# Schrödinger operators with oblique transmission conditions in
${\mathbb{R}}^{2}$

Jussi Behrndt  ,  Markus Holzmann  and  Georg Stenzel  Institut für Angewandte
Mathematik
Technische Universität Graz
Steyrergasse 30
8010 Graz
Austria<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

In this paper we study the spectrum of self-adjoint Schrödinger operators in
$L^{2}(\mathbb{R}^{2})$ with a new type of transmission conditions along a
smooth closed curve $\Sigma\subseteq\mathbb{R}^{2}$. Although these oblique
transmission conditions are formally similar to $\delta^{\prime}$-conditions
on $\Sigma$ (instead of the normal derivative here the Wirtinger derivative is
used) the spectral properties are significantly different: it turns out that
for attractive interaction strengths the discrete spectrum is always unbounded
from below. Besides this unexpected spectral effect we also identify the
essential spectrum, and we prove a Krein-type resolvent formula and a Birman-
Schwinger principle. Furthermore, we show that these Schrödinger operators
with oblique transmission conditions arise naturally as non-relativistic
limits of Dirac operators with electrostatic and Lorentz scalar
$\delta$-interactions justifying their usage as models in quantum mechanics.

##  1\. Introduction

In many quantum mechanical applications one considers particles moving in an
external potential field which is localized near a set $\Sigma$ of measure
zero. Such strongly localized fields can be modeled by singular potentials
that are supported on $\Sigma$ only; of particular importance in this regard
are $\delta$ and $\delta^{\prime}$-interactions. To be more precise, assume
that $\Sigma$ splits ${\mathbb{R}}^{2}$ into a bounded domain $\Omega_{+}$ and
an unbounded domain
$\Omega_{-}={\mathbb{R}}^{2}\setminus\overline{\Omega_{+}}$, and consider the
formal Schrödinger differential expressions

(1.1)
$\mathcal{H}_{\delta,\alpha}=-\Delta+\alpha\delta_{\Sigma}\quad\text{and}\quad\mathcal{H}_{\delta^{\prime},\alpha}=-\Delta+\alpha\delta^{\prime}_{\Sigma},\quad\alpha\in{\mathbb{R}}.$

These singular perturbations of the free Schrödinger operator $-\Delta$ are
characterized by certain transmission conditions along the interface $\Sigma$
for the functions in the operator domain. For $\delta$-interactions one
considers functions $f:{\mathbb{R}}^{2}\to{\mathbb{C}}$ such that the
restrictions $f_{\pm}=f\upharpoonright\Omega_{\pm}$ satisfy the transmission
conditions

(1.2)
$f_{+}=f_{-}\quad\text{and}\quad-\frac{\alpha}{2}\big{(}f_{+}+f_{-}\big{)}=\big{(}\partial_{\nu}f_{+}-\partial_{\nu}f_{-}\big{)}\quad\text{on
}\Sigma,$

while $\delta^{\prime}$-interactions are modeled by the transmission
conditions

(1.3)
$f_{+}-f_{-}=-\frac{\alpha}{2}\big{(}\partial_{\nu}f_{+}+\partial_{\nu}f_{-}\big{)}\quad\text{and}\quad\partial_{\nu}f_{+}=\partial_{\nu}f_{-}\quad\text{on
}\Sigma;$

here $\partial_{\nu}f_{\pm}$ is the normal derivative and
$\nu=(\nu_{1},\nu_{2})$ the unit normal vector field on $\Sigma$ pointing
outwards of $\Omega_{+}$. The spectra and resonances of the self-adjoint
realizations associated with the formal expressions (1.1) in
$L^{2}({\mathbb{R}}^{2})$ are well understood, see, e.g., [8, 9, 12, 13, 15,
16, 17, 18, 24]. In particular, the essential spectrum is given by
$[0,\infty)$ and the discrete spectrum consists of at most finitely many
points for every interaction strength $\alpha<0$, while there is no negative
spectrum if $\alpha\geq 0$.

In contrast to the transmission conditions (1.2) and (1.3) we are interested
in a new type of transmission conditions of the form

(1.4)
$(\nu_{1}+i\nu_{2})\big{(}f_{+}-f_{-}\big{)}=-\alpha\big{(}\partial_{\overline{z}}f_{+}+\partial_{\overline{z}}f_{-}\big{)}\quad\text{and}\quad\partial_{\overline{z}}f_{+}=\partial_{\overline{z}}f_{-}\quad\text{on
}\Sigma,$

where $\alpha\in{\mathbb{R}}$ and
$\partial_{\overline{z}}=\frac{1}{2}(\partial_{1}+i\partial_{2})$ is the
Wirtinger derivative. In the sequel such jump conditions will be referred to
as oblique transmission conditions. Note that the conditions (1.4) can be
rewritten as

(1.5)
$f_{+}-f_{-}=-\frac{\alpha}{2}\big{(}\partial_{\nu}f_{+}+\partial_{\nu}f_{-}+i\partial_{t}f_{+}+i\partial_{t}f_{-}\big{)}\quad\text{and}\quad\partial_{\overline{z}}f_{+}=\partial_{\overline{z}}f_{-}\quad\text{on
}\Sigma,$

where $\partial_{t}$ denotes the tangential derivative. Thus, on a formal
level there is some analogy to the $\delta^{\prime}$-transmission conditions
in (1.3), but it will turn out that the properties of the corresponding self-
adjoint realization in $L^{2}({\mathbb{R}}^{2})$ differ significantly from
those of Schrödinger operators with $\delta^{\prime}$-interactions.

To make matters mathematically rigorous, assume that the curve $\Sigma$ is the
boundary of a bounded and simply connected $C^{\infty}$-domain $\Omega_{+}$
with open complement
$\Omega_{-}={\mathbb{R}}^{2}\setminus\overline{\Omega_{+}}$, denote the
$L^{2}$-based Sobolev space of first order by $H^{1}$, let
$\gamma_{D}^{\pm}:H^{1}(\Omega_{\pm})\rightarrow L^{2}(\Sigma)$ be the
Dirichlet trace operators, and define for $\alpha\in{\mathbb{R}}$ the
Schrödinger operator with oblique transmission conditions by

(1.6) $\begin{split}T_{\alpha}f&=\left(-\Delta f_{+}\right)\oplus\left(-\Delta
f_{-}\right),\\\ {\rm dom\,}T_{\alpha}&=\Big{\\{}f\in H^{1}(\Omega_{+})\oplus
H^{1}(\Omega_{-})\>\big{|}\>\partial_{\overline{z}}f_{+}\oplus\partial_{\overline{z}}f_{-}\in
H^{1}({\mathbb{R}}^{2}),\\\ &\hskip
42.67912pt(\nu_{1}+i\nu_{2})\bigl{(}\gamma_{D}^{+}f_{+}-\gamma_{D}^{-}f_{-}\bigr{)}=-\alpha\bigl{(}\gamma_{D}^{+}(\partial_{\overline{z}}f_{+})+\gamma_{D}^{-}(\partial_{\overline{z}}f_{-})\bigr{)}\Big{\\}}.\end{split}$

The next theorem is the main result in this paper. We discuss the spectral
properties of the Schrödinger operators $T_{\alpha}$ and, in particular, we
show in item (ii) that for every $\alpha<0$ the operator $T_{\alpha}$ is
necessarily unbounded from below and the discrete spectrum in $(-\infty,0)$ is
infinite and accumulates to $-\infty$. In items (iii) and (iv) we shall make
use of the potential operator $\Psi_{\lambda}:L^{2}(\Sigma)\to
L^{2}({\mathbb{R}}^{2})$ and the single layer boundary integral operator
$S(\lambda):L^{2}(\Sigma)\to L^{2}(\Sigma)$ defined in (2.2) and (2.4),
respectively.

######  Theorem 1.1.

For any $\alpha\in{\mathbb{R}}$ the operator $T_{\alpha}$ is self-adjoint in
$L^{2}({\mathbb{R}}^{2})$ and the essential spectrum is given by

$\sigma_{\text{\rm ess}}(T_{\alpha})=[0,\infty).$

Furthermore, the following statements hold:

  1. (i)

If $\alpha\geq 0$, then $\sigma_{\text{\rm disc}}(T_{\alpha})=\emptyset$ and
$T_{\alpha}$ is a nonnegative operator in $L^{2}({\mathbb{R}}^{2})$.

  2. (ii)

If $\alpha<0$, then $\sigma_{\text{\rm disc}}(T_{\alpha})$ is infinite,
unbounded from below, and does not accumulate to $0$. Moreover, for every
fixed $n\in\mathbb{N}$ the $n$-th discrete eigenvalue
$\lambda_{n}\in\sigma_{\text{\rm disc}}(T_{\alpha})$ (ordered non-
increasingly) admits the asymptotic expansion

$\lambda_{n}=-\frac{4}{\alpha^{2}}+\mathcal{O}(1)\quad\text{for }\alpha\to
0^{-},$

where the dependence on $n$ appears in the $\mathcal{O}(1)$-term.

  3. (iii)

For $\lambda\in{\mathbb{C}}\setminus[0,\infty)$ the Birman-Schwinger principle
is valid:

$\lambda\in\sigma_{\rm p}(T_{\alpha})\quad\Longleftrightarrow\quad
1\in\sigma_{\rm p}\big{(}\alpha\lambda S(\lambda)\big{)}.$

  4. (iv)

For
$\lambda\in\rho(T_{\alpha})={\mathbb{C}}\setminus([0,\infty)\cup\sigma_{\rm
p}(T_{\alpha}))$ the operator $I-\alpha\lambda S(\lambda)$ is boundedly
invertible in $L^{2}(\Sigma)$ and the resolvent formula

$(T_{\alpha}-\lambda)^{-1}=(-\Delta-\lambda)^{-1}+\alpha\Psi_{\lambda}\bigl{(}I-\alpha\lambda
S(\lambda)\bigr{)}^{-1}\Psi_{\overline{\lambda}}^{\ast}$

holds, where $-\Delta$ is the free Schrödinger operator defined on
$H^{2}({\mathbb{R}}^{2})$.

To illustrate the significance of Theorem 1.1 we show that Schrödinger
operators with oblique transmission conditions arise naturally as non-
relativistic limits of Dirac operators with electrostatic and Lorentz scalar
$\delta$-interactions. To motivate this, consider one-dimensional Dirac
operators with $\delta^{\prime}$-interactions of strength
$\alpha\in\mathbb{R}$ supported in $\Sigma=\\{0\\}$. These are first order
differential operators in $L^{2}(\mathbb{R})^{2}$ and the singular interaction
is modeled by transmission conditions for functions in the operator domain,
which for sufficiently smooth $f=(f_{1},f_{2})\in L^{2}(\mathbb{R})^{2}$ are
given by

(1.7) $f_{1}(0+)-f_{1}(0-)=i\frac{\alpha
c}{2}\big{(}f_{2}(0+)+f_{2}(0-)\big{)}\quad\text{and}\quad
f_{2}(0+)=f_{2}(0-),$

where $c>0$ is the speed of light. It is known that the associated self-
adjoint Dirac operators converge in the non-relativistic limit to a
Schrödinger operator with a $\delta^{\prime}$-interaction of strength
$\alpha$; cf. [2, 19] and also [10, 11] for generalizations. It is not
difficult to see that (1.7) can be rewritten as the transmission conditions
associated with a Dirac operator with a combination of an electrostatic and a
Lorentz scalar $\delta$-interaction of strengths $\eta=-\frac{\alpha
c^{2}}{2}$ and $\tau=\frac{\alpha c^{2}}{2}$, respectively, as they were
studied in dimension one recently in [7] and in higher space dimensions in,
e.g., [3, 5, 6, 7].

