# Ground States for Infrared Renormalized Translation-Invariant Non-
Relativistic QED

David Hasler111Institut für Mathematik, Friedrich-Schiller-Universität Jena,
Ernst-Abbe-Platz 2, 07743 Jena, Germany<EMAIL_ADDRESS>Oliver
Siebert222FB Mathematik, Universität Tübingen, Auf der Morgenstelle 10, 72076
Tübingen, Germany<EMAIL_ADDRESS>
###### Abstract

We consider a translation-invariant Pauli-Fierz model describing a non-
relativistic charged quantum mechanical particle interacting with the
quantized electromagnetic field. The charged particle may be spinless or have
spin one half. We decompose the Hamiltonian with respect to the total momentum
into a direct integral of so called fiber Hamiltonians. We perform an infrared
renormalization, in the sense of norm resolvent convergence, for each fiber
Hamiltonian, which has the physical interpretation of removing an infinite
photon cloud. We show that the renormalized fiber Hamiltonians have a ground
state for almost all values for the total momentum with modulus less than one.

##  1 Introduction

The infrared problem is a technical obstacle in describing scattering of
charged particles with a quantized radiation field, which has been thoroughly
investigated in the literature over the last decades. As already pointed out
in the seminal paper by Bloch and Nordsieck [BN37] it is an artifact of the
emergence of a cloud of infinitely many photons with a finite amount of total
energy in the asymptotic scattering states. While there appeared several
perturbative methods like the one by Fadeev and Kulish [KF70] for the
computation of divergence-free scattering transition probabilities, a
mathematical precise theory was not available for a long time. A first
rigorous description of so called dressed one-electron states, i.e., the
‘plane-wave’ states of an electron traveling with a cloud of photons at some
fixed total momentum, was given by Kibble [Kib68] based on ideas of Chung
[Chu65]. The basic principle behind it is, that due to the infinite number of
photons, the states cease to exist in the original Hilbert space of the Fock
representation. Instead, one has to use non-Fock coherent representations of
the canonical commutation relations (CCR). A significant problem is that
different total momenta lead to different inequivalent representations, making
it hard to construct scattering states, cf. [Frö73] and for recent work on
this subject [CD19].

As a first step towards a more realistic model Fröhlich discussed so called
dressed one-electron states in the Nelson model [Frö73, Frö74], where a
charged particle like an electron interacts with a bosonic field via a
coupling term which is linear in the creation and annihilation operators. He
shows that it is impossible to implement one-electron states for different
total momenta in a joint Hilbert space, and he gave a non-constructive proof
for the existence of the corresponding coherent states as a weak-* limit where
the infrared regularization is removed. Later Pizzo accomplished an explicit
perturbative construction of such states in a Hilbert space by means of a
dressing transformation and an iterative algorithm [Piz03], which he then used
for the construction of asymptotic scattering states [Piz05]. Related results
for the weak coupling model of non-relativistic QED, which is technically more
involved due to the interaction being quadratic in the creation and
annihilation operators were obtained in [CF07, CFP10, CFP09, Bac+07, Che08].
Furthermore, in the last years there appeared a series of papers for Coloumb
scattering in the Nelson model beetween two electrons starting with [DP14] and
between an electron and atom [DP22] in the presence of a quantized radiation
field. Recently, there was also a novel approach developed for the
construction of scattering states in the Nelson model without any infrared
limiting procedure [BDG21].

Due to its significance for scattering theory, a lot of progress was made to
investigate such one-electron states without infrared regularization in
different models, e.g., apart from the Nelson model and non-relativistic QED
also for the UV-renormalized Nelson model [BDP12] and the semi-relativistic
Pauli-Fierz model [KM14]. Those states can be described mathematically as
ground states of a (renormalized) fiber Hamiltonian corresponding to fixed
total momentum.

In this paper we show the existence of one-electron states in the non-
relativistic Pauli-Fierz model for almost all momenta with modulus less than
one and for all values of the coupling constant. The charged particle may be
spinless or have spin one half, where in the latter case have to assume an
energy inequality, cf. (2.12). Moreover, we show that they are ground states
of fiber-wise renormalized Hamiltonians. A similar result has been obtained
for the UV renormalized translation invariant Nelson model in [BDP12]
following the strategy developed in [Piz03]. Moreover, we want to mention a
related result for the Nelson model in [Ara01]. We use a compactness argument,
originally tracing back to [GLL01], in the same way as in [HS20, HHS21], where
we showed the existence of ground states at zero momentum. The proof is non-
constructive but also non-perturbative. The infrared renormalization is done
via a Gross transformation similarly as performed in [Nel64, Piz03, Piz05] for
the Nelson model or for non-relativistic QED in [CF07].

The main results are presented in Section 3. In Section 4 we outline the
strategy of the proof. The renormalization of the fiber Hamiltonians is
performed in Section 5. Key elements of the proof of the main result, stated
in Theorem 3.5, are an argument based on second order perturbation theory in
the total momentum and a photon number bound together with a bound on
derivatives of the photon momenta, which can be found in Section 6 and Section
7, respectively. Finally, Section 8 contains the proof of the main result,
where the compactness argument is presented and previous estimates are used to
establish indeed compactness.

##  2 Model

Let $\mathfrak{h}$ be a complex Hilbert space. We define the symmetric Fock
space

$\mathcal{F}(\mathfrak{h}):=\bigoplus_{n=0}^{\infty}\mathfrak{h}^{(n)},$

with $\mathfrak{h}^{(0)}:={\mathbb{C}}$ and
$\mathfrak{h}^{(n)}:=\mathcal{P}_{n}(\bigotimes_{k=1}^{n}\mathfrak{h})$,
$n\in{\mathbb{N}}$, with $\mathcal{P}_{n}$ denoting the orthogonal projection
onto the subspace of totally symmetric tensors. Thus we can identify
$\psi\in\mathcal{F}(\mathfrak{h})$ with the sequence
$(\psi_{(n)})_{n\in{\mathbb{N}}_{0}}$ with $\psi_{(n)}\in\mathfrak{h}^{(n)}$.
The vacuum is the vector $\Omega:=(1,0,0,\ldots)\in\mathcal{F}(\mathfrak{h})$.
For a linear subspace $\mathfrak{v}$ of the Hilbert space $\mathfrak{h}$ we
define
$\mathcal{F}_{\operatorname{fin}}(\mathfrak{v})\subset\mathcal{F}(\mathfrak{h})$
as the vector space of finite linear combinations of $\Omega$ and vectors of
the the form $\mathcal{P}_{n}(v_{1}\otimes\cdots\otimes v_{n})$ with
$v_{1},...,v_{n}\in\mathfrak{v}$ and $n\in{\mathbb{N}}$.

We define for $f\in\mathfrak{h}$ the creation operator $a^{*}(f)$ acting on
vectors $\psi\in\mathcal{F}(\mathfrak{h})$ by

$(a^{*}(f)\psi)_{(n)}=\sqrt{n}\mathcal{P}_{n}(f\otimes\psi_{(n-1)})$

with domain
$D(a^{*}(f)):=\\{\psi\in\mathcal{F}(\mathfrak{h}):a^{*}(f)\psi\in\mathcal{F}(\mathfrak{h})\\}$.
This yields a densely defined closed operator. For $f\in\mathfrak{h}$ we
define the annihilation $a(f)$ as the adjoint of $a^{*}(f)$, i.e.,

$a(f)=\left[a^{*}(f)\right]^{*}.$

It follows from the definition that $a(f)$ is anti-linear, and $a^{*}(f)$ is
linear in $f$. Creation and annihilation operators are well known to satisfy
the so called canonical commutation relations

$\displaystyle[a^{*}(f),a^{*}(g)]=0\quad,\quad[a(f),a(g)]=0\quad,\quad[a(f),a^{*}(g)]=\langle{f,g}\rangle_{\mathfrak{h}}\;,$
(2.1)

where $f,g\in\mathfrak{h}$, $[\cdot,\cdot]$ stands for the commutator, and
$\langle{f,g}\rangle_{\mathfrak{h}}$ denotes the inner product of
$\mathfrak{h}$. For $f\in\mathfrak{h}$ we introduce the following notation for
the field operator and the conjugate field operator

$\displaystyle\phi(f)$ $\displaystyle=\text{closure of
}\quad\frac{1}{\sqrt{2}}(a(f)+a^{*}(f)),$ $\displaystyle\pi(f)$
$\displaystyle=\text{closure of
}\quad\frac{-\mathrm{i}}{\sqrt{2}}(a(f)-a^{*}(f)).$

We note that $\pi(f)=\phi(\mathrm{i}f)$.

For a self-adjoint operator $A$ in $\mathfrak{h}$ we define the operator
$\mathsf{d}\Gamma(A)$ as follows. In $\mathfrak{h}^{(n)}$ we set

$A^{(n)}:=A\otimes\mathds{1}\otimes\cdots\otimes\mathds{1}+\mathds{1}\otimes
A\otimes\cdots\otimes\mathds{1}+\cdots+\mathds{1}\otimes\cdots\otimes\mathds{1}\otimes
A,\,n\in{\mathbb{N}},$

in the sense of [RS81, VIII.10] and $A^{(0)}:=0$. By definition
$\psi\in\mathcal{F}(\mathfrak{h})$ is in the domain of $\mathsf{d}\Gamma(A)$
if $\psi_{(n)}\in D(A^{(n)})$ for all $n\in{\mathbb{N}}_{0}$ and

$\displaystyle(\mathsf{d}\Gamma(A)\psi)_{(n)}=A^{(n)}\psi_{(n)},\quad
n\in{\mathbb{N}}_{0},$ (2.2)

is a vector in $\mathcal{F}(\mathfrak{h})$, in which case
$\mathsf{d}\Gamma(A)\psi$ is defined by (2.2). The operator
$\mathsf{d}\Gamma(A)$ is self-adjoint, see for example [RS81, VIII.10].

Henceforth, we shall consider specifically

$\mathfrak{h}:=L^{2}({\mathbb{Z}}_{2}\times{\mathbb{R}}^{3})\cong
L^{2}({\mathbb{R}}^{3};{\mathbb{C}}^{2})$ (2.3)

and write $\mathcal{F}$ for $\mathcal{F}(\mathfrak{h})$. The Hilbert space
$\mathfrak{h}$ describes so called transversally polarized photons. By
physical interpretation the variable
$(\lambda,k)\in{\mathbb{Z}}_{2}\times{\mathbb{R}}^{3}$ consists of the wave
vector $k$ and the polarization label $\lambda$. Because of (2.3), the
elements $\psi\in\mathcal{F}_{0}$ can be identified with sequences
$(\psi_{(n)})_{n=0}^{\infty}$ of so called $n$-photon wave functions,
$\psi_{(n)}\in
L^{2}_{\mathrm{sym}}(({\mathbb{Z}}_{2}\times{\mathbb{R}}^{3})^{n})$, where the
subscript “${\mathrm{sym}}$” stands for the subspace of functions wich are
totally symmetric in their $n$ arguments. Henceforth, we shall make use of
this identification without mention. The Fock space inherits a scalar product
from $\mathfrak{h}$, explicitly

$\displaystyle\langle{\psi,\varphi}\rangle$
$\displaystyle=\overline{\psi}_{(0)}\varphi_{(0)}+\sum_{n=1}^{\infty}\sum_{\begin{subarray}{c}\lambda_{1},\ldots,\lambda_{n}\\\
\in\\{1,2\\}\end{subarray}}\int\overline{\psi_{(n)}(\lambda_{1},{k}_{1},\ldots,\lambda_{n},{k}_{n})}\varphi_{(n)}(\lambda_{1},{k}_{1},\ldots,\lambda_{n},{k}_{n})d{k}_{1}\ldots
d{k}_{n}\;.$

For $m\geq 0$ the field energy operator denoted by $H_{\mathrm{f},m}$ is
defined by

$H_{\mathrm{f},m}=\mathsf{d}\Gamma(\omega_{m}),$

where $\omega_{m}\colon{\mathbb{Z}}_{2}\times{\mathbb{R}}^{3}\to{\mathbb{R}}$,
$\omega_{m}(\lambda,k):=\omega_{m}(k):=\sqrt{m^{2}+k^{2}}$. The operator of
momentum $P_{\mathrm{f}}$ is defined as a three dimensional vector of
operators, where the $j$-th component is given by

$(P_{\mathrm{f}})_{j}:=\mathsf{d}\Gamma(K_{j}),$

with $K_{j}\colon{\mathbb{Z}}_{2}\times{\mathbb{R}}^{3}\to{\mathbb{R}}$,
$K_{j}(\lambda,k)=k_{j}$.

The Hilbert space describing the system composed of a charged particle with
spin $s\in\\{0,\frac{1}{2}\\}$ and the quantized field is

$\mathcal{H}_{\mathrm{full}}:=L^{2}({\mathbb{R}}^{3}\times{\mathbb{Z}}_{2s+1})\otimes\mathcal{F}\;.$
(2.4)

We consider the Hamiltonian

$H_{m}=\frac{1}{2}\left(p+eA(x)\right)^{2}+eS\cdot B(x)+H_{\mathrm{f},m}\;,$

with

$\displaystyle A_{j}(x)=\phi(f_{A,j}E_{x}),\quad B_{j}(x)$
$\displaystyle=\phi(f_{B,j}E_{x}),\quad j=1,2,3,$ (2.5)

where we defined the following functions
$E_{y}\colon{\mathbb{R}}^{3}\to{\mathbb{R}}^{3}$ by $k\mapsto
e^{-\mathrm{i}k\cdot y}$ for $y\in{\mathbb{R}}^{3}$,

$\displaystyle f_{A,j}(k,\lambda)$
$\displaystyle:=\frac{\rho(k)}{\sqrt{\left|k\right|}}{\varepsilon}_{\lambda}(k),\quad
f_{B,j}(k,\lambda):=-\mathrm{i}[k\wedge f_{A}(k,\lambda)]_{j},\quad
k\in{\mathbb{R}}^{3},$

where the $\varepsilon_{\lambda}(k)\in{\mathbb{R}}^{3}$ are so called
polarization vectors, depending measurably on $\widehat{k}=k/|k|$, such that
$(\widehat{k},\varepsilon_{1}(k),\varepsilon_{2}(k))$ forms an orthonormal
basis. For the proof we shall make an explicit choice of the polarization
vectors in (7.21), below. We note that the $x$ in (2.5) denotes the operator
of multiplication with the position coordinates of the first component of the
tensor product (2.4). Mathematically this amounts to the following definition.
By means of the unitary isomorphism $\mathcal{H}_{\mathrm{full}}\cong
L^{2}({\mathbb{R}}^{3}\times{\mathbb{Z}}_{2s+1};\mathcal{F})$ one has the
identity

$\displaystyle[(A_{j}(x)\psi](x,s)=\phi(f_{A,j}E_{x})\psi(x,s),\quad(x,s)\in{\mathbb{R}}^{3}\times{\mathbb{Z}}_{2s+1},$
(2.6)

for $\psi\in L^{2}({\mathbb{R}}^{3}\times{\mathbb{Z}}_{2s+1};\mathcal{F})$
such that $\psi(x,s)\in D(\phi(f_{A,j}E_{x}))$ for all
$(x,s)\in{\mathbb{R}}^{3}\times{\mathbb{Z}}_{2s+1}$ and (2.6) defines again an
element in $L^{2}({\mathbb{R}}^{3}\times{\mathbb{Z}}_{2s+1};\mathcal{F})$.
Likewise we define $B_{j}(x)$. In the physics literature the operators (2.5)
are often written in terms of so called operator valued distributions. This is
outlined in the following remark.

######  Remark 2.1.

Introducing operator valued distributions $a_{\lambda}(k)$ and
$a_{\lambda}^{*}(k)$ satisfying the so called canonical commutation relations,
[RS75, X.7], one can write

$\displaystyle A(x)$
$\displaystyle=\sum_{\lambda=1,2}\int\frac{\varepsilon_{\lambda}(k)}{\sqrt{2|k|}}\left(\overline{\rho(k)}a_{\lambda}(k)e^{\mathrm{i}k\cdot
x}+\rho(k)a_{\lambda}^{*}(k)e^{-\mathrm{i}k\cdot x}\right)dk\;,$ (2.7)
$\displaystyle B(x)$
$\displaystyle=\sum_{\lambda=1,2}\int\frac{\mathrm{i}k\wedge\varepsilon_{\lambda}(k)}{\sqrt{2|k|}}\left(\overline{\rho(k)}a_{\lambda}(k)e^{\mathrm{i}k\cdot
x}-\rho(k)a_{\lambda}^{*}(k)e^{-\mathrm{i}k\cdot x}\right)dk\;,$ (2.8)

where the integrals are understood as weak integrals on a suitable dense
subspace.

We shall adopt the standard convention that for $v=(v_{1},v_{2},v_{3})$ we
write $v^{2}:=\sum_{j=1}^{3}v_{j}v_{j}$. By $x$ we denote the position of the
electron and its canonically conjugate momentum by $p=-\mathrm{i}\nabla_{x}$.
If $s=1/2$, let $S=(\sigma_{1},\sigma_{2},\sigma_{3})$ denote the vector of
Pauli-matrices. If $s=0$, let $S=0$. The number $e\in{\mathbb{R}}$ is called
the coupling constant. The so called form factor
$\rho\colon{\mathbb{R}}^{3}\to{\mathbb{C}}$ is a measurable function for which
we shall assume the following hypothesis for the main theorem. We assume that
for some $\Lambda\in(0,\infty)$ we have

$\rho(k)=\frac{1}{(2\pi)^{3/2}}{\mathbf{1}}_{[0,\Lambda]}(|k|)\;,\quad
k\in{\mathbb{R}}^{3}.$ (2.9)

where ${\mathbf{1}}_{[0,\Lambda]}$ denotes the characteristic function of the
set $[0,\Lambda]$. One can show that $H_{m}$ is self-adjoint on
$D(\Delta\otimes\mathds{1})\cap D(\mathds{1}\otimes H_{\mathrm{f},m})$, cf.
[HH08a, Hir02].

The Hamiltonian is translation invariant and commutes with the generator of
translations, i.e., the operator of total momentum

$P_{\mathrm{tot}}=p+P_{\mathrm{f}}\;.$

Let

$W=\exp(\mathrm{i}x\cdot P_{\mathrm{f}})\;.$

Note $WP_{\mathrm{tot}}W^{*}=p$ so that in the new representation $p$ is the
total momentum. One easily computes

$WH_{m}W^{*}=\frac{1}{2}\left(p-P_{\mathrm{f}}+eA\right)^{2}+eS\cdot
B+H_{\mathrm{f},m}\;,$

where we set $A:=A(0)$ and $B:=B(0)$. Let $F$ be the Fourier transform in the
electron variable $x$, i.e., on $L^{2}({\mathbb{R}}^{3})$,

$\displaystyle(F\psi)(\xi)=\frac{1}{(2\pi)^{3/2}}\int_{{\mathbb{R}}^{3}}e^{-\mathrm{i}\xi\cdot
x}\psi(x)dx\;.$ (2.10)

Then the composition $U=FW$ is a unitary operator

$\displaystyle U\colon\mathcal{H}_{\mathrm{full}}\to
L^{2}({\mathbb{R}}^{3}\times{\mathbb{Z}}_{2s+1})\otimes\mathcal{F}\cong
L^{2}({\mathbb{R}}^{3};{\mathbb{C}}^{2s+1}\otimes\mathcal{F})=\int_{{\mathbb{R}}^{3}}^{\oplus}{\mathbb{C}}^{2s+1}\otimes\mathcal{F}d\xi,$

yielding the so called fiber decomposition of the Hamiltonian,

$\displaystyle UH_{m}U^{*}=\int_{{\mathbb{R}}^{3}}^{\oplus}H_{m}(\xi)d\xi,$

where

$H_{m}(\xi)=\frac{1}{2}(\xi-P_{\mathrm{f}}+eA)^{2}+eS\cdot B+H_{\mathrm{f},m}$
(2.11)

is an operator in the so called reduced Hilbert space

$\displaystyle\mathcal{H}:={\mathbb{C}}^{2s+1}\otimes\mathcal{F},$

cf. [RS78, Spo04]. The following result shows that the operator (2.11) is well
defined, cf. [Hir07], [HL08], [LMS07] and [HS20].

