# Special potentials for relativistic Laplacians I
Fractional Rollnik-class

Giacomo Ascione, Atsuhide Ishida and József Lőrinczi  Giacomo Ascione, Scuola
Superiore Meridionale
80138 Napoli, Italy<EMAIL_ADDRESS>Atsuhide Ishida, Katsushika
Division, Institute of Arts and Sciences
Tokyo University of Science, Tokyo, 125-8585, Japan
and
Alfréd Rényi Institute of Mathematics
Reáltanoda utca 13-15, 1053 Budapest, Hungary<EMAIL_ADDRESS>József
Lőrinczi, Alfréd Rényi Institute of Mathematics
Reáltanoda utca 13-15, 1053 Budapest, Hungary<EMAIL_ADDRESS>
###### Abstract.

We propose a counterpart of the classical Rollnik-class of potentials for
fractional and massive relativistic Laplacians, and describe this space in
terms of appropriate Riesz potentials. These definitions rely on precise
resolvent estimates. We show that Coulomb-type potentials are elements of
fractional Rollnik-class up to but not including the critical singularity of
the Hardy potential. For the operators with fractional exponent $\alpha=1$
there exists no fractional Rollnik potential, however, in low dimensions we
make sense of these classes as limiting cases by using $\Gamma$-convergence.
In a second part of the paper we derive detailed results on the self-
adjointness and spectral properties of relativistic Schrödinger operators
obtained under perturbations by fractional Rollnik potentials. We also define
an extended fractional Rollnik-class which is the maximal space for the
Hilbert-Schmidt property of the related Birman-Schwinger operators.

_Key-words_ : fractional and relativistic Laplacians, Riesz potential,
resolvent estimates, $\Gamma$-convergence, Coulomb potentials, subcriticality

2010 MS Classification: Primary 46N50, 81Q10; Secondary 42B20

###### Contents

  1. 1 Introduction
  2. 2 Preliminaries
    1. 2.1 Massive and massless relativistic Laplacians
    2. 2.2 Lizorkin space and Riesz potential
    3. 2.3 Resolvent estimates
  3. 3 Fractional Rollnik class
    1. 3.1 Definitions and basic properties
      1. 3.1.1 Fractional Rollnik norm
      2. 3.1.2 Restrictions on the fractional exponent
    2. 3.2 Comparison properties
    3. 3.3 Specific cases
      1. 3.3.1 Potentials with compact support
      2. 3.3.2 Asymmetric fractional Rollnik potentials
      3. 3.3.3 Rotationally symmetric potentials
      4. 3.3.4 Coulomb-type potentials
    4. 3.4 Extended fractional Rollnik-class
  4. 4 Perturbations by fractional Rollnik class potentials
    1. 4.1 Self-adjointness
    2. 4.2 Spectral properties
      1. 4.2.1 Essential spectrum
      2. 4.2.2 Discrete spectrum
      3. 4.2.3 Non-existence of negative and zero eigenvalues
  5. 5 The case $\alpha=1$
  6. 6 Appendix: Proofs of resolvent estimates
    1. 6.1 Massless case
      1. 6.1.1 Proof of estimate (2.14) in the $m=0$ case
      2. 6.1.2 Proof of estimate (2.18)
    2. 6.2 Massive case
      1. 6.2.1 The function $\kappa_{m}$
      2. 6.2.2 Proof of estimate (2.14) in the $m>0$ case
      3. 6.2.3 Proof of estimate (2.15)
      4. 6.2.4 Proof of estimate (2.16)
      5. 6.2.5 Proof of estimate (2.19)

##  1\. Introduction

The aim of this paper is to define and explore a relativistic counterpart of
the classical Rollnik-class of potentials, which plays a fundamental role in
mathematical quantum theory. We call this fractional Rollnik-class, and also
set ourselves the task to describe the properties of relativistic Schrödinger
operators obtained as perturbations of the (free) relativistic operators by
this class of potentials. Since both the massive and massless cases involve
singular integral operators, in our approach the Riesz potential will play a
role to a far-reaching extent. In particular, we will make use of it to a
detailed description of Coulomb-type potentials belonging to fractional
Rollnik-class and investigate their boundaries on the verge of the critical
Hardy potential.

In the paper [26] Rollnik identified the class of potentials

${\mathcal{R}}=\left\\{V:\mathbb{R}^{3}\to\mathbb{R}:\,\|V\|^{2}_{{\mathcal{R}}}:=\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}\frac{|V(x)||V(y)|}{|x-y|^{2}}dxdy<\infty\right\\},$
(1.1)

which later turned out to play a special role in the classical theory of
Schrödinger operators on the Hilbert space $L^{2}(\mathbb{R}^{3})$. He noticed
that for an eigenvalue $\lambda<0$ generated by a negative potential $V$, the
eigenvalue equation $H\varphi:=(-\frac{1}{2}\Delta+V)\varphi=\lambda\varphi$
can equivalently be written as
$K_{\lambda}\psi:=\sqrt{-V}(-\frac{1}{2}\Delta-\lambda)^{-1}\sqrt{-V}\psi=\psi$,
with $\psi=\sqrt{-V}\varphi$, so the original Schrödinger operator $H$ has a
eigenvalue $\lambda$ exactly when the related operator $K_{\lambda}$ has
eigenvalue 1. Thereby he appears to have anticipated by a five years (if we
count by the publication dates) what is now known as the Birman-Schwinger
principle. In fact, Rollnik studied mostly the case $\lambda=0$ and required
his operator $K_{0}$ (in his notation $W(0)$) to be compact, that he satisfied
by turning its kernel square-integrable, which results in (1.1).

The integral condition (1.1) has appeared in subsequent work by Schwinger
[30], Scadron, Weinberg and Wright [29], Coester [9], and in the series of
papers by Grossmann and Wu [18]. A later development is Simon’s mathematical
work in the paper [31] and the monograph [32], in which he made a
comprehensive analysis of Rollnik potentials in the non-relativistic case.

Our goal here is to construct and analyse a counterpart of Rollnik-class for
the massive and massless relativistic Laplacians

$L_{m,\alpha}=(-\Delta+m^{2/\alpha})^{\alpha/2}-m,\quad 0<\alpha<2,\;m\geq 0,$

on $L^{2}(\mathbb{R}^{d})$, $d\geq 1$, and study relativistic Schrödinger
operators obtained as perturbations of $L_{m,\alpha}$ by these potentials. For
the massless case $m=0$ the operator $L_{m,\alpha}$ reduces to the fractional
Laplacian with exponent $\frac{\alpha}{2}$, and for $\alpha=1$ and $m>0$ it is
the physical relativistic operator widely used in mathematical physics. For
the purposes of understanding fractional Rollnik-class it will be useful to
keep the full range $\alpha\in(0,2)$ in order to appreciate how our results
(qualitatively) depend on the details of the operators of different fractional
exponent. Our guiding principle is to keep to the original idea that the
Birman-Schwinger operators associated with the non-local Schrödinger operators
we consider should have the Hilbert-Schmidt property, however, for the rest we
will have to trod a new path due to qualitative differences between the
Laplacian and the fractional operators requiring very different techniques.

In the following we summarise our approach and the main results of this paper.

(1) Our first task is to find an appropriate concept of a fractional Rollnik-
class holding to this principle and investigate its structure. The
construction will depend on precise estimates of the resolvent kernels of the
operators $L_{m,\alpha}$. Making use of estimate (2.14) and assuming

$\alpha<d<2\alpha,$ (1.2)

we introduce the fractional Rollnik-class

$\mathcal{R}_{d,\alpha}=\left\\{V:\mathbb{R}^{d}\rightarrow\mathbb{R}:\,\|V\|^{2}_{\mathcal{R}_{d,\alpha}}=\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\frac{|V(x)||V(y)|}{|x-y|^{2(d-\alpha)}}dxdy<\infty\right\\}$

in the massless case and

$\mathcal{R}^{m}_{\alpha}=\mathcal{R}_{d,\alpha}\cap\mathcal{R}$

in the massive case (see Definition 3.1). This definition is consistent with
the classical case as it reproduces $\mathcal{R}$ for $d=3$ and $\alpha=2$
(see also Corollary 3.14 below). Observing that the Fourier transform of the
Riesz potential $\mathcal{I}_{\beta}$ of functions on Lizorkin space can be
computed (Theorem 2.2), we show that the fractional Rollnik norm of $V$ can be
expressed in the form

$\left\|V\right\|_{\mathcal{R}_{d,\alpha}}=c(d,\alpha)\left\|\mathcal{I}_{\alpha-\frac{d}{2}}|V|\right\|_{2}$
(1.3)

where $c(d,\alpha)$ is a constant, we show that the left-hand side above is
indeed a norm (Theorem 3.4), turning $\mathcal{R}_{d,\alpha}$ into a Banach
space (Theorem 3.6). Furthermore, as it will be seen, fractional Rollnik-class
contains a variety of $L^{p}$ subspaces (Propositions 3.11-3.13).

To get a closer idea of fractional Rollnik potentials, we consider several
specific cases (Section 3.4). In particular, we consider potentials having
Coulomb-type singularities and by using (1.3) and formula (3.15) allowing to
compute Riesz potentials of radial functions, we show that
$V(x)=|x|^{-\beta_{1}}\mathbf{1}_{\\{|x|<1\\}}+|x|^{-\beta_{2}}\mathbf{1}_{\\{|x|\geq
1\\}}$ belongs to $\mathcal{R}_{d,\alpha}$ if and only if $\beta_{1}<\alpha$
and $\beta_{2}\in(\alpha,d)$ (see Propositions 3.19-3.22 and Corollary 3.24).
Since for the fractional Laplacian $(-\Delta)^{\alpha/2}$ the critical
potential is the Hardy potential $V(x)=|x|^{-\alpha}$, we see that the
Coulomb-type fractional Rollnik-class potentials are subcritical with possible
singularities arbitrarily approaching but not attaining the critical exponent.
Also, we show (Theorem 3.18) that the Hardy exponent marks the boundary to
Coulomb-potentials in fractional Rollnik class. We also show that
$V(x)=|x|^{-\alpha}(-\log|x|)^{-\gamma}\mathbf{1}_{\\{|x|<1/2\\}}$, with
$\alpha<d$, belongs to $\mathcal{R}_{d,\alpha}$ exactly when
$\gamma>\frac{1}{2}$ (see Proposition 3.20). Apart from Coulomb singularities,
we also discuss potentials with compact support and show that their fractional
Rollnik norm increases under symmetric rearrangement and it can be bounded by
the fractional perimeter of the supporting set (Section 3.4.1), and asymmetric
potentials whose non-radial symmetry deficit we can bound by the $L^{1}$-norm
of the difference between the potential and a family of rotationally symmetric
potentials (Section 3.4.2).

Using another crucial resolvent estimate (see (2.16)-(2.17) and (2.18)-(2.20)
below) we also introduce extended fractional Rollnik-classes
${\overline{\mathcal{R}}}_{d,\alpha}$ and
${\overline{\mathcal{R}}}^{m}_{d,\alpha}$ (Definition 3.25), containing
$\mathcal{R}_{d,\alpha}$ as a proper subset (Proposition 3.26 and Remark 3.29
below). We will discuss the role of the extended fractional Rollnik-class
further below.

The restriction (1.2) has the implication that according to our definition
above, in case $d\geq 2\alpha$ no non-trivial continuous potential exists in
$\mathcal{R}_{d,\alpha}$ (Theorem 3.9). Also, there exists no fractional
Rollnik-class potential for the physical case $\alpha=1$ in any dimension.
However, in low dimensions we can make sense of them by further developing
Definition 3.1. For $d=1$ it is covered by the extended fractional Rollnik-
class, and for $d=2$ we obtain it as a limiting case. Proposition 3.10 shows,
however, that the required limit is not straightforward. Since, as it turns
out, with a sequence $(\alpha_{n})_{n\in\mathbb{N}}$ accumulating at
$\alpha=1$ the corresponding Schrödinger operators with
$\mathcal{R}_{d,\alpha_{n}}$-potentials have a limit in $\Gamma$-convergence
sense (see Theorem 5.3), we arrive to define it by
$\mathcal{R}_{d,1}^{p}:=L^{p}(\mathbb{R}^{d})\cap\liminf_{\alpha\downarrow
1}\mathcal{R}_{d,\alpha}$ with $p>1$.

(2) Having defined fractional Rollnik-class, our interest next is to study the
properties of relativistic Schrödinger operators of the form
$H_{m,\alpha}=L_{m,\alpha}+V$ with $V$ in $\mathcal{R}^{m}_{d,\alpha}$ or
${\overline{\mathcal{R}}}^{m}_{d,\alpha}$, for $m\geq 0$. First we consider
the self-adjointness of $H_{m,\alpha}$. To obtain this it suffices to have the
operator norm of
$K^{m}_{\lambda}=|V|^{1/2}(\lambda+L_{m,\alpha})^{-1}|V|^{1/2}$ vanish as
$\lambda\to\infty$ (Lemma 4.1). To ensure this, we show that $K^{m}_{\lambda}$
is a Hilbert-Schmidt operator at least for a large enough $\lambda$. In
particular, we show that for potentials in $\mathcal{R}_{d,\alpha}$ the
operator $H_{m,\alpha}$ is self-adjoint (Theorem 4.3). Also, we show that if a
potential is in the extended fractional Rollnik-class
${\overline{\mathcal{R}}}_{d,\alpha}$, then the related $K^{m}_{\lambda}$ is a
Hilbert-Schmidt operator, and conversely, if $K^{m}_{\lambda}$ is a Hilbert-
Schmidt operator for a potential, then the potential is in
${\overline{\mathcal{R}}}_{d,\alpha}$ (Theorem 4.4), thus the extended
fractional Rollnik-class is the maximal space for the Hilbert-Schmidt property
to hold for the operators we consider.

Next we study the spectrum of $H_{m,\alpha}$. Since fractional Rollnik
potentials are small in an aggregate sense (without being subject to any
specific fall-off rate), a perturbation by such potentials does not change the
essential spectrum of $L_{m,\alpha}$ (see Theorem 4.7). However, contrary to
the classical case, we can not make use of Fourier methods and develop an
other argument to show this. In case $H_{m,\alpha}$ has a non-empty discrete
spectrum, we can estimate the number of negative eigenvalues (Theorem 4.8);
since this number is controlled above by the fractional Rollnik norm, the
discrete spectrum is finite in both the massless and massive cases. We obtain
similar results for ${\overline{\mathcal{R}}}_{d,\alpha}$-class potentials
(Theorems 4.9-4.10), where now the control is made by the quantity
$[V]_{\overline{\mathcal{R}}_{d,\alpha}}$, which defines the extended class.
By using the extended Birman-Schwinger principle, we can further derive
conditions under which no bound states exist and zero (the spectral edge) is
not an embedded eigenvalue. This occurs when then fractional Rollnik norm is
too small (Theorem 4.13).

(3) The main estimates on the resolvent
$R^{m}_{\lambda}=(\lambda+L_{m,\alpha})^{-1}$, $\lambda>0$, (uniform in
$\lambda$, pointwise upper and lower for large $\lambda$ and for $\lambda=0$
both for the massive and massless cases) on which our constructions are built
are stated in the expressions (2.14)-(2.15), (2.16)-(2.17) and (2.19)-(2.20).
Detailed proofs will be presented in the Appendix. We note that these results
have an independent interest, in areas such as analytic and probabilistic
potential theory.

##  2\. Preliminaries

###  2.1. Massive and massless relativistic Laplacians

Let $\alpha\in(0,2)$, $m\geq 0$, $d\in\mathbb{N}$, and consider the family of
functions $\Psi_{m,\alpha}(u)=(u+m^{2/\alpha})^{\alpha/2}-m$ for every $u\geq
0$. Using these functions, we define the operators

$\displaystyle L_{m,\alpha}\\!\\!\\!$ $\displaystyle=$
$\displaystyle\\!\\!\\!\Psi_{m,\alpha}(-\Delta)=(-\Delta+m^{2/\alpha})^{\alpha/2}-m\qquad\mbox{if
$m>0$}$ $\displaystyle L_{0,\alpha}\\!\\!\\!$ $\displaystyle=$
$\displaystyle\\!\\!\\!\Psi_{0,\alpha}(-\Delta)=(-\Delta)^{\alpha/2}\hskip
85.35826pt\mbox{if $m=0$}$

and generically combine the notation into just $L_{m,\alpha}$, $m\geq 0$,
whenever suitable. These operators can be defined in several different ways.
We define them through the Fourier multipliers

$\widehat{(L_{m,\alpha}f)}(y)=\Psi_{m,\alpha}(|y|^{2})\widehat{f}(y),\quad
y\in\mathbb{R}^{d},\;f\in\operatorname{Dom}(L_{m,\alpha}),$

with domain

$\operatorname{Dom}(L_{m,\alpha})=\Big{\\{}f\in
L^{2}(\mathbb{R}^{d}):\Psi_{m,\alpha}(|\cdot|^{2})\widehat{f}\in
L^{2}(\mathbb{R}^{d})\Big{\\}},\quad m\geq 0.$

In the $m=0$ case we have more specifically
$\operatorname{Dom}(L_{0,\alpha})=H^{\alpha}(\mathbb{R}^{d})$, i.e., the
fractional Sobolev space of order $0<\alpha<2$.

For $f\in C^{\infty}_{\rm c}(\mathbb{R}^{d})$ we have the expression

$L_{m,\alpha}f(x)=-\lim_{\varepsilon\downarrow
0}\int_{|y-x|>\varepsilon}\left(f(y)-f(x)\right)\nu_{m,\alpha}(dy)$

with the Lévy jump measures as follows. For $m>0$ we have

$\nu_{m,\alpha}(dx)=j_{m,\alpha}(|x|)dx=\frac{2^{\frac{\alpha-d}{2}}m^{\frac{d+\alpha}{2\alpha}}\alpha}{\pi^{d/2}\Gamma(1-\frac{\alpha}{2})}\frac{K_{(d+\alpha)/2}(m^{1/\alpha}|x|)}{|x|^{(d+\alpha)/2}}\,dx,\quad
x\in\mathbb{R}^{d}\setminus\\{0\\},$ (2.1)

and for $m=0$ we have

$\nu_{0,\alpha}(dx)=j_{0,\alpha}(|x|)dx=\frac{2^{\alpha}\Gamma(\frac{d+\alpha}{2})}{\pi^{d/2}|\Gamma(-\frac{\alpha}{2})|}\frac{dx}{|x|^{d+\alpha}},\quad
x\in\mathbb{R}^{d}\setminus\\{0\\}.$ (2.2)

Here

$K_{\rho}(z)=\frac{1}{2}\left(\frac{z}{2}\right)^{\rho}\int_{0}^{\infty}t^{-\rho-1}e^{-t-\frac{z^{2}}{4t}}dt,\quad
z>0,\;\rho>-\frac{1}{2},$

is the standard modified Bessel function of the second (sometimes also called
third) kind. Below we will make use of the asymptotic formulae

$K_{\rho}(z)\sim\sqrt{\frac{\pi}{2|z|}}\,e^{-|z|}\quad\mbox{as
$|z|\to\infty$}\qquad\mbox{and}\qquad
K_{\rho}(z)\sim\frac{\Gamma(\rho)}{2}\left(\frac{2}{z}\right)^{\rho}\quad\mbox{as
$|z|\downarrow 0$}.$ (2.3)

A consequence of the second expression is that $j_{m,\alpha}\to j_{0,\alpha}$
as $m\downarrow 0$, thus the operator $L_{0,\alpha}$ is the massless limit of
$L_{m,\alpha}$.

The operators $L_{m,\alpha}$, $m\geq 0$, are positive with spectrum
$\operatorname{Spec}L_{m,\alpha}=\operatorname{Spec}_{\rm
ess}L_{m,\alpha}=[0,\infty)$, and self-adjoint with core $C^{\infty}_{\rm
c}(\mathbb{R}^{d})$, for every $0<\alpha<2$ and $m\geq 0$. In case $m>0$, the
operator $L_{m,\alpha}$ is called _relativistic fractional Laplacian with
exponent $\frac{\alpha}{2}$ and rest mass $m$_, while for $m=0$ it is the
_fractional Laplacian with exponent $\frac{\alpha}{2}$_, but for simplicity we
will refer to them as the massive and massless relativistic Laplacians,
respectively.

Below we will use the following generic notations. $L^{p}(\mathbb{R}^{d})$
norms will be denoted by $\|\cdot\|_{p}$ and $L^{p}(E)$,
$E\subsetneq\mathbb{R}^{d}$, norms by $\|\cdot\|_{L^{p}(E)}$. A ball centred
in $x\in\mathbb{R}^{d}$ with radius $r>0$ will be denoted by $B_{r}(x)$, and
simply $B_{r}$ when $x=0$. We write $\omega_{d}$ for the Lebesgue measure of
$B_{1}$ and $\sigma_{d}$ for the surface measure of $\partial B_{1}$, so that
$\sigma_{d}=d\omega_{d}$. When convenient, we will also use the notations
$a\wedge b=\min\\{a,b\\}$ and $a\vee b=\max\\{a,b\\}$.

###  2.2. Lizorkin space and Riesz potential

Consider the negative Laplacian $-\Delta$ and for $\beta\in(0,d)$ let

$\gamma_{d}(\beta)=\frac{2^{\beta}\pi^{d/2}\Gamma\big{(}\frac{\beta}{2}\big{)}}{\Gamma\big{(}\frac{d-\beta}{2}\big{)}}.$

The Riesz potential of order $0<\beta<d$ of a measurable function $f$ is
defined as

$\mathcal{I}_{\beta}f(x)=(-\Delta)^{\beta/2}f(x)=\frac{1}{\gamma_{d}(\beta)}\int_{\mathbb{R}^{d}}\frac{f(y)}{|x-y|^{d-\beta}}dy=(k_{\beta}\ast
f)(x).$

The requirement $\beta>0$ is necessary in order to guarantee that the Riesz
kernel $k_{\beta}$ is at least $L^{1}_{\rm loc}(\mathbb{R}^{d})$. It is clear
that the integral defining $\mathcal{I}_{\beta}f$ is absolutely convergent
almost everywhere if and only if $f$ is taken from

${\mathfrak{X}}_{\beta}^{d}:=\Big{\\{}f:\mathbb{R}^{d}\to\mathbb{C}:\mbox{$f$
is measurable and}\,\int_{|x|<1}|f(x)|dx+\int_{|x|\geq
1}\frac{|f(x)|}{|x|^{d-\beta}}dx<\infty\Big{\\}}.$ (2.4)

If $f$ belongs to the Schwartz space $\mathcal{S}(\mathbb{R}^{d})$ of rapidly
decreasing functions, its Riesz potential $\mathcal{I}_{\beta}f$ is well-
defined since $\mathcal{S}(\mathbb{R}^{d})\subset{\mathfrak{X}}_{\beta}^{d}$.
However, it is not guaranteed that $\mathcal{I}_{\beta}f\in
L^{1}(\mathbb{R}^{d})+L^{2}(\mathbb{R}^{d})$. To ensure this, we need to
consider $f$ in a suitable subspace of $\mathcal{S}(\mathbb{R}^{d})$. Let
$\mathbb{N}_{0}^{d}$, where $\mathbb{N}_{0}=\\{0\\}\cup\mathbb{N}$, be the set
of all multi-indices and for all $\mathbf{j}\in\mathbb{N}^{d}$ denote by
$D^{\mathbf{j}}h$ the partial derivative of $h$ taken $j_{i}$-times with
respect to the $x_{i}$ variable, for all $i=1,\dots,d$. Following the lines of
[28, Ch. 2.2], we introduce the spaces

$\mathcal{S}_{0}(\mathbb{R}^{d}):=\left\\{h\in\mathcal{S}(\mathbb{R}^{d}):\
D^{\mathbf{j}}h(0)=0,\ \forall\mathbf{j}\in\mathbb{N}_{0}^{d}\right\\}$

and

$\mathcal{FS}_{0}(\mathbb{R}^{d}):=\left\\{h\in\mathcal{S}(\mathbb{R}^{d}):\
h=\mathcal{F}[g],\ g\in\mathcal{S}_{0}\right\\},$

where $\mathcal{F}$ denotes Fourier transform. The set
$\mathcal{FS}_{0}(\mathbb{R}^{d})$ is called Lizorkin space. Since
$\mathcal{F}$ maps $\mathcal{S}(\mathbb{R}^{d})$ into itself, it is clear that
$\mathcal{FS}_{0}(\mathbb{R}^{d})\subset\mathcal{S}(\mathbb{R}^{d})$. In
particular, it is not difficult to check that
$\mathcal{FS}_{0}(\mathbb{R}^{d})$ is formed by all (complex-valued) functions
in $\mathcal{S}(\mathbb{R}^{d})$ that are orthogonal to all polynomials. We
will make use of the following denseness result, which was shown in [28, Th.
2.7].