To find a counterpart of the above result in dimension two, consider a Dirac
operator with electrostatic and Lorentz scalar $\delta$-shell interactions of
strength $\eta$ and $\tau$, respectively, supported on $\Sigma$, which is
formally given by

(1.8) $\mathcal{A}_{\eta,\tau}=A_{0}+\left(\eta
I_{2}+\tau\sigma_{3}\right)\delta_{\Sigma};$

here $A_{0}$ is the unperturbed Dirac operator, $I_{2}$ is the $2\times
2$-identity matrix and $\sigma_{3}\in\mathbb{C}^{2\times 2}$ is given in
(3.1). The differential expression $\mathcal{A}_{\eta,\tau}$ gives rise to a
self-adjoint operator $A_{\eta,\tau}$ in $L^{2}(\mathbb{R}^{2})^{2}$, see
(3.3). If one chooses, as above, $\eta=-\frac{\alpha c^{2}}{2}$ and
$\tau=\frac{\alpha c^{2}}{2}$ and computes the non-relativistic limit, then
instead of a Schrödinger operator with a $\delta^{\prime}$-interaction one
gets the somewhat unexpected limit $T_{\alpha}$. Of course, this is compatible
with the one-dimensional result described above, as the one-dimensional
counterparts of (1.3) and (1.5) coincide, since there are no tangential
derivatives in $\mathbb{R}$. However, in higher dimensions Schrödinger
operators with oblique transmission conditions should be viewed as the non-
relativistic counterparts of Dirac operators with transmission conditions
generalizing (1.7). Related results on nonrelativistic limits of three-
dimensional Dirac operators with singular interactions can be found in [4, 5,
21]. The precise result about the non-relativistic limit described above is
stated in the following theorem and shown in Section 3.

######  Theorem 1.2.

Let $\alpha\in{\mathbb{R}}$. Then for all
$\lambda\in\mathbb{C}\setminus\mathbb{R}$ one has

$\lim_{c\rightarrow\infty}\left(A_{-\alpha c^{2}/2,\alpha
c^{2}/2}-(\lambda+c^{2}/2)\right)^{-1}=\left(\begin{array}[]{cc}\left(T_{\alpha}-\lambda\right)^{-1}&0\\\
0&0\end{array}\right),$

where the convergence is in the operator norm and the convergence rate is
$\mathcal{O}\left(\frac{1}{c}\right)$.

Notations. Throughout this paper $\Omega_{+}\subseteq{\mathbb{R}}^{2}$ is a
bounded and simply connected $C^{\infty}$-domain and
$\Omega_{-}={\mathbb{R}}^{2}\setminus\overline{\Omega_{+}}$ is the
corresponding exterior domain with boundary
$\Sigma=\partial\Omega_{-}=\partial\Omega_{+}$. The unit normal vector field
on $\Sigma$ pointing outwards of $\Omega_{+}$ is denoted by $\nu$. Moreover,
for $z\in{\mathbb{C}}\setminus[0,\infty)$ we choose the square root $\sqrt{z}$
such that $\text{Im}\sqrt{z}>0$ holds. The modified Bessel function of order
$j\in{\mathbb{N}}_{0}$ is denoted by $K_{j}$.

For $s\geq 0$ the spaces $H^{s}({\mathbb{R}}^{2})^{n}$,
$H^{s}(\Omega_{\pm})^{n}$, and $H^{s}(\Sigma)^{n}$ are the standard
$L^{2}$-based Sobolev spaces of ${\mathbb{C}}^{n}$-valued functions defined on
${\mathbb{R}}^{2}$, $\Omega_{\pm}$, and $\Sigma$, respectively. If $n=1$ we
simply write $H^{s}({\mathbb{R}}^{2})$, $H^{s}(\Omega_{\pm})$, and
$H^{s}(\Sigma)$. For negative $s<0$ we define the spaces
$H^{s}({\mathbb{R}}^{2})^{n}$ and $H^{s}(\Sigma)^{n}$ as the anti-dual spaces
of $H^{-s}({\mathbb{R}}^{2})^{n}$ and $H^{-s}(\Sigma)^{n}$, respectively. We
denote the restrictions of functions
$f:\mathbb{R}^{2}\rightarrow\mathbb{C}^{n}$ onto $\Omega_{\pm}$ by $f_{\pm}$;
in this sense we write
$H^{1}(\mathbb{R}^{2}\setminus\Sigma)^{n}=H^{1}(\Omega_{+})^{n}\oplus
H^{1}(\Omega_{-})^{n}$ and identify $f\in
H^{1}(\mathbb{R}^{2}\setminus\Sigma)^{n}$ with $f_{+}\oplus f_{-}$, where
$f_{\pm}\in H^{1}(\Omega_{\pm})^{n}$. The Dirichlet trace operators are
denoted by $\gamma_{D}^{\pm}:H^{1}(\Omega_{\pm})\rightarrow L^{2}(\Sigma)$ and
we shall write $\gamma_{D}:H^{1}(\mathbb{R}^{2})\rightarrow L^{2}(\Sigma)$ for
the trace on $H^{1}(\mathbb{R}^{2})$; sometimes these trace operators are also
viewed as bounded mappings to $H^{1/2}(\Sigma)$.

For a Hilbert space $\mathcal{H}$ we write $\mathcal{L}(\mathcal{H})$ for the
space of all everywhere defined, linear, and bounded operators on
$\mathcal{H}$. Furthermore, the domain, kernel, and range of a linear operator
$T$ from a Hilbert space $\mathcal{G}$ to $\mathcal{H}$ are denoted by
$\text{dom}\,T$, $\text{ker}\,T$, and $\text{ran}\,T$, respectively. The
resolvent set, the spectrum, the essential spectrum, the discrete spectrum,
and the point spectrum of a self-adjoint operator $T$ are denoted by
$\rho(T)$, $\sigma(T)$, $\sigma_{\text{ess}}(T)$, $\sigma_{\text{disc}}(T)$,
and $\sigma_{\textup{p}}(T)$. The eigenvalues of compact self-adjoint
operators $K\in\mathcal{L}(\mathcal{H})$ are denoted by $\mu_{n}(K)$ and are
ordered by their absolute values.

### Acknowledgement.

We are indebted to the referee for a very careful reading of our manuscript
and various helpful suggestions to improve the text. Jussi Behrndt and Markus
Holzmann gratefully acknowledge financial support by the Austrian Science Fund
(FWF): P33568-N. This publication is based upon work from COST Action CA 18232
MAT-DYN-NET, supported by COST (European Cooperation in Science and
Technology), www.cost.eu.

##  2\. Proof of Theorem 1.1

In this section the main result of this paper will be proved. For this, some
families of integral operators are used. Define for
$\lambda\in{\mathbb{C}}\setminus[0,\infty)$ the function $L_{\lambda}$ by

(2.1)
$L_{\lambda}(x)=\frac{\sqrt{\lambda}}{2\pi}K_{1}\bigl{(}-i\sqrt{\lambda}|x|\bigr{)}\frac{x_{1}-ix_{2}}{|x|},\quad
x=(x_{1},x_{2})\in\mathbb{R}^{2}\setminus\\{0\\},$

and the operator $\Psi_{\lambda}:L^{2}(\Sigma)\to L^{2}({\mathbb{R}}^{2})$ by

(2.2)
$\Psi_{\lambda}\varphi(x)=\int_{\Sigma}L_{\lambda}(x-y)\varphi(y)d\sigma(y),\quad\varphi\in
L^{2}(\Sigma),~{}x\in{\mathbb{R}}^{2}\setminus\Sigma.$

Moreover, for $\lambda\in{\mathbb{C}}\setminus[0,\infty)$ we make use of the
single layer potential $SL(\lambda):L^{2}(\Sigma)\rightarrow
H^{1}(\mathbb{R}^{2})$ and the single layer boundary integral operator
$S(\lambda):L^{2}(\Sigma)\to L^{2}(\Sigma)$ associated with $-\Delta-\lambda$
that are defined by

(2.3)
$SL(\lambda)\varphi(x)=\int_{\Sigma}\frac{1}{2\pi}K_{0}\bigl{(}-i\sqrt{\lambda}|x-y|\bigr{)}\varphi(y)d\sigma(y),\quad\varphi\in
L^{2}(\Sigma),\quad x\in\mathbb{R}^{2}\setminus\Sigma,$

and

(2.4)
$S(\lambda)\varphi(x)=\int_{\Sigma}\frac{1}{2\pi}K_{0}\bigl{(}-i\sqrt{\lambda}|x-y|\bigr{)}\varphi(y)d\sigma(y),\quad\varphi\in
L^{2}(\Sigma),~{}x\in\Sigma.$

It is known that $SL(\lambda)$ and $S(\lambda)$ are bounded and
$\text{ran}\,S(\lambda)\subseteq H^{1}(\Sigma)$; cf. [25, Theorem 6.12 and
Theorem 7.2]. In particular, $S(\lambda)$ gives rise to a compact operator in
$H^{s}(\Sigma)$ for every $s\in[0,1]$. Furthermore, $S(\lambda)$ is self-
adjoint and positive for $\lambda<0$ (see Step 1 in the proof of Proposition
2.2). Some properties of $\Psi_{\lambda}$ and $S(\lambda)$ that are important
in the proof of Theorem 1.1 are summarized in the following two propositions;
cf. Appendix A for the proof of Proposition 2.1 and Proposition 2.2.

######  Proposition 2.1.

Let $\lambda\in{\mathbb{C}}\setminus[0,\infty)$ and let $\Psi_{\lambda}$ be
given by (2.2). Then

(2.5) $\Psi_{\lambda}=-2i\partial_{z}SL(\lambda):L^{2}(\Sigma)\to
L^{2}({\mathbb{R}}^{2})$

is bounded and the following is true:

  1. (i)

$\Psi_{\lambda}$ gives rise to a bijective mapping
$\Psi_{\lambda}:H^{1/2}(\Sigma)\to\mathcal{H}_{\lambda}$, where

$\mathcal{H}_{\lambda}:=\big{\\{}f\in
H^{1}(\mathbb{R}^{2}\setminus\Sigma)\>|\>\partial_{\overline{z}}f_{+}\oplus\partial_{\overline{z}}f_{-}\in
H^{1}({\mathbb{R}}^{2}),\left(-\Delta-\lambda\right)f_{\pm}=0\text{ on
}\Omega_{\pm}\\}.$

  2. (ii)

$\Psi_{\lambda}^{\ast}:L^{2}({\mathbb{R}}^{2})\to L^{2}(\Sigma)$ is compact,
$\Psi_{\lambda}^{*}=-2i\gamma_{D}\partial_{\overline{z}}(-\Delta-\overline{\lambda})^{-1}$,
and ${\rm ran\,}\Psi_{\lambda}^{\ast}\subseteq H^{1/2}(\Sigma)$.

  3. (iii)

For all $\varphi\in H^{1/2}(\Sigma)$ the jump relations

$\begin{split}i(\nu_{1}+i\nu_{2})\big{(}\gamma_{D}^{+}(\Psi_{\lambda}\varphi)_{+}-\gamma_{D}^{-}(\Psi_{\lambda}\varphi)_{-}\big{)}&=\varphi,\\\
-i\big{(}\gamma_{D}^{+}\partial_{\overline{z}}(\Psi_{\lambda}\varphi)_{+}+\gamma_{D}^{-}\partial_{\overline{z}}(\Psi_{\lambda}\varphi)_{-}\big{)}&=\lambda
S(\lambda)\varphi,\end{split}$

hold.