######  Theorem 2.2.

For all $m\geq 0$, $\xi\in{\mathbb{R}}^{3}$, and $e\in{\mathbb{R}}$ the
operator $H_{m}(\xi)$ is bounded from below and self-adjoint on the natural
domain of $P_{\mathrm{f}}^{2}+H_{\mathrm{f},m}$.

We call

$E_{m}(\xi):=\inf\sigma(H_{m}(\xi)).$

the ground state energy, which in general does not need to be an eigenvalue.
Let us collect a few elementary properties of the ground state energy as a
function of $\xi$.

######  Lemma 2.3.

The following holds.

  * (i)

The function $t_{m}\colon\xi\mapsto\frac{\xi^{2}}{2}-E_{m}(\xi)$ is convex.

  * (ii)

The function $\xi\mapsto E_{m}(\xi)$ is almost everywhere differentiable.

  * (iii)

The function $\xi\mapsto E_{m}(\xi)$ is rotationally invariant.

  * (iv)

If $E_{m}$ is differentiable in $\xi$ we have $|\nabla E_{m}(\xi)|\leq|\xi|$

###### Proof.

(i) This follows since the supremum of convex functions is convex. (ii) Convex
functions are locally Lipschitz, see for example Lemma C.2 in the Appendix.
Lipschitz functions are almost everywhere differentiable by Rademacher’s
theorem, see for example Theorem C.4 in the Appendix. So (ii) follows from
(i). (iii) Follows from well known transformation properties of the field
operators and the rotation invariance of $\rho$. (iv) By restricting the
function $t_{m}$ onto a straight line through the origin the claim follows
from Lemma C.3 in the appendix, where the symmetry assumption follows from
(iii). ∎

Henceforth, we shall write $H_{\mathrm{f}}$, $H$, $H(\xi)$, and $E(\xi)$ for
$H_{\mathrm{f},0}$, $H_{0}$, $H_{0}(\xi)$, and $E_{0}(\xi)$, respectively. For
massless photons, wich corresponds to the case $m=0$, the fiber Hamiltonian
does not have a ground state for momenta $\xi$ for which $\nabla E(\xi)\neq
0$. This is the content of the following theorem shown in [HH08].

######  Theorem 2.4.

Let $e\neq 0$. If $E(\cdot)$ is differentiable at $\xi$ and has a nonzero
derivative, then $H(\xi)$ does not have a ground state.

The physical interpretation of Theorem 2.4 is that charged particles with
nonzero velocity $\nabla E(\xi)\neq 0$, aquire an infinite photon cloud which
ceases to be square integrable. This will be made mathematically more precise
below and can be viewed as a manifestation of the so called infrared
catastrophe. On the other hand, if one considers the case with zero momentum
$\xi=0$ one can show in fact that the fiber Hamiltonian has a ground state.
This has been established for small values of the coupling constant in [Che08]
and recently [HS20] for all values of the coupling constant under an energy
inequality assumption (2.12), which is discussed below.

To control the infrared divergence for $\xi\neq 0$ one typically introduces an
infrared regularization. One way to achieve this, is to introduce a positive
photon mass, i.e., one considers $H_{m}(\xi$) for $m>0$. One can shown that
the operator $H_{m}(\xi)$ has a ground state for all $m>0$ and $|\xi|$
sufficiently small. Such a result has been obtained in various situations
[Frö74, Frö73, Spo04, GLL01, LMS07]. In the following we will work with the
result from [HS20], since it suits our specific situation. To formulate it, we
will use the following energy inequality, which reads as follows. We say that
the energy inequality holds for $e\in{\mathbb{R}}$ and $m\geq 0$ if

$\displaystyle E_{m}(\xi)\geq
E_{m}(0),\quad\quad\forall\xi\in{\mathbb{R}}^{3}.$ (2.12)

This inequality has been intensively investigated in the literature. In the
spinless case $s=0$ it has been shown to hold for all values $m\geq 0$ and
$e\in{\mathbb{R}}$ using functional integration, [Gro72, Spo04, Hir07, LMS07].
This is the content of the following theorem.

######  Theorem 2.5.

In the spin-less case, $s=0$, (2.12) holds for all $e\in{\mathbb{R}}$ and
$m\geq 0$.

In case of spin one half it has been shown to hold, but only in a limited
range of parameters. That is, for $s=1/2$ Inequality (2.12) has to the best of
our knowledge not yet been shown by means of functional integration. For
$s=1/2$ Inequality (2.12) follows for small $|e|$ from the main theorem stated
in [Che08], which in turn is based on perturbative arguments. Now let us state
the result about existence of ground states for positive photon mass from
[HS20], which will be used in this paper, see also [Frö74, Frö73, Spo04,
GLL01, LMS07].

######  Theorem 2.6.

Let $e\in{\mathbb{R}}$ and $m>0$ and suppose the energy inequality (2.12)
holds. If $|\xi|\leq 1$, then $E_{m}(\xi)$ is an eigenvalue of $H_{m}(\xi)$
isolated from the essential spectrum.

Henceforth, we let $\psi_{m}(\xi)$ denote a normalized eigenvector of
$H_{m}(\xi)$ with eigenvalue $E_{m}(\xi)$, whose existence is granted by
Theorem 2.6. In Proposition 4.1, in Section 4, we will show that $E_{m}(\xi)$
is monotonically decreasing as $m\downarrow 0$ and

$E_{m}(\xi)\to E(\xi).$ (2.13)

As the positive mass is removed, the ground state $\psi_{m}(\xi)$ acquires an
infinite photon cloud which ceases to be square integrable and hence does not
converge in the Hilbert space. However, if one removes this diverging photon
cloud by a Bogoluibov transformation, one can show that the transformed ground
states converge to a nonzero vector in Hilbert space. Such a picture has been
established in various situations [Frö73, Frö74, Piz05, CF07, CFP09]. To
describe this in more detail we define for $m\geq 0$ the function

$h_{m,\xi}(\lambda,k)=e{\varepsilon}_{\lambda}(k)\cdot\nabla_{\xi}E_{m}(\xi)\frac{\rho(k)}{\sqrt{\left|k\right|}}\frac{1}{\omega_{m}(k)-k\cdot\nabla_{\xi}E_{m}(\xi)},$
(2.14)

where $(\lambda,k)\in\\{1,2\\}\times{\mathbb{R}}^{3}$. In Remark 2.7 we sketch
a heuristic argument for (2.14).

######  Remark 2.7.

Following [Piz03, Section 2] and references therein we give a heuristic and
formal argument for the choice (2.14). Assuming that $\psi_{m}(\xi)$ should be
“coherent in the infrared region” we make the following “Ansatz” as $k\to 0$

$\displaystyle a_{\lambda}(k)\psi_{m}(\xi)\approx-
g_{m,\xi}(\lambda,k)\psi_{m}(\xi)$ (2.15)

for some function $g_{m,\xi}$ with values in the linear maps of
${\mathbb{C}}^{2s+1}$. By a formal application of the virial theorem we have

$\displaystyle\langle{\psi_{m}(\xi),[H_{m}(\xi),a_{\lambda}(k)]\psi_{m}(\xi)}\rangle=0.$
(2.16)

Calculating the formal commutator $[H_{m}(\xi),a_{\lambda}(k)]$ by means of
the so called pull-through formula, c.f. [BFS98, Lemma A.1], inserting the
ansatz (2.15) into (2.16), using the Feynman-Hellmann formula, c.f. Lemma 6.1,
and solving for $g_{m,\xi}$ we find

$g_{m,\xi}(\lambda,k)\approx e\left(\nabla_{\xi}E_{m}(\xi)\cdot
f_{A}(\lambda,k)+S\cdot
f_{B}(\lambda,k)\right)\frac{1}{\omega_{m}(k)-k\cdot\nabla_{\xi}E_{m}(\xi)+\frac{1}{2}k^{2}}$

as $k\to 0$. Dropping the higher order terms involving $k^{2}$ and $f_{B}$ we
arrive at (2.14).

From Lemma 2.3 we see that the derivative of $E_{m}$ exists almost everywhere.
Now if the derivative exists, then $h_{m,\xi}\in\mathfrak{h}$ whenever $m>0$
and $|\nabla E_{m}(\xi)|<1$, where the latter inequality always holds if
$|\xi|<1$ by Lemma 2.3. However in case $m=0$ we have
$h_{0,\xi}\notin\mathfrak{h}$. Whenever $h_{m,\xi}\in\mathfrak{h}$, we can
define

$\displaystyle U_{m}(\xi)=e^{\mathrm{i}\pi(h_{m,\xi})}.$ (2.17)

Working with $U_{m}(\xi)$ will require control of the derivatives of the
ground state energies. In Proposition 4.2 at the end of Section 4 we show that
for any seqeunce $(m_{j})_{j\in{\mathbb{N}}}$ of nonnegative numbers which
tend to zero, we have

$\displaystyle\nabla E_{m_{j}}(\xi)\overset{j\to\infty}{\longrightarrow}\nabla
E(\xi)$ (2.18)

for almost all $\xi\in{\mathbb{R}}^{3}$. This will allow us to study strong
limits of $U_{m}(\xi)$ as $m\downarrow 0$.

##  3 Statement of Main Results

As a first result we will show in Section 8 the following theorem, which
describes explicitly the removal of the divergent photon cloud of the ground
state as the positive photon mass tends to zero.

######  Theorem 3.1.

Let $e\in{\mathbb{R}}$ and suppose there exists an $m_{0}>0$ such that the
energy inequality (2.12) holds for all $m\in(0,m_{0})$. Then for almost all
$\xi\in{\mathbb{R}}^{3}$ with $|\xi|<1$, there exists a sequence
$(m_{j})_{j\in{\mathbb{N}}}$ of positive numbers converging to zero such that
$U_{m_{j}}(\xi)\psi_{m_{j}}(\xi)$ converges to a nonzero vector,
$\widehat{\psi}_{0}(\xi)$, in the Hilbert space $\mathcal{H}$.

######  Remark 3.2.

We note that Theorem 3.1 is similar to a result obtained in [CFP09], where
convergence is shown for a spinless charged particle. The proof given in that
paper is constructive, and holds for all $\xi$ with $|\xi|<1/3$ and small
values of $|e|$. Theorem 3.1 extends that result for almost all $\xi$ with
$|\xi|<1$ to all values of the coupling constant $e\in{\mathbb{R}}$, see also
Remark 3.6.

Next we address the question whether the nonzero vector
$\widehat{\psi}_{0}(\xi)$ given in Theorem 3.1 can be expressed as the ground
state of an infrared renormalized Hamiltonian. The main result of this paper
is, that this is indeed the case. For this we will show in Theorem 3.4, below,
that as $m_{j}$ tends to zero the operators

$\displaystyle U_{m_{j}}(\xi)H(\xi)U_{m_{j}}(\xi)^{*}$ (3.1)

coverge in resolvent sense for almost all $\xi$ with $|\xi|<1$, to a self-
adjoint operator, which will be given explicitly in (3.3) below, and which we
will refer to as the infrared renormalized fiber Hamiltonian. To capture this
limit in explicit terms we introduce first for measurable functions
$f,g\colon{\mathbb{Z}}_{2}\times{\mathbb{R}}^{3}\to{\mathbb{C}}$ with
$\overline{f}g\in L^{1}({\mathbb{Z}}_{2}\times{\mathbb{R}}^{3})$ the
sesquilinear form

$\mathfrak{s}(f,g)=\frac{1}{2}\sum_{\lambda=1}^{2}\int\overline{f}(\lambda,k)g(\lambda,k)dk.$

Second for a measurable function
$f\colon{\mathbb{Z}}_{2}\times{\mathbb{R}}^{3}\to{\mathbb{R}}$ we define the
following formal operator

$\displaystyle T_{m}(f;\xi)$
$\displaystyle:=\frac{1}{2}\sum_{j=1}^{3}\left(\xi_{j}-P_{f,j}+eA_{j}-\phi({K_{j}}f)-2e\operatorname{Re}\mathfrak{s}(f,f_{A,j})-\mathfrak{s}(f,{K_{j}}f)\right)^{2}$
$\displaystyle\quad+e\sum_{j=1}^{3}S_{j}\left(B_{j}-2\operatorname{Re}\mathfrak{s}(f,f_{B,j})\right)+H_{\mathrm{f},m}+\phi(\omega_{m}f)+\mathfrak{s}(f,\omega_{m}f).$
(3.2)

We will show in in Section 5, Lemma 5.1, that this operator is
$P_{\mathrm{f}}^{2}+H_{\mathrm{f},m}$ bounded, provided the function $f$
satisfies certain properties. Specifically, using the next lemma, following
from Proposition 5.5 in Section 5, we can give an explicit definition of the
infrared renormalized fiber Hamiltonian.

######  Lemma 3.3.

Let $E$ be differentiable in $\xi$ and suppose $|\nabla_{\xi}E(\xi)|<1$. Then

$\displaystyle\widehat{H}(\xi):=T_{0}(h_{0,\xi};\xi)$ (3.3)

can be realized as a selfadjoint operator with domain
$D(P_{\mathrm{f}}^{2}+H_{\mathrm{f}})$.

We now state the first main result of this paper, which follows from
Proposition 5.5 in Section 5 as well.

######  Theorem 3.4.

Let $E$ be differentiable in $\xi$ and suppose $|\nabla_{\xi}E(\xi)|<1$. Let
$(m_{j})_{j\in{\mathbb{N}}}$ be a sequence of positive numbers converging to
zero and $\nabla E_{m_{j}}(\xi)\to\nabla E(\xi)$. Then the operators

$U_{m_{j}}(\xi)H(\xi)U_{m_{j}}(\xi)^{*},$

converge as $j\to\infty$ in norm resolvent sense to $\widehat{H}(\xi)$.

We note that in view of Lemma 2.3 and (2.18) the assumptions of Theorem 3.4
are valid for almost all $\xi\in{\mathbb{R}}^{3}$ with $|\xi|<1$. We now state
the second main result of this paper, which will be shown in Section 8, and
which covers charged particles without spin and charged particles with spin
one half.

######  Theorem 3.5.

Let $e\in{\mathbb{R}}$. Suppose that one of the following two assumptions is
satisfied:

  * (i)

$s=0$,

  * (ii)

$s=1/2$ and there exists an $m_{0}>0$ such that the energy inequality (2.12)
holds for all $m\in(0,m_{0})$.

Then for almost all $\xi\in{\mathbb{R}}^{3}$ with $\left|\xi\right|<1$ the
following holds.

  * (a)

The function $E$ is differentiable at $\xi$, $|\nabla E(\xi)|<1$, and the
operator $\widehat{H}(\xi)$ has a ground state, i.e., $E(\xi)$ is an
eigenvalue of $\widehat{H}(\xi)$.

  * (b)

There exists a sequence $(m_{j})_{j\in{\mathbb{N}}}$ of positive numbers
converging to zero such that $U_{m_{j}}(\xi)\psi_{m_{j}}(\xi)$ converges to
the ground state.

The fact that the assertion of Theorem 3.5 only holds for almost all $\xi$
with length less than one, does not affect the fiber direct integral, since
sets of measure zero do not play a role in integration.

######  Remark 3.6.

We will determine in Theorem 8.1, below, the set of $\xi$’s, for which the
assertions of Theorem 3.1 as well as Theorem 3.5 can be shown, more precisely.
That set only depends on regularity properties of the energies $E_{m}$. We
note that regularity properties for such models have been studied using
various methods. In this regard we want to mention well established methods
from renormalization, iterated perturbation, or statistical mechanics, cf.
[Che08, Bac+07, Piz03, AH12].

##  4 Outline of the Proofs

Theorem 3.4 will be shown in Section 5 using well established estimates
involving creation and annihilation operators. To prove the existence of a
ground state for the renormalized Hamiltonian $\widehat{H}(\xi)$, i.e.,
Theorem 3.5, we follow closely ideas given in [GLL01] and [LMS07] combined
with a regularization procedure used in [Frö73, Frö74, Piz05, CFP09].

The basic idea of the proof is to regularize the Hamiltonian by adding a
positive mass to the photons and to study ground state properties when this
regularization is removed. That is, we consider the Hamiltonian $H_{m}(\xi)$
for $m>0$. Recall that the operator $H_{m}(\xi)$ has a normalized ground state
for all $m>0$ and $|\xi|$ sufficiently small, see Theorem 2.6, which we denote
by $\psi_{m}(\xi)$. In Proposition 4.1 we will show that $E_{m}(\xi)$ is
monotonically decreasing as $m\downarrow 0$ and

$E_{m}(\xi)\to E(\xi).$

Working with $U_{m}(\xi)$ requires control of the derivatives of the ground
state energies. In Proposition 4.2 at the end of this section we show that for
any seqeunce $(m_{j})_{j\in{\mathbb{N}}}$ of nonnegative numbers which
converges to zero, there exists a set $D$ such that ${\mathbb{R}}^{3}\setminus
D$ has Lebesgue measure zero, such that $E_{m_{j}}$, $j\in{\mathbb{N}}$, and
$E$ are differentiable at all points in $D$, and for all $\xi\in D$ we have

$\displaystyle\nabla E_{m_{j}}(\xi)\overset{j\to\infty}{\longrightarrow}\nabla
E(\xi).$ (4.1)

Using (4.1) we will show in Proposition 5.5 in Section 5 that

$U_{m_{j}}(\xi)H(\xi)U_{m_{j}}(\xi)^{*}\to\widehat{H}(\xi)$

in norm resolvent sense for almost all $\xi$ with $|\xi|<1$. We then show in
Proposition 5.6 using (2.13) and standard estimates involving creation and
annihilation operators that $U_{m_{j}}(\xi)\psi_{m_{j}}(\xi)$ is a minimizing
sequence for $\widehat{H}(\xi)$, i.e.,

$\displaystyle
0\leq\langle{U_{m_{j}}(\xi)\psi_{m_{j}}(\xi),(\widehat{H}(\xi)-E(\xi))U_{m_{j}}(\xi)\psi_{m_{j}}(\xi)}\rangle\to
0.$ (4.2)

The key idea for proving the existence of a ground state is to show that
$U_{m_{j}}(\xi)\psi_{m_{j}}(\xi)$ has a convergent subsequence. This is
achieved by showing that the sequence of these vectors lies in a compact
subset of the Hilbert space. To this end, we make use of the so called pull-
through formula, which is shown in Lemma 7.2,

$a_{\lambda}(k)\psi_{m}(\xi)=\frac{e\rho(k)}{\sqrt{2\left|k\right|}}R_{m,\xi}(k)\left(-{\varepsilon}_{\lambda}(k)\cdot
v(\xi)+S\cdot(\mathrm{i}k\wedge{\varepsilon}_{\lambda}(k))\right)\psi_{m}(\xi),$
(4.3)

where we defined

$\displaystyle R_{m,\xi}(k)$
$\displaystyle:=(H_{m}(\xi-k)+\omega_{m}(k)-E_{m}(\xi))^{-1},$ (4.4)
$\displaystyle v(\xi)$ $\displaystyle:=\xi-P_{\mathrm{f}}+eA.$ (4.5)

To estimate the resolvent occuring on the right hand side of (4.3), we will
relate it to the second order derivative of the ground state energy as a
function of $\xi$. In Proposition 4.2 estimates on the second order
derivatives of the ground state energy are established using convexity
properties, where it is shown that for almost all $\xi$ the second order
derivatives $\partial_{l}^{2}E_{m_{j}}(\xi)$ exist and

$\liminf_{j}(-\partial_{l}^{2}E_{m_{j}}(\xi))<\infty$

for every $l=1,2,3$. In Section 6 it will be shown that these bounds on the
second order derivative of the energy yield bounds on the resolvent. Using
these resolvent bounds in the pull-through formula (4.3) we will derive in
Propositions 7.4 and 7.5 of Section 7 estimates on

$a_{\lambda}(k)U_{m_{j}}(\xi)\psi_{m_{j}}(\xi)\text{ and
}\nabla_{k}a_{\lambda}(k)U_{m_{j}}(\xi)\psi_{m_{j}}(\xi).$

With the help of these estimates we will prove a fractional derivative bound
in Lemma 7.6. In Section 8 we use this fractional derivative bound to show
that the vectors $U_{m_{j}}(\xi)\psi_{m_{j}}(\xi)$, $j\in{\mathbb{N}}$, lie
indeed in a compact set. Thus by compactness there exists a strongly
convergent subsequence which converges to a nonzero vector, say
$\widehat{\psi}_{0}(\xi)$. We recall that this is the content of Theorem 3.1 ,
which will be proven in Section 8.