######  Theorem 2.1.

The Lizorkin space $\mathcal{FS}_{0}(\mathbb{R}^{d})$ is dense in
$L^{p}(\mathbb{R}^{d})$ for all $p\in(1,\infty)$.

For our purposes below the following connection between Lizorkin space and the
Riesz potential will be important, for details we refer to [28, Lem. 2.15, Th.
2.16].

######  Theorem 2.2.

Let $\beta\in(0,d)$. If $h\in\mathcal{FS}_{0}(\mathbb{R}^{d})$, then
$\mathcal{I}_{\beta}h\in\mathcal{F}\mathcal{S}_{0}(\mathbb{R}^{d})$ and

$\mathcal{F}[\mathcal{I}_{\beta}h](z)=|z|^{-\beta}\mathcal{F}[h](z).$

In the special case of radially symmetric functions
$f:\mathbb{R}^{d}\to\mathbb{C}$ written in the form $f(x)=g(|x|)$ with some
$g:[0,\infty)\to\mathbb{C}$, it has been shown in [34, Th. 1] that the Riesz
potential $\mathcal{I}_{\beta}f$ is radially symmetric and for $d\geq 2$ the
expression

$\mathcal{I}_{\beta}f(x)=\frac{1}{\gamma_{d}(\beta)}\int_{0}^{\infty}r^{\beta-1}F_{\beta}\left(\frac{|x|}{r}\right)g(r)dr$
(2.5)

holds, where

$F_{\beta}(r)=\sigma_{d-1}\int_{-1}^{1}\frac{(1-t^{2})^{\frac{d-3}{2}}}{(1+r^{2}-2rt)^{\frac{d-\beta}{2}}}\,dt,\quad
r>0.$ (2.6)

The asymptotic behaviour of the function $F_{\beta}$ is known. First, it is
continuous for $r\in(0,1)\cup(1,\infty)$, and for $\beta>1$ it is continuous
also in $r=1$ with value

$F_{\beta}(1)=\sigma_{d-1}2^{\beta-1}B\left(\frac{d-1}{2},\frac{\beta-1}{2}\right),$

where $B$ is the Beta-function. Also, if $\beta=1$, then

$\lim_{r\to 1}\frac{F_{\beta}(r)}{-\sigma_{d-1}\log(|r-1|)}=1$

while if $\beta\in(0,1)$, then

$\lim_{r\to
1}\frac{2F_{\beta}(r)}{\sigma_{d-1}B\left(\frac{d-1}{2},\frac{1-\beta}{2}\right)|r-1|^{\beta-1}}=1.$

For its behaviour at zero we have

$F_{\beta}(r)=\sigma_{d}+O(r),\ \mbox{ as }r\to 0$ (2.7)

while at infinity

$F_{\beta}(r)=\sigma_{d}r^{\beta-d}+O(r^{\beta-d-1}),\ \mbox{ as }r\to\infty.$
(2.8)

We note that in [28, Eq. (2.10)] the same formula has been obtained in terms
of hypergeometric functions. However, the above representation is preferable
for our purposes since usually the analytic extension of the hypergeometric
function is considered on the cut complex plane
$\mathbb{C}\setminus(1,\infty)$. This is due to the fact that $z=1$ and
$z=\infty$ are branch points of this function.

The $d=1$ case is not covered by (2.6), however, we can derive a similar type
of representation. Let $f:\mathbb{R}\to\mathbb{C}$ be even. First observe that

$\displaystyle\mathcal{I}_{\beta}f(-x)=\frac{1}{\gamma_{1}(\beta)}\int_{\mathbb{R}}\frac{f(y)}{|x+y|^{1-\beta}}\,dy=\frac{1}{\gamma_{1}(\beta)}\int_{\mathbb{R}}\frac{f(y)}{|x-y|^{1-\beta}}\,dy=\mathcal{I}_{\beta}(x),$

thus $\mathcal{I}_{\beta}f$ is also even. Next, notice that for $x>0$

$\displaystyle\mathcal{I}_{\beta}f(x)$
$\displaystyle=\frac{1}{\gamma_{1}(\beta)}\int_{\mathbb{R}}\frac{f(y)}{|x-y|^{1-\beta}}\,dy=\frac{1}{\gamma_{1}(\beta)}\left(\int_{0}^{\infty}\frac{f(y)}{|x-y|^{d-\beta}}\,dy+\int_{-\infty}^{0}\frac{f(y)}{|x-y|^{1-\beta}}\,dy\right)$
$\displaystyle=\frac{1}{\gamma_{1}(\beta)}\left(\int_{0}^{\infty}\frac{f(y)}{|x-y|^{1-\beta}}\,dy+\int_{0}^{\infty}\frac{f(y)}{|x+y|^{1-\beta}}\,dy\right)$
$\displaystyle=\frac{1}{\gamma_{1}(\beta)}\int_{0}^{\infty}y^{\beta-1}f(y)\left(\frac{1}{\left|\frac{x}{y}-1\right|^{1-\beta}}+\frac{1}{\left|\frac{x}{y}+1\right|^{1-\beta}}\right)\,dy.$

This means that (2.5) holds for $d=1$ with

$F_{\beta}(r)=\frac{1}{\left|r-1\right|^{1-\beta}}+\frac{1}{\left|r+1\right|^{1-\beta}}.$

For its the asymptotic behaviour, we still have (2.7)-(2.8), while

$\lim_{r\to 1}\frac{F(r)}{|r-1|^{\beta-1}}=1.$

Below we will also make use of the Hardy-Littlewood-Sobolev inequality, which
gives conditions under which $\mathcal{I}_{\beta}$ is a bounded operator
between specific $L^{p}$ spaces. Let $\beta\in(0,d)$,
$p_{\beta}:=\frac{dp}{dp-\beta}$ be the Sobolev exponent, and $p>1$. Then

$\left\|\mathcal{I}_{\beta}f\right\|_{p_{\beta}}\leq
C_{p,\beta}\left\|f\right\|_{p}$ (2.9)

holds. A direct consequence is the inequality

$\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\frac{|f(x)||g(x)|}{|x-y|^{\delta}}\,dx\,dy\leq
C_{p,\delta,d}\left\|f\right\|_{p}\left\|g\right\|_{q},$ (2.10)

where $\delta\in(0,d)$, $f\in L^{p}(\mathbb{R}^{d})$, $g\in
L^{q}(\mathbb{R}^{d})$ and $\frac{1}{p}+\frac{1}{q}=2-\frac{\delta}{d}$.

###  2.3. Resolvent estimates

For every operator $L_{m,\alpha}$, $m\geq 0$, introduced in Section 2.1 above,
the one-parameter operator semigroup $\\{e^{-tL_{m,\alpha}}:t\geq 0\\}$ can be
defined by functional calculus, each of whose elements is an integral operator
for $t>0$. The integral kernel $p(t,x,y)=e^{-tL_{m,\alpha}}(x,y)$, $t>0$,
which we call heat kernel as in the literature, satisfies the property
$p(t,x,y)=p(t,|x-y|,0)$, and we use the simpler conventional notation
$p_{t}(x)$. It is well-known that for $m=0$ the semigroup is transient
whenever $\alpha<d$, and for $m>0$ whenever $d\geq 3$, while in the
complementary cases the semigroups are recurrent. In our considerations below
we will occasionally have to make a distinction between transient and
recurrent cases. Results relying on the $0$-resolvent operator will apply only
in the transient case, since otherwise $0$-resolvents are not well-defined. On
the other hand, results relying on $\lambda$-resolvents with $\lambda>0$ can
be stated also in the recurrent case.

The following heat kernel estimates are known, for a proof see [3], [7, Th.
1.2], [8, Th. 4.1].

######  Proposition 2.3.

  1. (1)

Let $p_{t}(x)$ be the integral kernel of the operator $e^{-tL_{0,\alpha}}$.
Then

$p_{t}(x)\,\,\asymp\,\,t|x|^{-d-\alpha}\wedge t^{-d/\alpha}$ (2.11)

holds for all $x\in\mathbb{R}^{d}$ and $t>0$.

  2. (2)

Let $p^{m}_{t}(x)$ be the integral kernel of the operator
$e^{-tL_{m,\alpha}}$, $m>0$. Then

$\displaystyle p^{m}_{t}(x)\asymp\begin{cases}\ t^{-d/\alpha}\wedge
t|x|^{-(d+\alpha)/2}\kappa_{m}(|x|)&{\rm if}\quad 0<t\leq\frac{1}{m}\\\ \\\ \
m^{d/\alpha-d/2}t^{-d/2}e^{-c\left(m^{1/\alpha}|x|\wedge
m^{2/\alpha-1}|x|^{2}/t\right)}&{\rm if}\quad t>\frac{1}{m}\end{cases}$ (2.12)

hold with $c>0$ and

$\kappa_{m}(r)=m^{(d+\alpha)/(2\alpha)}K_{(d+\alpha)/2}(m^{1/\alpha}r).$
(2.13)

For every $\lambda>0$, let $R^{m}_{\lambda}=(\lambda+L_{m,\alpha})^{-1}$,
$m\geq 0$, be the resolvent of the relativistic operator, with integral kernel
$R^{m}_{\lambda}(x-y)$. We have then

$R^{m}_{\lambda}(x-y)=\int_{0}^{\infty}e^{-\lambda t}p^{m}_{t}(x-y)dt,\quad
m\geq 0,$

and similarly to the heat kernel write just $R^{m}_{\lambda}(x)$. For the
massless case we drop the superscripts for all these objects.

Our results below will largely depend on detailed resolvent estimates for the
fractional and relativistic Laplace operators. We will need uniform (in
$\lambda$), and pointwise upper and lower estimates on the resolvent kernel
$R^{m}_{\lambda}$ separately for large $\lambda$ and for $\lambda=0$, both for
the fractional ($m=0$) and the relativistic ($m>0)$ Laplacians. These results
stand also on their own, and can be applied in independent directions such as
potential theory.

We have the following uniform resolvent kernel estimates in the transient
massless ($d>\alpha$) and transient massive ($d>2$) cases. For every
$\lambda\geq 0$

$\displaystyle
R^{m}_{\lambda}(x)\leq\begin{cases}C_{d,\alpha}|x|^{\alpha-d}&\mbox{if \,
$m=0$}\\\ &\\\
C_{d,\alpha}\big{(}|x|^{\alpha-d}+m^{\frac{2}{\alpha}-1}{|x|^{2-d}}\big{)}&\mbox{if
\, $m>0$}\end{cases}$ (2.14)

holds with appropriate constant $C_{d,\alpha}>0$. In fact, for $\lambda=0$ the
expression

$R_{0}(x)=C_{d,\alpha}|x|^{\alpha-d},$

holds [21, Prop. 4.293], while for the massive case we have

$R^{m}_{0}(x)\asymp C_{d,\alpha,m}\left(|x|^{\alpha-d}\mathbf{1}_{\\{|x|\leq
1\\}}+|x|^{2-d}\mathbf{1}_{\\{|x|>1\\}}\right).$ (2.15)

Throughout the paper we will also make use of the non-uniform estimate

$\displaystyle R^{m}_{\lambda}(x)\leq
C_{d,\alpha,\lambda,m}H^{0}_{d,\alpha}(|x|),$ (2.16)

which holds for every $\alpha\in(0,2)$, $d\geq 1$, $m\geq 0$ and $\lambda>0$
with

$H_{d,\alpha}^{0}(r):=\begin{cases}r^{-(d-\alpha)}\mathbf{1}_{\\{r\leq
1\\}}+2^{\frac{d+\alpha}{2}-1}\Gamma\left(\frac{d+\alpha}{2}\right)r^{-(d+\alpha)}\mathbf{1}_{\\{r>1\\}}&\mbox{if
\, $\alpha<d$}\\\ &\\\ \log\left(1+\frac{1}{r}\right)\mathbf{1}_{\\{r\leq
1\\}}+2^{\frac{d+\alpha}{2}-1}\Gamma\left(\frac{d+\alpha}{2}\right)r^{-2}\mathbf{1}_{\\{r>1\\}}&\mbox{if
\, $\alpha=d=1$}\\\ &\\\ \mathbf{1}_{\\{r\leq
1\\}}+2^{\frac{d+\alpha}{2}-1}\Gamma\left(\frac{d+\alpha}{2}\right)r^{-(1+\alpha)}\mathbf{1}_{\\{r>1\\}}&\mbox{if
\, $\alpha>d=1$}.\end{cases}$ (2.17)

Again, for $m=0$ we actually have

$\displaystyle R_{\lambda}(x)\asymp
C_{d,\alpha,\lambda}H^{0}_{d,\alpha}(|x|).$ (2.18)

In the massive case we do not have this equivalence. However, for $m>0$ and
$\lambda>m$, we prove that

$\displaystyle R^{m}_{\lambda}(x)\asymp
C_{d,\alpha,\lambda,m}H^{m}_{d,\alpha}(|x|),$ (2.19)

where

$H^{m}_{d,\alpha}(r):=\begin{cases}r^{-(d-\alpha)}\mathbf{1}_{\\{r\leq
1\\}}+r^{-\frac{d+\alpha}{2}}\kappa_{m}(r)\mathbf{1}_{\\{r>0\\}}&\mbox{if \,
$\alpha<d$}\\\ &\\\ \log\left(1+\frac{1}{r}\right)\mathbf{1}_{\\{r\leq
1\\}}+r^{-\frac{d+\alpha}{2}}\kappa_{m}(r)\mathbf{1}_{\\{r>0\\}}&\mbox{if \,
$\alpha=d=1$}\\\ &\\\ \mathbf{1}_{\\{r\leq
1\\}}+r^{-\frac{d+\alpha}{2}}\kappa_{m}(r)\mathbf{1}_{\\{r>0\\}}&\mbox{if \,
$\alpha>d=1$}.\end{cases}$ (2.20)

We note that

$\displaystyle R^{m}_{\lambda}(x)\geq
C_{d,\alpha,\lambda,m}H^{m}_{d,\alpha}(|x|)$ (2.21)

holds in fact for all $m,\lambda>0$. The proof of these resolvent estimates is
given in the Appendix.

##  3\. Fractional Rollnik class

###  3.1. Definitions and basic properties

####  3.1.1. Fractional Rollnik norm

Making use of (2.14) we introduce the following class of potentials for the
operators $L_{m,\alpha}$, $m\geq 0$.

######  Definition 3.1.

Let $d\in\mathbb{N}$, $0<\alpha<2$ and assume that

$0<\alpha<d<2\alpha.$ (3.1)

  1. (1)

Let $m=0$. We call

$\mathcal{R}_{d,\alpha}=\left\\{V:\mathbb{R}^{d}\rightarrow\mathbb{R}:\,\|V\|^{2}_{\mathcal{R}_{d,\alpha}}=\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\frac{|V(x)||V(y)|}{|x-y|^{2(d-\alpha)}}dxdy<\infty\right\\}$

_fractional Rollnik-class of order $\alpha$ for the massless Laplacian
$L_{0,\alpha}$_.

  2. (2)

Let $d=3$ and $m>0$. We define the _fractional Rollnik-class of order $\alpha$
for the massive Laplacian $L_{m,\alpha}$_ by
$\mathcal{R}^{m}_{\alpha}=\mathcal{R}_{d,\alpha}\cap\mathcal{R}$, where
$\mathcal{R}$ is the classical Rollnik class.

It is seen that the study of the massive case reduces to the study of the
massless case. First therefore we focus on the class $\mathcal{R}_{d,\alpha}$
and as we progress we will consider extensions and which will also accommodate
cases of recurrent semigroups.

We begin by showing that the expression $\|\cdot\|_{\mathcal{R}_{d,\alpha}}$
used in the definition of $\mathcal{R}_{d,\alpha}$ is indeed a norm and that
$\mathcal{R}_{d,\alpha}$ is a Banach space with respect to this norm. Before
that, we note the following two properties which will be used below.

######  Lemma 3.2.

We have that $L^{1}_{\rm loc}(\mathbb{R}^{d})\subset\mathcal{R}_{d,\alpha}$
and

$\left\|V\right\|_{\mathcal{R}_{d,\alpha}}^{2}\geq\frac{1}{(2r(K))^{2d-2\alpha}}\left\|V\right\|_{L^{1}(K)}^{2},$

where $K\subset\mathbb{R}^{d}$ is any compact set and $r(K)=\inf\\{r\geq 0:\
K\subset B_{r}\\}$.

###### Proof.

Without loss of generality, it will be sufficient to show this for every ball
$B_{r}\subset\mathbb{R}^{d}$. Then we have

$\left\|V\right\|_{\mathcal{R}_{d,\alpha}}^{2}\geq\int_{B_{r}}\int_{B_{r}}|V(x)||V(y)||x-y|^{2\alpha-2d}\,dx\,dy\geq(2r)^{2\alpha-2d}\left\|V\right\|_{L^{1}(B_{r})}^{2}.$

∎

A straightforward but useful consequence of the Hardy-Littlewood-Sobolev
inequality is the following inclusion property.

######  Proposition 3.3.

$L^{d/\alpha}(\mathbb{R}^{d})$ is continuously embedded in
$\mathcal{R}_{d,\alpha}$, i.e.,

$\left\|V\right\|_{\mathcal{R}_{d,\alpha}}\leq
C_{d,\alpha}\left\|V\right\|_{d/\alpha}.$

###### Proof.

It is immediate by (2.10) with $\sigma=2(d-\alpha)$ and
$p=q=\frac{d}{\alpha}$. ∎

Apart from $\left\|\cdot\right\|_{\mathcal{R}_{d,\alpha}}$, we can also define
the quantity

${\left|\kern-1.07639pt\left|\kern-1.07639pt\left|V\right|\kern-1.07639pt\right|\kern-1.07639pt\right|}_{\mathcal{R}_{d,\alpha}}^{2}=\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}V(x)\overline{V(y)}|x-y|^{2\alpha-2d}\,dx\,dy$

on $\mathcal{R}_{d,\alpha}$, where we allow $V$ to be possibly complex-valued.
Next we prove that both $\left\|\cdot\right\|_{\mathcal{R}_{d,\alpha}}$ and
${\left|\kern-1.07639pt\left|\kern-1.07639pt\left|\cdot\right|\kern-1.07639pt\right|\kern-1.07639pt\right|}_{\mathcal{R}_{d,\alpha}}$
are norms on $\mathcal{R}_{d,\alpha}$; we take inspiration from a strategy
proposed in [22, Sect. 4] for $d=1$ only, with substantial modifications.
Recall the factor $\gamma_{d}$ from (2.2).

######  Theorem 3.4.

The functionals $\left\|\cdot\right\|_{\mathcal{R}_{d,\alpha}}$ and
${\left|\kern-1.07639pt\left|\kern-1.07639pt\left|\cdot\right|\kern-1.07639pt\right|\kern-1.07639pt\right|}_{\mathcal{R}_{d,\alpha}}$
are norms on $\mathcal{R}_{d,\alpha}$. Furthermore, for all
$V\in\mathcal{R}_{d,\alpha}$ we have

${\left|\kern-1.07639pt\left|\kern-1.07639pt\left|V\right|\kern-1.07639pt\right|\kern-1.07639pt\right|}_{\mathcal{R}_{d,\alpha}}^{2}=\gamma_{d}(2\alpha-d)\left\|\mathcal{I}_{\alpha-\frac{d}{2}}V\right\|_{2}^{2}\quad\mbox{
and
}\quad\left\|V\right\|_{\mathcal{R}_{d,\alpha}}^{2}=\gamma_{d}(2\alpha-d)\left\|\mathcal{I}_{\alpha-\frac{d}{2}}|V|\right\|_{2}^{2}.$
(3.2)

###### Proof.

The proof will proceed through several steps.

_Step 1:_ First we show (3.2), starting with assuming that
$V\in\mathcal{F}\mathcal{S}_{0}(\mathbb{R}^{d})$. Clearly, by the definition
of Lizorkin space, we know that $V\in\mathcal{S}(\mathbb{R}^{d})$ and then
$V\in L^{2}(\mathbb{R}^{d})\cap L^{d/\alpha}(\mathbb{R}^{d})$, which in turn
implies $V\in\mathcal{R}_{d,\alpha}$. Furthermore, also
$\mathcal{I}_{2\alpha-d}V\in\mathcal{F}\mathcal{S}_{0}(\mathbb{R}^{d})\subset
L^{2}(\mathbb{R}^{d})$ by Theorem 2.2. Hence, by an application of Fubini’s
theorem and Plancherel’s identity we obtain

$\displaystyle{\left|\kern-1.07639pt\left|\kern-1.07639pt\left|V\right|\kern-1.07639pt\right|\kern-1.07639pt\right|}_{\mathcal{R}_{d,\alpha}}^{2}$
$\displaystyle=\gamma_{d}(2\alpha-d)\left(V,\mathcal{I}_{2\alpha-d}V\right)_{L^{2}(\mathbb{R}^{d})}$
$\displaystyle=\gamma_{d}(2\alpha-d)\left(\mathcal{F}[V],|\cdot|^{d-2\alpha}\mathcal{F}[V]\right)_{L^{2}(\mathbb{R}^{d})}$
$\displaystyle=\gamma_{d}(2\alpha-d)\left(|\cdot|^{\frac{d}{2}-\alpha}\mathcal{F}[V],|\cdot|^{\frac{d}{2}-\alpha}\mathcal{F}[V]\right)_{L^{2}(\mathbb{R}^{d})}=\gamma_{d}(2\alpha-d)\left\|\mathcal{I}_{\alpha-\frac{d}{2}}V\right\|_{2}^{2}.$

_Step 2:_ Next assume that $V\in L^{d/\alpha}(\mathbb{R}^{d})$. Since
$\mathcal{F}\mathcal{S}_{0}(\mathbb{R}^{d})$ is dense in
$L^{d/\alpha}(\mathbb{R}^{d})$ by Theorem 2.1, we may choose a sequence
$(V_{n})_{n\in\mathbb{N}}\subset\mathcal{F}\mathcal{S}_{0}(\mathbb{R}^{d})$
such that $V_{n}\to V\in L^{d/\alpha}(\mathbb{R}^{d})$. Since $V_{n}-V\in
L^{d/\alpha}(\mathbb{R}^{d})$, by Proposition 3.3 we have

$\displaystyle\left|\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}V_{n}(x)V_{n}(y)|x-y|^{2\alpha-2d}dx\,dy-\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}V(x)V(y)|x-y|^{2\alpha-2d}dx\,dy\right|$
$\displaystyle\qquad\leq\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}|V_{n}(x)-V(x)||V_{n}(y)||x-y|^{2\alpha-2d}dx\,dy$
$\displaystyle\qquad\qquad+\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}|V(x)||V_{n}(y)-V(y)||x-y|^{2\alpha-2d}dx\,dy$
$\displaystyle\qquad\leq
C_{d,\alpha}\left\|V_{n}-V\right\|_{d/\alpha}\left(\left\|V_{n}\right\|_{{d/\alpha}}+\left\|V\right\|_{{d/\alpha}}\right)\to
0\quad\mbox{as $n\to\infty$}.$

On the other hand, by the triangle inequality and (2.9),

$\left|\left\|\mathcal{I}_{\alpha-\frac{d}{2}}V_{n}\right\|_{2}-\left\|\mathcal{I}_{\alpha-\frac{d}{2}}V\right\|_{2}\right|\leq\left\|\mathcal{I}_{\alpha-\frac{d}{2}}(V_{n}-V)\right\|_{2}\leq
C_{d,\alpha}\left\|V_{n}-V\right\|_{d/\alpha}\to 0.$

This proves the first equality in (3.2) for $V\in
L^{d/\alpha}(\mathbb{R}^{d})$. The second equality follows by applying the
first on $|V|\in L^{d/\alpha}(\mathbb{R}^{d})$.

