# Integral operators defined “up to a polynomial”

Serena Dipierro  University of Western Australia, Department of Mathematics
and Statistics, 35 Stirling Highway, WA6009 Crawley, Australia
<EMAIL_ADDRESS>,  Aleksandr Dzhugan  Università di Bologna,
Dipartimento di Matematica, Piazza di Porta San Donato 5, 40126 Bologna, Italy
<EMAIL_ADDRESS>and  Enrico Valdinoci  University of Western
Australia, Department of Mathematics and Statistics, 35 Stirling Highway,
WA6009 Crawley, Australia<EMAIL_ADDRESS>
###### Abstract.

We introduce a suitable notion of integral operators (comprising the
fractional Laplacian as a particular case) acting on functions with minimal
requirements at infinity. For these functions, the classical definition would
lead to divergent expressions, thus we replace it with an appropriate
framework obtained by a cut-off procedure. The notion obtained in this way
quotients out the polynomials which produce the divergent pattern once the
cut-off is removed.

We also present results of stability under the appropriate notion of
convergence and compatibility results between polynomials of different orders.
Additionally, we address the solvability of the Dirichlet problem.

The theory is developed in general in the pointwise sense. A viscosity
counterpart is also presented under the additional assumption that the
interaction kernel has a sign, in conformity with the maximum principle
structure.

###### Key words and phrases:

Integral operators, growth at infinity, fractional equations

###### 2020 Mathematics Subject Classification:

45H05, 26A33, 35R11

##  1\. Introduction

A classical line of investigation in mathematical analysis and mathematical
physics consists in the study of integro-differential operators. The
motivations for this stream of research come both from theoretical mathematics
(such as harmonic analysis, singular integral theory, fractional calculus,
etc.) and concrete problems in applied sciences (with questions related to
water waves, crystal dislocations, finance, optimization, minimal surfaces,
etc.): see e.g. the introduction in [BOOKCA] and the references therein for a
number of explicit motivations and examples.

A special focus of this stream of research deals with linear integro-
differential operators of the form

(1.1)
$Au(x)={\mbox{P.V.}}\int_{\mathbb{R}^{n}}(u(x)-u(y))K(x,y)\,dy=\lim_{\varepsilon\searrow
0}\int_{\mathbb{R}^{n}\setminus B_{\varepsilon}(x)}(u(x)-u(y))K(x,y)\,dy.$

As customary, the notation $B_{\varepsilon}(x)$ denotes the open ball of
radius $\varepsilon$ centered at the point $x$ (when $x$ is the origin, one
simply uses the notation $B_{\varepsilon}$ for short). The notation “P.V.”
above (which will be omitted in the rest of this paper for the sake of
simplicity) means “in the principal value sense” and takes into account
possible integral cancellations. The action of such operator is to “weight”
the oscillations of the function $u$ according to the kernel $K$. To make
sense of the expression above, two types of assumptions need to be accounted
for:

  * ✓

if the kernel $K$ is singular when $x=y$, the function $u$ needs to be regular
enough near the point $x$ (to allow integral cancellations and take advantage
of the principal value in (1.1)),

  * ✓

the function $u$ needs to be sufficiently well-behaved at infinity (namely,
its growth has to be balanced by the kernel $K$ to obtain in (1.1) a
convergent integral at infinity).

Roughly speaking, these two conditions correspond to the request that the
integral in (1.1) converges both in the vicinity of the given point $x$ and at
infinity. With respect to this, the regularity condition is necessary for the
local convergence of the integral and it is common to differential (rather
than integral) operators: in a sense, for differential problems the regularity
of $u$ ensures that incremental quotients converge to derivatives and,
somewhat similarly, for integral problems the regularity of $u$ allows the
increment inside the integral to compensate the possible singularity of the
kernel. Instead, the second assumption on the behavior of $u$ at infinity is
needed only to guarantee the tail convergence, it is a merely nonlocal feature
and has no counterpart for the case of differential operators.

Conditions “at infinity” are also technically more difficult to deal with.
First of all, they are more expensive to be computed, since they need to
account for virtually all the values of the given function (while regularity
ones deal with the values in an arbitrarily small region). Furthermore, these
conditions are typically lost after one analyzes the problem at a small scale
(since blow-up procedures alter the behavior of the solutions at infinity,
with the aim of detecting the local patterns). Moreover, it is sometimes
difficult to detect optimal assumptions for nonlocal problems even in very
basic and fundamental questions (see e.g. the open problem after Theorem 3.2
in [MR3916700]), hence any theory based only on “essential” assumptions is
doomed to have promising future developments.

It would be therefore very desirable to develop a theory of integral operators
that does not heavily rely on the conditions at infinity (in spite of the
striking fact that these conditions are needed even in the definition of the
operator itself!). To this end, a theory of “fractional Laplacian operators up
to polynomials” has been developed in [MR3988080, DSV19] to address the case
of functions with polynomial growth (see also [MR1421222] for related
approaches; see e.g. [MR2707618, MR2944369, MR3613319, MR3967804, MR4303657]
and the references therein for the basics on the fractional Laplace
operators). The gist of this method is to consider the family of cut-offs

(1.2) $\chi_{R}(x):=\begin{cases}1,&{\mbox{ if }}x\in B_{R};\\\
0,&\mbox{otherwise}\end{cases}$

and apply the operator to the function $\chi_{R}u$. Of course, in general, it
is not possible to send $R\to+\infty$, since the operator is not well-defined
on $u$, nevertheless it is still possible to perform such an operation once an
appropriate polynomial is “taken out” from the equation. Given the “rigidity”
of the space of polynomials (which is finite dimensional and easily
computable) the method is flexible and solid, it produces interesting results
and can be efficiently combined with blow-up procedures, see [MR3770173,
MR4038144].

The goal of this note is twofold: on the one hand, we review and extend the
theory developed in [MR3988080, DSV19], on the other hand, we generalize the
previous setting in order to include much more general classes of kernels (in
particular, kernels which are not necessarily scaling invariant). Besides its
interest in pure mathematics, this generalization has a concrete impact on the
study of interaction potentials of interatomic type arising in molecular
mechanics and materials science, such as the Morse potential [MORS]

(1.3) $K(x,y)=e^{-2(|x-y|-1)}-e^{-(|x-y|-1)},$

the Buckingham potential [BUCK]

(1.4) $K(x,y)=e^{-|x-y|}-\frac{1}{|x-y|^{6}},$

as well as their desingularized forms obtained by setting
$K_{\varepsilon}(x,y):=\min\left\\{\frac{1}{\varepsilon},\,K(x,y)\right\\}$.
Other classical potentials arising in probability and modelization include
also the Gauss kernel

(1.5) $K(x,y)=e^{-|x-y|^{2}},$

the Abel kernel

(1.6) $K(x,y)=e^{-|x-y|},$

the mollification kernel

(1.7) $K(x,y)=\begin{cases}e^{-\frac{1}{1-|x-y|^{2}}}&{\mbox{ if }}|x-y|<1,\\\
0&{\mbox{ if }}|x-y|\geqslant 1\end{cases}$

and the class of kernels comparable to that of the fractional Laplacian

(1.8) $\frac{\lambda}{|x-y|^{n+2s}}\leqslant
K(x,y)\leqslant\frac{\Lambda}{|x-y|^{n+2s}}$

for $s\in(0,1)$ and $\Lambda\geqslant\lambda>0$.

The theory of integral operators that we develop is broad enough to include
the kernels above (and others as well) into a unified setting. The operators
will be suitably defined “up to a polynomial”, in a sense that will be made
precise in Definition 2.3. This framework relies on a suitable decomposition
of the integral operator with respect to cut-off functions that is showcased
in Theorem 2.1. This setting is stable under the appropriate notion of
convergence, as it will be detailed in Proposition 2.11, and it presents nice
compatibility results between polynomials of different orders, as it will be
pointed out in Corollary 2.13 and Lemma 2.14. We also stress that the
generalized notion of operators that we deal with is “as good as the classical
one” in terms of producing solutions for the associated Dirichlet problem:
indeed, as it will be clarified in Theorem 3.1, the solvability of the
classical Dirichlet problem in the class of functions with nice behavior at
infinity is sufficient to ensure the solvability of the generalized Dirichlet
problem for the operator defined “up to a polynomial”.

To develop this theory, we mainly focused on the case of sufficiently smooth
(though not necessarily well-behaved at infinity) functions. This choice was
dictated by three main reasons. First of all, we aimed at developing the core
of the theory by focusing on its essential features, rather than complicating
it by additional difficulties. Moreover, we intended to split the
complications arising from the possible lack of smoothness of the solutions
with those produced by their behavior at infinity, consistently with the
initial discussion presented right after (1.1). Additionally, we stress that
the generality of kernels addressed by our theory goes far beyond the ones of
“elliptic” type, therefore a comprehensive regularity theory does not hold in
such an extensive framework.

However, one can also recast our theory in terms of viscosity solutions. For
this, since viscosity theory relates to maximum principles, one needs the
additional assumption that the kernel has a sign. In particular, in this
context one can obtain a viscosity definition of operators “up to a
polynomial” (see Definition 4.2) and discuss its stability properties under
uniform convergence (see Lemma 4.7) and the consistency properties with
respect of polynomials of different degree (see Corollary 4.8 and Lemma 4.9).
When the structure is compatible with both settings, the pointwise framework
and the viscosity one are essentially equivalent (see Lemma 4.6). Furthermore,
for kernels comparable with that of the fractional Laplacian a complete
solvability of the Dirichlet problem can be obtained (see Theorem 4.13).

In the forthcoming Section 2 we introduce the main definitions for integral
operators “up to a polynomial” and present their fundamental properties. The
corresponding Dirichlet problem will be discussed in Section 3.

While Sections 2 and 3 focus on the pointwise definition of this generalized
notion of operators, we devote Section 4 to the corresponding viscosity
theory.

##  2\. Definitions and main properties of operators “up to a polynomial”

The mathematical setting in which we work is the following. For every
$\vartheta\in[0,2]$, we define ${\mathcal{C}}_{\vartheta}$ as the set of
functions $u\in L^{1}_{{\rm{loc}}}(\mathbb{R}^{n})$ such that

(2.1) $u\in\begin{cases}C(B_{4})\cap L^{\infty}(B_{4})&{\mbox{ if
}}\vartheta=0,\\\ C^{\vartheta}(B_{4})&{\mbox{ if }}\vartheta\in(0,1),\\\
C^{0,1}(B_{4})&{\mbox{ if }}\vartheta=1,\\\ C^{1,\vartheta-1}(B_{4})&{\mbox{
if }}\vartheta\in(1,2),\\\ C^{1,1}(B_{4})&{\mbox{ if
}}\vartheta=2.\end{cases}$

Furthermore, for all $m\in\mathbb{N}_{0}$ and all $\vartheta\in[0,2]$, we
introduce ${\mathcal{K}}_{m,\vartheta}$ as the space of kernels $K=K(x,y)$
such that111As customary, in this paper we denote by $\Omega^{c}$ the
complementary set $\mathbb{R}^{n}\setminus\Omega$ for a given
$\Omega\subseteq\mathbb{R}^{n}$.

(2.2) for all $y\in B_{3}^{c}$, the map $x\in B_{1}\mapsto K(x,y)$ is
$C^{m}(B_{1})$ (2.3) and
$\displaystyle\int_{\mathbb{R}^{n}}\min\\{|x-y|^{\vartheta},1\\}\,|K(x,y)|\,dy<+\infty\quad{\mbox{
for all $x\in B_{1}$.}}$

If $\vartheta\in(1,2]$ we require additionally that every
$K\in{\mathcal{K}}_{m,\vartheta}$ satisfies222We observe that these
assumptions are satisfied by the kernel in (1.3) for every $n$, by the kernel
in (1.4) with $n=5$ and by all the corresponding desingularized kernels for
every $n$. The kernels in (1.5), (1.6) and (1.7) also satisfy these
assumptions for every $n$. The kernel in (1.8) satisfies (2.3) for every $n$
and every $\vartheta\in(2s,2]$.

(2.4) $K(x,x+z)=K(x,x-z)\quad{\mbox{ for all\leavevmode\nobreak\ $x$, $z\in
B_{1}$}}.$

Given $K\in{\mathcal{K}}_{m,\vartheta}$, we consider the space
${\mathcal{C}}_{\vartheta,K}$ of all the functions
$u\in{\mathcal{C}}_{\vartheta}$ for which

(2.5) $\displaystyle\sum_{|\alpha|\leqslant m-1}\int_{B_{R}\setminus
B_{3}}|u(y)|\,|\partial^{\alpha}_{x}K(x,y)|\,dy<+\infty\quad{\mbox{ for
all\leavevmode\nobreak\ $R>3$ and $x\in B_{1}$}}$ (2.6) and
$\displaystyle\int_{B^{c}_{3}}|u(y)|\sup_{{|\alpha|=m}\atop{x\in
B_{1}}}|\partial^{\alpha}_{x}K(x,y)|\,dy<+\infty.$

In this setting, we have the following results that show the role played by a
cut-off function in the computation of the operator in (1.1) on functions in
${\mathcal{C}}_{\vartheta,K}$:

######  Theorem 2.1.

Let333In this paper, we use the notation $\mathbb{N}_{0}$ to denote the set of
natural numbers including zero, that is $\\{0,1,2,3,\dots\\}$. The notation
$\mathbb{N}$ is instead reserved for the strictly positive natural numbers
$\\{1,2,3,\dots\\}$, therefore $\mathbb{N}_{0}=\mathbb{N}\cup 0$. For $m=0$,
the result in Theorem 2.1 holds true, simply by taking $P_{u,\tau}:=0$ and
$\psi(x,y):=K(x,y)$. $m\in\mathbb{N}_{0}$, $\vartheta\in[0,2]$,
$K\in{\mathcal{K}}_{m,\vartheta}$ and $u\in{\mathcal{C}}_{\vartheta,K}$. Let
also $\tau:\mathbb{R}^{n}\to[0,1]$ be compactly supported and such that
$\tau=1$ in $B_{3}$.

Then, there exist a function $f_{u,\tau}:\mathbb{R}^{n}\to\mathbb{R}$ and a
polynomial $P_{u,\tau}$ of degree at most $m-1$ such that

(2.7) $A(\tau u)=P_{u,\tau}+f_{u,\tau}$

in $B_{1}$. In addition, $f_{u,\tau}$ can be written in the following form:
there exists $\psi:B_{1}\times B_{3}^{c}\to\mathbb{R}$, with444Notice that the
right hand side of (2.8) may be infinite and in this case (2.8) is obviously
true. On the contrary, if the right hand side of (2.8) is finite, then the
quantity on the left hand side of (2.8) is bounded as well.

(2.8) $\sup_{x\in B_{1}}|\partial^{\gamma}_{x}\psi(x,y)|\leqslant
C\,\sup_{{x\in B_{1}}\atop{m\leqslant|\eta|\leqslant
m+|\gamma|}}|\partial^{\eta}_{x}K(x,y)|,$

for every $\gamma\in{\mathbb{N}}^{n}$ and for a suitable constant $C>0$
depending on $m$, $n$ and $|\gamma|$, such that

(2.9) $f_{u,\tau}=f_{1,u}+f_{2,u}+f^{*}_{u,\tau},$

where

(2.10)
$\begin{split}&f_{1,u}(x):=\int_{B_{3}}\big{(}u(x)-u(y)\big{)}K(x,y)\,dy,\\\
&f_{2,u}(x):=u(x)\int_{B^{c}_{3}}K(x,y)\,dy\\\ {\mbox{and
}}\qquad&f^{*}_{u,\tau}(x):=\int_{B^{c}_{3}}\tau(y)u(y)\psi(x,y)\,dy.\end{split}$

###### Proof.

We observe that

(2.11) $f_{1,u}$ and $f_{2,u}$ are well-defined and finite for every $x\in
B_{1}$.