Using lower semicontinuity of nonnegative quadratic forms [Sim78] (or
alternatively the spectral theorem and Fatou’s Lemma), we will conclude from
(4.2) and Theorem 3.1 that for almost all $\xi$ with $|\xi|<1$

$\displaystyle 0$
$\displaystyle\leq\langle{\widehat{\psi}_{0}(\xi),(\widehat{H}(\xi)-E(\xi))\widehat{\psi}_{0}(\xi)}\rangle$
$\displaystyle\qquad\leq\liminf_{j\to\infty}\langle{U_{m_{j}}(\xi)\psi_{m_{j}}(\xi),(\widehat{H}(\xi)-E(\xi))U_{m_{j}}(\xi)\psi_{m_{j}}}\rangle=0,$

i.e., that $\widehat{\psi}_{0}(\xi)$ is a ground state of $\widehat{H}(\xi)$.
As outlined in Section 8 this will then establish Theorem 3.5.

In the remaining part of this section we derive further properties of the
ground state energies $E_{m}(\xi)$.

######  Proposition 4.1.

Let $\xi\in{\mathbb{R}}^{3}$. Whenever $m_{1}\geq m_{2}\geq 0$ we have
$E_{m_{1}}(\xi)\geq E_{m_{2}}(\xi)\geq E(\xi)$, and

$\displaystyle E(\xi)=\lim_{m\downarrow 0}E_{m}(\xi)\;.$ (4.6)

###### Proof.

For $0\leq m_{2}\leq m_{1}$ we have
$\omega\leq\omega_{m_{2}}\leq\omega_{m_{1}}$ and hence $H(\xi)\leq
H_{m_{2}}(\xi)\leq H_{m_{1}}(\xi)$.This implies the first statement. Moreover,
it implies the existence of the limit

$E(\xi)\leq\lim_{m\downarrow 0}E_{m}(\xi).$ (4.7)

To show the opposite inequality we argue as follows. From Theorem 2.2 it
follows that any core for $P_{\mathrm{f}}^{2}+H_{\mathrm{f}}$ is a core for
$H(0)$. Thus for any ${\varepsilon}>0$, there exists a vector $\phi\in
D(N)\cap D(P_{\mathrm{f}}^{2}+H_{\mathrm{f}})$ such that

$\langle{\phi,H(\xi)\phi}\rangle\leq E(\xi)+{\varepsilon}\;.$

On the other hand, since $H_{m}(\xi)\leq
H(\xi)+m\mathsf{d}\Gamma(\mathds{1})$, it follows that for any $m$,

$E_{m}(\xi)\leq\langle{\phi,H_{m}(\xi)\phi}\rangle\leq\langle{\phi,H(\xi)\phi}\rangle+m\langle{\phi,\mathsf{d}\Gamma(\mathds{1})\phi}\rangle\leq
E(\xi)+{\varepsilon}+m\langle{\phi,\mathsf{d}\Gamma(\mathds{1})\phi}\rangle\;.$

Hence

$\lim_{m\downarrow 0}E_{m}(\xi)\leq E(\xi)+{\varepsilon}.$ (4.8)

Since ${\varepsilon}>0$ is arbitrary, (4.6) follows from (4.7) and (4.8). ∎

######  Proposition 4.2.

Let $(m_{j})_{j\in{\mathbb{N}}}$ be a sequence of nonnegative numbers which
converges to zero. Then there exists a set $D$ such that
${\mathbb{R}}^{3}\setminus D$ has Lebesgue measure zero and the following
holds. The functions $E_{m_{j}}$, $j\in{\mathbb{N}}$, and $E$ are
differentiable at all points in $D$, and for all $\xi\in D$ we have

  * (a)

$\nabla E_{m_{j}}(\xi)\overset{j\to\infty}{\longrightarrow}\nabla E(\xi)$,

  * (b)

the second partial derivatives $\partial_{l}^{2}E_{m_{j}}(\xi)$ exist and
satisfy for every $l=1,2,3$ that
$\liminf_{j}(-\partial_{l}^{2}E_{m_{j}}(\xi))<\infty$ .

###### Proof.

We note that $t_{m}\colon\xi\mapsto\frac{1}{2}\xi^{2}-E_{m}(\xi)$ are convex
functions by Lemma 2.3. Together with Proposition 4.1, we see that (a) and (b)
follow from Lemmas C.6 and C.7 in the appendix, respectively. ∎

##  5 Transformation of the Hamiltonian

We will now look at unitary transformations of the form

$e^{\mathrm{i}\pi(f)}H_{m}(\xi)e^{-\mathrm{i}\pi(f)},\quad f\in\mathfrak{h},$

and limits of such objects. Some background is collected in Appendix B.

Before we state the lemma we define

$\|f\|_{(m)}=\|f\|+\|\omega_{m}^{-1/2}f\|$

for all measurable functions
$f\colon{\mathbb{R}}^{3}\times{\mathbb{Z}}_{2}\to{\mathbb{C}}$, where
$\|\cdot\|$ denotes the $L^{2}$-norm in
$L^{2}({\mathbb{Z}}_{2}\times{\mathbb{R}}^{3})$. We write $f\in
L^{2}_{(m)}({\mathbb{Z}}_{2}\times{\mathbb{R}}^{3})$ if $\|f\|_{(m)}<\infty$.

######  Lemma 5.1.

Let $m\geq 0$ and
$f\colon{\mathbb{Z}}_{2}\times{\mathbb{R}}^{3}\to{\mathbb{R}}$ be a measurable
function, with

$\displaystyle K_{j}f,K_{j}^{2}f,\omega_{m}f\in
L^{2}_{(m)}({\mathbb{Z}}_{2}\times{\mathbb{R}}^{3}),$ (5.1)
$\displaystyle\omega_{m}|f|^{2},K_{j}|f|^{2},ef\overline{f}_{A,j},ef\overline{f}_{B,j}\in
L^{1}({\mathbb{Z}}^{2}\times{\mathbb{R}}^{3}).$ (5.2)

Then $T_{m}(f;\xi)$, given in (3.2), defines a
$P_{\mathrm{f}}^{2}+H_{\mathrm{f},m}$ bounded operator on
$D(P_{\mathrm{f}}^{2}+H_{\mathrm{f},m})$. In particular, there exits a
constant $C$ such that for all $\psi\in
D(P_{\mathrm{f}}^{2}+H_{\mathrm{f},m})$ we have

$\displaystyle\|T_{m}(f,\xi)\psi\|\leq$ $\displaystyle
C\sum_{j}\big{\\{}\|ef_{B,j}\|_{(m)}+\|e\overline{f}f_{B,j}\|_{1}+\|\omega_{m}f\|_{(m)}+\||f|^{2}\omega_{m}\|_{1}$
(5.3) $\displaystyle+(\|ef_{A,j}\|_{(m)}+\|K_{j}f\|_{(m)}+|c_{j}(f)|+1)^{2}$
$\displaystyle+\|K_{j}^{2}f\|_{(m)}+\|eK_{j}f_{A,j}\|_{(m)})\big{\\}}\|(P_{\mathrm{f}}^{2}+H_{\mathrm{f},m}+1)\psi\|,$

where

$\displaystyle c_{j}(h)$
$\displaystyle:=\xi_{j}-2e{\operatorname{Re}}\mathfrak{s}(h,f_{A,j})-\mathfrak{s}(h,K_{j}h).$
(5.4)

###### Proof.

From the assumptions (5.2) it follows that all expressions involving the
sesquilinear form $\mathfrak{s}$ are finite. Multiplying out the square in
(3.2) we will show that each term is $P_{\mathrm{f}}^{2}+H_{\mathrm{f},m}$
bounded. For
$\psi\in\mathcal{F}_{\operatorname{fin}}(C_{\mathrm{c}}({\mathbb{Z}}_{2}\times{\mathbb{R}}^{3}))$
we find that

$\displaystyle T_{m}(f;\xi)\psi$ (5.5)
$\displaystyle=\left(\frac{1}{2}P_{\mathrm{f}}^{2}+H_{\mathrm{f},m}\right)\psi$
$\displaystyle-\frac{1}{2}\sum_{j}P_{\mathrm{f},j}\left(\phi(ef_{A,j}-{K_{j}}f)+c_{j}(f))\right)\psi$
(5.6)
$\displaystyle-\frac{1}{2}\sum_{j}\left(\phi(ef_{A,j}-{K_{j}}f)+c_{j}(f)\right)P_{\mathrm{f},j}\psi$
(5.7)
$\displaystyle+\frac{1}{2}\sum_{j}\left(\phi(ef_{A,j}-{K_{j}}f)+c_{j}(f)\right)^{2}\psi$
(5.8)
$\displaystyle+e\sum_{j}S_{j}\left(\phi(f_{B,j})-2{\operatorname{Re}}\mathfrak{s}(f,f_{B,j})\right)\psi+(\phi(\omega_{m}f)+\mathfrak{s}(f,\omega_{m}f))\psi.$
(5.9)

From the first identity in Lemma A.2, see that the terms in (5.9) are
$H_{\mathrm{f},m}$ bounded, since $\omega_{m}f,f_{B,j}\in L^{2}_{(m)}$, i.e.,

$\displaystyle\|\eqref{eq:pfsqhfm00-4}\|\leq
C\big{(}\sum_{j}(\|ef_{B,j}\|_{(m)}+\|e\overline{f}f_{B,j}\|_{1}+\|\omega_{m}f\|_{(m)}+\||f|^{2}\omega_{m}\|_{1}\big{)}\|(H_{\mathrm{f},m}+1)^{1/2}\psi\|.$

Now let us estimate the terms in (5.8). Clearly $(\phi(K_{j}f)+eA_{j})^{2}$ is
$H_{\mathrm{f},m}$ bounded by Lemma A.2, since $K_{j}f,f_{A,j}\in
L^{2}_{(m)}({\mathbb{Z}}_{2}\times{\mathbb{R}}^{3})$ and so

$\displaystyle\|\eqref{eq:pfsqhfm00-3}\|\leq
C\sum_{j=1}^{3}(\|ef_{A,j}\|_{(m)}+\|K_{j}f\|_{(m)}+|c_{j}(f)|+1)^{2}\|(H_{\mathrm{f},m}+1)\psi\|.$

Similarly, the term in (5.7) can be bounded by Lemma A.2 as

$\displaystyle\|\eqref{eq:pfsqhfm00-2}\|\leq
C\sum_{j}\big{(}\|ef_{A,j}\|_{(m)}+\|K_{j}f\|_{(m)}+|c_{j}(f)|\big{)}\|(H_{\mathrm{f},m}+1)^{1/2}P_{\mathrm{f},j}\psi\|.$

For (5.6) observe that Lemma B.1 yields on
$\mathcal{F}_{\operatorname{fin}}(C_{\mathrm{c}}({\mathbb{Z}}_{2}\times{\mathbb{R}}^{3}))$

$\displaystyle
P_{f,j}\phi(K_{j}f+ef_{A,j})=\phi(K_{j}f+ef_{A,j})P_{f,j}-\mathrm{i}\phi(\mathrm{i}K_{j}^{2}f+\mathrm{i}eK_{j}f_{A,j}).$
(5.10)

Now each term in (5.10) is estimated using Lemma A.2 and we find for
$\psi\in\mathcal{F}_{\operatorname{fin}}(C_{\mathrm{c}}({\mathbb{Z}}_{2}\times{\mathbb{R}}^{3}))$

$\displaystyle\|\eqref{eq:pfsqhfm00-1}\|$ $\displaystyle\leq
C\sum_{j}\big{(}\|ef_{A,j}\|_{(m)}+\|K_{j}f\|_{(m)}+|c_{j}(f)|\big{)}\|(H_{\mathrm{f},m}+1)^{1/2}P_{\mathrm{f},j}\psi\|$
$\displaystyle\quad+C\sum_{j}(\|K_{j}^{2}f\|_{(m)}+\|eK_{j}f_{A,j}\|_{(m)})\|(H_{\mathrm{f},m}+1)\psi\|.$
(5.11)

Then (5.11) extends to all $\psi\in D(P_{\mathrm{f}}^{2}+H_{\mathrm{f},m})$,
since
$\mathcal{F}_{\operatorname{fin}}(C_{\mathrm{c}}({\mathbb{Z}}_{2}\times{\mathbb{R}}^{3}))$
is a core for $P_{\mathrm{f}}^{2}+H_{\mathrm{f},m}$. Finally, collecting above
estimates shows (5.3) and hence the claim follows.

∎

We note that if the assumptions of Lemma 5.1 hold, we will define
$T_{m}(f;\xi)$ according to that lemma as the operator with domain
$D(P_{\mathrm{f}}^{2}+H_{\mathrm{f},m})$. The following lemma will be used for
the proof of Proposition 5.3, below, to establish self-adjointness. We note
that the self-adjointness could alternatively be shown directly using
alternative methods, cf. [HS20, LMS07, Hir02] and references therein.

######  Lemma 5.2.

There exists a constant $c_{0}$ such that for $m\geq 0$ and $f\in\mathfrak{h}$
with $\omega^{2}f\in L^{2}_{(m)}({\mathbb{Z}}_{2}\times{\mathbb{R}}^{3})$ we
have

$\displaystyle\left\|\left(\frac{1}{2}P_{\mathrm{f}}^{2}+H_{\mathrm{f},m}\right)e^{\mathrm{i}\pi(f)}\psi\right\|\leq
C(f)\|(P_{\mathrm{f}}^{2}+H_{\mathrm{f},m}+1)\psi\|,$ (5.12)

where

$\displaystyle C(f)$
$\displaystyle=c_{0}\big{(}\big{[}(1+\|\omega_{m}^{1/2}f\|+\|\omega_{m}f\|_{(m)}\big{]}^{2}+\|\omega_{m}^{2}f\|_{(m)}\big{)}.$

In particular,
$e^{\mathrm{i}\pi(f)}D(P_{\mathrm{f}}^{2}+H_{\mathrm{f},m})=D(P_{\mathrm{f}}^{2}+H_{\mathrm{f},m})$.

###### Proof.

We apply Lemma B.5 to the operators $P_{\mathrm{f},j}$ and $H_{\mathrm{f},m}$,
and find on
$\mathcal{F}_{\operatorname{fin}}(C_{\mathrm{c}}({\mathbb{Z}}_{2}\times{\mathbb{R}}^{3}))\subset
D(P_{\mathrm{f}}^{2}+H_{\mathrm{f},m})$

$\displaystyle
e^{-\mathrm{i}\pi(f)}(P_{\mathrm{f}}^{2}+H_{\mathrm{f},m})e^{\mathrm{i}\pi(f)}$
(5.13)
$\displaystyle\quad=\frac{1}{2}e^{-\mathrm{i}\pi(f)}P_{\mathrm{f}}e^{\mathrm{i}\pi(f)}\cdot
e^{-\mathrm{i}\pi(f)}P_{\mathrm{f}}e^{\mathrm{i}\pi(f)}+e^{-\mathrm{i}\pi(f)}H_{\mathrm{f},m}e^{\mathrm{i}\pi(f)}$
$\displaystyle\quad=\frac{1}{2}\sum_{j=1}^{3}(P_{f,j}+\phi(K_{j}f)+\frac{1}{2}\mathfrak{s}(f,K_{j}f))^{2}+H_{\mathrm{f},m}+\phi(\omega_{m}f)+\frac{1}{2}\mathfrak{s}(f,\omega_{m}f).$

Now the bound (5.12) follows from Lemma 5.1 with $e=0$, $\xi=0$, and $s=0$
using the fact that
$\mathcal{F}_{\operatorname{fin}}(C_{\mathrm{c}}({\mathbb{Z}}_{2}\times{\mathbb{R}}^{3}))$
is a core for $P_{\mathrm{f}}^{2}+H_{\mathrm{f},m}$. This implies that
$e^{\mathrm{i}\pi(f)}D(P_{\mathrm{f}}^{2}+H_{\mathrm{f},m})\subset
D(P_{\mathrm{f}}^{2}+H_{\mathrm{f},m})$. Finally, the unitarity of
$e^{\mathrm{i}\pi(f)}$ implies that we have in fact equality. ∎

######  Proposition 5.3.

Let $m\geq 0$, and let $f\in\mathfrak{h}$ satisfy $K_{j}^{2}f\in
L^{2}_{(m)}({\mathbb{Z}}_{2}\times{\mathbb{R}}^{3})$, $j=1,2,3$. Then we have
on $D(P_{\mathrm{f}}^{2}+H_{\mathrm{f},m})$

$\displaystyle e^{-\mathrm{i}\pi(f)}$ $\displaystyle
H_{m}(\xi)e^{\mathrm{i}\pi(f)}=T_{m}(f;\xi)$ (5.14)

and the operator is self-adjoint on $D(P_{\mathrm{f}}^{2}+H_{\mathrm{f},m})$.