_Step 3:_ To complete this part of the proof, consider now any
$V\in\mathcal{R}_{d,\alpha}$. Let $M\in\mathbb{N}$ and $\eta_{M}$ be a
$C^{\infty}_{\rm c}(\mathbb{R}^{d})$ function such that $\eta_{M}(x)=1$ for
every $x\in B_{M}$, $\eta_{M}(x)\in[0,1]$ and ${\rm supp}(\eta_{M})\subset
B_{M+1}$. Then $V_{M}:=\min\\{|V|,M\\}\eta_{M}\in
L^{{d/\alpha}}(\mathbb{R}^{d})$ and (3.2) holds for this function. Next, since
$V_{M}\leq V_{M+1}$ for all $M\in\mathbb{N}$, by the monotone convergence
theorem we have

$\lim_{M\to\infty}\left\|V_{M}\right\|_{\mathcal{R}_{d,\alpha}}^{2}=\left\|V\right\|_{\mathcal{R}_{d,\alpha}}^{2}\quad\mbox{
and
}\quad\lim_{M\to\infty}\left\|\mathcal{I}_{\alpha-\frac{d}{2}}V_{M}\right\|_{2}^{2}=\left\|\mathcal{I}_{\alpha-\frac{d}{2}}|V|\right\|_{2}^{2}.$

This proves the second equality in (3.2). To prove the first, notice that we
have in fact shown that if $V\in\mathcal{R}_{d,\alpha}$, then
$\left\|\mathcal{I}_{\alpha-\frac{d}{2}}|V|\right\|_{2}<\infty$. Hence by
setting $V_{M}=\max\\{\min\\{V,M\\},-M\\}\eta_{M}$, we can use the dominated
convergence theorem to obtain

$\lim_{M\to\infty}{\left|\kern-1.07639pt\left|\kern-1.07639pt\left|V_{M}\right|\kern-1.07639pt\right|\kern-1.07639pt\right|}_{\mathcal{R}_{d,\alpha}}^{2}={\left|\kern-1.07639pt\left|\kern-1.07639pt\left|V\right|\kern-1.07639pt\right|\kern-1.07639pt\right|}_{\mathcal{R}_{d,\alpha}}^{2}\quad\mbox{
and
}\quad\lim_{M\to\infty}\left\|\mathcal{I}_{\alpha-\frac{d}{2}}V_{M}\right\|_{2}^{2}=\left\|\mathcal{I}_{\alpha-\frac{d}{2}}V\right\|_{2}^{2}.$

_Step 4:_ Finally we show the norm property. Identity (3.2) guarantees that
for all $V\in\mathcal{R}_{d,\alpha}$, both
$\left\|V\right\|_{\mathcal{R}_{d,\alpha}}$ and
${\left|\kern-1.07639pt\left|\kern-1.07639pt\left|V\right|\kern-1.07639pt\right|\kern-1.07639pt\right|}_{\mathcal{R}_{d,\alpha}}$
are non-negative. Furthermore, note that directly by the definition of the
Riesz potential, the facts $V_{1},V_{2}\geq 0$ and $V_{1}\leq V_{2}$, imply
$\left\|\mathcal{I}_{\alpha-\frac{d}{2}}V_{1}\right\|_{2}\leq\left\|\mathcal{I}_{\alpha-\frac{d}{2}}V_{2}\right\|_{2}$.
Hence we have

$\displaystyle\left\|V_{1}+V_{2}\right\|_{\mathcal{R}_{d,\alpha}}$
$\displaystyle=$
$\displaystyle\sqrt{\gamma(2\alpha-d)}\left\|\mathcal{I}_{\alpha-\frac{d}{2}}|V_{1}+V_{2}|\right\|_{2}\leq\sqrt{\gamma(2\alpha-d)}\left\|\mathcal{I}_{\alpha-\frac{d}{2}}|V_{1}|+\mathcal{I}_{\alpha-\frac{d}{2}}|V_{2}|\right\|_{2}$
$\displaystyle\leq$
$\displaystyle\sqrt{\gamma(2\alpha-d)}\left\|\mathcal{I}_{\alpha-\frac{d}{2}}|V_{1}|\right\|_{2}+\sqrt{\gamma(2\alpha-d)}\left\|\mathcal{I}_{\alpha-\frac{d}{2}}|V_{2}|\right\|_{2}$
$\displaystyle=$
$\displaystyle\left\|V_{1}\right\|_{\mathcal{R}_{d,\alpha}}+\left\|V_{2}\right\|_{\mathcal{R}_{d,\alpha}}.$

The fact that $\left\|V\right\|_{\mathcal{R}_{d,\alpha}}=0$ implies $V=0$ a.e.
follows by injectivity of the Riesz potential $\mathcal{I}_{2\alpha-d}$. The
same argument can be repeated for
${\left|\kern-1.07639pt\left|\kern-1.07639pt\left|\cdot\right|\kern-1.07639pt\right|\kern-1.07639pt\right|}_{\mathcal{R}_{d,\alpha}}$,
which proves that both are norms. ∎

######  Remark 3.5.

Notice that
${\left|\kern-1.07639pt\left|\kern-1.07639pt\left|\cdot\right|\kern-1.07639pt\right|\kern-1.07639pt\right|}_{\mathcal{R}_{d,\alpha}}$
is induced by the scalar product

$(V_{1},V_{2})_{\mathcal{R}_{d,\alpha}}=\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}V_{1}(x)\overline{V_{2}(y)}|x-y|^{2\alpha-2d}\,dx\,dy.$

We can prove the following completeness result with respect to the norm
$\left\|\cdot\right\|_{\mathcal{R}_{d,\alpha}}$.

######  Theorem 3.6.

The fractional Rollnik class $\mathcal{R}_{d,\alpha}$ equipped with the norm
$\|\cdot\|_{\mathcal{R}_{d,\alpha}}$ is a Banach space.

###### Proof.

Let $(V_{n})_{n\in\mathbb{N}}$ be a Cauchy sequence in
$\mathcal{R}_{d,\alpha}$. By Lemma 3.2 it is a Cauchy sequence also in
$L^{1}(B_{m})$ for all $m\in\mathbb{N}$. By a simple diagonal argument we know
that there exists a subsequence $(V_{n_{k}})_{k\in\mathbb{N}}$ and a function
$V$ such that $V_{n_{k}}(x)\to V(x)$ for a.e. $x\in\mathbb{R}^{d}$. By Fatou’s
lemma we have that

$\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}|V(x)-V_{n_{k}}(x)||V(y)-V_{n_{k}}(y)||x-y|^{2\alpha-2d}\,dx\,dy\leq\liminf_{j\to\infty}\left\|V_{n_{j}}-V_{n_{k}}\right\|_{\mathcal{R}_{d,\alpha}}^{2}.$

Fix $\varepsilon>0$. Since $(V_{n})_{n\in\mathbb{N}}$ is a Cauchy sequence in
$\mathcal{R}_{d,\alpha}$, there exists a label $k_{0}\in\mathbb{N}$ such that
$\left\|V_{n_{j}}-V_{n_{k}}\right\|_{\mathcal{R}_{d,\alpha}}<\varepsilon$
holds for all $j,k\geq k_{0}$. Taking the limit inferior, this means that for
all $k\geq k_{0}$

$\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}|V(x)-V_{n_{k}}(x)||V(y)-V_{n_{k}}(y)||x-y|^{2\alpha-2d}\,dx\,dy\leq\liminf_{j\to\infty}\left\|V_{n_{j}}-V_{n_{k}}\right\|_{\mathcal{R}_{d,\alpha}}^{2}<\varepsilon^{2}.$

This proves that $V-V_{n_{k}}\in\mathcal{R}_{d,\alpha}$, for all $k\geq
k_{0}$, and thus also $V\in\mathcal{R}_{d,\alpha}$. Furthermore, for all
$k\geq k_{0}$ we have
$\left\|V-V_{n_{k}}\right\|_{\mathcal{R}_{d,\alpha}}<\varepsilon$. Since
$\varepsilon>0$ is arbitrary, we get $V_{n_{k}}\to V$ in
$\mathcal{R}_{d,\alpha}$. This is sufficient to conclude that $V_{n}\to V$ in
$\mathcal{R}_{d,\alpha}$ as $n\to\infty$. ∎

It is natural to ask whether $\mathcal{R}_{d,\alpha}$ is a Hilbert space when
equipped with the norm
${\left|\kern-1.07639pt\left|\kern-1.07639pt\left|\cdot\right|\kern-1.07639pt\right|\kern-1.07639pt\right|}{\mathcal{R}_{d,\alpha}}$.
Adapting an argument in [22] originally only for $d=1$, we see that this is
not the case.

######  Theorem 3.7.

The fractional Rollnik class $\mathcal{R}_{d,\alpha}$ is not complete with
respect to the norm
${\left|\kern-1.07639pt\left|\kern-1.07639pt\left|\cdot\right|\kern-1.07639pt\right|\kern-1.07639pt\right|}_{\mathcal{R}_{d,\alpha}}$.

###### Proof.

Assume, to the contrary, that $\mathcal{R}_{d,\alpha}$ is complete with
respect to
${\left|\kern-1.07639pt\left|\kern-1.07639pt\left|\cdot\right|\kern-1.07639pt\right|\kern-1.07639pt\right|}_{\mathcal{R}_{d,\alpha}}$.
Consider any function $\phi\in L^{2}(\mathbb{R}^{d})$ such that
$\phi\not\in\operatorname{Dom}((-\Delta)^{(2\alpha-d)/4})$. Since
$\mathcal{F}\mathcal{S}_{0}(\mathbb{R}^{d})$ is dense in
$L^{2}(\mathbb{R}^{d})$, there exists a sequence
$(\phi_{n})_{n\in\mathbb{N}}\subset\mathcal{F}\mathcal{S}_{0}(\mathbb{R}^{d})$
such that $\phi_{n}\to\phi$ in $L^{2}(\mathbb{R}^{d})$. Define the sequence of
functions

$f_{n}=\mathcal{F}^{-1}[|\cdot|^{\frac{d}{2}-\alpha}\mathcal{F}[\phi_{n}]]\,\in\,\mathcal{F}\mathcal{S}_{0}(\mathbb{R}^{d})\,\subset\,\mathcal{R}_{d,\alpha}.$

Since $\phi_{n}\to\phi$ in $L^{2}(\mathbb{R}^{d})$, also
$\mathcal{I}_{\alpha-\frac{d}{2}}f_{n}\to\phi$ in $L^{2}(\mathbb{R}^{d})$,
hence $(\mathcal{I}_{\alpha-\frac{d}{2}}f_{n})_{n\in\mathbb{N}}$ is a Cauchy
sequence in $L^{2}(\mathbb{R}^{d})$. By the first equality in (3.2), this
implies that $(f_{n})_{n\in\mathbb{N}}$ is a Cauchy sequence in
$(\mathcal{R}_{d,\alpha},{\left|\kern-1.07639pt\left|\kern-1.07639pt\left|\cdot\right|\kern-1.07639pt\right|\kern-1.07639pt\right|}_{\mathcal{R}_{d,\alpha}})$.
Since by assumption this space is complete, there exists a function $f$ such
that $f_{n}\to f$ in
$(\mathcal{R}_{d,\alpha},{\left|\kern-1.07639pt\left|\kern-1.07639pt\left|\cdot\right|\kern-1.07639pt\right|\kern-1.07639pt\right|}_{\mathcal{R}_{d,\alpha}})$.
Again by (3.2), we know that $\mathcal{I}_{\alpha-\frac{d}{2}}f\in
L^{2}(\mathbb{R}^{d})$ and
$\mathcal{I}_{\alpha-\frac{d}{2}}f_{n}\to\mathcal{I}_{\alpha-\frac{d}{2}}f$.
However, this is impossible since then we would have
$\phi=\mathcal{I}_{\alpha-\frac{d}{2}}f\in\operatorname{Dom}((-\Delta)^{(2\alpha-d)/4})$.
∎

######  Remark 3.8.

We note that if $V_{1},V_{2}$ are two potentials such that
$|V_{1}(x)|\leq|V_{2}(x)|$ a.e. and $V_{2}\in\mathcal{R}_{d,\alpha}$, then
clearly also $V_{1}\in\mathcal{R}_{d,\alpha}$. Furthermore, by Lemma 3.2, it
is clear that if $V\in\mathcal{R}_{d,\alpha}$, then for all $R>0$

$\left\|V\right\|_{L^{1}(B_{R})}\leq
2^{d-\alpha}\left\|V\right\|_{\mathcal{R}_{d,\alpha}}R^{d-\alpha}.$ (3.3)

This means that fractional Rollnik-class potentials tend to concentrate their
mass at infinity.

####  3.1.2. Restrictions on the fractional exponent

Next we turn to discussing the role of condition (3.1). A straightforward
calculation shows that $|x|^{-2(d-\alpha)}\in L^{1}_{\rm loc}(\mathbb{R}^{d})$
only when $\alpha<2d$. This implies that if $d\geq 2\alpha$ there exist no
continuous potentials $V\in\mathcal{R}_{d,\alpha}$ apart from $V\equiv 0$. In
fact, we can prove the following more general property.

######  Theorem 3.9.

Let $k:[0,\infty)\to[0,\infty)$ be a function such that for all $\delta>0$

$\int_{0}^{\delta}r^{d-1}k(r)=\infty.$ (3.4)

Suppose that $V:\mathbb{R}^{d}\to\mathbb{R}$ is a measurable function such
that

$\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}|V(x)||V(y)|k(|x-y|)dxdy<\infty.$
(3.5)

Then for any choice of $\varepsilon,r>0$ and $x\in\mathbb{R}^{d}$ there is no
ball $B_{r}(x)$ such that $|V(y)|>\varepsilon$ for a.e. $y\in B_{r}(x)$. In
particular, if $V$ is continuous, then $V\equiv 0$.

###### Proof.

Let $V:\mathbb{R}^{d}\to\mathbb{R}$ be satisfying (3.5) and assume, to the
contrary, that there exist $\varepsilon^{\ast},r^{\ast}>0$ and
$x^{\ast}\in\mathbb{R}^{d}$ such that $|V(y)|>\varepsilon^{\ast}$ for all
$y\in B_{r^{\ast}}(x^{\ast})$. Then

$\displaystyle\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}|V(x)||V(y)|k(|x-y|)\,dx\,dy$
$\displaystyle\geq\int_{B_{r^{\ast}}(x^{\ast})}\int_{B_{r^{\ast}}(x^{\ast})}|V(x)||V(y)|k(|x-y|)\,dx\,dy$
$\displaystyle\geq\varepsilon^{2}\int_{B_{r^{\ast}}(x^{\ast})}\int_{B_{r^{\ast}}(x^{\ast})}k(|x-y|)\,dx\,dy,$

which implies
$\int_{B_{r^{\ast}}(x^{\ast})}\int_{B_{r^{\ast}}(x^{\ast})}k(|x-y|)\,dx\,dy<\infty$.
However, by (3.4) we also have

$\int_{B_{r^{\ast}}(x^{\ast})}\int_{B_{r^{\ast}}(x^{\ast})}k(|x-y|)\,dx\,dy\geq\int_{B_{r^{\ast}}(x^{\ast})}\int_{B_{r^{\ast}-|x-x^{\ast}|}(x)}k(|x-y|)\,dx\,dy=\infty,$

which is impossible. ∎

In fact, a fractional Rollnik-class cannot be obtained even in the limit
$\alpha\uparrow\frac{d}{2}$.

######  Proposition 3.10.

Let $d=1,2$ and suppose $V:\mathbb{R}^{d}\to\mathbb{R}$ is a measurable
function such that
$\liminf\limits_{\alpha\downarrow\frac{d}{2}}\left\|V\right\|_{\mathcal{R}_{d,\alpha}}<\infty$.
Then for any choice of $\varepsilon,r>0$ and $x\in\mathbb{R}^{d}$ there is no
ball $B_{r}(x)$ such that $|V(y)|>\varepsilon$ for a.e. $y\in B_{r}(x)$. In
particular, if $V$ is continuous, then $V\equiv 0$.

###### Proof.

Let $V:\mathbb{R}^{d}\to\mathbb{R}$ be such that
$\liminf\limits_{\alpha\downarrow\frac{d}{2}}\left\|V\right\|_{\mathcal{R}_{d,\alpha}}<\infty$
and assume, to the contrary, that there exists $\varepsilon^{*},r^{*}$ and
$x^{*}\in\mathbb{R}^{d}$ such that $|V(y)|>\varepsilon^{*}$ for almost every
$y\in B_{r^{*}}(x^{*})$. Let also $\alpha_{k}$ be such that
$\alpha_{k}\downarrow\frac{d}{2}$ and
$\lim_{k\to\infty}\left\|V\right\|_{\mathcal{R}_{d,\alpha_{k}}}=\liminf\limits_{\alpha\downarrow\frac{d}{2}}\left\|V\right\|_{\mathcal{R}_{d,\alpha}}$.
Without loss of generality, we assume that
$\left\|V\right\|_{\mathcal{R}_{d,\alpha_{k}}}<\infty$ for all
$k\in\mathbb{N}$. Observe that for all $k\in\mathbb{N}$

$\infty>\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}|V(x)||V(y)||x-y|^{2\alpha_{k}-2d}dxdy\geq\varepsilon^{2}\int_{B_{r^{\ast}}(x^{\ast})}\int_{B_{r^{\ast}}(x^{\ast})}|x-y|^{2\alpha_{k}-2d}dxdy.$
(3.6)

On the other hand,

$\int_{B_{r^{\ast}}(x^{\ast})}\int_{B_{r^{\ast}}(x^{\ast})}|x-y|^{2\alpha_{k}-2d}dxdy\geq\int_{B_{r^{\ast}}(x^{\ast})}\int_{B_{r^{\ast}-|x-x^{\ast}|}(x)}|x-y|^{2\alpha_{k}-2d}dydx.$

We can evaluate the inner integral for $x\in B_{r^{\ast}}(x^{\ast})$ as

$\int_{B_{r^{\ast}-|x-x^{\ast}|}(x)}|x-y|^{2\alpha_{k}-2d}dy=\sigma_{d}\int_{0}^{r^{\ast}-|x-x^{\ast}|}\rho^{2\alpha_{k}-d-1}d\rho=\frac{\sigma_{d}}{2\alpha_{k}-d}(r^{\ast}-|x-x^{\ast}|)^{2\alpha_{k}-d}$

and thus

$\displaystyle\int_{B_{r^{\ast}}(x^{\ast})}\int_{B_{r^{\ast}-|x-x^{\ast}|}(x)}|x-y|^{2\alpha_{k}-2d}dydx$
$\displaystyle=\frac{\sigma_{d}^{2}}{2\alpha_{k}-d}\int_{0}^{r^{\ast}}(r^{\ast}-\rho)^{2\alpha_{k}-d}\rho^{d-1}d\rho$
$\displaystyle=\frac{\sigma_{d}^{2}}{2\alpha_{k}-d}(r^{\ast})^{2\alpha_{k}}B(2\alpha_{k}-d+1;d),$

where $B(x,y)$ is the Beta function. Taking the limit as $k\to\infty$ we get

$\int_{B_{r^{\ast}}(x^{\ast})}\int_{B_{r^{\ast}-|x-x^{\ast}|}(x)}|x-y|^{2\alpha_{k}-2d}dydx=\infty,$

which is in contradiction with (3.6). ∎

Clearly, by (3.1) we have the possible ranges $D_{\alpha}$ of $\alpha$ as
follows:

$D_{\alpha}=\begin{cases}(\frac{1}{2},1)&\mbox{if \, $d=1$}\\\ (1,2)&\mbox{if
\, $d=2$}\\\ (\frac{3}{2},2)&\mbox{if \, $d=3$}\\\ \emptyset&\mbox{if \,
$d\geq 4$}.\end{cases}$

This just excludes the case $\alpha=1$, which will be dealt with below. In
Section 3.4 we introduce an extended fractional Rollnik-class which is subject
only to the constraint $d<2\alpha$, in particular, the case $\alpha=1$ is
covered by it in the $d=1$ case. In $d=2$ we will obtain it in Section 5 below
we as a limiting case. As Proposition 3.10 suggests, this limit is not
straightforward and we arrive at

$\mathcal{R}_{d,1}^{p}:=L^{p}(\mathbb{R}^{d})\,\cap\,\liminf_{\alpha\downarrow
1}\mathcal{R}_{d,\alpha},\quad p\geq 1.$

For later use below, we define more generally for $\alpha=\frac{d}{2}$,
$d=1,2,3$, the class

$\mathcal{R}_{d,\frac{d}{2}}^{p}:=L^{p}(\mathbb{R}^{d})\,\cap\,\liminf_{\alpha\downarrow\frac{d}{2}}\mathcal{R}_{d,\alpha},\quad
p\geq 1,$ (3.7)

where the limit inferior means
$\liminf_{\alpha\downarrow\frac{d}{2}}\mathcal{R}_{d,\alpha}=\cup_{\beta\in\left(\frac{d}{2},d\right)}\cap_{\alpha\in\left(\frac{d}{2},\beta\right)}\mathcal{R}_{d,\alpha}$.
The intersection with the $L^{p}$ spaces in the definition will be justified
by Theorem 5.3 below.

###  3.2. Comparison properties

We have already seen in Proposition 3.3 that
$L^{d/\alpha}(\mathbb{R}^{d})\subset\mathcal{R}_{d,\alpha}$. A corollary to
this result is the following.

######  Proposition 3.11.

If $p\leq\frac{d}{\alpha}\leq q$, then $L^{p}(\mathbb{R}^{d})\cap
L^{q}(\mathbb{R}^{d})\subset\mathcal{R}_{d,\alpha}$.

###### Proof.

By interpolation it is immediate that $L^{p}(\mathbb{R}^{d})\cap
L^{q}(\mathbb{R}^{d})\subset
L^{d/\alpha}(\mathbb{R}^{d})\subset\mathcal{R}_{d,\alpha}$. ∎

######  Proposition 3.12.

If $\alpha>\frac{d}{2}$, then $L^{1}(\mathbb{R}^{d})\cap
L^{2}(\mathbb{R}^{d})\subset\mathcal{R}_{d,\alpha}$ and the estimate

$\|V\|_{\mathcal{R}_{d,\alpha}}\leq\left(\frac{\sigma_{d}}{2(d-\alpha)}\right)^{(d-\alpha)/d}\sqrt{\frac{d}{2\alpha-d}}\,\|V\|_{1}^{(2\alpha-d)/d}\|V\|_{2}^{2(d-\alpha)/d}$
(3.8)

holds, where $\sigma_{d}$ is the surface area of the $d$-dimensional unit
sphere.