To check this, we first consider the case in which $\vartheta\in[0,1]$. In
this case, for every $x\in B_{1}$ and $y\in B_{3}$,

$|u(x)-u(y)|\leqslant C|x-y|^{\vartheta},$

for some $C>0$, and thus

$\displaystyle|f_{1,u}(x)|\leqslant\int_{B_{3}}\big{|}u(x)-u(y)\big{|}\,|K(x,y)|\,dy\leqslant
C\int_{B_{3}}|x-y|^{\vartheta}\,|K(x,y)|\,dy$ $\displaystyle\qquad\leqslant
C\left(\int_{B_{3}\cap
B_{1}(x)}|x-y|^{\vartheta}\,|K(x,y)|\,dy+\int_{B_{3}\setminus
B_{1}(x)}|K(x,y)|\,dy\right),$

up to renaming $C>0$, and this shows that $f_{1,u}$ is well-defined and
finite, thanks to (2.3).

If instead $\vartheta\in(1,2]$ we claim that, since
$u\in{\mathcal{C}}_{\vartheta}$, there exists a constant $L>0$ such that for
all $|z|$ sufficiently small (say $z\in B_{1}$) we have that

(2.12) $|2u(x)-u(x+z)-u(x-z)|\leqslant L|z|^{\vartheta}.$

Indeed, in this case we know that $u\in C^{1,\vartheta-1}(B_{4})$ and thus

(2.13) $\begin{split}&|2u(x)-u(x+z)-u(x-z)|=|(u(x)-u(x+z))+(u(x)-u(x-z))|\\\
&\qquad=\left|-\int_{0}^{1}\nabla u(x+tz)\cdot z\,dt+\int_{0}^{1}\nabla
u(x-tz)\cdot z\,dt\right|\\\ &\qquad\leqslant\int_{0}^{1}|\nabla
u(x+tz)-\nabla u(x-tz)|\,|z|\,dt\\\ &\qquad\leqslant
C\int_{0}^{1}|x+tz-(x-tz)|^{\vartheta-1}|z|\,dt=C\int_{0}^{1}t^{\vartheta-1}|z|^{\vartheta-1}|z|\,dt=C|z|^{\vartheta},\end{split}$

up to relabeling $C$ at every step. This establishes (2.12).

Now, we notice that

(2.14)
$\begin{split}&f_{1,u}(x)=\int_{B_{3}}\big{(}u(x)-u(y)\big{)}K(x,y)\,dy\\\
&\qquad=\int_{B_{1}(x)}\big{(}u(x)-u(y)\big{)}K(x,y)\,dy+\int_{B_{3}\setminus
B_{1}(x)}\big{(}u(x)-u(y)\big{)}K(x,y)\,dy\\\
&\qquad=:I_{1}+I_{2}.\end{split}$

Using (2.4) and (2.12), we obtain that

$\displaystyle\left|\int_{B_{1}(x)}\big{(}u(x)-u(y)\big{)}K(x,y)\,dy\right|$
$\displaystyle\quad=\frac{1}{2}\left|\int_{B_{1}}\big{(}u(x)-u(x+z)\big{)}K(x,x+z)\,dz+\int_{B_{1}}\big{(}u(x)-u(x-z)\big{)}K(x,x-z)\,dz\right|$
$\displaystyle\quad=\frac{1}{2}\left|\int_{B_{1}}\big{(}2u(x)-u(x+z)-u(x-z)\big{)}K(x,x+z)\,dz\right|$
$\displaystyle\quad\leqslant\frac{1}{2}\int_{B_{1}}\big{|}2u(x)-u(x+z)-u(x-z)\big{|}\,|K(x,x+z)|\,dz$
$\displaystyle\quad\leqslant\frac{L}{2}\int_{B_{1}}|z|^{\vartheta}\,|K(x,x+z)|\,dz$
$\displaystyle\quad\leqslant\frac{L}{2}\int_{B_{1}}|z|^{\vartheta}\,|K(x,x+z)|\,dz.$

As a consequence,

(2.15)
$|I_{1}|\leqslant\frac{L}{2}\int_{B_{1}}|z|^{\vartheta}\,|K(x,x+z)|\,dz,$

which is finite, thanks to (2.3).

Furthermore,

$\displaystyle|I_{2}|\leqslant\int_{B_{3}\setminus
B_{1}(x)}\big{(}|u(x)|+|u(y)|\big{)}\,|K(x,y)|\,dy\leqslant
2\|u\|_{L^{\infty}(B_{4})}\int_{B_{3}\setminus B_{1}(x)}|K(x,y)|\,dy,$

which is finite, in light of (2.3). This, together with (2.14) and (2.15),
proves that $f_{1,u}$ is well-defined and finite in the case
$\vartheta\in(1,2]$.

Also, $f_{2,u}$ is well-defined and finite for every $\vartheta\in[0,2]$,
thanks to (2.3). These observations establish (2.11).

As a consequence, for any $x\in B_{1}$, we can write

(2.16) $\begin{split}&A(\tau u)(x)\\\ =\;&\int_{B_{3}}\big{(}(\tau u)(x)-(\tau
u)(y)\big{)}K(x,y)\,dy+\int_{B^{c}_{3}}\big{(}(\tau u)(x)-(\tau
u)(y)\big{)}K(x,y)\,dy\\\
=\;&\int_{B_{3}}\big{(}u(x)-u(y)\big{)}K(x,y)\,dy+u(x)\int_{B^{c}_{3}}K(x,y)\,dy-\int_{B^{c}_{3}}(\tau
u)(y)K(x,y)\,dy\\\ =\;&f_{1,u}(x)+f_{2,u}(x)-\int_{B^{c}_{3}}(\tau
u)(y)K(x,y)\,dy.\end{split}$

Now, in light of the assumption in (2.2), we are allowed to use Proposition
5.34 in [chierchia] (see also e.g. Theorem 4 on page 461 of [MR2033094]) and
we find that

$K(x,y)=\sum_{|\alpha|\leqslant
m-1}\partial^{\alpha}_{x}K(0,y)\frac{x^{\alpha}}{\alpha!}-\psi(x,y),$

where

(2.17)
$\psi(x,y):=-\sum_{|\alpha|=m}\frac{m\,x^{\alpha}}{\alpha!}\int_{0}^{1}(1-t)^{m-1}\partial^{\alpha}_{x}K(tx,y)\,dt.$

As a consequence,

(2.18) $\begin{split}&\int_{B^{c}_{3}}(\tau
u)(y)K(x,y)\,dy=\int_{B^{c}_{3}}(\tau u)(y)\left(\sum_{|\alpha|\leqslant
m-1}\partial^{\alpha}_{x}K(0,y)\frac{x^{\alpha}}{\alpha!}-\psi(x,y)\right)\,dy\\\
&\qquad=\sum_{|\alpha|\leqslant m-1}\left(\int_{B^{c}_{3}}(\tau
u)(y)\partial^{\alpha}_{x}K(0,y)\,dy\right)\frac{x^{\alpha}}{\alpha!}-\int_{B^{c}_{3}}(\tau
u)(y)\psi(x,y)\,dy.\end{split}$

Now, we set, for every $|\alpha|\leqslant m-1$,

(2.19) $\theta_{\tau,\alpha}:=\int_{B^{c}_{3}}(\tau
u)(y)\frac{\partial^{\alpha}_{x}K(0,y)}{\alpha!}\,dy.$

Suppose that the support of $\tau$ is contained in some ball $B_{R}$ with
$R>3$, and thus

(2.20) $|\theta_{\tau,\alpha}|\leqslant\int_{B_{R}\setminus B_{3}}\left|(\tau
u)(y)\frac{\partial^{\alpha}_{x}K(0,y)}{\alpha!}\right|\,dy\leqslant\frac{1}{\alpha!}\int_{B_{R}\setminus
B_{3}}|u(y)|\,\big{|}\partial^{\alpha}_{x}K(0,y)\big{|}\,dy.$

We stress that the coefficients $\theta_{\tau,\alpha}$ are well-defined,
thanks to (2.5).

Hence, setting

(2.21) $P_{u,\tau}(x):=-\sum_{|\alpha|\leqslant
m-1}\theta_{\tau,\alpha}x^{\alpha}$

we have that $P_{u,\tau}$ is a polynomial in $x$ of degree at most $m-1$.
Plugging this information into (2.18), we obtain that

$\int_{B^{c}_{3}}(\tau u)(y)K(x,y)\,dy=-P_{u,\tau}(x)-\int_{B^{c}_{3}}(\tau
u)(y)\psi(x,y)\,dy.$

Now, we notice that, for all $x\in B_{1}$ and all $y\in B_{3}^{c}$,

$|(\tau u)(y)\psi(x,y)|\leqslant C|u(y)|\sup_{{|\alpha|=m}\atop{z\in
B_{1}}}|\partial^{\alpha}_{x}K(z,y)|,$

for some $C>0$, possibly depending on $m$ and $n$. The last function lies in
$L^{1}(B_{3}^{c})$, thanks to (2.6), and therefore, recalling the definition
of $f^{*}_{u,\tau}$ in (2.10), we have that

(2.22) $f^{*}_{u,\tau}$ is well-defined and finite.

With this setting, we have that

$\int_{B^{c}_{3}}(\tau u)(y)K(x,y)\,dy=-P_{u,\tau}(x)-f^{*}_{u,\tau}(x),$

and therefore, plugging this information into (2.16) and recalling (2.9), we
obtain (2.7).

Hence, to complete the proof of Theorem 2.1, it remains to check (2.8). For
this, recalling the definition of $\psi$ in (2.17), we have that, for all
$x\in B_{1}$ and all $y\in B_{3}^{c}$,

$\displaystyle\partial^{\gamma}_{x}\psi(x,y)$ $\displaystyle=$
$\displaystyle\sum_{|\alpha|=m}\int_{0}^{1}c_{\alpha}(t)\,\partial^{\gamma}_{x}\left(x^{\alpha}\,\partial^{\alpha}_{x}K(tx,y)\right)\,dt$
$\displaystyle=$
$\displaystyle\sum_{|\alpha|=m}\int_{0}^{1}c_{\alpha}(t)\,\sum_{\beta\leqslant\gamma}\binom{\gamma}{\beta}\partial^{\beta}_{x}(x^{\alpha})\,\partial_{x}^{\gamma-\beta}(\partial^{\alpha}_{x}K(tx,y))\,dt$
$\displaystyle=$
$\displaystyle\sum_{|\alpha|=m}\int_{0}^{1}c_{\alpha}(t)\,\sum_{\beta\leqslant\gamma}\binom{\gamma}{\beta}\partial^{\beta}_{x}(x^{\alpha})\,t^{|\gamma-\beta|}\,\partial^{\alpha+\gamma-\beta}_{x}K(tx,y)\,dt,$

where $c_{\alpha}(t):=\frac{m}{\alpha!}(1-t)^{m-1}$. Here
$\beta\leqslant\gamma$ means that $\beta_{1}\leqslant\gamma_{1}$, $\cdots$,
$\beta_{n}\leqslant\gamma_{n}$ and we used the notation

$\binom{\gamma}{\beta}=\binom{\gamma_{1}}{\beta_{1}}\times\cdots\times\binom{\gamma_{n}}{\beta_{n}}.$

Hence,

$|\partial^{\gamma}_{x}\psi(x,y)|\leqslant C\sup_{{z\in
B_{1}}\atop{m\leqslant|\eta|\leqslant
m+|\gamma|}}|\partial_{x}^{\eta}K(z,y)|,$

for some $C>0$ depending on $m$, $n$ and $|\gamma|$. This establishes (2.8). ∎

######  Corollary 2.2.

Let $m\in\mathbb{N}_{0}$, $\vartheta\in[0,2]$,
$K\in{\mathcal{K}}_{m,\vartheta}$, $u\in\mathcal{C}_{\vartheta,K}$ and $R>3$.
Let $\tau_{R}:\mathbb{R}^{n}\to[0,1]$ be supported in $B_{R}$, with
$\tau_{R}=1$ in $B_{3}$, and such that

(2.23) $\lim_{R\to+\infty}\tau_{R}=1\mbox{ a.e. in }\mathbb{R}^{n}.$

Then, there exist a function $f_{u}:\mathbb{R}^{n}\to\mathbb{R}$ and a family
of polynomials $P_{u,\tau_{R}}$, which have degree at most $m-1$, such that

(2.24) $\lim_{R\to+\infty}[A(\tau_{R}u)(x)-P_{u,\tau_{R}}(x)]=f_{u}(x)$

for any $x\in B_{1}$. More precisely, we have that

(2.25) $f_{u}=f_{1,u}+f_{2,u}+f_{3,u},$

where $f_{1,u}$ and $f_{2,u}$ are as in (2.10) and

(2.26) $f_{3,u}(x):=\int_{B^{c}_{3}}u(y)\psi(x,y)\,dy.$

###### Proof.

We apply Theorem 2.1 with $\tau:=\tau_{R}$ for any fixed $R$, and then send
$R\to+\infty$. Indeed, by (2.8) (used here with $\gamma:=0$), for any $x\in
B_{1}$ and $y\in B_{3}^{c}$ we have

$\big{|}(\tau_{R}u)(y)\psi(x,y)\big{|}\leqslant
C|u(y)|\sup_{{|\eta|=m}\atop{z\in B_{1}}}|\partial^{\eta}_{x}K(z,y)|,$

for some $C>0$ and the latter function of $y$ lies in $L^{1}(B_{3}^{c})$,
thanks to (2.6).

Consequently, we use (2.10), (2.23) and the Dominated Convergence Theorem,
thus obtaining that

$\lim_{R\to+\infty}f^{*}_{u,\tau_{R}}=\lim_{R\to+\infty}\int_{B_{3}^{c}}(\tau_{R}u)(y)\psi(x,y)\,dy=\int_{B_{3}^{c}}u(y)\psi(x,y)\,dy=f_{3,u}(x).$

Accordingly, taking the limit in (2.7) we obtain (2.24). Also, the claims in
(2.25) and (2.26) follow555It is also interesting to point out that, when
$\tau_{R}:=\chi_{R}$, the limit in (2.24) is uniform for $x\in B_{1}$. Indeed,
by (2.7) and (2.8), for all $R_{2}>R_{1}>4$,  $\displaystyle\sup_{x\in
B_{1}}\Big{|}[A(\tau_{R_{1}}u)(x)-P_{u,\tau_{R_{1}}}(x)]-[A(\tau_{R_{2}}u)(x)-P_{u,\tau_{R_{2}}}(x)]\Big{|}$
$\displaystyle\leqslant$ $\displaystyle\sup_{x\in
B_{1}}\int_{B_{R_{2}}\setminus B_{R_{1}}}|u(y)|\,|\psi(x,y)|\,dy$
$\displaystyle\leqslant$ $\displaystyle C\int_{B_{R_{2}}\setminus
B_{R_{1}}}|u(y)|\sup_{{|\eta|=m}\atop{z\in
B_{1}}}|\partial^{\eta}_{x}K(z,y)|,$ which is as small as we wish, owing to
(2.6). This observation will be further expanded in Lemma 2.8. from (2.10). ∎

We are now ready to introduce the formal setting to deal with general
operators defined “up to a polynomial”:

######  Definition 2.3.

Let $m\in\mathbb{N}_{0}$, $\vartheta\in[0,2]$,
$K\in{\mathcal{K}}_{m,\vartheta}$, $u\in\mathcal{C}_{\vartheta,K}$ and
$f:B_{1}\to\mathbb{R}$ be bounded and continuous. We say that

$Au\stackrel{{\scriptstyle m}}{{=}}f\quad{\mbox{ in }}B_{1}$

if there exist a family of polynomials $P_{R}$, with $\deg P_{R}\leqslant
m-1$, and functions $f_{R}:B_{1}\to\mathbb{R}$ such that

(2.27) $A(\chi_{R}u)=f_{R}+P_{R}$

in $B_{1}$, with

(2.28) $\lim_{R\to+\infty}f_{R}(x)=f(x).$

######  Remark 2.4.

We observe that (2.27) is considered here in the pointwise sense. This is
possible, since the setting in (2.1) suffices for writing the equation
pointwise (recall (2.11) and (2.22)). A viscosity theory is also possible by
appropriate modifications of the setting (in particular, to pursue a viscosity
theory, to be consistent with the elliptic framework, one would need the
additional assumption that the kernel is nonnegative). For instance, for
fractional elliptic equations a viscosity approach is useful to establish
existence results by the Perron method, which combined with fractional
elliptic regularity theory for viscosity solutions often provides the
existence of nice solutions for the Dirichlet problem (see e.g. [MR2707618]).
The viscosity setting will be briefly discussed in Section 4.