###### Proof.

We apply Lemma B.4 to the field operators $A$ and $B$ and Lemma B.5 to the
operators $P_{\mathrm{f},j}$ and $H_{\mathrm{f},m}$, and find on
$\mathcal{F}_{\operatorname{fin}}(C_{\mathrm{c}}({\mathbb{Z}}_{2}\times{\mathbb{R}}^{3}))\subset
D(P_{\mathrm{f}}^{2}+H_{\mathrm{f},m})$

$\displaystyle e^{-\mathrm{i}\pi(f)}H_{m}(\xi)e^{\mathrm{i}\pi(f)}$ (5.15)
$\displaystyle\quad=\frac{1}{2}e^{-\mathrm{i}\pi(f)}(\xi-
P_{\mathrm{f}}+eA)e^{\mathrm{i}\pi(f)}\cdot e^{-\mathrm{i}\pi(f)}(\xi-
P_{\mathrm{f}}+eA)e^{\mathrm{i}\pi(f)}+eS\cdot
e^{\mathrm{i}\pi(f)}Be^{\mathrm{i}\pi(f)}$
$\displaystyle\qquad\qquad+e^{-\mathrm{i}\pi(f)}H_{\mathrm{f},m}e^{\mathrm{i}\pi(f)}$
$\displaystyle\quad=\frac{1}{2}\sum_{j=1}^{3}\left(\xi_{j}-P_{f,j}+eA_{j}-\phi({K_{j}}f)-2e{\operatorname{Re}}\mathfrak{s}(f,f_{A,j})-\mathfrak{s}(f,{K_{j}}f)\right)^{2}$
$\displaystyle\quad\qquad+e\sum_{j=1}^{3}S_{j}\left(B_{j}-2{\operatorname{Re}}\mathfrak{s}(f,f_{B,j})\right)+H_{\mathrm{f},m}+\phi(\omega_{m}f)+\mathfrak{s}(f,\omega_{m}f)$
$\displaystyle=T_{m}(f;\xi),$

where the last identity is simply the definition (3.2) of $T_{m}(f;\xi)$. By
Theorem 2.2 we know that $H_{m}(\xi)$ is self-adjoint on
$D(P_{\mathrm{f}}^{2}+H_{\mathrm{f}})$. By unitarity of
$e^{-\mathrm{i}\pi(f)}$ if follows that
$e^{-\mathrm{i}\pi(f)}H_{m}(\xi)e^{\mathrm{i}\pi(f)}$ is self-adjoint on the
domain $e^{-\mathrm{i}\pi(f)}D(P_{\mathrm{f}}^{2}+H_{\mathrm{f}})$. But this
domain equals $D(P_{\mathrm{f}}^{2}+H_{\mathrm{f}})$ by Lemma 5.2. ∎

Next we want to consider limits of expressions of the form (3.1). For this we
prove the following lemma.

######  Lemma 5.4.

Suppose $f$ satisfies the assumptions of Lemma 5.1 and let $g\in\mathfrak{h}$.
Then for all $\psi\in D(P_{\mathrm{f}}^{2}+H_{\mathrm{f}})$ we have

$\left\|(T_{0}(f;\xi)-e^{-\mathrm{i}\pi(g)}H(\xi)e^{\mathrm{i}\pi(g)})\psi\right\|\leq
C(g,f;\xi)\left\|(H_{\mathrm{f}}+\frac{1}{2}P_{\mathrm{f}}^{2}+1)\psi\right\|,$
(5.16)

where

$\displaystyle C(g,f;\xi)$
$\displaystyle=|D(f,g)|+\sqrt{2}\|\omega(f-g)\|_{(0)}$ (5.17)
$\displaystyle+\sum_{j=1}^{3}\big{\\{}(|d_{j}(f,g)|+\sqrt{2}\|K_{j}(f-g)\|_{(0)})$
$\displaystyle\times(2+\sqrt{2}(\|ef_{A,j}-K_{j}f\|_{(0)}+\|ef_{A,j}-K_{j}g\|_{(0)})+|c_{j}(f)|+|c_{j}(g)|)\big{\\}}$
$\displaystyle+\sum_{j=1}^{3}\sqrt{2}\|K_{j}^{2}(f-g)\|_{(0)},$

with

$\displaystyle c_{j}(h)$
$\displaystyle:=\xi_{j}-2e{\operatorname{Re}}\mathfrak{s}(h,f_{A,j})-\mathfrak{s}(h,K_{j}h),$
$\displaystyle D(f,g)$
$\displaystyle:=-2\sum_{j=1}^{3}e\S_{j}{\operatorname{Re}}\mathfrak{s}(f-g,f_{B,j})+\mathfrak{s}(f-g,\omega
f)+\mathfrak{s}(g,\omega(f-g)),$ $\displaystyle d_{j}(f,g)$
$\displaystyle:={\mathrm{c}}_{j}(f)-c_{j}(g)=-2e{\operatorname{Re}}\mathfrak{s}(f-g,f_{A,j})-\mathfrak{s}(f-g,K_{j}f)-\mathfrak{s}(g,K_{j}(f-g)).$

###### Proof.

First we take the difference of the operators given in (5.14). Using
Proposition 5.3 and (3.2) we find on $D(P_{\mathrm{f}}^{2}+H_{\mathrm{f}})$

$\displaystyle T_{0}(f;\xi)-e^{-\mathrm{i}\pi(g)}H(\xi)e^{\mathrm{i}\pi(g)}$
$\displaystyle\quad=T_{0}(f;\xi)-T_{0}(g;\xi)$
$\displaystyle\quad=\frac{1}{2}\sum_{j=1}^{3}\left\\{F_{j}(f)^{2}-F_{j}(g))^{2}\right\\}+\Phi(\omega(f-g))+D(f,g)$
$\displaystyle\quad=\frac{1}{2}\sum_{j=1}^{3}\left\\{F_{j}(f)(F_{j}(f)-F_{j}(g))+(F_{j}(f)-F_{j}(g))F_{j}(g)\right\\}$
(5.18) $\displaystyle\qquad+\Phi(\omega(f-g))+D(f,g)$ (5.19)

where we defined for $j=1,2,3$,

$\displaystyle F_{j}(h)$
$\displaystyle:=-P_{\mathrm{f},j}+eA_{j}-\Phi(K_{j}h)+c_{j}(h).$

Now the terms in the line (5.19) are easily estimated using Lemma A.2 and
contribute (5.17) to the constant. Now let us estimate the term in line
(5.18). Taking the difference we find

$\displaystyle F_{j}(f)-F_{j}(g)=-\Phi(K_{j}(f-g))+d_{j}(f,g).$

Recall the notation $A_{j}=\Phi(f_{A,j})$. First we estimate the first term in
the sum in line (5.18). By the triangle inequality

$\displaystyle\|F_{j}(f)(F_{j}(f)-F_{j}(g))\psi\|$
$\displaystyle\quad\leq\|P_{\mathrm{f},j}(F_{j}(f)-F_{j}(g))\psi\|$ (5.20)
$\displaystyle\quad+\|\Phi(ef_{A,j}-K_{j}f)(F_{j}(f)-F_{j}(g))\psi\|$ (5.21)
$\displaystyle\quad+\|c_{j}(f)(F_{j}(f)-F_{j}(g))\psi\|.$ (5.22)

Now we estimate the terms on the right hand side. Using Lemma A.2 and Lemma
B.1 we find

(5.20)
$\displaystyle\leq\left\\{|d_{j}(f,g)|+\sqrt{2}\|K_{j}(f-g)\|_{(0)}\right\\}\|(H_{\mathrm{f}}+1)^{1/2}P_{f,j}\psi\|$
$\displaystyle\quad+\sqrt{2}\|K_{j}^{2}(f-g)\|_{(0)}\|(H_{\mathrm{f}}+1)^{1/2}\psi\|.$
(5.23)

Again by Lemma A.2 we have

(5.21)
$\displaystyle\leq(|d_{j}(f,g)|+\sqrt{2}\|K_{j}(f-g)\|_{(0)})\sqrt{2}\|ef_{A,j}-K_{j}g\|_{(0)}\|(H_{\mathrm{f}}+1)\psi\|,$
(5.22)
$\displaystyle\leq(|d_{j}(g,f)|+\sqrt{2}\|K_{j}(f-g)\|_{(0)})|c_{j}(f)|\|(H_{\mathrm{f}}+1)^{1/2}\psi\|.$

The second term in the sum of (5.18) is estimated analogously with $g$ and $f$
interchanged, with the only difference that we do not need the term in line
(5.23). The lemma now follows by collecting estimates and observing that
$d_{j}(g,f)=-d_{j}(f,g)$. ∎

Now Lemma 3.3 will follow as an immediate consequence of the following
proposition. Furthermore, we will show below that it also implies Theorem 3.4.

######  Proposition 5.5.

Let $E$ be differentiable in $\xi$ and $|\nabla E(\xi)|<1$. If $m>0$ and
$|\nabla E_{m}(\xi)|<1$, then $h_{m,\xi}\in\mathfrak{h}$. Furthermore, assume
$(m_{j})_{j\in{\mathbb{N}}}$ is a sequence of positive numbers converging to
zero and $\nabla E_{m_{j}}(\xi)\to\nabla E(\xi)$. Then

$\left(e^{-\mathrm{i}\pi(h_{m_{j},\xi})}H(\xi)e^{\mathrm{i}\pi(h_{m_{j},\xi})}-T_{0}(h_{0,\xi};\xi)\right)(P_{\mathrm{f}}^{2}+H_{\mathrm{f}}+1)^{-1}\to
0,\quad(j\to\infty).$ (5.24)

###### Proof.

For $m>0$ and $|\nabla E_{m}(\xi)|<1$ we have $h_{m,\xi}\in\mathfrak{h}$,
since $\omega_{m}(k)-k\nabla E_{m}(\xi)>\omega_{m}(k)(1-|\nabla E_{m}(\xi)|)$.
Statement (5.24) follows by inserting $h_{0,\xi}$ for $f$ and $h_{m,\xi}$,
$m>0$, for $g$ in Lemma 5.4 and by observing that the constant
$C(h_{m,\xi},h_{0,\xi};\xi)$ given in (5.17) tends to zero as $m\downarrow 0$.
This can be seen by looking at the explicit expressions and using dominated
convergence. ∎

###### Proof of Lemma 3.3.

The operators
$e^{-\mathrm{i}\pi(h_{m,\xi})}H(\xi)e^{\mathrm{i}\pi(h_{m,\xi})}$ are self-
adjoint on $D(P_{\mathrm{f}}^{2}+H_{\mathrm{f}})$ by Proposition 5.3. On the
other hand $T_{0}(h_{0,\xi};\xi)$ is $P_{\mathrm{f}}^{2}+H_{\mathrm{f}}$
bounded by Lemma 5.1. So the self-adjointness of $T_{0}(h_{0,\xi};\xi)$
follows in view of Kato-Rellich, and the following estimate

$\displaystyle\|(e^{-\mathrm{i}\pi(h_{m,\xi})}H(\xi)e^{\mathrm{i}\pi(h_{m,\xi})}-T_{0}(h_{0,\xi};\xi))(e^{-\mathrm{i}\pi(h_{m,\xi})}H(\xi)e^{\mathrm{i}\pi(h_{m,\xi})}+\mathrm{i})^{-1}\|$
(5.25)
$\displaystyle\leq\|(e^{-\mathrm{i}\pi(h_{m,\xi})}H(\xi)e^{\mathrm{i}\pi(h_{m,\xi})}-T_{0}(h_{0,\xi};\xi))(P_{\mathrm{f}}^{2}+H_{\mathrm{f}}+1)^{-1}\|$
$\displaystyle\quad\times\|(P_{\mathrm{f}}^{2}+H_{\mathrm{f}}+1)(e^{-\mathrm{i}\pi(h_{m,\xi})}H(\xi)e^{\mathrm{i}\pi(h_{m,\xi})}+\mathrm{i})^{-1}\|$
$\displaystyle\leq\|(e^{-\mathrm{i}\pi(h_{m,\xi})}H(\xi)e^{\mathrm{i}\pi(h_{m,\xi})}-T_{0}(h_{0,\xi};\xi))(P_{\mathrm{f}}^{2}+H_{\mathrm{f}}+1)^{-1}\|$
$\displaystyle\quad\times\|e^{\mathrm{i}\pi(h_{m,\xi})}(P_{\mathrm{f}}^{2}+H_{\mathrm{f}}+1)e^{-\mathrm{i}\pi(h_{m,\xi})}(H(\xi)+\mathrm{i})^{-1}\|$
$\displaystyle\leq\|(e^{\mathrm{i}\pi(h_{m,\xi})}H(\xi)e^{\mathrm{i}\pi(h_{m,\xi})}-T_{0}(h_{0,\xi};\xi))(P_{\mathrm{f}}^{2}+H_{\mathrm{f}}+1)^{-1}\|$
$\displaystyle\quad\times\|(P_{\mathrm{f}}^{2}+H_{\mathrm{f}}+1)e^{-\mathrm{i}\pi(h_{m,\xi})}(P_{\mathrm{f}}^{2}+H_{\mathrm{f}}+1)^{-1}\|\|(P_{\mathrm{f}}^{2}+H_{\mathrm{f}}+1)(H(\xi)+\mathrm{i})^{-1}\|.$

Now the last term on the right hand side of (5.25) is bounded by Theorem 2.2,
the second term on the right hand side is bounded uniformly in $m\geq 0$ in
view of the estimate (5.12) in Lemma 5.2, and the first term on the right hand
side of (5.25) tends to zero as $m\downarrow 0$ by Proposition 5.5. ∎

###### Proof of Theorem 3.4.

By Proposition 5.5 (a) the resolvent $(T_{0}(h_{0,\xi};\xi)+\mathrm{i})^{-1}$
is well-defined and we obtain

$\displaystyle\|(e^{-\mathrm{i}\pi(h_{m,\xi})}H(\xi)e^{\mathrm{i}\pi(h_{m,\xi})}-T_{0}(h_{0,\xi};\xi))(T_{0}(h_{0,\xi};\xi)+\mathrm{i})^{-1}\|$
$\displaystyle\leq\|(e^{-\mathrm{i}\pi(h_{m,\xi})}H(\xi)e^{\mathrm{i}\pi(h_{m,\xi})}-T_{0}(h_{0,\xi};\xi))(H_{\mathrm{f}}+P_{\mathrm{f}}^{2}+1)^{-1}\|$
$\displaystyle\quad\times\|(H_{\mathrm{f}}+P_{\mathrm{f}}^{2}+1)(T_{0}(h_{0,\xi};\xi)+\mathrm{i})^{-1}\|.$
(5.26)

Now the right hand side of (5.26) tends to zero as $m\downarrow 0$ by (b) of
Proposition 5.5. This implies norm resolvent convergence. We recall that for a
sequence of self-adjoint operators $(A_{n})_{n\in{\mathbb{N}}}$ to converge in
norm resolvent sense to a selfadjoint operator $A$, it suffices to show that
$\|(A_{n}-A)(A+\mathrm{i})^{-1}\|\to 0$ as $n\to\infty$, which can be seen
from [RS81, Theorem VIII.19] and the first resolvent identity. ∎

In the following proposition we will establish that
$U_{m_{j}}(\xi)\psi_{m_{j}}(\xi)$ is a minimizing sequence of
$\widehat{H}(\xi)$.

######  Proposition 5.6.

Suppose $|\xi|<1$. Furthermore, assume $(m_{j})$ is a sequence of positive
numbers converging to zero and $\nabla E_{m_{j}}(\xi)\to\nabla E(\xi)$. Then

$0\leq\langle{U_{m_{j}}(\xi)\psi_{m_{j}}(\xi),(\widehat{H}(\xi)-E(\xi))U_{m_{j}}(\xi)\psi_{m_{j}}(\xi)}\rangle\to
0$

in the limit $j\to\infty$.

###### Proof.

We recall that the assumption $|\xi|<1$ guarantees by Lemma 2.3 that always
$|\nabla E_{m}(\xi)|<1$ and that, for $m>0$, $U_{m}(\xi)$ is a well defined
unitary operator. Therefore we know that $h_{m,\xi}\in\mathfrak{h}$. For the
proof we define

$\widehat{H}_{m}(\xi):=U_{m}(\xi)H(\xi)U_{m}(\xi)^{*}.$

Then we have by definition

$\displaystyle\left<U_{m}(\xi)\psi_{m},(\widehat{H}(\xi)-E(\xi))U_{m_{j}}\psi_{m_{j}}\right>$
$\displaystyle=\left<U_{m}(\xi)\psi_{m},(\widehat{H}(\xi)-\widehat{H}_{m}(\xi)+\widehat{H}_{m}(\xi)-E(\xi))U_{m}(\xi)\psi_{m}\right>$
$\displaystyle=\left<U_{m}(\xi)\psi_{m},(\widehat{H}(\xi)-\widehat{H}_{m}(\xi))U_{m}(\xi)\psi_{m}\right>+\left<\psi_{m},({H}(\xi)-E(\xi))\psi_{m}\right>.$
(5.27)

Now we use Proposition 5.5 to control the first term of (5.27) along the
sequence $(m_{j})$

$\displaystyle(\widehat{H}(\xi)-\widehat{H}_{m_{j}}(\xi))^{*}(\widehat{H}(\xi)-\widehat{H}_{m_{j}}(\xi))$
$\displaystyle\leq C_{j}^{2}(H_{\mathrm{f}}+P_{\mathrm{f}}^{2}+1)^{2},$

with $C_{j}\to 0$ as $j\to\infty$. Taking the square root yields

$\displaystyle\widehat{H}(\xi)-\widehat{H}_{m_{j}}(\xi)\leq|\widehat{H}(\xi)-\widehat{H}_{m_{j}}(\xi)|$
$\displaystyle\leq|C_{j}|(H_{\mathrm{f}}+P_{\mathrm{f}}^{2}+1).$

Inserting this above we find

$\displaystyle\left<U_{m_{j}}(\xi)\psi_{m_{j}},(\widehat{H}(\xi)-\widehat{H}_{m_{j}}(\xi))U_{m_{j}}(\xi)\psi_{m_{j}}\right>\leq|C_{j}|\langle
U_{m_{j}}(\xi)\psi_{m_{j}},(H_{\mathrm{f}}+P_{\mathrm{f}}^{2}+1)U_{m_{j}}(\xi)\psi_{m_{j}}\rangle.$

The second term in (5.27) can be bounded by means of $H(\xi)\leq H_{m}(\xi)$

$\displaystyle\left<\psi_{m},({H}(\xi)-E(\xi))\psi_{m}\right>\leq\left<\psi_{m},({H}_{m}(\xi)-E(\xi))\psi_{m}\right>=E_{m}(\xi)-E(\xi).$

Now, as $H(\xi)$ is closed on $D(P_{\mathrm{f}}^{2}+H_{\mathrm{f}})$ we find
with Lemma 5.2 for some constant $C$ independent of $m\geq 0$

$\displaystyle\langle
U_{m}(\xi)\psi_{m},(H_{\mathrm{f}}+P_{\mathrm{f}}^{2}+1)U_{m}(\xi)\psi_{m}\rangle$
$\displaystyle\leq
C\langle\psi_{m},(H_{\mathrm{f}}+P_{\mathrm{f}}^{2}+1)\psi_{m}\rangle$
$\displaystyle\leq C\langle\psi_{m},(H(\xi)+1)\psi_{m}\rangle$
$\displaystyle\leq
C\langle\psi_{m},(H_{m}(\xi)+1)\psi_{m}\rangle=C(E_{m}(\xi)+1),$

which is in fact bounded by Proposition 4.1. Collecting estimates and using
that $E_{m}(\xi)\to E(\xi)$ by Proposition 4.1, the claim follows. ∎

##  6 Perturbation Theory

In this section we use perturbation theory to obtain resolvent bounds. More
precisely, we use second order perturbation theory to determine the second
derivative of the energy in terms of the resolvent. Then using bounds on the
second derivative of the energy, we find bounds on the resolvent. These bounds
will be used to obtain infrared bounds for the ground states of the massive
Hamiltonians.