For $\lambda<0$ denote by $\mu_{n}(S(\lambda))$ the discrete eigenvalues of
the positive self-adjoint operator $S(\lambda)$ order non-increasingly and
with multiplicities taken into account.

######  Proposition 2.2.

Let $S(\lambda)$ be defined by (2.4) and let $n\in\mathbb{N}$ be fixed. Then
the following holds:

  1. (i)

The function $(-\infty,0)\ni\lambda\mapsto\lambda\mu_{n}(S(\lambda))$ is
continuous, strictly monotonically increasing and

$\lim_{\lambda\to
0^{-}}\lambda\mu_{n}(S(\lambda))=0\quad\text{and}\quad\lim_{\lambda\to-\infty}\lambda\mu_{n}(S(\lambda))=-\infty.$

  2. (ii)

For $a<0$ the unique solution $\lambda_{n}(a)\in(-\infty,0)$ of
$\lambda\mu_{n}(S(\lambda))=a$ (see (i)) admits the asymptotic expansion
$\lambda_{n}(a)=-4a^{2}+\mathcal{O}(1)$ for $a\to-\infty$, where the
dependence on $n$ appears in the $\mathcal{O}(1)$-term.

###### Proof of Theorem 1.1.

_Step 1:_ We verify that $T_{\alpha}$ is symmetric in
$L^{2}({\mathbb{R}}^{2})$. Observe first that for $f\in{\rm dom\,}T_{\alpha}$
we have $\partial_{\overline{z}}f_{\pm}\in H^{1}(\Omega_{\pm})$ and $\Delta
f_{\pm}=4\partial_{z}\partial_{\overline{z}}f_{\pm}\in L^{2}(\Omega_{\pm})$,
and hence $T_{\alpha}$ is well-defined. Moreover, as
$C_{0}^{\infty}(\mathbb{R}^{2}\setminus\Sigma)\subseteq{\rm dom\,}T_{\alpha}$
it is also clear that $\text{dom}\,T_{\alpha}$ is dense. In order to show that
$T_{\alpha}$ is symmetric, we note that integration by parts in $\Omega_{\pm}$
yields for $f,g\in{\rm dom\,}T_{\alpha}$

(2.6) $\begin{split}(-\Delta
f_{\pm},g_{\pm}&)_{L^{2}(\Omega_{\pm})}=(-4\partial_{z}\partial_{\overline{z}}f_{\pm},g_{\pm})_{L^{2}(\Omega_{\pm})}\\\
&=4(\partial_{\overline{z}}f_{\pm},\partial_{\overline{z}}g_{\pm})_{L^{2}(\Omega_{\pm})}\mp
2\big{(}(\nu_{1}-i\nu_{2})\gamma_{D}^{\pm}(\partial_{\overline{z}}f_{\pm}),\gamma_{D}^{\pm}g_{\pm}\big{)}_{L^{2}(\Sigma)}\\\
&=4(\partial_{\overline{z}}f_{\pm},\partial_{\overline{z}}g_{\pm})_{L^{2}(\Omega_{\pm})}\mp
2\big{(}\gamma_{D}^{\pm}(\partial_{\overline{z}}f_{\pm}),(\nu_{1}+i\nu_{2})\gamma_{D}^{\pm}g_{\pm})\big{)}_{L^{2}(\Sigma)}.\end{split}$

Now, consider (2.6) for $f=g$ and add the equations for $\Omega_{+}$ and
$\Omega_{-}$. Then, using
$\gamma_{D}^{+}(\partial_{\overline{z}}f_{+})=\gamma_{D}^{-}(\partial_{\overline{z}}f_{-})$
and the transmission condition for $f\in{\rm dom\,}T_{\alpha}$, one finds that

(2.7)
$\begin{split}\big{(}T_{\alpha}f,f&\big{)}_{L^{2}({\mathbb{R}}^{2})}=4\big{(}\|\partial_{\overline{z}}f_{+}\|^{2}_{L^{2}(\Omega_{+})}+\|\partial_{\overline{z}}f_{-}\|^{2}_{L^{2}(\Omega_{-})}\big{)}\\\
&\quad-\big{(}\gamma_{D}^{+}(\partial_{\overline{z}}f_{+})+\gamma_{D}^{-}(\partial_{\overline{z}}f_{-}),(\nu_{1}+i\nu_{2})(\gamma_{D}^{+}f_{+}-\gamma_{D}^{-}f_{-})\big{)}_{L^{2}(\Sigma)}\\\
&=4\|\partial_{\overline{z}}f_{+}\oplus\partial_{\overline{z}}f_{-}\|^{2}_{L^{2}(\mathbb{R}^{2})}+\alpha\|\gamma_{D}^{+}(\partial_{\overline{z}}f_{+})+\gamma_{D}^{-}(\partial_{\overline{z}}f_{-})\|^{2}_{L^{2}(\Sigma)}\in\mathbb{R}.\end{split}$

Since this holds for all $f\in{\rm dom\,}T_{\alpha}$, we conclude that
$T_{\alpha}$ is symmetric.

_Step 2:_ Proof of the Birman-Schwinger principle in (iii): To show the first
implication, assume that $\lambda\in{\mathbb{C}}\setminus[0,\infty)$ with
$1\in\sigma_{\textup{p}}(\alpha\lambda S(\lambda))$ is given and choose
$\varphi\in{{\rm ker\,}\,}(I-\alpha\lambda S(\lambda))\setminus\\{0\\}$. Then
it follows from the mapping properties of $S(\lambda)$ that
$\varphi=\alpha\lambda S(\lambda)\varphi\in H^{1/2}(\Sigma)$ holds. Therefore,
Proposition 2.1 (i) implies that
$f:=\Psi_{\lambda}\varphi\in\mathcal{H}_{\lambda}$ fulfils $f\neq 0$, $f\in
H^{1}(\mathbb{R}^{2}\setminus\Sigma)$,
$\partial_{\overline{z}}f_{+}\oplus\partial_{\overline{z}}f_{-}\in
H^{1}({\mathbb{R}}^{2})$ and, as $\varphi\in{{\rm ker\,}\,}(1-\alpha\lambda
S(\lambda))\setminus\\{0\\}$, Proposition 2.1 (iii) implies

$i(\nu_{1}+i\nu_{2})\big{(}\gamma_{D}^{+}f_{+}-\gamma_{D}^{-}f_{-}\big{)}=\varphi=\alpha\lambda
S(\lambda)\varphi=-i\alpha\big{(}\gamma_{D}^{+}(\partial_{\overline{z}}f_{+})+\gamma_{D}^{-}(\partial_{\overline{z}}f_{-})\big{)}.$

Hence, $f\in{\rm dom\,}T_{\alpha}$. Moreover, as $f\in\mathcal{H}_{\lambda}$,
we conclude $f\in{{\rm ker\,}\,}(T_{\alpha}-\lambda)\setminus\\{0\\}$ and
hence $\lambda\in\sigma_{\textup{p}}(T_{\alpha})$.

To show the second implication, assume that
$\lambda\in\sigma_{\textup{p}}(T_{\alpha})$ is given and choose $f\in{{\rm
ker\,}\,}(T_{\alpha}-\lambda)\setminus\\{0\\}$. Then, by Proposition 2.1 (i)
there exists a unique $\varphi\in H^{1/2}(\Sigma)$ such that
$f=\Psi_{\lambda}\varphi$. Moreover, using $f\in{\rm dom\,}T_{\alpha}$ and
Proposition 2.1 (iii) one finds that

$0=i(\nu_{1}+i\nu_{2})\big{(}\gamma_{D}^{+}f_{+}-\gamma_{D}^{-}f_{-}\big{)}+i\alpha\big{(}\gamma_{D}^{+}(\partial_{\overline{z}}f_{+})+\gamma_{D}^{-}(\partial_{\overline{z}}f_{-})\big{)}=(I-\alpha\lambda
S(\lambda))\varphi.$

Since $\varphi\neq 0$, we conclude $1\in\sigma_{\textup{p}}(\alpha\lambda
S(\lambda))$.

_Step 3:_ Next, we prove that $T_{\alpha}$ is a self-adjoint operator and the
resolvent formula in (iv). Let
$\lambda\in{\mathbb{C}}\setminus([0,\infty)\cup\sigma_{\textup{p}}(T_{\alpha}))$
be fixed. First, we show that $I-\alpha\lambda S(\lambda)$ gives rise to a
bijective map in $H^{s}(\Sigma)$ for every $s\in[0,1]$. Recall that
$S(\lambda)$ is compact in $H^{s}(\Sigma)$. Since $I-\alpha\lambda S(\lambda)$
is injective for our choice of $\lambda$ by the Birman-Schwinger principle in
(iii), Fredholm’s alternative shows that $I-\alpha\lambda S(\lambda)$ is
indeed bijective.

Recall that $T_{\alpha}$ is symmetric; cf. Step 1. Hence, to show that
$T_{\alpha}$ is self-adjoint, it suffices to verify that
$\text{ran}(T_{\alpha}-\lambda)=L^{2}(\mathbb{R}^{2})$ holds for
$\lambda\in{\mathbb{C}}\setminus([0,\infty)\cup\sigma_{\textup{p}}(T_{\alpha}))$.
Fix such a $\lambda$, let $f\in L^{2}(\mathbb{R}^{2})$, and define

(2.8) $g=(-\Delta-\lambda)^{-1}f+\alpha\Psi_{\lambda}(I-\alpha\lambda
S(\lambda))^{-1}\Psi_{\overline{\lambda}}^{\ast}f,$

which is well-defined by the considerations above. Since
$\Psi_{\overline{\lambda}}^{\ast}f\in H^{1/2}(\Sigma)$ by Proposition 2.1 (ii)
and $(I-\alpha\lambda S(\lambda))^{-1}$ is bijective in $H^{1/2}(\Sigma)$, we
conclude with Proposition 2.1 (i) that $\Psi_{\lambda}(I-\alpha\lambda
S(\lambda))^{-1}\Psi_{\overline{\lambda}}^{\ast}f\in\mathcal{H}_{\lambda}\subseteq
H^{1}(\mathbb{R}^{2}\setminus\Sigma)$. In particular, with
$(-\Delta-\lambda)^{-1}f\in H^{2}(\mathbb{R}^{2})$ this implies that $g\in
H^{1}(\mathbb{R}^{2}\setminus\Sigma)$ and
$\partial_{\overline{z}}g_{+}\oplus\partial_{\overline{z}}g_{-}\in
H^{1}({\mathbb{R}}^{2})$. Moreover, with Proposition 2.1 (ii)–(iii) we obtain
that

$\begin{split}i(\nu_{1}+i\nu_{2})&\big{(}\gamma_{D}^{+}g_{+}-\gamma_{D}^{-}g_{-}\big{)}+i\alpha\big{(}\gamma_{D}^{+}(\partial_{\overline{z}}g_{+})+\gamma_{D}^{-}(\partial_{\overline{z}}g_{-})\big{)}\\\
&=\alpha(I-\alpha\lambda
S(\lambda))^{-1}\Psi_{\overline{\lambda}}^{\ast}f-\alpha\Psi_{\overline{\lambda}}^{\ast}f-\alpha^{2}\lambda
S(\lambda)(I-\alpha\lambda
S(\lambda))^{-1}\Psi_{\overline{\lambda}}^{\ast}f\\\ &=\alpha(I-\alpha\lambda
S(\lambda))(I-\alpha\lambda
S(\lambda))^{-1}\Psi_{\overline{\lambda}}^{\ast}f-\alpha\Psi_{\overline{\lambda}}^{\ast}f=0\end{split}$

and hence, $g\in{\rm dom\,}T_{\alpha}$. As $\Psi_{\lambda}(I-\alpha\lambda
S(\lambda))^{-1}\Psi_{\overline{\lambda}}^{\ast}f\in\mathcal{H}_{\lambda}$ by
Proposition 2.1 (i), we conclude

$\begin{split}\left(-\Delta-\lambda\right)g_{\pm}&=\left(-\Delta-\lambda\right)\big{(}(-\Delta-\lambda)^{-1}f\big{)}_{\pm}+\alpha\left(-\Delta-\lambda\right)\big{(}\Psi_{\lambda}(I-\alpha\lambda
S(\lambda))^{-1}\Psi_{\overline{\lambda}}^{\ast}f\big{)}_{\pm}\\\
&=\left(-\Delta-\lambda\right)\big{(}(-\Delta-\lambda)^{-1}f\big{)}_{\pm}=f_{\pm},\end{split}$

i.e. $(T_{\alpha}-\lambda)g=f$. Since $f\in L^{2}({\mathbb{R}}^{2})$ was
arbitrary, we conclude that ${\rm
ran\,}(T_{\alpha}-\lambda)=L^{2}({\mathbb{R}}^{2})$ and that $T_{\alpha}$ is
self-adjoint. Moreover, the resolvent formula in item (iv) follows from (2.8).

