder I” with the
operators $\lvert S\rvert^{\alpha}$ and $\lvert S\rvert^{1-\alpha}$ and the
properties “Square” and “Powers”. “Hilbert–Schmidt kernel” can be seen as a
corollary of Lemma 2.2 in [13]. “Orthogonality” is based on the observation,
that if $S^{*}T=0$, we have $(S+T)^{*}(S+T)=S^{*}S+T^{*}T$, “Monotonicity II”,
and “Adjoint” to replace the condition $S^{*}T=0$ by the non-equivalent
condition $ST^{*}=0$. ∎

######  Definition .

We say a densely defined operator $T$ on $\mathrm{L}^{2}(\mathbb{R}^{2})$ has
the integral kernel $t\colon\mathbb{R}^{2}\times\mathbb{R}^{2}\to\mathbb{C}$,
if for any $f\in C^{0}_{c}(\mathbb{R}^{2})$, the identity

$\displaystyle(Tf)(x)=\int_{\mathbb{R}^{2}}t(x,y)f(y)dy$ (2.5)

holds for almost all $x\in\mathbb{R}^{2}$. In this case, we define

$\displaystyle\operatorname{iker}T(x,y)\coloneqq t(x,y).$ (2.6)

We say, that $t$ is _nice_ , or respectively, that $T$ is a _nice_ integral
operator, if for any fixed $x$, the functions
$t(x,{\makebox[4.30554pt]{{$\cdot$}}})$ and
$t({\makebox[4.30554pt]{{$\cdot$}}},x)$ are in
$\mathrm{L}^{1}(\mathbb{R}^{2})$ with a norm bounded independently of $x$.

######  Corollary .

Let $T$ be a _nice_ integral operator. Then $T$ is a bounded operator on
$\mathrm{L}^{2}(\mathbb{R}^{2})$.

###### Proof.

The expression $\left\lVert\left(x\mapsto\lVert
t(x,{\cdot})\rVert_{\mathrm{L}^{1}(\mathbb{R}^{2})}\right)\right\rVert_{\mathrm{L}^{\infty}(\mathbb{R}^{2})}$
is finite and an upper bound for the operator norm of $T$ as an operator on
$\mathrm{L}^{\infty}(\mathbb{R}^{2})$. On the other hand, the expression
$\left\lVert\left(y\mapsto\lVert
t({\cdot},y)\rVert_{\mathrm{L}^{1}(\mathbb{R}^{2})}\right)\right\rVert_{\mathrm{L}^{\infty}(\mathbb{R}^{2})}$
is finite and an upper bound for the operator norm of $T$ as a bounded
operator on $\mathrm{L}^{1}(\mathbb{R}^{2})$. Hence, by the Riesz-Thorin
interpolation theorem, the operator $T$ is bounded on
$\mathrm{L}^{2}(\mathbb{R}^{2})$ with an operator norm bounded by the square
root of the product of both of these expressions. ∎

######  Lemma .

Let $S,T$ be _nice_ integral operators on $\mathrm{L}^{2}(\mathbb{R}^{2})$
with integral kernels $s,t$. Let $x,z\in\mathbb{R}^{2}$. Then we have the
identities

$\displaystyle\operatorname{iker}(S+T)(x,z)=$ $\displaystyle(s+t)(x,z),$ (2.7)
$\displaystyle\operatorname{iker}(ST)(x,z)=$
$\displaystyle\int_{\mathbb{R}^{2}}s(x,y)t(y,z)dy.$ (2.8)

In particular, $S+T$ and $ST$ are _nice_ integral operators.

The first statement is trivial and the second follows by Fubini to interchange
the integral over $y$ with the one over $z$, for any test function
$f\in\mathrm{L}^{1}(\mathbb{R}^{2})\cap\mathrm{L}^{\infty}(\mathbb{R}^{2})$.

######  Definition .

Let $\gamma,d\in\mathbb{N}$ and $\lambda\in\mathbb{[}0,\infty)$. Then we
define the space
$W^{\gamma,\infty}_{(\lambda)}(\mathbb{R}^{2},\mathbb{C}^{d})$ as the subspace
of the Sobolev space $W^{\gamma,\infty}(\mathbb{R}^{2},\mathbb{C}^{d})$, where
the norm

$\displaystyle\lVert
u\rVert_{W^{\gamma,\infty}_{(\lambda)}(\mathbb{R}^{2},\mathbb{C}^{d})}\coloneqq\sum_{\gamma^{\prime}\leq\gamma}\sup_{x\in\mathbb{R}^{2}}\left\lVert\left(1+\lVert
x\rVert\right)^{\lambda}\left(\nabla^{\otimes\gamma^{\prime}}u\right)(x)\right\rVert$
(2.9)

is finite. The supremum in this definition refers to the almost everywhere
supremum. This is a Banach space. The limit space

$\displaystyle
W^{\gamma,\infty}_{(\infty)}(\mathbb{R}^{2},\mathbb{C}^{d})\coloneqq\bigcap_{\lambda\geq
0}W^{\gamma,\infty}_{(\lambda)}(\mathbb{R}^{2},\mathbb{C}^{d})$ (2.10)

is only a vector space equipped with the inverse limit topology associated to
the intersection (a set is open, if and only if it is open in each space for
finite $\lambda$.).

These spaces are motivated by Schwartz semi-norms.

##  3\. Setting and main result

We introduce the Landau Hamilton operator $H_{0}$ with a constant magnetic
field $B_{0}>0$, defined on (a suitable subspace of)
$\mathrm{L}^{2}(\mathbb{R}^{2})$, with magnetic gauge $A_{0}$ given by

$\displaystyle A_{0}(x)\coloneqq$ $\displaystyle\frac{B_{0}}{2}Jx,$ (3.1)
$\displaystyle H_{0}\coloneqq$ $\displaystyle(-i\nabla-A_{0})^{2}.$ (3.2)

The spectrum of $H_{0}$, $\sigma(H_{0})$, equals $B_{0}(2\mathbb{N}+1)$. Let
$P_{l}$ be the projection onto the eigenspace with eigenvalue $B_{0}(2l+1)$
for $l\in\mathbb{N}$.

Furthermore, for any $\alpha>0$, we introduce the $\alpha$-Rényi entropy
functions $h_{\alpha}\colon[0,1]\to[0,\ln(2)]$,

$\displaystyle
h_{\alpha}(x)\coloneqq\begin{cases}\frac{1}{1-\alpha}\ln\left(x^{\alpha}+(1-x)^{\alpha}\right)&\text{
for }\alpha\not=1,\\\ -x\ln x-(1-x)\ln(1-x)&\text{ for }\alpha=1,\end{cases}$
(3.3)

for $x\in(0,1)$ and $h_{\alpha}(0)=h_{\alpha}(1)=0$. Throughout this paper,
let $\Lambda\subset\mathbb{R}^{2}$ be a bounded open set with Lipschitz-
boundary.

Let $\mu\in\mathbb{R}\setminus B_{0}(2\mathbb{N}+1)$. We define
$1_{\leq\mu}(H_{0})$ as the spectral projection associated to $H_{0}$ and
$\mu$. We are interested in how the leading order asymptotic expansion of the
local entropy,

$\displaystyle
S_{\alpha}(L\Lambda)\coloneqq\operatorname{tr}h_{\alpha}\left(1_{L\Lambda}1_{\leq\mu}(H_{0})1_{L\Lambda}\right),$
(3.4)

as $L\to\infty$ changes under slight perturbations of $H_{0}$. The trace is
defined as the usual Hilbert space trace of trace class operators on
$\mathrm{L}^{2}(\mathbb{R}^{2})$. This quantity is the local entropy or
entanglement entropy of the ground state restricted to $L\Lambda$. Under the
assumption that $\Lambda$ has $C^{3}$ boundary, the leading term of order $L$
for the operator $H=H_{0}$ has been calculated by Leschke, Sobolev and Spitzer
in [11]. This allows us to focus on bounding the error term that arises, as we
introduce a perturbation to $H_{0}$. Our main result is Section 3 and relies
on the exact calculations of the leading term for $H=H_{0}$, see [11], and the
estimates we will prove in this paper.

The following condition is needed to state our main results and a lot of
results along the way. Throughout this paper, we fix $0<\varepsilon<1$.

######  Definition .

Let $\gamma$ be a natural number. We call a magnetic field
$B_{\varepsilon}\colon\mathbb{R}^{2}\to\mathbb{R}$ and a potential
$V_{\varepsilon}\colon\mathbb{R}^{2}\to\mathbb{R}$ $(\gamma,\varepsilon)$
tame, if $V_{\varepsilon}\in
W^{\gamma,\infty}_{(\varepsilon)}(\mathbb{R}^{2},\mathbb{R})$ and
$B_{\varepsilon}\in
W^{\gamma,\infty}_{(1+\varepsilon)}(\mathbb{R}^{2},\mathbb{R})$.

######  Remark.

All the following estimates will depend on $B_{\varepsilon},V_{\varepsilon}$
only through $\varepsilon,\gamma$ and the norms of
$B_{\varepsilon},V_{\varepsilon}$ in the spaces
$W^{\gamma,\infty}_{(1+\varepsilon)}(\mathbb{R}^{2},\mathbb{R})$ and
$W^{\gamma,\infty}_{(\varepsilon)}(\mathbb{R}^{2},\mathbb{R})$. Maybe somewhat
counter-intuitively, small values of $\varepsilon$ correspond to slowly
decaying $B_{\varepsilon},V_{\varepsilon}$.

To define the perturbed Hamiltonian $H$, we need to choose a gauge
$A_{\varepsilon}$ of the magnetic field $B_{\varepsilon}$. We choose the
convolution, which is given by

$\displaystyle
A_{\varepsilon}(x)\coloneqq\left(B_{\varepsilon}*\frac{J{\makebox[4.30554pt]{{$\cdot$}}}}{2\pi\lVert{\makebox[4.30554pt]{{$\cdot$}}}\rVert^{2}}\right)(x)=\int_{\mathbb{R}^{2}}B_{\varepsilon}(x-y)\frac{Jy}{2\pi\lVert
y\rVert^{2}}dy$ (3.5)

for any $x\in\mathbb{R}^{2}$. Its relevant properties are summed up in the
following Lemma.

######  Lemma .

Let $\gamma\in\mathbb{N}$, $f\in
W_{(1+\varepsilon)}^{\gamma,\infty}(\mathbb{R}^{2},\mathbb{R})$ and define
$g\in\operatorname{Maps}(\mathbb{R}^{2},\mathbb{R}^{2})$ as the convolution

$\displaystyle g\coloneqq
f*\frac{J{\makebox[4.30554pt]{{$\cdot$}}}}{2\pi\lVert{\makebox[4.30554pt]{{$\cdot$}}}\rVert^{2}}.$
(3.6)

Then, for any $x\in\mathbb{R}^{2}$, we have the identities

$\displaystyle\nabla_{x}\times g(x)$ $\displaystyle=f(x),$ (3.7)
$\displaystyle\nabla_{x}\cdot g(x)$ $\displaystyle=0.$ (3.8)

Furthermore, we have $g\in
W^{\gamma,\infty}_{(\varepsilon)}(\mathbb{R}^{2},\mathbb{R}^{2})$.

######  Remark.

A gauge satisfying (3.8) is commonly referred to as a Coulomb gauge. The
restriction to $\varepsilon<1$ is necessary to get the described decay. A
value of $\varepsilon>1$ will only achieve a $(1+\lVert x\rVert)^{-1}$ decay
in $A_{\varepsilon}$.

The proof can be found in Appendix B.

Now we define the perturbed gauge $A$ and the perturbed Hamiltonian $H$ by

$\displaystyle A$ $\displaystyle\coloneqq A_{0}-A_{\varepsilon},$ (3.9)
$\displaystyle H$ $\displaystyle\coloneqq(-i\nabla-A)^{2}+V_{\varepsilon}.$
(3.10)

As we can see, this gauge corresponds to the magnetic field
${B_{0}}-B_{\varepsilon}$, that is, $\nabla_{x}\times
A(x)=B_{0}-B_{\varepsilon}(x)$. The operator $H$ is self-adjoint and its
domain agrees with the domain of $H_{0}$, which we will see in Section 4.

We need the following $p$-Schatten quasi-norm estimate, which will be proven
in the next section.

######  Theorem 3.1.

Let $l\in\mathbb{N},\gamma\in\mathbb{Z}^{+}$. Let
$B_{\varepsilon},V_{\varepsilon}$ be $(\gamma,\varepsilon)$ tame and let
$1\geq p>\frac{2}{\gamma+3}$. Let $a,b\in\mathbb{R}\setminus
B_{0}(2\mathbb{N}+1)$ with $a<b$. Then we have the estimates

$\displaystyle\left\lVert
1_{L\Lambda}1_{[a,b]}(H)1_{L\Lambda^{\complement}}\right\rVert_{p}^{p}$
$\displaystyle\leq CL,$ (3.11) $\displaystyle\left\lVert
1_{L\Lambda}\left(1_{[a,b]}(H)-1_{[a,b]}(H_{0})\right)1_{L\Lambda^{\complement}}\right\rVert_{p}^{p}$
$\displaystyle\leq CL^{1-p\varepsilon}.$ (3.12)

The constants $C$ depend on
$\gamma,a,b,\Lambda,p,\varepsilon,B_{\varepsilon},V_{\varepsilon}$.

Finally, we need the following statement due to Aleksandrov and Peller, which
is a Corollary of Theorem 5.11 in [1] and the inclusion
$C_{c}^{\infty}(\mathbb{R})\subset B_{\infty,1}^{1}(\mathbb{R})$, where the
latter refers to the Besov space as used by Aleksandrov and Peller.

######  Proposition  (based on Theorem 5.11 in [1]).

Let $f\in C^{\infty}_{c}(\mathbb{R})$. Then there is a constant $C<\infty$,
such that for any self-adjoint bounded operators $A,B$, such that $A-B$ is
trace class, we have the estimate

$\displaystyle\lVert f(A)-f(B)\rVert_{1}\leq C\lVert A-B\rVert_{1}\,.$ (3.13)

Now we state the key result of this paper, which is proved below.

######  Theorem 3.2.

Let $\alpha>0$ and choose $\beta=\min(0.5,\alpha)$. Define $\gamma$ as the
smallest integer, such that $\gamma>\frac{1}{\beta}-3$. Let
$B_{\varepsilon},V_{\varepsilon}$ be $(\gamma,\varepsilon)$ tame. Let
$a,b\in\mathbb{R}\setminus{B_{0}}(2\mathbb{N}+1),a<b$ and $I\coloneqq[a,b]$.
Then we have

$\displaystyle\operatorname{tr}\left(h_{\alpha}\left(1_{L\Lambda}1_{I}(H)1_{L\Lambda}\right)-h_{\alpha}\left(1_{L\Lambda}1_{I}(H_{0})1_{L\Lambda}\right)\right)=o(L),$
(3.14)

as $L\to\infty$.

######  Remark.

The choice of $\beta=0.5$ for $\alpha\geq 0.5$ delivers the optimal value for
$\gamma$, namely $0$. For $\alpha>\frac{1}{3}$, we can get away with a non-
differentiable $B_{\varepsilon},V_{\varepsilon}$.

The assumption that $a,b\not\in B_{0}(2\mathbb{N}+1)$ cannot be dropped, as
the following counter example illustrates. Let $B_{\varepsilon}=0,a=0$ and
$b=B_{0}$. By Section 4, the spectrum of $H$ has an accumulation point at
$B_{0}$. If we assume $V_{\varepsilon}>0$ pointwise, then all eigenvalues of
$H$ are strictly larger than $B_{0}$ and hence $1_{I}(H)=0$. But Theorem 8 in
[11], which we will elaborate on shortly, states, that the leading order
asymptotic expansion of
$\operatorname{tr}h_{\alpha}\left(1_{L\Lambda}1_{I}(H_{0})1_{L\Lambda}\right)$
for large $L$ is of order $O(L)$ and does not vanish. On the other hand, if we
assume that $-B_{0}<V_{\varepsilon}<0$ pointwise, there is a spectral gap of
the form $(B_{0},2B_{0})$ in the spectrum of $H$. Hence, we can move $b$ to
$1.5B_{0}$ without changing the operators. Now we can apply our Theorem 3.2.
Hence under our general assumptions, it is possible to get both one-sided
limits, when $b=B_{0}$. We expect similar results, whenever $a$ or $b$ are in
the spectrum of $H_{0}$. It is, however, a little more complicated to see,
whether the leading order expansion for $H_{0}$ changes, when we add or remove
a single Landau level from the interval $I$.

The following corollary is our main result. It combines Theorem 8 in [11],
which can be stated as the corollary for the case
$B_{\varepsilon}=V_{\varepsilon}=0$, with our Theorem 3.2.

######  Corollary .

Let $\alpha>0$ and choose $\beta=\min(\alpha,0.5)$. Define $\gamma$ as the
smallest positive integer, such that $\gamma>\frac{1}{\beta}-3$. Let
$B_{\varepsilon},V_{\varepsilon}$ be $(\gamma,\varepsilon)$ tame. Let
$\mu\not\in\sigma(H_{0})$ and define $\nu$ as the largest integer, such that
$B_{0}(2\nu+1)<\mu$. Assume that the boundary $\partial\Lambda$ is
$C^{3}$-smooth. Then

$\displaystyle
S_{\alpha}(L\Lambda)=\operatorname{tr}(h_{\alpha}(1_{L\Lambda}1_{\leq\mu}(H)1_{L\Lambda})=L\sqrt{B_{0}}\lvert\partial\Lambda\rvert
M_{\leq\nu}(h_{\alpha})+o(L),$ (3.15)

as $L\to\infty$ with $0<M_{\leq\nu}(h_{\alpha})<\infty$ as described in [11]
for $\nu\geq 0$ and $M_{\leq\nu}(h_{\alpha})\coloneqq 0$ for $\nu<0$.

In the case $\mu<B_{0}$, the projection is finite dimensional and the entropy
has an order at most $O(1)$ in $L$ as $L\to\infty$.

###### Proof of Theorem 3.2.

We define the function $g_{\alpha}\colon[0,1]\to[0,\ln(2)]$ by the identity

$\displaystyle g_{\alpha}(4x(1-x))=h_{\alpha}(x).$ (3.16)

The symmetry of $h_{\alpha}$ guarantees the existence of $g_{\alpha}$. We have

$\displaystyle g_{\alpha}(t)=h_{\alpha}\left(\frac{1-\sqrt{1-t}}{2}\right).$
(3.17)

Let $\varepsilon_{0}>0$. We choose a smooth cut-off function
$\varphi\colon[0,1]\to[0,1]$ with $\varphi(x)=1$, if $x\leq\varepsilon_{0}$,
and $\varphi(x)=0,$ if $x\geq 2\varepsilon_{0}$. Now we write

$\displaystyle
g_{\alpha}(t)=(1-\varphi(t))g_{\alpha}(t)+\varphi(t)g_{\alpha}(t).$ (3.18)

The advantage of this decomposition is that the first summand is smooth, and
the second summand is small. The second summand can be bounded using the fact,
that $h_{\alpha}$ is $\beta$-Hölder continuous on $[0,1]$ and smooth on
$(0,1)$. As $h_{\alpha}$ is symmetric around $t=\frac{1}{2}$ and analytic on
$(0,1)$, its Taylor expansion at that point contains only even powers of
$(t-\frac{1}{2})$. Thus, we see that $g_{\alpha}$ is analytic at $t=1$. Hence
$g_{\alpha}\in C^{\infty}((0,1])$ and it is $\beta$-Hölder continuous on
$[0,1]$, as $\beta=\min(\alpha,0.5)$.

We choose $\beta^{\prime}<\beta\leq\frac{1}{2}$, such that
$\gamma>\frac{1}{\beta^{\prime}}-3$. Hence, we have

$\displaystyle\varphi(t)g_{\alpha}(t)\leq
C\varepsilon_{0}^{\beta-\beta^{\prime}}t^{\beta^{\prime}}.$ (3.19)

We define $P,P^{\prime}$ as the spectral projections,

$\displaystyle P\coloneqq$ $\displaystyle 1_{I}(H_{0}),$ (3.20) $\displaystyle
P^{\prime}\coloneqq$ $\displaystyle 1_{I}(H).$ (3.21)

We observe

$\displaystyle
h_{\alpha}(1_{L\Lambda}P^{(^{\prime})}1_{L\Lambda})=g_{\alpha}(4\lvert
1_{L\Lambda^{\complement}}P^{(^{\prime})}1_{L\Lambda}\rvert^{2}).$ (3.22)

We can now apply Section 3. Thus,

$\displaystyle\left\lVert\left((1-\varphi)g_{\alpha}\right)\left(4\lvert
1_{L\Lambda^{\complement}}P^{\prime}1_{L\Lambda}\rvert^{2}\right)-\left((1-\varphi)g_{\alpha}\right)\left(4\lvert
1_{L\Lambda^{\complement}}P1_{L\Lambda}\rvert^{2}\right)\right\rVert_{1}$
(3.23) $\displaystyle\leq$ $\displaystyle C\left\lVert\lvert
1_{L\Lambda^{\complement}}P^{\prime}1_{L\Lambda}\rvert^{2}-\lvert
1_{L\Lambda^{\complement}}P1_{L\Lambda}\rvert^{2}\right\rVert_{1}$ (3.24)
$\displaystyle\leq$ $\displaystyle C\left\lVert
1_{L\Lambda^{\complement}}(P^{\prime}-P)1_{L\Lambda}\right\rVert_{1}$ (3.25)
$\displaystyle\leq$ $\displaystyle CL^{1-\varepsilon}\,.$ (3.26)

Note that the last constant $C$ depends on $\varepsilon_{0}$, but not on $L$.
In the second step we used the identity $\lvert A\rvert^{2}-\lvert
B\rvert^{2}=A^{*}(A-B)+(A^{*}-B^{*})B$. In the last step, we used Theorem 3.1
with $p=1$.