If $|\xi|\leq 1$, then we know by Theorem 2.6 that $E_{m}(\xi)$ is an
eigenvalue of $H_{m}(\xi)$ isolated from the essential spectrum. Let
$P_{m}(\xi)$ denote the projection onto the kernel of $H_{m}(\xi)$, which is
finite dimensional. We recall the definition of $v(\xi)$ in (4.5).

######  Lemma 6.1.

Let $e\in{\mathbb{R}}$ und $m>0$, and suppose (2.12) holds. Let $|\xi|\leq 1$
and suppose that all first order partial derivatives of $E_{m}(\cdot)$ exist
at $\xi$. Then the following holds.

  1. (a)

$P_{m}v(\xi)P_{m}=\nabla_{\xi}E_{m}(\xi)P_{m}$.

  2. (b)

$E_{m}(\cdot)$ is twice partially differentiable at $\xi$ and for $j=1,2,3$,

$\displaystyle\partial_{j}^{2}E_{m}(\xi)$ $\displaystyle\ \leq
1-2P_{m}(\xi)(v(\xi)-\nabla
E_{m}(\xi))_{j}\frac{1}{H_{m}(\xi)-E_{m}(\xi)}(v(\xi)-\nabla
E_{m}(\xi))_{j}P_{m}(\xi)$

as inequality in the sense of operators on $\operatorname{ran}P_{m}(\xi)$.

######  Remark 6.2.

We note that by (a) the vectors in $\operatorname{ran}(v(\xi)-\nabla
E_{m}(\xi))_{j}P_{m}(\xi)$ are orthogonal to $\operatorname{ran}P_{m}(\xi)$
and hence the resolvent in (b) is well defined.

###### Proof.

For the proof we use analytic perturbation theory. For details we refer the
reader to [Kat80, RS78]. On ${\mathbb{C}}^{3}$ the operator valued function
$\zeta\mapsto H_{m}(\xi+\zeta):=H_{m}(\xi)+\zeta\cdot
v(\xi)+\frac{1}{2}\zeta^{2}$ is an analytic family of type (A) in each
component. By Theorem 2.6 we know that $E_{m}(\xi)$ is an eigenvalue isolated
from the essential spectrum, and thus with finite multiplicity
$n(\xi)=\dim\operatorname{ran}P_{m}(\xi)$.
By degenerate perturbation theory, cf. [RS78, Theorem XII.13], we know that
there exist $n(\xi)$ complex analytic functions $e_{s}$, $s=1,\ldots,n(\xi)$,
in a neighborhood of zero such that

$e_{s}(0)=E_{m}(\xi),\quad s=1,\ldots,n(\xi),$ (6.1)

and $e_{s}(\zeta)$ is an eigenvalue of $H_{m}(\xi+\zeta)$. The functions
$e_{s}$, $s=1,\ldots,n(\xi)$, are real for real $\zeta$, and

$\partial_{j}e_{s}(0),\quad s=1,\ldots,n(\zeta),$

are the eigenvalues of

$P_{m}(\xi)v_{j}(\xi)P_{m}(\xi),$

which can be seen from a power series expansion and comparison of
coefficients. Since $x\mapsto E_{m}(\xi+x)=\inf_{s=1,\ldots,n}e_{s}(x)$ is
differentiable at $x=0$, by assumption, and $e_{s}(0)=E_{m}(\xi)$, it can be
seen from a first order Taylor expansion that all derivatives must be equal at
the origin and equal $\partial_{j}E(\xi)$, i.e.,

$\partial_{j}e_{s}(0)=\partial_{j}E(\xi),\quad s=1,\ldots,n(\xi).$ (6.2)

Thus, $\nabla E_{m}(\xi)P_{m}(\xi)=P_{m}(\xi)v(\xi)P_{m}(\xi)$. This shows
(a). Now by second order degenerate perturbation theory, we find

$\partial_{j}^{2}e_{s}(0),\quad s=1,\ldots,n(\xi),$

are the eigenvalues of

$P_{m}(\xi)-2P_{m}(\xi)(v(\xi)-\nabla
E(\xi))(H_{m}(\xi)-E_{m}(\xi))^{-1}(v(\xi)-\nabla E(\xi))P_{m}(\xi),$

which can again be seen from a power series expansion. By (6.2) and (6.1) and
by the definition of $E_{m}(\xi)$ as an infimum, it can be seen from a second
order Taylor expansion that $E_{m}$ is twice differentiable at $\xi$ and that

$\partial_{j}^{2}E_{m}(\xi)=\inf_{s}\partial_{j}^{2}e_{s}(0).$

This shows (b). ∎

##  7 Infrared Bounds

In this section we derive infrared bounds. These will then be used in the next
section to show that infrared regularized ground states for a fixed momentum
lie in a compact set. For $\xi\in{\mathbb{R}}^{3}$ and $m\geq 0$ we define the
following quantity

$\Delta_{m}(\xi)=\inf_{k\in{\mathbb{R}}^{3}}\\{E_{m}(\xi-k)-E_{m}(\xi)+\omega_{m}(k)\\},$

which we shall use to estimate the resolvent of the Hamiltonian. We will use
the notation

$\displaystyle v:=v(0)=-P_{\mathrm{f}}+eA.$

The following proposition tells us when this resolvent defined in (4.4) is
well defined.

######  Proposition 7.1.

Let $e\in{\mathbb{R}}$ and $m>0$, and suppose the energy Inequality (2.12)
holds. Then the following holds.

  * (a)

We have $\Delta_{m}(\xi)>0$, whenever $|\xi|\leq 1$.

  * (b)

For $|\xi|\leq 1$ and all $k\in{\mathbb{R}}^{3}$ we have

$H_{m}(\xi-k)+\omega_{m}(k)-E_{m}(\xi)>0.$

  * (c)

For $|\xi|<1$ and all $k\in{\mathbb{R}}^{3}$

$\left\|R_{m,\xi}(k)\right\|\leq\frac{1}{1-|\xi|}\frac{1}{\left|k\right|}.$
(7.1)

###### Proof.

First we show that the function $\xi\mapsto E_{m}(\xi)$ satisfies the
assumptions (i)–(iii) of Proposition C.1 in the Appendix. Property (i) is
simply the energy inequality (2.12). From Lemma 2.3 we know on the one hand
that $t_{m}(\xi)=\frac{1}{2}\xi^{2}-E_{m}(\xi)$ is convex, i.e., (iii) and on
the other hand that $E_{m}(-\xi)=E_{m}(\xi)$, which implies
$t_{m}(-\xi)=t_{m}(\xi)$. Thus by convexity $t_{m}(0)\leq t_{m}(\xi)$, which
implies (ii). It therefore follows from Proposition C.1

$E_{m}(\xi-k)-E_{m}(\xi)\geq\left\\{\begin{array}[]{ll}-|k||\xi|+\frac{1}{2}k^{2}&\text{,
if }|k|\leq|\xi|,\\\ -\frac{1}{2}\xi^{2}&\text{, if
}|k|\geq|\xi|.\end{array}\right.$ (7.2)

(a) We find using $\omega(k)>|k|$ in (7.2)

$\displaystyle E_{m}(\xi-k)-E_{m}(\xi)+\omega(k)$
$\displaystyle>\left\\{\begin{array}[]{ll}-|k||\xi|+|k|&\text{, if
}|k|\leq|\xi|,\\\ -\frac{1}{2}\xi^{2}+|\xi|&\text{, if
}|k|\geq|\xi|,\end{array}\right.$ $\displaystyle\geq 0,$

provided $|\xi|\leq 1$.
(b) From (a) we have $H_{m}(\xi-k)+\omega_{m}(k)-E_{m}(\xi)\geq
E_{m}(\xi-k)+\omega_{m}(k)-E_{m}(\xi)\geq\Delta_{m}(\xi)>0.$
(c) From (7.2) we find

$\displaystyle H_{m}(\xi-k)+\omega_{m}(k)-E_{m}(\xi)$ $\displaystyle\geq
E_{m}(\xi-k)-E_{m}(\xi)+|k|$
$\displaystyle\geq\begin{cases}|k|(1-|\xi|)&\text{, if
}\left|k\right|\leq\left|\xi\right|,\\\ |k|-|\xi|^{2}/2&\text{, if
}\left|k\right|\geq\left|\xi\right|.\end{cases}$

Observing that $|k|-|\xi|^{2}/2\geq|k|-\frac{|\xi||k|}{2}\geq(1-|\xi|)|k|$ if
$\left|k\right|\geq\left|\xi\right|$ shows the claim. ∎

Now we use the following relation, which we shall refer to as the pull-through
resolvent identity. To formulate it we shall make use of the physics notation
of the pointwise annihilation operator. One defines for
$(\lambda,k)\in{\mathbb{Z}}_{2}\times{\mathbb{R}}^{3}$ and
$\psi\in\mathcal{F}$

$\displaystyle[a_{\lambda}(k)\psi_{(n)}](\lambda_{1},{k}_{1},\ldots,\lambda_{n},{k}_{n})$
$\displaystyle=\sqrt{n+1}\psi_{(n+1)}(\lambda,k,\lambda_{1},{k}_{1},\ldots,\lambda_{n},{k}_{n}),\,n\in{\mathbb{N}}_{0}.$
(7.3)

Note that by the Fubini-Tonelli theorem
$a_{\lambda}(k)\psi_{(n+1)}\in\mathfrak{h}^{(n)}$ for almost every $k$. If
$\psi\in D(H_{\mathrm{f}})$ and $f\in\mathfrak{h}$ with $\|f\|_{(0)}<\infty$,
then a straightforward calculation using the definitions shows that in the
sense of weak integrals

$\displaystyle a(f)\psi$ $\displaystyle=\sum_{\lambda=1,2}\int
d{k}\overline{f(\lambda,{k})}a_{\lambda}({k})\psi.$ (7.4)

In fact, the integral on the right hand side (7.4) exists as a Bochner
integral. This can be seen from the following estimate

$\displaystyle\sum_{\lambda=1,2}\int\|\overline{f(\lambda,k)}a_{\lambda}(k)\psi\|dk$
$\displaystyle=\sum_{\lambda=1,2}\int
dk|\overline{f(\lambda,k)}\omega(k)^{-1/2}|\|\omega(k)^{1/2}a_{\lambda}(k)\psi\|dk$
$\displaystyle\leq\|\omega^{-1/2}f\|\left(\sum_{\lambda=1,2}\int
dk\langle{a_{\lambda}(k)\psi,\omega(k)a_{\lambda}(k)\psi}\rangle\right)^{1/2}$
$\displaystyle=\|\omega^{-1/2}f\|\|H_{\mathrm{f}}^{1/2}\psi\|.$

######  Lemma 7.2 (Pull-through resolvent identity).

Let $e\in{\mathbb{R}}$ and $m>0$, and suppose the energy Inequality (2.12)
holds. Suppose $|\xi|<1$. Then for all $k\in{\mathbb{R}}^{3}$ and
$\lambda\in\\{1,2\\}$ we have

$\displaystyle
a_{\lambda}(k)\psi_{m}(\xi)=\frac{e\rho(k)}{\sqrt{2\left|k\right|}}R_{m,\xi}(k)\left(-{\varepsilon}_{\lambda}(k)v(\xi)+S\cdot(\mathrm{i}k\wedge{\varepsilon}_{\lambda}(k))\right)\psi_{m}(\xi).$
(7.5)

###### Sketch of proof.

For details we refer to [CF07, Lemma 6.1] and [HH08, Lemma 7]. As $\psi_{m}\in
D(H_{\mathrm{f}})$, we have

$\sum_{n=0}^{\infty}\sum_{\lambda}\int|k|\left\|(a_{\lambda}(k)\psi_{m})_{(n)}\right\|^{2}dk=\langle\psi_{m},H_{\mathrm{f}}\psi_{m}\rangle<\infty\;,$

which implies $\left\|a_{\lambda}(k)\psi_{m}\right\|<\infty$ almost
everywhere.

The pull-through formula yields, e.g. [BFS98, Lemma A.1], for a.e.
$k\in{\mathbb{R}}^{3}$, $\lambda\in\\{1,2\\}$, in the sense of measurable
fucntions

$\displaystyle a_{\lambda}(k)H_{m}(\xi)\psi$
$\displaystyle=\left(\frac{1}{2}(\xi-k-P_{\mathrm{f}}+eA)^{2}+eS\cdot
B+H_{\mathrm{f},m}+\omega_{m}(k)\right)a_{\lambda}(k)\psi$
$\displaystyle\qquad+(ef_{A}(k)\cdot v(\xi)+eS\cdot f_{B}(k))\psi$

for any $\psi\in\mathcal{F}$. Therefore,

$\displaystyle a_{\lambda}(k)E_{m}(\xi)\psi_{m}(\xi)$
$\displaystyle=a_{\lambda}(k)H_{m}(\xi)\psi_{m}(\xi)$
$\displaystyle=((H_{m}(\xi-k)+\omega_{m}(k))a_{\lambda}(k)+ef_{A}(k)\cdot
v(\xi)+eS\cdot f_{B}(k))\psi_{m}(\xi),$

or in other words,

$\displaystyle((H_{m}(\xi-k)-E_{m}(\xi)+\omega_{m}(k))a_{\lambda}(k)\psi_{m}(\xi)=-(ef_{A}(k)\cdot
v(\xi)+eS\cdot f_{B}(k))\psi_{m}(\xi).$

Multiplying with $R_{m,\xi}(k)$ yields the desired formula. ∎

The next lemma is needed to estimate the resolvent occurring in (7.5). It uses
the estimate obtained by second order perturbation theory stated in Lemma 6.1.

######  Lemma 7.3.

There exists a constant $C$ such that the following holds. Let
$e\in{\mathbb{R}}$ and $m>0$, and suppose the energy Inequality (2.12) holds.
Then for all $\xi\in{\mathbb{R}}^{3}$ with $|\xi|<1$, such that $\nabla
E_{m}(\xi)$ exists, and all $k\in{\mathbb{R}}^{3}$ and $i=1,2,3$, we have

  1. (a)

$\left\|R_{m,\xi}(k)(v_{i}(\xi)-\partial_{i}E_{m}(\xi))\psi_{m}(\xi)\right\|\leq\frac{C|\frac{1}{2}-\partial_{i}^{2}E_{m}(\xi)|^{1/2}(1+|E_{m}(\xi)|)^{1/2}}{1-|\xi|}(|k|^{-1/2}+1)$,

  2. (b)

$\left\|R_{m,\xi}(k)v_{i}(\xi)\right\|\leq
C(1-|\xi|)^{-1}(1+|E_{m}(\xi)|)^{1/2}(|k|^{-1}+1)$.

###### Proof.

To simplify the notation we drop the $\xi$ label in $\psi_{m}(\xi)$,
$R_{m,\xi}(k)$, and $v(\xi)$, and we write

$\displaystyle h_{m}:=H_{m}(\xi),\qquad e_{m}:=E_{m}(\xi).$

(a) We start with the product inequality

$\displaystyle\|R_{m}(k)v_{j}\psi_{m}\|\leq\|R_{m}(k)(h_{m}-e_{m})^{1/2}\|\|(h_{m}-e_{m})^{-1/2}(v_{j}-\partial_{j}E_{m}(\xi))\psi_{m}\|.$
(7.6)

By Lemma 6.1 the second factor on the right hand side can be estimated using

$\|(h_{m}-e_{m})^{-1/2}(v_{j}-\partial_{j}E_{m}(\xi))\psi_{m}\|\leq\left|\frac{1}{2}-\partial_{j}^{2}E_{m}(\xi)\right|^{1/2}\;.$

It remains to estimate the first factor in (7.6). First we use the trivial
identity

$\displaystyle h_{m}-e_{m}$
$\displaystyle=\frac{1}{2}(v-k)^{2}+H_{f,m}-e_{m}+\frac{1}{2}k^{2}+(v-k)\cdot
k.$

Estimating the last term using

$(v-k)\cdot k\leq\frac{1}{2}|k|+\frac{1}{2}|k|(v-k)^{2},$

we find with $\frac{1}{2}(v-k)^{2}\leq H_{m}(\xi-k)$ that

$\displaystyle h_{m}-e_{m}$
$\displaystyle\leq(1+|k|)(H_{m}(\xi-k)+\omega_{m}(k)-e_{m})+\frac{1}{2}(|k|+k^{2})+|k|e_{m}.$
(7.7)

Now multiplying this inequality on both sides with the self-adjoint operator
$R_{m}(k)$ we obtain

$\displaystyle
R_{m}(k)(h_{m}-e_{m})R_{m}(k)\leq(1+|k|)R_{m}(k)+\left(\frac{1}{2}(|k|+k^{2})+|k|e_{m}\right)R_{m}(k)^{2}.$

Using this, we estimate

$\displaystyle\|R_{m}(k)(h_{m}-e_{m})^{1/2}\|^{2}$ (7.8)
$\displaystyle\qquad\leq\|(h_{m}-e_{m})^{1/2}R_{m}(k)\|^{2}$
$\displaystyle\qquad=\sup_{\|\phi\|=1}\langle{\phi,R_{m}(k)(h_{m}-e_{m})R_{m}(k)\phi}\rangle$
$\displaystyle\qquad\leq(1+|k|)\|R_{m}(k)\|+\left(\frac{1}{2}(|k|+k^{2})+|k|e_{m}\right)\|R_{m}(k)\|^{2}.$
(7.9)

Now using Inequality (7.1) this implies the bound stated in (a).
(b) Using that $v^{2}\leq h_{m}$ we see from (7.7) that

$\displaystyle v^{2}$
$\displaystyle\leq(1+|k|)(H_{m}(\xi-k)+\omega_{m}(k)-e_{m})+\frac{1}{2}(|k|+k^{2})+(1+|k|)e_{m}.$

This implies

$\displaystyle\left\|R_{m}(k)v_{i}\right\|^{2}\leq(1+|k|)\|R_{m}(k)\|+\left(\frac{1}{2}(|k|+k^{2})+(|k|+1)e_{m}\right)\|R_{m}(k)\|^{2}.$

Now using Inequality (7.1) this implies the bound in (b). ∎

We note that the assumption $|\xi|<1$ guarantees by Lemma 2.3 that $|\nabla
E_{m}(\xi)|<1$ always holds, provided the derivative exists. Therefore we know
that $h_{m,\xi}\in\mathfrak{h}$ and the following two expressions are well
defined and positive,

$\displaystyle D_{m,\xi}(k)$ $\displaystyle:=(\omega_{m}(k)-k\cdot\nabla
E_{m}(\xi))^{-1},$ (7.10) $\displaystyle\widehat{D}_{m,\xi}(k)$
$\displaystyle:=(1-\omega_{m}(k)^{-1}k\cdot\nabla E_{m}(\xi))^{-1}.$ (7.11)

To write the following estimates in compact notation we define

$F_{m,\xi}:=\sum_{i=1}^{3}\left(1+\left|\frac{1}{2}-\partial_{i}^{2}E_{m}(\xi)\right|^{1/2}\right)(1+|E_{m}(\xi)|)^{1/2}.$
(7.12)

Recall that if $\nabla E_{m}(\xi)$ exists, then also the second derivative,
cf. Lemma 6.1.