_Step 4:_ Next, we show $\sigma_{\text{ess}}(T_{\alpha})=[0,\infty)$. Let
$\lambda\in{\mathbb{C}}\setminus\mathbb{R}$. Since
$\Psi_{\overline{\lambda}}^{\ast}:L^{2}({\mathbb{R}}^{2})\to L^{2}(\Sigma)$ is
compact by Proposition 2.1 (ii), the resolvent formula in (iv) implies that
$(T_{\alpha}-\lambda)^{-1}-(-\Delta-\lambda)^{-1}$ is a compact operator in
$L^{2}({\mathbb{R}}^{2})$. Consequently, Weyl’s Theorem [27, Theorem XIII.14]
yields that
$\sigma_{\textup{ess}}(T_{\alpha})=\sigma_{\textup{ess}}(-\Delta)=[0,\infty)$.

_Step 5:_ Proof of (i): Let $\alpha\geq 0$. Then, (2.7) implies that
$T_{\alpha}$ is non-negative and hence, $\sigma(T_{\alpha})\subset[0,\infty)$.
Since the latter set coincides with $\sigma_{\textup{ess}}(T_{\alpha})$, see
Step 4, we conclude $\sigma_{\textup{disc}}(T_{\alpha})=\emptyset$.

_Step 6:_ Proof of (ii): Let $\alpha<0$. Since
$\sigma_{\text{ess}}(T_{\alpha})=[0,\infty)$, it follows from the Birman-
Schwinger principle in (iii) that

$\sigma_{\text{disc}}(T_{\alpha})=\left\\{\lambda_{n}\>|\>n\in{\mathbb{N}}\right\\}=\left\\{\lambda<0\>|\>\exists
n\in{\mathbb{N}}\text{ such that
}\lambda\mu_{n}(S(\lambda))=\alpha^{-1}\right\\}$

holds. Note that by Proposition 2.2 the equation
$\lambda\mu_{n}(S(\lambda))=\alpha^{-1}$ has a unique solution $\lambda_{n}$
for all $n\in\mathbb{N}$. Moreover, for any $n\in{\mathbb{N}}$ there cannot be
infinitely many $k\neq n$ with $\lambda_{n}=\lambda_{k}$, since otherwise
$\alpha^{-1}<0$ would be an eigenvalue with infinite multiplicity of the self-
adjoint and compact operator $\lambda_{n}S(\lambda_{n})$. Thus
$\sigma_{\text{disc}}(T_{\alpha})$ is indeed an infinite set. Furthermore, as
$S(\lambda)$ is a positive self-adjoint operator in $L^{2}(\Sigma)$; cf. Step
1 in the proof of Proposition 2.2, we have by definition
$\mu_{n}(S(\lambda))\geq\mu_{n+1}(S(\lambda))$ implying
$\lambda\mu_{n}(S(\lambda))\leq\lambda\mu_{n+1}(S(\lambda))$. Therefore, the
monotonicity of the map $\lambda\mapsto\lambda\mu_{n}(S(\lambda))$ from
Proposition 2.2 yields $\lambda_{n+1}\leq\lambda_{n}$ for all
$n\in{\mathbb{N}}$. This shows that $0$ cannot be an accumulation point of the
sequence $(\lambda_{n})_{n\in{\mathbb{N}}}$ and as
$\sigma_{\text{ess}}(T_{\alpha})\cap(-\infty,0)=\emptyset$ the sequence
$(\lambda_{n})_{n\in{\mathbb{N}}}$ has no finite accumulation points, that is,
$\sigma_{\text{disc}}(T_{\alpha})$ must be unbounded from below.

It remains to prove the asymptotic expansion in item (ii). By the above
considerations $\lambda_{n}$ is determined as the unique solution of
$\lambda\mu_{n}(S(\lambda))=\alpha^{-1}$. Clearly, if $\alpha\rightarrow
0^{-}$, then $a:=\alpha^{-1}\rightarrow-\infty$. Hence, it follows from
Proposition 2.2 (ii) with $a=\alpha^{-1}$ that
$\lambda_{n}=-\frac{4}{\alpha^{2}}+\mathcal{O}(1)$ for $\alpha\rightarrow
0^{-}$ and that the dependence on $n$ appears in the $\mathcal{O}(1)$-term. ∎

##  3\. Proof of Theorem 1.2

In this section we show that $T_{\alpha}$ is the nonrelativistic limit of a
family of Dirac operators with electrostatic and Lorentz scalar $\delta$-shell
potentials formally given by (1.8), whose interaction strengths are suitably
scaled. First, we recall the rigorous definition of the operator
$A_{\eta,\tau}$ associated with (1.8), see [5, 6, 7] for details. Let

(3.1) $\sigma_{1}=\left(\begin{array}[]{cc}0&1\\\ 1&0\\\
\end{array}\right),\hskip 8.53581pt\hskip
8.53581pt\sigma_{2}=\left(\begin{array}[]{cc}0&-i\\\ i&0\\\
\end{array}\right),\hskip 8.53581pt\text{ and }\hskip
8.53581pt\sigma_{3}=\left(\begin{array}[]{cc}1&0\\\ 0&-1\\\
\end{array}\right),$

be the Pauli spin matrices and denote the $2\times 2$ identity matrix by
$I_{2}$. Furthermore, for $x=(x_{1},x_{2})\in{\mathbb{R}}^{2}$ we will use the
abbreviations

(3.2) $\sigma\cdot
x=\sigma_{1}x_{1}+\sigma_{2}x_{2}\quad\text{and}\quad\sigma\cdot\nabla=\sigma_{1}\partial_{1}+\sigma_{2}\partial_{2}.$

We define Dirac operators with electrostatic and Lorentz scalar $\delta$-shell
interactions of strengths $\eta,\tau\in{\mathbb{R}}$ in
$L^{2}({\mathbb{R}}^{2})^{2}$ by

(3.3)
$\begin{split}A_{\eta,\tau}f&=\left(-ic(\sigma\cdot\nabla)+\frac{c^{2}}{2}\sigma_{3}\right)f_{+}\oplus\left(-ic(\sigma\cdot\nabla)+\frac{c^{2}}{2}\sigma_{3}\right)f_{-},\\\
{\rm dom\,}A_{\eta,\tau}&=\Big{\\{}f\in H^{1}(\Omega_{+})^{2}\oplus
H^{1}(\Omega_{-})^{2}\>\big{|}\\\
&~{}~{}ic(\sigma\cdot\nu)\left(\gamma_{D}^{+}f_{+}-\gamma_{D}^{-}f_{-}\right)+\frac{1}{2}(\eta
I_{2}+\tau\sigma_{3})\left(\gamma_{D}^{+}f_{+}+\gamma_{D}^{-}f_{-}\right)=0\Big{\\}}.\end{split}$

It is shown in [6, 7] that $A_{\eta,\tau}$ is self-adjoint in
$L^{2}({\mathbb{R}}^{2})^{2}$, whenever $\eta^{2}-\tau^{2}\neq 4c^{2}$, and as
in [5] one sees that these operators are the self-adjoint realisations of the
formal differential expression (1.8). In the above definition we are using
units such that $\hbar=1$ and consider the mass $m=\frac{1}{2}$, but we keep
the speed of light $c$ as a parameter for the discussion of the non-
relativistic limit $c\to\infty$.

Throughout this section we make use of the self-adjoint free Dirac operator
$A_{0}$, which coincides with $A_{0,0}$ given in (3.3) and which is defined on
$H^{1}(\mathbb{R}^{2})^{2}$. For
$\lambda\in\rho(A_{0})=\mathbb{C}\setminus((-\infty,-\frac{c^{2}}{2}]\cup[\frac{c^{2}}{2},\infty))$
the integral kernel of the resolvent of $A_{0}$ is given by
$G_{\lambda}(x-y)$, where $G_{\lambda}(x)$ is defined for
$x\in{\mathbb{R}}^{2}\setminus\\{0\\}$ by

(3.4) $G_{\lambda}(x)=\frac{1}{2\pi
c}\sqrt{\frac{\lambda^{2}}{c^{2}}-\frac{c^{2}}{4}}K_{1}\left(-i\sqrt{\frac{\lambda^{2}}{c^{2}}-\frac{c^{2}}{4}}|x|\right)\frac{1}{|x|}(\sigma\cdot
x)\\\ +\frac{1}{2\pi
c}K_{0}\left(-i\sqrt{\frac{\lambda^{2}}{c^{2}}-\frac{c^{2}}{4}}|x|\right)\left(\frac{\lambda}{c}I_{2}+\frac{c}{2}\sigma_{3}\right);$

cf. [6, equation (3.2)]. With this function we define the two families of
integral operators

(3.5)
$\begin{split}\Phi_{\lambda}\varphi(x)&=\int_{\Sigma}G_{\lambda}(x-y)\varphi(y)\mathrm{d}\sigma(y),\quad\varphi\in
L^{2}(\Sigma)^{2},~{}x\in{\mathbb{R}}^{2}\setminus\Sigma,\\\
\mathcal{C}_{\lambda}\varphi(x)&=\lim_{\varepsilon\to
0^{+}}\int\limits_{\Sigma\setminus
B(x,\varepsilon)}G_{\lambda}(x-y)\varphi(y)\mathrm{d}\sigma(y),\quad\varphi\in
L^{2}(\Sigma)^{2},~{}x\in\Sigma,\end{split}$

where $B(x,\varepsilon)$ is the ball of radius $\varepsilon$ centered at $x$.
Both operators $\Phi_{\lambda}:L^{2}(\Sigma)^{2}\to
L^{2}({\mathbb{R}}^{2})^{2}$ and $\mathcal{C}_{\lambda}:L^{2}(\Sigma)^{2}\to
L^{2}(\Sigma)^{2}$ are well-defined and bounded; cf. [6, Proposition 3.3 and
equation (3.7)].

In the following lemma, which is a preparation for the proof of Theorem 1.2,
we will use the matrices

$M_{1}=\left(\begin{array}[]{cc}1&0\\\ 0&0\end{array}\right),\quad
M_{2}=\left(\begin{array}[]{cc}0&1\\\
0&0\end{array}\right),\quad\text{and}\quad
M_{3}=\left(\begin{array}[]{cc}0&0\\\ 0&1\end{array}\right);$

products of scalar operators and matrices are understood componentwise, e.g.