###### Proof.

Let $V\in L^{1}(\mathbb{R}^{d})\cap L^{2}(\mathbb{R}^{d})$ and $\delta>0$. We
compute

$\iint_{|x-y|\geq\delta}\frac{|V(x)||V(y)|}{|x-y|^{2(d-\alpha)}}dxdy\leq\frac{1}{\delta^{2(d-\alpha)}}\iint_{\mathbb{R}^{2d}}|V(x)||V(y)|dxdy=\frac{\|V\|_{1}^{2}}{\delta^{2(d-\alpha)}}.$

By the Schwarz inequality, we also have

$\displaystyle\iint_{|x-y|<\delta}\frac{|V(x)||V(y)|}{|x-y|^{2(d-\alpha)}}dxdy$
$\displaystyle\leq$
$\displaystyle\iint_{|x-y|<\delta}\frac{|V(x)|^{2}}{|x-y|^{2(d-\alpha)}}dxdy$
(3.9) $\displaystyle\leq$
$\displaystyle\|V\|_{2}^{2}\int_{|x|<\delta}\frac{dx}{|x|^{2(d-\alpha)}}=\frac{\sigma_{d}\delta^{2\alpha-d}}{2\alpha-d}\|V\|_{2}^{2},$

and thus

$\|V\|_{\mathcal{R}_{d,\alpha}}^{2}\leq\frac{\|V\|_{1}^{2}}{\delta^{2(d-\alpha)}}+\frac{\sigma_{d}\delta^{2\alpha-d}}{2\alpha-d}\|V\|_{2}^{2}.$

Optimizing over $\delta$ by the choice

$\delta=\left(\frac{(2(d-\alpha)\|V\|_{1}^{2}}{\sigma_{d}\|V\|_{2}^{2}}\right)^{1/d}$

leads to (3.8). ∎

######  Proposition 3.13.

The following inclusions hold:

  1. (1)

Let $\alpha\leq\alpha_{1}\leq d$. Then $\mathcal{R}_{d,\alpha}\cap
L^{1}(\mathbb{R}^{d})\subset\mathcal{R}_{d,\alpha_{1}}$.

  2. (2)

Let $\alpha_{2}\leq\alpha$ and $\frac{d}{\alpha_{2}}<\min\\{p,q\\}$ for
$\frac{1}{p}+\frac{1}{q}=1$. Then $\mathcal{R}_{d,\alpha}\cap
L^{p}(\mathbb{R}^{d})\cap
L^{q}(\mathbb{R}^{d})\subset\mathcal{R}_{d,\alpha_{2}}$. In particular,
$\mathcal{R}_{d,\alpha}\cap
L^{2}(\mathbb{R}^{d})\subset\mathcal{R}_{d,\alpha_{2}}$.

  3. (3)

Let $\alpha_{1}\leq\alpha\leq\alpha_{2}$. Then
$\mathcal{R}_{d,\alpha_{1}}\cap\mathcal{R}_{d,\alpha_{2}}\subset\mathcal{R}_{d,\alpha}$.

###### Proof.

(1) Let $V\in\mathcal{R}_{d,\alpha}\cap L^{1}(\mathbb{R}^{d})$. As in
Proposition 3.12, we compute

$\iint_{|x-y|\geq
1}\frac{|V(x)||V(y)|}{|x-y|^{2d-2\alpha_{1}}}dxdy\leq\|V\|_{1}^{2}.$

We clearly have

$\iint_{|x-y|<1}\frac{|V(x)||V(y)|}{|x-y|^{2d-2\alpha_{1}}}dxdy\leq\iint_{\mathbb{R}^{2d}}\frac{|V(x)||V(y)|}{|x-y|^{2d-2\alpha}}dxdy=\|V\|_{\mathcal{R}_{d,\alpha}}^{2}$

using that $2d-2\alpha_{1}\leq 2d-2\alpha$.

(2) Let $V\in\mathcal{R}_{d,\alpha}\cap L^{p}(\mathbb{R}^{d})\cap
L^{q}(\mathbb{R}^{d})$. We have

$\iint_{|x-y|\geq
1}\frac{|V(x)||V(y)|}{|x-y|^{2d-2\alpha_{2}}}dxdy\leq\iint_{\mathbb{R}^{2d}}\frac{|V(x)||V(y)|}{|x-y|^{2d-2\alpha}}dxdy=\|V\|_{\mathcal{R}_{d,\alpha}}^{2}$

since $2d-2\alpha_{2}\geq 2d-2\alpha$. On the other hand, as in (3.9), we
compute

$\iint_{|x-y|<1}\frac{|V(x)|^{p}}{|x-y|^{pd-p\alpha_{2}}}dxdy\leq\frac{\sigma_{d}}{p\alpha_{2}-pd+d}\|V\|_{p}^{p}$

noting that $p\alpha_{2}-pd+d>0$ is equivalent to $d/\alpha_{2}<q$. Therefore,
by the Hölder inequality we obtain

$\iint_{|x-y|<1}\frac{|V(x)||V(y)|}{|x-y|^{2d-2\alpha_{2}}}dxdy\leq\left(\frac{\sigma_{d}}{p\alpha_{2}-pd+d}\right)^{1/p}\left(\frac{\sigma_{d}}{q\alpha_{2}-qd+d}\right)^{1/q}\|V\|_{p}\|V\|_{q}.$

(3) We have

$\displaystyle\left\|V\right\|_{\mathcal{R}_{d,\alpha}}^{2}$
$\displaystyle=\iint_{|x-y|<1}\frac{|V(x)||V(y)|}{|x-y|^{2d-2\alpha}}dx\,dy+\iint_{|x-y|\geq
1}\frac{|V(x)||V(y)|}{|x-y|^{2d-2\alpha}}dx\,dy$
$\displaystyle\leq\iint_{|x-y|<1}\frac{|V(x)||V(y)|}{|x-y|^{2d-2\alpha_{1}}}dx\,dy+\iint_{|x-y|\geq
1}\frac{|V(x)||V(y)|}{|x-y|^{2d-2\alpha_{2}}}dx\,dy\leq\left\|V\right\|_{\mathcal{R}_{d,\alpha_{1}}}^{2}+\left\|V\right\|_{\mathcal{R}_{d,\alpha_{2}}}^{2}.$

∎

By extension of the definition of $\mathcal{R}_{d,\alpha}$ to $\alpha=2$ we
consider the space $\mathcal{R}_{3,2}$ which coincides with the classical
Rollnik-class given by (1.1). A consequence of part (3) in Proposition 3.13
and the dominated convergence theorem is the following result.

######  Corollary 3.14.

Let $d=3$ and $V\in\mathcal{R}_{3,\alpha_{0}}\cap\mathcal{R}_{3,2}$ for some
$\alpha_{0}\in\left(\frac{3}{2},2\right)$. Then the function
$\alpha\in[\alpha_{0},2]\mapsto\left\|V\right\|_{\mathcal{R}_{d,\alpha}}$ is
continuous.

######  Remark 3.15.

We see that for $\mathcal{R}_{d,2}$ the only possibility permitted by (3.1) is
$d=3$ and $\mathcal{R}_{3,2}=\mathcal{R}$. Corollary 3.14 shows, moreover,
continuity as $\alpha\uparrow 2$. Also, for $d=3$ and $\alpha=2$, estimate
(3.8) gives

$\|V\|_{\mathcal{R}}\leq\sqrt{3}(2\pi)^{1/3}\|V\|_{1}^{1/3}\|V\|_{2}^{2/3},$

which coincides with the standard norm estimate for the classical Rollnik-
class.

Next we compare the fractional Rollnik-class with the domain of the Riesz
potential. Recall the definition in (2.4).

######  Proposition 3.16.

The following properties hold.

  1. (1)

$\mathcal{R}_{d,\alpha}\subset{\mathfrak{X}}^{d}_{\alpha-\frac{d}{2}}$.

  2. (2)

If $V\in\mathcal{R}_{d,\alpha}$ satisfies $|\\{V(x)=0\\}|=0$, then
$V\in{\mathfrak{X}}^{d}_{2\alpha-d}$.

###### Proof.

To prove (1), just observe that by (3.2) it follows that if
$V\in\mathcal{R}_{d,\alpha}$, then $\mathcal{I}_{\alpha-\frac{d}{2}}|V|\in
L^{2}(\mathbb{R}^{d})$, in particular it is a.e. finite. Hence
$|V|\in{\mathfrak{X}}^{d}_{\alpha-\frac{d}{2}}$ and then also
$V\in{\mathfrak{X}}^{d}_{\alpha-\frac{d}{2}}$. Next consider (2) and define
the measure $\mu_{V}(dx)=|V(x)|dx$. Since $|\\{V(x)=0\\}|=0$, the measure
$\mu_{V}$ is equivalent to Lebesgue measure. Since
$V\in\mathcal{R}_{d,\alpha}$, we obtain

$\int_{\mathbb{R}^{d}}\mathcal{I}_{2\alpha-d}|V|(x)\mu_{V}(dx)=\int_{\mathbb{R}^{d}}|V(x)|\mathcal{I}_{2\alpha-d}|V|(x)dx<\infty,$

hence $\mathcal{I}_{2\alpha-d}|V|$ is $\mu_{V}$-a.e. finite. Since $\mu_{V}$
and Lebesgue measure are equivalent, $\mathcal{I}_{2\alpha-d}|V|$ is a.e.
finite and then $|V|\in{\mathfrak{X}}^{d}_{2\alpha-d}$, which in turn implies
$V\in{\mathfrak{X}}^{d}_{2\alpha-d}$. ∎

###  3.3. Specific cases

####  3.3.1. Potentials with compact support

Let $V_{K}=-\mathbf{1}_{K}$ for a compact set $K\subset\mathbb{R}^{d}$. Then
clearly $V_{K}\in L^{d/\alpha}(\mathbb{R}^{d})\subset\mathcal{R}_{d,\alpha}$.
We compare the Rollnik norm of $V_{K}$ with the norm of $V_{B^{K}}$, where
$B^{K}$ is a ball centred in the origin having the same volume as $K$. Recall
the fractional perimeter of index $s\in(0,1)$ of a Borel set
$E\subset\mathbb{R}^{d}$ defined by

$\operatorname{Per}_{s}(E)=\frac{s2^{s-1}\Gamma\left(\frac{d+s}{2}\right)}{\pi^{d/2}\Gamma\left(1-\frac{s}{2}\right)}\int_{E}\int_{\mathbb{R}^{d}\setminus
E}\frac{dxdy}{|x-y|^{d+s}}.$

A sufficient condition for a set $E$ to have a finite fractional perimeter of
order $s$ is that its indicator function $\mathbf{1}_{E}$ belongs to the
fractional Sobolev space $W^{\frac{s}{p},p}(\mathbb{R}^{d})$. In particular,
$E$ has finite fractional perimeter if it has finite classical perimeter
$\operatorname{Per}(E)$. The inequality

$\operatorname{Per}_{s}(E)\geq\operatorname{Per}_{s}(B^{E}),\ s\in(0,1]$
(3.10)

is well-known. Indeed, the case $s=1$ is the classical isoperimetric
inequality, while for $s\in(0,1)$ it can be obtained as a consequence of [14,
Th. 4.1]. For quantitative versions of (3.10) we refer to [15, 16].

A straightforward consequence of the Riesz rearrangement inequality [25, 5]
gives

$\left\|V_{K}\right\|_{\mathcal{R}_{d,\alpha}}\leq\left\|V_{B^{K}}\right\|_{\mathcal{R}_{d,\alpha}},\quad
0<\alpha<d<2\alpha.$ (3.11)

Combining the two inequalities above, we get for any compact set $K$ with
finite $s$-perimeter

$\left\|V_{K}\right\|_{\mathcal{R}_{d,\alpha}}^{2}\leq\left\|V_{B^{K}}\right\|_{\mathcal{R}_{d,\alpha}}^{2}=\frac{\left\|V_{B^{K}}\right\|_{\mathcal{R}_{d,\alpha}}^{2}}{\operatorname{Per}_{s}(B^{K})}\operatorname{Per}_{s}(B^{K})\leq\frac{\left\|V_{B^{K}}\right\|_{\mathcal{R}_{d,\alpha}}^{2}}{\operatorname{Per}_{s}(B^{K})}\operatorname{Per}_{s}(K),$

where the constant
$\left\|V_{B^{K}}\right\|_{\mathcal{R}_{d,\alpha}}/\operatorname{Per}_{s}(B^{K})$
is sharp in the sense that all the inequalities become equalities if
$K=B^{K}$. The constant can be exactly evaluated. Indeed, by [12, Prop. 2.3]
we have

$\operatorname{Per}_{s}(B^{K})=\begin{cases}\left(\frac{|K|}{\omega_{d}}\right)^{\frac{d-s}{d}}\frac{\left(\Gamma\left(\frac{d+s}{2}\right)\right)^{2}}{2\pi\Gamma\left(1-\frac{s}{2}\right)\Gamma\left(\frac{d-s}{2}+1\right)}\sigma_{d}&s\in(0,1)\\\
\left(\frac{|K|}{\omega_{d}}\right)^{\frac{d-1}{d}}\sigma_{d}&s=1.\end{cases}$

On the other hand, by using [12, Prop. 7.5] we obtain

$\left\|V_{B^{K}}\right\|_{\mathcal{R}_{d,\alpha}}^{2}=\left(\frac{|K|}{\omega_{d}}\right)^{\frac{d+\alpha}{d}}\frac{\sigma_{d}}{\alpha(d+\alpha)}\mu_{2\alpha-d},$

where

$\mu_{2\alpha-d}=\begin{cases}2^{1+2\alpha-d}\pi^{\frac{d-1}{2}}\frac{\Gamma\left(\frac{1+2\alpha-d}{2}\right)}{\Gamma(\alpha)}&\alpha\not=\frac{d+1}{2}\\\
\frac{16\pi^{\frac{d-1}{2}}}{\Gamma\left(\frac{d+1}{2}\right)}&\alpha=\frac{d+1}{2}.\end{cases}$

####  3.3.2. Asymmetric fractional Rollnik potentials

By (3.11) symmetric rearrangement increases the fractional Rollnik norm, and
since the kernel in the Rollnik integral is spherically symmetric, the
inequality reduces to equality only for spherically symmetric potentials. We
can use the fractional Rollnik norm to estimate the symmetry deficit of a non-
radial $L^{\infty}(\mathbb{R}^{d})\cap L^{1}(\mathbb{R}^{d})$ potential from a
rotationally symmetric potential. We discuss this by comparing with a
spherical potential well.

Let $V\in L^{1}(\mathbb{R}^{d})\cap L^{\infty}(\mathbb{R}^{d})$ with $V\leq 0$
a.e. Clearly, $V\in
L^{d/\alpha}(\mathbb{R}^{d})\subset\mathcal{R}_{d,\alpha}$. Denote by $B^{V}$
the ball centered in the origin with volume
$\left\|V\right\|_{1}/\left\|V\right\|_{\infty}$, and consider the family of
potential wells $V_{B,a}:=-\left\|V\right\|_{\infty}\mathbf{1}_{B^{V}+a}$, for
$a\in\mathbb{R}^{d}$. As a direct consequence of [13, Th. 4] we get

$\left\|V_{B,0}\right\|_{\mathcal{R}_{d,\alpha}}^{2}-\left\|V\right\|_{\mathcal{R}_{d,\alpha}}^{2}\geq\frac{C_{d,\alpha}}{2}\left(\frac{\left\|V\right\|_{\infty}}{\left\|V\right\|_{1}}\right)^{\frac{2(d-\alpha)}{d}}\inf_{a\in\mathbb{R}^{d}}\left\|V-V_{B,a}\right\|_{1}^{2},$

i.e., the difference of the fractional Rollnik norms of $V$ and the spherical
potential well $V_{B}$ controls from above an $L^{1}$ distance of $V$ from the
family of spherical potential wells with the same $L^{\infty}$ and $L^{1}$
norms.

####  3.3.3. Rotationally symmetric potentials

Consider a generic potential $V$ and denote by $V^{\ast}$ the symmetric
decreasing rearrangement of $|V|$. Then the following property holds.

######  Proposition 3.17.

Let $d\in\mathbb{N}$ and $\alpha\in(0,2)$ such that $\alpha<d<2\alpha$. If
$V^{\ast}\in\mathcal{R}_{d,\alpha}$, then $V\in\mathcal{R}_{d,\alpha}$ and

$\left\|V\right\|_{\mathcal{R}_{d,\alpha}}\leq\left\|V^{\ast}\right\|_{\mathcal{R}_{d,\alpha}}.$
(3.12)

Equality holds if and only if there exists a matrix $A\in\mathbb{R}^{d\times
d}$ with $|{\rm det}(A)|=1$ and a vector $a\in\mathbb{R}^{d}$ such that

$|V(x)|=V^{\ast}(Ax+a).$ (3.13)

Furthermore, if $V$ is continuous, then equality holds if and only if there
exists a matrix $A\in\mathbb{R}^{d\times d}$ with $|{\rm det}(A)|=1$ and a
vector $a\in\mathbb{R}^{d}$ such that

$\mbox{either}\quad V(x)=V^{\ast}(Ax+a),\quad\mbox{ or }\quad
V(x)=-V^{\ast}(Ax+a).$ (3.14)

###### Proof.

The inequality (3.12) follows by a direct application of Riesz rearrangement
inequality [25]. If equality holds in (3.12), then we are under the
assumptions of [20, Lem. 3] and (3.13) holds. If, moreover, $V$ is continuous,
then $V^{\ast}(x)=0$ for some $x$ with $|x|=R$ implies $V^{\ast}(x)=0$ for all
$x\in\mathbb{R}^{d}$ with $|x|\geq R$. Thus either $V^{\ast}(Ax+a)>0$ for all
$x\in\mathbb{R}^{d}$, or there exists a bounded open set
$K\subset\mathbb{R}^{d}$ such that $V^{\ast}(Ax+a)=0$ exactly when $x\not\in
K^{c}$. By (3.13) this means that $V$ cannot change sign and so (3.14) holds.
∎

We focus now on the case when $V$ is a non-negative radially symmetric
potential, with s non-increasing profile (or, equivalently, on $V^{\ast}$ for
any potential $V$). Let this profile be $v:[0,\infty)\to[0,\infty]$ such that
$v(t)=V(x)$ for all $x\in\mathbb{R}^{d}$ with $|x|=t$. Making use of (3.3) we
show a bound on $v$ near the origin.

######  Theorem 3.18.

Let $d\in\mathbb{N}$ and $\alpha\in(0,2)$ with $\alpha<d<2\alpha$. Suppose
that $V\in\mathcal{R}_{d,\alpha}$ is non-negative, radially symmetric, with
non-increasing profile $v$. Then there exists $t_{0}>0$ such that

$v(t)<\frac{2^{d-\alpha}\left\|V\right\|_{\mathcal{R}_{d,\alpha}}}{\omega_{d}}t^{-\alpha},\quad
t\in(0,t_{0}).$

###### Proof.

Consider the function

$U(r)=\int_{0}^{r}t^{d-1}v(t)dt\leq\frac{2^{d-\alpha}\left\|V\right\|_{\mathcal{R}_{d,\alpha}}}{\sigma_{d}}r^{d-\alpha},$

where the bound follows by (3.3). Assume, to the contrary, that there exists a
sequence $(t_{n})_{n\in\mathbb{N}}$ such that $t_{n}\downarrow 0$ and
$v(t_{n})\geq\frac{2^{d-\alpha}\left\|V\right\|_{\mathcal{R}_{d,\alpha}}}{\omega_{d}}t_{n}^{-\alpha}$.
Then we have

$U(t_{0})=\sum_{n=0}^{\infty}\int_{t_{n+1}}^{t_{n}}t^{d-1}v(t)dt\geq\frac{2^{d-\alpha}\left\|V\right\|_{\mathcal{R}_{d,\alpha}}}{\sigma_{d}}\sum_{n=0}^{\infty}t_{n}^{-\alpha}(t_{n}^{d}-t_{n+1}^{d})=\frac{2^{d-\alpha}\left\|V\right\|_{\mathcal{R}_{d,\alpha}}}{\sigma_{d}}\sum_{n=0}^{\infty}(t_{n}^{d-\alpha}-t_{n}^{-\alpha}t_{n+1}^{d}).$

Since $t_{n}>t_{n+1}$ and thus $t_{n}^{-\alpha}<t_{n+1}^{-\alpha}$, in the end
we have

$U(t_{0})>\frac{2^{d-\alpha}\left\|V\right\|_{\mathcal{R}_{d,\alpha}}}{\sigma_{d}}\sum_{n=0}^{\infty}(t_{n}^{d-\alpha}-t_{n+1}^{d-\alpha})=\frac{2^{d-\alpha}\left\|V\right\|_{\mathcal{R}_{d,\alpha}}}{\sigma_{d}}t_{0}^{d-\alpha},$

which is impossible. ∎

####  3.3.4. Coulomb-type potentials

We give an alternative expression of the fractional Rollnik norm for radial
potentials, which will be useful below. Let $V(x)=v(|x|)$ with a function
$v:\mathbb{R}_{0}^{+}\to\mathbb{C}$. Then, by making use of (2.5) we have

$\displaystyle\begin{split}\left\|V\right\|_{\mathcal{R}_{d,\alpha}}^{2}&=\gamma_{d}(2\alpha-d)\int_{\mathbb{R}^{d}}V(x)\mathcal{I}_{2\alpha-d}V(x)dx\\\
&=\int_{\mathbb{R}^{d}}v(|x|)\int_{0}^{\infty}r^{2\alpha-d-1}F_{2\alpha-d}\left(\frac{|x|}{r}\right)v(r)dr\,dx\\\
&=\sigma_{d}\int_{0}^{\infty}\int_{0}^{\infty}s^{d-1}r^{2\alpha-d-1}v(s)F_{2\alpha-d}\left(\frac{s}{r}\right)v(r)dr\,ds.\end{split}$
(3.15)

######  Proposition 3.19.

The potential $V(x)=|x|^{-\beta}\mathbf{1}_{\\{|x|<1\\}}$ belongs to
$\mathcal{R}_{d,\alpha}$ if and only if $\beta<\alpha$.

###### Proof.

By (3.15) we get

$\displaystyle\left\|V\right\|_{\mathcal{R}_{d,\alpha}}^{2}$
$\displaystyle=\sigma_{d}\int_{0}^{1}\int_{0}^{1}s^{d-1-\beta}r^{2\alpha-d-1-\beta}F_{2\alpha-d}\left(\frac{s}{r}\right)ds\,dr$
$\displaystyle=\sigma_{d}\int_{0}^{1}\int_{0}^{1/r}z^{d-\beta-1}r^{2(\alpha-\beta)-1}F_{2\alpha-d}(z)dz\,dr$
$\displaystyle=\sigma_{d}\int_{0}^{1}\int_{0}^{1}z^{d-\beta-1}r^{2(\alpha-\beta)-1}F_{2\alpha-d}(z)dz\,dr+\sigma_{d}\int_{0}^{1}\int_{1}^{1/r^{2}}z^{d-\beta-1}r^{2(\alpha-\beta)-1}F_{2\alpha-d}(z)dz\,dr$
$\displaystyle=\sigma_{d}\int_{0}^{1}\int_{0}^{1}z^{d-\beta-1}r^{2(\alpha-\beta)-1}F_{2\alpha-d}(z)dr\,dz+\sigma_{d}\int_{1}^{\infty}\int_{0}^{1/z}z^{d-\beta-1}r^{2(\alpha-\beta)-1}F_{2\alpha-d}(z)dr\,dz$
$\displaystyle=\frac{\sigma_{d}}{2(\alpha-\beta)}\int_{0}^{1}z^{d-\beta-1}F_{2\alpha-d}(z)dz+\frac{\sigma_{d}}{2(\alpha-\beta)}\int_{1}^{\infty}z^{d-2\alpha+\beta-1}F_{2\alpha-d}(z)dz,$

where integration with respect to $r$ was possible since $\beta<\alpha$, while
otherwise the integral is divergent. The first integral is finite since
$z^{d-\beta-1}F_{2\alpha-d}(z)\sim\sigma_{d}z^{d-\beta-1}$ as $z\to 0$, which
is integrable near zero, while $F_{2\alpha-d}$ is always integrable near $1$.
For the second integral observe that
$z^{d-2\alpha+\beta-1}F_{2\alpha-d}(z)\sim\sigma_{d}z^{\beta-1-d}$ as
$z\to\infty$, which is integrable at infinity since $\beta<d$. ∎

In case $\beta=\alpha$, we can make a finer analysis including logarithmic
corrections.