We can also apply Theorem 3.1 for the remaining term, after using (3.19),
$1\geq 2\beta^{\prime}>\frac{2}{\gamma+3}$ and that $H=H_{0}$ is admissible
for Theorem 3.1.

$\displaystyle\left\lVert\left(\varphi g_{\alpha}\right)\left(4\lvert
1_{L\Lambda^{\complement}}P^{(^{\prime})}1_{L\Lambda}\rvert^{2}\right)\right\rVert_{1}\leq
C\varepsilon_{0}^{\beta-\beta^{\prime}}\left\lVert\lvert
1_{L\Lambda^{\complement}}P^{(^{\prime})}1_{L\Lambda}\rvert\right\rVert_{2\beta^{\prime}}^{2\beta^{\prime}}\leq
C\varepsilon_{0}^{\beta-\beta^{\prime}}L.$ (3.27)

Hence,

$\displaystyle\left\lvert\operatorname{tr}h_{\alpha}(1_{L\Lambda}P^{\prime}1_{L\Lambda})-\operatorname{tr}h_{\alpha}(1_{L\Lambda}P1_{L\Lambda})\right\rvert\leq
C(\varepsilon_{0})L^{1-\varepsilon}+C\varepsilon_{0}^{\beta-\beta^{\prime}}L.$
(3.28)

Note that the first constant $C(\varepsilon_{0})$ depends on $\varepsilon_{0}$
while the second one does not. This term is in $o(L)$, as for any
$\varepsilon>0$ we can choose $L$ large enough to let the first term be less
than $\varepsilon_{0}L$. This proves that the leading term expansion of the
$\alpha$-Rényi entropy for the perturbed Landau Hamiltonian $H$ agrees with
the main term in the same expansion for the Landau Hamiltonian $H_{0}$. This
finishes the proof. ∎

######  Remark.

We can actually pick $\varepsilon_{0}$ dependent on $L$, which does lead to a
smaller error term, if we bound the constant $C(\varepsilon_{0})$ more
precisely. This does however not lead to an improved error term in Section 3,
as the known error term for the constant magnetic field is too large. Hence, I
did not include the details here.

##  4\. The Ansatz for the proof of Theorem 3.1

The goal of this section is to explain how to prove Theorem 3.1 and, to reduce
it to two more technical statements. The general approach has been inspired by
[16].

We define

$\displaystyle H_{\varepsilon}$ $\displaystyle\coloneqq H-H_{0},$ (4.1)

where $H$ and $H_{0}$ were defined in (3.10) and (3.2).

We expand $H_{\varepsilon}$ as

$\displaystyle H_{\varepsilon}$ $\displaystyle=H-H_{0}$ (4.2)
$\displaystyle=\left(-i\nabla-A\right)^{2}-\left(-i\nabla-
A_{0}\right)^{2}+V_{\varepsilon}$ (4.3)
$\displaystyle=(A_{0}-A)\cdot\left(-i\nabla-A+A_{0}-A_{0}\right)+\left(-i\nabla-
A_{0}\right)\cdot(A_{0}-A)+V_{\varepsilon}$ (4.4)
$\displaystyle=2A_{\varepsilon}\cdot\left(-i\nabla-
A_{0}\right)+A_{\varepsilon}^{2}+V_{\varepsilon}.$ (4.5)

We used the identity $a^{2}-b^{2}=(a-b)a+b(a-b)$ in the third step and (3.8),
which is equivalent to $\nabla\cdot
A_{\varepsilon}=A_{\varepsilon}\cdot\nabla$, in the last step. We now
introduce the pseudo potential

$\displaystyle W_{\varepsilon}\coloneqq A_{\varepsilon}^{2}+V_{\varepsilon}.$
(4.6)

We introduce a few more operators. Let $I\subset\mathbb{N}$ be cofinite,
$\zeta\in\mathbb{C}$ and $\zeta\not\in B_{0}(2I+1)$. Then we define the
bounded operator

$\displaystyle M_{I,\zeta}\coloneqq\sum_{l\in
I}\frac{P_{l}}{B_{0}(2l+1)-\zeta}.$ (4.7)

It satisfies $M_{I,\zeta}^{*}=M_{I,\overline{\zeta}}$. For
$\zeta\not\in\sigma(H_{0})$, we have the identity

$\displaystyle M_{\mathbb{N},\zeta}=\frac{1}{H_{0}-\zeta}.$ (4.8)

There are some results describing the kernel of the resolvent operator, but we
also need the special case

$\displaystyle T_{l}\coloneqq
M_{\mathbb{N}\setminus\\{l\\},{B_{0}}(2l+1)}=\sum_{k\not=l}\frac{P_{k}}{2{B_{0}}(k-l)}.$
(4.9)

Hence it is more convenient to deal with the operator $M_{I,\zeta}$ in this
generality.

We define $n_{0}$ as the smallest integer such that

$\displaystyle n_{0}>\frac{1}{2\varepsilon}.$ (4.10)

The following lemma will be proved in Section 6 after some preparations.

######  Lemma .

Let $B_{\varepsilon}$ and $V_{\varepsilon}$ be $(0,\varepsilon)$ tame. Then
for any $I\subset N$ cofinite and any
$\zeta\in\mathbb{C}\setminus{B_{0}}(2I+1)$, the operator
$H_{\varepsilon}M_{I,\zeta}$ is in the $4n_{0}$-Schatten class, and the
$4n_{0}$-Schatten norm is in
$\mathrm{L}^{\infty}_{loc}(\mathbb{C}\setminus\sigma(H_{0}))$ as a function of
$\zeta$. The upper bound for the norm depends on ${B_{0}}$.

As $p$-Schatten class operators are compact, we now know that
$H_{\varepsilon}$ is relatively $H_{0}$-compact. This implies

######  Corollary .

The essential spectrum of $H$ agrees with the essential spectrum of $H_{0}$
which is ${B_{0}}(2\mathbb{N}+1)$.

######  Remark.

The statement is also true if $V=0$ and $B$ is smooth and converges to $B_{0}$
as $\lVert x\rVert\to\infty$ (at any rate), see [7]. They state smoothness of
$B$ as a condition, but I think it is not required. However, their algebraic
proof does not imply that the eigenspaces of $H_{0}$ and $H$ are at all
related.

As $\sigma(H_{0})$ is discrete, this implies, that
$\sigma(H)=\overline{\sigma_{p}(H)}$ and that the continuous part of the
spectrum of $H$ vanishes. We continue with the Riesz integral representation.

######  Fact .

For any path $\Gamma$ in $\mathbb{C}$ that intersects $\mathbb{R}$ in exactly
two points $\lambda_{1}<\lambda_{2}$, does not intersect
$\sigma(H)\subset\mathbb{R}$ and has winding number $+1$ around
$(\lambda_{1}+\lambda_{2})/2$, we have the identity

$\displaystyle-\frac{1}{2\pi
i}\int_{\Gamma}\frac{d\zeta}{H-\zeta}=1_{\lambda_{1}<E<\lambda_{2}}(H).$
(4.11)

With the resolvent identity, we can write

$\displaystyle\frac{1}{H-\zeta}=$
$\displaystyle\frac{1}{H_{0}-\zeta}-\frac{1}{H-\zeta}H_{\varepsilon}\frac{1}{H_{0}-\zeta}$
(4.12) $\displaystyle=$
$\displaystyle\frac{1}{H_{0}-\zeta}-\frac{1}{H_{0}-\zeta}H_{\varepsilon}\frac{1}{H_{0}-\zeta}+\frac{1}{H_{0}-\zeta}H_{\varepsilon}\frac{1}{H-\zeta}H_{\varepsilon}\frac{1}{H_{0}-\zeta}\,.$
(4.13)

By induction, this leads to

######  Corollary .

For any $n\in\mathbb{Z}^{+},\zeta\not\in\sigma(H)\cup\sigma(H_{0})$, we have

$\displaystyle\frac{1}{H-\zeta}=\sum_{k=0}^{2n-1}\frac{(-1)^{k}}{H_{0}-\zeta}\left(H_{\varepsilon}\frac{1}{H_{0}-\zeta}\right)^{k}+\left(\frac{1}{H_{0}-\zeta}H_{\varepsilon}\right)^{n}\frac{1}{H-\zeta}\left(H_{\varepsilon}\frac{1}{H_{0}-\zeta}\right)^{n},$
(4.14)

where $H_{\varepsilon}=H-H_{0}$, as in (4.1).

For the summands in Section 4 except the last summand, we can resolve the path
integral over some paths.

######  Lemma .

Let $l,k\in\mathbb{N}$ and $\Gamma$ be the path along the circle $\partial
D_{B_{0}}({B_{0}}(2l+1))$ that rotates in positive direction. Then we have

$\displaystyle-\frac{1}{2\pi
i}\int_{\Gamma}\frac{1}{H_{0}-\zeta}\left(H_{\varepsilon}\frac{1}{H_{0}-\zeta}\right)^{k}d\zeta$
$\displaystyle=\sum_{m=0}^{k}(T_{l}H_{\varepsilon})^{m}P_{l}(H_{\varepsilon}T_{l})^{k-m},$
(4.15)

where $H_{\varepsilon}=H-H_{0}$, as in (4.1).

###### Proof.

Let $N>2l$ and either $I=\mathbb{N}$ and $\zeta\in\Gamma$ or
$I=\mathbb{N}\setminus\\{l\\}$ and $\zeta={B_{0}}(2l+1)$. We introduce
$P_{\leq N}\coloneqq\sum_{n\leq N}P_{n}$ and $P_{>N}\coloneqq 1-P_{\leq N}$.
We continue with the identity

$\displaystyle P_{\leq N}M_{I,\zeta}=\sum_{j\in I,j\leq
N}\frac{P_{j}}{{B_{0}}(2j+1)-\zeta}.$ (4.16)

There is a constant $C$, independent of $N$ and $\zeta$, such that the
estimate $\left\lVert{P_{>N}}M_{I,\zeta}\right\rVert_{\infty}\leq\frac{C}{N}$
holds (see Appendix A). Furthermore, by Section 4 and as the $4n_{0}$-Schatten
norm is an upper bound for the operator norm, we have the estimate
$\left\lVert H_{\varepsilon}M_{I,\zeta}\right\rVert_{\infty}<C$ with a
constant $C$ independent of $\zeta$ (and $N$). We use the telescope sum
$b(ab)^{k}-c(ac)^{k}=\sum_{k^{\prime}=0}^{k}(ba)^{k^{\prime}}(b-c)(ac)^{k-k^{\prime}}$,
which holds in any ring, and the triangle inequality to get

$\displaystyle\left\lVert
M_{I,\zeta}\left(H_{\varepsilon}M_{I,\zeta}\right)^{k}-P_{\leq
N}M_{I,\zeta}\left(H_{\varepsilon}P_{\leq
N}M_{I,\zeta}\right)^{k}\right\rVert_{\infty}$ (4.17) $\displaystyle\leq$
$\displaystyle\sum_{k^{\prime}=0}^{k}\left\lVert\left(M_{I,\zeta}H_{\varepsilon}\right)^{k^{\prime}}P_{>N}M_{I,\zeta}\left(H_{\varepsilon}P_{\leq
N}M_{I,\zeta}\right)^{k-k^{\prime}}\right\rVert_{\infty}\leq\frac{C}{N},$
(4.18)

where $C$ is independent of $N$ and $\zeta$. The second step relies on the
submultiplicativity of the norm, and the identity $M_{I,\zeta}P_{\leq
N}=P_{\leq N}M_{I,\zeta}$. Thus, we have

$\displaystyle-\frac{1}{2\pi
i}\int_{\Gamma}\frac{1}{H_{0}-\zeta}\left(H_{\varepsilon}\frac{1}{H_{0}-\zeta}\right)^{k}d\zeta$
(4.19) $\displaystyle=$ $\displaystyle-\frac{1}{2\pi
i}\lim_{N\to\infty}\int_{\Gamma}P_{\leq
N}M_{\mathbb{N},\zeta}\left(H_{\varepsilon}P_{\leq
N}M_{\mathbb{N},\zeta}\right)^{k}d\zeta$ (4.20) $\displaystyle=$
$\displaystyle-\frac{1}{2\pi
i}\lim_{N\to\infty}\int_{\Gamma}\sum_{\sigma\in\\{0,\dots,N\\}^{k+1}}\frac{P_{\sigma_{0}}\prod_{j=1}^{k}H_{\varepsilon}P_{\sigma_{j}}}{\prod_{j=0}^{k}({B_{0}}(2\sigma_{j}+1)-\zeta)}d\zeta$
(4.21) $\displaystyle=$
$\displaystyle\lim_{N\to\infty}\sum_{\sigma\in\\{0,\dots,N\\}^{k+1}}P_{\sigma_{0}}\left(\prod_{j=1}^{k}H_{\varepsilon}P_{\sigma_{j}}\right)\begin{cases}\prod_{\sigma_{j}\not=l}\frac{1}{2{B_{0}}(\sigma_{j}-l)}&\mathrm{if}\\#\\{j\mid\sigma_{j}=l\\}=1\\\
0&\mathrm{else}\end{cases}$ (4.22) $\displaystyle=$
$\displaystyle\lim_{N\to\infty}\sum_{m=0}^{k}(P_{\leq
N}T_{l}H_{\varepsilon})^{m}P_{l}(H_{\varepsilon}P_{\leq N}T_{l})^{k-m}$ (4.23)
$\displaystyle=$
$\displaystyle\sum_{m=0}^{k}(T_{l}H_{\varepsilon})^{m}P_{l}(H_{\varepsilon}T_{l})^{k-m}.$
(4.24)

In the first step, we used that (4.18) holds uniformly in $\zeta\in\Gamma$ for
$I=\mathbb{N}$. In the second step, we inserted (4.16) $k+1$ times and
multiplied out all terms in order to get a finite sum. We then exchanged this
finite sum with the complex path integral and resolved this complex-valued
integral. The fourth step uses (4.16) in reverse. The final step follows by
(4.18) for $I=\mathbb{N}\setminus\\{l\\}$ and $\zeta={B_{0}}(2l+1)$. This
finishes the proof. ∎

We will prove the following theorem at the end of Section 6.

######  Theorem 4.1.

Let $k,l,m,\gamma\in\mathbb{N}$ with $k\geq m$. Let
$B_{\varepsilon},V_{\varepsilon}$ be $(\gamma,\varepsilon)$ tame and let
$1\geq p>\frac{2}{\gamma+3}$. Then there is a constant $C>0$ and a
$\lambda>0$, such that for any $R\geq 0$, we have the upper bound

$\displaystyle\left\lVert
1_{[0,1]^{2}+x_{0}}(T_{l}H_{\varepsilon})^{m}P_{l}(H_{\varepsilon}T_{l})^{k-m}1_{D^{\complement}_{R}(x_{0})}\right\rVert_{p}\leq
C\frac{\exp\left(-\lambda R^{2}\right)}{(1+\lVert
x_{0}\rVert)^{k\varepsilon}},$ (4.25)

for any $x_{0}\in\mathbb{R}^{2}$. The constant $C$ depends on
${B_{0}},l,k,m,\gamma,p,\varepsilon,B_{\varepsilon},V_{\varepsilon}$, but is
independent of $R$ and $x_{0}$.

######  Remark.

For $k=m=0$, this is Lemma 12 in [11].

We will now follow Theorem 13 in [11]. But we go a slightly different
direction with the proof111We replace a sum by an integral..

######  Theorem 4.2.

Let $k,l,m,\gamma\in\mathbb{N}$ with $k\geq m$, let
$B_{\varepsilon},V_{\varepsilon}$ be $(\gamma,\varepsilon)$ tame and let
$1\geq p>\frac{2}{\gamma+3}$. Then for any $L>1$ we have

$\displaystyle\left\lVert
1_{L\Lambda}(T_{l}H_{\varepsilon})^{m}P_{l}(H_{\varepsilon}T_{l})^{k-m}1_{L\Lambda^{\complement}}\right\rVert_{p}^{p}\leq
CL^{1-pk\varepsilon}.$ (4.26)

The constant $C$ depends on
$\Lambda,{B_{0}},l,k,m,\gamma,p,\varepsilon,B_{\varepsilon},V_{\varepsilon}$.

###### Proof.

We define

$\displaystyle
T\coloneqq(T_{l}H_{\varepsilon})^{m}P_{l}(H_{\varepsilon}T_{l})^{k-m}.$ (4.27)

We choose an $h_{0}\in[0,1)^{2}$. We will now use the $p$-Schatten norm
property we called orthogonality in the first and forth step, and the
$p$-triangle inequality in the second step. Hence,

$\displaystyle\left\lVert
1_{L\Lambda}T1_{L\Lambda^{\complement}}\right\rVert_{p}^{p}$ (4.28)
$\displaystyle\leq$
$\displaystyle\left\lVert\sum_{z\in\mathbb{Z}^{2},z+h_{0}\in
D_{\sqrt{2}}(L\Lambda)}1_{[0,1)^{2}+z+h_{0}}T1_{L\Lambda^{\complement}}\right\rVert_{p}^{p}$
(4.29) $\displaystyle\leq$ $\displaystyle\sum_{z\in\mathbb{Z}^{2},z+h_{0}\in
D_{\sqrt{2}}(L\Lambda)}\left\lVert
1_{[0,1)^{2}+z+h_{0}}T1_{L\Lambda^{\complement}}\right\rVert_{p}^{p}$ (4.30)
$\displaystyle=$ $\displaystyle\sum_{z\in\mathbb{Z}^{2},z+h_{0}\in
D_{\sqrt{2}}(L\Lambda)}\left\lVert
1_{[0,1]^{2}+z+h_{0}}T1_{L\Lambda^{\complement}}\right\rVert_{p}^{p}$ (4.31)
$\displaystyle\leq$ $\displaystyle\sum_{z\in\mathbb{Z}^{2},z+h_{0}\in
D_{\sqrt{2}}(L\Lambda)}\left\lVert
1_{[0,1]^{2}+z+h_{0}}T1_{D_{\operatorname{dist}(z+h_{0},L\Lambda^{\complement})}^{\complement}(z+h_{0})}\right\rVert_{p}^{p}$
(4.32) $\displaystyle\leq$ $\displaystyle\sum_{z\in\mathbb{Z}^{2},z+h_{0}\in
D_{\sqrt{2}}(L\Lambda)}C\frac{\exp\left(-p\lambda\operatorname{dist}(z+h_{0},L\Lambda^{\complement})^{2}\right)}{(1+\lVert
z+h_{0}\rVert)^{pk\varepsilon}}\,.$ (4.33)

The last step follows by Theorem 4.1. The constant $C$ is independent of
$z,h_{0}$. Now we can integrate this upper bound over $h_{0}\in[0,1)^{2}$.
This integral can be resolved by Appendix A. Hence, we have

$\displaystyle\left\lVert
1_{L\Lambda}T1_{L\Lambda^{\complement}}\right\rVert_{p}^{p}\leq$
$\displaystyle\int_{[0,1)^{2}}dh_{0}\sum_{z\in\mathbb{Z}^{2},z+h_{0}\in
D_{\sqrt{2}}(L\Lambda)}C\frac{\exp\left(-p\lambda\operatorname{dist}(z+h_{0},L\Lambda^{\complement})^{2}\right)}{(1+\lVert
z+h_{0}\rVert)^{pk\varepsilon}}$ (4.34) $\displaystyle=$
$\displaystyle\int_{D_{\sqrt{2}}(L\Lambda)}\frac{\exp\left(-p\lambda\operatorname{dist}(x,L\Lambda^{\complement})^{2}\right)}{(1+\lVert
x\rVert)^{pk\varepsilon}}dx$ (4.35) $\displaystyle=$ $\displaystyle
L^{2}\int_{D_{\frac{\sqrt{2}}{L}}(\Lambda)}\frac{\exp\left(-p\lambda
L^{2}\operatorname{dist}(x^{\prime},\Lambda^{\complement})^{2}\right)}{(1+L\lVert
x^{\prime}\rVert)^{pk\varepsilon}}dx^{\prime}$ (4.36) $\displaystyle\leq$
$\displaystyle CL^{2}\left(\int_{\Lambda}\frac{\exp\left(-p\lambda
L^{2}\operatorname{dist}(x^{\prime},\Lambda^{\complement})^{2}\right)}{(1+L\lVert
x^{\prime}\rVert)^{pk\varepsilon}}dx^{\prime}+L^{-pk\varepsilon}\left\lvert
D_{\frac{\sqrt{2}}{L}}(\Lambda)\setminus\Lambda\right\rvert\right).$ (4.37)

The constant $C$ does not depend on $L$. We are left to show, that the term
behind $CL^{2}$ is bounded by $CL^{-1-pk\varepsilon}$.