$(-\Delta-\lambda)^{-1}M_{1}=\left(\begin{array}[]{cc}(-\Delta-\lambda)^{-1}&0\\\
0&0\end{array}\right):L^{2}({\mathbb{R}}^{2})^{2}\to
L^{2}({\mathbb{R}}^{2})^{2}.$

######  Lemma 3.1.

Let $\lambda\in{\mathbb{C}}\setminus{\mathbb{R}}$. Then there exists a
constant $K>0$, depending only on $\lambda$ and $\Sigma$, such that the
estimates

(3.6a)
$\displaystyle\big{\|}\left(A_{0}-(\lambda+c^{2}/2)\right)^{-1}-\left(-\Delta-\lambda\right)^{-1}M_{1}\big{\|}$
$\displaystyle\leq\frac{K}{c},$ (3.6b)
$\displaystyle\big{\|}c\Phi_{\lambda+c^{2}/2}M_{3}-\Psi_{\lambda}M_{2}\big{\|}$
$\displaystyle\leq\frac{K}{c},$ (3.6c)
$\displaystyle\big{\|}cM_{3}\Phi_{\lambda+c^{2}/2}^{\ast}-M_{2}^{\top}\Psi_{\lambda}^{\ast}\big{\|}$
$\displaystyle\leq\frac{K}{c},$ (3.6d)
$\displaystyle\big{\|}c^{2}M_{3}\mathcal{C}_{\lambda+c^{2}/2}M_{3}-\lambda
S(\lambda)M_{3}\big{\|}$ $\displaystyle\leq\frac{K}{c},$

are valid for all sufficiently large $c>0$.

###### Proof.

We use a similar strategy as in the proof of [4, Proposition 5.2]. In the
following let $\lambda\in{\mathbb{C}}\setminus{\mathbb{R}}$ be fixed. Then
$\lambda+\frac{c^{2}}{2}\in{\mathbb{C}}\setminus{\mathbb{R}}$ and hence all
operators in (3.6a)–(3.6d) are well-defined. One verifies by direct
calculation that for sufficiently large $c>0$ and all $t\in[0,1]$

(3.7)
$0<\frac{1}{2}\left|\sqrt{\lambda}\right|\leq\left|\sqrt{\lambda+t\frac{\lambda^{2}}{c^{2}}}\right|\leq\frac{3}{2}\left|\sqrt{\lambda}\right|\quad\text{and}\quad\frac{1}{2}\text{Im}\sqrt{\lambda}\leq\text{Im}\sqrt{\lambda+t\frac{\lambda^{2}}{c^{2}}}$

hold. With the well-known asymptotic expansions of the modified Bessel
functions and $K_{1}^{\prime}(z)=-K_{0}(z)-\frac{1}{z}K_{1}(z)$, (see [1]) one
shows that there exist constants $\widehat{K},\kappa,R>0$, depending only on
$\lambda$, such that

(3.8)
$\left|K_{j}\left(-i\sqrt{\lambda+t\frac{\lambda^{2}}{c^{2}}}|x|\right)\right|\leq\widehat{K}\left\\{\begin{array}[]{ll}|x|^{-1},&\text{for
}|x|<R,\\\ e^{-\kappa|x|},&\text{for }|x|\geq R,\end{array}\right.$

and

(3.9)
$\left|K_{1}^{\prime}\left(-i\sqrt{\lambda+t\frac{\lambda^{2}}{c^{2}}}|x|\right)\right|\leq\widehat{K}\left\\{\begin{array}[]{ll}|x|^{-2},&\text{for
}|x|<R,\\\ e^{-\kappa|x|},&\text{for }|x|\geq R,\end{array}\right.$

hold for all $x\in{\mathbb{R}}^{2}\setminus\\{0\\}$, $j\in\\{0,1\\}$,
$t\in[0,1]$, and sufficiently large $c>0$.

Next, with $G_{\lambda+c^{2}/2}$ defined by (3.4) we find

(3.10) $G_{\lambda+c^{2}/2}(x)=\frac{1}{2\pi
c}\sqrt{\lambda+\frac{\lambda^{2}}{c^{2}}}K_{1}\left(-i\sqrt{\lambda+\frac{\lambda^{2}}{c^{2}}}|x|\right)\frac{1}{|x|}(\sigma\cdot
x)\\\ +\frac{1}{2\pi
c}K_{0}\left(-i\sqrt{\lambda+\frac{\lambda^{2}}{c^{2}}}|x|\right)\left(\frac{\lambda}{c}I_{2}+cM_{1}\right).$

Let

$U_{\lambda}(x)=\frac{1}{2\pi}K_{0}\bigl{(}-i\sqrt{\lambda}|x|\bigr{)},\quad
x\in{\mathbb{R}}^{2}\setminus\\{0\\},$

be the integral kernel of the resolvent of the free Laplace operator; cf. [28,
Chapter 7.5]. Then

$G_{\lambda+c^{2}/2}(x)-U_{\lambda}(x)M_{1}=t_{1}(x)+t_{2}(x)+t_{3}(x)$

holds, where the matrix-valued functions $t_{1},t_{2}$, and $t_{3}$ are given
by

$\begin{split}t_{1}(x)&=\frac{1}{2\pi
c}\sqrt{\lambda+\frac{\lambda^{2}}{c^{2}}}K_{1}\left(-i\sqrt{\lambda+\frac{\lambda^{2}}{c^{2}}}|x|\right)\frac{\sigma\cdot
x}{|x|},\\\
t_{2}(x)&=\frac{1}{2\pi}\left(K_{0}\left(-i\sqrt{\lambda+\frac{\lambda^{2}}{c^{2}}}|x|\right)-K_{0}\bigl{(}-i\sqrt{\lambda}|x|\bigr{)}\right)M_{1},\\\
t_{3}(x)&=\frac{\lambda}{2\pi
c^{2}}K_{0}\left(-i\sqrt{\lambda+\frac{\lambda^{2}}{c^{2}}}|x|\right)I_{2}.\end{split}$

With (3.7) and (3.8) applied with $t=1$ one finds that there exist constants
$k_{1},\kappa,R>0$, depending only on $\lambda$, such that for $j\in\\{1,3\\}$
and sufficiently large $c>0$ one has

$\left|t_{j}(x)\right|\leq\frac{k_{1}}{c}\left\\{\begin{array}[]{ll}|x|^{-1},&\text{for
}|x|<R,\\\ e^{-\kappa|x|},&\text{for }|x|\geq R.\end{array}\right.$

To estimate $t_{2}$, we use $K_{0}^{\prime}=-K_{1}$ and obtain with the
fundamental theorem of calculus, (3.7), and (3.8)

(3.11)
$\begin{split}\Bigg{|}K_{0}\left(-i\sqrt{\lambda+\frac{\lambda^{2}}{c^{2}}}|x|\right)&-K_{0}\left(-i\sqrt{\lambda}|x|\right)\Bigg{|}\leq\int_{0}^{1}\left|\frac{d}{dt}K_{0}\left(-i\sqrt{\lambda+t\frac{\lambda^{2}}{c^{2}}}|x|\right)\right|dt\\\
&=\int_{0}^{1}\frac{|\lambda|^{2}|x|}{\left|\sqrt{\lambda+t\frac{\lambda^{2}}{c^{2}}}\right|}\frac{1}{2c^{2}}\left|K_{1}\left(-i\sqrt{\lambda+t\frac{\lambda^{2}}{c^{2}}}|x|\right)\right|dt\\\
&\leq\frac{k_{2}}{c^{2}}\left\\{\begin{array}[]{ll}1,&\text{for }|x|<R,\\\
e^{-\frac{\kappa}{2}|x|},&\text{for }|x|\geq R,\end{array}\right.\end{split}$

with a constant $k_{2}$ which depends only on $\lambda$. Thus, if we define
$k_{3}=2k_{1}+\frac{k_{2}R}{2\pi}$, then

$\left|G_{\lambda+c^{2}/2}(x)-U_{\lambda}(x)M_{1}\right|\leq\frac{k_{3}}{c}\left\\{\begin{array}[]{ll}|x|^{-1},&\text{for
}|x|<R,\\\ e^{-\frac{\kappa}{2}|x|},&\text{for }|x|\geq R.\end{array}\right.$

This estimation for the integral kernel yields with the Schur test; cf. [4,
Proposition A.3] for a similar argument,

$\big{\|}\left(A_{0}-(\lambda+c^{2}/2)\right)^{-1}-\left(-\Delta-\lambda\right)^{-1}M_{1}\big{\|}\leq\frac{K}{c}$

for all sufficiently large $c>0$, which is the first claimed estimate (3.6a).

Next, we prove (3.6b). Recall that the integral kernel $L_{\lambda}$ of
$\Psi_{\lambda}$ is given by (2.1). Using that $\sigma_{1}M_{3}=M_{2}$,
$\sigma_{2}M_{3}=-iM_{2}$, and $M_{1}M_{3}=0$, we obtain with (3.10) the
decomposition

$cG_{\lambda+c^{2}/2}(x)M_{3}-L_{\lambda}(x)M_{2}=\tau_{1}(x)+\tau_{2}(x)+\tau_{3}(x)$

with

$\begin{split}\tau_{1}(x)&=\frac{1}{2\pi}\left(\sqrt{\lambda+\frac{\lambda^{2}}{c^{2}}}-\sqrt{\lambda}\right)K_{1}\left(-i\sqrt{\lambda+\frac{\lambda^{2}}{c^{2}}}|x|\right)\frac{x_{1}-ix_{2}}{|x|}M_{2},\\\
\tau_{2}(x)&=\frac{\sqrt{\lambda}}{2\pi}\left(K_{1}\left(-i\sqrt{\lambda+\frac{\lambda^{2}}{c^{2}}}|x|\right)-K_{1}\bigl{(}-i\sqrt{\lambda}|x|\bigr{)}\right)\frac{x_{1}-ix_{2}}{|x|}M_{2},\\\
\tau_{3}(x)&=\frac{\lambda}{2\pi
c}K_{0}\left(-i\sqrt{\lambda+\frac{\lambda^{2}}{c^{2}}}|x|\right)M_{3}.\end{split}$

Similar as above it can be shown that there exists a $k_{4}>0$, depending only
on $\lambda$, such that for all $j\in\\{1,2,3\\}$

$\left|\tau_{j}(x)\right|\leq\frac{k_{4}}{c}\left\\{\begin{array}[]{ll}|x|^{-1},&\text{for
}|x|<R,\\\ e^{-\frac{\kappa}{2}|x|},&\text{for }|x|\geq R;\end{array}\right.$

to see the estimate for $\tau_{2}$ one has to use (3.9). With the help of the
Schur test the estimate (3.6b) follows (see also [4, Proposition A.4] for a
similar argument); the constant $k_{4}$ depends in this case on $\lambda$ and
$\Sigma$. The estimate in (3.6c) follows by taking adjoints.