######  Proposition 3.20.

The potential
$V(x)=|x|^{-\alpha}(-\log|x|)^{-\gamma}\mathbf{1}_{\\{|x|<1/2\\}}$,
$\alpha<d$, belongs to $\mathcal{R}_{d,\alpha}$ if and only if
$\gamma>\frac{1}{2}$.

###### Proof.

By (3.15) we obtain

$\displaystyle\left\|V\right\|_{\mathcal{R}_{d,\alpha}}^{2}$
$\displaystyle=\sigma_{d}\int_{0}^{1/2}\int_{0}^{1/2}s^{d-1-\alpha}r^{\alpha-d-1}(-\log
s)^{-\gamma}(-\log r)^{-\gamma}F_{2\alpha-d}\left(\frac{s}{r}\right)ds\,dr$
$\displaystyle=\sigma_{d}\int_{0}^{1/2}\int_{0}^{1/2r}z^{d-\alpha-1}(-\log
z-\log r)^{-\gamma}r^{-1}(-\log r)^{-\gamma}F_{2\alpha-d}(z)dz\,dr$
$\displaystyle=\sigma_{d}\int_{0}^{1/2}\int_{0}^{1}z^{d-\alpha-1}(-\log z-\log
r)^{-\gamma}r^{-1}(-\log r)^{-\gamma}F_{2\alpha-d}(z)dz\,dr$
$\displaystyle\qquad+\sigma_{d}\int_{0}^{1/2}\int_{1}^{1/2r}z^{d-\alpha-1}(-\log
z-\log r)^{-\gamma}r^{-1}(-\log r)^{-\gamma}F_{2\alpha-d}(z)dz\,dr$
$\displaystyle=\sigma_{d}\int_{0}^{1}\int_{0}^{1/2}z^{d-\alpha-1}(-\log z-\log
r)^{-\gamma}r^{-1}(-\log r)^{-\gamma}F_{2\alpha-d}(z)dr\,dz$
$\displaystyle\qquad+\sigma_{d}\int_{1}^{\infty}\int_{0}^{1/2z}z^{d-\alpha-1}(-\log
z-\log r)^{-\gamma}r^{-1}(-\log r)^{-\gamma}F_{2\alpha-d}(z)dr\,dz$
$\displaystyle=\sigma_{d}(I_{1}+I_{2}).$

If $\gamma>\frac{1}{2}$, then

$\displaystyle I_{1}$
$\displaystyle=\int_{0}^{1}z^{d-\alpha-1}F_{2\alpha-d}(z)\int_{\log
2}^{\infty}x^{-\gamma}\left(x-\log z\right)^{-\gamma}dx\,dz$
$\displaystyle=\int_{0}^{1}z^{d-\alpha-1}F_{2\alpha-d}(z)\int_{\log 2-\log
z}^{\infty}(x+\log z)^{-\gamma}x^{-\gamma}dx\,dz$
$\displaystyle=\int_{0}^{1}z^{d-\alpha-1}F_{2\alpha-d}(z)(-\log
z)^{-2\gamma}\int_{\log 2-\log z}^{\infty}(1-\frac{-\log
z}{x})^{-\gamma}\left(\frac{x}{-\log z}\right)^{-2\gamma}dx\,dz$
$\displaystyle=\int_{0}^{1}z^{d-\alpha-1}F_{2\alpha-d}(z)(-\log
z)^{1-2\gamma}\int_{0}^{\frac{-\log z}{\log 2-\log
z}}(1-x)^{-\gamma}x^{-2\gamma-2}dx\,dz$
$\displaystyle=\int_{0}^{1}z^{d-\alpha-1}F_{2\alpha-d}(z)(-\log
z)^{1-2\gamma}\int_{0}^{\frac{-\log z}{\log 2-\log z}}B\left(\frac{-\log
z}{\log 2-\log z};2\gamma-1,1-\gamma\right)\,dz,$

with the incomplete Beta-function $B(\cdot;\cdot,\cdot)$. In the limit
$z\downarrow 0$ we have

$B\left(\frac{-\log z}{\log 2-\log z};2\gamma-1,1-\gamma\right)\to
B(2\gamma-1,1-\gamma),$

hence

$z^{d-\alpha-1}\log^{1-2\gamma}(z)F_{2\alpha-d}(z)B\left(\frac{-\log z}{\log
2-\log z};2\gamma-1,1-\gamma\right)\sim
B(1-\gamma,2\gamma-1)\sigma_{d}z^{d-\alpha-1}\log^{1-2\gamma}(z),$

where the right hand side is integrable in a neighbourhood of zero. As to the
behaviour at $1$, recall that the incomplete Beta-function satisfies

$B\left(z;2\gamma-1,1-\gamma\right)\sim\frac{z^{2\gamma-1}}{2\gamma-1}\quad\mbox{
as $z\downarrow 0$}.$

Indeed, by [17, eq. (8.391)], we have

$B\left(z;2\gamma-1,1-\gamma\right)=\frac{z^{2\gamma-1}}{2\gamma-1}{}_{2}F_{1}(2\gamma-1,\gamma;2\gamma;z),$

where ${}_{2}F_{1}(a,b;c,z)$ is Gauss’ hypergeometric function, so it is clear
that

$\lim_{z\to
0^{+}}\frac{(2\gamma-1)B\left(z;2\gamma-1,1-\gamma\right)}{z^{2\gamma-1}}={}_{2}F_{1}(2\gamma-1,\gamma;2\gamma;0)=1.$

Hence we get, in general,

$z^{d-\alpha-1}\log^{1-2\gamma}(z)F_{2\alpha-d}(z)B\left(\frac{-\log z}{\log
2-\log
z};2\gamma-1,1-\gamma\right)\sim\frac{F_{2\alpha-d}(z)}{2\gamma-1}\quad\mbox{
as $z\downarrow 1$},$

where the right-hand side is integrable near $1$. This proves that
$I_{1}<\infty$. Next we estimate $I_{2}$ and write

$\displaystyle I_{2}$
$\displaystyle=\int_{1}^{\infty}z^{d-\alpha-1}F_{2\alpha-d}(z)\int_{\log(2z)}^{\infty}x^{-\gamma}(x-\log
z)^{-\gamma}dx\,dz$
$\displaystyle=\int_{1}^{\infty}z^{d-\alpha-1}\log^{1-2\gamma}(z)F_{2\alpha-d}(z)\int_{0}^{\log
z/(\log 2+\log z)}x^{2\gamma-2}\left(1-x\right)^{-\gamma}dx\,dz$
$\displaystyle=\int_{1}^{\infty}z^{d-\alpha-1}\log^{1-2\gamma}(z)F_{2\alpha-d}(z)B\left(\frac{\log
z}{\log 2+\log z};2\gamma-1,1-\gamma\right),$

with the incomplete Beta-function $B(\cdot;\cdot,\cdot)$. We have

$B\left(\frac{\log z}{\log 2+\log z};2\gamma-1,1-\gamma\right)\to
B(1-\gamma,2\gamma-1)$

as $z\to\infty$, hence

$z^{d-\alpha-1}\log^{1-2\gamma}(z)F_{2\alpha-d}(z)B\left(\frac{\log z}{\log
2+\log z};2\gamma-1,1-\gamma\right)\sim
B(1-\gamma,2\gamma-1)\sigma_{d}z^{\alpha-d-1}\log^{1-2\gamma}(z),$

which is integrable near infinity since $\alpha<d$. Again, to handle the
behaviour at $1$, note that

$z^{d-\alpha-1}\log^{1-2\gamma}(z)F_{2\alpha-d}(z)B\left(\frac{\log z}{\log
2+\log
z};2\gamma-1,1-\gamma\right)\sim\frac{F_{2\alpha-d}(z)}{2\gamma-1}\quad\mbox{
as $z\downarrow 1$},$

where the right-hand side is integrable near $1$.

Finally, consider the case $\gamma\leq\frac{1}{2}$. Then

$I_{1}\geq\sigma_{d}\int_{0}^{1}z^{d-\alpha-1}F_{2\alpha-d}(z)\int_{1}^{\infty}\frac{1}{(-\log
z+x)^{2\gamma}}dx\,dz=\infty$

due to $2\gamma\leq 1$. ∎

######  Remark 3.21.

Notice that for $V$ as in Proposition 3.20 we have

$\displaystyle\int_{\mathbb{R}^{d}}|V(x)|^{d/\alpha}dx$ $\displaystyle=$
$\displaystyle\int_{B_{1/2}}|x|^{-d}(-\log(|x|))^{-\frac{\gamma d}{\alpha}}dx$
$\displaystyle=$ $\displaystyle\sigma_{d}\int_{0}^{1/2}r^{-1}(-\log
r)^{-\frac{\gamma d}{\alpha}}dr=\sigma_{d}\int_{\log
2}^{\infty}x^{-\frac{\gamma d}{\alpha}}dx,$

which is finite if and only if $\gamma>\frac{\alpha}{d}$. If $d=3$ and
$\alpha>\frac{3}{2}$ or if $d=2$ and $\alpha>1$, then
$\frac{\alpha}{d}>\frac{1}{2}$. Hence, under these conditions, whenever
$\gamma\in\left(\frac{1}{2},\frac{\alpha}{d}\right)$, we have
$V\in\mathcal{R}_{d,\alpha}$ but $V\not\in L^{d/\alpha}(\mathbb{R}^{d})$.

Next we study the possible decay at infinity of fractional Rollnik potentials.

######  Proposition 3.22.

The potential $V(x)=|x|^{-\beta}\mathbf{1}_{\\{|x|\geq 1\\}}$ belongs to
$\mathcal{R}_{d,\alpha}$ if and only if $\beta\in(\alpha,d)$.

###### Proof.

By (3.15) we get

$\displaystyle\left\|V\right\|_{\mathcal{R}_{d,\alpha}}^{2}$
$\displaystyle=\sigma_{d}\int_{1}^{\infty}\int_{1}^{\infty}s^{d-1-\beta}r^{2\alpha-d-1-\beta}F_{2\alpha-d}\left(\frac{s}{r}\right)ds\,dr$
$\displaystyle=\sigma_{d}\int_{1}^{\infty}\int_{1/r}^{\infty}z^{d-\beta-1}r^{2(\alpha-\beta)-1}F_{2\alpha-d}(z)dz\,dr$
$\displaystyle=\sigma_{d}\int_{1}^{\infty}\int_{1/r}^{1}z^{d-\beta-1}r^{2(\alpha-\beta)-1}F_{2\alpha-d}(z)dz\,dr+\sigma_{d}\int_{1}^{\infty}\int_{1}^{\infty}z^{d-\beta-1}r^{2(\alpha-\beta)-1}F_{2\alpha-d}(z)dz\,dr$
$\displaystyle=\sigma_{d}\int_{1}^{\infty}\int_{1/z}^{\infty}z^{d-\beta-1}r^{2(\alpha-\beta)-1}F_{2\alpha-d}(z)dr\,dz+\sigma_{d}\int_{1}^{\infty}\int_{1}^{\infty}z^{d-\beta-1}r^{2(\alpha-\beta)-1}F_{2\alpha-d}(z)dr\,dz$
$\displaystyle=\frac{\sigma_{d}}{2(\beta-\alpha)}\int_{1}^{\infty}z^{d-2\alpha+\beta-1}F_{2\alpha-d}(z)dzdz+\frac{\sigma_{d}}{2(\beta-\alpha)}\int_{1}^{\infty}z^{d-\beta-1}F_{2\alpha-d}(z)dz,$

where again integration on $r$ makes since since $\beta>\alpha$ and else
diverge. For the former integral notice that $F_{2\alpha-d}$ is integrable
near $1$, and

$z^{d-2\alpha+\beta-1}F_{2\alpha-d}\sim\sigma_{d}z^{\beta-1-d}\mbox{ as
}z\to\infty,$

which is integrable if and only if $\beta<d$. For the latter integral, observe
again that $F_{2\alpha-d}$ is integrable near $1$, while

$z^{d-\beta-1}F_{2\alpha-d}\sim\sigma_{d}z^{2\alpha-d-\beta-1}\mbox{ as
}z\to\infty,$

which is integrable since $2\alpha-d<\alpha<\beta$. ∎

######  Proposition 3.23.

The potential $V(x)=|x|^{-\alpha}\log^{-\gamma}(x)\mathbf{1}_{\\{|x|\geq
2\\}}$ belongs to $\mathcal{R}_{d,\alpha}$ if and only if
$\gamma>\frac{1}{2}$.

###### Proof.

By (3.15) we get

$\displaystyle\left\|V\right\|_{\mathcal{R}_{d,\alpha}}^{2}$
$\displaystyle=\sigma_{d}\int_{2}^{\infty}\int_{2}^{\infty}s^{d-1-\alpha}r^{\alpha-d-1}\log^{-\gamma}(s)\log^{-\gamma}(r)F_{2\alpha-d}\left(\frac{s}{r}\right)ds\,dr$
$\displaystyle=\sigma_{d}\int_{2}^{\infty}\int_{2/r}^{\infty}z^{d-\alpha-1}r^{-1}(\log(r)+\log
z)^{-\gamma}\log^{-\gamma}(r)F_{2\alpha-d}(z)dz\,dr$
$\displaystyle=\sigma_{d}\int_{2}^{\infty}\int_{2/r}^{1}z^{d-\alpha-1}r^{-1}(\log(r)+\log
z)^{-\gamma}\log^{-\gamma}(r)F_{2\alpha-d}(z)\,dr$
$\displaystyle\quad+\sigma_{d}\int_{2}^{\infty}\int_{1}^{\infty}z^{d-\alpha-1}r^{-1}(\log(r)+\log
z)^{-\gamma}\log^{-\gamma}(r)F_{2\alpha-d}(z)dz\,dr$
$\displaystyle=\sigma_{d}\int_{0}^{1}\int_{2/z}^{\infty}z^{d-\alpha-1}r^{-1}(\log(r)+\log
z)^{-\gamma}\log^{-\gamma}(r)F_{2\alpha-d}(z)dr\,dz$
$\displaystyle\quad+\sigma_{d}\int_{1}^{\infty}\int_{2}^{\infty}z^{d-\alpha-1}r^{-1}(\log(r)+\log
z)^{-\gamma}\log^{-\gamma}(r)F_{2\alpha-d}(z)dr\,dz$
$\displaystyle=\sigma_{d}(I_{1}+I_{2}).$

First consider $I_{1}$ with $\gamma>\frac{1}{2}$. We have

$\displaystyle I_{1}$
$\displaystyle=\int_{0}^{1}z^{d-\alpha-1}F_{2\alpha-d}(z)\int_{\log(2/z)}^{\infty}(x+\log
z)^{-\gamma}x^{-\gamma}dx\,dz$
$\displaystyle=\int_{0}^{1}z^{d-\alpha-1}F_{2\alpha-d}(z)(-\log
z)^{1-2\gamma}\int_{0}^{\frac{-\log z}{\log 2-\log
z}}\left(1-x\right)^{-\gamma}x^{2\gamma-2}dx\,dz$
$\displaystyle=\int_{0}^{1}z^{d-\alpha-1}F_{2\alpha-d}(z)(-\log
z)^{1-2\gamma}B\left(\frac{-\log z}{\log 2-\log
z};2\gamma-1,1-\gamma\right)dz.$

As in Proposition 3.20, we have

$z^{d-\alpha-1}F_{2\alpha-d}(z)(-\log z)^{1-2\gamma}B\left(\frac{-\log z}{\log
2-\log
z};2\gamma-1,1-\gamma\right)\sim\frac{F_{2\alpha-d}(z)}{2\gamma-1}\mbox{ as
$z\uparrow 1$},$

where the right-hand side is integrable near $1$. Concerning the behaviour at
$0$, observe that

$z^{d-\alpha-1}F_{2\alpha-d}(z)(-\log z)^{1-2\gamma}B\left(\frac{-\log z}{\log
2-\log z};2\gamma-1,1-\gamma\right)\\\ \sim\sigma_{d}z^{d-\alpha-1}(-\log
z)^{1-2\gamma}B(2\gamma-1,\gamma-1)\mbox{ as $z\downarrow 0$},$

which is integrable near $0$ since $\alpha<d$. Hence $I_{1}<\infty$. Next
consider $I_{2}$. We can rewrite

$\displaystyle I_{2}$
$\displaystyle=\int_{1}^{\infty}z^{d-\alpha-1}F_{2\alpha-d}(z)\int_{\log
2}^{\infty}(x+\log z)^{-\gamma}x^{-\gamma}\,dx\,dz$
$\displaystyle=\int_{1}^{\infty}z^{d-\alpha-1}F_{2\alpha-d}(z)\int_{\log
2+\log z}^{\infty}x^{-\gamma}(x-\log z)^{-\gamma}\,dx\,dz$
$\displaystyle=\int_{1}^{\infty}z^{d-\alpha-1}F_{2\alpha-d}(z)\log^{-2\gamma}(z)\int_{\log
2+\log z}^{\infty}\left(\frac{x}{\log z}\right)^{-2\gamma}\left(1-\frac{\log
z}{x}\right)^{-\gamma}\,dx\,dz$
$\displaystyle=\int_{1}^{\infty}z^{d-\alpha-1}F_{2\alpha-d}(z)\log^{1-2\gamma}(z)\int_{0}^{\frac{\log
z}{\log 2+\log z}}x^{2\gamma-2}\left(1-x\right)^{-\gamma}\,dx\,dz$
$\displaystyle=\int_{1}^{\infty}z^{d-\alpha-1}F_{2\alpha-d}(z)\log^{1-2\gamma}(z)B\left(\frac{\log
z}{\log 2+\log z};2\gamma-1,1-\gamma\right)\,dz.$

Like before, we thus have

$z^{d-\alpha-1}F_{2\alpha-d}(z)\log^{1-2\gamma}(z)B\left(\frac{\log z}{\log
2+\log
z};2\gamma-1,1-\gamma\right)\sim\frac{F_{2\alpha-d}(z)}{2\gamma-1}\mbox{ as
$z\downarrow 1$},$

where the right-hand side is integrable near $1$. Concerning the behaviour at
$\infty$, observe that

$z^{d-\alpha-1}F_{2\alpha-d}(z)\log^{1-2\gamma}(z)B\left(\frac{-\log z}{\log
2-\log
z};2\gamma-1,1-\gamma\right)\sim\sigma_{d}z^{\alpha-d-1}\log^{1-2\gamma}(z)B(2\gamma-1,\gamma-1)\mbox{
as $z\uparrow\infty$},$

which is integrable near $\infty$ since $\alpha<d$. This proves that
$I_{2}<\infty$.

For the case $\gamma\leq\frac{1}{2}$, note that

$I_{2}\geq\int_{1}^{\infty}z^{d-\alpha-1}F_{2\alpha-d}(z)\int_{1}^{\infty}(x+\log
z)^{-2\gamma}dx\,dz=\infty.$

∎

######  Corollary 3.24.

The potential
$V(x)=|x|^{-\beta_{1}}\mathbf{1}_{\\{|x|<1\\}}+|x|^{-\beta_{2}}\mathbf{1}_{\\{|x|\geq
1\\}}$ belongs to $\mathcal{R}_{d,\alpha}$ if and only if $\beta_{1}<\alpha$
and $\beta_{2}\in(\alpha,d)$.

###### Proof.

By Propositions 3.19 and 3.22 it follows that if $\beta_{1}<\alpha$ and
$\beta_{2}\in(\alpha,d)$, then $V\in\mathcal{R}_{d,\alpha}$. Conversely, if
$V\in\mathcal{R}_{d,\alpha}$, then it follows by Remark 3.8 that both
$V_{1}(x)=|x|^{-\beta_{1}}\mathbf{1}_{\\{|x|<1\\}}$ and
$V_{2}(x)=|x|^{-\beta_{2}}\mathbf{1}_{\\{|x|\geq 1\\}}$ belong to
$\mathcal{R}_{d,\alpha}$, which implies $\beta_{1}<\alpha$ and
$\beta_{2}\in(\alpha,d)$ again by Propositions 3.19 and 3.22. ∎

###  3.4. Extended fractional Rollnik-class

Next we introduce another class of potentials related to the fractional
Rollnik-class, constructed on the resolvent estimates (2.17) and (2.20). This
class will be justified by the behaviours we will discuss in Section 4 below.

######  Definition 3.25.

Let $\alpha\in(0,2)$, $d\in\mathbb{N}$, $m\geq 0$, and assume that
$d<2\alpha$. We call

$\overline{\mathcal{R}}^{m}_{d,\alpha}:=\left\\{V:\mathbb{R}^{d}\to\mathbb{R}:[V]_{\overline{\mathcal{R}}_{d,\alpha}}^{2}:=\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}H^{m}_{d,\alpha}(|x-y|)^{2}|V(x)||V(y)|dx\,dy<\infty\right\\}$

extended fractional Rollnik-class. If $m=0$, we omit the superscript.

First we settle the relationship between the fractional Rollnik-classes and
their extensions.

######  Proposition 3.26.

Let $\alpha\in(0,2)$, $d\in\mathbb{N}$ and assume that $d<2\alpha$.

  1. (1)

If $d>\alpha$, we have
$\mathcal{R}_{d,\alpha}\subset\overline{\mathcal{R}}_{d,\alpha}$. In
particular, there exists a constant $C_{d,\alpha}>0$ such that

$[V]_{\overline{\mathcal{R}}_{d,\alpha}}\leq
C_{d,\alpha}\left\|V\right\|_{\mathcal{R}_{d,\alpha}}$

for every $V\in\mathcal{R}_{d,\alpha}$.

  2. (2)

For every $m>0$ we have
$\overline{\mathcal{R}}_{d,\alpha}\subset\overline{\mathcal{R}}^{m}_{d,\alpha}$.
In particular, there exists a constant $C_{d,\alpha}>0$ such that

$[V]_{\overline{\mathcal{R}}^{m}_{d,\alpha}}\leq
C_{d,\alpha}[V]_{\overline{\mathcal{R}}_{d,\alpha}}$

for every $V\in\mathcal{R}_{d,\alpha}$.

###### Proof.

(1) easily follows since $r\geq 1$ trivially implies $r^{-2(d+\alpha)}\leq
r^{-2(d-\alpha)}$, and thus

$\displaystyle[V]_{\overline{\mathcal{R}}_{d,\alpha}}^{2}$
$\displaystyle=\iint_{|x-y|\leq
1}\frac{|V(x)||V(y)|}{|x-y|^{2(d-\alpha)}}dx\,dy+2^{d+\alpha-2}\Gamma^{2}\Big{(}\frac{d+\alpha}{2}\Big{)}\iint_{|x-y|\geq
1}\frac{|V(x)||V(y)|}{|x-y|^{2(d+\alpha)}}dx\,dy$
$\displaystyle\leq\iint_{|x-y|\leq
1}\frac{|V(x)||V(y)|}{|x-y|^{2(d-\alpha)}}dx\,dy+2^{d+\alpha-2}\Gamma^{2}\Big{(}\frac{d+\alpha}{2}\Big{)}\iint_{|x-y|\geq
1}\frac{|V(x)||V(y)|}{|x-y|^{2(d-\alpha)}}dx\,dy$ $\displaystyle\leq
C_{d,\alpha}\left\|V\right\|_{\mathcal{R}_{d,\alpha}}^{2}.$

To obtain (2) note that by (6.6) there exists a constant $C_{d,\alpha}>0$ such
that $H^{m}_{d,\alpha}(r)\leq C_{d,\alpha}H_{d,\alpha}(r)$, for all $r>0$. ∎

The above definitions yield a consistent behaviour in the massless limit.