As $L\geq 1$, by (A.28), we have

$\displaystyle\left\lvert
D_{\frac{\sqrt{2}}{L}}(\Lambda)\setminus\Lambda\right\rvert\leq\frac{C}{L},$
(4.38)

because we can ignore the $\frac{1}{L^{2}}$ part. The constant depends on
$\Lambda$ and this is the desired estimate.

To estimate the remaining integral, we first use Appendix A and then once more
Appendix A to estimate the integral over the enumerator. Thus,

$\displaystyle\int_{\Lambda}\exp\left(-p\lambda
L^{2}\operatorname{dist}(x^{\prime},\Lambda^{\complement})^{2}\right)dx^{\prime}$
(4.39) $\displaystyle=$ $\displaystyle\int_{\mathbb{R}}{p\lambda
L^{2}h}\exp\left(-p\lambda
L^{2}h^{2}\right)\left\lvert\\{x^{\prime}\in\Lambda\mid\operatorname{dist}(x^{\prime},\Lambda^{\complement})\leq
h\\}\right\rvert dh$ (4.40) $\displaystyle\leq$
$\displaystyle\int_{0}^{\infty}{p\lambda L^{2}h}\exp\left(-p\lambda
L^{2}h^{2}\right)Chdh$ (4.41) $\displaystyle=$
$\displaystyle\int_{0}^{\infty}C\exp\left(-(h^{\prime})^{2}\right)(h^{\prime})^{2}\frac{dh^{\prime}}{L}$
(4.42) $\displaystyle=$ $\displaystyle\frac{C}{L}\,.$ (4.43)

In the second to last step, we used the substitution
$(h^{\prime})^{2}=p\lambda L^{2}h^{2}$. The constant $C$ depends on
$p,\lambda$ and in turn on $p,l,k,m,\gamma,{B_{0}}$ and the decay of
$B_{\varepsilon},V_{\varepsilon}$.

To deal with the denominator in (4.37), we use $0\in\Lambda$. Hence there is
an $r>0$, such that $D_{2r}(0)\subset\Lambda$. For the integral over
$\Lambda\cap D_{r}(0)^{\complement}$, we can bound the denominator from below
by $CL^{pk\varepsilon}$ and use the integral estimate above for the
enumerator. For the integral over $D_{r}(0)$ we estimate the enumerator by
$Ce^{-L}$ and the denominator by $1$. This finishes the proof.

∎

Now, we need to consider the final summand in Section 4. For that, we need the
following theorem, which will be proven in Section 6.

######  Theorem 4.3.

Let $\gamma\in\mathbb{N},B_{\varepsilon},V_{\varepsilon}$ be
$(\gamma,\varepsilon)$ tame, $\Gamma$ be a (finite-length) path in
$\mathbb{C}\setminus\sigma(H)$, $\nu>0$ and let $1\geq p>\frac{2}{\gamma+3}$.
Then there is an $n\in\mathbb{N}$ and a $C>0$, such that we have the following
upper bound for any $x_{0}\in\mathbb{R}^{2}$ and $L>1$:

$\displaystyle\left\lVert\int_{\Gamma}1_{[0,1]^{2}+x_{0}}(M_{\mathbb{N},\zeta}H_{\varepsilon})^{n}\frac{1}{H-\zeta}(H_{\varepsilon}M_{\mathbb{N},\zeta})^{n}1_{L\Lambda^{\complement}}d\zeta\right\rVert_{p}\leq
C(1+\lVert x_{0}\rVert)^{\gamma}L^{-\nu}.$ (4.44)

The constant $C$ depends on
${B_{0}},\gamma,\varepsilon,B_{\varepsilon},V_{\varepsilon}$, but is
independent of $x_{0}$.

By the $p$-triangle inequality, the covering of $L\Lambda$ by translated unit
boxes, like in the proof of Theorem 4.2, and choosing $\nu$ sufficiently
large, we arrive at

######  Corollary .

Let $\gamma\in\mathbb{N}$, $B_{\varepsilon},V_{\varepsilon}$ be
$(\gamma,\varepsilon)$ tame, let $\Gamma$ be a (finite-length) path in
$\mathbb{C}\setminus\sigma(H)$ and let $1\geq p>\frac{2}{\gamma+3}$. Then
there is an $n\in\mathbb{N}$ and a $C>0$, such that for any $L>1$ we have

$\displaystyle\left\lVert\int_{\Gamma}1_{L\Lambda}(M_{\mathbb{N},\zeta}H_{\varepsilon})^{n}\frac{1}{H-\zeta}(H_{\varepsilon}M_{\mathbb{N},\zeta})^{n}1_{L\Lambda^{\complement}}d\zeta\right\rVert_{p}^{p}\leq
C.$ (4.45)

The constant $C$ depends on
$\Lambda,{B_{0}},\gamma,p,\varepsilon,B_{\varepsilon},V_{\varepsilon}$.

We can now conclude the

###### Proof of Theorem 3.1.

We assume that $a,b\not\in\sigma(H)$. We begin with a fixed Landau level,
meaning we even assume $B_{0}(2l-1)<a<B_{0}(2l+1)<b<B_{0}(2l+3)$ for some
$l\in\mathbb{N}$. We choose $\Gamma$ as a path along the circle through $a,b$
with centre $\frac{a+b}{2}$. We choose $n\in\mathbb{N}$, as in Section 4. Now
we use Section 4. Hence, for any $\zeta\in\operatorname{im}\Gamma$, we have

$\displaystyle\frac{1}{H-\zeta}=\sum_{k=0}^{2n-1}(-1)^{k}\frac{1}{H_{0}-\zeta}\left(H_{\varepsilon}\frac{1}{H_{0}-\zeta}\right)^{k}+\left(\frac{1}{H_{0}-\zeta}H_{\varepsilon}\right)^{n}\frac{1}{H-\zeta}\left(H_{\varepsilon}\frac{1}{H_{0}-\zeta}\right)^{n}.$
(4.46)

The path integral over every summand for $0\leq k\leq 2n-1$ can be resolved by
Section 4 and then bounded by Theorem 4.2. Hence, we have

$\displaystyle\left\lVert-\frac{1}{2\pi
i}\int_{\Gamma}(-1)^{k}1_{L\Lambda}\frac{1}{H_{0}-\zeta}\left(H_{\varepsilon}\frac{1}{H_{0}-\zeta}\right)^{k}1_{L\Lambda^{\complement}}\right\rVert_{p}^{p}$
(4.47) $\displaystyle=$
$\displaystyle\left\lVert(-1)^{k}\sum_{m=0}^{k}1_{L\Lambda}(T_{l}H_{\varepsilon})^{m}P_{l}(H_{\varepsilon}T_{l})^{k-m}1_{L\Lambda^{\complement}}\right\rVert_{p}^{p}$
(4.48) $\displaystyle\leq$ $\displaystyle CL^{1-pk\varepsilon}.$ (4.49)

In particular, we realize that $P_{l}$ is the integral over the summand for
$k=0$ and hence this summand is cancelled in (3.12). Section 4 tells us that
the path integral over the final summand is even bounded in the $p$-Schatten
norm independently of $L$. Another application of the $p$-triangle inequality
finishes the proof for a fixed Landau level.

For every $l\in\mathbb{N}$, such that $a<{B_{0}}(2l+1)<b$, we choose a circle
path, such that the last one hits $\mathbb{R}$ at $b$, each two neighbouring
paths hit $\mathbb{R}$ at one common point not in $\sigma(H)$, the first path
hits $\mathbb{R}$ at $a$ and every circle has a real-valued centre. Then we
apply the estimate for a single Landau level and the $p$-triangle inequality.

If there is no Landau eigenvalue between $a$ and $b$, the associated
projections are finite dimensional and will lead to an $O(1)$ term with
respect to $L$. This also solves the case, where $a\in\sigma(H)$ or
$b\in\sigma(H)$. Thus, it finishes the proof. ∎

##  5\. Kernel estimates

In this section we establish several properties of the Landau Hamilton
operator $H_{0}$ and the operators $P_{l},M_{I,\zeta}$ and in particular,
their integral kernels. At the end of this section, we will also include an
important integral bound.

We introduce the Laguerre polynomials and their generating function. For any
$l\in\mathbb{N}$, the Laguerre polynomials $\mathcal{L}_{l}$ is given by

$\displaystyle\mathcal{L}_{l}\colon[0,\infty)\to\mathbb{R},\quad
t\mapsto\sum_{k=0}^{l}\binom{l}{k}\frac{(-1)^{k}}{k!}t^{k}.$ (5.1)

For any $s\in[0,\infty)$, $-1<t<1$, their generating function is given by

$\displaystyle\sum_{l\in\mathbb{N}}t^{l}\mathcal{L}_{l}(s)=\frac{1}{1-t}\exp\left(\frac{-ts}{1-t}\right).$
(5.2)

Let $x,y\in\mathbb{R}^{2}$. For $l\in\mathbb{N}$, we define $p_{l}$ as the
integral kernel of $P_{l}$,

$\displaystyle
p_{l}(x,y)\coloneqq\frac{B_{0}}{2\pi}\exp\left(-\frac{B_{0}}{4}\lVert
x-y\rVert^{2}+i\frac{B_{0}}{2}\langle x\mid
Jy\rangle\right)\mathcal{L}_{l}\left(B_{0}\lVert x-y\rVert^{2}/2\right).$
(5.3)

Furthermore, for $0<t<1$, we define the operator
$Q_{t}\coloneqq\sum_{l}t^{l}P_{l}$. Its integral kernel is given by

$\displaystyle q_{t}(x,y)\coloneqq$ $\displaystyle\sum_{l}t^{l}p_{l}(x,y)$
(5.4) $\displaystyle=$
$\displaystyle\frac{B_{0}}{2\pi(1-t)}\exp\left(-\frac{B_{0}}{4}\lVert
x-y\rVert^{2}+i\frac{B_{0}}{2}\langle x\mid
Jy\rangle-\frac{B_{0}t}{2-2t}\lVert x-y\rVert^{2}\right)$ (5.5)
$\displaystyle=$
$\displaystyle\frac{B_{0}}{2\pi(1-t)}\exp\left(-\frac{B_{0}(1+t)}{4(1-t)}\lVert
x-y\rVert^{2}+i\frac{B_{0}}{2}\langle x\mid Jy\rangle\right).$ (5.6)

We easily calculate

$\displaystyle\left(-i\nabla_{x}-\frac{B_{0}}{2}Jx\right)q_{t}(x,y)$
$\displaystyle=\left(\frac{iB_{0}(1+t)}{2(1-t)}(x-y)-\frac{B_{0}}{2}J(x-y)\right)q_{t}(x,y),$
(5.7)

and

$\displaystyle\left(-i\nabla_{x}-\frac{B_{0}}{2}Jx\right)^{\otimes
2}q_{t}(x,y)$ (5.8) $\displaystyle=$
$\displaystyle\left(\left(\frac{iB_{0}(1+t)}{2(1-t)}(x-y)-\frac{B_{0}}{2}J(x-y)\right)^{\otimes
2}+\left(\frac{B_{0}(1+t)}{2(1-t)}\begin{pmatrix}1&0\\\
0&1\end{pmatrix}-\frac{B_{0}}{2}J\right)\right)q_{t}(x,y).$ (5.9)

######  Lemma .

For any $j\in\mathbb{N}$, there are $C,a>0$, independent of $l,B_{0}$, such
that for any $x,y\in\mathbb{R}^{2}$

$\displaystyle\left\lVert(-i\nabla_{x}-A_{0}(x))^{\otimes
j}p_{l}(x,y)\right\rVert$ $\displaystyle\leq
B_{0}^{1+0.5j}Ca^{l}\exp\left(\frac{-B_{0}\lVert x-y\rVert^{2}}{8}\right).$
(5.10)

The norm on the left-hand side is the 2-norm on $\mathbb{C}^{2^{j}}$.

###### Proof.

Using the explicit formula for the Laguerre polynomials, for any $t\geq
0,j^{\prime}\in\mathbb{N},0<\delta<1$, we bound the $j^{\prime}$th
differential as follows:

$\displaystyle\lVert\mathcal{L}_{l}^{(j^{\prime})}(t)\rVert$
$\displaystyle\leq\sum_{k=0}^{l-j^{\prime}}2^{l}\frac{t^{k}}{k!}$ (5.11)
$\displaystyle=\sum_{k=0}^{l-j^{\prime}}\left(\frac{2}{\delta}\right)^{l}\frac{(\delta
t)^{k}}{k!}$ (5.12)
$\displaystyle\leq\left(\frac{2}{\delta}\right)^{l}\exp(\delta t).$ (5.13)

Each of the $j$ differential operators have to be resolved with the product
rule, where we apply the $-i\nabla_{x}$ to the polynomial, which is resolved
by chain rule, and $-i\nabla_{x}-A_{0}(x)$ to the exponential. This will
always be the exponential times a polynomial expression in $x-y$, taking
values in $\mathbb{C}^{2^{j}}$. This leads to the first bound, with a constant
$C$ depending only on $j$, as the dependency on $l$ is encoded entirely in the
polynomial $L_{l}$ and its differentials. Thus, we have

$\displaystyle\left\lVert(-i\nabla_{x}-A_{0}(x))^{\otimes
j}p_{l}(x,y)\right\rVert$ (5.14) $\displaystyle\leq$ $\displaystyle
C\sum_{j^{\prime}=0}^{j}\left\lVert\mathcal{L}_{l}^{(j^{\prime})}\left(\frac{B_{0}}{2}\lVert
x-y\rVert^{2}\right)\right\rVert\left(1+\sqrt{B_{0}}\lVert
x-y\rVert\right)^{j}B_{0}^{1+0.5j}\exp\left(-\frac{B_{0}}{4}\lVert
x-y\rVert^{2}\right).$ (5.15)

By setting $t=B_{0}\lVert x-y\rVert^{2}/2$ and $\delta=\frac{1}{8}$ in (5.13),
we can finally estimate

$\displaystyle\left\lVert(-i\nabla_{x}-A_{0}(x))^{\otimes
j}p_{l}(x,y)\right\rVert$ (5.16) $\displaystyle\leq$ $\displaystyle
CB_{0}^{1+0.5j}\sum_{j^{\prime}=0}^{j}\left\lVert\mathcal{L}_{l}^{(j^{\prime})}\left(\frac{B_{0}}{2}\lVert
x-y\rVert^{2}\right)\right\rVert\left(1+\sqrt{B_{0}}\lVert
x-y\rVert\right)^{j}\exp\left(-\frac{B_{0}}{4}\lVert x-y\rVert^{2}\right)$
(5.17) $\displaystyle\leq$ $\displaystyle
CB_{0}^{1+0.5j}16^{l}\exp\left(\frac{B_{0}}{16}\lVert
x-y\rVert^{2}\right)\left(1+\sqrt{B_{0}}\lVert
x-y\rVert\right)^{j}\exp\left(-\frac{B_{0}}{4}\lVert x-y\rVert^{2}\right)$
(5.18) $\displaystyle\leq$ $\displaystyle
CB_{0}^{1+0.5j}16^{l}\exp\left(\frac{B_{0}}{8}\lVert
x-y\rVert^{2}\right)\exp\left(-\frac{B_{0}}{4}\lVert x-y\rVert^{2}\right)$
(5.19) $\displaystyle\leq$ $\displaystyle
CB_{0}^{1+0.5j}16^{l}\exp\left(-\frac{B_{0}}{8}\lVert x-y\rVert^{2}\right).$
(5.20)

In the second to last step, we used that polynomials can be bounded by
exponentials. The constant $C$ changed, but still only depends on $j$. ∎

######  Lemma .

Let $I\subset\mathbb{N}$ be cofinite, $\zeta\in\mathbb{C},\zeta\not\in
B_{0}(2I+1)$ and $l_{0}\in\mathbb{N}$, such that
$l_{0}\geq\max(I^{\complement}\cup\\{\Re\frac{\zeta+B_{0}}{2B_{0}}\\})$. Then
we have the identity

$\displaystyle B_{0}M_{I,\zeta}=$
$\displaystyle\int_{0}^{1}t^{-\zeta/B_{0}}\left(Q_{t^{2}}-\sum_{l\leq
l_{0}}t^{2l}P_{l}\right)dt+\sum_{l\in I,l\leq
l_{0}}\frac{P_{l}}{(2l+1-\zeta/B_{0})}.$ (5.21)

###### Proof.

The idea of this proof is the formal identity

$\displaystyle\int_{0}^{1}\sum_{l\in I}t^{2l-\zeta/B_{0}}P_{l}dt$
$\displaystyle=\sum_{l\in I}\frac{1}{1+2l-\zeta/B_{0}}P_{l}.$ (5.22)

Now we need to establish the precise meaning of this identity. First, we note
that $t^{-\zeta/B_{0}}=\exp(-\zeta/B_{0}\ln(t))$ is well defined, as $t>0$. If
$\Re(\zeta)/B_{0}\geq 2l+1$, then the integral of the summands for $l$ will
not exist, which is the reason we introduced $l_{0}$. We bounded the real part
of $\zeta$ a little stronger than necessary to make the proof easier. Hence,
we have

$\displaystyle\int_{0}^{1}\sum_{l>l_{0}}t^{2l-\zeta/B_{0}}P_{l}dt=\sum_{l>l_{0}}\frac{1}{1+2l-\zeta/B_{0}}P_{l}.$
(5.23)

For any single $l>l_{0}$, the integral exists as a Bochner integral with
respect to the operator norm. Appendix A finishes the proof. ∎

We will deal with a few integral kernels that have a singularity at the
diagonal. To describe such a singularity, for any $s\in\mathbb{R}$, we
introduce

$\displaystyle
b_{s}\colon\mathbb{R}^{2}\to[0,\infty),\quad(x,y)\mapsto\begin{cases}-1_{D_{\frac{1}{\sqrt{B_{0}}}}(0)}(x-y)\ln(\sqrt{B_{0}}\lVert
x-y\rVert)&s=0,\\\ 1_{D_{\frac{1}{\sqrt{B_{0}}}}(0)}(x-y)\frac{1}{\lVert
x-y\rVert^{s}}&s\not=0.\end{cases}$ (5.24)

######  Lemma .

Let $I\subset\mathbb{N}$ be cofinite. Then there is a function
$F\in\mathrm{L}^{\infty}_{loc}(\mathbb{C}\setminus(2I+1))$, such that the
following pointwise upper bounds hold for all $x,y\in\mathbb{R}^{2},x\not=y$
and $\zeta\in\mathbb{C}\setminus{B_{0}}(2I+1)$:

$\displaystyle\lvert\operatorname{iker}M_{I,\zeta}(x,y)\rvert\leq$
$\displaystyle
F\left(\frac{\zeta}{B_{0}}\right)\left(b_{0}(x,y)+\exp\left(-\frac{B_{0}}{8}\lVert
x-y\rVert^{2}\right)\right),$ (5.25)
$\displaystyle\lVert\operatorname{iker}(-i\nabla-
A_{0})M_{I,\zeta}(x,y)\rVert\leq$ $\displaystyle
F\left(\frac{\zeta}{B_{0}}\right)\left(b_{1}(x,y)+\sqrt{B_{0}}\exp\left(-\frac{B_{0}}{8}\lVert
x-y\rVert^{2}\right)\right),$ (5.26)
$\displaystyle\lVert(-i\nabla_{x}-A_{0}(x))^{\otimes
2}\operatorname{iker}M_{I,\zeta}(x,y)\rVert\leq$ $\displaystyle
F\left(\frac{\zeta}{B_{0}}\right)\left(b_{2}(x,y)+{B_{0}}\exp\left(-\frac{B_{0}}{8}\lVert
x-y\rVert^{2}\right)\right).$ (5.27)

######  Remark.

The last inequality is structurally different, because the implied operator
$(-i\nabla-A_{0})^{\otimes 2}M_{I,\zeta}$ does not have a _nice_ integral
kernel. The differential of the integral kernel can still be considered but is
not $\mathrm{L}^{1}$ with respect to $y$ for any fixed $x$ and hence not a
_nice_ integral kernel. In general, this kernel does not fully describe the
operator.

###### Proof.

The set $I\subset\mathbb{N}$ is fixed throughout the proof.