It remains to prove (3.6d). Taking $M_{3}\sigma\cdot xM_{3}=0$, which holds
for any $x\in\mathbb{R}^{2}$, and (3.11) into account we obtain that

$\begin{split}\big{|}c^{2}M_{3}G_{\lambda+c^{2}/2}(x)M_{3}&-\lambda
U_{\lambda}(x)M_{3}\big{|}\\\
&=\frac{|\lambda|}{2\pi}\left|K_{0}\left(-i\sqrt{\lambda+\frac{\lambda^{2}}{c^{2}}}|x|\right)-K_{0}\bigl{(}-i\sqrt{\lambda}|x|\bigr{)}\right|\leq\frac{k_{5}}{c^{2}}\end{split}$

holds for all $x\in{\mathbb{R}}^{2}\setminus\\{0\\}$. Using the dominated
convergence theorem, one sees that

$\big{(}c^{2}M_{3}\mathcal{C}_{\lambda+c^{2}/2}M_{3}f\big{)}(x)=\int_{\Sigma}c^{2}M_{3}G_{\lambda+c^{2}/2}(x-y)M_{3}f(y)d\sigma(y)$

holds for all $f\in L^{2}(\Sigma)^{2}$ and $x\in\Sigma$, i.e. the integral
does not have to be understood as principal value. Thus we obtain with the
Schur test [23, III. Example 2.4] that

$\|c^{2}M_{3}\mathcal{C}_{\lambda+mc^{2}}M_{3}-\lambda
S(\lambda)M_{3}\|\leq\frac{K}{c^{2}}.$

In this case, the constant $K$ depends on $\lambda$ and $\Sigma$. This yields
(3.6d) and finishes the proof of this lemma. ∎

Now we are prepared to prove Theorem 1.2 and show that $A_{-\alpha
c^{2}/2,\alpha c^{2}/2}$ converges in the nonrelativistic limit to
$T_{\alpha}$ defined in (1.6).

###### Proof of Theorem 1.2.

Let $\lambda\in{\mathbb{C}}\setminus{\mathbb{R}}$ be fixed. Then, by [7, Lemma
5.4, Proposition 5.5, Theorem 5.6, and Lemma 5.9] (see also [6, Theorem 4.6])
the operator $I_{2}-\alpha
c^{2}M_{3}\mathcal{C}_{\lambda+c^{2}/2}:L^{2}(\Sigma)^{2}\to
L^{2}(\Sigma)^{2}$ is boundedly invertible and the resolvent of $A_{-\alpha
c^{2}/2,\alpha c^{2}/2}-\frac{c^{2}}{2}$ is given by

(3.12) $\begin{split}\big{(}A_{-\alpha c^{2}/2,\alpha
c^{2}/2}-&(\lambda+c^{2}/2)\big{)}^{-1}=\left(A_{0}-(\lambda+c^{2}/2)\right)^{-1}\\\
&+\Phi_{\lambda+c^{2}/2}\left(I-\alpha
c^{2}M_{3}\mathcal{C}_{\lambda+c^{2}/2}\right)^{-1}\alpha
c^{2}M_{3}\Phi_{\overline{\lambda}+c^{2}/2}^{\ast}.\end{split}$

Because of $M_{3}=M_{3}^{2}$ it follows from [26, Proposition 2.1.8] that

$\sigma\big{(}M_{3}\mathcal{C}_{\lambda+c^{2}/2}\big{)}\cup\\{0\\}=\sigma\big{(}M_{3}\mathcal{C}_{\lambda+c^{2}/2}M_{3}\big{)}\cup\\{0\\}.$

In particular, this yields that the operator $I-\alpha
c^{2}M_{3}\mathcal{C}_{\lambda+c^{2}/2}M_{3}$ is boundedly invertible in
$L^{2}(\Sigma)^{2}$ for all $c>0$ and a direct calculation shows

(3.13) $(I-\alpha
c^{2}M_{3}\mathcal{C}_{\lambda+c^{2}/2})^{-1}M_{3}=M_{3}(I-\alpha
c^{2}M_{3}\mathcal{C}_{\lambda+c^{2}/2}M_{3})^{-1}.$

Recall that for $\lambda\in\mathbb{C}\setminus\mathbb{R}$ also
$I-\alpha\lambda S(\lambda)$ is boundedly invertible in $L^{2}(\Sigma)$; cf.
Theorem 1.1 (iv). Hence, we obtain from Lemma 3.1 and [23, IV. Theorem 1.16]
that

(3.14) $\|(I-\alpha
c^{2}M_{3}\mathcal{C}_{\lambda+c^{2}/2}M_{3})^{-1}-(I-\alpha\lambda
S(\lambda)M_{3})^{-1}\|\leq\frac{K}{c}$

holds for all sufficiently large $c>0$ with a constant $K>0$ which depends
only on $\lambda$, $\alpha$, and $\Sigma$.

To conclude, note that (3.12) and (3.13) yield

$\begin{split}\big{(}A_{-\alpha c^{2}/2,\alpha
c^{2}/2}-&(\lambda+c^{2}/2)\big{)}^{-1}=\big{(}A_{0}-(\lambda+c^{2}/2)\big{)}^{-1}\\\
&+c\Phi_{\lambda+c^{2}/2}M_{3}(I-\alpha
c^{2}M_{3}\mathcal{C}_{\lambda+c^{2}/2}M_{3})^{-1}\alpha
cM_{3}\Phi_{\overline{\lambda}+c^{2}/2}^{\ast},\end{split}$

while Theorem 1.1 (iv) and $M_{2}M_{3}M_{2}^{\top}=M_{1}$ show

$\begin{split}(T_{\alpha}-\lambda)^{-1}M_{1}&=(-\Delta-\lambda)^{-1}M_{1}+\Psi_{\lambda}(I-\alpha\lambda
S(\lambda))^{-1}\alpha\Psi_{\overline{\lambda}}^{\ast}M_{1}\\\
&=(-\Delta-\lambda)^{-1}M_{1}+\Psi_{\lambda}M_{2}(I-\alpha\lambda
S(\lambda)M_{3})^{-1}\alpha
M_{2}^{\top}\Psi_{\overline{\lambda}}^{\ast}.\end{split}$

Using Lemma 3.1 and (3.14) the last two displayed formulae finally lead to the
claimed convergence result and it also follows that the order of convergence
is $\mathcal{O}(\frac{1}{c})$. ∎

##  Appendix A Proof of Propositions 2.1 and 2.2

Recall that for $\lambda\in\mathbb{C}\setminus[0,\infty)$ the operators
$\Psi_{\lambda}$, $SL(\lambda)$, and $S(\lambda)$ are defined by (2.2), (2.3),
and (2.4), respectively. First, we collect some properties of the single layer
potential $SL(\lambda)$ that are needed in the following. It is well-known
that $SL(\lambda):H^{1/2}(\Sigma)\to H^{2}(\mathbb{R}^{2}\setminus\Sigma)$
gives rise to a bounded operator, that $(-\Delta-\lambda)SL(\lambda)\varphi=0$
in $\mathbb{R}^{2}\setminus\Sigma$, and that for $\varphi\in H^{1/2}(\Sigma)$
the jump relations

(A.1)
$\gamma_{D}^{+}(SL(\lambda)\varphi)_{+}=\gamma_{D}^{-}(SL(\lambda)\varphi)_{-}\quad\text{and}\quad\partial_{\nu}(SL(\lambda)\varphi)_{+}-\partial_{\nu}(SL(\lambda)\varphi)_{-}=\varphi$

hold; cf. [25] or [22, Section 3.3]. Furthermore, for the single layer
boundary integral operator $S(\lambda)$ from (2.4) we have
$S(\lambda)=\gamma_{D}SL(\lambda)$ and for all $\varphi\in L^{2}(\Sigma)$ the
representations

(A.2)
$SL(\lambda)\varphi=(-\Delta-\lambda)^{-1}\gamma_{D}^{\prime}\varphi\quad\text{and}\quad
S(\lambda)\varphi=\gamma_{D}(-\Delta-\lambda)^{-1}\gamma_{D}^{\prime}\varphi$

hold (see [22, 25]); here $\gamma_{D}:H^{1}(\mathbb{R}^{2})\rightarrow
L^{2}(\Sigma)$ and $\gamma_{D}^{\prime}:L^{2}(\Sigma)\rightarrow
H^{-1}(\mathbb{R}^{2})$ is the anti-dual operator.

###### Proof of Proposition 2.1.

First, we prove item (ii). For $\lambda\in{\mathbb{C}}\setminus[0,\infty)$
define the operator

(A.3)
$\widehat{\Psi}_{\lambda}:=-2i\gamma_{D}\partial_{\overline{z}}(-\Delta-\overline{\lambda})^{-1}.$

Since $(-\Delta-\overline{\lambda})^{-1}:L^{2}(\mathbb{R}^{2})\rightarrow
H^{2}(\mathbb{R}^{2})$ and $\gamma_{D}:H^{1}(\mathbb{R}^{2})\rightarrow
H^{1/2}(\Sigma)$ are bounded, we get that
$\widehat{\Psi}_{\lambda}:L^{2}({\mathbb{R}}^{2})\to H^{1/2}(\Sigma)$ is well-
defined and bounded. Furthermore, as $H^{1/2}(\Sigma)$ is compactly embedded
in $L^{2}(\Sigma)$ by Rellich’s embedding theorem, the operator
$\widehat{\Psi}_{\lambda}:L^{2}({\mathbb{R}}^{2})\to L^{2}(\Sigma)$ is
compact. Note that $\widehat{\Psi}_{\lambda}$ is an integral operator with
integral kernel

$\begin{split}k(x,y)&=-2i\partial_{\overline{z}}\frac{1}{2\pi}K_{0}\bigl{(}-i\sqrt{\overline{\lambda}}|x-y|\bigr{)}\\\
&=\frac{\sqrt{\overline{\lambda}}}{2\pi}K_{1}\bigl{(}-i\sqrt{\overline{\lambda}}|x-y|\bigr{)}\frac{x_{1}-y_{1}+i(x_{2}-y_{2})}{|x-y|}\\\
&=\overline{L_{\lambda}(y-x)},\end{split}$

where we used $K_{0}^{\prime}=-K_{1}$ in the second step and
$\overline{\sqrt{\lambda}}=-\sqrt{\overline{\lambda}}$ in the last step
(recall that $\textup{Im}\sqrt{\omega}>0$ for
$\omega\in\mathbb{C}\setminus[0,\infty)$). Hence, we conclude that

$\Psi_{\lambda}=\widehat{\Psi}_{\lambda}^{*}:L^{2}(\Sigma)\rightarrow
L^{2}(\mathbb{R}^{2})$

is bounded and that all claims in item (ii) are true.

Next, we show (2.5). Let $\varphi\in L^{2}(\Sigma)$ and $f\in
H^{1}(\mathbb{R}^{2})$. Since
$\Delta=4\partial_{\overline{z}}\partial_{z}=4\partial_{z}\partial_{\overline{z}}$,
we see that
$\partial_{\overline{z}}(-\Delta-\overline{\lambda})^{-1}f=(-\Delta-\overline{\lambda})^{-1}\partial_{\overline{z}}f$.
Hence, item (ii) and (A.2) imply

$\begin{split}(\Psi_{\lambda}\varphi,f)_{L^{2}(\Sigma)}&=\big{(}\varphi,-2i\gamma_{D}\partial_{\overline{z}}(-\Delta-\overline{\lambda})^{-1}f\big{)}_{L^{2}(\mathbb{R}^{2})}\\\
&=\big{(}\varphi,-2i\gamma_{D}(-\Delta-\overline{\lambda})^{-1}\partial_{\overline{z}}f\big{)}_{L^{2}(\mathbb{R}^{2})}\\\
&=\big{(}-2i\partial_{z}SL(\lambda)\varphi,f\big{)}_{L^{2}(\mathbb{R}^{2})}.\end{split}$

Since $H^{1}(\mathbb{R}^{2})$ is dense in $L^{2}(\mathbb{R}^{2})$, we conclude
that (2.5) is true. In particular, this and the properties of the single layer
potential mentioned at the beginning of this appendix imply that

(A.4) $\Psi_{\lambda}:H^{1/2}(\Sigma)\rightarrow
H^{1}(\mathbb{R}^{2}\setminus\Sigma)$

is bounded and for $\varphi\in H^{1/2}(\Sigma)$ we have

(A.5)
$i\partial_{\overline{z}}\left(\Psi_{\lambda}\varphi\right)_{\pm}=2\partial_{\overline{z}}\partial_{z}\left(SL(\lambda)\varphi\right)_{\pm}=\frac{1}{2}\Delta\left(SL(\lambda)\varphi\right)_{\pm}=-\frac{\lambda}{2}\left(SL(\lambda)\varphi\right)_{\pm}.$

Since $SL(\lambda)\varphi\in H^{1}({\mathbb{R}}^{2})$, we obtain
$\partial_{\overline{z}}\left(\Psi_{\lambda}\varphi\right)_{+}\oplus\partial_{\overline{z}}\left(\Psi_{\lambda}\varphi\right)_{-}\in
H^{1}({\mathbb{R}}^{2})$ for any $\varphi\in H^{1/2}(\Sigma)$.