######  Proposition 3.27.

Let $V\in\overline{\mathcal{R}}_{d,\alpha}$. Then $\lim_{m\downarrow
0}[V]_{\overline{\mathcal{R}}^{m}_{d,\alpha}}=[V]_{\overline{\mathcal{R}}_{d,\alpha}}$.

###### Proof.

By Remark 6.16 in the Appendix we have $\lim_{m\downarrow
0}H_{d,\alpha}^{m}(r)=H_{d,\alpha}(r)$, and then the claim follows by the
dominated convergence theorem. ∎

Concerning the relationship between $L^{p}$-spaces and
$\overline{\mathcal{R}}_{d,\alpha}^{m}$, we can prove the following continuous
embedding property.

######  Proposition 3.28.

Let $m\geq 0$. Then
$L^{2}(\mathbb{R}^{d})\subset\overline{\mathcal{R}}_{d,\alpha}^{m}$. In
particular,

$[V]_{\overline{\mathcal{R}}_{d,\alpha}^{m}}\leq\left\|H_{d,\alpha}^{m}\right\|_{2}\left\|V\right\|_{2}.$
(3.16)

###### Proof.

Note that if $d<2\alpha$ then $H_{d,\alpha}^{m}\in L^{2}(\mathbb{R}^{d})$ for
all $m\geq 0$ and thus (3.16) follows by a direct application of Young’s
convolution inequality. ∎

######  Remark 3.29.

Proposition 3.28 guarantees that for $\beta>\frac{d}{2}$ the potential
$V(x)=|x|^{-\beta}\mathbf{1}_{\\{|x|\geq 1\\}}$ belongs to
$\overline{\mathcal{R}}_{d,\alpha}^{m}$ for all $m\geq 0$. However, since
$d<2\alpha$, we can take $\beta\in\left(\frac{d}{2},\alpha\right)$ such that
$V\not\in\mathcal{R}_{d,\alpha}$ due to Proposition 3.22, and thus
$V\in\overline{\mathcal{R}}_{d,\alpha}\setminus\mathcal{R}_{d,\alpha}$.

Similarly to (3.7) we define $\overline{\mathcal{R}}_{d,\frac{d}{2}}^{m,p}$
for some $p>1$ through of a limiting argument by

$\overline{\mathcal{R}}_{d,\frac{d}{2}}^{m,p}:=L^{p}(\mathbb{R}^{d})\,\cap\,\liminf_{\alpha\downarrow\frac{d}{2}}\overline{\mathcal{R}}_{d,\alpha}^{m}.$

By Proposition 3.28, for $p=2$ we have
$\overline{\mathcal{R}}_{d,\frac{d}{2}}^{m,2}=L^{2}(\mathbb{R}^{d})$.
Furthermore, Proposition 3.26 implies that for every $p>1$

$\mathcal{R}_{d,\frac{d}{2}}^{p}\subset\overline{\mathcal{R}}_{d,\frac{d}{2}}^{m,p}.$

##  4\. Perturbations by fractional Rollnik class potentials

###  4.1. Self-adjointness

In what follows our main interest is to study the properties of fractional
Schrödinger operators for $m\geq 0$

$H=L_{m,\alpha}+V$

with fractional Rollnik-class potentials $V$. First we show that $H$ can be
defined as a self-adjoint linear operator for $V\in{\mathcal{R}}_{d,\alpha}$.

Define

$K^{m}_{\lambda}=|V|^{1/2}(\lambda+L_{m,\alpha})^{-1}|V|^{1/2},$

which is an integral operator with kernel

$K^{m}_{\lambda}(x-y)=\sqrt{|V(x)|}R^{m}_{\lambda}(x-y)\sqrt{|V(y)|}.$

First we give a criterion for self-adjointness of the operator $H$.

######  Lemma 4.1.

Fix $m\geq 0$ and assume that
$\lim_{\lambda\to\infty}\left\|K_{\lambda}^{m}\right\|_{\mathscr{B}(L^{2}(\mathbb{R}^{d}))}=0$.
Then $H$ is self-adjoint on $L^{2}(\mathbb{R}^{d})$.

###### Proof.

By the assumption we have directly that

$\|\sqrt{|V|}(\lambda+L_{m,\alpha})^{-1/2}\|_{\mathscr{B}(L^{2})}\rightarrow
0$

as $\lambda\rightarrow\infty$. Define the function

$\widetilde{V}^{1/2}(x)=\begin{cases}V(x)/\sqrt{|V(x)|}&\mbox{if}\
V(x)\not=0,\\\ 0&\mbox{if}\ V(x)=0.\end{cases}$

Then $|\widetilde{V}^{1/2}|=\sqrt{|V|}$ and $\tilde{V}^{1/2}\sqrt{|V|}=V$
hold. For $\phi\in L^{2}(\mathbb{R}^{d})$ we compute

$\displaystyle\vspace{-3cm}\|(\lambda+L_{m,\alpha})^{-1/2}V(\lambda+L_{m,\alpha})^{-1/2}\phi\|^{2}_{2}$
$\displaystyle=$
$\displaystyle(\tilde{V}^{1/2}(\lambda+L_{m,\alpha})^{-1}V(\lambda+L_{m,\alpha})^{-1/2}\phi,\sqrt{|V|}(\lambda+(-\Delta)^{\alpha/2})^{-1/2}\phi)_{L^{2}}$
$\displaystyle\leq$
$\displaystyle\|K^{m}_{\lambda}\|_{\mathscr{B}(L^{2})}\|\sqrt{|V|}(\lambda+L_{m,\alpha})^{-1/2}\|_{\mathscr{B}(L^{2})}^{2}\|\phi\|_{2}^{2}.$

Thus we obtain

$\|(\lambda+L_{m,\alpha})^{-1/2}V(\lambda+L_{m,\alpha})^{-1/2}\|_{\mathscr{B}(L^{2})}\rightarrow
0$ (4.1)

as $\lambda\rightarrow\infty$. It follows then for
$\phi\in\operatorname{Dom}(L_{m,\alpha})$ that

$|(V\phi,\phi)_{L^{2}}|\leq\varepsilon_{\lambda}|((\lambda+L_{m,\alpha})\phi,\phi)_{L^{2}}|$

with $\varepsilon_{\lambda}\rightarrow 0$ as $\lambda\rightarrow\infty$. An
application of the KLMN theorem [23, Th. X.17] completes the proof. ∎

Using Lemma 4.1 it remains to show that under specific assumptions on $V$ it
follows that
$\lim_{\lambda\to\infty}\left\|K_{\lambda}^{m}\right\|_{\mathscr{B}(L^{2}(\mathbb{R}^{d}))}=0$.
This can be guaranteed if for a sufficiently large $\lambda_{0}>0$ the
operator $K^{m}_{\lambda_{0}}$ is a Hilbert-Schmidt operator.

######  Lemma 4.2.

Suppose that there exists $\lambda_{0}>0$ such that $K^{m}_{\lambda_{0}}$ is a
Hilbert-Schmidt operator. Then $K^{m}_{\lambda}$ is a Hilbert-Schmidt operator
for all $\lambda>\lambda_{0}$ and

$\lim_{\lambda\to\infty}\left\|K_{\lambda}^{m}\right\|_{\mathscr{B}(L^{2}(\mathbb{R}^{d}))}=\lim_{\lambda\to\infty}\left\|K_{\lambda}^{m}\right\|_{\rm
HS}=0,$

where $\left\|\cdot\right\|_{\rm HS}$ is the Hilbert-Schmidt norm.

###### Proof.

Assume that $K^{m}_{\lambda_{0}}$ is a Hilbert-Schmidt operator and let
$\lambda>\lambda_{0}$. Then we have

$R_{\lambda}^{m}(x)=\int_{0}^{\infty}e^{-\lambda
t}p^{m}_{t}(x)\,dt\leq\int_{0}^{\infty}e^{-\lambda_{0}t}p^{m}_{t}(x)\,dt=R_{\lambda_{0}}^{m}(x).$

Furthermore, by the dominated convergence theorem, it is clear that
$\lim_{\lambda\to\infty}R_{\lambda}(x)=0$. We get

$\displaystyle\left\|K_{\lambda}^{m}\right\|_{\rm HS}^{2}$ $\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{2d}}|V(x)|\,R_{\lambda}^{m}(|x-y|)^{2}\,|V(y)|dx\,dy\leq\int_{\mathbb{R}^{2d}}|V(x)|\,R_{\lambda_{0}}^{m}(|x-y|)^{2}\,|V(y)|dx\,dy$
$\displaystyle=$ $\displaystyle\left\|K_{\lambda_{0}}^{m}\right\|_{\rm
HS}^{2}<\infty.$

Again by dominated convergence, it follows that
$\lim_{\lambda\to\infty}\left\|K_{\lambda}^{m}\right\|_{\rm HS}=0$. ∎

Once this established, our goal is determine conditions under which
$K_{\lambda}^{m}$ is a Hilbert-Schmidt operator for some $\lambda>0$. First
consider the case of the fractional Rollnik-class.

######  Theorem 4.3.

Let $\alpha\in(0,2)$, $d\geq 1$, $m\geq 0$ and assume that $\alpha<d<2\alpha$.
If $V\in\mathcal{R}_{d,\alpha}$, then $K^{m}_{\lambda}$ is a Hilbert-Schmidt
operator for every $\lambda>0$. Furthermore, there exists a constant
$C_{d,\alpha}>0$ such that

$\left\|K^{m}_{\lambda}\right\|_{\rm
HS}\leq\sqrt{C_{d,\alpha}}\Big{(}1+\frac{me^{-\lambda/m}}{\lambda}\Big{)}\left\|V\right\|_{\mathcal{R}_{d,\alpha}}.$

(The bracket equals 1 if $m=0$.) In particular, if
$V\in\mathcal{R}_{d,\alpha}$, then $H$ is a self-adjoint operator.

###### Proof.

By Lemmas 6.1 and 6.10, it is straightforward that

$\iint_{\mathbb{R}^{2d}}K^{m}_{\lambda}(x-y)^{2}dxdy=\iint_{\mathbb{R}^{2d}}R^{m}_{\lambda}(x-y)^{2}|V(x)||V(y)|dxdy\leq
C_{d,\alpha}^{2}\left(1+\frac{me^{-\frac{\lambda}{m}}}{\lambda}\right)^{2}\|V\|_{\mathcal{R}_{d,\alpha}}^{2}$
(4.2)

for every $\lambda>0$ and $m\geq 0$. Lemma 4.1 then implies that $H$ is self-
adjoint on $L^{2}(\mathbb{R}^{d})$. ∎

We can, however, provide a finer argument to characterize the class of
potentials $V$ such that $K_{\lambda}^{m}$ is a Hilbert-Schmidt operator, at
least for $\lambda>m$.

######  Theorem 4.4.

Let $\alpha\in(0,2)$, $d\geq 1$, $m\geq 0$ and $\lambda>0$ and assume that
$d<2\alpha$. Then the following properties hold:

  1. (1)

If $V\in\overline{\mathcal{R}}_{d,\alpha}$, then $K_{\lambda}^{m}$ is a
Hilbert-Schmidt operator and there exists a constant $C_{d,\alpha,\lambda,m}$,
independent of $V$, such that

$\left\|K_{\lambda}^{m}\right\|_{\rm HS}\leq
C_{d,\alpha,\lambda,m}[V]_{\overline{\mathcal{R}}_{d,\alpha}}.$

  2. (2)

If $K_{\lambda}^{m}$ is a Hilbert-Schmidt operator, then
$V\in\overline{\mathcal{R}}_{d,\alpha}^{m}$ and there exists a constant
$C_{d,\alpha,\lambda,m}$, independent of $V$, such that

$[V]_{\overline{\mathcal{R}}_{d,\alpha}^{m}}\leq
C_{d,\alpha,\lambda,m}\left\|K_{\lambda}^{m}\right\|_{\rm HS}.$

  3. (3)

If $\lambda>m$, then $K_{\lambda}^{m}$ is a Hilbert-Schmidt operator if and
only if $V\in\overline{\mathcal{R}}_{d,\alpha}^{m}$, and there exists a
constant $C_{d,\alpha,\lambda,m}>1$ independent of $V$ such that

$\frac{1}{C_{d,\alpha,\lambda,m}}[V]_{\overline{\mathcal{R}}_{d,\alpha}^{m}}\leq\left\|K_{\lambda}^{m}\right\|_{\rm
HS}\leq C_{d,\alpha,\lambda,m}[V]_{\overline{\mathcal{R}}_{d,\alpha}^{m}}.$

###### Proof.

To obtain (1), let $V\in\overline{\mathcal{R}}_{d,\alpha}$. Then by Corollary
6.13 we have

$\left\|K_{\lambda}^{m}\right\|_{\rm
HS}^{2}=\iint_{\mathbb{R}^{2d}}|V(x)|\,R^{m}_{\lambda}(|x-y|)^{2}\,|V(y)|dx\,dy\\\
\leq
C_{d,\alpha,\lambda,m}\iint_{\mathbb{R}^{2d}}|V(x)|\,H_{d,\alpha}^{0}(|x-y|)^{2}\,|V(y)|dx\,dy=C_{d,\alpha,\lambda,m}[V]_{\overline{\mathcal{R}}_{d,\alpha}}^{2}.$

Next we show (2). Assume that $K_{\lambda}^{m}$ is a Hilbert-Schmidt operator.
By Lemma 6.17 we then have

$[V]_{\overline{\mathcal{R}}_{d,\alpha}^{m}}^{2}=\iint_{\mathbb{R}^{2d}}|V(x)|\,H_{d,\alpha}^{m}(|x-y|)^{2}\,|V(y)|dx\,dy\\\
\leq
C_{d,\alpha,\lambda,m}\iint_{\mathbb{R}^{2d}}|V(x)|\,R^{m}_{\lambda}(|x-y|)^{2}\,|V(y)|dx\,dy=C_{d,\alpha,\lambda,m}\left\|K_{\lambda}^{m}\right\|_{\rm
HS}^{2}.$

Finally, to show (3) note that one implication has already been shown in (2).
If we assume further that $\lambda>m$ and
$V\in\overline{\mathcal{R}}_{d,\alpha}^{m}$, then by Corollary 6.15 we get

$\left\|K_{\lambda}^{m}\right\|_{\rm
HS}^{2}=\iint_{\mathbb{R}^{2d}}|V(x)|\,R^{m}_{\lambda}(|x-y|)^{2}\,|V(y)|dx\,dy\\\
\leq
C_{d,\alpha,\lambda,m}\iint_{\mathbb{R}^{2d}}|V(x)|\,H_{d,\alpha}^{m}(|x-y|)^{2}\,V(y)|dx\,dy=C_{d,\alpha,\lambda,m}[V]_{\overline{\mathcal{R}}^{m}_{d,\alpha}}^{2}.$

∎

######  Remark 4.5.

We note that by the Kato-Rellich theorem the relativistic Schrödinger
operators with Coulomb potentials $|x|^{-\gamma}$, $\gamma>0$, are self-
adjoint if $\gamma<\frac{d}{2}$ for $d\leq 3$ by [19, Prop. 1.5]. On the other
hand, a combination of Theorem 4.4 and Corollary 3.24 improve this to
$\gamma<\alpha$.

Finally we note that if $d\geq 2\alpha$, then $K^{m}_{\lambda}$ cannot be a
Hilbert-Schmidt operator for continuous potentials $V$.

######  Proposition 4.6.

Let $\alpha\in(0,2)$, $d\geq 1$, $\lambda>0$, $m\geq 0$ and assume $d\geq
2\alpha$. If there exist $\varepsilon^{\ast},r^{\ast}>0$ and
$x^{\ast}\in\mathbb{R}^{d}$ such that $|V(y)|>\varepsilon^{\ast}$ for all
$y\in B_{r^{\ast}}(x^{\ast})$, then $K_{\lambda}^{m}$ is not a Hilbert-Schmidt
operator. In particular, if $V$ is continuous, then $K_{\lambda}^{m}$ is not a
Hilbert-Schmidt operator, unless $V\equiv 0$.

###### Proof.

By Lemma 6.17 we see that the function $r\in[0,\infty)\mapsto
R_{\lambda}^{m}(r\mathbf{e}_{1})^{2}\in[0,\infty)$ satisfies (3.4), since
clearly $H^{m}_{d,\alpha}$ does not belong to $L^{2}(B_{\delta})$ for any ball
$B_{\delta}$ centred in $0$. The statement follows then by Theorem 3.9. ∎

###  4.2. Spectral properties

####  4.2.1. Essential spectrum

In this section we discuss the spectrum of non-local Schrödinger operators
with fractional Rollnik-class potentials. In the classical case, for
$V\in\mathcal{R}$ Fourier transform gives that ${\operatorname{Spec}}_{\rm
ess}(-\Delta+V)={\operatorname{Spec}}_{\rm ess}(-\Delta)$ without any further
requirements [32]. However, if $(d,\alpha)\not=(3,2)$, it appears to be
difficult to construct a Fourier theory of $\mathcal{R}_{d,\alpha}$. Therefore
we need other ways to prove the stability of the essential spectrum.

######  Theorem 4.7.

Let $\alpha\in(0,2)$, $d\geq 1$, $m\geq 0$, and assume that $d<2\alpha$. If
the potential
$V\in\overline{\mathcal{R}}^{m}_{d,\alpha}\cap(L^{1}(\mathbb{R}^{d})+L^{2}(\mathbb{R}^{d}))$,
then

${\operatorname{Spec}}_{\rm ess}H={\operatorname{Spec}}_{\rm
ess}L_{m,\alpha}=[0,\infty)$

holds for $m\geq 0$.

###### Proof.

Due to (4.1) we may choose $\lambda>m$ sufficiently large such that

$\|(\lambda+L_{m,\alpha})^{-1/2}V(\lambda+L_{m,\alpha})^{-1/2}\|_{\mathscr{B}(L^{2})}<1.$
(4.3)

Write

$G_{m,\lambda}=(\lambda+L_{m,\alpha})^{-1/2}V(\lambda+L_{m,\alpha})^{-1/2}.$

By the Neumann expansion we have

$\displaystyle(\lambda+H)^{-1}$ $\displaystyle=$
$\displaystyle(\lambda+L_{m,\alpha})^{-1/2}(1+G_{m,\lambda})^{-1}(\lambda+L_{m,\alpha})^{-1/2}$
$\displaystyle=$
$\displaystyle(\lambda+L_{m,\alpha})^{-1/2}\sum_{j=0}^{\infty}(-1)^{j}G_{m,\lambda}^{j}(\lambda+L_{m,\alpha})^{-1/2}.$

and $\lambda$ that satisfies (4.3) belongs to the resolvent set of $H$. We
thus obtain

$(\lambda+H)^{-1}-(\lambda+L_{m,\alpha})^{-1}=-A_{m,\lambda}+B_{m,\lambda},$

where

$\displaystyle A_{m,\lambda}$ $\displaystyle=$
$\displaystyle(\lambda+L_{m,\alpha})^{-1}V(\lambda+L_{m,\alpha})^{-1},$
$\displaystyle B_{m,\lambda}$ $\displaystyle=$
$\displaystyle(\lambda+L_{m,\alpha})^{-1/2}\sum_{j=2}^{\infty}(-1)^{j}G_{m,\lambda}^{j}(\lambda+L_{m,\alpha})^{-1/2}.$

By Theorem 4.4 $K_{\lambda}^{m}$ is a Hilbert-Schmidt operator and thus
compact. Therefore $B_{m,\lambda}$ is also compact since the operator-norm
limit of a sequence of compact operators is compact. We now prove compactness
of $A_{m,\lambda}$. To do this, we split $V=V_{1}+V_{2}$, where $V_{1}\in
L^{1}(\mathbb{R}^{d})$ and $V_{2}\in L^{2}(\mathbb{R}^{d})$. It is clear that

$A_{m,\lambda}=(\lambda+L_{m,\alpha})^{-1}V_{1}(\lambda+L_{m,\alpha})^{-1}+(\lambda+L_{m,\alpha})^{-1}V_{2}(\lambda+L_{m,\alpha})^{-1}=:A_{m,\lambda}^{1}+A_{m,\lambda}^{2}.$

Since $V_{1}\in L^{1}(\mathbb{R}^{d})$, we have by Corollary 6.15,

$\int_{\mathbb{R}^{2d}}R^{m}_{\lambda}(x-y)|V_{1}(y)|dxdy\leq
C_{d,\alpha,\lambda,m}\|V_{1}\|_{1}\int_{\mathbb{R}^{d}}H_{d,\alpha}^{m}(|x|)dx<\infty.$
(4.4)

This bound implies that $(\lambda+L_{m,\alpha})^{-1/2}|V_{1}|^{1/2}$ is a
Hilbert-Schmidt operator. Hence $A_{m,\lambda}^{1}$ is compact. On the other
hand, we prove that $(\lambda+L_{m,\alpha})^{-1/2}V_{2}$ is a Hilbert-Schmidt
operator by replacing $|V_{1}(y)|$ and $\|V_{1}\|_{1}$ in (4.4) with
$|V_{2}(y)|^{2}$ and $\|V_{2}\|_{2}^{2}$. Therefore $A^{2}_{m,\lambda}$ is
compact. Hence $A_{m,\lambda}$ is compact since it is a sum of compact
operators. The Weyl theorem then completes the proof. ∎

####  4.2.2. Discrete spectrum

Next we consider the discrete spectrum. First we provide a bound on the number
of negative eigenvalues, which in order to keep explicit, we only consider the
case $\alpha<d$ and $V\in\mathcal{R}_{d,\alpha}$.

######  Theorem 4.8.

Let $\alpha\in(0,2)$, $d\geq 1$, $m\geq 0$ and assume that $\alpha<d<2\alpha$.
Let the potential
$V\in\mathcal{R}_{d,\alpha}\cap(L^{1}(\mathbb{R}^{d})+L^{2}(\mathbb{R}^{d}))$.
Denote by $V_{-}$ its negative part and by $N(V)$ the number of the negative
eigenvalues of $H$. Then there exists $\lambda_{0}>0$ such that

$N(V)\leq
C_{d,\alpha}^{2}\left(1+\frac{me^{-\lambda_{0}/m}}{\lambda_{0}}\right)^{2}\|V_{-}\|_{\mathcal{R}_{d,\alpha}}^{2}$
(4.5)

holds (the quantity in the brackets is set to $1$ and is independent of
$\lambda_{0}$ if $m=0$), where $C_{d,\alpha}>0$ is same constant as in Lemma
6.10. In particular, the discrete spectrum of $H$ is finite for both the
massless and massive cases.