######  Proposition 7.4.

There exists a constant $C$ such that the following holds. Let
$e\in{\mathbb{R}}$ and $m>0$, and suppose the energy Inequality (2.12) holds.
Let $\xi\in{\mathbb{R}}^{3}$ with $|\xi|<1$, such that $\nabla E_{m}(\xi)$
exists. Then for a.e. $k\in{\mathbb{R}}^{3}$

$\left\|a_{\lambda}(k)U_{m}(\xi)\psi_{m}(\xi)\right\|\leq\frac{C|e\rho(k)|F_{m,\xi}}{1-|\xi|}(\widehat{D}_{m,\xi}(k)+1)(|k|^{-1}+|k|^{-1/2}).$

###### Proof.

Using the convergence of (7.4) as a Bochner integral and Lemma B.4 we find for
any $f\in\mathfrak{h}$ with $\|f\|_{(0)}<\infty$ and
$\varphi\in\mathcal{F}_{\operatorname{fin}}(\mathfrak{h})$ that

$\displaystyle\sum_{\lambda=1,2}\int
dk\overline{f(\lambda,k)}\langle{\varphi,U_{m}(\xi)^{*}a_{\lambda}(k)U_{m}(\xi)\psi_{m}(\xi)}\rangle$
(7.13)
$\displaystyle=\langle{\varphi,U_{m}(\xi)^{*}a(f)U_{m}(\xi)\psi_{m}(\xi)}\rangle$
$\displaystyle=\sum_{\lambda=1,2}\int
dk\overline{f(\lambda,k)}\langle{\varphi,a_{\lambda}(k)\psi_{m}(\xi)}\rangle$
$\displaystyle\quad+\sum_{\lambda=1,2}\int
dk\overline{f(\lambda,k)}e{\varepsilon}_{\lambda}(k)\cdot\nabla_{\xi}E_{m}(\xi)\frac{\rho(k)}{\sqrt{2\left|k\right|}}\frac{1}{\omega_{m}(k)-k\nabla_{\xi}\cdot
E_{m}(\xi)}\langle{\varphi,\psi_{m}(\xi)}\rangle.$

Since $f\in\mathfrak{h}$ with $\|f\|_{(0)}<\infty$ and
$\varphi\in\mathcal{F}_{\operatorname{fin}}(\mathfrak{h})$ are arbitrary it
follows from (7.13) and density of
$\mathcal{F}_{\operatorname{fin}}(\mathfrak{h})$ that for a.e.
$k\in{\mathbb{R}}^{3}$

$\displaystyle U_{m}(\xi)^{*}a_{\lambda}(k)U_{m}(\xi)\psi_{m}(\xi)$
$\displaystyle=a_{\lambda}(k)\psi_{m}(\xi)+e{\varepsilon}_{\lambda}(k)\nabla_{\xi}E_{m}(\xi)\frac{\rho(k)}{\sqrt{2\left|k\right|}}\frac{1}{\omega_{m}(k)-k\nabla_{\xi}E_{m}(\xi)}\psi_{m}(\xi).$
(7.14)

The goal is to show that as $k\to 0$ the leading order contribution of the
first term in (7.14) cancels the second term. To show this we use the pull-
through resolvent identity in Lemma 7.2 and can write

$a_{\lambda}(k)\psi_{m}(\xi)=(\mathrm{I})+(\mathrm{II}),$ (7.15)

where

$\displaystyle(\mathrm{I})$
$\displaystyle:=R_{m,\xi}(k)\frac{e\rho(k)}{\sqrt{2\left|k\right|}}S\cdot(\mathrm{i}k\wedge{\varepsilon}_{\lambda}(k))\psi_{m}(\xi),$
$\displaystyle(\mathrm{II})$
$\displaystyle:=-R_{m,\xi}(k)\frac{e\rho(k)}{\sqrt{2\left|k\right|}}{\varepsilon}_{\lambda}(k)\cdot
v(\xi)\psi_{m}(\xi).$

The first term, (I), can be estimated using (7.1)

$\displaystyle\left\|(\mathrm{I})\right\|\leq\frac{|e\rho(k)|}{1-|\xi|}\frac{1}{\sqrt{2\left|k\right|}}.$

Furthermore, we write the second term

$(\mathrm{II})=(\mathrm{II})_{1}+(\mathrm{II})_{2}$

by dividing it into the following two parts

$\displaystyle(\mathrm{II})_{1}$
$\displaystyle:=-R_{m,\xi}(k)\frac{e\rho(k)}{\sqrt{2\left|k\right|}}{\varepsilon}_{\lambda}(k)\cdot(v(\xi)-\nabla
E_{m}(\xi))\psi_{m}(\xi),$ $\displaystyle(\mathrm{II})_{2}$
$\displaystyle:=-R_{m,\xi}(k)\frac{e\rho(k)}{\sqrt{2\left|k\right|}}{\varepsilon}_{\lambda}(k)\cdot\nabla
E_{m}(\xi)\psi_{m}(\xi).$

We have

$\left\|(\mathrm{II})_{1}\right\|\leq
C\frac{|e\rho(k)|}{\sqrt{2\left|k\right|}}F_{m,\xi}(1-|\xi|)^{-1}(|k|^{-1/2}+1)$

by Lemma 7.3. To estimate $(\mathrm{II})_{2}$, we proceed as follows.
Similarly to [HH08] we introduce the operator

$\displaystyle R^{(0)}_{m,\xi}(k)$
$\displaystyle:=(H_{m}(\xi)+\omega_{m}(k)-E_{m}(\xi))^{-1}.$

Using the second resolvent identity, we find

$\displaystyle R_{m,\xi}(k)\psi_{m}(\xi)$
$\displaystyle=R_{m,\xi}^{(0)}(k)\psi_{m}+R_{m,\xi}(k)(-\frac{1}{2}k^{2}+k\cdot
v(\xi))R_{m,\xi}^{(0)}(k)\psi_{m}(\xi)$
$\displaystyle=\omega_{m}(k)^{-1}\psi_{m}+\omega_{m}(k)^{-1}R_{m,\xi}(k)(k\cdot
v(\xi)-\frac{1}{2}k^{2})\psi_{m}(\xi)$
$\displaystyle=\omega_{m}(k)^{-1}\left(\psi_{m}+k\cdot\nabla
E_{m}(\xi)R_{m,\xi}(k)\psi_{m}(\xi)+({\mathrm{III}})_{1}+({\mathrm{III}})_{2}\right),$
(7.16)

where we defined

$\displaystyle({\mathrm{III}})_{1}$
$\displaystyle:=-\frac{1}{2}k^{2}R_{m,\xi}(k)\psi_{m}(\xi),$
$\displaystyle(\mathrm{III})_{2}$
$\displaystyle:=R_{m,\xi}(k)k\cdot(v(\xi)-\nabla E(\xi))\psi_{m}(\xi).$

Hence, multiplying out (7.16) and bringing the term
$\omega_{m}(k)^{-1}k\cdot\nabla E_{m}(\xi)R_{m,\xi}(k)\psi_{m}(\xi)$ to the
left, we arrive at

$\displaystyle
R_{m,\xi}(k)\psi_{m}(\xi)=D_{m,\xi}(k)(\psi_{m}(\xi)+(\mathrm{III})_{1}+(\mathrm{III})_{2}).$
(7.17)

By Inequality (7.1) and by Lemma 7.3, respectively, we find

$\displaystyle\left\|(\mathrm{III})_{1}\right\|$
$\displaystyle\leq\frac{C|k|}{1-|\xi|},$ (7.18)
$\displaystyle\left\|(\mathrm{III})_{2}\right\|$
$\displaystyle\leq\frac{CF_{m,\xi}}{1-|\xi|}(|k|^{1/2}+|k|).$ (7.19)

Now inserting the above, we arrive at

$\displaystyle(\mathrm{II})_{2}$
$\displaystyle=-{\varepsilon}_{\lambda}(k)\cdot\nabla_{\xi}E_{m}(\xi)\frac{e\rho(k)}{\sqrt{2|k|}}\frac{1}{\omega_{m}(k)-k\cdot\nabla_{\xi}E_{m}(\xi)}\psi_{m}(\xi)$
$\displaystyle\qquad-{\varepsilon}_{\lambda}(k)\cdot\nabla_{\xi}E_{m}(\xi)\frac{e\rho(k)}{\sqrt{2|k|}}\frac{1}{\omega_{m}(k)-k\cdot\nabla_{\xi}E_{m}(\xi)}((\mathrm{III})_{1}+(\mathrm{III})_{2}).$
(7.20)

Using (7.18) and (7.19) we can estimate the second term on the right hand side
of (7.20) and find that it is of order $|k|^{-1}$. The claim now follows by
collecting estimates and using that the first term on the right hand side of
(7.20) exactly cancels the second term on the right hand side of (7.14). ∎

We still need an estimate involving derivatives. To this end, we shall
henceforth make an explicit choice of the polarization vectors. After a
possible unitary transformation on Fock space we can always achieve that the
polarization vectors are given by

$\displaystyle\varepsilon_{1}(k)=\frac{(k_{2},-k_{1},0)}{\sqrt{k_{1}^{2}+k_{2}^{2}}}\quad\mathrm{and}\quad\varepsilon_{2}(k)=\frac{k}{|k|}\wedge\varepsilon_{1}(k)\;.$
(7.21)

######  Proposition 7.5.

There exists a constant $C$ (depending on $\Lambda$) such that the following
holds. Let $e\in{\mathbb{R}}$ and $m>0$, and suppose the energy Inequality
(2.12) holds. Let $\xi\in{\mathbb{R}}^{3}$ with $|\xi|<1$ such that $\nabla
E_{m}(\xi)$ exists. Then for almost all $k$ with $|k|<\Lambda$,

$\displaystyle\left\|\nabla_{k}(a_{\lambda}(k)U_{m}(\xi)\psi_{m}(\xi))\right\|\leq\frac{CF_{m,\xi}^{2}|e\rho(k)|(\widehat{D}_{m,\xi}(k)+1)}{(1-|\xi|)^{2}|k|\sqrt{k_{1}^{2}+k_{2}^{2}}}.$

###### Proof.

First we note that for all nonzero $k\in{\mathbb{R}}^{3}$ and $\lambda=1,2$,
we have

$\left|\frac{\partial}{\partial
k_{j}}\varepsilon_{\lambda}(k)\right|\leq\frac{1}{\sqrt{k_{1}^{2}+k_{2}^{2}}},\quad
j=1,2,3.$ (7.22)

Moreover, as $\rho$ is constant for $|k|<\Lambda$, we do not need to take
derivatives of $\rho$ into account. Since $|k|<\Lambda$ it suffices to
consider the leading order contributions as $|k|$ is small. From
differentiating (7.14) we find

$\displaystyle
U_{m}(\xi)^{*}\nabla_{k}a_{\lambda}(k)U_{m}(\xi)\psi_{m}(\xi)=\nabla_{k}U_{m}(\xi)^{*}a_{\lambda}(k)U_{m}(\xi)\psi_{m}(\xi)$
(7.23)
$\displaystyle\quad=\nabla_{k}\left(a_{\lambda}(k)\psi_{m}(\xi)+{\varepsilon}_{\lambda}(k)\cdot\nabla_{\xi}E_{m}(\xi)\frac{e\rho(k)}{\sqrt{2\left|k\right|}}\frac{1}{\omega_{m}(k)-k\nabla_{\xi}E_{m}(\xi)}\psi_{m}(\xi)\right),$

where the derivative is understood in the strong sense. First we calculate the
derivative of the first term using the pull-through resolvent identity in
Lemma 7.2.

For $k$ with $|k|<\Lambda$ we find by the product rule

$\displaystyle\nabla_{k}a_{\lambda}(k)\psi_{m}(\xi)$ $\displaystyle\
=\nabla_{k}\left(-\frac{e\rho(k)}{\sqrt{2\left|k\right|}}R_{m,\xi}(k)\left({\varepsilon}_{\lambda}(k)\cdot
v(\xi)+\sigma\cdot(k\wedge{\varepsilon}_{\lambda}(k)\right)\psi_{m}(\xi)\right)$
$\displaystyle\ =J_{1}+J_{2},$ (7.24)

where we introduced

$\displaystyle J_{1}$
$\displaystyle:=R_{m,\xi}(k)(k-v(\xi)+\nabla_{k}\omega_{m}(k))R_{m,\xi}(k)$
$\displaystyle\quad\times\frac{e\rho(k)}{\sqrt{2|k|}}\left({\varepsilon}_{\lambda}(k)\cdot
v(\xi)+\sigma\cdot(k\wedge{\varepsilon}_{\lambda}(k)\right)\psi_{m}(\xi)$

and

$J_{2}:=J_{2,1}+J_{2,2}$

with

$\displaystyle J_{2,1}$
$\displaystyle:=-R_{m,\xi}(k)\nabla_{k}\left(\frac{e\rho(k)}{\sqrt{2|k|}}{\varepsilon}_{\lambda}(k)\cdot
v(\xi)\psi_{m}(\xi)\right),$ $\displaystyle J_{2,2}$
$\displaystyle:=-R_{m,\xi}(k)\nabla_{k}\left(\frac{e\rho(k)}{\sqrt{2|k|}}S\cdot(k\wedge{\varepsilon}_{\lambda}(k))\psi_{m}(\xi)\right).$

To estimate $J_{2,2}$ we use (7.1) and (7.22), which yields

$\displaystyle\|J_{2,2}\|\leq\frac{C|e\rho(k)|}{1-|\xi|}\frac{1}{|k|\sqrt{k_{1}^{2}+k_{2}^{2}}}.$
(7.25)

Next we consider $J_{2,1}$ and we write

$v(\xi)=\nabla E_{m}(\xi)+(v(\xi)-\nabla E_{m}(\xi))$ (7.26)

and use Lemma 7.3 (a) and (7.22) to find

$\displaystyle\left\|J_{2,1}+R_{m,\xi}(k)\nabla_{k}\left(\frac{e\rho(k)}{\sqrt{2|k|}}{\varepsilon}_{\lambda}(k)\cdot\nabla
E_{m}(\xi)\psi_{m}(\xi)\right)\right\|\leq\frac{CF_{m,\xi}|e\rho(k)|}{(1-|\xi|)|k|\sqrt{k_{1}^{2}+k_{2}^{2}}}.$

Now we use (7.17)–(7.19) to replace the resolvent $R_{m,\xi}$ by $D_{m,\xi}$
and get (noting that $\left|\nabla E_{m}(\xi)\right|<1$)

$\displaystyle\left\|J_{2,1}+D_{m,\xi}(k)\nabla_{k}\left(\frac{e\rho(k)}{\sqrt{2|k|}}{\varepsilon}_{\lambda}(k)\cdot\nabla
E_{m}(\xi)\psi_{m}(\xi)\right)\right\|\leq\frac{CF_{m,\xi}|e\rho(k)|(\widehat{D}_{m,\xi}(k)+1)}{(1-|\xi|)|k|\sqrt{k_{1}^{2}+k_{2}^{2}}}.$
(7.27)

Finally let us consider $J_{1}$. The term involving the spin has a lower order
singularity as well as the term involving the $-k$. Thus we find with (7.1)
and Lemma 7.3 (b)

$\displaystyle\left\|J_{1}-R_{m,\xi}(k)(\nabla_{k}\omega_{m}(k)-v(\xi))R_{m,\xi}(k)\frac{e\rho(k)}{\sqrt{2\left|k\right|}}{\varepsilon}_{\lambda}(k)\cdot
v(\xi)\psi_{m}(\xi)\right\|\leq\frac{CF_{m,\xi}|e\rho(k)|}{(1-|\xi|)^{2}|k|^{3/2}}.$

Using again (7.26) to replace the second $v(\xi)$ by $\nabla E_{m}(\xi)$, we
find with Lemma 7.3 (a) and Lemma 7.3 (b) that

$\displaystyle\left\|J_{1}-R_{m,\xi}(k)(\nabla_{k}\omega_{m}(k)-v(\xi))R_{m,\xi}(k)\frac{e\rho(k)}{\sqrt{2\left|k\right|}}{\varepsilon}_{\lambda}(k)\cdot\nabla
E_{m}(\xi)\psi_{m}(\xi)\right\|\leq\frac{CF_{m,\xi}^{2}|e\rho(k)|}{(1-|\xi|)^{2}|k|^{2}}.$

Now we use as before (7.17)–(7.19) to replace the second resolvent $R_{m,\xi}$
by $D_{m,\xi}$

$\displaystyle\left\|J_{1}-R_{m,\xi}(k)(\nabla_{k}\omega_{m}(k)-v(\xi))D_{m,\xi}(k){\varepsilon}_{\lambda}(k)\cdot\frac{e\rho(k)}{\sqrt{2|k|}}\nabla
E(\xi)\psi_{m}(\xi)\right\|$
$\displaystyle\qquad\leq\frac{CF_{m,\xi}^{2}|e\rho(k)|(\widehat{D}_{m,\xi}(k)+1)}{(1-|\xi|)^{2}|k|^{2}}.$
(7.28)

Repeating the above, that is, using again first (7.26) to replace $v(\xi)$ by
$\nabla E_{m}(\xi)$ and then (7.17)–(7.19) to replace the first resolvent
$R_{m,\xi}$ by $D_{m,\xi}$, we arrive at

$\displaystyle\left\|J_{1}-(\nabla_{k}\omega_{m}(k)-\nabla
E_{m}(\xi))[D_{m,\xi}(k)]^{2}{\varepsilon}_{\lambda}(k)\cdot\frac{e\rho(k)}{\sqrt{2|k|}}\nabla
E(\xi)\psi_{m}(\xi)\right\|\leq\text{r.h.s. of }\eqref{eq:estofderest}.$
(7.29)

Finally, note that the derivative of the second term in (7.23) is

$\displaystyle\nabla_{k}$
$\displaystyle\left({\varepsilon}_{\lambda}(k)\cdot\nabla_{\xi}E_{m}(\xi)\frac{e\rho(k)}{\sqrt{2\left|k\right|}}D_{m,\xi}(k)\psi_{m}(\xi)\right)$
$\displaystyle=D_{m,\xi}(k)\nabla_{k}\left(\frac{e\rho(k)}{\sqrt{2|k|}}{\varepsilon}_{\lambda}(k)\cdot\nabla
E_{m}(\xi)\psi_{m}(\xi)\right)$ (7.30) $\displaystyle\qquad-
D_{m,\xi}(k)^{2}(\nabla_{k}\omega_{m}(k)-\nabla
E_{m}(\xi))\frac{e\rho(k)}{\sqrt{2|k|}}{\varepsilon}_{\lambda}(k)\cdot\nabla
E_{m}(\xi)\psi_{m}(\xi).$ (7.31)

Thus, considering the decomposition (7.24) for the first term and combining
(7.27) with (7.30), noting (7.25), as well as (7.29) with (7.31) we obtain the
desired upper bound.

∎

######  Lemma 7.6 ($y$-Bound).

Let $e\in{\mathbb{R}}$. Suppose there exists an $m_{0}>0$ such that (2.12)
holds for all $m\in(0,m_{0})$. Let $|\xi|<1$. Then for every $M>0$ there
exists a constant $C$, and a $\delta>0$ such that for all $m\in(0,m_{0})$ and
all $n\in{\mathbb{N}}$,

$\sum_{\lambda_{1},\ldots,\lambda_{n}}\int\sum_{i=1}^{n}n^{-1}|y_{i}|^{\delta}\|F(U_{m}(\xi)\psi_{m}(\xi))_{(n)}(\lambda_{1},{y}_{1},\ldots,\lambda_{n},y_{n})\|^{2}\,dy_{1}\ldots
dy_{n}\leq C\;,$

whenever $|E_{m}(\xi)|<M$ and
$\max\\{|\partial_{i}^{2}E_{m}(\xi)|:i=1,2,3\\}<M$. Here
$F(U_{m}(\xi)\psi_{m}(\xi))_{(n)}$ denotes the Fourier transform of the
$n$-photon component of $U_{m}(\xi)\psi_{m}(\xi)$.