For any $t\in[0,1),l\in\mathbb{N},j\in\\{0,1,2\\}$, we define

$\displaystyle q_{t,j}(x,y)$
$\displaystyle\coloneqq(-i\nabla_{x}-A_{0}(x))^{\otimes j}q_{t}(x,y),$ (5.28)
$\displaystyle p_{l,j}(x,y)$
$\displaystyle\coloneqq(-i\nabla_{x}-A_{0}(x))^{\otimes j}p_{l}(x,y).$ (5.29)

As $q_{t,j},p_{l,j}$ are _nice_ integral kernels, we can apply dominated
convergence and see that

$\displaystyle q_{t,j}(x,y)$
$\displaystyle=\operatorname{iker}\left((-i\nabla-A_{0})^{\otimes
j}Q_{t}\right)(x,y),$ (5.30) $\displaystyle p_{l,j}(x,y)$
$\displaystyle=\operatorname{iker}\left((-i\nabla-A_{0})^{\otimes
j}P_{l}\right)(x,y).$ (5.31)

We choose $l_{0}\in\mathbb{N}$ minimal, such that $(2l_{0}-1){B_{0}}>\Re\zeta$
and $l_{0}\geq\max(I^{\complement})$. Now we use the representation
established in Section 5. To prove, that for $j\in\\{0,1\\}$, the operators
have integral kernels, we want to use Appendix A. Hence, we only need to show,
that the following inequality holds, in order to finish the proof for $j=0,1$:

$\displaystyle\int_{0}^{1}\left\lVert
t^{-\zeta/{B_{0}}}\left(q_{t^{2},j}(x,y)-\sum_{l\leq
l_{0}}t^{2l}p_{l,j}(x,y)\right)\right\rVert dt+\sum_{l\in I,l\leq
l_{0}}\frac{\left\lVert p_{l,j}(x,y)\right\rVert}{(2l+1-\zeta/{B_{0}})}$
(5.32) $\displaystyle\leq$ $\displaystyle
B_{0}F\left(\frac{\zeta}{B_{0}}\right)\left(b_{j}(x,y)+{B_{0}}\exp\left(-\frac{B_{0}}{8}\lVert
x-y\rVert^{2}\right)\right).$ (5.33)

For $j=2$, however, we need to consider, that as the integrand is smooth on
$(0,1)$ and the summands at the end are smooth, we can try to exchange the
integral with the differential operator $(-i\nabla-A_{0})$. This will work, if
the absolute value of the differential is integrable, by dominated
convergence. Hence above integral bound also covers the case $j=2$ and we will
now proceed to bound all terms at the same time by choosing $j\in\\{0,1,2\\}$.
We want to use Section 5 to bound the first integral on the interval
$(0,t_{0})$ and the sums. Hence,

$\displaystyle\int_{0}^{t_{0}}\left\lVert
t^{-\zeta/{B_{0}}}\left(q_{t^{2},j}(x,y)-\sum_{l\leq
l_{0}}t^{2l}p_{l,j}(x,y)\right)\right\rVert dt$ (5.34) $\displaystyle\leq$
$\displaystyle\int_{0}^{t_{0}}\sum_{l>l_{0}}t^{2l-\Re(\zeta)/{B_{0}}}\lvert
p_{l,j}(x,y)\rvert dt$ (5.35) $\displaystyle\leq$
$\displaystyle\int_{0}^{t_{0}}\sum_{l>l_{0}}t^{2l-\Re(\zeta)/{B_{0}}}C{B_{0}}^{1+0.5j}a^{l}\exp\left(-\frac{B_{0}}{8}\lVert
x-y\rVert^{2}\right)dt$ (5.36) $\displaystyle=$ $\displaystyle
C{B_{0}}^{1+0.5j}\sum_{l>l_{0}}\frac{(t_{0}^{2}a)^{l}t_{0}}{(2l+1-\Re(\zeta)/{B_{0}})t_{0}^{\Re(\zeta)/{B_{0}}}}\exp\left(-\frac{B_{0}}{8}\lVert
x-y\rVert^{2}\right)$ (5.37) $\displaystyle\leq$ $\displaystyle
F(\zeta/{B_{0}}){B_{0}}^{1+0.5j}\exp\left(-\frac{B_{0}}{8}\lVert
x-y\rVert^{2}\right).$ (5.38)

The last step holds, if $t_{0}^{2}a<1$, so we fix such a $t_{0}$
now222Actually $a=16$, so we could choose for example $t_{0}=0.1$, but the
value is not relevant.. The function $F_{0}$ is in
$\mathrm{L}^{\infty}_{loc}(\mathbb{C}\setminus B_{0}(2I+1))$, as $l_{0}$ is
chosen locally bounded in $\zeta/{B_{0}}$. For fixed $l_{0}$ the function
$F_{0}$ is continuous. The next step is bounding the remaining finite sum
terms. Here, we will use, that $l\leq l_{0}$ and hence $a^{l}\leq C$. Thus,

$\displaystyle\int_{t_{0}}^{1}\left\lVert\sum_{l\leq
l_{0}}t^{2l-\zeta/{B_{0}}}p_{l,j}(x,y)\right\rVert dt+\sum_{l\in I,l\leq
l_{0}}\frac{\left\lVert p_{l,j}(x,y)\right\rVert}{(2l+1-\zeta/{B_{0}})}$
(5.39) $\displaystyle\leq$ $\displaystyle C{B_{0}}^{1+0.5j}\left(\sum_{l\leq
l_{0}}\left(\int_{t_{0}}^{1}t^{2l-\Re(\zeta)/{B_{0}}}dt\right)+\sum_{l\in
I,l\leq l_{0}}\left(\frac{1}{\lVert
2l+1-\zeta/{B_{0}}\rVert}\right)\right)\exp\left(-\frac{B_{0}}{8}\lVert
x-y\rVert^{2}\right)$ (5.40) $\displaystyle\leq$ $\displaystyle
F(\zeta/{B_{0}}){B_{0}}^{1+0.5j}\exp\left(-\frac{B_{0}}{8}\lVert
x-y\rVert^{2}\right).$ (5.41)

The function $F_{1}$ is in $\mathrm{L}^{\infty}_{loc}(\mathbb{C}\setminus
B_{0}(2I+1))$ by the same argumentation as $F_{0}$. We will now turn our
attention to the last remaining term. It is given by

$\displaystyle\int_{t_{0}}^{1}\left\lVert
q_{t^{2},j}(x,y)t^{-\zeta/B_{0}}\right\rVert dt.$ (5.42)

The integrand is given by (5.7) for $j=1$ and by (5.9) for $j=2$. Only in the
following lines, we denote by $j\mapsto\delta(j,2)$ the function, that is $1$,
if $j=2$ and $0$ otherwise. We introduce the parameter
$h\coloneqq\sqrt{B_{0}}\lVert x-y\rVert$ and estimate

$\displaystyle\int_{t_{0}}^{1}\left\lVert
t^{-\zeta/B_{0}}q_{t^{2},j}(x,y)\right\rVert dt$ (5.43) $\displaystyle\leq$
$\displaystyle\left(t_{0}^{-\Re(\zeta)/{B_{0}}}+1\right)\int_{t_{0}}^{1}\frac{C{B_{0}}}{1-t^{2}}\left(\left(\frac{\sqrt{B_{0}}h}{2(1-t^{2})}\right)^{j}+\frac{\delta(j,2){B_{0}}}{2(1-t^{2})}\right)\exp\left(-\frac{1+t^{2}}{4(1-t^{2})}h^{2}\right)dt$
(5.44) $\displaystyle\leq$
$\displaystyle\int_{0}^{1}\frac{F(\zeta/{B_{0}}){B_{0}}}{1-t}\left(\left(\frac{\sqrt{B_{0}}h}{2(1-t)}\right)^{j}+\frac{\delta(j,2){B_{0}}}{2(1-t)}\right)\exp\left(-\frac{1}{5(1-t)}h^{2}\right)dt.$
(5.45)

In the last step, we used the fact, that $\frac{1+t^{2}}{1+t}\geq
2\sqrt{2}-2>\frac{4}{5}$ to bound the factor in the exponential. The function
$F_{2}$ is just continuous on $\mathbb{C}$.

We want to do a change of variables to $s\coloneqq\frac{h^{2}}{5(1-t)}$. The
interval is changed to $(h^{2}/5,\infty)$ and the determinant is
$h^{2}/(5s^{2})$. In total we have

$\displaystyle\int_{t_{0}}^{1}\left\lVert
t^{-\zeta/B_{0}}q_{t^{2},j}(x,y)\right\rVert dt$ (5.46) $\displaystyle\leq$
$\displaystyle
F\left(\frac{\zeta}{B_{0}}\right){B_{0}}^{1+0.5j}\int_{h^{2}/5}^{\infty}\frac{s}{h^{2}}\left(\left(\frac{s}{h}\right)^{j}+\frac{\delta(j,2)s}{h^{2}}\right)\exp\left(-s\right)\frac{h^{2}}{s^{2}}ds$
(5.47) $\displaystyle\leq$ $\displaystyle
F\left(\frac{\zeta}{B_{0}}\right){B_{0}}^{1+0.5j}\int_{h^{2}/5}^{\infty}\frac{1}{s}\left(\left(\frac{s}{h}\right)^{j}+\frac{\delta(j,2)s}{h^{2}}\right)\exp\left(-s\right)ds\eqqcolon\Theta.$
(5.48)

If $h>1$, we can bound the integrand by $C\exp(-\frac{5}{8}s)$. The reduction
in the exponent takes care of the factor $s$, that appears in the case $j=2$.
Negative powers of $h$ can be bounded by one. The integral can then be
resolved and we have

$\displaystyle\Theta\leq$ $\displaystyle
CF\left(\frac{\zeta}{B_{0}}\right){B_{0}}^{1+0.5j}\exp\left(-\frac{5}{8}\frac{h^{2}}{5}\right)$
(5.49) $\displaystyle=$ $\displaystyle
CF\left(\frac{\zeta}{B_{0}}\right){B_{0}}^{1+0.5j}\exp\left(-\frac{B_{0}}{8}\lVert
x-y\rVert^{2}\right).$ (5.50)

This is the desired upper bound.

If $h\leq 1$, $j>0$, we can set the lower interval limit to $0$ and get an
integrable function in $s$ multiplied by $h^{-j}$. This gives us

$\displaystyle\Theta\leq
CF\left(\frac{\zeta}{B_{0}}\right){B_{0}}^{1+0.5j}h^{-j}\leq$ $\displaystyle
CF_{2}\left(\frac{\zeta}{B_{0}}\right){B_{0}}b_{j}(x,y),$ (5.51)

which is the desired upper bound.

Finally, if $h\leq 1$, $j=0$, we get a constant from the integral starting at
$\frac{1}{5}$. For the integral up to $\frac{1}{5}$, we can bound the
integrand by $\frac{1}{s}$. Hence, the remaining integral is bounded by
$C(1-\ln(h^{2}))\leq C(1+b_{0}(x,y))$. Once again, this is the desired result.
∎

We need one very important bound, which will have multiple uses later.

######  Lemma .

Let $u_{1},u_{2},u_{3}\colon\mathbb{R}^{2}\to\mathbb{R}^{+}$ be functions,
such that $\ln\circ u_{j}$ is Lipschitz with Lipschitz constant $C_{lip}>0$.
Let $0\leq s_{1},s_{2}<2$ and $\lambda>0$ be real numbers. Then there is a
constant $C>0$, depending only on ${B_{0}},s_{1},s_{2},\lambda$ and $C_{lip}$,
such that for all $x,y\in\mathbb{R}^{2},x\not=y$ we have the estimate

$\displaystyle\int_{\mathbb{R}^{2}}\frac{\left(b_{s_{1}}(x,y)+\exp(-{B_{0}}\lambda\lVert
x-y\rVert^{2})\right)\left(b_{s_{2}}(y,z)+\exp(-{B_{0}}\lambda\lVert
y-z\rVert^{2})\right)}{u_{1}(x)u_{2}(y)u_{3}(z)}dy$ (5.52) $\displaystyle\leq$
$\displaystyle\frac{Cb_{s_{1}+s_{2}-2}(x,z)+C\exp\left(-\frac{{B_{0}}\lambda}{3}\lVert
x-z\rVert^{2}\right)}{u_{1}(x)u_{2}(x)u_{3}(x)}\,.$ (5.53)

If $1/(u_{1}u_{2}u_{3})\in\mathrm{L}^{2}(\mathbb{R}^{2})$ and $s_{1}+s_{2}<3$,
then the integral kernel is Hilbert–Schmidt.

This is to be used together with Section 5 with $\lambda=\frac{1}{8}$. The
general $\lambda$ is included to be able to chain more resolvents inductively.
As all summands in the integral are positive, we may assume that they have the
same constants in front.

###### Proof.

We first need two minor results. Let $a,b\in\mathbb{R}^{2},j\in\\{1,2,3\\}$.
Then for any $\delta>0$, we have

$\displaystyle\frac{u_{j}(a)}{u_{j}(b)}=$ $\displaystyle\exp\left(\ln\circ
u_{j}(a)-\ln\circ u_{j}(b)\right)$ (5.54) $\displaystyle\leq$
$\displaystyle\exp\left(C_{lip}\lVert a-b\rVert\right)$ (5.55)
$\displaystyle\leq$ $\displaystyle\exp\left(\delta\lVert
a-b\rVert^{2}+\frac{C_{lip}^{2}}{4\delta}\right)$ (5.56) $\displaystyle=$
$\displaystyle C(C_{lip},\delta)\exp\left(\delta\lVert a-b\rVert^{2}\right).$
(5.57)

We used the Young inequality. Furthermore (for any $x,y,z\in\mathbb{R}^{2}$)
we have the identity

$\displaystyle\lVert x-y\rVert^{2}+\lVert y-z\rVert^{2}=\frac{1}{2}\lVert
x-z\rVert^{2}+2\left\lVert y-\frac{x+z}{2}\right\rVert^{2}.$ (5.58)

We write $R\coloneqq\frac{1}{\sqrt{B_{0}}}$. Let us begin with the left-hand
side of (5.53) and just write out most of the Hölder estimates. Hence,

$\displaystyle LHS\leq$ $\displaystyle\frac{C}{u_{1}(x)u_{3}(z)}$ (5.59)
$\displaystyle\Bigg{(}\int_{D_{R}(x)}b_{s_{1}}(x,y)b_{s_{2}}(y,z)dy\left\lVert\frac{1}{u_{2}({\makebox[4.30554pt]{{$\cdot$}}})}\right\rVert_{\mathrm{L}^{\infty}\left(D_{R}(x)\right)}$
(5.60) $\displaystyle\quad+\exp\left(-\frac{{B_{0}}\lambda}{2}\lVert
x-z\rVert^{2}\right)\int_{\mathbb{R}^{2}}\frac{\exp\left(-2{B_{0}}\lambda\left\lVert
y-\frac{x+z}{2}\right\rVert^{2}\right)}{u_{2}(y)}dy$ (5.61)
$\displaystyle\quad+\left\lVert
b_{s_{1}}(x,{\makebox[4.30554pt]{{$\cdot$}}})\right\rVert_{\mathrm{L}^{1}\left(D_{R}(x)\right)}\left\lVert\exp\left(-{B_{0}}\lambda\lVert{\makebox[4.30554pt]{{$\cdot$}}}-z\rVert^{2}\right)\right\rVert_{\mathrm{L}^{\infty}\left(D_{R}(x)\right)}\left\lVert\frac{1}{u_{2}({\makebox[4.30554pt]{{$\cdot$}}})}\right\rVert_{\mathrm{L}^{\infty}\left(D_{R}(x)\right)}$
(5.62) $\displaystyle\quad+\left\lVert
b_{s_{2}}({\makebox[4.30554pt]{{$\cdot$}}},z)\right\rVert_{\mathrm{L}^{1}\left(D_{R}(z)\right)}\left\lVert\exp\left(-{B_{0}}\lambda\lVert
x-{\makebox[4.30554pt]{{$\cdot$}}}\rVert^{2}\right)\right\rVert_{\mathrm{L}^{\infty}\left(D_{R}(z)\right)}\left\lVert\frac{1}{u_{2}({\makebox[4.30554pt]{{$\cdot$}}})}\right\rVert_{\mathrm{L}^{\infty}\left(D_{R}(z)\right)}\Bigg{)}.$
(5.63)

The $\mathrm{L}^{\infty}$ norms of the non-exponential terms can be bounded by
a constant times the function evaluated at the centre, where the constant is
given by (5.55), using $a$ as the centre of the ball and $b$ as any point in
the ball. For the $\mathrm{L}^{\infty}$ norms of the exponential terms, we use
Appendix A with $x_{0}\coloneqq y-z$. We are left to estimate the four
$\mathrm{L}^{1}$ norms, some of which are written as integrals. The last two
$\mathrm{L}^{1}$ norms can be bounded by a constant and that is sufficient.
For the exponential integral, we first use (5.57) with $\delta=B_{0}$ to
replace the $u_{2}(y)$ in the denominator by $u_{2}((x+z)/2)$, getting a
different Gaussian in the numerator, and then we can just bound its integral.
With all of these, we get

$\displaystyle LHS\leq$ $\displaystyle\frac{C}{u_{1}(x)u_{3}(z)}\times$ (5.64)
$\displaystyle\Bigg{(}\int_{D_{R}(x)}b_{s_{1}}(x,y)b_{s_{2}}(y,z)dy\frac{1}{u_{2}(x)}+\frac{\exp\left(-{{B_{0}}\lambda}/2\lVert
x-z\rVert^{2}\right)}{u_{2}((x+z)/2)}$ (5.65)
$\displaystyle\quad+\exp\left(-\frac{{B_{0}}\lambda}{2}\lVert
x-z\rVert^{2}\right)\frac{1}{u_{2}(x)}+\exp\left(-\frac{{B_{0}}\lambda}{2}\lVert
x-z\rVert^{2}\right)\frac{1}{u_{2}(z)}\Bigg{)}.$ (5.66)

If we apply (5.57) again, we can get the desired bound for the last three
summands. So, we only need to get the same bound for the first summand. If
$\lVert x-z\rVert>2R$, the first summand vanishes. Otherwise, the term
$1/u_{3}(z)$ can be bounded by $C/u_{3}(x)$ by (5.57). In the case
$2R\geq\lVert x-z\rVert\geq R/2$, we just bound the integral by a constant
depending on $R$, which can then be bounded by a constant times the Gaussian.
We are left to consider the case $\lVert x-z\rVert<R/2$. So, we are left to
bound the integral

$\displaystyle\int_{D_{R}(x)}b_{s_{1}}(x,y)b_{s_{2}}(y,z)dy.$ (5.67)

We have $b_{s}(x,{\makebox[4.30554pt]{{$\cdot$}}})\in\mathrm{L}^{p}$ for any
$1\leq p<2/s$ and $b_{s}$ is symmetric in $x,y$. Hence, if $s_{1}+s_{2}<2$, we
can bound this by a constant (independent of $x,z$) using Hölder. This can
then by bounded by the Gaussian, as $\lVert x-z\rVert\leq 2R$. We are left
with the case $s_{1}+s_{2}\geq 2$, where we want to bound the integral by
$b_{s_{1}+s_{2}-2}(x,z)+C$. As $s_{1},s_{2}<2$, we have $s_{1},s_{2}>0$. Let
$e_{1}\in\mathbb{R}^{2}$ be the standard unit vector and let
$D_{r_{1},r_{2}}(0)$ be the annulus between the two radii $r_{1}\leq r_{2}$.
Then we have

$\displaystyle\int_{D_{R}(x)}b_{s_{1}}(x,y)b_{s_{2}}(y,z)dy$ (5.68)
$\displaystyle\leq$ $\displaystyle\int_{D_{R}(x)}\frac{1}{\lVert
x-y\rVert^{s_{1}}\lVert y-z\rVert^{s_{2}}}dy$ (5.69) $\displaystyle=$
$\displaystyle\int_{D_{R}(0)}\frac{1}{\lVert y\rVert^{s_{1}}\lVert
y-(z-x)\rVert^{s_{2}}}dy$ (5.70) $\displaystyle=$
$\displaystyle\int_{D_{R\lVert x-z\rVert^{-1}}(0)}\frac{\lVert
x-z\rVert^{2-s_{1}-s_{2}}}{\lVert y\rVert^{s_{1}}\lVert
y-e_{1}\rVert^{s_{2}}}dy$ (5.71) $\displaystyle\leq$
$\displaystyle\frac{\int_{D_{2}(0)}\lVert y\rVert^{-s_{1}}\lVert
y-e_{1}\rVert^{-s_{2}}dy}{\lVert x-z\rVert^{s_{1}+s_{2}-2}}+\int_{D_{2,R\lVert
x-z\rVert^{-1}}(0)}\frac{\lVert x-z\rVert^{2-s_{1}-s_{2}}}{\lVert
y\rVert^{s_{1}}\lVert y-e_{1}\rVert^{s_{2}}}dy$ (5.72) $\displaystyle\leq$
$\displaystyle\frac{C}{\lVert x-z\rVert^{s_{1}+s_{2}-2}}+\int_{D_{2,R\lVert
x-z\rVert^{-1}}(0)}\frac{C\lVert x-z\rVert^{2-s_{1}-s_{2}}}{\lVert
y\rVert^{s_{1}+s_{2}}}dy$ (5.73) $\displaystyle\leq$ $\displaystyle
Cb_{s_{1}+s_{2}-2}(x,z).$ (5.74)

In the final step, we have to consider the case $s_{1}+s_{2}=2$ separately. In
this case, the integral at the end yields the term $b_{0}(x,z)$ up to a
constant. In the case $s_{1}+s_{2}>2$, the integral over $\lVert
y\rVert^{-s_{1}-s_{2}}$ can be bounded by a constant, independent of $x,z$ and
we are left with the correct singularity at the diagonal. This finishes the
proof of the upper bound.