######  Remark 2.5.

From Definition 2.3 one immediately sees that for all $j\in\mathbb{N}$ and
$K\in{\mathcal{K}}_{m,\vartheta}\cap{\mathcal{K}}_{m+j,\vartheta}$,

$\mbox{if }Au\stackrel{{\scriptstyle m}}{{=}}f,\mbox{ then
}Au\stackrel{{\scriptstyle m+j}}{{=}}f,$

in $B_{1}$, since polynomials of degree at most $m-1$ are also polynomials of
degree at most $m+j-1$.

######  Remark 2.6.

From Definition 2.3 and Corollary 2.2 (used here with $\tau_{R}:=\chi_{R}$, in
the notation of (1.2)), we can write $Au\stackrel{{\scriptstyle m}}{{=}}f_{u}$
in $B_{1}$ for any $K\in\mathcal{K}_{m,\vartheta}$ and
$u\in\mathcal{C}_{\vartheta,K}$.

######  Remark 2.7.

We observe that from Definition 2.3 it follows that any polynomial of degree
less than or equal to $m-1$ can be arbitrarily added to $f_{R}$ and subtracted
from $P_{R}$ in (2.27), hence, for any polynomial $P$ with $\deg P\leqslant
m-1$ we have that

$\mbox{if }Au\stackrel{{\scriptstyle m}}{{=}}f,\mbox{ then
}Au\stackrel{{\scriptstyle m}}{{=}}f+P$

in $B_{1}$.

We now investigate in further detail the convergence properties of the
approximating source term $f_{R}$.

######  Lemma 2.8.

Let $m\in\mathbb{N}_{0}$, $\vartheta\in[0,2]$ and
$K\in{\mathcal{K}}_{m,\vartheta}$. Let $u\in\mathcal{C}_{\vartheta,K}$, $f$
and $f_{R}$ be as in Definition 2.3.

Then, if $R^{\prime}>R>4$ we have that

(2.29)
$\inf\|f_{R^{\prime}}-f_{R}-P\|_{L^{\infty}(B_{1})}\leqslant\int_{B_{R}^{c}}|u(y)|\sup_{{|\alpha|=m}\atop{x\in
B_{1}}}|\partial^{\alpha}_{x}K(x,y)|\,dy,$

where the $\inf$ in (2.29) is taken over all the polynomials $P$ with degree
at most $m-1$.

###### Proof.

We define $v:=(1-\chi_{4})u$. In this way $v=0$ in $B_{4}$ and
$|v|\leqslant|u|$, so

(2.30) $v\in\mathcal{C}_{\vartheta,K}.$

Moreover, if $R>4$,

$(\chi_{R}-\chi_{4})u=(\chi_{R}-\chi_{4})v.$

Hence, from (2.27),

(2.31)
$A((\chi_{R}-\chi_{4})v)=A((\chi_{R}-\chi_{4})u)=f_{R}-f_{4}+P_{R}-P_{4}=f_{R}-f_{4}+\tilde{P}_{R},$

where $\tilde{P}_{R}:=P_{R}-P_{4}$ is a polynomial of degree at most $m-1$.

We also remark that, due to (2.30), we can use Theorem 2.1 here on the
function $v$. More specifically, using Theorem 2.1 on the function $v$ (twice,
once with $\tau:=\chi_{R}$ and once with $\tau:=\chi_{4}$), we obtain that

(2.32)
$\begin{split}A((\chi_{R}-\chi_{4})v)&=P_{v,\chi_{R}}-P_{v,\chi_{4}}+f_{v,\chi_{R}}-f_{v,\chi_{4}}\\\
&=\bar{P}_{v,\chi_{R}}+(f_{1,v}+f_{2,v}+f^{*}_{v,\chi_{R}})-(f_{1,v}+f_{2,v}+f^{*}_{v,\chi_{4}})\\\
&=\bar{P}_{v,\chi_{R}}+f^{*}_{v,\chi_{R}}-f^{*}_{v,\chi_{4}}\\\
&=\bar{P}_{v,\chi_{R}}+\int_{B_{R}\setminus
B_{4}}u(y)\psi(x,y)\,dy\end{split}$

in $B_{1}$, where $\bar{P}_{v,\chi_{R}}:=P_{v,\chi_{R}}-P_{v,\chi_{4}}$ is a
polynomial of degree at most $m-1$. Comparing the right hand sides of (2.31)
and (2.32), we obtain that in $B_{1}$

$f_{R}=f_{4}+P^{*}_{R}+\int_{B_{R}\setminus B_{3}}u(y)\psi(x,y)\,dy,$

where $P_{R}^{*}:=\bar{P}_{v,\chi_{R}}-\tilde{P}_{R}$ is a polynomial of
degree at most $m-1$.

Therefore, for any $R^{\prime}>R$,

$\displaystyle f_{R^{\prime}}-P^{*}_{R^{\prime}}-f_{R}+P^{*}_{R}$
$\displaystyle=$ $\displaystyle\left(f_{4}+\int_{B_{R^{\prime}}\setminus
B_{3}}u(y)\psi(x,y)\,dy\right)-\left(f_{4}+\int_{B_{R}\setminus
B_{3}}u(y)\psi(x,y)\,dy\right)$ $\displaystyle=$
$\displaystyle\int_{B_{R^{\prime}}\setminus B_{R}}u(y)\psi(x,y)\,dy$

and, as a consequence,

(2.33)
$\|f_{R^{\prime}}-P^{*}_{R^{\prime}}-f_{R}+P^{*}_{R}\|_{L^{\infty}(B_{1})}=\|\Psi_{R^{\prime},R}\|_{L^{\infty}(B_{1})},$

where

$\Psi_{R^{\prime},R}(x):=\int_{B_{R^{\prime}}\setminus
B_{R}}u(y)\psi(x,y)\,dy.$

From (2.8) and Remark 2.6, we know that

$\begin{split}&\|\Psi_{R^{\prime},R}\|_{L^{\infty}(B_{1})}\leqslant\sup_{x\in
B_{1}}\int_{B_{R^{\prime}}\setminus B_{R}}|u(y)||\psi(x,y)|\,dy\\\
&\qquad\qquad\leqslant\int_{B_{R^{\prime}}\setminus
B_{R}}|u(y)|\sup_{{|\alpha|=m}\atop{x\in
B_{1}}}|\partial^{\alpha}_{x}K(x,y)|\,dy\leqslant\int_{B^{c}_{R}}|u(y)|\sup_{{|\alpha|=m}\atop{x\in
B_{1}}}|\partial^{\alpha}_{x}K(x,y)|\,dy.\end{split}$

This and (2.33) imply that

$\|f_{R^{\prime}}-P^{*}_{R^{\prime}}-f_{R}+P^{*}_{R}\|_{L^{\infty}(B_{1})}\leqslant\int_{B^{c}_{R}}|u(y)|\sup_{{|\alpha|=m}\atop{x\in
B_{1}}}|\partial^{\alpha}_{x}K(x,y)|\,dy,$

which gives (2.29). ∎

Next result deals with the stability of the equation under uniform convergence
(and this can be seen as an adaptation to our setting of the result contained
e.g. in Lemma 5 of [CS11]).

######  Lemma 2.9.

Let $\vartheta\in[0,2]$ and $K\in{\mathcal{K}}_{0,\vartheta}$. For every
$k\in\mathbb{N}$, let $u_{k}\in\mathcal{C}_{\vartheta,K}$ and $f_{k}$ be
bounded and continuous in $B_{1}$. Assume that

(2.34) $Au_{k}=f_{k}$

in $B_{1}$, that

(2.35) $f_{k}$ converges uniformly in $B_{1}$ to some function $f$ as
$k\to+\infty$,

that

$\begin{split}&{\mbox{$u_{k}$ converges in\leavevmode\nobreak\ $B_{4}$ to some
function\leavevmode\nobreak\ $u$ as\leavevmode\nobreak\ $k\to+\infty$}}\\\
&{\mbox{in the topology of }}\begin{cases}L^{\infty}(B_{4})&{\mbox{ if
}}\vartheta=0,\\\ C^{\vartheta}(B_{4})&{\mbox{ if }}\vartheta\in(0,1),\\\
C^{0,1}(B_{4})&{\mbox{ if }}\vartheta=1,\\\ C^{1,\vartheta-1}(B_{4})&{\mbox{
if }}\vartheta\in(1,2),\\\ C^{1,1}(B_{4})&{\mbox{ if
}}\vartheta=2.\end{cases}\end{split}$

and that666We observe that condition (2.36) cannot be dropped from Lemma 2.9.
Indeed, if $s\in(0,1)$ and  $\mathbb{R}\ni x\mapsto
u_{k}(x):=-\frac{\chi_{(k,k^{2})}(x)\,x^{2s}}{\log k}$ we have that $u_{k}\to
0=:u$ locally uniformly and that, for each $x\in(-1,1)$,
$\displaystyle\int_{\mathbb{R}}\frac{u_{k}(x)-u_{k}(y)}{|x-y|^{1+2s}}\,dy=\frac{1}{\log
k}\int_{k}^{k^{2}}\frac{y^{2s}}{(y-x)^{1+2s}}\,dy=:f_{k}(x).$ We stress that,
if $x\in(-1,1)$ and $y>k$,  $y-x\geqslant
y-1=\frac{k-1}{k}\,y+\frac{y}{k}-1\geqslant\frac{k-1}{k}\,y$ and
$y-x\leqslant y+1=\frac{k+1}{k}\,y-\frac{y}{k}+1\leqslant\frac{k+1}{k}\,y.$ As
a result, if $x\in(-1,1)$,
$f_{k}(x)\leqslant\frac{k^{1+2s}}{(k-1)^{1+2s}}\,\frac{1}{\log
k}\int_{k}^{k^{2}}\frac{y^{2s}}{y^{1+2s}}\,dy=\frac{k^{1+2s}}{(k-1)^{1+2s}}$
and  $f_{k}(x)\geqslant\frac{k^{1+2s}}{(k+1)^{1+2s}}\,\frac{1}{\log
k}\int_{k}^{k^{2}}\frac{y^{2s}}{y^{1+2s}}\,dy=\frac{k^{1+2s}}{(k+1)^{1+2s}},$
thus $f_{k}\to 1=:f$ uniformly in $(-1,1)$. This example shows that
$(-\Delta)^{s}u_{k}=f_{k}\to f=1\not=0=(-\Delta)^{s}u.$

(2.36) $\lim_{k\to+\infty}\int_{\mathbb{R}^{n}\setminus
B_{3}}\big{|}u(y)-u_{k}(y)\big{|}|K(x,y)|\,dy=0,$

for every $x\in B_{1}$.

Then,

$Au=f$

in $B_{1}$.

###### Proof.

Let $x_{0}\in B_{1}$ and $\rho>0$ such that $B_{\rho}(x_{0})\Subset B_{1}$. We
claim that

(2.37)
$\lim_{k\to+\infty}\int_{B_{\rho}(x_{0})}\big{(}u_{k}(x_{0})-u_{k}(y)\big{)}K(x_{0},y)\,dy=\int_{B_{\rho}(x_{0})}\big{(}u(x_{0})-u(y)\big{)}K(x_{0},y)\,dy.$

For this, we distinguish two cases. If $\vartheta\in[0,1]$, we observe that

$\displaystyle\left|\int_{B_{\rho}(x_{0})}\big{(}u_{k}(x_{0})-u_{k}(y)\big{)}K(x_{0},y)\,dy-\int_{B_{\rho}(x_{0})}\big{(}u(x_{0})-u(y)\big{)}K(x_{0},y)\,dy\right|$
$\displaystyle\leqslant$
$\displaystyle\int_{B_{\rho}(x_{0})}\big{|}(u_{k}-u)(x_{0})-(u_{k}-u)(y)\big{|}\,|K(x_{0},y)|\,dy$
$\displaystyle\leqslant$
$\displaystyle\|u_{k}-u\|_{C^{\vartheta}(B_{4})}\int_{B_{\rho}(x_{0})}|x_{0}-y|^{\vartheta}\,|K(x_{0},y)|\,dy$
$\displaystyle\leqslant$ $\displaystyle C\,\|u_{k}-u\|_{C^{\vartheta}(B_{4})}$

for some $C>0$, thanks to (2.3), and this proves (2.37) in this case.

Hence, to complete the proof of (2.37), we now assume that
$\vartheta\in(1,2]$. In this case, we recall (2.4) and we see that, for $k$
sufficiently large,

$\displaystyle\int_{B_{\rho}(x_{0})}\big{(}u_{k}(x_{0})-u_{k}(y)\big{)}K(x_{0},y)\,dy$
$\displaystyle=$
$\displaystyle\frac{1}{2}\int_{B_{\rho}}\big{(}u_{k}(x_{0})-u_{k}(x_{0}+z)\big{)}K(x_{0},x_{0}+z)\,dz+\frac{1}{2}\int_{B_{\rho}}\big{(}u_{k}(x_{0})-u_{k}(x_{0}-z)\big{)}K(x_{0},x_{0}-z)\,dz$
$\displaystyle=$
$\displaystyle\frac{1}{2}\int_{B_{\rho}}\big{(}2u_{k}(x_{0})-u_{k}(x_{0}+z)-u_{k}(x_{0}-z)\big{)}K(x_{0},x_{0}+z)\,dz,$

and a similar computation holds with $u$ instead of $u_{k}$. Consequently,
recalling also (LABEL:siw74b8437654693) (used here with $u_{k}-u$ in place of
$u$),

$\displaystyle\left|\int_{B_{\rho}(x_{0})}\big{(}u_{k}(x_{0})-u_{k}(y)\big{)}K(x_{0},y)\,dy-\int_{B_{\rho}(x_{0})}\big{(}u(x_{0})-u(y)\big{)}K(x_{0},y)\,dy\right|$
$\displaystyle\leqslant$
$\displaystyle\frac{1}{2}\int_{B_{\rho}}\big{|}2(u_{k}-u)(x_{0})-(u_{k}-u)(x_{0}+z)-(u_{k}-u)(x_{0}-z)\big{|}\,|K(x_{0},x_{0}+z)|\,dz$
$\displaystyle\leqslant$
$\displaystyle\frac{\|u_{k}-u\|_{C^{1,\vartheta-1}(B_{4})}}{2}\int_{B_{\rho}}|z|^{\vartheta}\,|K(x_{0},x_{0}+z)|\,dz$
$\displaystyle\leqslant$ $\displaystyle
C\,\|u_{k}-u\|_{C^{1,\vartheta-1}(B_{4})},$

for some $C>0$, thanks to (2.3), and this completes the proof of (2.37).