###### Proof.

The argument is based on the Birman-Schwinger principle, for details see [21,
Sect. 4.9.7] and [24, Th. XIII.10]. We may assume that $V\leq 0$, otherwise
apply the proof to the negative part of $V$. Let
$H(\gamma)=L_{m,\alpha}+\gamma V$ with $\gamma>0$. By Theorem 4.7 we have
${\operatorname{Spec}}_{\rm ess}(H(\gamma))=[0,\infty)$. If
$-\lambda\in{\operatorname{Spec}}_{\rm p}(H(\gamma))$, that is,
$H(\gamma)\phi=-\lambda\phi$ for $\phi\in L^{2}(\mathbb{R}^{d})$ holds, then
we have

$(\lambda+L_{m,\alpha})^{-1}(-V)\phi=\frac{1}{\gamma}\phi.$

This implies that $1/\gamma\in{\operatorname{Spec}}_{\rm p}(K_{\lambda}^{m})$
and $\sqrt{-V}\phi$ is the corresponding eigenfunction. Denote by
$\mu_{n}(\gamma)$ the $n$th negative eigenvalue of $H(\gamma)$, counting
multiplicity. Since the negative eigenvalues of $H$ make a discrete set, we
can choose $\lambda_{0}>0$ sufficiently small so that
$\mu_{1}(1)<-\lambda_{0}<0$ holds. Denote

$N(V)=\\#\left\\{n\in\mathbb{N}:\mu_{n}(1)<-\lambda_{0}\right\\}.$

In a similar way as for classical Schrödinger operators, see [24, Lem.
XIII.3.C], it can be shown that $\mu_{n}(\gamma)$ is monotone decreasing and
continuous in $\gamma$. For every $1\leq n\leq N(V)$ the relations
$\mu_{n}(0)=0$ and $\mu_{n}(1)<-\lambda_{0}$ imply that there exists a
$0<\gamma_{n}<1$ such that $\mu_{n}(\gamma_{n})=-\lambda_{0}$. We thus compute

$N(V)=\\#\left\\{n\in\mathbb{N}:\mu_{n}(\gamma_{n})=-\lambda_{0}\right\\}\leq\sum_{n=1}^{N(V)}\frac{1}{\gamma_{n}^{2}}\leq\sum_{\nu\in{\operatorname{Spec}}_{\rm
p}(K_{\lambda_{0}})}\nu^{2}=\|K^{m}_{\lambda_{0}}\|_{\rm HS}^{2}.$ (4.6)

By (4.2) and (4.6) we obtain (4.5). ∎

By the same proof we can show the following result for extended fractional
Rollnik class.

######  Theorem 4.9.

Let $\alpha\in(0,2)$, $d\geq 1$, $m\geq 0$ and assume that $d<2\alpha$. Let
the potential
$V\in\overline{\mathcal{R}}^{0}_{d,\alpha}\cap(L^{1}(\mathbb{R}^{d})+L^{2}(\mathbb{R}^{d}))$.
Denote by $V_{-}$ its negative part and by $N(V)$ the number of the negative
eigenvalues of $H$ for $m\geq 0$. Then there exists $\lambda_{0}>0$ such that

$N(V)\leq
C_{d,\alpha,\lambda_{0},m}^{2}[V_{-}]_{\overline{\mathcal{R}}^{0}_{d,\alpha}}^{2}$

holds, where $C_{d,\alpha,\lambda_{0},m}>0$ is the same constant as in
Corollary 6.13. In particular, the discrete spectrum of $H$ is finite for both
the massless and massive operators.

In case $m>0$ and $V\in\overline{\mathcal{R}}^{m}_{d,\alpha}$, we cannot
directly provide a bound on the number of negative eigenvalues, however, we
can prove that they are only finitely many.

######  Theorem 4.10.

Let $\alpha\in(0,2)$, $d\geq 1$, $m>0$ and assume that $d<2\alpha$. Let
$V\in\overline{\mathcal{R}}^{0}_{d,\alpha}\cap(L^{1}(\mathbb{R}^{d})+L^{2}(\mathbb{R}^{d}))$.
Denote by $V_{-}$ its negative part and by $N_{m}(V)$ the number of
eigenvalues of $H$, counting multiplicity, that are less then $-m$. Then there
exists $\lambda_{0}>0$ such that

$N_{m}(V)\leq
C_{d,\alpha,\lambda_{0},m}^{2}[V_{-}]_{\overline{\mathcal{R}}^{m}_{d,\alpha}}^{2}$

holds, where $C_{d,\alpha,\lambda_{0},m}>0$ is the same constant as in
Corollary 6.15. In particular, the discrete spectrum of $H$ is finite.

###### Proof.

By a similar argument to the proof of Theorem 4.8, we may assume that $V\leq
0$ and define $H(\gamma)=L_{m,\alpha}+\gamma V$ with $\gamma>0$, so that if
$-\lambda\in{\rm Spec}_{p}(H(\gamma))$ then
$1/\gamma\in\operatorname{Spec}_{p}(K_{\lambda})$. Since $\gamma
V\in\overline{\mathcal{R}}^{m}_{d,\alpha}$, by Theorem 4.7 we have
$\operatorname{Spec}_{\rm ess}(H(\gamma))=[0,\infty)$. Let $N(V)$ be the
number of negative eigenvalues of $H$ (counting multiplicity). For fixed
$\gamma>0$, the sequence $(\mu_{n}(\gamma))_{0\leq n\leq N(V)}$ of negative
eigenvalues of $H(\gamma)$, counting multiplicity, where
$\mu_{n}(\gamma)=-\infty$ if $n>N(V)$, is monotone, decreasing and continuous
in $\gamma$. Furthermore, $\operatorname{Spec}_{p}(H)$ is continuous, hence
there is a finite number of eigenvalues $\mu_{n}(1)$ lying in the interval
$[-m,0]$. Thus in case $N_{m}(V)=0$, the statement follows. In the contrary
case $N_{m}(V)>0$ let

$n_{\star}:=\min\\{n\geq 1:\ \mu_{n}(1)<-m\\}.$

Hence for all $n\geq n_{\star}$ we have $\mu_{n}(1)<-m$. Choose
$\lambda_{0}>m$ such that for all $n\geq n_{\star}$ we have
$\mu_{n}(1)\leq\mu_{n_{\star}}(1)<-\lambda_{0}$. On the other hand,
$\mu_{n}(0)=0$ thus due to the fact that $\mu_{n}(\gamma)$ is continuous in
$\gamma$, there exists a sequence $(\gamma_{n})_{n_{\star}\leq n\leq
n_{\star}+N_{m}(V)}$ such that $\mu_{n}(\gamma_{n})=-\lambda_{0}$ and
$\gamma_{n}\in(0,1)$. In particular, $1/\gamma_{n}\in\operatorname{Spec}_{\rm
p}(K_{\lambda_{0}}^{m})$ for all $n_{\star}\leq n\leq n_{\star}+N_{m}(V)$ and

$N_{m}(V)=\\#\\{n\in\mathbb{N}:\
\mu_{n+n_{\star}}(\gamma_{n+n_{\star}})=-\lambda_{0}\\}\leq\sum_{n=1}^{N_{m}(V)}\frac{1}{\gamma_{n+n_{\star}}^{2}}\leq\sum_{\nu\in\operatorname{Spec}_{\rm
p}(K_{\lambda_{0}}^{m})}\nu^{2}=\left\|K_{\lambda_{0}}^{m}\right\|_{\rm
HS}^{2}$

holds. Since $\lambda_{0}>m$, by part (3) of Theorem 4.4 it follows that
$K_{\lambda_{0}}^{m}$ is a Hilbert-Schmidt operator, and the same result then
completes the proof. ∎

####  4.2.3. Non-existence of negative and zero eigenvalues

It is known, see [33], [21, Th. 4.308], that whenever the semigroup
$\\{e^{-tL_{m,\alpha}}:t\geq 0\\}$ is recurrent, the operator
$H=L_{m,\alpha}+V$ has at least one bound state whenever $V$ is a non-positive
potential. However, when $\\{e^{-tL_{m,\alpha}}:t\geq 0\\}$ is transient, a
bound state exists only when the potential is strong enough in a specific
sense. By making use of the extended Birman-Schwinger principle, we can show
that for too small Rollnik-norm there exist no bound states and as a bonus we
can also rule out embedded eigenvalues at zero. We will use the following
result, for a proof see [21, Cor. 4.150, Cor. 4.151, Prop. 4.309].

######  Theorem 4.11.

Let $\alpha<d$. Suppose that $V\leq 0$ is form-compact with respect to
$L_{m,\alpha}$ and the operator norm $\|K^{m}_{0}\|<1$. Then $H$ has no non-
positive eigenvalues, in particular, it has no bound states at negative
eigenvalues and zero is not an embedded eigenvalue.

In what follows, we derive suitable upper estimates on the norm of $K_{0}^{m}$
which will permit to use this observation. We will separately consider the
cases $m=0$ and $m>0$, which will require different treatment.

First we note that the map
$\lambda\in[0,\infty)\mapsto\left\|K^{m}_{\lambda}\right\|_{\mathscr{B}(L^{2}(\mathbb{R}^{d}))}$
is monotone and continuous.

######  Lemma 4.12.

Let $m\geq 0$ and suppose that
$K^{m}_{\lambda_{0}}\in\mathscr{B}(L^{2}(\mathbb{R}^{d}))$ for some
$\lambda_{0}>0$. Then $K^{m}_{\lambda}\in\mathscr{B}(L^{2}(\mathbb{R}^{d}))$
for all $\lambda\geq\lambda_{0}$ and
$\left\|K^{m}_{\lambda}\right\|_{\mathscr{B}(L^{2}(\mathbb{R}^{d}))}\leq\left\|K^{m}_{\lambda_{0}}\right\|_{\mathscr{B}(L^{2}(\mathbb{R}^{d}))}$.
Furthermore, if
$\left\|K^{m}_{\lambda_{0}}\right\|_{\mathscr{B}(L^{2}(\mathbb{R}^{d}))}<\infty$,
then
$\lim_{\lambda\downarrow\lambda_{0}}\left\|K^{m}_{\lambda}\right\|_{\mathscr{B}(L^{2}(\mathbb{R}^{d}))}=\left\|K^{m}_{\lambda_{0}}\right\|_{{\mathscr{B}}(L^{2})}$.

###### Proof.

Since $K_{\lambda}$ is positivity preserving, we have

$\left\|K_{\lambda}\right\|_{\mathscr{B}(L^{2}(\mathbb{R}^{d}))}=\sup_{\begin{subarray}{c}f\in
L^{2}(\mathbb{R}^{d})\\\ f\geq 0,\ f\not\equiv
0\end{subarray}}\frac{\left\|K_{\lambda}f\right\|_{2}}{\left\|f\right\|_{2}}.$
(4.7)

Indeed, choosing $C$ to be the right-hand side of (4.7), for all $f\in
L^{2}(\mathbb{R}^{d})$ with positive and negative parts $f^{+},f^{-}$
respectively, we get

$\left\|K_{\lambda}f\right\|_{2}\leq\sqrt{\left\|K_{\lambda}f^{+}\right\|_{2}^{2}+\left\|K_{\lambda}f^{-}\right\|_{2}^{2}}\leq
C\sqrt{\left\|f^{+}\right\|_{2}^{2}+\left\|f^{-}\right\|_{2}^{2}}=C\left\|f\right\|_{2}.$

Once this is established, the statement clearly follows by the fact that
$0\leq R_{\lambda}(x)\leq R_{\lambda_{0}}(x)$. By the monotone convergence
theorem, the same relation guarantees that
$\left\|K_{\lambda}f\right\|_{2}\to\left\|K_{\lambda_{0}}f\right\|_{2}$ for
any $f\in L^{2}(\mathbb{R}^{d})$ with $f\geq 0$, and so
$\lim_{\lambda\to\lambda_{0}}\left\|K_{\lambda}\right\|_{\mathscr{B}(L^{2}(\mathbb{R}^{d}))}\leq\left\|K_{\lambda_{0}}\right\|_{\mathscr{B}(L^{2}(\mathbb{R}^{d}))}$.
To prove that in fact equality holds, for every $\varepsilon>0$ consider
$f_{\varepsilon}\in L^{2}(\mathbb{R}^{d})$ with $f_{\varepsilon}\geq 0$ such
that

$\left|\frac{\left\|K_{\lambda_{0}}f_{\varepsilon}\right\|_{2}}{\left\|f_{\varepsilon}\right\|_{2}}-\left\|K_{\lambda_{0}}\right\|_{\mathscr{B}(L^{2}(\mathbb{R}^{d}))}\right|<\varepsilon.$

There also exists $\delta>0$ such that if $0<\lambda-\lambda_{0}<\delta$ then

$\left|\frac{\left\|K_{\lambda}f_{\varepsilon}\right\|_{2}}{\left\|f_{\varepsilon}\right\|_{2}}-\frac{\left\|K_{\lambda_{0}}f_{\varepsilon}\right\|_{2}}{\left\|f_{\varepsilon}\right\|_{2}}\right|<\varepsilon.$

Then for all $\lambda_{0}<\lambda<\lambda_{0}+\delta$ it follows that

$\left|\frac{\left\|K_{\lambda}f_{\varepsilon}\right\|_{2}}{\left\|f_{\varepsilon}\right\|_{2}}-\left\|K_{\lambda_{0}}\right\|_{\mathscr{B}(L^{2}(\mathbb{R}^{d}))}\right|<2\varepsilon,$

which in turn implies

$\left\|K_{\lambda}\right\|_{\mathscr{B}(L^{2}(\mathbb{R}^{d}))}\geq\frac{\left\|K_{\lambda}f_{\varepsilon}\right\|_{2}}{\left\|f_{\varepsilon}\right\|_{2}}\geq\left\|K_{\lambda_{0}}\right\|_{\mathscr{B}(L^{2}(\mathbb{R}^{d}))}-2\varepsilon.$

Taking $\varepsilon\to 0$ the claim follows. ∎

Now we focus on estimating
$\left\|K_{0}^{m}\right\|_{\mathscr{B}(L^{2}(\mathbb{R}^{d}))}$. In fact, we
will provide a necessary and sufficient condition for $K_{0}^{m}$ to be a
Hilbert-Schmidt operator.

######  Theorem 4.13.

Let $\alpha\in(0,2)$ with $0<\alpha<d<2\alpha$ and $m\geq 0$.

  1. (1)

If $m=0$, then $K_{0}$ is a Hilbert-Schmidt operator if and only if
$V\in\mathcal{R}_{d,\alpha}$. Furthermore, there exists a constant
$C_{d,\alpha}>0$ such that

$\left\|K_{0}\right\|_{\rm
HS}=C_{d,\alpha}\left\|V\right\|_{\mathcal{R}_{d,\alpha}}$ (4.8)

for all $V\in\mathcal{R}_{d,\alpha}$. In particular, whenever $V\leq 0$ such
that

$\left\|V\right\|_{\mathcal{R}_{d,\alpha}}<\frac{1}{C_{d,\alpha}},$ (4.9)

the operator $H=L_{0,\alpha}+V$ has no bound states at negative eigenvalues
and zero is not an embedded eigenvalue.

  2. (2)

If $m>0$, let $d=3$. Then $K_{0}^{m}$ is a Hilbert-Schmidt operator if and
only if $V\in\mathcal{R}_{3,\alpha}\cap\mathcal{R}$ and there exist two
constants $C^{1}_{\alpha},C^{2}_{\alpha,m}>0$ such that

$C^{2}_{\alpha,m}\sqrt{\left\|V\right\|_{\mathcal{R}_{3,\alpha}}^{2}+\left\|V\right\|_{\mathcal{R}}^{2}}\leq\left\|K_{0}^{m}\right\|_{\rm
HS}\leq
C^{1}_{\alpha}\sqrt{\left\|V\right\|_{\mathcal{R}_{3,\alpha}}^{2}+m^{\frac{4}{\alpha}-2}\left\|V\right\|_{\mathcal{R}}^{2}}$
(4.10)

for all $V\in\mathcal{R}_{3,\alpha}\cap\mathcal{R}$ In particular, whenever
$V\leq 0$ is such that

$\sqrt{\left\|V\right\|_{\mathcal{R}_{3,\alpha}}^{2}+m^{\frac{4}{\alpha}-2}\left\|V\right\|_{\mathcal{R}}^{2}}<\frac{1}{C_{\alpha}},$

the operator $H=L_{m,\alpha}+V$ has no bound states at negative eigenvalues
and zero is not an embedded eigenvalue.

###### Proof.

To prove (1) observe that by (2.3)

$\iint_{\mathbb{R}^{2d}}K_{0}(x-y)^{2}dxdy=\iint_{\mathbb{R}^{2d}}R_{0}(x-y)^{2}|V(x)||V(y)|dxdy=C_{d,\alpha}^{2}\|V\|_{\mathcal{R}_{d,\alpha}}^{2}.$

Note that (4.8) and (4.9) directly imply that $V$ is form-compact with respect
to $L_{0,\alpha}$ with relative bound strictly less than one, thus by Theorem
4.11 the operator $H=L_{0,\alpha}+V$ has no bound states at negative
eigenvalues and zero is not an embedded eigenvalue. To get (2) first we show
that if $V\in\mathcal{R}_{3,\alpha}\cap\mathcal{R}$, then $K_{0}^{m}$ is a
Hilbert-Schmidt operator and that the second inequality in (4.10) holds. By
Lemma 6.7

$\displaystyle\iint_{\mathbb{R}^{6}}K^{m}_{0}(x-y)^{2}dxdy=\iint_{\mathbb{R}^{6}}R^{m}_{0}(x-y)^{2}|V(x)||V(y)|dxdy$
$\displaystyle\leq
C_{\alpha}\left(\iint_{\mathbb{R}^{6}}|x-y|^{-(6-2\alpha)}|V(x)||V(y)|dxdy+m^{\frac{4}{\alpha}-2}\iint_{\mathbb{R}^{6}}|x-y|^{-2}|V(x)||V(y)|dxdy\right)$
$\displaystyle=C_{\alpha}(\left\|V\right\|_{\mathcal{R}_{3,\alpha}}^{2}+m^{\frac{4}{\alpha}-2}\left\|V\right\|_{\mathcal{R}}^{2}).$

If $V\in\mathcal{R}_{3,\alpha}\cap\mathcal{R}$, we restate Lemma 6.8 for any
$\alpha\in(0,2)$, $m>0$ and $d\geq 3$ as follows

$R_{0}^{m}(x)\geq
C_{d,\alpha,m}\max\\{|x|^{-(d-\alpha)},|x|^{-(d-2)}\\}\geq\frac{C_{d,\alpha,m}}{2}(|x|^{-(d-\alpha)}+|x|^{-(d-2)}).$

Hence, for $d=3$ we get

$\displaystyle\iint_{\mathbb{R}^{6}}|K^{m}_{0}(x-y)|^{2}dxdy=\iint_{\mathbb{R}^{6}}R^{m}_{0}(x-y)^{2}|V(x)||V(y)|dxdy$
$\displaystyle\geq
C_{\alpha,m}\left(\iint_{\mathbb{R}^{6}}|x-y|^{-(6-2\alpha)}|V(x)||V(y)|dxdy+\iint_{\mathbb{R}^{6}}|x-y|^{-2}|V(x)||V(y)|dxdy\right)$
$\displaystyle=C_{\alpha,m}(\left\|V\right\|_{\mathcal{R}_{3,\alpha}}^{2}+\left\|V\right\|_{\mathcal{R}}^{2}).$

The remaining part of the claim follows in the same way as in part (1). ∎

##  5\. The case $\alpha=1$

In this section we study the $\alpha\to 1$ limit and, more generally, consider
$\alpha\to\frac{d}{2}$ for $d=1,2,3$, using the spaces defined in (3.7). As
seen in Proposition 3.10, this limit taken directly does not lead to a
sensible set of potentials. Here we show, however, that the limit of
Schrödinger operators with Rollnik-class potentials in $\Gamma$-convergence
sense behaves much better, and as a bonus we even obtain self-adjointness of
the limit operator.

Recall the Lévy jump measures (2.1)-(2.2) and let

$[u]_{m,\alpha}^{2}:=\frac{1}{2}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}|u(x+h)-u(x)|^{2}j_{m,\alpha}(|h|)dxdh$

be a Gagliardo-type seminorm on $L^{2}(\mathbb{R}^{d})$ and rewrite fractional
Sobolev space

$H^{\alpha/2}(\mathbb{R}^{d}):=\Big{\\{}u\in
L^{2}(\mathbb{R}^{d}):[u]_{0,\alpha}<\infty\Big{\\}}$

equivalently in its terms. For a given potential $V$, we define the quadratic
form associated to the operator formally written as $H=L_{m,\alpha}+V$ by

$\mathcal{A}_{m,\alpha}^{V}[u]:=[u]_{m,\alpha}+\int_{\mathbb{R}^{d}}V(x)u^{2}(x)dx,\quad
u\in L^{2}(\mathbb{R}^{d}).$

We denote
$\mathcal{D}(\mathcal{A}_{m,\alpha}^{V})=H^{\alpha/2}(\mathbb{R}^{d})\cap
L^{2}(|V|)$ and set $\mathcal{A}_{m,\alpha}^{V}[u]=\infty$ if
$u\not\in\mathcal{D}(\mathcal{A}_{m,\alpha}^{V})$. We will show self-
adjointness of the limit Schrödinger operator obtained on varying $\alpha$ by
means of the $\Gamma$-convergence of the associated quadratic forms.

First we recall the concept of $\Gamma$-convergence.

######  Definition 5.1.

Let $\mathscr{M}$ be a metric space, $(F_{n})_{n\in\mathbb{N}}$ a sequence of
functionals $F_{n}:\mathscr{M}\to\overline{\mathbb{R}}$, and
$F:\mathscr{M}\to\overline{\mathbb{R}}$ be another functional. We say that the
sequence is _$\Gamma$ -convergent_ to $F$ as $n\to\infty$, denoted
$F=\Gamma-\lim_{n\to\infty}F_{n}$, if

  1. (1)

for all sequences $(x_{n})_{n\in\mathbb{N}}\subset\mathscr{M}$ convergent to
$x\in\mathscr{M}$ the property $F(x)\leq\liminf_{n\to\infty}F_{n}(x_{n})$
holds

  2. (2)

for all $x\in\mathscr{M}$ there exists a sequence $(x_{n})_{n\in\mathbb{N}}\to
x$ in $\mathscr{M}$ such that $F(x)\geq\limsup_{n\to\infty}F_{n}(x_{n})$.

Furthermore, if $\mathscr{M}=L^{2}(\mathbb{R}^{d})$, we say that
$(F_{n})_{n\in\mathbb{N}}$ is _Mosco-convergent_ to $F$ as $n\to\infty$,
denoted $F={\rm M}-\lim_{n\to\infty}F_{n}$, if property (2) above holds and

  * $(1_{w})$

for all sequences $(x_{n})_{n\in\mathbb{N}}\subset L^{2}(\mathbb{R}^{d})$
weakly convergent to $x\in L^{2}(\mathbb{R}^{d})$ it follows that
$F(x)\leq\liminf_{n\to\infty}F_{n}(x_{n})$.

It is clear that if $F={\rm M}-\lim_{n\to\infty}F_{n}$, then also
$F=\Gamma-\lim_{n\to\infty}F_{n}$ in the strong topology of
$L^{2}(\mathbb{R}^{d})$.

We will make use of the following decomposition, obtained in [27, Lem. 2], see
also [2, Prop. 2.7]. For every $m>0$, $\alpha\in(0,2)$ and $d\geq 1$ there
exists a function $\Sigma_{m,\alpha}:\mathbb{R}^{d}\to\mathbb{R}^{+}$ with the
property that

$\int_{\mathbb{R}^{d}}\Sigma_{m,\alpha}(|x|)dx=m,$ (5.1)

such that

$j_{0,\alpha}(r)=j_{m,\alpha}(r)+\Sigma_{m,\alpha}(r).$ (5.2)

This leads to the following domination property.

######  Lemma 5.2.

Let $d\geq 1$, $m>0$ and $\alpha\in(0,2)$. Then

$[u]^{2}_{m,\alpha}\leq[u]^{2}_{0,\alpha}\leq[u]_{m,\alpha}^{2}+2m\left\|u\right\|_{2}^{2}$
(5.3)

for every $u\in L^{2}(\mathbb{R}^{d})$. In particular, if $u\in
H^{\alpha/2}(\mathbb{R}^{d})$, then $[u]_{m,\alpha}<\infty$ for all $m>0$.
Conversely, if $[u]_{m,\alpha}<\infty$ for some $m>0$, then $u\in
H^{\alpha/2}(\mathbb{R}^{d})$.