If $1/(u_{1}u_{2}u_{3})\in\mathrm{L}^{2}$ and $s_{1}+s_{2}<3$, we get

$\displaystyle
C\int_{\mathbb{R}^{2}}dx\int_{\mathbb{R}^{2}}dz\left(\frac{b_{s_{1}+s_{2}-2}(x,z)+\exp\left(-\frac{{B_{0}}\lambda}{3}\lVert
x-z\rVert^{2}\right)}{u_{1}(x)u_{2}(x)u_{3}(x)}\right)^{2}$ (5.75)
$\displaystyle=$ $\displaystyle
C\int_{\mathbb{R}^{2}}dx\int_{\mathbb{R}^{2}}d(x-z)\left(\frac{b_{s_{1}+s_{2}-2}(x,z)+\exp\left(-\frac{{B_{0}}\lambda}{3}\lVert
x-z\rVert^{2}\right)}{u_{1}(x)u_{2}(x)u_{3}(x)}\right)^{2}$ (5.76)
$\displaystyle\leq$
$\displaystyle\int_{\mathbb{R}^{2}}dx\left(\frac{C}{\left(u_{1}(x)u_{2}(x)u_{3}(x)\right)^{2}}\right)\leq
C.$ (5.77)

Hence the integral kernel is Hilbert–Schmidt. ∎

######  Corollary .

Let $n\in\mathbb{N}$ and for any $0\leq i\leq n$, let there be an operator
$K_{i}$ with integral kernel $k_{i}$ on $\mathrm{L}^{2}(\mathbb{R}^{2})$, log-
Lipschitz functions $u_{i},v_{i}\colon\mathbb{R}^{2}\to\mathbb{R}^{+}$,
$\lambda_{i}>0$, and $0\leq s_{i}<2$. Assume the integral kernels $k_{i}$
satisfy the upper bound

$\displaystyle u_{i}(x)\lvert k_{i}(x,y)\rvert v_{i}(y)\leq
Cb_{s_{i}}(x,y)+C\exp\left(-\lambda_{i}\lVert x-y\rVert^{2}\right),$ (5.78)

for any $x\not=y$. Define $K\coloneqq\prod_{i=0}^{n}K_{i}$ and let

$\displaystyle s=-2n+\sum_{i=0}^{n}s_{i}.$ (5.79)

Then $K$ has an integral kernel $k$ and there are $\lambda>0,C>0$, such that
for any $x\not=y$, we have the inequalities

$\displaystyle\lvert
k(x,y)\rvert\leq\frac{Cb_{s}(x,y)+C\exp\left(-\lambda\lVert
x-y\rVert^{2}\right)}{\prod_{i=0}^{n}u_{i}(x)v_{i}(x)},$ (5.80)
$\displaystyle\lvert
k(x,y)\rvert\leq\frac{Cb_{s}(x,y)+C\exp\left(-\lambda\lVert
x-y\rVert^{2}\right)}{\prod_{i=0}^{n}u_{i}(y)v_{i}(y)}.$ (5.81)

For $s<0$, we can replace $b_{s}$ by $0$ in (5.80) and (5.81), as $b_{s}$ is
bounded and can be absorbed in the Gaussian.

###### Proof.

The case $n=0$ follows by (5.57). We continue with the case $n=1$. By Section
5, we only have to show that $K_{0}K_{1}$ has is an integral operator and that
for any $x,z\in\mathbb{R}^{2}$ with $x\neq z$, we have

$\displaystyle\operatorname{iker}K_{0}K_{1}(x,z)=\int_{\mathbb{R}^{2}}dy\operatorname{iker}K_{0}(x,y)\operatorname{iker}K_{1}(y,z).$
(5.82)

To do so, it is sufficient to find a function space $\mathcal{Y}\supset
C^{0}_{c}(\mathbb{R}^{2})$, on which $K_{0}$ and $K_{1}$ are continuous. We
claim the topological vector space

$\displaystyle\mathcal{Y}\coloneqq\bigcap_{\lambda\in\mathbb{R}}\left\\{f({\cdot})\exp(\lambda\lVert{\cdot}\rVert)\in\mathrm{L}^{\infty}(\mathbb{R}^{2})\right\\}$
(5.83)

does the trick.

We observe that any log-Lipschitz function
$u\colon\mathbb{R}^{2}\to\mathbb{R}^{+}$ with log-Lipschitz constant $C_{Lip}$
satisfies for any $x\in\mathbb{R}^{2}$ that

$\displaystyle u(0)\exp(-C_{Lip}\lVert x\rVert)\leq u(x)\leq
u(0)\exp(C_{Lip}\lVert x\rVert).$ (5.84)

Hence, such a function defines a continuous multiplication operator on
$\mathcal{Y}$. By the assumption (5.78), the operators $K_{i}$ can each be
written as a product of two such multiplication operators and a nice integral
operator $K_{i}^{\prime}$ satisfying the kernel estimate

$\displaystyle\operatorname{iker}K_{i}^{\prime}(x,y)\leq
b_{s_{i}}(x,y)+\exp\left(-\lambda_{i}\lVert x-y\rVert^{2}\right)$ (5.85)

for any $x,y\in\mathbb{R}^{2}$ with $x\neq y$. We observe that such an
integral operator is bounded on $\mathcal{Y}$. Now, by Fubini we can conclude
(5.82). This finishes the case $n=1$ with
$\lambda=\frac{1}{3}\min\\{\lambda_{1},\lambda_{2}\\}$.

As the resulting estimate for $K_{0}K_{1}$ is of the same form as the required
estimate in (5.78), the induction over $n$ follows trivially. ∎

##  6\. Proof of Theorem 4.1 and Theorem 4.3

We will first briefly summarize the approach for both proofs. We will start by
conjugating with the unitary operator $U_{x_{0}}$, as defined in Appendix A.
Then, we can show that the operators we produce this way are Hilbert–Schmidt
operators from $\mathrm{L}^{2}(\mathbb{R}^{2})$ to $H^{\gamma+2}([0,1]^{2})$
using the quasi isometry $D_{\gamma+2}$, that we have constructed in Appendix
A and some commutator relations to move the differentials around. The proofs
will conclude with Appendix A.

We denote by $t_{l}$ the integral kernel of $T_{l}$. By Section 5, we can only
apply one full differential in $x$ or $y$ to $t_{l}$, before we get a
function, that is not a _nice_ integral kernel anymore. However, the operator
$P_{l}$ has a smooth integral kernel, which is why we would like to move
differentials over to it. We will see that we can apply two differentials
after $M_{I,\zeta}$ and still remain with a bounded operators, that is (in
general) not an integral operator in Section 6.

We also still need to prove Section 4. However, it is convenient to prove a
more general integral kernel bound along with it. For that, we need to
introduce some new notation.

For any $x_{0}\in\mathbb{R}^{2}$, $j\in\mathbb{N}$, we introduce the three
multiplication operators, which are given for any $x\in\mathbb{R}^{2}$ by

$\displaystyle B_{\varepsilon,x_{0}}^{(j)}(x)$
$\displaystyle\coloneqq(-i)^{j}\left(\left(\prod_{h=1}^{j}\partial_{\theta_{h}}\right)B_{\varepsilon}(x+x_{0})\right)_{\theta\in\\{1,2\\}^{j}},$
(6.1) $\displaystyle A_{\varepsilon,x_{0}}^{(j)}(x)$ $\displaystyle\coloneqq
B_{\varepsilon}^{(j)}({\makebox[4.30554pt]{{$\cdot$}}}+x_{0})*\frac{J{\makebox[4.30554pt]{{$\cdot$}}}}{2\pi\lVert{\makebox[4.30554pt]{{$\cdot$}}}\rVert^{2}}(x),$
(6.2) $\displaystyle W^{(j)}_{\varepsilon,x_{0}}(x)$
$\displaystyle\coloneqq(-i)^{j}\left(\left(\prod_{h=1}^{j}\partial_{\theta_{h}}\right)W_{\varepsilon}(x+x_{0})\right)_{\theta\in\\{1,2\\}^{j}},$
(6.3) $\displaystyle H_{\varepsilon,x_{0}}^{(j)}$ $\displaystyle\coloneqq
A^{(j)}_{\varepsilon,x_{0}}\cdot\left(-i\nabla-
A_{0}\right)+W^{(j)}_{\varepsilon,x_{0}}.$ (6.4)

The last equation defines a non-multiplication operator. [We have not defined
$A_{x_{0}}$ and $H_{x_{0}}$, as this may lead to confusion with $A_{0}$ and
$H_{0}$, if we set $x_{0}=0$.] The scalar product in the definition of
$H^{(j)}_{\varepsilon,x_{0}}$ reduces the final component of
$A^{(j)}_{\varepsilon,x_{0}}$, which originates from the convolution with the
$\mathbb{R}^{2}$ valued function
$\frac{J{\cdot}}{2\pi\lVert{\cdot\rVert^{2}}}$. We will write
$X_{\varepsilon,x_{0}}$ for $X^{(0)}_{\varepsilon,x_{0}}$ for
$X\in\\{A,W,B\\}$.

We observe that by Section 3 for $f=B_{\varepsilon}^{(j)}$, we have

$\displaystyle A^{(j)}_{\varepsilon,x_{0}}\cdot\left(-i\nabla-
A_{0}\right)=\left(-i\nabla-A_{0}\right)\cdot A^{(j)}_{\varepsilon,x_{0}},$
(6.5)

where the scalar product on both sides reduces the final component of
$A^{(j)}_{\varepsilon,x_{0}}$, which originates from the convolution with the
$\mathbb{R}^{2}$ valued function
$\frac{J{\cdot}}{2\pi\lVert{\cdot\rVert^{2}}}$, as above. Hence, we have, with
the same scalar product,

$\displaystyle H_{\varepsilon,x_{0}}^{(j)}$ $\displaystyle=\left(-i\nabla-
A_{0}\right)\cdot A^{(j)}_{\varepsilon,x_{0}}+W^{(j)}_{\varepsilon,x_{0}}.$
(6.6)

The idea behind these definitions is, as we hinted at in the introduction to
this section, that by conjugating with the unitary operator
$U_{x_{0}}\colon\mathrm{L}^{2}(\mathbb{R}^{2})\to\mathrm{L}^{2}(\mathbb{R}^{2})$,
as defined in Appendix A, we observe the identity

$\displaystyle\left\lVert
1_{[0,1]^{2}-x_{0}}(T_{l}H_{\varepsilon})^{m}P_{l}(H_{\varepsilon}T_{l})^{k-m}1_{D^{\complement}_{R}(x_{0})}\right\rVert_{p}=$
$\displaystyle\left\lVert
1_{[0,1]^{2}}(T_{l}H_{\varepsilon,x_{0}})^{m}P_{l}(H_{\varepsilon,x_{0}}T_{l})^{k-m}1_{D^{\complement}_{R}(0)}\right\rVert_{p},$
(6.7)

as the $p$-Schatten norm is unitarily invariant. Something similar applies for
the proof of Theorem 4.3.

It is now time to prove Section 4. However, as we will need a more general
statement, we will prove that instead.

######  Lemma .

Let $\gamma\in\mathbb{N}$, $V_{\varepsilon},B_{\varepsilon}$ be
$(\gamma,\varepsilon)$ tame, $I\subset\mathbb{N}$ cofinite, and let
$\zeta\in\mathbb{C}\setminus(2I+1){B_{0}}$. Furthermore, let
$x_{0}\in\mathbb{R}^{2}$ and $\mathbb{N}\ni d\leq\gamma$. Then there is a
function $F\in L^{\infty}_{loc}(\mathbb{C}\setminus(2I+1){B_{0}})$ and a real
number $\lambda>0$, such that for any $x,y\in\mathbb{R}^{2}$ with $x\neq y$,
we have the upper bound

$\displaystyle\left\lVert\operatorname{iker}H_{\varepsilon,x_{0}}^{(d)}M_{I,\zeta}(x,y)\right\rVert\leq
F(\zeta)\frac{b_{1}(x,y)+\exp(-\lambda\lVert x-y\rVert^{2})}{(1+\lVert
x+x_{0}\rVert)^{\varepsilon}}.$ (6.8)

In particular, this is a nice integral kernel and the operator norm of
$H_{\varepsilon,x_{0}}^{(d)}M_{I,\zeta}$ is bounded independently of $x_{0}$.
Furthermore, we have the estimate

$\displaystyle\left\lVert H_{\varepsilon}M_{I,\zeta}\right\rVert_{4n_{0}}\leq
F(\zeta),$ (6.9)

where $n_{0}$ is the smallest integer such that $2n_{0}\varepsilon>1$.

This lemma generalizes Section 4. The operators $H_{\varepsilon,x_{0}}^{(d)}$
and $M_{I,\zeta}$ have been defined in (6.4) and (4.7). The function $b_{s}$
has been defined in (5.24).

###### Proof.

We can estimate pointwise for $x\not=y$, using the assumption that
$B_{\varepsilon},V_{\varepsilon}$ are $(\gamma,\varepsilon)$ tame, Section 3,
and Section 5. Thus, we have

$\displaystyle\left\lVert\operatorname{iker}A_{\varepsilon,x_{0}}^{(d)}(-i\nabla-
A_{0})M_{I,\zeta}(x,y)\right\rVert$ (6.10) $\displaystyle\leq$
$\displaystyle\frac{C}{(1+\lVert
x+x_{0}\rVert)^{\varepsilon}}\left\lVert\operatorname{iker}(-i\nabla-
A_{0})M_{I,\zeta}(x,y)\right\rVert$ (6.11) $\displaystyle\leq$ $\displaystyle
F\left(\frac{\zeta}{B_{0}}\right)\frac{b_{1}(x,y)+\sqrt{B_{0}}\exp\left(-\frac{B_{0}}{8}\lVert
x-y\rVert^{2}\right)}{(1+\lVert x+x_{0}\rVert)^{\varepsilon}}.$ (6.12)

And now for the other part, we observe that
$W_{\varepsilon}=A_{\varepsilon}^{2}+V_{\varepsilon}\in
W^{\gamma,\infty}_{(\varepsilon)}(\mathbb{R}^{2},\mathbb{R})$ and can then use
Section 5 to see

$\displaystyle\left\lVert\operatorname{iker}W_{\varepsilon,x_{0}}^{(d)}M_{I,\zeta}(x,y)\right\rVert$
(6.13) $\displaystyle\leq$ $\displaystyle\frac{C}{(1+\lVert
x+x_{0}\rVert)^{\varepsilon}}\lVert\operatorname{iker}M_{I,\zeta}(x,y)\rVert$
(6.14) $\displaystyle\leq$ $\displaystyle
F\left(\frac{\zeta}{B_{0}}\right)\frac{b_{0}(x,y)+\exp\left(-\frac{B_{0}}{8}\lVert
x-y\rVert^{2}\right)}{(1+\lVert x+x_{0}\rVert)^{\varepsilon}}$ (6.15)
$\displaystyle\leq$ $\displaystyle
F\left(\frac{\zeta}{B_{0}}\right)\frac{b_{1}(x,y)+\sqrt{B_{0}}\exp\left(-\frac{B_{0}}{8}\lVert
x-y\rVert^{2}\right)}{(1+\lVert x+x_{0}\rVert)^{\varepsilon}}.$ (6.16)

In the last step we used $b_{0}\leq Cb_{1}$ and $1=C\sqrt{B_{0}}$. This shows
the first claim.

We use properties we denoted as powers and Hilbert–Schmidt kernel of the
$p$-Schatten norms. Hence the $4n_{0}$-Schatten norm of $T$ can be calculated
as the $4n_{0}$th root of the square integral of the integral kernel of
$(TT^{*})^{n_{0}}$. We note, that $u(x)\coloneqq(1+\lVert
x\rVert)^{\varepsilon}$ is log-Lipschitz. We want to use Section 5. Hence, we
define for $0\leq i\leq 2n_{0}-1$

$\displaystyle K_{i}\coloneqq\begin{cases}H_{\varepsilon}M_{I,\zeta}&i\text{
even},\\\ \left(H_{\varepsilon}M_{I,\zeta}\right)^{*}&i\text{
odd}.\end{cases}$ (6.17)

For even $i$, we choose $u_{i}(x)=u(x),v_{i}(x)=1$ and for odd $i$, we choose
$v_{i}(x)=u(x),u_{i}(x)=1$. We always have $s_{i}=1$. Now we can apply Section
5 and get for any $x\not=y$ that

$\displaystyle\left\lvert\operatorname{iker}\left(\left(H_{\varepsilon}M_{I,\zeta}\right)\left(H_{\varepsilon}M_{I,\zeta}\right)^{*}\right)^{n_{0}}(x,y)\right\rvert$
(6.18) $\displaystyle\leq$ $\displaystyle
F\left(\zeta\right)\frac{b_{0}(x,y)+\exp\left(-\lambda{B_{0}}\lVert
x-y\rVert^{2}\right)}{(1+\lVert x\rVert)^{2n_{0}\varepsilon}}.$ (6.19)

The function $F$ is in
$\mathrm{L}^{\infty}_{loc}(\mathbb{C}\setminus(2I+1){B_{0}})$. This integral
kernel is in $\mathrm{L}^{2}$, as $2n_{0}\varepsilon>1$. The $b_{0}$ term only
appears for $n_{0}=1$, as for $n_{0}>1$, we get $s<0$, which corresponds to a
bounded $b_{s}$. ∎

We will now prove some useful methods to deal with the differentials we will
have to apply in order to use Appendix A. We will first see that, in a way,
$M_{I,\zeta}$ can take two differentials, and then we will see how to move
further differentials past $M_{I,\zeta}$ and $H_{\varepsilon,x_{0}}$.

Let $j_{1},j_{2}\in\\{1,2\\}$ and $h\in\\{\pm 1\\}$. Then we observe the
commutator relation

$\displaystyle\left[\left(-i\nabla-hA_{0}\right)_{j_{1}},\left(-i\nabla-
A_{0}\right)_{j_{2}}\right]$
$\displaystyle=i\frac{B_{0}}{2}\left([\nabla_{j_{1}},(JX)_{j_{2}}]+h[(JX)_{j_{1}},\nabla_{j_{2}}\right)$
(6.20)
$\displaystyle=i\frac{B_{0}}{2}\left(J_{j_{2}j_{1}}-hJ_{j_{1}j_{2}}\right).$
(6.21)

Here, $X$ refers to the multiplication operator associated to the identity on
$\mathbb{R}^{2}$. As the matrix $J$ is skew-symmetric, this states that the so
called covariant derivative $-i\nabla+A_{0}$ commutes with $-i\nabla-A_{0}$
and hence it commutes with the operators $H_{0},P_{l},M_{I,\zeta},T_{l}$ for
any $l\in\mathbb{N}$, cofinite subset $I\subset\mathbb{N}$ and any
$\zeta\in\mathbb{C}\setminus{B_{0}}(2I+1)$.

For $h=+1$, however, it motivates the definition of the annihilation and
construction operators. They are defined by

$\displaystyle a_{\pm}\coloneqq\frac{1}{\sqrt{B_{0}}}\left(\left(-i\nabla-
A_{0}\right)_{1}\pm i\left(-i\nabla-A_{0}\right)_{2}\right).$ (6.22)

Using (6.21) for $j_{1}=1,j_{2}=2,h=1$ and (2.3), we observe

$\displaystyle{B_{0}}a_{-}a_{+}=H_{0}+B_{0},\quad{B_{0}}a_{+}a_{-}=H_{0}-B_{0},\quad
a_{+}^{*}=a_{-}.$ (6.23)

This implies that $a_{-}$ is surjective and $a_{+}$ is injective. Let
$l\in\mathbb{N}$. Then we have

$\displaystyle(H_{0}+B_{0})a_{-}P_{l}={B_{0}}a_{-}a_{+}a_{-}P_{l}=a_{-}(H_{0}-B_{0})P_{l}=2lB_{0}a_{-}P_{l}.$
(6.24)

This states that $a_{-}P_{l}$ maps into the eigenspace of $H_{0}$ with
eigenvalue $(2l-1)B_{0}$, which is the image of $P_{l-1}$. If $l>0$, as
$a_{-}$ is surjective, it has to map the image of $P_{l}$ onto the image of
$P_{l-1}$. With an analogous computation for $a_{-}^{*}=a_{+}$, we arrive at

$\displaystyle P_{l-1}a_{-}=P_{l-1}a_{-}P_{l}=a_{-}P_{l}.$ (6.25)

We recall that the operator $M_{i,\zeta}$ has been defined in (4.7).