We now claim that

(2.38) $\lim_{k\to+\infty}\int_{B_{3}\setminus
B_{\rho}(x_{0})}\big{(}u_{k}(x_{0})-u_{k}(y)\big{)}K(x_{0},y)\,dy=\int_{B_{3}\setminus
B_{\rho}(x_{0})}\big{(}u(x_{0})-u(y)\big{)}K(x_{0},y)\,dy.$

To prove it, we use (2.3) to conclude that

$\displaystyle\left|\int_{B_{3}\setminus
B_{\rho}(x_{0})}\big{(}u_{k}(x_{0})-u_{k}(y)\big{)}K(x_{0},y)\,dy-\int_{B_{3}\setminus
B_{\rho}(x_{0})}\big{(}u(x_{0})-u(y)\big{)}K(x_{0},y)\,dy\right|$
$\displaystyle\leqslant$ $\displaystyle
C\,\|u_{k}-u\|_{L^{\infty}(B_{4})}\int_{B_{3}\setminus
B_{\rho}(x_{0})}|K(x_{0},y)|\,dy$ $\displaystyle\leqslant$ $\displaystyle
C\,\|u_{k}-u\|_{L^{\infty}(B_{4})}$

up to renaming $C>0$ from line to line, and this establishes (2.38).

Furthermore, using (2.3),

$\displaystyle\left|\int_{\mathbb{R}^{n}\setminus
B_{3}}\big{(}u_{k}(x_{0})-u_{k}(y)\big{)}K(x_{0},y)\,dy-\int_{\mathbb{R}^{n}\setminus
B_{3}}\big{(}u(x_{0})-u(y)\big{)}K(x_{0},y)\,dy\right|$
$\displaystyle\leqslant$ $\displaystyle\int_{\mathbb{R}^{n}\setminus
B_{3}}\big{|}u_{k}(x_{0})-u(x_{0})\big{|}\,|K(x_{0},y)|\,dy+\int_{\mathbb{R}^{n}\setminus
B_{3}}\big{|}u_{k}(y)-u(y)\big{|}|K(x_{0},y)|\,dy$ $\displaystyle\leqslant$
$\displaystyle
C\,\|u_{k}-u\|_{L^{\infty}(B_{4})}+\int_{\mathbb{R}^{n}\setminus
B_{3}}\big{|}u_{k}(y)-u(y)\big{|}|K(x_{0},y)|\,dy,$

which is infinitesimal thanks to (2.36).

Gathering together this, (2.37) and (2.38), we conclude that $Au_{k}(x_{0})\to
Au(x_{0})$ as $k\to+\infty$. From this, (2.34) and (2.35) we obtain the
desired result. ∎

A natural question is whether the stability result in Lemma 2.9 carries over
directly to the setting introduced in Definition 2.3. The answer is in general
negative, as pointed out by the following counterexample:

######  Proposition 2.10.

Let $k\in\mathbb{N}$. Let

$u_{k}(x):=\begin{cases}0&{\mbox{ if }}x\in(-\infty,k],\\\ kx&{\mbox{ if
}}x\in(k,+\infty)\end{cases}$

and

$f_{k}(x):=\frac{kx}{k-x}+k\log\frac{k}{k-x}.$

Then,

(2.39) $\displaystyle\sqrt{-\Delta}u_{k}\stackrel{{\scriptstyle
1}}{{=}}f_{k}{\mbox{ in }}(-1,1),$ (2.40) $u_{k}$ converges to zero locally
uniformly, (2.41)
$\displaystyle\lim_{k\to+\infty}\int_{\mathbb{R}\setminus(-1,1)}\frac{|u_{k}(y)|}{|y|^{2+a}}\,dy=0\,{\mbox{
for all }}a>1,$ (2.42) $f_{k}(x)$ converges to $2x$ uniformly in $[-1,1]$.

###### Proof.

Let $R>k$ and $u_{k,R}(x):=u_{k}(x)\chi_{(-R,R)}(x)$. Then, if $x\in(-1,1)$,

$\displaystyle\int_{\mathbb{R}}\frac{u_{k,R}(x)-u_{k,R}(y)}{|x-y|^{2}}\,dy=k\int_{k}^{R}\frac{y}{(y-x)^{2}}\,dy=\frac{kx}{k-x}-\frac{kx}{R-x}+k\log(R-x)-k\log(k-x)$
$\displaystyle\qquad=\frac{kx}{k-x}-\frac{kx}{R-x}+k\log\frac{R-x}{R}+k\log
R+k\log\frac{k}{k-x}-k\log k$
$\displaystyle\qquad=f_{k}(x)+k\log\frac{R-x}{R}-\frac{kx}{R-x}+k\log R-k\log
k.$

Since the term $k\log R-k\log k$ is a constant in $x$ (hence a polynomial of
degree zero) and the function $k\log\frac{R-x}{R}-\frac{kx}{R-x}$ goes to zero
as $R\to+\infty$, the identity above proves (2.39).

Additionally, the claim in (2.40) is obvious and, if $a>1$,

$\displaystyle\lim_{k\to+\infty}\int_{\mathbb{R}\setminus(-1,1)}\frac{|u_{k}(y)|}{|y|^{2+a}}\,dy=\lim_{k\to+\infty}k\int_{k}^{+\infty}\frac{dy}{y^{1+a}}=\frac{1}{a}\lim_{k\to+\infty}\frac{1}{k^{a-1}}=0,$

thus establishing (2.41).

Furthermore, for every $x\in[-1,1]$, if $k$ is large enough,

$\displaystyle|f_{k}(x)-2x|\leqslant\left|\frac{kx}{k-x}-x\right|+\left|k\log\frac{k}{k-x}-x\right|=\left|\frac{x^{2}}{k-x}\right|+\left|k\int_{1}^{\frac{k}{k-x}}\frac{dt}{t}-x\right|$
$\displaystyle\qquad\leqslant\frac{1}{k-1}+\left|k\int_{1}^{1+\frac{x}{k-x}}\frac{dt}{t}-\frac{kx}{k-x}\right|+\left|\frac{kx}{k-x}-x\right|$
$\displaystyle\qquad\leqslant\frac{2}{k-1}+k\,\left|\int_{1}^{1+\frac{x}{k-x}}\frac{dt}{t}-\int_{1}^{1+\frac{x}{k-x}}dt\right|\leqslant\frac{2}{k-1}+k\,\int_{1-\frac{1}{k-1}}^{1+\frac{1}{k-1}}\frac{|1-t|}{t}\,dt$
$\displaystyle\qquad\leqslant\frac{6}{k-1},$

which gives (2.42).∎

Concerning the example in Proposition 2.10, notice in particular that, in
$(-1,1)$,

$\sqrt{-\Delta}u_{k}\stackrel{{\scriptstyle
1}}{{=}}f_{k}\stackrel{{\scriptstyle\mbox{{\tiny{$k\to+\infty$}}}}}{{-\\!\\!\\!-\\!\\!\\!-\\!\\!\\!\longrightarrow}}2x\stackrel{{\scriptstyle
1}}{{\neq}}0=\sqrt{-\Delta}0,$

showing that some care is necessary to pass Definition 2.3 to the limit and
additional assumptions are needed to exchange the order in which different
limits are taken.

From the positive side, as an affirmative counterpart of the counterexample in
Proposition 2.10, we provide the following stability result for the setting of
Definition 2.3:

######  Proposition 2.11.

Let $m\in\mathbb{N}_{0}$, $\vartheta\in[0,2]$ and
$K\in{\mathcal{K}}_{m,\vartheta}$. For every $k\in\mathbb{N}$, let
$u_{k}\in\mathcal{C}_{\vartheta,K}$ and $f_{k}$ be bounded and continuous in
$B_{1}$. Assume that

(2.43) $Au_{k}\stackrel{{\scriptstyle m}}{{=}}f_{k}\quad{\mbox{
in\leavevmode\nobreak\ $B_{1}$,}}$

that

$f_{k}$ converges uniformly in $B_{1}$ to some function $f$ as $k\to+\infty$,

that

(2.44) $\begin{split}&{\mbox{$u_{k}$ converges in\leavevmode\nobreak\ $B_{4}$
to some function\leavevmode\nobreak\ $u$ as\leavevmode\nobreak\
$k\to+\infty$}}\\\ &{\mbox{in the topology of
}}\begin{cases}L^{\infty}(B_{4})&{\mbox{ if }}\vartheta=0,\\\
C^{\vartheta}(B_{4})&{\mbox{ if }}\vartheta\in(0,1),\\\ C^{0,1}(B_{4})&{\mbox{
if }}\vartheta=1,\\\ C^{1,\vartheta-1}(B_{4})&{\mbox{ if
}}\vartheta\in(1,2),\\\ C^{1,1}(B_{4})&{\mbox{ if
}}\vartheta=2,\end{cases}\end{split}$

that

(2.45) $\lim_{k\to+\infty}\sup_{{x\in B_{1}}\atop{R>4}}\int_{B_{R}\setminus
B_{1}(x)}\big{(}(u-u_{k})(x)-(u-u_{k})(y)\big{)}K(x,y)\,dy=0$

and that

(2.46) $\lim_{R\to+\infty}\sup_{k\in\mathbb{N}}\int_{\mathbb{R}^{n}\setminus
B_{R}}|u_{k}(y)|\sup_{{|\eta|=m}\atop{z\in
B_{1}}}|\partial^{\eta}_{x}K(z,y)|\,dy=0.$

Then,

(2.47) $Au\stackrel{{\scriptstyle m}}{{=}}f\quad{\mbox{ in\leavevmode\nobreak\
$B_{1}$.}}$

To prove Proposition 2.11, we establish a uniqueness result in the spirit of
Lemma 1.2 of [MR3988080]:

######  Lemma 2.12.

Let $m\in\mathbb{N}_{0}$, $\vartheta\in[0,2]$,
$K\in{\mathcal{K}}_{m,\vartheta}$ and $u\in\mathcal{C}_{\vartheta,K}$. Let
$f_{1}$ and $f_{2}$ be bounded and continuous in $B_{1}$. Suppose that

(2.48) $Au\stackrel{{\scriptstyle m}}{{=}}f_{1}$ and
$Au\stackrel{{\scriptstyle m}}{{=}}f_{2}$ in $B_{1}$.

Then, there exists a polynomial of degree at most $m-1$ such that
$f_{1}-f_{2}=P$.

###### Proof.

In light of (2.48) and Definition 2.3, we have that there exist two families
of polynomials $P_{R}^{1}$ and $P^{2}_{R}$, with degree at most $m-1$, such
that, for every $x\in B_{1}$,

$\displaystyle\lim_{R\to+\infty}\big{(}A(\chi_{R}u)(x)-P_{R}^{1}(x)\big{)}=f_{1}(x)$
and
$\displaystyle\lim_{R\to+\infty}\big{(}A(\chi_{R}u)(x)-P_{R}^{2}(x)\big{)}=f_{2}(x).$

As a consequence, for every $x\in B_{1}$,

$\displaystyle
f_{1}(x)-f_{2}(x)=\lim_{R\to+\infty}\big{(}A(\chi_{R}u)(x)-P_{R}^{1}(x)\big{)}-\lim_{R\to+\infty}\big{(}A(\chi_{R}u)(x)-P_{R}^{2}(x)\big{)}$
$\displaystyle\qquad\qquad=\lim_{R\to+\infty}\big{(}P_{R}^{2}(x)-P_{R}^{1}(x)\big{)}.$

We remark that $P_{R}^{2}-P_{R}^{1}$ is a polynomial of degree at most $m-1$.
Accordingly, we can use Lemma 2.1 of [MR3988080] to conclude that
$f_{1}-f_{2}$ is a polynomial of degree at most $m-1$. This establishes the
desired result. ∎

###### Proof of Proposition 2.11.

We exploit the setting of Corollary 2.2 with $\tau_{R}:=\chi_{R}$. In this
way, for each $k$, we find a function $f_{u_{k}}:\mathbb{R}^{n}\to\mathbb{R}$
and a family of polynomials $P_{u_{k},R}$, which have degree at most $m-1$,
such that, in $B_{1}$,

(2.49) $\lim_{R\to+\infty}[A(\chi_{R}u_{k})(x)-P_{u_{k},R}(x)]=f_{u_{k}}(x).$

As a matter of fact (recall the footnote on page 5), we see that, for every
$x\in B_{1}$ and every $k\in\mathbb{N}$,

(2.50) $\Big{|}A(\chi_{R}u_{k})(x)-P_{u_{k},R}(x)-f_{u_{k}}(x)\Big{|}\leqslant
C\int_{\mathbb{R}^{n}\setminus B_{R}}|u_{k}(y)|\sup_{{|\eta|=m}\atop{z\in
B_{1}}}|\partial^{\eta}_{x}K(z,y)|\leqslant\frac{1}{R},$

as long as $R$ is sufficiently large, thanks to (2.46).

Comparing (2.49) with Definition 2.3, we thus conclude that
$Au_{k}\stackrel{{\scriptstyle m}}{{=}}f_{u_{k}}$ in $B_{1}$. This and (2.43),
together with the uniqueness result in Lemma 2.12, yield that
$f_{u_{k}}=f_{k}+\tilde{P}_{k}$ for a suitable polynomial $\tilde{P}_{k}$ of
degree at most $m-1$.

Consequently, setting $\tilde{P}_{u_{k},R}:=P_{u_{k},R}+\tilde{P}_{k}$, we
have that, by (2.50),

(2.51)
$\Big{|}A(\chi_{R}u_{k})(x)-\tilde{P}_{u_{k},R}(x)-f_{k}(x)\Big{|}\leqslant\frac{1}{R}.$

Now, we claim that, given $R>4$,

(2.52)
$\lim_{k\to+\infty}\sup_{B_{1}}\Big{|}A\big{(}(u-u_{k})\chi_{R}\big{)}\Big{|}=0.$

To this end, we calculate that, for each $x\in B_{1}$,

$\displaystyle\Big{|}A\big{(}(u-u_{k})\chi_{R}\big{)}(x)\Big{|}$
$\displaystyle=$
$\displaystyle\left|\int_{B_{1}(x)}\big{(}(u-u_{k})(x)-(u-u_{k})(y)\big{)}K(x,y)\,dy+\int_{B_{R}\setminus
B_{1}(x)}\big{(}(u-u_{k})(x)-(u-u_{k})(y)\big{)}K(x,y)\,dy\right.$
$\displaystyle\qquad\qquad\left.+\int_{\mathbb{R}^{n}\setminus
B_{R}}(u-u_{k})(x)\,K(x,y)\,dy\right|$ $\displaystyle\leqslant$
$\displaystyle\left|\int_{B_{1}(x)}\big{(}(u-u_{k})(x)-(u-u_{k})(y)\big{)}K(x,y)\,dy\right|+\left|\int_{B_{R}\setminus
B_{1}(x)}\big{(}(u-u_{k})(x)-(u-u_{k})(y)\big{)}K(x,y)\,dy\right|$
$\displaystyle\qquad\qquad+\|u-u_{k}\|_{L^{\infty}(B_{1})}\int_{\mathbb{R}^{n}\setminus
B_{R}}|K(x,y)|\,dy$

and hence (2.52) follows from (2.3), (2.37) (used here with $\rho:=1$; notice
that we can use (2.37) in this setting in light of (2.44)) and (2.45).

Thus, in light of (2.52), given $R>4$ we can find $k_{R}\in\mathbb{N}$ such
that

(2.53)
$\sup_{B_{1}}\Big{|}A\big{(}(u-u_{k_{R}})\chi_{R}\big{)}\Big{|}\leqslant\frac{1}{R}.$

We define $f_{R}:=f_{k_{R}}$ and $P_{R}:=\tilde{P}_{u_{k_{R}},R}$. We stress
that $P_{R}$ is a polynomial of degree at most $m-1$. Moreover, for every
$x\in B_{1}$,

$\displaystyle\Big{|}A(\chi_{R}u)(x)-f_{R}(x)-P_{R}(x)\Big{|}\leqslant\Big{|}A\big{(}(u-u_{k_{R}})\chi_{R}\big{)}(x)\Big{|}+\Big{|}A(\chi_{R}u_{k_{R}})(x)-f_{R}(x)-P_{R}(x)\Big{|}$
$\displaystyle\qquad\leqslant\frac{1}{R}+\Big{|}A(\chi_{R}u_{k_{R}})(x)-f_{k_{R}}(x)-\tilde{P}_{k_{R}}(x)\Big{|}\leqslant\frac{2}{R}$

thanks to (2.50) and (2.53), which proves (2.47). ∎

A consequence of Lemmata 2.8 and 2.9 is the following equivalence result:

######  Corollary 2.13.

Let $\vartheta\in[0,2]$, $K\in{\mathcal{K}}_{0,\vartheta}$ and
$u\in\mathcal{C}_{\vartheta,K}$. Let $f$ be bounded and continuous in $B_{1}$.