Now, we show (iii). Let $\varphi\in H^{1/2}(\Sigma)$. With (A.5) we see that

$-i(\gamma_{D}^{+}\partial_{\overline{z}}(\Psi_{\lambda}\varphi)_{+}+\gamma_{D}^{-}\partial_{\overline{z}}(\Psi_{\lambda}\varphi)_{-})=\lambda\gamma_{D}SL(\lambda)\varphi=\lambda
S(\lambda)\varphi$

holds. Moreover, we obtain with $SL(\lambda)\varphi\in
H^{2}(\mathbb{R}^{2}\setminus\Sigma)$

$i(\nu_{1}+i\nu_{2})\gamma^{\pm}_{D}\left(-2i\partial_{z}SL(\lambda)\varphi\right)_{\pm}=\partial_{\nu}\left(SL(\lambda)\varphi\right)_{\pm}-i\partial_{t}\left(SL(\lambda)\varphi\right)_{\pm},$

where $\partial_{t}$ is the tangential derivative on $\Sigma$. As
$SL(\lambda)\varphi\in H^{1}(\mathbb{R}^{2})$, one has the relation
$\partial_{t}\left(SL(\lambda)\varphi\right)_{+}=\partial_{t}\left(SL(\lambda)\varphi\right)_{-}$
and consequently with (A.1)

$i(\nu_{1}+i\nu_{2})\left(\gamma^{+}_{D}(\Psi_{\lambda}\varphi)_{+}-\gamma^{-}_{D}(\Psi_{\lambda}\varphi)_{-}\right)=\partial_{\nu}\left(SL(\lambda)\varphi\right)_{+}-\partial_{\nu}\left(SL(\lambda)\varphi\right)_{-}=\varphi.$

This finishes the proof of (iii).

It remains to prove item (i). By applying the Wirtinger derivative
$\partial_{z}$ to (A.5) one gets with (2.5) that

$-\Delta\left(\Psi_{\lambda}\varphi\right)_{\pm}=-4\partial_{z}\partial_{\overline{z}}\left(\Psi_{\lambda}\varphi\right)_{\pm}=-2i\lambda\partial_{z}\left(SL(\lambda)\varphi\right)_{\pm}=\lambda\left(\Psi_{\lambda}\varphi\right)_{\pm}$

holds for all $\varphi\in L^{2}(\Sigma)$ in the distributional sense. This,
(A.4), (A.5), and the properties of $SL(\lambda)$ described at the beginning
of this appendix show that $\Psi_{\lambda}\varphi\in\mathcal{H}_{\lambda}$ for
all $\varphi\in H^{1/2}(\Sigma)$ and therefore the mapping
$\Psi_{\lambda}:H^{1/2}(\Sigma)\to\mathcal{H}_{\lambda}$ is well-defined.
Moreover, it follows from (iii) that this mapping is injective. To prove that
$\Psi_{\lambda}:H^{1/2}(\Sigma)\to\mathcal{H}_{\lambda}$ is surjective, let
$f\in\mathcal{H}_{\lambda}$ be fixed. Define the function
$\varphi=i(\nu_{1}+i\nu_{2})\big{(}\gamma_{D}^{+}f_{+}-\gamma_{D}^{-}f_{-}\big{)}\in
H^{1/2}(\Sigma)$ and $g=\Psi_{\lambda}\varphi\in\mathcal{H}_{\lambda}$. By
(iii) we have that

$\begin{split}\gamma_{D}^{+}(f-g)_{+}-\gamma_{D}^{-}(f-g)_{-}&=\gamma_{D}^{+}f_{+}-\gamma_{D}^{-}f_{-}+i(\nu_{1}-i\nu_{2})\varphi=0.\end{split}$

This shows $f-g\in H^{1}({\mathbb{R}}^{2})$. Moreover, due to
$f,g\in\mathcal{H}_{\lambda}$ we have that $\partial_{\overline{z}}(f-g)\in
H^{1}({\mathbb{R}}^{2})$, which implies $f-g\in H^{2}(\mathbb{R}^{2})$.
Combining this with $f,g\in\mathcal{H}_{\lambda}$ we find that
$f-g\in\text{ker}\left(-\Delta-\lambda\right)=\\{0\\}$, i.e.
$f=g=\Psi_{\lambda}\varphi$. Thus
$\Psi_{\lambda}:H^{1/2}(\Sigma)\to\mathcal{H}_{\lambda}$ is also surjective
and all claims in assertion (i) are shown. ∎

###### Proof of Proposition 2.2.

The proof of item (i) is divided into 3 steps. In Step 1 we show that the map
$(-\infty,0)\ni\lambda\mapsto\mu_{n}(S(\lambda))\in(0,\infty)$ is continuous
and strictly monotonically increasing, in Step 2 we show that the same is true
for the map
$(-\infty,0)\ni\lambda\mapsto\lambda\mu_{n}(S(\lambda))\in(-\infty,0)$. Using
these results, we complete the proof of assertion (i) in Step 3.

Step 1: Let $n\in\mathbb{N}$. We show that the map
$(-\infty,0)\ni\lambda\mapsto\mu_{n}(S(\lambda))\in(0,\infty)$ is continuous
and strictly monotonically increasing. To verify that $\mu_{n}(S(\lambda))>0$
for $\lambda\in(-\infty,0)$, it suffices to prove that $S(\lambda)$ is a
positive self-adjoint operator. From the definition of $S(\lambda)$ in (2.4)
it follows that $S(\lambda)$ is self-adjoint. Next, let $\varphi\in
L^{2}(\Sigma)$ with $\varphi\neq 0$ and set $f:=SL(\lambda)\varphi$. Using the
properties of $SL(\lambda)$ described at the beginning of this appendix one
finds that $f\neq 0$ and

$\begin{split}\big{(}S(\lambda)\varphi&,\varphi\big{)}_{L^{2}(\Sigma)}=\big{(}\gamma_{D}f,\partial_{\nu}f_{+}-\partial_{\nu}f_{-}\big{)}_{L^{2}(\Sigma)}\\\
&=\big{(}f_{+},\Delta f_{+}\big{)}_{L^{2}(\Omega_{+})}+\|\nabla
f_{+}\|_{L^{2}(\Omega_{+})}^{2}+\big{(}f_{-},\Delta
f_{-}\big{)}_{L^{2}(\Omega_{-})}+\|\nabla f_{-}\|_{L^{2}(\Omega_{-})}^{2}\\\
&\geq\big{(}f_{+},\Delta f_{+}\big{)}_{L^{2}(\Omega_{+})}+\big{(}f_{-},\Delta
f_{-}\big{)}_{L^{2}(\Omega_{-})}=-\lambda\|f\|_{L^{2}({\mathbb{R}}^{2})}^{2}>0.\end{split}$

Therefore, $\mu_{n}(S(\lambda))>0$ must be true.

Next, we show that
$(-\infty,0)\ni\lambda\mapsto\mu_{n}(S(\lambda))\in(0,\infty)$ is
monotonically increasing and continuous. With (A.2) one sees that
$S(\cdot):\mathbb{C}\setminus[0,\infty)\rightarrow\mathcal{L}(L^{2}(\Sigma))$
is holomorphic and that
$\frac{d}{d\lambda}S(\lambda)=\gamma_{D}(-\Delta-\lambda)^{-2}\gamma_{D}^{\prime}$
holds. In particular, for any $\varphi\in L^{2}(\Sigma)$ the function
$(-\infty,0)\ni\lambda\mapsto(S(\lambda)\varphi,\varphi)_{L^{2}(\Sigma)}$ is
continuously differentiable and

$\begin{split}\frac{d}{d\lambda}\big{(}S(\lambda)\varphi,\varphi\big{)}_{L^{2}(\Sigma)}=\big{(}(-\Delta-\lambda)^{-1}\gamma_{D}^{\prime}\varphi,(-\Delta-\lambda)^{-1}\gamma_{D}^{\prime}\varphi\big{)}_{L^{2}(\Sigma)}&\\\
=\|SL(\lambda)\varphi\|_{L^{2}({\mathbb{R}}^{2})}^{2}&\geq 0\end{split}$

is true. Thus, the min-max principle implies that the map
$(-\infty,0)\ni\lambda\mapsto\mu_{n}(S(\lambda))$ is monotonically increasing
for every $n\in{\mathbb{N}}$. Furthermore, due to the holomorphy of
$S(\cdot):\mathbb{C}\setminus[0,\infty)\rightarrow\mathcal{L}(L^{2}(\Sigma))$
and the estimate

$|\mu_{n}(S(\eta))-\mu_{n}(S(\lambda))|\leq\|S(\eta)-S(\lambda)\|,\quad\eta,\lambda<0,$

(see [29, Satz 3.17]), we find that
$(-\infty,0)\ni\lambda\mapsto\mu_{n}(S(\lambda))$ is continuous for
$n\in{\mathbb{N}}$.

It remains to show that the latter map is strictly monotonically increasing.
Define for $\alpha\in{\mathbb{R}}\setminus\\{0\\}$ the operator-valued
function
$\mathcal{B}_{1}:{\mathbb{C}}\setminus[0,\infty)\to\mathcal{L}(L^{2}(\Sigma))$
by $\mathcal{B}_{1}(\lambda)=I-\alpha S(\lambda)$. By the properties of
$S(\lambda)$ it is easy to see that $\mathcal{B}_{1}$ is holomorphic and
$\mathcal{B}_{1}(\lambda)$ is a Fredholm operator with index $0$ for any fixed
$\lambda$, since $S(\lambda)$ is compact in $L^{2}(\Sigma)$. Moreover, by [18,
Theorem 1.2] there exists a constant $K>0$ such that

(A.6)
$\|S(\lambda)\|\leq\frac{K}{\sqrt{2+|\lambda|}}\ln\sqrt{2+\frac{1}{|\lambda|}},\quad\lambda\in{\mathbb{C}}\setminus[0,\infty).$

Hence, there exists $\lambda_{0}<0$ such that $\|S(\lambda)\|<|\alpha|^{-1}$
is valid for all $\lambda<\lambda_{0}$ . This implies that
$\mathcal{B}_{1}(\lambda)$ is boundedly invertible for every
$\lambda<\lambda_{0}$. Therefore, by [20, Chapter XI., Corollary 8.4] the set

$\mathcal{M}_{\alpha,1}=\left\\{\lambda\in{\mathbb{C}}\setminus[0,\infty)\>\big{|}\>\mathcal{B}_{1}(\lambda)=I-\alpha
S(\lambda)\text{ is not invertible}\right\\}$

is at most countable and does not have an accumulation point in
${\mathbb{C}}\setminus[0,\infty)$. Now assume that $\lambda_{1}<\lambda_{2}<0$
satisfy $\mu_{n}(S(\lambda_{1}))=\mu_{n}(S(\lambda_{2}))=:\alpha^{-1}$ for
some $n\in{\mathbb{N}}$. Then it follows from the monotonicity of
$\lambda\mapsto\mu_{n}(S(\lambda))$ that
$[\lambda_{1},\lambda_{2}]\subseteq\mathcal{M}_{\alpha,1}$, which is a
contradiction to the fact that $\mathcal{M}_{\alpha,1}$ is at most countable.
Therefore, the mapping $(-\infty,0)\ni\lambda\mapsto\mu_{n}(S(\lambda))$ is
continuous and strictly monotonically increasing for $n\in{\mathbb{N}}$.