######  Lemma .

For any $I\subset\mathbb{N}$ cofinite and $j_{1},j_{2}\in\\{1,2\\}$, there is
an $F\in L^{\infty}\left(\mathbb{C}\setminus B_{0}(2I+1)\right)$, such that
for any $\zeta\in\mathbb{C}\setminus B_{0}(2I+1)$, we have the estimates

$\displaystyle\left\lVert\left(-i\nabla-
A_{0}\right)_{j_{1}}M_{I,\zeta}\left(-i\nabla-
A_{0}\right)_{j_{2}}\right\rVert_{\infty}\leq F(\zeta),$ (6.26)
$\displaystyle\left\lVert\left(-i\nabla-A_{0}\right)_{j_{1}}\left(-i\nabla-
A_{0}\right)_{j_{2}}M_{I,\zeta}\right\rVert_{\infty}\leq F(\zeta).$ (6.27)

###### Proof.

We will only prove the first claim, as the second follows completely
analogous. As both components of $(-i\nabla-A_{0})$ are linear combinations of
$a_{+},a_{-}$, it suffices to show that for any $h_{1},h_{2}\in\\{+,-\\}$, we
have the required estimate for the operator $a_{h_{1}}M_{I,\zeta}a_{h_{2}}$.
Let $l\in\mathbb{N}$. We consider the operator

$\displaystyle S_{l}\coloneqq a_{h_{1}}M_{I,\zeta}a_{h_{2}}P_{l}.$ (6.28)

For any $k\in\mathbb{N}$, we define $k++=k+1$ and $k+-=k-1$. Using (6.25), we
see

$\displaystyle a_{h_{1}}M_{I,\zeta}a_{h_{2}}P_{l}$
$\displaystyle=a_{h_{1}}M_{I,\zeta}P_{l+h_{2}}a_{h_{2}}$ (6.29)
$\displaystyle=\frac{1_{I}(l+h_{2})}{(2(l+h_{2})+1)B_{0}-\zeta}a_{h_{1}}P_{l+h_{2}}a_{h_{1}}$
(6.30)
$\displaystyle=\frac{1_{I}(l+h_{2})}{(2(l+h_{2})+1)B_{0}-\zeta}P_{l+h_{1}+h_{2}}a_{h_{1}}a_{h_{2}}.$
(6.31)

We use the convention $\frac{0}{0}=0$ in this proof. Hence, the family of
operators $S_{l}$ satisfy the conditions of Appendix A. So we just need to
bound the norm of $S_{l}$. Using (6.23), we observe that for any
$h\in\\{+,-\\}$ and $k\in\mathbb{N}$,

$\displaystyle\lVert a_{h}P_{k}\rVert^{2}=\lVert
P_{k}a_{h}^{*}a_{h}P_{k}\rVert=2k+1+h.$ (6.32)

Using (6.30), this leads to

$\displaystyle\lVert S_{l}\rVert$
$\displaystyle=\frac{1_{I}(l+h_{2})}{\lvert(2(l+h_{2})+1)B_{0}-\zeta\rvert}\lVert
a_{h_{1}}P_{l+h_{2}}\rVert\lVert P_{l+h_{2}}a_{h_{2}}P_{l}\rVert$ (6.33)
$\displaystyle=\frac{1_{I}(l+h_{2})}{\lvert(2(l+h_{2})+1)B_{0}-\zeta\rvert}\sqrt{(2(l+h_{2})+1+h_{1})(2l+1+h_{2})}\leq
F(\zeta).$ (6.34)

This finishes the proof. ∎

######  Lemma .

Let $\gamma,n\in\mathbb{N}$ and assume that
$(B_{\varepsilon},V_{\varepsilon})$ are $(\gamma,\varepsilon)$ tame. Let
$\mathbb{N}\ni\gamma^{\prime}\leq\gamma$. Then there is a set of matrices
$\left(N_{\mu}\in\operatorname{Lin}\left((\mathbb{C}^{2^{\gamma^{\prime}}},\mathbb{C}^{2^{\gamma^{\prime}}}\right)\right)_{\mu\in\mathbb{N}^{n+1},\lvert\mu\rvert=\gamma^{\prime}}$,
such that for any admissible $I,\zeta,x_{0}$, the identity

$\displaystyle\left(-i\nabla+A_{0}\right)^{\otimes\gamma^{\prime}}\left(M_{I,\zeta}H_{\varepsilon,x_{0}}\right)^{n}$
$\displaystyle=\sum_{\mu\in\mathbb{N}^{n+1},\lvert\mu\rvert=\gamma^{\prime}}N_{\mu}\left(\bigotimes_{j=1}^{n}M_{I,\zeta}H_{\varepsilon,x_{0}}^{(\mu_{j})}\right)\otimes\left(-i\nabla+A_{0}\right)^{\otimes\mu_{n+1}}$
(6.35)

holds in the sense that both operators agree as continuous operators from the
space $W^{\gamma^{\prime},\infty}_{(\infty)}(\mathbb{R}^{2},\mathbb{C})$ to
the space
$W^{0,\infty}_{(\infty)}\left(\mathbb{R}^{2},\mathbb{C}^{2^{\gamma^{\prime}}}\right)$.

###### Proof.

Let $d\in\mathbb{N},h\in\\{1,2\\}$ with $0\leq d<\gamma$. We recall (6.1) to
(6.4), and (6.6). We have $A_{\varepsilon,x_{0}}^{(d)}\in
W^{\gamma-d,\infty}_{(\varepsilon)}(\mathbb{R}^{2},\mathbb{C}^{2^{d+1}})$ and
$W_{\varepsilon,x_{0}}^{(d)}\in
W^{\gamma-d,\infty}_{(\varepsilon)}(\mathbb{R}^{2},\mathbb{C}^{2^{d}})$ by the
assumptions and Section 3. Hence, by the product rule, we have for any
$\mathbb{N}\ni\gamma^{\prime}\leq\gamma-d$ that the multiplication operators
$A_{\varepsilon,x_{0}}^{(d)}$ and $W_{\varepsilon,x_{0}}^{(d)}$ are continuous
operators from
$W^{\gamma^{\prime},\infty}_{(\infty)}(\mathbb{R}^{2},\mathbb{C})$ to the
spaces
$W^{\gamma^{\prime},\infty}_{(\infty)}(\mathbb{R}^{2},\mathbb{C}^{2^{d+1}})$
respectively
$W^{\gamma^{\prime},\infty}_{(\infty)}(\mathbb{R}^{2},\mathbb{C}^{2^{d}})$.
Furthermore, the operator $(-i\nabla+A_{0})$ obviously maps
$W^{\gamma^{\prime}+1,\infty}_{(\infty)}(\mathbb{R}^{2},\mathbb{C})$ to
$W^{\gamma^{\prime},\infty}_{(\infty)}(\mathbb{R}^{2},\mathbb{C}^{2})$
continuously for any $\gamma^{\prime}\in\mathbb{N}$. Finally, by Section 5 and
the fact that the covariant derivative $-i\nabla+A_{0}$ commutes with
$M_{I,\zeta}$ by (6.21), for any $\gamma^{\prime}\in\mathbb{N}$, the operators
$M_{I,\zeta}$ and $(-i\nabla-A_{0})M_{I,\zeta}$ are continuous from
$W^{\gamma^{\prime},\infty}_{(\infty)}(\mathbb{R}^{2},\mathbb{C})$ to the
spaces $W^{\gamma^{\prime},\infty}_{(\infty)}(\mathbb{R}^{2},\mathbb{C})$,
respectively
$W^{\gamma^{\prime},\infty}_{(\infty)}(\mathbb{R}^{2},\mathbb{C}^{2})$. These
statements guarantee that every composition of operators we consider is well-
defined in the claimed sense.

Now, by (6.21) and (6.6), we have

$\displaystyle\left(-i\nabla+A_{0}\right)_{h}M_{I,\zeta}H^{(d)}_{\varepsilon,x_{0}}$
(6.36) $\displaystyle=$ $\displaystyle M_{I,\zeta}\left(\left(-i\nabla-
A_{0}\right)\cdot\left(-i\nabla+A_{0}\right)_{h}A^{(d)}_{\varepsilon,x_{0}}+\left(-i\nabla+A_{0}\right)_{h}W^{(d)}_{\varepsilon,x_{0}}\right)$
(6.37) $\displaystyle=$ $\displaystyle M_{I,\zeta}\Big{(}\left(-i\nabla-
A_{0}\right)\cdot
A^{(d)}_{\varepsilon,x_{0}}\left(-i\nabla+A_{0}\right)_{h}-i\left(-i\nabla-
A_{0}\right)\cdot\partial_{h}A^{(d)}_{\varepsilon,x_{0}}$ (6.38)
$\displaystyle\phantom{M_{I,\zeta}\Big{(}}\quad+W^{(d)}_{\varepsilon,x_{0}}\left(-i\nabla+A_{0}\right)_{h}-i\partial_{h}W^{(d)}_{\varepsilon,x_{0}}\Big{)}$
(6.39) $\displaystyle=$ $\displaystyle
M_{I,\zeta}H^{(d)}_{\varepsilon,x_{0}}\left(-i\nabla+A_{0}\right)_{h}+M_{I,\zeta}\left(e_{h}\cdot
H^{(d+1)}_{\varepsilon,x_{0}}\right).$ (6.40)

The scalar product $e_{h}\cdot H^{(d+1)}_{\varepsilon,x_{0}}$ reduces the
first component of the tensor product
$\left(\mathbb{C}^{2}\right)^{\otimes(d+1)}$.

Let
$N^{\prime}_{d}\colon\mathbb{C}^{2^{d}}\otimes\mathbb{C}^{2}\to\mathbb{C}^{2}\otimes\mathbb{C}^{2^{d}}$
that swaps the tensor factors ($u\otimes v\mapsto v\otimes u$). Then we have

$\displaystyle\left(-i\nabla+A_{0}\right)\otimes
M_{I,\zeta}H^{(d)}_{\varepsilon,x_{0}}=N^{\prime}_{d}M_{I,\zeta}H^{(d)}_{\varepsilon,x_{0}}\otimes\left(-i\nabla+A_{0}\right)+M_{I,\zeta}H^{(d+1)}_{\varepsilon,x_{0}}.$
(6.41)

The case $n=0$ or $\gamma^{\prime}=0$ is tautological. The case
$n=\gamma^{\prime}=1$ follows, if we set $d=0$ above. Now we consider $n=1$
and the step $\gamma^{\prime}\mapsto\gamma^{\prime}+1\leq\gamma$,

$\displaystyle\left(-i\nabla+A_{0}\right)^{\otimes(\gamma^{\prime}+1)}M_{I,\zeta}H_{\varepsilon,x_{0}}$
(6.42) $\displaystyle=$
$\displaystyle\left(-i\nabla+A_{0}\right)\otimes\sum_{\mu\in\mathbb{N}^{2},\mu_{1}+\mu_{2}=\gamma^{\prime}}N_{\mu}M_{I,\zeta}H_{\varepsilon,x_{0}}^{(\mu_{1})}\otimes\left(-i\nabla+A_{0}\right)^{\otimes\mu_{2}}$
(6.43) $\displaystyle=$
$\displaystyle\sum_{\mu\in\mathbb{N}^{2},\mu_{1}+\mu_{2}=\gamma^{\prime}}\left(Id_{\mathbb{C}^{2}}\otimes
N_{\mu}\right)\left(-i\nabla+A_{0}\right)\otimes
M_{I,\zeta}H_{\varepsilon,x_{0}}^{(\mu_{1})}\otimes\left(-i\nabla+A_{0}\right)^{\otimes\mu_{2}}$
(6.44) $\displaystyle=$
$\displaystyle\sum_{\mu\in\mathbb{N}^{2},\mu_{1}+\mu_{2}=\gamma^{\prime}}\left(Id_{\mathbb{C}^{2}}\otimes
N_{\mu}\right)\left(N^{\prime}_{\mu_{1}}M_{I,\zeta}H_{\varepsilon,x_{0}}^{(\mu_{1})}\otimes\left(-i\nabla+A_{0}\right)+M_{I,\zeta}H_{\varepsilon,x_{0}}^{(\mu_{1}+1)}\right)\otimes\left(-i\nabla+A_{0}\right)^{\otimes\mu_{2}}$
(6.45) $\displaystyle=$
$\displaystyle\sum_{\mu\in\mathbb{N}^{2},\mu_{1}+\mu_{2}=\gamma^{\prime}+1}N_{\mu}M_{I,\zeta}H_{\varepsilon,x_{0}}^{(\mu_{1})}\otimes\left(-i\nabla+A_{0}\right)^{\otimes\mu_{2}}.$
(6.46)

In the last step, we used the inductive definition

$\displaystyle N_{(\mu_{1},\mu_{2})}\coloneqq\left(Id_{\mathbb{C}^{2}}\otimes
N_{(\mu_{1},\mu_{2}-1)}\right)\left(N^{\prime}_{\mu_{1}}\otimes
Id_{\mathbb{C}^{\mu_{2}}}\right)+\left(Id_{\mathbb{C}^{2}}\otimes
N_{(\mu_{1}-1,\mu_{2})}\right).$ (6.48)

To conclude the proof, we do an induction on $n$ over the statement of the
lemma. The idea is to use the induction hypothesis and then the case $n=1$. We
omit the details, as it works pretty similar to the induction on
$\gamma^{\prime}$. The only annoying part is creating a recursive description
for the $N_{\gamma}$s. But we have no use for such a description. ∎

We can now prove Theorem 4.3.

###### Proof of Theorem 4.3.

We begin by conjugating with the unitary operator $U_{x_{0}}$, that we have
defined in Appendix A. Hence, as the $p$-Schatten quasi norm is unitarily
equivalent, we have

$\displaystyle\left\lVert\int_{\Gamma}1_{[0,1]^{2}-x_{0}}\left(M_{\mathbb{N},\zeta}H_{\varepsilon}\right)^{n}\frac{1}{H-\zeta}\left(H_{\varepsilon}M_{\mathbb{N},\zeta}\right)^{n}1_{L\Lambda^{\complement}}d\zeta\right\rVert_{p}$
(6.49) $\displaystyle=$
$\displaystyle\left\lVert\int_{\Gamma}1_{[0,1]^{2}}\left(M_{\mathbb{N},\zeta}H_{\varepsilon,x_{0}}\right)^{n}U_{x_{0}}\frac{1}{H-\zeta}\left(H_{\varepsilon}M_{\mathbb{N},\zeta}\right)^{n}1_{L\Lambda^{\complement}}U_{x_{0}}^{-1}d\zeta\right\rVert_{p}$
(6.50)

Let $q$ satisfy $\frac{1}{q}+\frac{1}{2}=\frac{1}{p}$. As
$p>\frac{2}{\gamma+3}$, we have $q>\frac{2}{\gamma+2}$. Hence, we can apply
Appendix A with $\gamma+2$ and the property Hölder I (see Section 2) to get
the upper bound

$\displaystyle\eqref{final summand conjugated}\leq$ $\displaystyle
C\left\lVert\int_{\Gamma}\left(M_{\mathbb{N},\zeta}H_{\varepsilon,x_{0}}\right)^{n}U_{x_{0}}\frac{1}{H-\zeta}\left(H_{\varepsilon}M_{\mathbb{N},\zeta}\right)^{n}1_{L\Lambda^{\complement}}U_{x_{0}}^{-1}d\zeta\right\rVert_{S_{2}(\mathrm{L}^{2}(\mathbb{R}^{2}),H^{\gamma+2}([0,1]^{2}))}$
(6.51) $\displaystyle\leq$ $\displaystyle
C\int_{\Gamma}\left\lVert\left(M_{\mathbb{N},\zeta}H_{\varepsilon,x_{0}}\right)^{n}U_{x_{0}}\frac{1}{H-\zeta}\left(H_{\varepsilon}M_{\mathbb{N},\zeta}\right)^{n}1_{L\Lambda^{\complement}}U_{x_{0}}^{-1}\right\rVert_{S_{2}(\mathrm{L}^{2}(\mathbb{R}^{2}),H^{\gamma+2}([0,1]^{2}))}d\zeta.$
(6.52)

The last step relies on the fact that the Hilbert–Schmidt norm (2-Schatten
norm) is a norm and not just a quasi-norm. Now it suffices to bound the
integrand uniformly on the integration path. For this, we first use the quasi-
isometry $D_{\gamma+2}$ constructed in Appendix A. Hence, we have

$\displaystyle\left\lVert\left(M_{\mathbb{N},\zeta}H_{\varepsilon,x_{0}}\right)^{n}U_{x_{0}}\frac{1}{H-\zeta}\left(H_{\varepsilon}M_{\mathbb{N},\zeta}\right)^{n}1_{L\Lambda^{\complement}}U_{x_{0}}^{-1}\right\rVert_{S_{2}(\mathrm{L}^{2}(\mathbb{R}^{2}),H^{\gamma+2}([0,1]^{2}))}$
(6.53) $\displaystyle\leq$ $\displaystyle
C\sum_{\gamma^{\prime}=-2}^{\gamma}\left\lVert\left(-i\nabla+A_{0}\right)^{\otimes(\gamma^{\prime}+2)}\left(M_{\mathbb{N},\zeta}H_{\varepsilon,x_{0}}\right)^{n}U_{x_{0}}\frac{1}{H-\zeta}\left(H_{\varepsilon}M_{\mathbb{N},\zeta}\right)^{n}1_{L\Lambda^{\complement}}U_{x_{0}}^{-1}\right\rVert_{S_{2}(\mathrm{L}^{2}(\mathbb{R}^{2}),\mathrm{L}^{2}([0,1]^{2})}$
(6.54) $\displaystyle=$ $\displaystyle
C\sum_{\gamma^{\prime}=-2}^{\gamma}\left\lVert
1_{[0,1]^{2}}\left(-i\nabla+A_{0}\right)^{\otimes(\gamma^{\prime}+2)}\left(M_{\mathbb{N},\zeta}H_{\varepsilon,x_{0}}\right)^{n}U_{x_{0}}\frac{1}{H-\zeta}\left(H_{\varepsilon}M_{\mathbb{N},\zeta}\right)^{n}1_{L\Lambda^{\complement}}U_{x_{0}}^{-1}\right\rVert_{2}$
(6.55) $\displaystyle\leq$ $\displaystyle
C\sum_{\gamma^{\prime}=-2}^{\gamma}\left\lVert
1_{[0,1]^{2}}\left(-i\nabla+A_{0}\right)^{\otimes(\gamma^{\prime}+2)}\left(M_{\mathbb{N},\zeta}H_{\varepsilon,x_{0}}\right)^{n}\right\rVert_{\infty}\left\lVert
U_{x_{0}}\right\rVert_{\infty}\left\lVert\frac{1}{H-\zeta}\left(H_{\varepsilon}M_{\mathbb{N},\zeta}\right)^{n}1_{L\Lambda^{\complement}}\right\rVert_{2}\left\lVert
U_{x_{0}}^{-1}\right\rVert_{\infty}$ (6.56) $\displaystyle=$ $\displaystyle
C\sum_{\gamma^{\prime}=-2}^{\gamma}\left\lVert
1_{[0,1]^{2}}\left(-i\nabla+A_{0}\right)^{\otimes(\gamma^{\prime}+2)}\left(M_{\mathbb{N},\zeta}H_{\varepsilon,x_{0}}\right)^{n}\right\rVert_{\infty}\left\lVert\frac{1}{H-\zeta}\right\rVert_{\infty}\left\lVert\left(H_{\varepsilon}M_{\mathbb{N},\zeta}\right)^{n}1_{L\Lambda^{\complement}}\right\rVert_{2}$
(6.57)

The third step relies on applications of Hölder I (see Section 2). The last
step uses that $U_{x_{0}}$ is unitary on $\mathrm{L}^{2}(\mathbb{R}^{2})$ and
another application of Hölder I. The conjugation with $U_{x_{0}}$ was only
needed fo the first term. It does make a difference there, as $U_{x_{0}}$ is
not unitary on $H^{\gamma+2}([0,1]^{2})$ and does not commute with
$D_{\gamma+2}$.

We begin with the last factor in (6.57). As we are still free to choose
$n\in\mathbb{N}$, we can assume $n>2$. We use the kernel estimate in Section 6
and Section 5, similar to the proof of the second result of Section 6 to
arrive at the following estimate for any $x,y\in\mathbb{R}^{2}$:

$\displaystyle\left\lvert\operatorname{iker}\left(H_{\varepsilon}M_{\mathbb{N},\zeta}\right)^{n}(x,y)\right\rvert\leq
F(\zeta)\frac{\exp\left(-\lambda\lVert x-y\rVert^{2}\right)}{(1+\lvert
y\rVert)^{n\varepsilon}}.$ (6.58)

Now we let $n\varepsilon>1+\nu$. Then using the Hilbert–Schmidt kernel
identity, we have

$\displaystyle\left\lVert\left(H_{\varepsilon}M_{\mathbb{N},\zeta}\right)^{n}1_{L\Lambda^{\complement}}\right\rVert_{2}^{2}=$
$\displaystyle\int_{\mathbb{R}^{2}}dx\int_{L\Lambda^{\complement}}dy\left\lvert\operatorname{iker}\left(H_{\varepsilon}M_{\mathbb{N},\zeta}\right)^{n}(x,y)\right\rvert^{2}$
(6.59) $\displaystyle\leq$ $\displaystyle
F(\zeta)\int_{L\Lambda^{\complement}}dy\frac{1}{(1+\lVert y\rVert)^{2+2\nu}}$
(6.60) $\displaystyle\leq$ $\displaystyle F(\zeta)L^{-2\nu}$ (6.61)

In the second step, we use that the Gauss kernel is integrable over $x$, that
the integral is independent of $y$, and that $n\varepsilon>1+\nu$. The third
step uses that there is some $r>0$ such that $D_{r}(0)\subset\Lambda$ and that
$\nu>0$.