Then

$Au=f{\mbox{ in }}B_{1}$

is equivalent to

$Au\stackrel{{\scriptstyle 0}}{{=}}f\mbox{ in the sense of Definition
\ref{def}}.$

###### Proof.

Suppose that $Au=f$ in $B_{1}$. Then, for $R>10$,

(2.54)
$A(\chi_{R/2}u)(x)=Au(x)-A((1-\chi_{R/2})u)(x)=f(x)+\int_{\mathbb{R}^{n}}(1-\chi_{R/2}(y))u(y)K(x,y)\,dy$

for every $x\in B_{1}$. Now, we set

$w:=(\chi_{R}-\chi_{R/2})u.$

We observe that $w=0$ in $B_{4}$, so we can exploit Theorem 2.1 to $w$
(applied here with $m=0$) and get that, for any $x\in B_{1}$,

(2.55) $\begin{split}A&((\chi_{R}-\chi_{R/2})u)(x)=Aw(x)\\\
&=f_{1,w}+f_{2,w}+f^{*}_{w,\chi_{R}}(x)=\int_{B_{R}\setminus
B_{3}}w(y)\psi(x,y)\,dy=\int_{B_{R}\setminus
B_{R/2}}u(y)\psi(x,y)\,dy.\end{split}$

Hence, from (2.54) and (2.55), we find that

(2.56)
$\begin{split}A(\chi_{R}u)(x)&=A((\chi_{R}-\chi_{R/2})u)(x)+A(\chi_{R/2}u)(x)\\\
&=\int_{B_{R}\setminus
B_{R/2}}u(y)\psi(x,y)\,dy+f(x)+\int_{\mathbb{R}^{n}}(1-\chi_{R/2}(y))u(y)\psi(x,y)\,dy\\\
&=:f_{R}(x)\end{split}$

for every $x\in B_{1}$. We remark that $f_{R}\to f$ in $B_{1}$ as
$R\to+\infty$, thanks to (2.8) (used here with $m=0$ and
$\gamma=(0,\ldots,0)$) and (2.6).

Now we recall Definition 2.3 (here with $m=0$ and $P_{R}=0$) and we conclude
that $Au\stackrel{{\scriptstyle 0}}{{=}}f$ in $B_{1}$, as desired.

Conversely, we now suppose that $Au\stackrel{{\scriptstyle 0}}{{=}}f$ in
$B_{1}$. From Definition 2.3 and the fact that $m=0$, we have that $P_{R}$ is
identically zero, and so we can write that $A(\chi_{R}u)=f_{R}$ in $B_{1}$,
with $f_{R}\to f$ in $B_{1}$ as $R\to+\infty$. We observe that $\chi_{R}u$
approaches $u$ locally uniformly in $\mathbb{R}^{n}$. Also, we can use here
Lemma 2.8: in this way, we find that

$\|f_{R^{\prime}}-f_{R}\|_{L^{\infty}(B_{1})}\leqslant\int_{B_{R}^{c}}|u(y)|\sup_{x\in
B_{1}}|K(x,y)|\,dy.$

Therefore, we send $R^{\prime}\to+\infty$ and obtain that, for any $x\in
B_{1}$,

$|f(x)-f_{R}(x)|=\lim_{R^{\prime}\to+\infty}|f_{R^{\prime}}(x)-f_{R}(x)|\leqslant\lim_{R^{\prime}\to+\infty}\|f_{R^{\prime}}-f_{R}\|_{L^{\infty}(B_{1})}\leqslant\int_{B_{R}^{c}}|u(y)|\sup_{x\in
B_{1}}|K(x,y)|\,dy.$

As a consequence, recalling (2.6) (here with $m=0$) we have that $f_{R}$
converges to $f$ uniformly in $B_{1}$ as $R\to+\infty$. From this, we can
exploit Lemma 2.9 and conclude that $Au=f$, as desired. ∎

A natural question deals with the consistency of the operator setting for
functions that are sufficiently well-behaved to allow definitions related to
two different indices: roughly speaking, in the best possible scenario, if we
know that $Au\stackrel{{\scriptstyle m}}{{=}}f$ and $j\leqslant m$, can we say
that $Au\stackrel{{\scriptstyle j}}{{=}}f$? Posed like this, the answer to
this question is negative, since, after all, in light of Lemma 2.12, the
function $f$ is uniquely defined only “up to a polynomial”. Nevertheless, the
answer becomes positive if we take into account this additional polynomial
normalization, as stated in the next result:

######  Lemma 2.14.

Let $j,m\in\mathbb{N}_{0}$, with $j\leqslant m$, $\vartheta\in[0,2]$ and
$K\in\mathcal{K}_{j,\vartheta}\cap\mathcal{K}_{m,\vartheta}$. Let $f$ be
bounded and continuous in $B_{1}$ and let $u\in\mathcal{C}_{\vartheta,K}$ such
that

(2.57) $Au\stackrel{{\scriptstyle m}}{{=}}f$

in $B_{1}$.

Then, there exist a function $\bar{f}$ and a polynomial $P$ of degree at most
$m-1$, such that $\bar{f}=f+P$ and $Au\stackrel{{\scriptstyle j}}{{=}}\bar{f}$
in $B_{1}$.

###### Proof.

Let $v:=(1-\chi_{4})u$ and $w:=\chi_{4}u$. We notice that $v$,
$w\in\mathcal{C}_{\vartheta,K}$. Hence, since $K\in\mathcal{K}_{j,\vartheta}$,
recalling Remark 2.6, (2.10), (2.25) and (2.26), we can write that

$Av\stackrel{{\scriptstyle j}}{{=}}f_{v}=\int_{B_{4}^{c}}u(y)\psi(x,y)$

in $B_{1}$. That is, by Definition 2.3,

(2.58)
$A(\chi_{R}v)=\int_{B_{4}^{c}}u(y)\psi(x,y)\,dy+\tilde{\varphi}_{R}+Q_{R},$

for some $\tilde{\varphi}_{R}$ such that $\tilde{\varphi}_{R}\to 0$ in $B_{1}$
as $R\to+\infty$ and a polynomial $Q_{R}$ of degree at most $j-1$.

Furthermore, by (2.57), and recalling Definition 2.3, we get that

(2.59) $A(\chi_{R}u)=f+\varphi_{R}+P_{R},$

for some $\varphi_{R}$ such that $\varphi_{R}\to 0$ if $R\to+\infty$ and a
polynomial $P_{R}$ with deg$P_{R}\leqslant m-1$. Therefore, subtracting (2.58)
from (2.59), we obtain

(2.60)
$f+\varphi_{R}+P_{R}-\int_{B_{4}^{c}}u(y)\psi(x,y)\,dy-\tilde{\varphi}_{R}-Q_{R}=A(\chi_{R}(u-v))=A(\chi_{R}w).$

We notice that, for every $x\in B_{1}$ and every $R>4$,

$\displaystyle
A(\chi_{R}w)(x)=\int_{\mathbb{R}^{n}}(\chi_{R}w(x)-\chi_{R}w(y))\,K(x,y)\,dy$
$\displaystyle\qquad\quad=\int_{B_{R}}(w(x)-w(y))\,K(x,y)\,dy+\int_{\mathbb{R}^{n}\setminus
B_{R}}w(x)\,K(x,y)\,dy$
$\displaystyle\qquad\quad=\int_{B_{4}}(w(x)-w(y))\,K(x,y)\,dy+\int_{\mathbb{R}^{n}\setminus
B_{4}}w(x)\,K(x,y)\,dy$
$\displaystyle\qquad\quad=\int_{\mathbb{R}^{n}}(\chi_{4}w(x)-\chi_{4}w(y))\,K(x,y)\,dy=A(\chi_{4}w)(x).$

As a consequence of this and (2.60), we have that, in $B_{1}$,

$f+\varphi_{R}+P_{R}-\int_{B_{4}^{c}}u(y)\psi(x,y)\,dy-\tilde{\varphi}_{R}-Q_{R}=A(\chi_{4}w).$

This shows that the limit

$\lim_{R\to+\infty}(\varphi_{R}+P_{R}-\tilde{\varphi}_{R}-Q_{R})$

exists. As a result, the limit

$\lim_{R\to+\infty}(P_{R}-Q_{R})$

exists. Then, exploiting Lemma 2.1 in [MR3988080] we conclude that

$\lim_{R\to+\infty}(P_{R}-Q_{R})=P,$

for some polynomial $P$ of degree at most $m-1$.

Now we set $\bar{f}:=f+P$ and $S_{R}:=\varphi_{R}+P_{R}-Q_{R}-P$, and we see
that $S_{R}\to 0$ as $R\to+\infty$. Thus, from (2.59) we obtain that

$A(\chi_{R}u)=\bar{f}+S_{R}+Q_{R}$

in $B_{1}$. Since the degree of $Q_{R}$ is at most $j-1$, this shows that
$Au\stackrel{{\scriptstyle j}}{{=}}\bar{f}$ in $B_{1}$, as desired. ∎

##  3\. The Dirichlet problem

In this section we consider the existence problem for equations involving
general operators that are defined “up to a polynomial”. The main result is
the following:

######  Theorem 3.1.

Let $m\in\mathbb{N}_{0}$, $\vartheta\in[0,2]$,
$K\in\mathcal{K}_{0,\vartheta}\cap{\mathcal{K}}_{m,\vartheta}$ and
$u\in\mathcal{C}_{\vartheta,K}$. Assume that $u_{0}\in L^{1}_{\rm
loc}(B_{1}^{c})$ satisfies (2.5) and (2.6) and $f:B_{1}\to\mathbb{R}$ is
bounded and continuous in $B_{1}$.

Additionally, suppose that for any $\tilde{f}:B_{1}\to\mathbb{R}$ which is
bounded and continuous in $B_{1}$ and any $\tilde{u}_{0}\in L^{1}(B_{1}^{c})$
there exists a unique solution $\tilde{u}\in{\mathcal{C}}_{\vartheta}$ to the
Dirichlet problem

(3.1) $\begin{cases}A\tilde{u}=\tilde{f}&{\mbox{ in }}B_{1},\\\
\tilde{u}=\tilde{u}_{0}&{\mbox{ in }}B_{1}^{c}.\end{cases}$

Then, there exists a function $u\in\mathcal{C}_{\vartheta,K}$ such that

(3.2) $\begin{cases}Au\stackrel{{\scriptstyle m}}{{=}}f&{\mbox{ in }}B_{1},\\\
u=u_{0}&{\mbox{ in }}B_{1}^{c}.\end{cases}$

Also, the solution to (3.2) is not unique, since the space of solutions of
(3.2) has dimention $N_{m}$, with

(3.3) $N_{m}:=\sum_{j=0}^{m-1}\binom{j+n-1}{n-1}.$

###### Proof.

To begin with, we prove the existence of solutions for (3.2). To do this, we
define

$u_{1}:=\chi_{B_{4}^{c}}\,u_{0}\qquad{\mbox{ and
}}\qquad\tilde{u}_{0}:=\chi_{B_{4}\setminus B_{1}}\,u_{0}.$

Since $u_{1}$ vanishes in $B_{4}$ and $K\in\mathcal{K}_{m,\vartheta}$, we can
write $Au_{1}\stackrel{{\scriptstyle m}}{{=}}f_{u_{1}}$ in $B_{1}$, for some
function $f_{u_{1}}$, due to Remark 2.6.

We now consider the solution of (3.1) with $\tilde{f}:=f-f_{u_{1}}$.
Therefore, using Remark 2.5 and Corollary 2.13 we obtain that

$\begin{cases}A\tilde{u}\stackrel{{\scriptstyle m}}{{=}}\tilde{f}&{\mbox{ in
}}B_{1},\\\ \tilde{u}=\tilde{u}_{0}&{\mbox{ in }}B_{1}^{c}.\end{cases}$

Then, we set $u:=u_{1}+\tilde{u}$ and we get that
$Au=Au_{1}+A\tilde{u}\stackrel{{\scriptstyle m}}{{=}}f_{u_{1}}+\tilde{f}=f$ in
$B_{1}$. Moreover, we have that $u=u_{1}+\tilde{u}_{0}=u_{0}$ in $B_{1}^{c}$,
that is $u\in{\mathcal{C}}_{\vartheta,K}$ is solution of (3.2). This
establishes the existence of solution for (3.2).

Now we prove that solutions of (3.2) are not unique and determine the
dimension of the corresponding linear space. For this, we notice that for any
polynomial $P$ with $\deg P\leqslant m-1$ there exists a unique solution
$\tilde{u}_{P}\in{\mathcal{C}}_{\vartheta}$ of the problem

(3.4) $\begin{cases}A\tilde{u}_{P}=P&{\mbox{ in }}B_{1},\\\
\tilde{u}_{P}=0&{\mbox{ in }}B_{1}^{c},\end{cases}$

due to the existence and uniqueness assumption for (3.1). This is equivalent
to say that $A\tilde{u}_{P}\stackrel{{\scriptstyle 0}}{{=}}P$ in $B_{1}$,
thanks to Corollary 2.13. Using Remark 2.5, we obtain that
$A\tilde{u}_{P}\stackrel{{\scriptstyle m}}{{=}}P$ in $B_{1}$. Thus, applying
Remark 2.7, we obtain that $\tilde{u}_{P}$ is a solution of

(3.5) $\begin{cases}A\tilde{u}_{P}\stackrel{{\scriptstyle m}}{{=}}0&{\mbox{ in
}}B_{1},\\\ \tilde{u}_{P}=0&{\mbox{ in }}B_{1}^{c}.\end{cases}$

From this it follows that if $u$ is a solution of (3.2), then
$u+\tilde{u}_{P}$ is also a solution of (3.2).

Viceversa, if $u$ and $v$ are two solutions of (3.2), then $w:=u-v$ is a
solution of

$\begin{cases}Aw\stackrel{{\scriptstyle m}}{{=}}0&{\mbox{ in }}B_{1},\\\
w=0&{\mbox{ in }}B_{1}^{c}.\end{cases}$

Here we can apply Lemma 2.14 with $j:=0$ thus obtaining that
$Aw\stackrel{{\scriptstyle 0}}{{=}}P$ in $B_{1}$, where $P$ is a polynomial of
$\deg P\leqslant m-1$. Using again Corollary 2.13, one deduces that

(3.6) $\begin{cases}Aw=P&{\mbox{ in }}B_{1},\\\ w=0&{\mbox{ in
}}B_{1}^{c}.\end{cases}$

Therefore, the uniqueness of the solution of (3.6), confronted with (3.4),
gives us that $w=\tilde{u}_{P}$, and thus $v=u+\tilde{u}_{P}$.

This reasoning gives that the space of solutions of (3.2) is isomorphic to the
space of polynomials with degree less than or equal to $m-1$, which has
exactly dimension $N_{m}$, given by (3.3) (see e.g. [2021arXiv210107941D]). ∎

##  4\. A viscosity approach

Up to now, we focused our attention on the case of equations defined
pointwise. In principle, this requires functions that are “sufficiently
regular” for the equation to be satisfied at every given point. However, a
less restrictive approach adopted in the classical theory of elliptic
equations is to consider weaker notions of solutions (and possibly recover the
pointwise setting via an appropriate regularity theory): in this spirit, a
convenient setting, which is also useful in case of fully nonlinear equations,
is that of viscosity solutions, which does not require a high degree of
regularity of the solution itself since the equation is computed pointwise
only at smooth functions touching from either below or above (see e.g. [CC95]
for a thorough discussion on viscosity solutions).