###### Proof.

The first bound in (5.3) is a straightforward consequence of (5.2) and the
fact that $\sigma_{m,\alpha}$ is positive. For the second we have

$\displaystyle[u]^{2}_{0,\alpha}$
$\displaystyle=\frac{1}{2}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}|u(x+h)-u(x)|^{2}j_{0,\alpha}(|h|)dxdh$
$\displaystyle\leq\frac{1}{2}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}|u(x+h)-u(x)|^{2}j_{m,\alpha}(|h|)dxdh+\frac{1}{2}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}|u(x+h)-u(x)|^{2}\Sigma_{m,\alpha}(|h|)dxdh$
$\displaystyle=:[u]^{2}_{m,\alpha}+I.$

Furthermore,

$\displaystyle I$
$\displaystyle\leq\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}(|u(y)|^{2}+|u(x)|^{2})\Sigma_{m,\alpha}(|x-y|)dxdy$
$\displaystyle=2\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}|u(y)|^{2}\Sigma_{m,\alpha}(|x-y|)dxdy=2\int_{\mathbb{R}^{d}}|u(y)|^{2}\left(\int_{\mathbb{R}^{d}}\sigma_{m,\alpha}(|x-y|)dx\right)dy=2m\left\|u\right\|_{2},$

where we used (5.1). ∎

Now we can prove our result on $\Gamma$-convergence yielding self-adjointness.

######  Theorem 5.3.

Let $d=1,2,3$, $m\geq 0$, and consider a potential
$V\in\overline{\mathcal{R}}^{m,p}_{d,\frac{d}{2}}$ for some $p\geq 4$ such
that $V_{-}\in L^{\infty}(\mathbb{R}^{d})$. Then the Schrödinger operator
$H=L_{m,\frac{d}{2}}+V$ is self-adjoint.

###### Proof.

By the definition of $\overline{\mathcal{R}}_{d,\frac{d}{2}}^{m,p}$ there
exists a sequence
$(\alpha_{k})_{k\in\mathbb{N}}\subset\left(\frac{d}{2},2\right)$ such that
$\alpha_{k}\downarrow\frac{d}{2}$ and
$V\in\overline{\mathcal{R}}_{d,\alpha_{k}}^{m}$. Consider the sequence of
Schrödinger operators $H_{k}:=L_{\alpha_{k},m}+V$. By Section 4.1 $H_{k}$ is
self-adjoint for all $k\in\mathbb{N}$. Hence we can consider the lower-
semicontinuous (in $L^{2}(\mathbb{R}^{d})$) quadratic forms
$\mathcal{A}_{m,\alpha_{k}}^{V}$. Define

$\displaystyle\mathcal{A}_{m,\frac{d}{2}}^{V}[u]:=[u]_{m,\alpha}+\int_{\mathbb{R}^{d}}V(x)u^{2}(x)dx,\quad
u\in L^{2}(\mathbb{R}^{d}).$

We show that
$\mathcal{A}_{m,\frac{d}{2}}^{V}=\Gamma-\lim_{k\to\infty}\mathcal{A}_{m,\alpha_{k}}^{V}$.
First note that $[\cdot]_{m,\frac{d}{2}}={\rm
M}-\lim_{k\to\infty}[\cdot]_{m,\alpha_{k}}$ by [6, Lem. 3.2]. We proceed in
two steps.

_Step 1._ We show that condition (1) in Definition 5.1 holds. Let $u_{k}\to u$
in $L^{2}(\mathbb{R}^{d})$. In case that
$\liminf_{k\to\infty}\mathcal{A}_{m,\alpha_{k}}^{V}[u_{k}]=\infty$, the
statement is trivial, thus we assume that
$\liminf_{k\to\infty}\mathcal{A}_{m,\alpha_{k}}^{V}[u_{k}]<\infty$. With no
loss of generality we assume that the sequence $u_{k}$ satisfies

$\displaystyle(1)\,\liminf_{k\to\infty}\mathcal{A}_{m,\alpha_{k}}^{V}[u_{k}]=\lim_{k\to\infty}\mathcal{A}_{m,\alpha_{k}}^{V}[u_{k}],$
$\displaystyle(2)\,\liminf_{k\to\infty}\,[u_{k}]_{m,\alpha_{k}}=\lim_{k\to\infty}[u_{k}]_{m,\alpha_{k}},$
$\displaystyle(3)\,\liminf_{k\to\infty}\int_{\mathbb{R}^{d}}(V(x)+C_{V})u_{k}^{2}(x)dx=\lim_{k\to\infty}\int_{\mathbb{R}^{d}}(V(x)+C_{V})u_{k}^{2}(x)dx,$
$\displaystyle(4)\,\lim_{k\to\infty}u_{k}(x)\to u(x)\ \mbox{ for a.e.
}x\in\mathbb{R}^{d},$

where $C_{V}:=\left\|V_{-}\right\|_{L^{\infty}}$. Note also that

$\lim_{k\to\infty}C_{V}\int_{\mathbb{R}^{d}}u_{k}^{2}(x)dx=C_{V}\int_{\mathbb{R}^{d}}u^{2}(x)dx.$

Then we have

$\lim_{k\to\infty}\mathcal{A}_{m,\alpha_{k}}^{V}[u_{k}]=\lim_{k\to\infty}[u_{k}]_{m,\alpha_{k}}+\lim_{k\to\infty}\int_{\mathbb{R}^{d}}(V(x)+C_{V})u_{k}^{2}(x)dx-
C_{V}\int_{\mathbb{R}^{d}}u^{2}(x)dx.$ (5.4)

By Mosco-convergence of $[\cdot]_{m,\alpha_{k}}$ to $[\cdot]_{m,\frac{d}{2}}$
it follows that

$\lim_{k\to\infty}[u_{k}]_{m,\alpha_{k}}\geq[u]_{m,\frac{d}{2}}.$

Furthermore, since $(V(x)+C_{V})u_{k}^{2}(x)\geq 0$, Fatou’s Lemma gives

$\lim_{k\to\infty}\int_{\mathbb{R}^{d}}(V(x)+C_{V})u_{k}^{2}(x)dx\geq\int_{\mathbb{R}^{d}}(V(x)+C_{V})u^{2}(x)dx.$

Hence, by (5.4) we get

$\displaystyle\lim_{k\to\infty}\mathcal{A}_{m,\alpha_{k}}^{V}[u_{k}]$
$\displaystyle=\lim_{k\to\infty}[u_{k}]_{m,\alpha_{k}}+\lim_{k\to\infty}\int_{\mathbb{R}^{d}}(V(x)+C_{V})u_{k}^{2}(x)dx-
C_{V}\int_{\mathbb{R}^{d}}u^{2}(x)dx$
$\displaystyle\geq[u]_{m,\alpha_{k}}+\int_{\mathbb{R}^{d}}(V(x)+C_{V})u^{2}(x)dx-
C_{V}\int_{\mathbb{R}^{d}}u^{2}(x)dx=\mathcal{A}_{m,\frac{d}{2}}^{V}[u].$

_Step 2._ Next we show that condition (2) in Definition 5.1 also holds. Let
$u\in L^{2}(\mathbb{R}^{d})$. If
$u\not\in\mathcal{D}(\mathcal{A}_{m,\frac{d}{2}}^{V})$, then the statement is
trivial with the sequence $u_{k}=u$ for all $k\in\mathbb{N}$. If
$u\in\mathcal{D}(\mathcal{A}_{m,\frac{d}{2}}^{V})$, then consider the sequence
$(u_{k})_{k\in\mathbb{N}}$ such that $u_{k}\to u$ strongly in
$L^{2}(\mathbb{R}^{d})$ and
$[u]_{m,\frac{d}{2}}=\lim_{k\to\infty}[u_{k}]_{m,\alpha_{k}}$. Without loss of
generality, we can assume that $u_{k}(x)\to u(x)$ for a.e.
$x\in\mathbb{R}^{d}$ and that there exists a constant $M_{1}$ (possibly
dependent on $m$, $d$ and $u$) such that
$\left\|u_{k}\right\|_{2}+[u_{k}]_{m,\alpha_{k}}\leq M_{1}$ for all
$k\in\mathbb{N}$. By (5.3) we have $[u_{k}]_{0,\alpha_{k}}^{2}\leq M_{2}$ for
some constant $M_{2}$. Thus we obtain

$\displaystyle[u_{k}]_{0,\frac{d}{2}}^{2}$
$\displaystyle=\frac{d2^{\frac{d}{2}-3}}{\pi^{\frac{d}{2}}}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}|u_{k}(x+h)-u_{k}(x)|^{2}|h|^{-\frac{3d}{2}}dhdx$
$\displaystyle=\frac{d2^{\frac{d}{2}-1}}{\pi^{\frac{d}{2}}}\int_{\mathbb{R}^{d}}\int_{B_{1}}|u_{k}(x+h)-u_{k}(x)|^{2}|h|^{-\frac{3d}{2}}dhdx$
$\displaystyle\qquad+\frac{d2^{\frac{d}{2}-1}}{\pi^{\frac{d}{2}}}\int_{\mathbb{R}^{d}}\int_{B_{1}^{c}}|u_{k}(x+h)-u_{k}(x)|^{2}|h|^{-\frac{3d}{2}}dhdx=I_{1}^{k}+I_{2}^{k}.$

For the first integral we have

$I_{1}^{k}\leq\frac{d2^{\frac{d}{2}-1}}{\pi^{\frac{d}{2}}}\int_{\mathbb{R}^{d}}\int_{B_{1}}|u_{k}(x+h)-u_{k}(x)|^{2}|h|^{-d-\alpha_{k}}dhdx\leq
C_{0}[u_{k}]_{0,\alpha_{k}}^{2},$

where

$C_{0}:=\max_{\alpha\in\left[\frac{d}{2},\alpha_{1}\right]}\frac{d2^{\frac{d}{2}-\alpha+1}\Gamma\left(1-\frac{\alpha}{2}\right)}{\alpha_{k}\Gamma\left(\frac{d+\alpha}{2}\right)}.$

For the second,

$I_{2}^{k}\leq\frac{\sigma_{d}2^{\frac{d}{2}+1}}{\pi^{\frac{d}{2}}}\left\|u_{k}\right\|_{2}^{2}.$

Hence there exists a constant $C_{1}>0$, independent of $k$, such that

$[u_{k}]_{0,\frac{d}{2}}\leq
C_{1}([u_{k}]_{0,\alpha_{k}}+\left\|u_{k}\right\|_{2})\leq M_{3},$

where $M_{3}>0$ is a suitable constant. By the Sobolev inequality (see [11,
Th. 6.5]) we have

$\left\|u_{k}\right\|_{q}\leq M_{4}(q)$

for every $q\in\left[2,\frac{8}{3}\right]$, where $M_{4}(q)>0$ is a suitable
constant. Note that $\frac{8}{3}$ is the critical Sobolev exponent in
$H^{\frac{d}{4}}(\mathbb{R}^{d})$. This further implies that

$\left\|u_{k}^{2}\right\|_{q}\leq M_{5}(q)$

for a suitable constant $M_{5}(q)$, for all $q\in\left[1,\frac{4}{3}\right]$.
Hence $\left\|u_{k}^{2}\right\|_{p^{\prime}}\leq M_{5}(p^{\prime})$ due to
$p\geq 4$. Without loss of generality, we can then assume that
$u_{k}^{2}\rightharpoonup u^{2}$ weakly in $L^{p^{\prime}}(\mathbb{R}^{d})$.
Then

$\lim_{k\to\infty}\int_{\mathbb{R}^{d}}V(x)u_{k}^{2}(x)dx=\int_{\mathbb{R}^{d}}V(x)u^{2}(x)dx$

and so

$\lim_{k\to\infty}\mathcal{A}_{m,\alpha_{k}}^{V}[u_{k}]=\lim_{k\to\infty}[u_{k}]_{m,\alpha_{k}}+\lim_{k\to\infty}\int_{\mathbb{R}^{d}}V(x)u_{k}^{2}(x)dx=[u]_{m,\frac{d}{2}}+\int_{\mathbb{R}^{d}}V(x)u^{2}(x)dx=\mathcal{A}_{m,\frac{d}{2}}^{V}[u].$

Hence
$\mathcal{A}^{V}_{m,\frac{d}{2}}=\Gamma-\lim_{k\to\infty}\mathcal{A}_{m,\alpha_{k}}^{V}$.
By [4, Prop. 1.28] this implies that $\mathcal{A}_{m,\frac{d}{2}}^{V}$ is
lower semi-continuous. However, $\mathcal{A}_{m,\frac{d}{2}}$ is the quadratic
form associated with the operator $H$, hence by [10, Th. 12.13] the operator
$H$ is self-adjoint. ∎

##  6\. Appendix: Proofs of resolvent estimates

In this section we prove the resolvent estimates stated in Section 2.3. We
treat the massless ($m=0$) and massive ($m>0$) cases separately.

###  6.1. Massless case

####  6.1.1. Proof of estimate (2.14) in the $m=0$ case

######  Proposition 6.1.

Let $\alpha\in(0,2)$ and $d\geq 1$. There exists $C_{d,\alpha}>0$ such that
for all $\lambda>0$ and $x\in\mathbb{R}^{d}\setminus\\{0\\}$

$R_{\lambda}(x)\;\leq\;C_{d,\alpha}\begin{cases}|x|^{\alpha-d},&\alpha<d,\\\\[10.0pt]
\log\left(1+\frac{1}{\lambda|x|}\right)+1,&\alpha=d=1,\\\\[10.0pt]
|x|^{\alpha-1}+\lambda^{\frac{1-\alpha}{\alpha}},&\alpha>d=1.\end{cases}$

Furthermore, if $\alpha<d$, then the statement holds also for $\lambda=0$.

###### Proof.

From (2.11) it follows that there exists a constant $c_{1}>0$ such that

$p_{t}(x)\leq c_{1}t|x|^{-d-\alpha}\;\;\mbox{if
$t<|x|^{\alpha}$}\quad\mbox{and}\quad p_{t}(x)\leq
c_{1}t^{-d/\alpha}\;\;\mbox{if $t>|x|^{\alpha}$}.$ (6.1)

We write

$R_{\lambda}(x)=\left(\int_{0}^{|x|^{\alpha}}+\int_{|x|^{\alpha}}^{\infty}\right)e^{-\lambda
t}p_{t}(x)dt=I_{1}+I_{2}.$

We first estimate $I_{1}$ as

$I_{1}\leq c_{1}|x|^{-d-\alpha}\int_{0}^{|x|^{\alpha}}te^{-\lambda t}dt\leq
c_{1}|x|^{-d-\alpha}\int_{0}^{|x|^{\alpha}}tdt=\frac{c_{1}}{2}|x|^{-d+\alpha}$

In estimating $I_{2}$ we have three cases to consider. If $\alpha<d$, we have

$I_{2}\leq c_{1}\int_{|x|^{\alpha}}^{\infty}t^{-d/\alpha}e^{-\lambda t}dt\leq
c_{1}\int_{|x|^{\alpha}}^{\infty}t^{-d/\alpha}dt=\frac{c_{1}}{\frac{d}{\alpha}-1}|x|^{-d+\alpha}.$

Note that the same estimate holds for $\lambda=0$ in this case. If $d=1$ and
$\alpha>d$, we get

$I_{2}\leq c_{1}\int_{|x|^{\alpha}}^{\infty}t^{-1/\alpha}e^{-\lambda t}dt\leq
c_{1}\lambda^{\frac{1-\alpha}{\alpha}}\int_{0}^{\infty}t^{-1/\alpha}e^{-t}dt=c_{1}\lambda^{\frac{1-\alpha}{\alpha}}\Gamma\Big{(}1-\frac{1}{\alpha}\Big{)}.$

In case $\alpha=d=1$ we use the exponential integral function

$E_{1}(x)=\int_{x}^{\infty}t^{-1}e^{-t}dt,\ x>0$

and recall the following bracketing property (see [1, Ch. V, eq. 5.1.20])

$\frac{e^{-x}}{2}\log\Big{(}1+\frac{2}{x}\Big{)}\leq E_{1}(x)\leq
e^{-x}\log\Big{(}1+\frac{1}{x}\Big{)}.$ (6.2)

Hence we obtain

$I_{2}\leq c_{1}\int_{|x|}^{\infty}t^{-1}e^{-\lambda t}dt\leq
c_{1}\int_{\lambda|x|}^{\infty}t^{-1}e^{-t}dt=c_{1}E_{1}(\lambda|x|)\leq
c_{1}\log\Big{(}1+\frac{1}{\lambda|x|}\Big{)}.$

∎

####  6.1.2. Proof of estimate (2.18)

Below we will make use of the inequality

$s^{p}e^{-s}\leq p^{p}e^{-p},$ (6.3)

for $0\leq s\leq p$. Beside Proposition 6.1, we can also prove a further
estimate that is sharper for large values of $|x|$.

######  Lemma 6.2.

Let $\alpha\in(0,2)$ and $d\geq 1$. Then there exists a positive constant
$C_{d,\alpha}$ such that for all $\lambda>0$ and
$x\in\mathbb{R}^{d}\setminus\\{0\\}$,

$R_{\lambda}(x)\leq\frac{C_{d,\alpha}}{\lambda^{2}|x|^{d+\alpha}}.$

###### Proof.

We give another estimate of $I_{1}$ and $I_{2}$ in the proof of Lemma 6.1.
Concerning $I_{1}$, we have

$I_{1}\leq c_{1}|x|^{-d-\alpha}\int_{0}^{|x|^{\alpha}}te^{-\lambda
t}dt=c_{1}|x|^{-d-\alpha}\left(-|x|^{\alpha}\frac{e^{-\lambda|x|^{\alpha}}}{\lambda}-\frac{e^{-\lambda|x|^{\alpha}}}{\lambda^{2}}+\frac{1}{\lambda^{2}}\right)\leq\frac{c_{1}}{\lambda^{2}|x|^{d+\alpha}}$

Next, observe that

$\displaystyle I_{2}\leq
c_{1}\int_{|x|^{\alpha}}^{\infty}t^{-d/\alpha}e^{-\lambda
t}dt=c_{2}\lambda^{\frac{d}{\alpha}-1}\int_{\lambda|x|^{\alpha}}^{\infty}s^{-d/\alpha}e^{-s}ds.$

By (6.3), we clearly have

$\displaystyle\begin{split}I_{2}&\leq\frac{2c_{1}}{e}\lambda^{\frac{d}{\alpha}-1}\int_{\lambda|x|^{\alpha}}^{\infty}s^{-d/\alpha-1}e^{-\frac{s}{2}}ds\leq\frac{2c_{1}}{e}e^{-\frac{\lambda|x|^{\alpha}}{2}}\lambda^{\frac{d}{\alpha}-1}\int_{\lambda|x|^{\alpha}}^{\infty}s^{-d/\alpha-1}ds\\\
&=\frac{2\alpha
c_{1}}{de}e^{-\frac{\lambda|x|^{\alpha}}{2}}\lambda^{-1}|x|^{-d}\leq\frac{4\alpha
c_{1}}{d\lambda^{2}e^{2}|x|^{d+\alpha}}.\end{split}$

∎

Combining Proposition 6.1 and Lemma 6.2 we get the following non-uniform upper
bound.

######  Proposition 6.3.

Let $\alpha\in(0,2)$ and $d\geq 1$. Then for all $\lambda>0$ there exists a
constant $C_{d,\alpha,\lambda}>0$ such that

$R_{\lambda}(x)\leq C_{d,\alpha,\lambda}H_{d,\alpha}^{0}(|x|),$

for all $x\in\mathbb{R}^{d}\setminus\\{0\\}$, where $H_{d,\alpha}^{0}$ is
defined in (2.18).

Proposition 6.3 corresponds to the upper bound in (2.18). We only have to
prove the lower bound.

######  Proposition 6.4.

Let $\alpha\in(0,2)$ and $d\geq 1$. Then for all $\lambda>0$ there exists
$C_{d,\alpha,\lambda}>0$ such that for all
$x\in\mathbb{R}^{d}\setminus\\{0\\}$

$R_{\lambda}(x)\geq C_{d,\alpha,\lambda}H_{d,\alpha}^{0}(x),$

where $H_{d,\alpha}^{0}$ is defined in (2.17).

###### Proof.

First we give an estimate for $|x|>1$. By (2.11) there exists a constant
$c_{2}$ such that

$p_{t}(x)\geq c_{2}t|x|^{-d-\alpha}\;\;\mbox{if
$t<|x|^{\alpha}$}\quad\mbox{and}\quad p_{t}(x)\geq
c_{2}t^{-d/\alpha}\;\;\mbox{if $t>|x|^{\alpha}$}.$ (6.4)

Using (6.4) we obtain

$\displaystyle R_{\lambda}(x)$
$\displaystyle\geq\int_{0}^{|x|^{\alpha}}e^{-\lambda t}p_{t}(x)dt\geq
c_{2}|x|^{-d-\alpha}\int_{0}^{|x|^{\alpha}}te^{-\lambda t}dt$
$\displaystyle\geq c_{2}|x|^{-d-\alpha}\int_{0}^{|x|^{\alpha}}te^{-\lambda
t}dt\geq\frac{c_{2}}{e^{\lambda}\lambda^{2}}|x|^{-d-\alpha}\left(e^{\lambda}-1-\lambda\right).$

Now let $|x|\leq 1$ and assume $\alpha<d$. Then we have, again by (6.1),

$\displaystyle R_{\lambda}(x)$ $\displaystyle\geq
c_{2}|x|^{-d-\alpha}\int_{0}^{|x|^{\alpha}}te^{-\lambda t}dt\geq
c_{2}|x|^{-d-\alpha}e^{-\lambda}\int_{0}^{|x|^{\alpha}}tdt=\frac{c_{2}}{2}e^{-\lambda}|x|^{-(d-\alpha)}.$

If $\alpha=d=1$ we get instead

$\displaystyle R_{\lambda}(x)$ $\displaystyle\geq
c_{2}\int_{|x|}^{\infty}t^{-1}e^{-\lambda
t}=c_{2}E_{1}(\lambda|x|)\geq\frac{c_{2}}{2}e^{-\lambda}\log\left(1+\frac{2}{\lambda|x|}\right)\geq
C_{1,1,\lambda}\log\left(1+\frac{1}{|x|}\right),$

where we also used (6.2). Finally, for $\alpha>d=1$, notice that

$R_{\lambda}(x)\geq
c_{2}\int_{|x|^{\alpha}}^{\infty}t^{-\frac{1}{\alpha}}e^{-\lambda t}dt\geq
c_{2}\int_{1}^{\infty}t^{-\frac{1}{\alpha}}e^{-\lambda t}dt.$

∎

###  6.2. Massive case

####  6.2.1. The function $\kappa_{m}$

Next we turn to the massive case. First we state some basic properties of the
function $\kappa_{m}$ defined in (2.13).

######  Lemma 6.5.

Let $\alpha\in(0,2)$ and $d\geq 1$.

  1. (1)

We have

$\lim_{m\downarrow
0}\kappa_{m}(r)=2^{\frac{d+\alpha}{2}-1}\Gamma\left(\frac{d+\alpha}{2}\right)r^{-\frac{d+\alpha}{2}}.$
(6.5)

  2. (2)

There exists a constant $C_{d,\alpha}$ such that

$\kappa_{m}(r)\leq C_{d,\alpha}r^{-\frac{d+\alpha}{2}}$ (6.6)

###### Proof.

Part (1) directly follows by (2.3). Consider part (2). If $r\leq 1$ the
expression

$c_{1,d,\alpha}:=\sup_{r\leq
1}r^{\frac{d+\alpha}{2}}K_{\frac{d+\alpha}{2}}(r),$

is finite by the second asymptotic relation in (2.3). Hence $\kappa_{m}(r)\leq