For the second factor in (6.57), we observe

$\displaystyle\left\lVert\frac{1}{H-\zeta}\right\rVert_{\infty}=\frac{1}{\operatorname{dist}(\zeta,\sigma(H))},$
(6.62)

which is bounded along the path $\Gamma$.

For the first factor in (6.57), we first consider the case
$\gamma^{\prime}\geq 0$. Here, we start by using Section 6 with the parameters
$\gamma^{\prime}$ and $2$. Hence, we have

$\displaystyle\left\lVert
1_{[0,1]^{2}}\left(-i\nabla+A_{0}\right)^{\otimes(\gamma^{\prime}+2)}\left(M_{\mathbb{N},\zeta}H_{\varepsilon,x_{0}}\right)^{n}\right\rVert_{\infty}$
(6.63) $\displaystyle=$ $\displaystyle\left\lVert
1_{[0,1]^{2}}\left(-i\nabla+A_{0}\right)^{\otimes
2}\otimes\\!\sum_{\begin{subarray}{c}\mu\in\mathbb{N}^{3},\\\
\lvert\mu\rvert=\gamma^{\prime}\end{subarray}}\\!N_{\mu}M_{\mathbb{N},\zeta}H_{\varepsilon,x_{0}}^{(\mu_{1})}\otimes
M_{\mathbb{N},\zeta}H_{\varepsilon,x_{0}}^{(\mu_{2})}\otimes\left(-i\nabla+A_{0}\right)^{\otimes\mu_{3}}\left(M_{\mathbb{N},\zeta}H_{\varepsilon,x_{0}}\right)^{n-2}\right\rVert_{\infty}$
(6.64) $\displaystyle\leq$ $\displaystyle
C\sup_{\mu_{1},\mu_{2},\mu_{3}\leq\gamma}\left\lVert
1_{[0,1]^{2}}\left(-i\nabla+A_{0}\right)^{\otimes 2}\otimes
M_{\mathbb{N},\zeta}\right\rVert_{\infty}\left\lVert
H_{\varepsilon,x_{0}}^{(\mu_{1})}\otimes
M_{\mathbb{N},\zeta}H_{\varepsilon,x_{0}}^{(\mu_{2})}\right\rVert_{\infty}$
(6.65)
$\displaystyle\quad\times\left\lVert\left(-i\nabla+A_{0}\right)^{\otimes\mu_{3}}\left(M_{\mathbb{N},\zeta}H_{\varepsilon,x_{0}}\right)^{n-2}\right\rVert_{\infty}.$
(6.66)

In the last step, we also used that $\mu_{j}\leq\gamma^{\prime}\leq\gamma$.
Now we need to estimate these three factors. We begin with the first one.

By the proof of Appendix A, we conclude that the map $D_{2}^{\prime}\colon
H^{2}([0,1]^{2})\to\mathrm{L}^{2}([0,1]^{2},\mathbb{C}^{7})$ given by
$u\mapsto\left((-i\nabla-A_{0})^{\otimes j}u\right)_{j=0}^{2}$ is a quasi-
isometry. Hence, as the operators $M_{\mathbb{N},\zeta}$ and $(-i\nabla-
A_{0})M_{\mathbb{N},\zeta}$ are bounded by Section 5 , and the operator
$(-i\nabla-A_{0})^{\otimes 2}M_{\mathbb{N},\zeta}$ is bounded by Section 6, we
have

$\displaystyle\left\lVert 1_{[0,1]^{2}}\left(-i\nabla+A_{0}\right)^{\otimes
2}M_{\mathbb{N},\zeta}\right\rVert_{\infty}\leq$ $\displaystyle C\left\lVert
M_{\mathbb{N},\zeta}\right\rVert_{S_{\infty}(\mathrm{L}^{2}(\mathbb{R}^{2}),H^{2}([0,1]^{2})}$
(6.67) $\displaystyle\leq$ $\displaystyle
C\sum_{j=0}^{2}\left\lVert\left(-i\nabla-A_{0}\right)^{\otimes
j}M_{\mathbb{N},\zeta}\right\rVert_{\infty}\leq F(\zeta).$ (6.68)

For any $\mathbb{N}\ni d\leq\gamma$, the multiplication operators
$A_{\varepsilon,x_{0}}^{(d)},W_{\varepsilon,x_{0}}^{(d)}$ are bounded
operators with a norm not depending on $x_{0}$. Furthermore, by Section 5, the
operators $M_{\mathbb{N},\zeta},(-i\nabla-A_{0})M_{\mathbb{N},\zeta}$ and
$M_{\mathbb{N},\zeta}(-i\nabla-A_{0})=\left((-i\nabla-
a_{0})M_{I,\overline{\zeta}}\right)^{*}$ are bounded, and the operator
$(-i\nabla-A_{0})\otimes M_{\mathbb{N},\zeta}(-i\nabla-A_{0})$ is bounded by
Section 6. Now, we use (6.4) and (6.6) to conclude

$\displaystyle\left\lVert H_{\varepsilon,x_{0}}^{(\mu_{1})}\otimes
M_{\mathbb{N},\zeta}H_{\varepsilon,x_{0}}^{(\mu_{2})}\right\rVert_{\infty}$
(6.69) $\displaystyle=$
$\displaystyle\left\lVert\left(A_{\varepsilon,x_{0}}^{(\mu_{1})}\cdot(-i\nabla-
A_{0})+W_{\varepsilon,x_{0}}^{(\mu_{1})}\right)\otimes
M_{\mathbb{N},\zeta}\left((-i\nabla-A_{0})\cdot
A_{\varepsilon,x_{0}}^{(\mu_{2})}+W_{\varepsilon,x_{0}}^{(\mu_{2})}\right)\right\rVert_{\infty}\leq
F(\zeta).$ (6.70)

We are left to estimate the expression in (6.66). We rename $\mu_{3}$ to $d$
and do an induction over $d$ for $0\leq d\leq\gamma$. Let $e\in\mathbb{N}$ be
minimal with $e\varepsilon\geq 1$. The claim of our induction is that for
$n\geq d(e+2)+3$, we have the estimate

$\displaystyle\left\lVert\left(-i\nabla+A_{0}\right)^{\otimes
d}\left(M_{\mathbb{N},\zeta}H_{\varepsilon,x_{0}}\right)^{n-2}\right\rVert_{\infty}\leq
F(\zeta)(1+\lVert x_{0}\rVert)^{d},$ (6.71)

for some $F\in\mathrm{L}^{\infty}_{loc}(\mathbb{C}\setminus\sigma(H))$
depending on $n,d$. The induction start at $d=0$ only uses that $\lVert
M_{I,\zeta}H_{\varepsilon,x_{0}}\rVert_{\infty}=\left\lVert\left(H_{\varepsilon,x_{0}}M_{I,\overline{\zeta}}\right)^{*}\right\rVert_{\infty}\leq
F(\zeta)$ by Section 6 and that the product of bounded operators is bounded.
For the step $d\to d+1\leq\gamma$, we first use Section 6 with the parameters
$d$ and $e+2$. Hence, we have

$\displaystyle\left\lVert\left(-i\nabla+A_{0}\right)^{\otimes(d+1)}\left(M_{\mathbb{N},\zeta}H_{\varepsilon,x_{0}}\right)^{n-2}\right\rVert_{\infty}$
(6.72) $\displaystyle=$
$\displaystyle\left\lVert\left(-i\nabla+A_{0}\right)\otimes\sum_{\mu\in\mathbb{N}^{e+3},\lvert\mu\rvert=d}N_{\mu}\bigotimes_{j=1}^{e+2}\left(M_{I,\zeta}H_{\varepsilon,x_{0}}^{(\mu_{j})}\right)\otimes\left(-i\nabla+A_{0}\right)^{\otimes\mu_{e+3}}\left(M_{\mathbb{N},\zeta}H_{\varepsilon,x_{0}}\right)^{n-4-e}\right\rVert_{\infty}$
(6.73) $\displaystyle\leq$ $\displaystyle C\sup_{\mu\in\mathbb{N}_{\leq
d}^{e+3}}\left\lVert\left(-i\nabla+A_{0}\right)\otimes\bigotimes_{j=1}^{e+2}\left(M_{I,\zeta}H_{\varepsilon,x_{0}}^{(\mu_{j})}\right)\right\rVert_{\infty}\left\lVert\left(-i\nabla+A_{0}\right)^{\otimes\mu_{e+3}}\left(M_{\mathbb{N},\zeta}H_{\varepsilon,x_{0}}\right)^{n-4-e}\right\rVert_{\infty}$
(6.74) $\displaystyle\leq$ $\displaystyle C\sup_{\mu\in\mathbb{N}_{\leq
d}^{e+2}}\left\lVert\left(-i\nabla+A_{0}\right)M_{\mathbb{N},\zeta}\otimes\bigotimes_{j=1}^{e}\left(H_{\varepsilon,x_{0}}^{(\mu_{j})}M_{I,\zeta}\right)\right\rVert_{\infty}\left\lVert
H_{\varepsilon,x_{0}}^{(\mu_{e+1})}\otimes
M_{\mathbb{N},\zeta}H_{\varepsilon,x_{0}}^{(\mu_{e+2})}\right\rVert_{\infty}F(\zeta)(1+\lVert
x_{0}\rVert)^{d}$ (6.75) $\displaystyle\leq$ $\displaystyle F(\zeta)(1+\lVert
x_{0}\rVert)^{d}\sup_{\mu\in\mathbb{N}_{\leq
d}^{e+2}}\left\lVert\left(-i\nabla+A_{0}\right)M_{\mathbb{N},\zeta}\otimes\bigotimes_{j=1}^{e}\left(H_{\varepsilon,x_{0}}^{(\mu_{j})}M_{\mathbb{N},\zeta}\right)\right\rVert_{\infty}.$
(6.76)

In the third step, we used the induction hypothesis and in the last step we
used (6.70). The remaining operator is just a product of integral operators.
The kernel of $(-i\nabla+A_{0})M_{\mathbb{N},\zeta}$ can be bounded using
Section 5. Hence, we have

$\displaystyle\left\lVert\operatorname{iker}(-i\nabla+A_{0})M_{\mathbb{N},\zeta}(x,y)\right\rVert$
(6.77) $\displaystyle\leq$
$\displaystyle\left\lVert\operatorname{iker}(-i\nabla-
A_{0})M_{\mathbb{N},\zeta}(x,y)\right\rVert+C\lVert
x\rVert\lVert\operatorname{iker}M_{\mathbb{N},\zeta}(x,y)\rVert$ (6.78)
$\displaystyle\leq$ $\displaystyle F(\zeta)(1+\lVert
x\rVert)\left(b_{1}(x,y)+\exp(-\lambda\lVert x-y\rVert^{2})\right).$ (6.79)

We used $b_{0}\leq Cb_{1}$. We have estimated the integral kernels of the
operators $H_{\varepsilon,x_{0}}^{(\mu_{j})}M_{\mathbb{N},\zeta}$ in Section
6. Now, we can apply Section 5. As $\varepsilon<1$, we have $e>1$ and hence
there is no singularity on the diagonal (the $b_{s}$ term is bounded). Hence,
we have

$\displaystyle\left\lVert\operatorname{iker}\left(-i\nabla+A_{0}\right)M_{\mathbb{N},\zeta}\otimes\bigotimes_{j=1}^{e}\left(H_{\varepsilon,x_{0}}^{(\mu_{j})}M_{\mathbb{N},\zeta}\right)(x,y)\right\rVert$
(6.80) $\displaystyle\leq$ $\displaystyle F(\zeta)\frac{1+\lVert
x\rVert}{(1+\lVert x+x_{0}\rVert)^{e\varepsilon}}\exp(-\lambda\lVert
x-y\rVert^{2})$ (6.81) $\displaystyle\leq$ $\displaystyle F(\zeta)(1+\lVert
x_{0}\rVert)\exp(-\lambda\lVert x-y\rVert^{2}).$ (6.82)

The final step relies on the fact $e\varepsilon\geq 1$. Using Section 2, we
can conclude

$\displaystyle\left\lVert\left(-i\nabla+A_{0}\right)M_{\mathbb{N},\zeta}\otimes\bigotimes_{j=1}^{e}\left(H_{\varepsilon,x_{0}}^{(\mu_{j})}M_{\mathbb{N},\zeta}\right)\right\rVert_{\infty}\leq
F(\zeta)(1+\lVert x_{0}\rVert).$ (6.83)

This finishes the induction over $d$. Hence, we have proven (6.71) and can
continue the estimate in (6.66). Using (6.68) and (6.70), we observe that for
$0\leq\gamma^{\prime}\leq\gamma$, we have

$\displaystyle\left\lVert
1_{[0,1]^{2}}\left(-i\nabla+A_{0}\right)^{\otimes(\gamma^{\prime}+2)}\left(M_{\mathbb{N},\zeta}H_{\varepsilon,x_{0}}\right)^{n}\right\rVert_{\infty}\leq
F(\zeta)(1+\lVert x_{0}\rVert)^{\gamma}.$ (6.84)

Now we need to consider the case $\gamma^{\prime}\in\\{-2,-1\\}$. For these,
we estimate

$\displaystyle\left\lVert
1_{[0,1]^{2}}\left(-i\nabla+A_{0}\right)^{\otimes(\gamma^{\prime}+2)}\left(M_{\mathbb{N},\zeta}H_{\varepsilon,x_{0}}\right)^{n}\right\rVert_{\infty}$
(6.85) $\displaystyle\leq$ $\displaystyle\left\lVert
1_{[0,1]^{2}}\left(-i\nabla+A_{0}\right)^{\otimes(\gamma^{\prime}+2)}M_{\mathbb{N},\zeta}\right\rVert_{\infty}\left\lVert
H_{\varepsilon,x_{0}}M_{\mathbb{N},\zeta}H_{\varepsilon,x_{0}}\right\rVert_{\infty}\left\lVert
M_{\mathbb{N},\zeta}H_{\varepsilon,x_{0}}\right\rVert_{\infty}^{n-2}$ (6.86)
$\displaystyle\leq$ $\displaystyle F(\zeta)\leq F(\zeta)(1+\lVert
x_{0}\rVert)^{\gamma}.$ (6.87)

The operator $M_{\mathbb{N},\zeta}H_{\varepsilon
x_{0}}=H_{\varepsilon,x_{0}}M_{\mathbb{N},\overline{\zeta}}^{*}$ has an
operator norm $\leq F(\zeta)$ by Section 6, the middle factor is bounded by
(6.70), and the first factor is bounded by Section 5 for $\gamma^{\prime}=-2$
and by (6.79) for $\gamma^{\prime}=-1$, in both cases the operator norm is
$\leq F(\zeta)$.

Now we have suitable upper bounds for the all factors in (6.57). The other
factors are bounded by (6.61) and (6.62). Thus, we conclude

$\displaystyle\left\lVert\left(M_{\mathbb{N},\zeta}H_{\varepsilon,x_{0}}\right)^{n}U_{x_{0}}\frac{1}{H-\zeta}\left(H_{\varepsilon}M_{\mathbb{N},\zeta}\right)^{n}1_{L\Lambda^{\complement}}U_{x_{0}}^{-1}\right\rVert_{S_{2}(\mathrm{L}^{2}(\mathbb{R}^{2}),H^{\gamma+2}[0,1]^{2})}\leq
F(\zeta)(1+\lVert x_{0}\rVert)^{\gamma}L^{-\nu}.$ (6.88)

Using (6.52), we have now finished this proof. ∎

We need one more technical lemma to prove Theorem 4.1.

######  Lemma .

Let $d\in\mathbb{N},\kappa\in[0,\infty)$, and let $S$ be an integral operator
on $\mathrm{L}^{2}(\mathbb{R}^{2})$ satisfying for any $x,y\in\mathbb{R}^{2}$

$\displaystyle\lvert\operatorname{iker}S(x,y)\rvert\leq C\frac{(1+\lVert
x\rVert)^{d}}{(1+\lVert x+x_{0}\rVert)^{\kappa}}\exp\left(-\lambda\lVert
x-y\rVert^{2}\right).$ (6.89)

Furthermore, let $\Omega\subset\mathbb{R}^{2}$ be bounded. Then there are
constants $C,\lambda^{\prime}$ such that for any $R\in[0,\infty)$, we have the
estimate

$\displaystyle\left\lVert
1_{\Omega}S1_{D_{R}^{\complement}(0)}\right\rVert_{2}\leq
C\frac{\exp(-\lambda^{\prime}R^{2})}{(1+\lVert x_{0}\rVert)^{\kappa}}.$ (6.90)

###### Proof.

We use the Hilbert–Schmidt kernel property (see Section 2). Hence, by the
unitary equivalence of the $p$-Schatten norms, we have

$\displaystyle\left\lVert
1_{\Omega}S1_{D_{R}^{\complement}(0)}\right\rVert_{2}^{2}=$
$\displaystyle\int_{\Omega}dx\int_{D_{R}^{\complement}(0)}dy\left\lVert\operatorname{iker}S(x,y)\right\rVert^{2}$
(6.91) $\displaystyle\leq$ $\displaystyle
C\int_{\Omega}dx\int_{D_{R}^{\complement}(0)}dy\frac{(1+\lVert
x\rVert)^{2d}}{(1+\lVert x+x_{0}\rVert)^{2\kappa}}\exp\left(-2\lambda\lVert
x-y\rVert^{2}\right)$ (6.92) $\displaystyle\leq$ $\displaystyle
C\int_{\Omega}dx\int_{D_{R}^{\complement}(0)}dy\frac{1}{(1+\lVert
x_{0}\rVert)^{2\kappa}}\exp(-\lambda\lVert y\rVert^{2})$ (6.93)
$\displaystyle\leq$ $\displaystyle
C\frac{\exp(-2\lambda^{\prime}R^{2})}{(1+\lVert x_{0}\rVert)^{2\kappa}}$
(6.94)

The second step uses $x\in\Omega$ and Appendix A. Then we used Appendix A
again. This finishes the proof. ∎

###### Proof of Theorem 4.1.

We start off similarly to the proof of Theorem 4.3. In particular, we begin by
conjugating with the unitary operator $U_{x_{0}}$, as defined in Appendix A.
Hence, we have333We have already mentioned this equality in (6.7).

$\displaystyle\left\lVert
1_{[0,1]^{2}-x_{0}}(T_{l}H_{\varepsilon})^{m}P_{l}(H_{\varepsilon}T_{l})^{k-m}1_{D^{\complement}_{R}(x_{0})}\right\rVert_{p}=$
$\displaystyle\left\lVert
1_{[0,1]^{2}}(T_{l}H_{\varepsilon,x_{0}})^{m}P_{l}(H_{\varepsilon,x_{0}}T_{l})^{k-m}1_{D^{\complement}_{R}(0)}\right\rVert_{p}.$
(6.95)

Now, once again, let $q$ satisfy $\frac{1}{q}+\frac{1}{2}=\frac{1}{p}$. As
$p>\frac{2}{\gamma+3}$, we have $q>\frac{2}{\gamma+2}$. Hence, we can apply
Appendix A with $\gamma+2$ and the property Hölder I (see Section 2) to get
the upper bound

$\displaystyle\eqref{cube norm proof eq1}\leq$ $\displaystyle C\left\lVert
1_{[0,1]^{2}}(T_{l}H_{\varepsilon,x_{0}})^{m}P_{l}(H_{\varepsilon,x_{0}}T_{l})^{k-m}1_{D^{\complement}_{R}(0)}\right\rVert_{S_{2}(\mathrm{L}^{2}(\mathbb{R}^{2}),H^{\gamma+2}([0,1]^{2}))}$
(6.96) $\displaystyle\leq$ $\displaystyle
C\sum_{\gamma^{\prime}=-2}^{\gamma}\left\lVert
1_{[0,1]^{2}}(-i\nabla+A_{0})^{\otimes(\gamma^{\prime}+2)}(T_{l}H_{\varepsilon,x_{0}})^{m}P_{l}(H_{\varepsilon,x_{0}}T_{l})^{k-m}1_{D^{\complement}_{R}(0)}\right\rVert_{2}.$
(6.97)

We used the quasi-isometry $D_{\gamma+2}$ as constructed in Appendix A. We
will now establish two kernel estimates that will be needed to finish this
proof.