In this section, we recast the setting of general operators defined “up to a
polynomial” into the viscosity solution framework. To this end, we proceed as
follows. For all $m\in\mathbb{N}_{0}$ and $\vartheta\in[0,2]$, we define
$\mathcal{K}_{m,\vartheta}^{+}$ as the space of kernels $K=K(x,y)$ verifying
(2.2), (2.3) and (2.4), and such that

(4.1) $K(x,y)\geqslant 0{\mbox{ for all $x\in B_{1}$ and
$y\in\mathbb{R}^{n}$}}.$

Given $K\in\mathcal{K}_{m,\vartheta}^{+}$, we consider the space
${\mathcal{V}}_{K}$ of all the functions $u\in
L^{1}_{\rm{loc}}(\mathbb{R}^{n})\cap C(B_{4})\cap L^{\infty}(B_{4})$ for which

(4.2) $\displaystyle\sum_{|\alpha|\leqslant m-1}\int_{B_{R}\setminus
B_{3}}|u(y)|\,|\partial^{\alpha}_{x}K(x,y)|\,dy<+\infty\quad{\mbox{ for
all\leavevmode\nobreak\ $R>3$ and $x\in B_{1}$}}$ (4.3) and
$\displaystyle\int_{B^{c}_{3}}|u(y)|\sup_{{|\alpha|=m}\atop{x\in
B_{1}}}|\partial^{\alpha}_{x}K(x,y)|\,dy<+\infty.$

######  Remark 4.1.

Notice that if (4.1) holds true and $K\in\mathcal{K}_{m,\vartheta}$ for some
$m\in\mathbb{N}_{0}$ and some $\vartheta\in[0,2]$, then
$K\in\mathcal{K}_{m,\vartheta}^{+}$.

In the viscosity framework we introduce the following definition.

######  Definition 4.2.

Let $m\in\mathbb{N}_{0}$, $\vartheta\in[0,2]$,
$K\in\mathcal{K}_{m,\vartheta}^{+}$, $u\in\mathcal{V}_{K}$ and
$f:B_{1}\to\mathbb{R}$ be bounded and continuous. We say that

$Au\stackrel{{\scriptstyle m}}{{=}}f\quad\mbox{in $B_{1}$ in the viscosity
sense}$

if there exist a family of polynomials $P_{R}$, with $\deg P_{R}\leqslant
m-1$, and bounded and continuous functions $f_{R}:B_{1}\to\mathbb{R}$ such
that

(4.4) $A(\chi_{R}u)=f_{R}+P_{R}$

in $B_{1}$ in the viscosity sense, with

(4.5) $\lim_{R\to+\infty}f_{R}(x)=f(x)\quad{\mbox{ uniformly in }}B_{1}.$

######  Remark 4.3.

We point out that the limit in (4.5) is assumed to hold uniformly (this is a
stronger assumption than the one in (2.28) that was assumed for the pointwise
setting, and it is taken here to make the setting compatible with the
viscosity method, see e.g. the proof of the forthcoming Corollary 4.8). See
also [MR4038144] for related observations.

######  Remark 4.4.

We observe that, for all $j\in\mathbb{N}$ and
$K\in\mathcal{K}_{m,\vartheta}^{+}\cap\mathcal{K}_{m+j,\vartheta}^{+}$,

$\mbox{if }Au\stackrel{{\scriptstyle m}}{{=}}f,\mbox{ then
}Au\stackrel{{\scriptstyle m+j}}{{=}}f$

in $B_{1}$ in the viscosity sense of Definition 4.2.

######  Remark 4.5.

From Definition 4.2 it follows that any polynomial of degree less than or
equal to $m-1$ can be arbitrarily added to $f_{R}$ and subtracted from $P_{R}$
in (4.4), hence, for any polynomial $P$ with $\deg P\leqslant m-1$ we have
that

$\mbox{if }Au\stackrel{{\scriptstyle m}}{{=}}f,\mbox{ then
}Au\stackrel{{\scriptstyle m}}{{=}}f+P$

in $B_{1}$ in the viscosity sense of Definition 4.2.

We now establish that when the structure is compatible with the both the
settings in Definitions 2.3 and 4.2, the pointwise and viscosity frameworks
are equivalent:

######  Lemma 4.6.

Let $m\in\mathbb{N}_{0}$, $\vartheta\in[0,2]$,
$K\in{\mathcal{K}}_{m,\vartheta}$, $u\in\mathcal{C}_{\vartheta,K}$ and
$f:B_{1}\to\mathbb{R}$ be bounded and continuous. If (4.1) holds true and $u$
is a solution of

$Au\stackrel{{\scriptstyle m}}{{=}}f\mbox{ in the sense of
Definition\leavevmode\nobreak\ \ref{def}},$

then $K\in\mathcal{K}_{m,\vartheta}^{+}$, $u\in\mathcal{V}_{K}$ and it is a
solution of

$Au\stackrel{{\scriptstyle m}}{{=}}f\mbox{ in the viscosity sense of
Definition\leavevmode\nobreak\ \ref{defv}}.$

Conversely, let $m\in\mathbb{N}_{0}$, $\vartheta\in[0,2]$,
$K\in\mathcal{K}_{m,\vartheta}^{+}$, $u\in\mathcal{V}_{K}$ and
$f:B_{1}\to\mathbb{R}$ be bounded and continuous. If
$K\in\mathcal{K}_{m,\vartheta}$ and $u\in{\mathcal{C}}_{\vartheta,K}$ is a
solution of

$Au\stackrel{{\scriptstyle m}}{{=}}f\mbox{ in the viscosity sense of
Definition\leavevmode\nobreak\ \ref{defv}},$

then $u$ is a solution of

$Au\stackrel{{\scriptstyle m}}{{=}}f\mbox{ in the sense of
Definition\leavevmode\nobreak\ \ref{def}}.$

###### Proof.

Assume that $u$ is a solution of $Au\stackrel{{\scriptstyle m}}{{=}}f\mbox{ in
$B_{1}$ in the sense of Definition\leavevmode\nobreak\ \ref{def}}$. It follows
that there exist a family of polynomials $P_{R}$ with $\deg P_{R}\leqslant
m-1$ and functions $f_{R}$ such that

(4.6) $A(\chi_{R}u)=f_{R}+P_{R}$

pointwise in $B_{1}$. Since $u\in\mathcal{C}_{\vartheta,K}$ we have that also
$u\in\mathcal{V}_{K}$. Moreover, we have that
$K\in\mathcal{K}_{m,\vartheta}^{+}$, due to (4.1) and Remark 4.1.

Now we observe that if $v\in{\mathcal{C}}_{\vartheta}$ and vanishes outside
$B_{2}$, $g$ is a bounded and continuous function, and $Av=g$ pointwise in
$B_{1}$, then also

(4.7) $Av=g$ in $B_{1}$ in the viscosity sense.

To check this, let $\varphi$ be a smooth function touching $v$ from below at
some point $x_{0}\in B_{1}$. Then, we have that $v(y)-\varphi(y)\geqslant
0=v(x_{0})-\varphi(x_{0})$ for all $y\in\mathbb{R}^{n}$ and therefore, by
(4.1),

$A\varphi(x_{0})=\int_{\mathbb{R}^{n}}(\varphi(x_{0})-\varphi(y))K(x_{0},y)\,dy\geqslant\int_{\mathbb{R}^{n}}(v(x_{0})-v(y))K(x_{0},y)\,dy=Av(x_{0})=g(x_{0}).$

Similarly, if $\varphi$ touches $v$ from above, one obtains the opposite
inequality, and these observations complete the proof of (4.7).

As a consequence of (4.6) and (4.7), we obtain that

$A(\chi_{R}u)=f_{R}+P_{R}$

in $B_{1}$ in the viscosity sense.

Hence, to finish the first part of the proof, we show that

(4.8) $f_{R}\to f$ uniformly in $B_{1}$.

For this, we observe that, in light of Remark 2.6, we can write
$Au\stackrel{{\scriptstyle m}}{{=}}f_{u}$ in $B_{1}$ in the sense of
Definition 2.3. That is, recalling Corollary 2.2, there exist functions
$f_{u,R}$ and a family of polynomials $P_{u,\tau_{R}}$, which have degree at
most $m-1$, such that

(4.9) $A(\chi_{R}u)=f_{u,R}+P_{u,\chi_{R}}$

pointwise in $B_{1}$.

Furthermore, using Lemma 2.8, we have that, if $R^{\prime}>R>4$, there exists
a polynomial $P_{R,R^{\prime}}$ of degree at most $m-1$ such that

(4.10)
$\|f_{u,R^{\prime}}-f_{u,R}-P_{R,R^{\prime}}\|_{L^{\infty}(B_{1})}\leqslant\int_{B_{R}^{c}}|u(y)|\sup_{{|\alpha|=m}\atop{x\in
B_{1}}}|\partial^{\alpha}_{x}K(x,y)|\,dy.$

We now claim that

(4.11) $P_{R,R^{\prime}}=0.$

Indeed, by a careful inspection of the proof of Lemma 2.8, one can notice that
the polynomial $P_{R,R^{\prime}}$ is explicit and, denoting by
$v:=(1-\chi_{4})u$, it is equal to

$P_{v,\chi_{R^{\prime}}}-P_{u,\chi_{R^{\prime}}}-P_{v,\chi_{R}}+P_{u,\chi_{R}},$

where we have used the notation of Theorem 2.1. In particular, recalling the
notation in formulas (2.19) and (2.21), we have that the coefficients of the
polynomial $P_{R,R^{\prime}}$ are given by

$\displaystyle\int_{B^{c}_{3}}(\chi_{R^{\prime}}v)(y)\frac{\partial^{\alpha}_{x}K(0,y)}{\alpha!}\,dy-\int_{B^{c}_{3}}(\chi_{R^{\prime}}u)(y)\frac{\partial^{\alpha}_{x}K(0,y)}{\alpha!}\,dy$
$\displaystyle\qquad-\int_{B^{c}_{3}}(\chi_{R}v)(y)\frac{\partial^{\alpha}_{x}K(0,y)}{\alpha!}\,dy+\int_{B^{c}_{3}}(\chi_{R}u)(y)\frac{\partial^{\alpha}_{x}K(0,y)}{\alpha!}\,dy$
$\displaystyle=$ $\displaystyle\int_{B_{R^{\prime}}\setminus
B_{4}}u(y)\frac{\partial^{\alpha}_{x}K(0,y)}{\alpha!}\,dy-\int_{B_{R^{\prime}}\setminus
B_{3}}u(y)\frac{\partial^{\alpha}_{x}K(0,y)}{\alpha!}\,dy$
$\displaystyle\qquad-\int_{B_{R}\setminus
B_{4}}u(y)\frac{\partial^{\alpha}_{x}K(0,y)}{\alpha!}\,dy+\int_{B_{R}\setminus
B_{3}}u(y)\frac{\partial^{\alpha}_{x}K(0,y)}{\alpha!}\,dy$ $\displaystyle=$
$\displaystyle-\int_{B_{4}\setminus
B_{3}}u(y)\frac{\partial^{\alpha}_{x}K(0,y)}{\alpha!}\,dy+\int_{B_{4}\setminus
B_{3}}u(y)\frac{\partial^{\alpha}_{x}K(0,y)}{\alpha!}\,dy$ $\displaystyle=$
$\displaystyle 0,$

for every $|\alpha|\leqslant m-1$, which proves (4.11).

Hence, using the information of formula (4.11) into (4.10), we obtain that
$f_{u,R}$ converges to $f_{u}$ uniformly in $B_{1}$.

Now, as a consequence of (4.6) and (4.9), we have that

(4.12)
$\lim_{R\to+\infty}\big{(}P_{u,\chi_{R}}-P_{R}\big{)}=\lim_{R\to+\infty}\big{(}f_{R}-f_{u,R}\big{)}=f-f_{u}$

in $B_{1}$. Therefore, in light of Lemma 2.1 in [MR3988080], we have that the
convergence in (4.12) is uniform in $B_{1}$. Furthermore,

$\displaystyle\|f_{R}-f\|_{L^{\infty}(B_{1})}\leqslant\|f_{R}-f_{u,R}+f_{u}-f\|_{L^{\infty}(B_{1})}+\|f_{u,R}-f_{u}\|_{L^{\infty}(B_{1})}.$

These considerations prove (4.8) and therefore, the first part of the proof is
complete.

Now take $K\in\mathcal{K}_{m,\vartheta}^{+}$ and a solution
$u\in{\mathcal{C}}_{\vartheta,K}$ to $Au\stackrel{{\scriptstyle
m}}{{=}}f\mbox{ in $B_{1}$ in the viscosity sense of
Definition\leavevmode\nobreak\ \ref{defv}}$. We have that there exist a family
of polynomials $P_{R}$ with $\deg P_{R}\leqslant m-1$ and functions $f_{R}$
such that

(4.13) $A(\chi_{R}u)=f_{R}+P_{R}$ in $B_{1}$ in the viscosity sense.

Our objective is now to check that

(4.14) the equation in (4.13) holds true in the pointwise sense as well.

Indeed, once this is established, we can send $R\to+\infty$ and conclude that
$Au\stackrel{{\scriptstyle m}}{{=}}f$ in the pointwise sense of Definition
2.3. To prove (4.14), we use a convolution argument. We pick $\rho\in(0,1)$
and we define $v_{\varepsilon}$ to be the convolution of $\chi_{R}u$ against a
given mollifier $\eta_{\varepsilon}$. We also denote by $g_{\varepsilon}$ the
convolution of $f_{R}+P_{R}$ against $\eta_{\varepsilon}$ and we remark that,
if $\varepsilon$ is small enough, then $Av_{\varepsilon}=g_{\varepsilon}$ in
$B_{\rho}$ in the viscosity sense, and actually also in the pointwise sense,
since $v_{\varepsilon}$ is smooth and can be used itself as a test function in
the viscosity definition. Hence, we can take any point $x_{0}\in B_{\rho}$ and
conclude that

(4.15) $\lim_{\varepsilon\to
0}\int_{\mathbb{R}^{n}}(v_{\varepsilon}(x_{0})-v_{\varepsilon}(y))K(x_{0},y)\,dy=\lim_{\varepsilon\to
0}Av_{\varepsilon}(x_{0})=\lim_{\varepsilon\to
0}g_{\varepsilon}(x_{0})=f_{R}(x_{0})+P_{R}(x_{0}).$

We now claim that, for all $R>5$,

(4.16) $\lim_{\varepsilon\to
0}\int_{\mathbb{R}^{n}}(v_{\varepsilon}(x_{0})-v_{\varepsilon}(y))K(x_{0},y)\,dy=\int_{\mathbb{R}^{n}}(\chi_{R}(x_{0})u(x_{0})-\chi_{R}(y)u(y))K(x_{0},y)\,dy.$

We stress that, once this is proved, then (4.14) would follow directly from
(4.15). Hence, our goal now is to check (4.16). We perform the argument when
$\vartheta\in(1,2]$ (the argument when $\vartheta\in[0,1]$ being similar and
simpler, not requiring any additional symmetrization). We exploit (2.4) to see
that

(4.17)
$\begin{split}&2\int_{B_{1}(x_{0})}(v_{\varepsilon}(x_{0})-v_{\varepsilon}(y))K(x_{0},y)\,dy\\\
&\qquad=\int_{B_{1}}(v_{\varepsilon}(x_{0})-v_{\varepsilon}(x_{0}+z))K(x_{0},x_{0}+z)\,dz+\int_{B_{1}}(v_{\varepsilon}(x_{0})-v_{\varepsilon}(x_{0}-z))K(x_{0},x_{0}-z)\,dz\\\
&\qquad=\int_{B_{1}}\big{(}2v_{\varepsilon}(x_{0})-v_{\varepsilon}(x_{0}+z)-v_{\varepsilon}(x_{0}-z)\big{)}K(x_{0},x_{0}+z)\,dz.\end{split}$

Also, since $u\in{\mathcal{C}}_{\vartheta,K}$ (and we are supposing
$\vartheta\in(1,2]$), for all $z\in B_{1}$,

$\displaystyle|2v_{\varepsilon}(x_{0})-v_{\varepsilon}(x_{0}+z)-v_{\varepsilon}(x_{0}-z)|=\left|\int_{0}^{1}\Big{(}\nabla
v_{\varepsilon}(x_{0}+tz)-\nabla v_{\varepsilon}(x_{0}-tz)\Big{)}\cdot
z\,dt\right|$ $\displaystyle\qquad\leqslant|z|\int_{0}^{1}\Big{|}\nabla
v_{\varepsilon}(x_{0}+tz)-\nabla v_{\varepsilon}(x_{0}-tz)\Big{|}\,dt$
$\displaystyle\qquad=|z|\int_{0}^{1}\left|\int_{B_{\varepsilon}}\nabla(\chi_{R}u)(x_{0}+tz-\zeta)\eta_{\varepsilon}(\zeta)\,d\zeta-\int_{B_{\varepsilon}}\nabla(\chi_{R}u)(x_{0}-tz-\zeta)\eta_{\varepsilon}(\zeta)\,d\zeta\right|\,dt$
$\displaystyle\qquad\leqslant|z|\int_{0}^{1}\int_{B_{\varepsilon}}\left|\nabla(\chi_{R}u)(x_{0}+tz-\zeta)-\nabla(\chi_{R}u)(x_{0}-tz-\zeta)\right|\eta_{\varepsilon}(\zeta)\,d\zeta\,dt$
$\displaystyle\qquad\leqslant
C|z|^{\vartheta}\int_{0}^{1}\int_{B_{\varepsilon}}\eta_{\varepsilon}(\zeta)\,d\zeta\,dt$
$\displaystyle\qquad=C|z|^{\vartheta},$

for some $C>0$.