###### Proof.

We drop the subscript $m$. We write $\widehat{\psi}=U_{m}(\xi)\psi_{m}(\xi)$.
Thus, $F\widehat{\psi}_{(n)}$ denotes the Fourier transform of
$\widehat{\psi}_{(n)}$ in all its $n$-components. We define the functions

$\displaystyle\widehat{\psi}_{(n)}(k)$ $\displaystyle:$
$\displaystyle(\lambda,k_{1},\lambda_{1},\ldots,k_{n-1},\lambda_{n-1})\mapsto\widehat{\psi}_{(n)}(k,\lambda,k_{1},\lambda_{1},\ldots,k_{n-1},\lambda_{n-1}),$
$\displaystyle F\widehat{\psi}_{(n)}(y)$ $\displaystyle:$
$\displaystyle(\lambda,y_{1},\lambda_{1},\ldots,y_{n-1},\lambda_{n-1})\mapsto
F\widehat{\psi}_{(n)}(y,\lambda,y_{1},\lambda_{1},\ldots,y_{n-1},\lambda_{n-1})\;.$

Step 1: There exists a $\delta>0$ and a constant $C$ such that for all
$a\in{\mathbb{R}}^{3}$,

$\int|1-e^{-\mathrm{i}ay}|^{2}\|F\widehat{\psi}_{(n)}(y)\|^{2}dy\leq\left\\{\begin{array}[]{ll}C|a|^{\delta}&\text{if}\
|a|<\frac{1}{2}\Lambda,\\\ C&\text{if}\
|a|\geq\frac{1}{2}\Lambda.\end{array}\right.$

The claim follows easily for $|a|\geq\frac{1}{2}\Lambda$, since
$\widehat{\psi}$ is a normalized state in Fock space and
$|1-e^{-\mathrm{i}ay}|\leq 2$. Now lets consider the case
$|a|<\frac{1}{2}\Lambda$. By the Fourier transform, we have the identity

$\displaystyle\int|1-e^{-\mathrm{i}ay}|^{2}\|F\widehat{\psi}_{(n)}(y)\|^{2}dy$
$\displaystyle\quad=\int\|\widehat{\psi}_{(n)}(k+a)-\widehat{\psi}_{n}(k)\|^{2}dk$
$\displaystyle\quad=\int_{|k|<\Lambda-|a|}\|\widehat{\psi}_{(n)}(k+a)-\widehat{\psi}_{n}(k)\|^{2}dk+\int_{\Lambda-|a|\leq|k|}\|\widehat{\psi}_{(n)}(k+a)-\widehat{\psi}_{n}(k)\|^{2}dk.$
(7.32)

To estimate the second integral we use Proposition 7.4 and observe that the
integrand vanishes for $\left|k\right|>\Lambda+\left|a\right|$,

$\displaystyle\int_{\Lambda-\left|a\right|\leq\left|k\right|}\|\widehat{\psi}_{(n)}(k+a)-\widehat{\psi}_{n}(k)\|^{2}dk$
$\displaystyle\leq\text{const.}\int_{\Lambda-\left|a\right|\leq\left|k\right|\leq\Lambda+\left|a\right|}\left(\frac{1}{\left|k+a\right|^{2}}+\frac{1}{\left|k\right|^{2}}\right)dk$
$\displaystyle\leq\text{const.}\int_{\Lambda-2\left|a\right|\leq\left|k\right|\leq\Lambda+2\left|a\right|}\frac{1}{\left|k\right|^{2}}dk$
$\displaystyle\leq\text{const.}\left|a\right|,$ (7.33)

where const. denotes a numerical constant changing from line to line. Next we
estimate the first integral and assume
$\left|k\right|<\Lambda-\left|a\right|$. Using Proposition 7.5 we find

$\displaystyle\|\widehat{\psi}_{(n)}(k+a)-\widehat{\psi}_{(n)}(k)\|$
$\displaystyle=\left\|\int_{0}^{1}\left(\frac{d}{dt}\widehat{\psi}_{(n)}(k+ta)\right)dt\right\|$
$\displaystyle\leq|a|\int_{0}^{1}\|\nabla_{k}\widehat{\psi}_{(n)}(k+ta)\|dt$
$\displaystyle\leq\text{const.}|a|\int_{0}^{1}\frac{\rho(k+ta)}{|k+ta||\pi_{3}(k+ta)|}dt\;,$
(7.34)

where $\pi_{3}$ denotes the projection in ${\mathbb{R}}^{3}$ along the 3-axis
and const. denotes a finite constant independent of $n$. Let $\pi_{a}$ denote
the projection in ${\mathbb{R}}^{3}$ along the vector $a$ and let $\pi_{3,a}$
denote the projection in the $(1,2)$-plane along $\pi_{3}a$ (with convention
that $\pi_{3,a}=\pi_{3}$, if $\pi_{3}a=0$). We find from (7.34)

$\displaystyle\|\widehat{\psi}_{(n)}(k+a)-\widehat{\psi}_{(n)}(k)\|\leq\text{const.}\frac{|a|}{|\pi_{a}(k)||\pi_{3,a}(k)|}\;.$
(7.35)

On the other hand using Proposition 7.4 we obtain

$\displaystyle\|\widehat{\psi}_{(n)}(k+a)-\widehat{\psi}_{(n)}(k)\|\leq\text{const.}\left(\frac{\rho(k+a)}{|k+a|}+\frac{\rho(k)}{|k|}\right)\;.$
(7.36)

Introducing Inequalities (7.35) and (7.36) into the second integral of (7.32),
we find for any $\theta$ with $0\leq\theta\leq 1$,

$\displaystyle\int_{\left|k\right|<\Lambda-\left|a\right|}\|\widehat{\psi}_{(n)}(k+a)-\widehat{\psi}_{n}(k)\|^{2}dk$
$\displaystyle=\text{const.}|a|^{2\theta}\int_{|k|<\Lambda-|a|}\frac{1}{|\pi_{a}(k)|^{2\theta}|\pi_{3,a}(k)|^{2\theta}}\left(\frac{\rho(k+a)}{|k+a|}+\frac{\rho(k)}{|k|}\right)^{2(1-\theta)}\,dk.$

Now we use Young’s inequality: $bc\leq b^{p}/p+c^{q}/q$, whenever $p,q>1$ and
$p^{-1}+q^{-1}=1$; and the convexity of $x\mapsto x^{2(1-\theta)q}$ on
${\mathbb{R}}_{+}$, for $0<\theta<1/2$. Thus for $0<\theta<1/2$,

$\displaystyle\int_{\left|k\right|<\Lambda-\left|a\right|}\|\widehat{\psi}_{(n)}(k+a)-\widehat{\psi}_{n}(k)\|^{2}dk$
$\displaystyle\quad\leq|a|^{2\theta}\text{const.}\int_{|k|\leq\Lambda}\Bigg{(}\frac{1}{|\pi_{a}(k)|^{4\theta
p}}+\frac{1}{|\pi_{3,a}(k)|^{4\theta
p}}+\left[\frac{1}{|k+a|}\right]^{2(1-\theta)q}+\left[\frac{1}{|k|}\right]^{2(1-\theta)q}\Bigg{)}dk\;.$
(7.37)

For any $q$ with $1<q\leq 3/2$, we can choose $\theta>0$ sufficiently small
such that the right hand side is finite. Inserting (7.33) und (7.37) into
(7.32) we obtain the desired estimate.

Step 2: Step 1 implies the statement of the Lemma.

From Step 1 we know that there exists a finite constant $C$ such that

$\int\frac{|1-e^{-\mathrm{i}ay}|^{2}\|F\widehat{\psi}_{(n)}(y)\|^{2}}{|a|^{\delta/2}}dy\frac{da}{|a|^{3}}\leq
C\;.$

After interchanging the order of integration and a change of integration
variables $b=|y|a$, we find

$\displaystyle C$
$\displaystyle\geq\int\|F\widehat{\psi}_{(n)}(y)\|^{2}\int\frac{|1-e^{-\mathrm{i}ay}|^{2}}{|a|^{\delta/2}}\frac{da}{|a|^{3}}dy$
$\displaystyle=\int\|F\widehat{\psi}_{(n)}(y)\|^{2}|y|^{\delta/2}\underbrace{\int\frac{|1-e^{-\mathrm{i}by/|y|}|^{2}}{|b|^{\delta/2}}\frac{db}{|b|^{3}}}_{=:\
c}dy,$

where $c$ is nonzero and does not depend on $y$. ∎

##  8 Compactness Argument

Using a compactness argument we show Theorem 3.1. It states that for suitable
$\xi$ there exists a sequence $(m_{j})$ of positive numbers converging to zero
such that $U_{m_{j}}(\xi)\psi_{m_{j}}(\xi)$ converges to a vector in the
Hilbert space, say $\widehat{\psi}_{0}$. Then we prove the main Theorem 3.5,
which states that this vector $\widehat{\psi}_{0}$ is indeed the ground state
of the renormalized Hamiltonian using a semi-continuity argument of quadratic
forms.

######  Theorem 8.1.

Let $e\in{\mathbb{R}}$, and suppose there exists an $m_{0}>0$ such that the
energy Inequality (2.12) holds for all $m\in(0,m_{0})$. Let
$\xi\in{\mathbb{R}}^{3}$ with $\left|\xi\right|<1$ such that $E$ is
differentiable at $\xi$. Suppose there exists a sequence
$(m_{j})_{j\in{\mathbb{N}}}$ of positive numbers converging to zero such that

  * (i)

$E_{m_{j}}$ is differentiable at $\xi$,

  * (ii)

$\nabla E_{m_{j}}(\xi)\overset{j\to\infty}{\longrightarrow}\nabla E(\xi)$,

  * (iii)

for every $l=1,2,3$ the partial derivatives $\partial_{l}^{2}E_{m_{j}}(\xi)$
exist and satisfy

$\sup_{j}(-\partial_{l}^{2}E_{m_{j}}(\xi))<\infty.$

Then $E(\xi)$ is an eigenvalue of $\widehat{H}(\xi)$, and there exists a
subsequence of $(U_{m_{j}}(\xi)\psi_{m_{j}}(\xi))_{j\in{\mathbb{N}}}$
converging to the eigenvector.

###### Proof.

Step 1: The sequence of vectors
$\widehat{\psi}_{m_{j}}(\xi):=U_{m_{j}}(\xi)\psi_{m_{j}}(\xi)$,
$j\in{\mathbb{N}}$, lies in a compact subspace of the reduced Hilbert space
$\mathcal{H}={\mathbb{C}}^{2s+1}\otimes\mathcal{F}$.

Let $L$ be the self-adjoint operator associated to the nonnegative and closed
quadratic form $q$ in $\mathcal{H}$ defined by

$q(\phi):=\langle{\phi,\mathsf{d}\Gamma(\mathds{1})\phi}\rangle+\sum_{n=1}^{\infty}n^{-3}\langle{\widehat{\phi}_{(n)},\sum_{i=1}^{n}|y_{i}|^{\delta}\widehat{\phi}_{(n)}}\rangle+\langle{\phi,H_{\mathrm{f}}\phi}\rangle$

on the natural form domain $D(q)$. We choose $\delta>0$ such that Lemma 7.6
holds. By this and Propositions 7.4 and 4.2, there exists a finite $C$ such
that for all $m$ with $0<m<m_{0}$,

$\psi_{m}\in\mathcal{K}:=\\{\phi\in D(q):\|\phi\|\leq 1,q(\phi)\leq C\\}\;.$

The set $\mathcal{K}$ is a compact subset of $\mathcal{H}$, provided $L$ has
compact resolvent [RS78, Theorem XIII.64]. Hence it remains to show that $L$
has compact resolvent. The operator $L$ preserves the $n$-photon sectors. Let
$L_{n}$ denote the restriction of $L$ to the $n$-photon sector. From Rellich’s
criterion [RS78, Theorem XIII.65] it follows that $L_{n}$ has compact
resolvent. Therefore $\mu_{l}(L_{n})\to\infty$ as $l$ tends to infinity, where
$\mu_{l}$ denotes the $l$-th eigenvalue obtained by the min-max principle.
Moreover, since $\mu_{l}(L_{n})\geq n$ for all $l,n$, it follows that
$\mu_{l}(L)\to\infty$ as $l\to\infty$. Hence $L$ has a compact resolvent.

Step 2: The sequence in Step 1 has a subsequence which converges to a
normalized vector $\widehat{\psi}_{0}\in\mathcal{H}$.

This follows directly from Step 1 and the property of compact sets.

Step 3: The vector $\widehat{\psi}_{0}$ is an eigenvector of the renormalized
fiber Hamiltonian $\widehat{H}(\xi)$ with eigenvalue $E(\xi)$.

Using lower semicontinuity of nonnegative quadratic forms it follows from Step
2 and Proposition 5.6 that for almost all $\xi$ with $|\xi|<1$

$\displaystyle 0$
$\displaystyle\leq\langle{\widehat{\psi}_{0}(\xi),(\widehat{H}(\xi)-E(\xi))\widehat{\psi}_{0}(\xi)}\rangle$
$\displaystyle\qquad\leq\liminf_{i\to\infty}\langle{U_{m_{j}}(\xi)\psi_{m_{j}}(\xi),(\widehat{H}(\xi)-E(\xi))U_{m_{j}}(\xi)\psi_{m_{j}}}\rangle=0,$

i.e., that $\widehat{\psi}_{0}(\xi)$ is a ground state of $\widehat{H}(\xi)$.
∎

Now the above theorem implies together with Proposition 4.2 Theorems 3.5 and
3.1.

###### Proof of Theorems 3.5 and 3.1.

Let $e\in{\mathbb{R}}$. By Theorem 2.5 in case $s=0$ or by assumption in case
$s=1/2$, there exists an $m_{0}>0$, such that (2.12) holds for all
$m\in(0,m_{0})$. Let $(m_{j})_{j\in{\mathbb{N}}}$ be any sequence in
$(0,m_{0})$ which converges to zero. Then by Proposition 4.2 there exists a
set $D\subset{\mathbb{R}}^{3}$ of full Lebesgue measure with the following
property: For all $\xi\in D$ the functions $E_{m_{j}}$, $j\in{\mathbb{N}}$,
and $E$ are differerentiable and

  * (a)

$\nabla E_{m_{j}}(\xi)\overset{j\to\infty}{\longrightarrow}\nabla E(\xi)$,

  * (b)

the second partial derivatives $\partial_{l}^{2}E_{m_{j}}(\xi)$ exist and
satisfy for every $l=1,2,3$ that
$\liminf_{j}(-\partial_{l}^{2}E_{m_{j}}(\xi))<\infty$ .

Now pick any $\xi\in D$ with $|\xi|<1$ and fix it. Then the assumptions of
Theorem 8.1 hold. Therefore, Theorem 3.5 and Theorem 3.1 follow directly from
Theorem 8.1. ∎

##  Appendix A Energy Estimates

In this section we collect a few well known properties related to the
canonical commutation relations, which we need in particular in Section 5.

######  Lemma A.1.

Let $f_{1},\ldots,f_{n}\in
L_{(n)}^{2}({\mathbb{R}}^{3}\times{\mathbb{Z}}_{2})$. Then for any
$\psi\in\mathcal{F}$ we have

$\displaystyle\|a(f_{1})\cdots a(f_{n})\psi\|$
$\displaystyle\leq\left(\prod_{j=1}^{n}\|f_{j}\omega_{m}^{-1/2}\|\right)\|H_{\mathrm{f},m}^{n/2}\psi\|.$

###### Proof.

In the following we use the notation $\underline{k}=(\lambda,k)$ and
$\int(\cdots)d\underline{k}=\sum_{\lambda=1,2}\int(\cdots)dk$. By the
definition of the annihilation operator, by Cauchy-Schwarz and Fubini, we find

$\displaystyle\|a(f_{1})\cdots a(f_{n})\psi\|$
$\displaystyle\leq\int|f_{1}(\underline{k}_{1})\cdots
f_{n}(\underline{k}_{n})|\|a(\underline{k}_{1})\cdots
a(\underline{k}_{n})\psi\|d\underline{k}_{1}\cdots d\underline{k}_{n}$
$\displaystyle\leq\left(\prod_{j=1}^{n}\|f_{j}\omega_{m}^{-1/2}\|\right)\left(\int\omega_{1}(\underline{k}_{1})\cdots\omega_{n}(\underline{k}_{n})\|a(\underline{k}_{1})\cdots
a(\underline{k}_{n})\psi\|^{2}d\underline{k}_{1}\cdots
d\underline{k}_{n}\right)^{1/2}.$

To estimate the second factor we use

$\displaystyle\int\omega_{1}(\underline{k}_{1})\cdots\omega_{n}(\underline{k}_{n})\|a(\underline{k}_{1})\cdots
a(\underline{k}_{n})\psi\|^{2}d\underline{k}_{1}\cdots d\underline{k}_{n}$
$\displaystyle=\int\omega_{1}(\underline{k}_{2})\cdots\omega_{n}(\underline{k}_{n})\|H_{\mathrm{f},m}^{1/2}a(\underline{k}_{2})\cdots
a(\underline{k}_{n})\psi\|^{2}d\underline{k}_{2}\cdots d\underline{k}_{n}$
$\displaystyle=\int\omega_{1}(\underline{k}_{2})\cdots\omega_{n}(\underline{k}_{n})\|H_{\mathrm{f},m}^{1/2}a(\underline{k}_{2})\cdots
a(\underline{k}_{n})H_{\mathrm{f},m}^{-1/2}H_{\mathrm{f}}^{1/2}\psi\|^{2}d\underline{k}_{2}\cdots
d\underline{k}_{n}$
$\displaystyle=\int\omega_{1}(\underline{k}_{2})\cdots\omega_{n}(\underline{k}_{n})\|H_{\mathrm{f},m}^{1/2}(H_{\mathrm{f},m}+\sum_{j=2}^{n}\omega_{m}(k_{j}))^{-1/2}a(\underline{k}_{2})\cdots
a(\underline{k}_{n})H_{\mathrm{f},m}^{1/2}\psi\|^{2}d\underline{k}_{2}\cdots
d\underline{k}_{n}$
$\displaystyle\leq\int\omega_{1}(\underline{k}_{2})\cdots\omega_{n}(\underline{k}_{n})\|a(\underline{k}_{2})\cdots
a(\underline{k}_{n})H_{\mathrm{f},m}^{1/2}\psi\|^{2}d\underline{k}_{2}\cdots
d\underline{k}_{n}$ $\displaystyle\vdots$
$\displaystyle\leq\|H_{\mathrm{f},m}^{n/2}\psi\|^{2},$

where we used the pull-through formula, cf. [BFS98, Lemma A.1], and that
$H_{\mathrm{f}}\varphi=0$ implies $a(f)\varphi=0$.