Looking at (5.3), we observe that for any $d\in\mathbb{N}$ and
$h\in\\{0,1\\}$, there are $\lambda,C\in\mathbb{R}^{+}$, depending on
$B_{0},l,d,h$, such that for any $x,y\in\mathbb{R}^{2}$, we have the upper
bound

$\displaystyle\left\lVert\left(-i\nabla-A_{0}\right)^{\otimes
h}\otimes\left(-i\nabla_{x}+A_{0}(x)\right)^{\otimes
d}p_{l}(x,y)\right\rVert\leq C(1+\lVert x\rVert)^{d+h}\exp\left(-\lambda\lVert
x-y\rVert^{2}\right).$ (6.98)

Let $\mathbb{N}\ni j\leq\gamma$. Then, using (6.4), we observe

$\displaystyle\left\lVert\operatorname{iker}H_{\varepsilon,x_{0}}^{(j)}\left(-i\nabla+A_{0}\right)^{\otimes
d}P_{l}(x,y)\right\rVert\leq C\frac{(1+\lVert x\rVert)^{d+1}}{(1+\lVert
x+x_{0}\rVert)^{\varepsilon}}\exp\left(-\lambda\lVert x-y\rVert^{2}\right).$
(6.99)

Now we consider the case $m=0$. Here, we can use (6.98), the kernel estimate
for $H_{\varepsilon,x_{0}}T_{l}$, that is provided by Section 6, and Section 5
to arrive at

$\displaystyle\left\lVert\operatorname{iker}(-i\nabla+A_{0})^{\otimes(\gamma^{\prime}+2)}P_{l}\left(H_{\varepsilon,x_{0}}T_{l}\right)^{k}(x,y)\right\rVert\leq
C\frac{(1+\lVert x\rVert)^{\gamma^{\prime}+2}}{(1+\lVert
x+x_{0}\rVert)^{k\varepsilon}}\exp\left(-\lambda\lVert x-y\rVert^{2}\right).$
(6.100)

As the kernel of $(-i\nabla+A_{0})^{\otimes d}P_{l}$ has no singularity at the
diagonal, the term $b_{s}$ can be ignored. By Section 6, we have now finished
the case $m=0$.

Now we consider the case $m>0$ and $\gamma^{\prime}\in\\{-2,-1\\}$. Here, we
can use Section 5 to get (compare (6.79))

$\displaystyle\left\lVert\operatorname{iker}\left(-i\nabla+A_{0}\right)^{\otimes(\gamma^{\prime}+2)}T_{l}(x,y)\right\rVert\leq
C(1+\lVert x\rVert)^{\gamma^{\prime}+2}\left(b_{1}(x,y)+\exp(-\lambda\lVert
x-y\rVert^{2})\right).$ (6.101)

With this kernel estimate, the one in Section 6, and (6.99) with $d=j=0$, we
can employ Section 5 to get

$\displaystyle\left\lVert\operatorname{iker}\left(-i\nabla+A_{0}\right)^{\otimes(\gamma^{\prime}+2)}(T_{l}H_{\varepsilon,x_{0}})^{m}P_{l}(H_{\varepsilon,x_{0}}T_{l})^{k-m}(x,y)\right\rVert\leq
C\frac{(1+\lVert x\rVert)^{\gamma^{\prime}+3}}{(1+\lVert
x+x_{0}\rVert)^{k\varepsilon}}\exp\left(-\lambda\lVert x-y\rVert^{2}\right).$
(6.102)

Once again, as $p_{l}$ has no singularity at the diagonal, the term $b_{s}$
can be ignored and by Section 6, we have finished this case as well.

We are left with the case $m>0$ and $\gamma^{\prime}\geq 0$. Here, we first
apply Section 6 with the parameters $\gamma^{\prime}$ and $m$. Hence, we have

$\displaystyle\left\lVert
1_{[0,1]^{2}}(-i\nabla+A_{0})^{\otimes(\gamma^{\prime}+2)}(T_{l}H_{\varepsilon,x_{0}})^{m}P_{l}(H_{\varepsilon,x_{0}}T_{l})^{k-m}1_{D^{\complement}_{R}(0)}\right\rVert_{2}$
(6.103) $\displaystyle=$ $\displaystyle\left\lVert
1_{[0,1]^{2}}(-i\nabla+A_{0})^{\otimes
2}\otimes\sum_{\mu\in\mathbb{N}^{k},\lvert\mu\rvert=\gamma^{\prime}}N_{\mu}\bigotimes_{j=1}^{m}(T_{l}H_{\varepsilon,x_{0}}^{(\mu_{j})})\otimes\left(-i\nabla+A_{0}\right)^{\otimes\mu_{m+1}}P_{l}(H_{\varepsilon,x_{0}}T_{l})^{k-m}1_{D^{\complement}_{R}(0)}\right\rVert_{2}$
(6.104) $\displaystyle\leq$ $\displaystyle
C\sup_{\mu\in\mathbb{N}_{\leq\gamma}^{m+1}}\left\lVert
1_{[0,1]^{2}}(-i\nabla+A_{0})^{\otimes
2}\otimes\bigotimes_{j=1}^{m}(T_{l}H_{\varepsilon,x_{0}}^{(\mu_{j})})\otimes\left(-i\nabla+A_{0}\right)^{\otimes\mu_{m+1}}P_{l}(H_{\varepsilon,x_{0}}T_{l})^{k-m}1_{D^{\complement}_{R}(0)}\right\rVert_{2}$
(6.105) $\displaystyle=$ $\displaystyle
C\\!\sup_{\mu\in\mathbb{N}_{\leq\gamma}^{m+1}}\Bigg{\lVert}1_{[0,1]^{2}}(-i\nabla+A_{0})^{\otimes
2}T_{l}\otimes$ (6.106)
$\displaystyle\phantom{C\\!\sup_{\mu\in\mathbb{N}_{\leq\gamma}^{m+1}}\Bigg{\lVert}}\bigotimes_{j=1}^{m-1}(H_{\varepsilon,x_{0}}^{(\mu_{j})}T_{l})\otimes
H_{\varepsilon,x_{0}}^{(\mu_{m})}\left(-i\nabla+A_{0}\right)^{\otimes\mu_{m+1}}P_{l}(H_{\varepsilon,x_{0}}T_{l})^{k-m}1_{D^{\complement}_{R}(0)}\Bigg{\rVert}_{2}$
(6.107)

The operator $1_{[0,1]^{2}}(-i\nabla+A_{0})^{\otimes 2}T_{l}$ does not have a
nice integral kernel. This is why we cannot directly get a kernel bound from
this representation. Let $\varphi\in C^{\infty}_{c}(\mathbb{R}^{2})$ be a
smooth cutoff function satisfying $\varphi(x)=1$ for $x\in D_{1}(0)$,
$\varphi(x)=0$ for $x\in D_{2}^{\complement}(0)$, and $0\leq\varphi(x)\leq 1$
everywhere. We introduce the operators $T_{l,n}$ and $T_{l,f}$, which are
defined by the integral kernels given for any $x,y\in\mathbb{R}^{2}$ with
$x\neq y$ by

$\displaystyle t_{l,n}(x,y)\coloneqq$ $\displaystyle\varphi(x-y)t_{l}(x,y),$
(6.108) $\displaystyle t_{l,f}(x,y)\coloneqq$
$\displaystyle(1-\varphi(x-y))t_{l}(x,y).$ (6.109)

Obviously, $T_{l,n}+T_{l,f}=T_{l}$. Furthermore, for any $d\in\\{0,1,2\\}$,
the operator $(-i\nabla-A_{0})^{\otimes d}T_{l,f}$ has a nice integral kernel
satisfying

$\displaystyle\left\lVert\operatorname{iker}(-i\nabla-A_{0})^{\otimes
d}T_{l,f}(x,y)\right\rVert\leq C\exp(-\lambda\lVert x-y\rVert^{2})$ (6.110)

by Section 5. This implies the kernel estimate

$\displaystyle\left\lVert\operatorname{iker}(-i\nabla+A_{0})^{\otimes
2}T_{l,f}(x,y)\right\rVert\leq C(1+\lVert x\rVert)^{2}\exp(-\lambda\lVert
x-y\rVert^{2}).$ (6.111)

Hence, the operator $1_{[0,1]^{2}}(-i\nabla+A_{0})^{\otimes 2}T_{l,f}$ is
bounded. The operator $(-i\nabla-A_{0})^{\otimes d}T_{l}$ is bounded by
Section 5 for $d=0,1$ and by Section 6 for $d=2$. Hence, the operator
$1_{[0,1]^{2}}(-i\nabla+A_{0})^{\otimes 2}T_{l}$ is bounded. By the triangle
inequality, we can conclude that the operator
$1_{[0,1]^{2}}(-i\nabla+A_{0})^{\otimes 2}T_{l,n}$ is bounded. Furthermore, we
have the identity

$\displaystyle 1_{[0,1]^{2}}(-i\nabla+A_{0})^{\otimes
2}T_{l,n}=1_{[0,1]^{2}}(-i\nabla+A_{0})^{\otimes
2}1_{[-1,2]^{2}}T_{l,n}=1_{[0,1]^{2}}(-i\nabla+A_{0})^{\otimes
2}T_{l,n}1_{[-3,4]^{2}}.$ (6.112)

The value at $x\in[0,1]^{2}$ of $(-i\nabla+A_{0})f$ only depends on $f$ in an
arbitrary small neighbourhood of $x$, which proves the first identity. The
second identity follows by the construction of $T_{l,n}$ as an integral
operator with a kernel that vanishes if $\lVert x-y\rVert\geq 2$.

We will now estimate the kernel of the operator in (6.107), where we replace
the first $T_{l}$ by $T_{l,f}$. The kernels of the operators
$H_{\varepsilon,x_{0}}^{(\mu_{j})}T_{l}$ and $H_{\varepsilon,x_{0}}T_{l}$ can
be bounded by Section 6, the kernel of
$H_{\varepsilon,x_{0}}^{(\mu_{m})}\left(-i\nabla+A_{0}\right)^{\otimes\mu_{m+1}}P_{l}$
has been bounded in (6.99), and the kernel of $(-i\nabla+A_{0})^{\otimes
2}T_{l,f}$ has been bounded in (6.111). Hence, we can apply Section 5 to
arrive at

$\displaystyle\left\lVert\operatorname{iker}(-i\nabla+A_{0})^{\otimes
2}T_{l,f}\otimes\bigotimes_{j=1}^{m-1}(H_{\varepsilon,x_{0}}^{(\mu_{j})}T_{l})\otimes
H_{\varepsilon,x_{0}}^{(\mu_{m})}\left(-i\nabla+A_{0}\right)^{\otimes\mu_{m+1}}P_{l}(H_{\varepsilon,x_{0}}T_{l})^{k-m}(x,y)\right\rVert$
(6.113) $\displaystyle\leq$ $\displaystyle C\frac{(1+\lVert
x\rVert)^{\gamma+3}}{(1+\lVert
x+x_{0}\rVert)^{k\varepsilon}}\exp\left(-\lambda\lVert x-y\rVert^{2}\right).$
(6.114)

Once more, the $b_{s}$ term can be ignored as the operator
$H_{\varepsilon,x_{0}}^{(\mu_{m})}\left(-i\nabla+A_{0}\right)^{\otimes\mu_{m+1}}P_{l}$
has no singularity at the diagonal and by Section 6, this establishes the
required estimate.

We are only left with the term in (6.107), where we replace the first $T_{l}$
by $T_{l,n}$. Here, we can use (6.112) to see

$\displaystyle\left\lVert 1_{[0,1]^{2}}(-i\nabla+A_{0})^{\otimes
2}T_{l,n}\otimes\bigotimes_{j=1}^{m-1}(H_{\varepsilon,x_{0}}^{(\mu_{j})}T_{l})\otimes
H_{\varepsilon,x_{0}}^{(\mu_{m})}\left(-i\nabla+A_{0}\right)^{\otimes\mu_{m+1}}P_{l}(H_{\varepsilon,x_{0}}T_{l})^{k-m}1_{D^{\complement}_{R}(0)}\right\rVert_{2}$
(6.115) $\displaystyle=$
$\displaystyle\Bigg{\lVert}1_{[0,1]^{2}}(-i\nabla+A_{0})^{\otimes
2}T_{l,n}1_{[-3,4]^{2}}\otimes$ (6.116)
$\displaystyle\phantom{\Bigg{\lVert}}\quad\bigotimes_{j=1}^{m-1}(H_{\varepsilon,x_{0}}^{(\mu_{j})}T_{l})\otimes
H_{\varepsilon,x_{0}}^{(\mu_{m})}\left(-i\nabla+A_{0}\right)^{\otimes\mu_{m+1}}P_{l}(H_{\varepsilon,x_{0}}T_{l})^{k-m}1_{D^{\complement}_{R}(0)}\Bigg{\rVert}_{2}$
(6.117) $\displaystyle\leq$ $\displaystyle C\left\lVert
1_{[0,1]^{2}}(-i\nabla+A_{0})^{\otimes 2}T_{l,n}\right\rVert_{\infty}$ (6.118)
$\displaystyle\quad\times\left\lVert
1_{[-3,4]^{2}}\otimes\bigotimes_{j=1}^{m-1}(H_{\varepsilon,x_{0}}^{(\mu_{j})}T_{l})\otimes
H_{\varepsilon,x_{0}}^{(\mu_{m})}\left(-i\nabla+A_{0}\right)^{\otimes\mu_{m+1}}P_{l}(H_{\varepsilon,x_{0}}T_{l})^{k-m}1_{D^{\complement}_{R}(0)}\right\rVert_{2}.$
(6.119)

The operator $1_{[0,1]^{2}}(-i\nabla+A_{0})^{\otimes 2}T_{l,n}$ is bounded.
For the remaining part, we estimate the kernel. This is incredibly similar to
(6.114). The kernels of the operators $H_{\varepsilon,x_{0}}^{(\mu_{j})}T_{l}$
and $H_{\varepsilon,x_{0}}T_{l}$ can be bounded by Section 6 and the kernel of
$H_{\varepsilon,x_{0}}^{(\mu_{m})}\left(-i\nabla+A_{0}\right)^{\otimes\mu_{m+1}}P_{l}$
has been bounded in (6.99). Hence, we can apply Section 5 to arrive at

$\displaystyle\left\lVert\operatorname{iker}\bigotimes_{j=1}^{m-1}(H_{\varepsilon,x_{0}}^{(\mu_{j})}T_{l})\otimes
H_{\varepsilon,x_{0}}^{(\mu_{m})}\left(-i\nabla+A_{0}\right)^{\otimes\mu_{m+1}}P_{l}(H_{\varepsilon,x_{0}}T_{l})^{k-m}(x,y)\right\rVert$
(6.120) $\displaystyle\leq$ $\displaystyle C\frac{(1+\lVert
x\rVert)^{\gamma+1}}{(1+\lVert
x+x_{0}\rVert)^{k\varepsilon}}\exp\left(-\lambda\lVert x-y\rVert^{2}\right).$
(6.121)

For one final time, the $b_{s}$ term can be ignored as the operator
$H_{\varepsilon,x_{0}}^{(\mu_{m})}\left(-i\nabla+A_{0}\right)^{\otimes\mu_{m+1}}P_{l}$
has no singularity at the diagonal and by Section 6, this establishes the
required estimate.

This brings this proof to a close. ∎

## Appendix A

######  Lemma .

Let $x_{0}\in\mathbb{R}^{2}$. Then there is a unitary operator
$U_{x_{0}}\colon\mathrm{L}^{2}(\mathbb{R}^{2})\to\mathrm{L}^{2}(\mathbb{R}^{2})$,
such that the following identities hold for any
$f\in\mathrm{L}^{\infty}(\mathbb{R}^{2})$, any $I\subset\mathbb{N}$ cofinite
and any $\zeta\in\mathbb{C}\setminus B_{0}(2I+1)$:

$\displaystyle U_{x_{0}}f(X)U_{x_{0}}^{-1}=$ $\displaystyle f(X+x_{0}),$ (A.1)
$\displaystyle U_{x_{0}}\left(-i\nabla-A_{0}\right)U_{x_{0}}^{-1}=$
$\displaystyle\left(-i\nabla-A_{0}\right),$ (A.2) $\displaystyle
U_{x_{0}}H_{\varepsilon}U_{x_{0}}^{-1}=$ $\displaystyle
H_{\varepsilon,x_{0}}\,,$ (A.3) $\displaystyle
U_{x_{0}}M_{I,\zeta}U_{x_{0}}^{-1}=$ $\displaystyle M_{I,\zeta}\,.$ (A.4)

Here, $X$ refers to the multiplication operator with the identity on
$\mathbb{R}^{2}$ and $f(X)$ is defined by functional calculus and hence the
multiplication operator with the function $f$.

The operators $H_{\varepsilon},H_{\varepsilon,x_{0}}$, and $M_{I,\zeta}$ have
been defined in (4.1), (6.4), and (4.7).

###### Proof.

For any $x_{0}\in\mathbb{R}^{2}$, we define the three unitary operators
$U_{x_{0}1},U_{x_{0}2},U_{x_{0}}$ by

$\displaystyle\forall x\in\mathbb{R}^{2}\colon\quad$
$\displaystyle\left(U_{x_{0}1}\psi\right)(x)\,$ $\displaystyle\coloneqq$
$\displaystyle\,\psi(x+x_{0}),$ (A.5) $\displaystyle\forall
x\in\mathbb{R}^{2}\colon\quad$ $\displaystyle\left(U_{x_{0}2}\psi\right)(x)\,$
$\displaystyle\coloneqq$
$\displaystyle\,\psi(x)\exp\left(-i\frac{B_{0}}{2}\langle x\mid
Jx_{0}\rangle\right),$ (A.6) $\displaystyle U_{x_{0}}\coloneqq
U_{x_{0}1}U_{x_{0}2}.$ (A.7)

As we can see, these operators and their inverses preserve
$C^{\infty}_{c}(\mathbb{R}^{2})$. Hence, it is sufficient to show that the
claimed operator identites hold, when evaluated at a test function $\psi\in
C^{\infty}_{c}(\mathbb{R}^{2})$.

We have

$\displaystyle U_{x_{0}1}U_{x_{0}2}f(X)U_{x_{0}2}^{-1}U_{x_{0}1}^{-1}$
$\displaystyle=U_{x_{0}1}f(X)U_{x_{0}1}^{-1}$ (A.8)
$\displaystyle=f(X+x_{0}).$ (A.9)

Now, we need to check how $(-i\nabla-\frac{B_{0}}{2}JX)$ behaves under
conjugation with $U_{x_{0}}$. Hence, we get

$\displaystyle\left(U_{x_{0}2}\left(-i\nabla-\frac{B_{0}}{2}JX\right)U_{x_{0}2}^{-1}U_{x_{0}1}^{-1}\psi\right)(x)$
(A.10) $\displaystyle=$ $\displaystyle\exp\left(-i\frac{B_{0}}{2}\langle x\mid
Jx_{0}\rangle\right)\left(-i\nabla_{x}-\frac{B_{0}}{2}Jx\right)\exp\left(i\frac{B_{0}}{2}\langle
x\mid Jx_{0}\rangle\right)\psi(x-x_{0})$ (A.11) $\displaystyle=$
$\displaystyle\left(-i\nabla_{x}-\frac{B_{0}}{2}Jx\right)\psi(x-x_{0})+\psi(x-x_{0})\left(-i\nabla_{x}\right)\left(i\frac{B_{0}}{2}\langle
x\mid Jx_{0}\rangle\right)$ (A.12) $\displaystyle=$
$\displaystyle\left(-i\nabla_{x}-\frac{B_{0}}{2}J(x-x_{0})\right)\psi(x-x_{0})$
(A.13) $\displaystyle=$
$\displaystyle\left(U_{x_{0}1}^{-1}\left(-i\nabla-\frac{B}{2}JX\right)\psi\right)(x).$
(A.14)

In the second step, we used the product and chain rule and the exponentials
cancel. The interior derivative is then resolved in the next step.

In conclusion, we have

$\displaystyle U_{x_{0}}\left(-i\nabla-
A_{0}\right)U_{x_{0}}^{-1}=\left(-i\nabla-A_{0}\right).$ (A.15)

This implies

$\displaystyle U_{x_{0}}T_{l}U_{x_{0}}^{-1}=T_{l}.$ (A.16)

Together with (A.9), this implies the identity

$\displaystyle U_{x_{0}}H_{\varepsilon}U_{x_{0}}^{-1}=$ $\displaystyle
H_{\varepsilon,x_{0}}.$ (A.17)

This finishes the proof. ∎

######  Lemma .

Let $\Omega\subset\mathbb{R}^{n}$ be measurable and let
$f\colon\Omega\to\mathbb{C}$ be integrable. Then we have the identity

$\displaystyle\int_{[0,1)^{n}}\sum_{z\in\mathbb{Z}^{n},z+h_{0}\in\Omega}f(z+h_{0})dh_{0}=\int_{\Omega}f(x)dx.$
(A.18)

###### Proof.

We observe

$\displaystyle\int_{[0,1)^{n}}\sum_{z\in\mathbb{Z}^{n},z+h_{0}\in\Omega}f(z+h_{0})dh_{0}=$
$\displaystyle\int_{[0,1)^{n}}\sum_{z\in\mathbb{Z}^{n}}1_{\Omega}(z+h_{0})f(z+h_{0})dh_{0}$
(A.19) $\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{n}}1_{\Omega}(x)f(x)dx$ (A.20)
$\displaystyle=$ $\displaystyle\int_{\Omega}f(x)dx.$ (A.21)

In the second step we used Fubini with
$[0,1)^{n}\times\mathbb{Z}^{n}=\mathbb{R}^{n}$.

∎

######  Lemma .

Let $f\in C^{1}(\mathbb{R})$ with $f^{\prime}\leq 0$ and
$\lim_{t\to\infty}f(t)=0$, $h\in C^{0}(\mathbb{R}^{2},\mathbb{R})$ and
$\Lambda\subset\mathbb{R}^{2}$ be measurable.

Then we have