From this and the Dominated Convergence Theorem, recalling (2.3), we deduce
from (4.17) that

$\displaystyle\lim_{\varepsilon\searrow
0}2\int_{B_{1}(x_{0})}(v_{\varepsilon}(x_{0})-v_{\varepsilon}(y))K(x_{0},y)\,dy=\int_{B_{1}}\lim_{\varepsilon\searrow
0}\big{(}2v_{\varepsilon}(x_{0})-v_{\varepsilon}(x_{0}+z)-v_{\varepsilon}(x_{0}-z)\big{)}K(x_{0},x_{0}+z)\,dz$
$\displaystyle\qquad=\int_{B_{1}}\big{(}2(\chi_{R}u)(x_{0})-(\chi_{R}u)(x_{0}+z)-(\chi_{R}u)(x_{0}-z)\big{)}K(x_{0},x_{0}+z)\,dz$
$\displaystyle\qquad=2\int_{B_{1}(x_{0})}\big{(}(\chi_{R}u)(x_{0})-(\chi_{R}u)(y)\big{)}K(x_{0},y)\,dy.$

Using again the Dominated Convergence Theorem, one deduces (4.16) from the
previous equation, as desired. ∎

Next result shows the stability of the equation under the uniform convergence
in the viscosity sense. To this end, we will also assume other mild conditions
on the kernel. First of all, we assume a continuity hypothesis in the first
variable, that is we suppose that

(4.18) ${\mbox{for all\leavevmode\nobreak\ $y\in\mathbb{R}^{n}$ and
all\leavevmode\nobreak\ $x_{0}\in B_{3}\setminus\\{y\\}$, }}\;\lim_{x\to
x_{0}}K(x,y)=K(x_{0},y).$

Additionally, we assume a local integrability condition outside a possible
singularity of the kernel and a locally uniform version of condition (2.3),
namely we suppose that

(4.19) ${\mbox{for all $x_{0}\in B_{3}$ and all\leavevmode\nobreak\ $r>0$,
}}\;\int_{\mathbb{R}^{n}\setminus B_{r}(x_{0})}\;\sup_{x\in B_{3}\setminus
B_{r}(y)}|K(x,y)|\,dy<+\infty$

and

(4.20) $\int_{B_{3}}\sup_{x\in B_{1}}|x-y|^{2}\,|K(x,y)|\,dy<+\infty.$

######  Lemma 4.7.

Let $\vartheta\in[0,2]$. Let $K\in\mathcal{K}_{0,\vartheta}^{+}$ satisfying
(4.18), (4.19) and (4.20). For every $k\in\mathbb{N}$, let
$u_{k}\in\mathcal{V}_{K}$ and $f_{k}$ be bounded and continuous in $B_{1}$.
Assume that

(4.21) $Au_{k}=f_{k}$

in $B_{1}$ in the viscosity sense, that

$f_{k}$ converges uniformly in $B_{1}$ to some function $f$ as $k\to+\infty$,

that

(4.22) $u_{k}$ converges uniformly in $B_{4}$ to some function
$u\in\mathcal{V}_{K}$ as $k\to+\infty$

and that

(4.23) $\lim_{k\to+\infty}\int_{\mathbb{R}^{n}\setminus
B_{3}}\big{|}u(y)-u_{k}(y)\big{|}\sup_{x\in B_{1}}|K(x,y)|\,dy=0.$

Then,

$Au=f$

in $B_{1}$ in the viscosity sense.

###### Proof.

Let $x_{0}\in B_{1}$ and $\rho>0$ such that $B_{\rho}(x_{0})\Subset B_{1}$.
Let $\varphi\in C^{2}(\overline{B_{\rho}(x_{0})})$ with $\varphi=u$ outside
$B_{\rho}(x_{0})$. Suppose that $v:=\varphi-u$ has a local maximum at $x_{0}$.

We define, for every $k\in\mathbb{N}$,

$\varepsilon_{k}:=\|u-u_{k}\|_{L^{\infty}(B_{1})}+\frac{1}{k}$

and

$\varphi_{k}(x):=\begin{cases}\varphi(x)-\sqrt{\varepsilon_{k}}\,|x-x_{0}|^{2}&{\mbox{
in }}\overline{B_{\rho}(x_{0})},\\\ u_{k}(x)&{\mbox{ in
}}\mathbb{R}^{n}\setminus\overline{B_{\rho}(x_{0})}.\end{cases}$

We let $v_{k}:=\varphi_{k}-u_{k}$ and $x_{k}\in\overline{B_{\rho}(x_{0})}$ be
such that

$v_{k}(x_{k})=\max_{\overline{B_{\rho}(x_{0})}}v_{k}.$

We observe that

$\displaystyle|x_{k}-x_{0}|^{2}$ $\displaystyle=$
$\displaystyle\frac{\varphi(x_{k})-\varphi_{k}(x_{k})}{\sqrt{\varepsilon_{k}}}$
$\displaystyle=$
$\displaystyle\frac{v(x_{k})-v_{k}(x_{k})+u(x_{k})-u_{k}(x_{k})}{\sqrt{\varepsilon_{k}}}$
$\displaystyle\leqslant$
$\displaystyle\frac{v(x_{0})-v_{k}(x_{0})+\varepsilon_{k}}{\sqrt{\varepsilon_{k}}}$
$\displaystyle=$
$\displaystyle\frac{\varphi(x_{0})-\varphi_{k}(x_{0})-u(x_{0})+u_{k}(x_{0})+\varepsilon_{k}}{\sqrt{\varepsilon_{k}}}$
$\displaystyle=$
$\displaystyle\frac{-u(x_{0})+u_{k}(x_{0})+\varepsilon_{k}}{\sqrt{\varepsilon_{k}}}$
$\displaystyle\leqslant$ $\displaystyle 2\sqrt{\varepsilon_{k}}.$

Thus, since $\varepsilon_{k}$ is infinitesimal due to (4.22), we have that
$x_{k}$ converges to $x_{0}$ as $k\to+\infty$ and, in particular, the function
$v_{k}$ has an interior maximum at $x_{k}$. This and (4.21) give that

(4.24) $0\leqslant A\varphi_{k}(x_{k})-f_{k}(x_{k})\leqslant
A\varphi_{k}(x_{k})-f(x_{0})+|f(x_{0})-f(x_{k})|+\|f-f_{k}\|_{L^{\infty}(B_{1})}.$

Now we claim that

(4.25)
$\lim_{k\to+\infty}\int_{B_{\rho}(x_{0})}\big{(}\varphi_{k}(x_{k})-\varphi_{k}(y)\big{)}K(x_{k},y)\,dy=\int_{B_{\rho}(x_{0})}\big{(}\varphi(x_{0})-\varphi(y)\big{)}K(x_{0},y)\,dy.$

To this end, we first observe that

(4.26)
$\begin{split}\varphi_{k}(x_{k})-\varphi_{k}(y)\,=\,&\varphi(x_{k})-\varphi(y)+\sqrt{\varepsilon_{k}}|y-x_{0}|^{2}-\sqrt{\varepsilon_{k}}|x_{k}-x_{0}|^{2}\\\
=\,&\varphi(x_{k})-\varphi(y)+\sqrt{\varepsilon_{k}}(2x_{0}-x_{k}-y)\cdot(x_{k}-y).\end{split}$

We define

(4.27) $F(y):=\sup_{x\in B_{1}}|x-y|^{2}\,|K(x,y)|$

and we observe that $F\in L^{1}(B_{3})$, thanks to (4.20). Accordingly, by the
absolute continuity of the Lebesgue integrals, for all $\varepsilon>0$ there
exists $\delta>0$ such that if the Lebesgue measure of a set $Z\subset B_{3}$
is less than $\delta$, then

(4.28) $\int_{Z}F(y)\,dy\leqslant\varepsilon.$

We recall (2.4) and we see that, for $k$ sufficiently large,

$\displaystyle\int_{B_{\rho/2}(x_{k})}\big{(}\varphi_{k}(x_{k})-\varphi_{k}(y)\big{)}K(x_{k},y)\,dy$
$\displaystyle=$
$\displaystyle\frac{1}{2}\int_{B_{\rho/2}}\big{(}\varphi_{k}(x_{k})-\varphi_{k}(x_{k}+z)\big{)}K(x_{k},x_{k}+z)\,dz+\frac{1}{2}\int_{B_{\rho/2}}\big{(}\varphi_{k}(x_{k})-\varphi_{k}(x_{k}-z)\big{)}K(x_{k},x_{k}-z)\,dz$
$\displaystyle=$
$\displaystyle\frac{1}{2}\int_{B_{\rho/2}}\big{(}2\varphi_{k}(x_{k})-\varphi_{k}(x_{k}+z)-\varphi_{k}(x_{k}-z)\big{)}K(x_{k},x_{k}+z)\,dz$
$\displaystyle=$
$\displaystyle\frac{1}{2}\int_{B_{\rho}(x_{0})}\chi_{B_{\rho/2}(x_{k})}(y)\big{(}2\varphi_{k}(x_{k})-\varphi_{k}(y)-\varphi_{k}(2x_{k}-y)\big{)}K(x_{k},y)\,dy.$

Thus, for all $y\in B_{\rho}(x_{0})$ we define

$\zeta_{k}(y):=\frac{1}{2}\chi_{B_{\rho/2}(x_{k})}(y)\big{(}2\varphi_{k}(x_{k})-\varphi_{k}(y)-\varphi_{k}(2x_{k}-y)\big{)}K(x_{k},y)$

and we point out that

$\displaystyle|\zeta_{k}(y)|$ $\displaystyle\leqslant$
$\displaystyle\frac{1}{2}\Big{|}\big{(}\varphi_{k}(x_{k})-\varphi_{k}(y)\big{)}+\big{(}\varphi_{k}(x_{k})-\varphi_{k}(2x_{k}-y)\big{)}\Big{|}\,|K(x_{k},y)|$
$\displaystyle=$
$\displaystyle\frac{1}{2}\left|\left(\int_{0}^{1}\nabla\varphi_{k}\big{(}tx_{k}+(1-t)y\big{)}\,dt-\int_{0}^{1}\nabla\varphi_{k}\big{(}tx_{k}+(1-t)(2x_{k}-y)\big{)}\,dt\right)\cdot(x_{k}-y)\right|\,|K(x_{k},y)|$
$\displaystyle=$
$\displaystyle\Bigg{|}\int_{0}^{1}\left((1-t)\int_{0}^{1}D^{2}\varphi_{k}\Big{(}\tau\big{(}tx_{k}+(1-t)y\big{)}+(1-\tau)\big{(}tx_{k}+(1-t)(2x_{k}-y)\big{)}\Big{)}\,d\tau\right)\,dt$
$\displaystyle\qquad(x_{k}-y)\cdot(x_{k}-y)\Bigg{|}\,|K(x_{k},y)|$
$\displaystyle\leqslant$ $\displaystyle C\,|x_{k}-y|^{2}\,|K(x_{k},y)|$
$\displaystyle\leqslant$ $\displaystyle C\,F(y),$

where the notation in (4.27) was used. In particular, since $F\in
L^{1}(B_{3})$ by (4.20), we can exploit the absolute continuity of the
Lebesgue integrals (see (4.28)) and deduce that for all $\varepsilon>0$ there
exists $\delta>0$ such that if the Lebesgue measure of a set $Z\subset
B_{\rho}(x_{0})$ is less than $\delta$, then

$\int_{Z}|\zeta_{k}(y)|\,dy\leqslant C\varepsilon.$

Hence, recalling (4.18), we utilize the Vitali Convergence Theorem and obtain
that

$\displaystyle\lim_{k\to+\infty}\int_{B_{\rho}(x_{0})}\zeta_{k}(y)\,dy=\int_{B_{\rho}(x_{0})}\lim_{k\to+\infty}\zeta_{k}(y)\,dy$
$\displaystyle\qquad=\frac{1}{2}\int_{B_{\rho/2}(x_{0})}\big{(}2\varphi(x_{0})-\varphi(y)-\varphi(2x_{0}-y)\big{)}K(x_{0},y)$

which proves that

(4.29)
$\lim_{k\to+\infty}\int_{B_{\rho/2}(x_{0})}\big{(}\varphi_{k}(x_{k})-\varphi_{k}(y)\big{)}K(x_{k},y)\,dy=\int_{B_{\rho/2}(x_{0})}\big{(}\varphi(x_{0})-\varphi(y)\big{)}K(x_{0},y)\,dy.$

Now, for every $y\in\mathbb{R}^{n}\setminus B_{\rho/2}(x_{0})$ we define

$\eta_{k}(y):=\big{(}\varphi_{k}(x_{k})-\varphi_{k}(y)\big{)}K(x_{k},y).$

We observe that, if $k$ is sufficiently large, for every $y\in B_{3}\setminus
B_{\rho/2}(x_{0})$,

$|\eta_{k}(y)|\leqslant C\,\sup_{x\in B_{3}\setminus B_{\rho/4}(y)}|K(x,y)|,$

which belongs to $L^{1}(B_{\rho}(x_{0})\setminus B_{\rho/2}(x_{0}))$ due to
(4.19).

Consequently, by (4.18) and the Dominated Convergence Theorem,

$\begin{split}&\lim_{k\to+\infty}\int_{B_{\rho}(x_{0})\setminus
B_{\rho/2}(x_{0})}\big{(}\varphi_{k}(x_{k})-\varphi_{k}(y)\big{)}K(x_{k},y)\,dy=\lim_{k\to+\infty}\int_{B_{\rho}(x_{0})\setminus
B_{\rho/2}(x_{0})}\eta_{k}(y)\,dy\\\ &\qquad=\int_{B_{\rho}(x_{0})\setminus
B_{\rho/2}(x_{0})}\lim_{k\to+\infty}\eta_{k}(y)\,dy=\int_{B_{\rho}(x_{0})\setminus
B_{\rho/2}(x_{0})}\big{(}\varphi(x_{0})-\varphi(y)\big{)}K(x_{0},y)\,dy.\end{split}$

This and (4.29) show the validity of (4.25).