Step 2: To show the continuity and strict monotonicity of the map
$(-\infty,0)\ni\lambda\mapsto\lambda\mu_{n}(S(\lambda))$ for all
$n\in{\mathbb{N}}$, we note first that the continuity follows from the
continuity of the map $\lambda\mapsto\mu_{n}(S(\lambda))$ shown in Step 1. In
order to prove the monotonicity, we use again
$\frac{d}{d\lambda}S(\lambda)=\gamma_{D}(-\Delta-\lambda)^{-2}\gamma_{D}^{\prime}$
and compute for a fixed $\varphi\in L^{2}(\Sigma)$ and $\lambda\in(-\infty,0)$
with the help of (2.5) and (A.2)

$\begin{split}\frac{d}{d\lambda}\big{(}\lambda
S(\lambda)\varphi,\varphi)_{L^{2}(\Sigma)}&=\big{(}S(\lambda)\varphi+\lambda\gamma_{D}(-\Delta-\lambda)^{-2}\gamma_{D}^{\prime}\varphi,\varphi\big{)}_{L^{2}(\Sigma)}\\\
&=\big{(}-4\gamma_{D}(-\Delta-\lambda)^{-1}\partial_{\overline{z}}\partial_{z}(-\Delta-\lambda)^{-1}\gamma_{D}^{\prime}\varphi,\varphi\big{)}_{L^{2}(\Sigma)}\\\
&=\|\Psi_{\lambda}\varphi\|_{L^{2}({\mathbb{R}}^{2})}^{2}\geq 0.\end{split}$

Thus, the min-max principle yields the monotonicity of the mapping
$(-\infty,0)\ni\lambda\mapsto\lambda\mu_{n}(S(\lambda))$. To see the strict
monotonicity, we use a similar strategy as in Step 1 and define for
$\alpha\in\mathbb{R}\setminus\\{0\\}$ the holomorphic mapping
$\mathcal{B}_{2}:{\mathbb{C}}\setminus[0,\infty)\to\mathcal{L}(L^{2}(\Sigma))$
by $\mathcal{B}_{2}(\lambda)=I-\alpha\lambda S(\lambda)$. Again,
$\mathcal{B}_{2}(\lambda)$ is a Fredholm operator with index zero for any
fixed $\lambda$ and it follows from (A.6) that there exists $\lambda_{3}<0$
such that $\|\lambda S(\lambda)\|<|\alpha|^{-1}$ holds for all
$\lambda\in(\lambda_{3},0)$. In particular, $\mathcal{B}_{2}(\lambda)$ is
boundedly invertible for all $\lambda\in(\lambda_{3},0)$. It follows from [20,
Chapter XI., Corollary 8.4] that the set

$\mathcal{M}_{\alpha,2}=\left\\{\lambda\in{\mathbb{C}}\setminus[0,\infty)\>\big{|}\>\mathcal{B}_{2}(\lambda)=I-\alpha\lambda
S(\lambda)\text{ is not invertible}\right\\}$

is at most countable and does not have an accumulation point in
${\mathbb{C}}\setminus[0,\infty)$. Now the same argument as in Step 1 shows
that $(-\infty,0)\ni\lambda\mapsto\lambda\mu_{n}(S(\lambda))$ is strictly
monotonously increasing.

Step 3: To study the limiting behaviour of $\lambda\mu_{n}(S(\lambda))$ for
$\lambda\rightarrow 0$, note that (A.6) implies $\|\lambda
S(\lambda)\|\rightarrow 0$ for $\lambda\rightarrow 0^{-}$ and hence,

(A.7) $\lim_{\lambda\to 0^{-}}\lambda\mu_{n}(S(\lambda))=0,\quad
n\in{\mathbb{N}}.$

Next, we consider the limit of $\lambda\mu_{n}(S(\lambda))$ for
$\lambda\to-\infty$. For this purpose, results on Schrödinger operators with
$\delta$-interactions will be used. Define for $\alpha<0$ the sesquilinear
form

$\mathfrak{h}_{\delta,\alpha}[f,g]=\big{(}\nabla f,\nabla
g\big{)}_{L^{2}({\mathbb{R}}^{2})}+\alpha\big{(}\gamma_{D}f,\gamma_{D}g\big{)}_{L^{2}(\Sigma)},\quad
f,g\in\text{dom}\,\mathfrak{h}_{\delta,\alpha}=H^{1}({\mathbb{R}}^{2}).$

By [9, 14] the form $\mathfrak{h}_{\delta,\alpha}$ is semi-bounded and closed,
and one can show for the self-adjoint operator $H_{\delta,\alpha}$, which is
associated with $\mathfrak{h}_{\delta,\alpha}$ by the first representation
theorem, that $\sigma_{\text{ess}}(H_{\delta,\alpha})=[0,\infty)$, that its
discrete spectrum $\sigma_{\text{disc}}(H_{\delta,\alpha})$ is finite, and for
$\lambda\in(-\infty,0)$ one has that

(A.8)
$\lambda\in\sigma_{\textup{p}}(H_{\delta,\alpha})\quad\Longleftrightarrow\quad-1\in\sigma_{\textup{p}}(\alpha
S(\lambda));$

see for instance [9, Lemma 2.3 and Theorem 4.2] and [8, Theorems 3.5 and
3.14]. Recall that the eigenvalues $\mu_{n}(S(\lambda))$ are ordered non-
increasingly with multiplicities taken into account. If we order the discrete
eigenvalues of $H_{\delta,\alpha}$ in an increasing way then the strict
monotonicity of $\lambda\mapsto\mu_{n}(S(\lambda))$ implies that the $k$-th
discrete eigenvalue $E_{k}(\alpha)$ (if it exists) satisfies the equation
$-1=\alpha\mu_{k}(S(E_{k}(\alpha)))$.

Let $n\in{\mathbb{N}}$. Then by [14, Theorem 1] the operator
$H_{\delta,\alpha}$ has at least $n$ negative discrete eigenvalues (counted
with multiplicities) if $-\alpha>0$ is sufficiently large, and the $n$-th
discrete eigenvalue $E_{n}(\alpha)$ of $H_{\delta,\alpha}$ admits the
asymptotic expansion

(A.9)
$E_{n}(\alpha)=-\frac{\alpha^{2}}{4}+\mu_{n}(H)+\mathcal{O}(\alpha^{-1}\ln|\alpha|),\qquad\alpha\rightarrow-\infty.$

Here $H$ is a fixed semibounded differential operator on $\Sigma$ that is
independent of $\alpha$ and has purely discrete spectrum
$\mu_{1}(H)\leq\mu_{2}(H)\leq\dots$. Thus for $\alpha\to-\infty$ we obtain
with (A.8) that

(A.10) $\frac{\alpha}{4}+\frac{|\mu_{n}(H)|+1}{\alpha}\leq
E_{n}(\alpha)\mu_{n}(S(E_{n}(\alpha)))=-\frac{E_{n}(\alpha)}{\alpha}\leq\frac{\alpha}{4}-\frac{|\mu_{n}(H)|+1}{\alpha}.$

This shows

(A.11) $\lim_{\lambda\to-\infty}\lambda\mu_{n}(S(\lambda))=-\infty$

and finishes the proof of item (i).

To show item (ii), we note first that by (A.7), (A.11), and the strict
monotonicity and continuity of the mapping
$\lambda\mapsto\lambda\mu_{n}(S(\lambda))$ it is clear that for any $a<0$
there is a unique solution $\lambda_{n}(a)$ of $\lambda\mu_{n}(S(\lambda))=a$.
Let $\mu_{n}(H)$ be as in (A.9), define the numbers
$k_{\pm}=\pm(|\mu_{n}(H)|+1)$ and let

(A.12)
$\alpha_{\pm}=-2|a|\left(\sqrt{1+\frac{k_{\pm}}{a^{2}}}+1\right)=-4|a|-\frac{k_{\pm}}{|a|}+f_{\pm}(a)$

with some functions $f_{\pm}(a)=\mathcal{O}(a^{-3})$ for large $|a|>0$, where
the latter representation holds due to a Taylor series expansion. Then one has

(A.13) $a=\frac{\alpha_{\pm}}{4}-\frac{k_{\pm}}{\alpha_{\pm}}$

and it follows with (A.10) that

$E_{n}(\alpha_{+})\mu_{n}(S(E_{n}(\alpha_{+})))\leq\frac{\alpha_{+}}{4}-\frac{|\mu_{n}(H)|+1}{\alpha_{+}}=a=\lambda_{n}(a)\mu_{n}(S(\lambda_{n}(a))$

and

$\lambda_{n}(a)\mu_{n}(S(\lambda_{n}(a))=a=\frac{\alpha_{-}}{4}+\frac{|\mu_{n}(H)|+1}{\alpha_{-}}\leq
E_{n}(\alpha_{-})\mu_{n}(S(E_{n}(\alpha_{-}))).$

Since $\lambda\mapsto\lambda\mu_{n}(S(\lambda))$ is monotone we find

(A.14) $E_{n}(\alpha_{+})\leq\lambda_{n}(a)\leq E_{n}(\alpha_{-}).$

From (A.12) we obtain

$\frac{1}{4}\alpha_{\pm}^{2}=4a^{2}+2k_{\pm}+g_{\pm}(a)$

with functions $g_{\pm}(a)=\mathcal{O}(a^{-2})$ for large $|a|>0$ and hence
(A.9) implies

(A.15)
$\big{|}E_{n}(\alpha_{\pm})+4a^{2}+2k_{\pm}+g_{\pm}(a)-\mu_{n}(H)\big{|}\leq
C_{1}\big{|}\alpha_{\pm}^{-1}\ln|\alpha_{\pm}|\big{|}$

for some constant $C_{1}>0$. Note that there exist constants $C_{2},C_{3}>0$
such that $C_{2}|a|\leq\alpha_{\pm}\leq C_{3}|a|$ holds for large $|a|>0$.
With this we conclude from (A.15) that

(A.16)
$\begin{split}\big{|}E_{n}(\alpha_{\pm})+4a^{2}+2k_{\pm}-\mu_{n}(H)\big{|}\leq
C_{4}\big{|}a^{-1}\ln|a|\big{|}\end{split}$

holds for some constant $C_{4}>0$ and for large $|a|>0$. Taking (A.14) and
(A.16) into account, one concludes finally that

$|\lambda_{n}(a)+4a^{2}|\leq
3|\mu_{n}(H)|+2+\mathcal{O}(a^{-1}\ln|a|)=\mathcal{O}(1)\quad\text{for
}a\to-\infty.$

∎

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