∎

######  Lemma A.2.

For any $f\in L^{2}_{(m)}({\mathbb{Z}}_{2}\times{\mathbb{R}}^{3})$ and
$\psi\in\mathcal{F}$

$\displaystyle\|\phi(f)\psi\|$
$\displaystyle\leq\sqrt{2}\|f\|_{(m)}\|(H_{\mathrm{f},m}+1)^{1/2}\psi\|,$
$\displaystyle\|\phi(f_{1})\phi(f_{2})\psi\|$ $\displaystyle\leq
2\|f_{1}\|_{(m)}\|f_{2}\|_{(m)}\|(H_{\mathrm{f},m}+1)\psi\|.$

Note that the proof of Lemma A.2 can be found in [Ara18, Theorem 5.18], but
for the convenience of the reader we give a proof below.

###### Proof.

From the canonical commutation relations we get

$\displaystyle\|a(f)^{*}\psi\|^{2}$
$\displaystyle=\langle{a^{*}(f)\psi,a^{*}(f)\psi}\rangle=\|f\|^{2}\|\psi\|^{2}+\|a(f)\psi\|^{2}.$

Now the first identity follows from the triangle inequality and Lemma A.1.

For the second identity we use again the canonical commutation relations and
find

$\displaystyle\|a^{*}(f_{1})a^{*}(f_{2})\psi\|^{2}$
$\displaystyle\quad=\langle{a^{*}(f_{1})a^{*}(f_{2})\psi,a^{*}(f_{1})a^{*}(f_{2})\psi}\rangle$
$\displaystyle\quad\leq\|a(f_{1})a(f_{2})\psi\|^{2}+2\|f_{1}\|^{2}\|a(f_{2})\psi\|^{2}+2\|f_{2}\|^{2}\|a(f_{1})\psi\|^{2}+2\|f_{1}\|^{2}\|f_{2}\|^{2},$

and

$\displaystyle\|a^{*}(g_{1})a(g_{2})\psi\|^{2}$
$\displaystyle=\langle{a^{*}(g_{1})a(g_{2})\psi,a^{*}(g_{1})a(g_{2})\psi}\rangle=\|a(g_{1})a(g_{2})\psi\|^{2}+\|g_{1}\|^{2}\|a(g_{2})\psi\|^{2}$
$\displaystyle\|a(f_{1})a^{*}(f_{2})\psi\|^{2}$
$\displaystyle\leq(\|f_{1}\|\|f_{2}\|\|\psi\|+\|a^{*}(f_{2})a(f_{1})\psi\|)^{2}.$

The second inequality follows now by collecting estimates, the triangle
inequality and Lemma A.1.

∎

##  Appendix B Statements about CCR Algebras

######  Lemma B.1.

Let $A$ be a self-ajoint operator in $\mathfrak{h}$ and $f\in
D(A)\subset\mathfrak{h}$. Then we have the relations

$\displaystyle\mathsf{d}\Gamma(A)a^{*}(f)$
$\displaystyle=a^{*}(f)\mathsf{d}\Gamma(A)+a^{*}(Af),$
$\displaystyle\mathsf{d}\Gamma(A)a(f)$
$\displaystyle=a(f)\mathsf{d}\Gamma(A)-a(Af),$
$\displaystyle\mathsf{d}\Gamma(A)\phi(f)$
$\displaystyle=\phi(f)\mathsf{d}\Gamma(A)-\mathrm{i}\phi(\mathrm{i}Af)$

on $\mathcal{F}_{\operatorname{fin}}(D(A))$.

###### Proof.

The first identity follows directly from the definition of the creation
operators and $\mathsf{d}\Gamma(A)$. The second identity follows from the
first by taking adjoints. The last identity follows from the first two. For
details see for example [Ara18, Proposition 5.10]. ∎

######  Lemma B.2.

Let $f,g\in\mathfrak{h}$. Then for every $\psi\in
D(\mathsf{d}\Gamma(\mathds{1}))$ one has

$(\phi(f)\phi(g)-\phi(g)\phi(f))\psi=\mathrm{i}\operatorname{Im}\langle
f,g\rangle\psi.$

###### Proof.

This follows directly from the canonical commutation relations (2.1). For
details see for example [Ara18, Proposition 5.14]. ∎

For $f\in\mathfrak{h}$ we define

$W(f):=\exp(\mathrm{i}\phi(f))=\exp(\mathrm{i}\pi(\mathrm{i}f)).$

Recall that $\mathcal{F}_{\operatorname{fin}}(\mathfrak{h})$ denotes the
subspace of elements $\psi\in\mathcal{F}(\mathfrak{h})$ such that $\psi_{n}=0$
for all but finitely many $n$. For the definition of an analytic vector for an
operator we refer the reader to [RS75].

######  Lemma B.3.

Let $f\in\mathfrak{h}$ and
$\psi\in\mathcal{F}_{\operatorname{fin}}(\mathfrak{h})$. Then for all
$t\in{\mathbb{C}}$ we have

$\sum_{n=0}^{\infty}\frac{\|\phi(f)^{n}\psi\|}{n!}|t|^{n}<\infty.$

In particular, $\mathcal{F}_{\operatorname{fin}}(\mathfrak{h})$ is a dense set
of analytic vectors for $\phi(f)$.

###### Proof.

See the proof of [BR03, Proposition 5.2.3]. ∎

######  Lemma B.4.

Let $f,g\in\mathfrak{h}$.

  1. (a)

Then $W(f)D(\phi(g))=D(\phi(g))$ and

$W(f)\phi(g)W(f)^{*}=\phi(g)-\operatorname{Im}\langle f,g\rangle.$

  2. (b)

Then $W(f)D(a^{\\#}(g))=D(a^{\\#}(g))$ and

$\displaystyle W(f)a(g)W(f)^{*}$
$\displaystyle=a(g)-\mathrm{i}2^{-1/2}\langle{g,f}\rangle,$ $\displaystyle
W(f)a^{*}(g)W(f)^{*}$
$\displaystyle=a^{*}(g)+\mathrm{i}2^{-1/2}\langle{f,g}\rangle.$

The proof of the lemma before can be found for example in [BR03, Proposition
5.2.4] and [Ara18, Corollary 5.12]. For the convenience of the reader we
sketch a proof below.

###### Sketch of proof.

(a) By Lemma B.3 every $\psi\in\mathcal{F}_{\operatorname{fin}}(\mathfrak{h})$
is analytic for $\phi(f)$. Thus one can define $\phi(g)W(f)^{*}$ on $\psi$ by
a power series expansion, which yields the identity

$\phi(g)W(f)^{*}\psi=W(f)^{*}\phi(g)-\operatorname{Im}\langle f,g\rangle\psi.$

Since $\mathcal{F}_{\operatorname{fin}}(\mathfrak{h})$ is an operator core for
$\phi(g)$, the claim now follows since $\phi(g)$ is by definition a closed
operator. For details we refer the reader to [BR03, Proposition 5.2.4].

Part (b) is shown similarly. ∎

######  Lemma B.5.

Let $A$ be a self-adjoint operator in $\mathfrak{h}$ and $f\in D(A)$. Then we
have $W(f)D(\mathsf{d}\Gamma(A))\supset D(\mathsf{d}\Gamma(A))\cap
D(\phi(\mathrm{i}Af))$ and as an identity on the latter

$W(f)\mathsf{d}\Gamma(A)W(f)^{*}=\mathsf{d}\Gamma(A)-\phi(\mathrm{i}Af)+\frac{1}{2}\langle
f,Af\rangle.$ (B.1)

###### Proof.

First we note that by definition of $\mathsf{d}\Gamma(A)$ we have on
$\mathcal{F}_{\operatorname{fin}}(D(A))$

$e^{\mathrm{i}\mathsf{d}\Gamma(A)t}\phi(f)^{n}=\phi(e^{\mathrm{i}At}f)^{n}e^{\mathrm{i}\mathsf{d}\Gamma(A)t}.$

On the same domain we can differentiate with respect to $t$, and find at $t=0$

$\displaystyle\mathrm{i}\mathsf{d}\Gamma(A)\phi(f)^{n}$
$\displaystyle=\phi(f)^{n}\mathrm{i}\mathsf{d}\Gamma(A)+\sum_{l=0}^{n-1}\phi(f)^{l}\phi(\mathrm{i}Af)\phi(f)^{n-1-l}$
$\displaystyle=\phi(f)^{n}\mathrm{i}\mathsf{d}\Gamma(A)+n\phi(f)^{n-1}\phi(\mathrm{i}Af)-\mathrm{i}\frac{n(n-1)}{2}\phi^{n-2}(f)\langle
f,Af\rangle,$

where we used Lemma B.2 for the last identity. Multiplying with
$(-\mathrm{i})^{n}(n!)^{-1}$ and summation over $n\in{\mathbb{N}}_{0}$ yields
on $\mathcal{F}_{\operatorname{fin}}(D(A))$

$\mathrm{i}\mathsf{d}\Gamma(A)W(f)^{*}=\mathrm{i}W(f)^{*}\mathsf{d}\Gamma(A)-\mathrm{i}W(f)^{*}\phi(\mathrm{i}Af)+\mathrm{i}\frac{1}{2}W(f)^{*}\langle
f,Af\rangle.$

Thus, it follows that (B.1) holds on $\mathcal{F}_{\operatorname{fin}}(D(A))$.
If $\psi\in D(\Gamma(A))\cap D(\phi(\mathrm{i}Af))$, then $\psi_{n}=1_{N\leq
n}\psi\in\mathcal{F}_{\operatorname{fin}}(D(A))$. Now
$(\mathsf{d}\Gamma(A)-\phi(\mathrm{i}Af)+\frac{1}{2}\langle
f,Af\rangle)\psi_{n}\to(\mathsf{d}\Gamma(A)-\phi(\mathrm{i}Af)+\frac{1}{2}\langle
f,Af\rangle)\psi$. Therefore, $\psi\in
D(W(f)\mathsf{d}\Gamma(A)W(f)^{*})=W(f)D(\mathsf{d}\Gamma(A))$ and by
closedness (B.1) holds for $\psi$. ∎

##  Appendix C Some Statements about Convex Functions

######  Proposition C.1.

Let $F\colon{\mathbb{R}}^{n}\to{\mathbb{R}}$ be a function that satisfies the
following conditions:

  * (i)

$F(0)\leq F(x)$,

  * (ii)

$F(x)\leq\frac{x^{2}}{2}+F(0)$,

  * (iii)

$x\mapsto\frac{x^{2}}{2}-F(x)$ is a convex function.

Then we have

$F(x-k)-F(x)\geq\left\\{\begin{array}[]{ll}-|k||x|+k^{2}/2&\text{ , if
}|k|\leq|x|,\\\ -x^{2}/2&\text{ , if }|k|\geq|x|.\end{array}\right.$

###### Proof.

A proof is given in [LMS07, Appendix A]. ∎

######  Lemma C.2.

Let $g\colon(a,b)\to{\mathbb{R}}$ be convex. Then for any compact interval
$[c,d]\subset(a,b)$, the function $g$ is Lipschitz continuous on $[c,d]$ with
Lipschitz constant $K$ bounded by
$\max\\{{\varepsilon}^{-1}|g(c)-g(c-{\varepsilon})|,{\varepsilon}^{-1}|g(d+{\varepsilon})-g({\varepsilon})|\\}$
for any ${\varepsilon}>0$ such that
$[c-{\varepsilon},d+{\varepsilon}]\subset(a,b)$.

###### Proof.

If $F$ is convex, then for all $s,t,s^{\prime},t^{\prime}\in(a,b)$ such that
$s\leq s^{\prime}<t^{\prime}$ and $s<t\leq t^{\prime}$,

$\frac{F(t)-F(s)}{t-s}\leq\frac{F(t^{\prime})-F(s^{\prime})}{t^{\prime}-s^{\prime}},$

see for example [Fol99]. The claim now follows from the above inequality. ∎

######  Lemma C.3.

Let $f\colon{\mathbb{R}}\to{\mathbb{R}}$ be a convex function with
$f(x)=f(-x)$ for all $x\in{\mathbb{R}}$. Then $f$ has a global minimum in $0$
and in points $x\geq 0$ where $f$ is differentiable we have $f^{\prime}(x)\geq
0$.

###### Proof.

By convexity and symmetry
$f(0)=f(\frac{1}{2}{x}-\frac{1}{2}x)\leq\frac{1}{2}(f(x)+f(-x))=f(x)$. If $f$
is differentiable at $x>0$, then by elementary convexity we conclude that
$0\leq\frac{f(x)-f(0)}{x-0}\leq f^{\prime}(x)$. The case when $x=0$ follows
from the minmality property. ∎

Let us state the following theorem, which implies in view of the previous
lemma that every convex function is almost everywhere differentiable.

######  Theorem C.4 (Rademacher).

Let $U\subset{\mathbb{R}}^{n}$ be open and $f\colon U\to{\mathbb{R}}$ be
Lipschitz continuous. Then $f$ is almost everywhere in $U$ differentiable.

A proof can be found in [Eva18]. In fact, for convex functions the second
derivative exists almost everywhere. This is the statement of the so called
Alexandrov theorem, [Ale39].

######  Theorem C.5 (Alexandrov).

Let $U\subset{\mathbb{R}}^{n}$ be open and $f:U\to{\mathbb{R}}$ convex. Then
$f$ has a second derivative almost everywhere.

In the following two lemmas we study sequences of convex functions which
converge pointwise. This will be needed to contol properties of the ground
state energy as the mass regularization is removed.

######  Lemma C.6.

Let $f,(f_{n})_{n\in{\mathbb{N}}}$ be convex functions from ${\mathbb{R}}$ to
${\mathbb{R}}$ with $f_{n}\overset{n\longrightarrow\infty}{\longrightarrow}f$
pointwise. Then then there exists a set $D$ with ${\mathbb{R}}\setminus D$ a
Lebesgue null set such that on $D$ the functions $f$ and $f_{n}$,
$n\in{\mathbb{N}}$, are differentiable and pointwise $f_{n}^{\prime}\to
f^{\prime}$.

###### Proof.

By Lemma C.2 we see that each convex function is locally Lipschitz continuous,
hence almost everywhere differentiable by Rademacher’s theorem, Theorem C.4.
Hence there exists a set $D$, where $f_{n}$, $n\in{\mathbb{N}}$, and $f$ are
differentiable such that ${\mathbb{R}}^{d}\setminus D$ is a set of Lebesgue
measure zero. By pointwise convergence and again Lemma C.2 we see that $f_{n}$
is a family of uniformly equicontinuous functions on any compact interval. Now
uniform equicontinuity and pointwise convergences imply uniform convergence
(see for example [RS81, Theorem I.27]). So let $x\in D$ be arbitary. Then we
have that $\delta_{n}:=\sup_{y\in[x-1,x+1]}|f(y)-f_{n}(y)|$ satisfies
$\delta_{n}\overset{n\longrightarrow\infty}{\longrightarrow}0$. Let
$n_{0}\in{\mathbb{N}}$ be such that $\delta_{n}\leq 1$ for all $n\geq n_{0}$.
Since $f_{n}$ is convex and differentiable in $x\in D$, we have

$\displaystyle
f_{n}^{\prime}(x)\leq\frac{f_{n}(x+\sqrt{\delta}_{n})-f_{n}(x)}{\sqrt{\delta}_{n}}.$

Therefore, for $n\geq n_{0}$, we get

$\displaystyle
f_{n}^{\prime}(x)\leq\frac{f(x+\sqrt{\delta}_{n})-f(x)+2\delta_{n}}{\sqrt{\delta}_{n}}\overset{n\longrightarrow\infty}{\longrightarrow}f^{\prime}(x).$

This implies $\lim\sup_{n\to\infty}f_{n}^{\prime}(x)\leq f^{\prime}(x)$.
Analogously, we obtain $\lim\inf_{n\to\infty}f_{n}^{\prime}(x)\geq
f^{\prime}(x)$. This shows that
$f_{n}^{\prime}(x)\overset{n\longrightarrow\infty}{\longrightarrow}f^{\prime}(x)$.
∎

######  Lemma C.7.

Let $f$ and $f_{n}$, $n\in{\mathbb{N}}$, be convex functions from
${\mathbb{R}}$ to ${\mathbb{R}}$ with
$f_{n}\overset{n\longrightarrow\infty}{\longrightarrow}f$ pointwise. Then
there exists a set $D\subset{\mathbb{R}}$ such that ${\mathbb{R}}\setminus D$
has Lebesgue measure zero and the following holds.

  * (a)

For all $x\in D$ and $n\in{\mathbb{N}}$ the function $f_{n}$ is twice
differentiable in $x$ and $f_{n}^{\prime\prime}(x)\geq 0$.

  * (b)

For all $x\in D$ we have $\liminf_{n}f_{n}^{\prime\prime}(x)<\infty$.

######  Remark C.8.

We note that (a) follows directly from Alexandrov’s theorem, Theorem C.5.
However, in the proof of (b), which we will present below, Part (a) will
follow as an intermediate step.

###### Proof.

Assume that $g$ is a convex function. Then it is locally Lipschitz continuous
by Lemma C.2. Thus it is absolutely continuous and therefore by the
fundamental theorem of calculus for Lebesgue integrals (see for example
[Fol99] Theorem 3.35), $g$ is almost everywhere differentiable, $g^{\prime}$
is locally in $L^{1}$, and for every $x\in{\mathbb{R}}$ we have

$g(x)=\int_{0}^{x}g^{\prime}(\xi)d\xi.$

By convexity $g^{\prime}$ is monotone on the set of points where $g$ is
differentiable. By possibly extending $g^{\prime}$ to a monotone function on
${\mathbb{R}}$, we can assume without loss that $g^{\prime}$ is a monotone
function on ${\mathbb{R}}$. Since monotone functions have at most countably
many discontinuities (see for example [Fol99] Theorem 3.23), we can assume
furthermore that $g^{\prime}$ is right continuous. Thus, there exists a unique
measure $\mu$ on ${\mathbb{R}}$ such that for all $a,b\in{\mathbb{R}}$

$\mu((a,b])=g^{\prime}(b)-g^{\prime}(a)$

(see for example [Fol99, Theorems 1.16 and 1.18]). Let $\mu=hd\lambda+\rho$ be
its Lebesgue-Radon-Nykodim representation, where $\rho$ is mutually singular
to the Lebesgue measure $\lambda$ and $h\geq 0$ is Borel measurable (see for
example [Fol99, Theorem 3.8]). Then by taking suitable families shrinking
nicely to $x$ we see that (see for example [Fol99, Theorem 3.22]) $g^{\prime}$
is almost everywhere differentiable and $g^{\prime\prime}=h$. This shows (a).
Furthermore, we find

$g^{\prime}(b)-g^{\prime}(a)=\mu((a,b])=\int_{a}^{b}g^{\prime\prime}(x)dx+\rho((a,b])\geq\int_{a}^{b}g^{\prime\prime}(x)dx.$
(C.1)

By Lemma C.6 there exist for each $m\in{\mathbb{N}}$ two numbers
$a\leq-m,m\leq b$ for which $f_{n}^{\prime}(b)$ and $f_{n}^{\prime}(a)$
converge as $n\to\infty$. Thus, inserting $f_{n}$ into (C.1), and using the
Lemma of Fatou, we obtain

$\lim_{n}(f_{n}^{\prime}(b)-f_{n}^{\prime}(a))\geq\int_{a}^{b}\liminf_{n}f_{n}^{\prime\prime}(x)dx.$

It follows that $\liminf_{n}f_{n}^{\prime\prime}(x)<\infty$ almost everywhere.
This shows (b). ∎

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