Having completed the proof of (4.25), we now claim that

(4.30) $\lim_{k\to+\infty}\int_{\mathbb{R}^{n}\setminus
B_{\rho}(x_{0})}\big{(}\varphi_{k}(x_{k})-\varphi_{k}(y)\big{)}K(x_{k},y)\,dy=\int_{\mathbb{R}^{n}\setminus
B_{\rho}(x_{0})}\big{(}\varphi(x_{0})-\varphi(y)\big{)}K(x_{0},y)\,dy.$

Indeed, for all $y\in\mathbb{R}^{n}\setminus\overline{B_{\rho}(x_{0})}$ we set

$\mu_{k}(y):=\big{(}\varphi_{k}(x_{k})-\varphi_{k}(y)\big{)}K(x_{k},y)=\big{(}\varphi(x_{k})-\sqrt{\varepsilon_{k}}|x_{k}-x_{0}|^{2}-u_{k}(y)\big{)}K(x_{k},y).$

If $y\in B_{3}\setminus\overline{B_{\rho}(x_{0})}$ and $k$ is sufficiently
large, then, by (4.22),

$\displaystyle|\mu_{k}(y)|\leqslant C\,(1+|u(y)|)\,|K(x_{k},y)|\leqslant
C\,(1+\|u\|_{L^{\infty}(B_{3})})\,\sup_{x\in B_{3}\setminus
B_{\rho/2}(y)}|K(x,y)|$

and the latter function belongs to
$L^{1}(B_{3}\setminus\overline{B_{\rho}(x_{0})})$ owing to (4.19).

This, (4.18), (4.22) and the Dominated Convergence Theorem lead to

(4.31)
$\begin{split}&\lim_{k\to+\infty}\int_{B_{3}\setminus{B_{\rho}(x_{0})}}\big{(}\varphi_{k}(x_{k})-\varphi_{k}(y)\big{)}K(x_{k},y)\,dy\,\\\
&\qquad=\,\int_{B_{3}\setminus{B_{\rho}(x_{0})}}\lim_{k\to+\infty}\big{(}\varphi(x_{k})-\sqrt{\varepsilon_{k}}|x_{k}-x_{0}|^{2}-u_{k}(y)\big{)}K(x_{k},y)\,dy\\\
&\qquad=\,\int_{B_{3}\setminus{B_{\rho}(x_{0})}}\big{(}\varphi(x_{0})-u(y)\big{)}K(x_{0},y)\,dy.\end{split}$

In light of (4.3) and (4.23) we also remark that, for large $k$,

(4.32) $\begin{split}&\int_{\mathbb{R}^{n}\setminus
B_{3}}|u_{k}(y)|\sup_{{x\in B_{1}}}K(x,y)\,dy\\\
&\qquad\leqslant\int_{\mathbb{R}^{n}\setminus
B_{3}}\big{|}u(y)-u_{k}(y)\big{|}\sup_{x\in
B_{1}}K(x,y)\,dy+\int_{\mathbb{R}^{n}\setminus B_{3}}|u(y)|\sup_{{x\in
B_{1}}}K(x,y)\,dy\\\ &\qquad\leqslant 1+\int_{\mathbb{R}^{n}\setminus
B_{3}}|u(y)|\sup_{{x\in B_{1}}}K(x,y)\,dy\leqslant C,\end{split}$

up to renaming $C$ once again. Furthermore, if $y\in\mathbb{R}^{n}\setminus
B_{3}$ and
$\delta_{k}:=\big{|}\varphi(x_{k})-\varphi(x_{0})\big{|}+\sqrt{\varepsilon_{k}}$,
we have that $\delta_{k}$ is infinitesimal as $k\to+\infty$ and

$\displaystyle\Big{|}\big{(}\varphi_{k}(x_{k})-\varphi_{k}(y)\big{)}K(x_{k},y)-\big{(}\varphi(x_{0})-u(y)\big{)}K(x_{0},y)\Big{|}$
$\displaystyle\leqslant$
$\displaystyle\Big{(}\big{|}\varphi(x_{k})-\varphi(x_{0})\big{|}+\sqrt{\varepsilon_{k}}+|u_{k}(y)-u(y)|\Big{)}|K(x_{k},y)|+\big{|}\varphi(x_{0})-u(y)\big{|}\,\big{|}K(x_{0},y)-K(x_{k},y)\big{|}$
$\displaystyle\leqslant$ $\displaystyle\delta_{k}\sup_{x\in
B_{1}}|K(x,y)|+|u_{k}(y)-u(y)|\sup_{x\in
B_{1}}|K(x,y)|+C\,(1+|u(y)|)\,\big{|}K(x_{0},y)-K(x_{k},y)\big{|}$
$\displaystyle\leqslant$ $\displaystyle\delta_{k}\sup_{x\in B_{1}\setminus
B_{1}(y)}|K(x,y)|+|u_{k}(y)-u(y)|\sup_{x\in
B_{1}}|K(x,y)|+C\,(1+|u(y)|)\,\big{|}K(x_{0},y)-K(x_{k},y)\big{|}.$

Gathering this information, (4.19) and (4.23), we find that

(4.33) $\begin{split}&\lim_{k\to+\infty}\int_{\mathbb{R}^{n}\setminus
B_{3}}\Big{|}\big{(}\varphi_{k}(x_{k})-\varphi_{k}(y)\big{)}K(x_{k},y)-\big{(}\varphi(x_{0})-u(y)\big{)}K(x_{0},y)\Big{|}\,dy\\\
&\qquad\leqslant\lim_{k\to+\infty}C\int_{\mathbb{R}^{n}\setminus
B_{3}}(1+|u(y)|)\,\big{|}K(x_{0},y)-K(x_{k},y)\big{|}\,dy.\end{split}$

We also point out that, if $y\in\mathbb{R}^{n}\setminus B_{3}$,

$(1+|u(y)|)\,\big{|}K(x_{0},y)-K(x_{k},y)\big{|}\leqslant
2(1+|u(y)|)\,\sup_{x\in B_{1}\setminus B_{1}(y)}|K(x,y)|$

and the latter function belongs to $L^{1}(\mathbb{R}^{n}\setminus B_{3})$,
thanks to (4.3) and (4.19). The Dominated Convergence Theorem and (4.18)
thereby give that

$\lim_{k\to+\infty}\int_{\mathbb{R}^{n}\setminus
B_{3}}(1+|u(y)|)\,\big{|}K(x_{0},y)-K(x_{k},y)\big{|}\,dy=0.$

This and (4.33) yield that

$\lim_{k\to+\infty}\int_{\mathbb{R}^{n}\setminus
B_{3}}\Big{|}\big{(}\varphi_{k}(x_{k})-\varphi_{k}(y)\big{)}K(x_{k},y)-\big{(}\varphi(x_{0})-u(y)\big{)}K(x_{0},y)\Big{|}\,dy=0,$

which, combined with (4.31), proves (4.30).

By combining (4.25) and (4.30), we deduce that $A\varphi_{k}(x_{k})\to
A\varphi(x_{0})$ as $k\to+\infty$. As a result, by passing to the limit in
(4.24), we conclude that $A\varphi(x_{0})\geqslant f(x_{0})$.

Similarly, one sees that if the function $\varphi-u$ has a local minimum at
$x_{0}$ then $A\varphi(x_{0})\leqslant f(x_{0})$. ∎

######  Corollary 4.8.

Let $\vartheta\in[0,2]$. Let $K\in\mathcal{K}_{0,\vartheta}^{+}$ satisfying
(4.18), (4.19) and (4.20) and $u\in\mathcal{V}_{K}$. Let $f$ be bounded and
continuous in $B_{1}$.

Then

$Au=f{\mbox{ in }}B_{1}$

in viscosity sense is equivalent to

$Au\stackrel{{\scriptstyle 0}}{{=}}f{\mbox{ in }}B_{1}$

in the viscosity sense of Definition 4.2.

###### Proof.

Suppose first that $Au=f$ in $B_{1}$ in viscosity sense. For every $R>5$, we
define

$f_{R}(x):=f(x)+\int_{B_{R}^{c}}u(y)K(x,y)\,dy.$

Notice that

$\sup_{x\in B_{1}}|f_{R}(x)-f(x)|\leqslant\int_{B_{R}^{c}}|u(y)|\sup_{x\in
B_{1}}K(x,y)\,dy,$

which is infinitesimal, thanks to (4.3) (used here with $m:=0$), and thus
$f_{R}$ converges to $f$ uniformly in $B_{1}$.

Our objective is to prove that $A(\chi_{R}u)=f_{R}$ in $B_{1}$ in the
viscosity sense (from which we obtain that $Au\stackrel{{\scriptstyle
0}}{{=}}f$ in $B_{1}$ in the viscosity sense of Definition 4.2).

To check this claim, we pick a point $x_{0}\in B_{1}$ and touch $\chi_{R}u$
from below by a test function $\varphi$ at $x_{0}$, with $\varphi=\chi_{R}u$
outside $B_{2}$. We define $\psi:=\varphi+(1-\chi_{R})u$ and we observe that
$\psi$ touches $u$ by below at $x_{0}$ and that $\psi=u$ outside $B_{2}$. As a
result, $A\psi(x_{0})\geqslant f(x_{0})$ and therefore

$\displaystyle f_{R}(x_{0})$ $\displaystyle=$ $\displaystyle
f(x_{0})+\int_{B_{R}^{c}}u(y)K(x_{0},y)\,dy$ $\displaystyle\leqslant$
$\displaystyle A\psi(x_{0})+\int_{B_{R}^{c}}u(y)K(x_{0},y)\,dy$
$\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{n}}(\psi(x_{0})-\psi(y))K(x_{0},y)\,dy+\int_{B_{R}^{c}}u(y)K(x_{0},y)\,dy$
$\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{n}}(\varphi(x_{0})-\psi(y))K(x_{0},y)\,dy+\int_{B_{R}^{c}}u(y)K(x_{0},y)\,dy$
$\displaystyle=$
$\displaystyle\int_{B_{R}}(\varphi(x_{0})-\varphi(y))K(x_{0},y)\,dy+\int_{B_{R}^{c}}(\varphi(x_{0})-\psi(y))K(x_{0},y)\,dy+\int_{B_{R}^{c}}u(y)K(x_{0},y)\,dy$
$\displaystyle=$ $\displaystyle
A\varphi(x_{0})+\int_{B_{R}^{c}}(\varphi(y)-\psi(y))K(x_{0},y)\,dy+\int_{B_{R}^{c}}u(y)K(x_{0},y)\,dy$
$\displaystyle=$ $\displaystyle A\varphi(x_{0}).$

Similarly, if $\varphi$ touches $\chi_{R}u$ from above, then
$A\varphi(x_{0})\leqslant f_{R}(x_{0})$. These observations entail that
$A(\chi_{R}u)=f_{R}$ in $B_{1}$ in the viscosity sense.

This proves one of the implications of Corollary 4.8. To prove the other, we
assume now that $Au\stackrel{{\scriptstyle 0}}{{=}}f$ in $B_{1}$ in the
viscosity sense of Definition 4.2. Then, we find $f_{R}:B_{1}\to\mathbb{R}$
such that $A(\chi_{R}u)=f_{R}$ in $B_{1}$ in viscosity sense, with $f_{R}$
converging to $f$ uniformly in $B_{1}$. We remark that

$\displaystyle\lim_{R\to+\infty}\int_{\mathbb{R}^{n}\setminus
B_{3}}\big{|}u(y)-(\chi_{R}u)(y)\big{|}\sup_{x\in B_{1}}|K(x,y)|\,dy=0,$

thanks to (4.3) (used here with $m:=0$) and the Dominated Convergence Theorem.

We can therefore apply Lemma 4.7 and conclude that $Au=f$ in $B_{1}$ in the
sense of viscosity, as desired. ∎

In the next result we state the viscosity counterpart of Lemma 2.14 (its proof
is omitted since it is similar to the one of Lemma 2.14, just noticing that
the functions $v$ and $\chi_{R}w-\chi_{4}w$ vanish in $B_{4}$, hence the
viscous and pointwise setting would equally apply to them).

######  Lemma 4.9.

Let $j,m\in\mathbb{N}_{0}$, with $j\leqslant m$, $\vartheta\in[0,2]$ and
$K\in\mathcal{K}_{j,\vartheta}^{+}\cap\mathcal{K}_{m,\vartheta}^{+}$. Let $f$
be bounded and continuous in $B_{1}$ and let $u\in\mathcal{V}_{K}$ such that

(4.34) $Au\stackrel{{\scriptstyle m}}{{=}}f$

in $B_{1}$.

Then, there exist a function $\bar{f}$ and a polynomial $P$ of degree at most
$m-1$ such that $\bar{f}=f+P$ and $Au\stackrel{{\scriptstyle j}}{{=}}\bar{f}$
in $B_{1}$.

Next, as a possible application, we show a specific case in which the
existence of solution to a Dirichlet problem is guaranteed. For this, we
consider a family of kernels comparable to the fractional Laplace operator, as
follows. For any $s\in(0,1)$, given real numbers $\Lambda\geqslant\lambda>0$,
we consider the family of kernels $K$ as defined in (1.8). We suppose that

(4.35) there exists $m\in\mathbb{N}_{0}$ such that condition (2.2) is
satisfied.

We also assume that condition (2.4) holds true and that $K$ is translation
invariant, i.e.

(4.36) $K(x+z,y+z)=K(x,y)$ for any $x$, $y$, $z\in\mathbb{R}^{n}$.

With this, we have that $K$ belongs to $\mathcal{K}_{m,\vartheta}^{+}$, with
$m$ as in (4.35) and for every $\vartheta\in(2s,2]$. Moreover, it also
satisfies (4.19) and (4.20).

We introduce the fractional Sobolev space

$\mathbb{H}^{s}(B_{1}):=\left\\{u\in
L^{2}(B_{1}):\iint_{\mathbb{R}^{2n}\setminus(B_{1}^{c}\times
B_{1}^{c})}(u(x)-u(y))^{2}K(x,y)\,dy<+\infty\right\\}$

and, given $g\in\mathbb{H}^{s}(B_{1})$, the class

$J_{g}(B_{1}):=\Big{\\{}u\in\mathbb{H}^{s}(B_{1}):\;u=g\;{\mbox{ in
}}B_{1}^{c}\Big{\\}}.$

We use this class to seek solutions to the Dirichlet problem (see [P18]). More
precisely, the following result can be proved by using the Direct Methods of
the Calculus of Variations and the strict convexity of the functional.

######  Proposition 4.10.

Let $K$ be as in (1.8), (2.4), (4.35) and (4.36). Let $f\in L^{2}(B_{1})$ and
$g\in\mathbb{H}^{s}(B_{1})$. Then, there exists a unique minimizer of the
functional

(4.37)
$\mathcal{E}(u):=\frac{1}{4}\iint_{\mathbb{R}^{n}\times\mathbb{R}^{n}}(u(x)-u(y))^{2}K(x,y)\,dx\,dy-\int_{B_{1}}f(x)u(x)\,dx$

over $J_{g}(B_{1})$.

In addition, $u\in J_{g}(B_{1})$ is a minimizer of (4.37) over $J_{g}(B_{1})$
if and only if it is a weak solution of

(4.38) $\begin{cases}Au=f&{\mbox{ in }}B_{1},\\\ u=g&\mbox{ in
}B_{1}^{c},\end{cases}$

that is, for every $\phi\in C^{\infty}_{0}(B_{1})$,

$\frac{1}{2}\iint_{\mathbb{R}^{n}\times\mathbb{R}^{n}}\big{(}u(x)-u(y)\big{)}\big{(}\phi(x)-\phi(y)\big{)}\,K(x,y)\,dx\,dy=\int_{B_{1}}f(x)\phi(x)\,dx.$

The next result is a generalization of Theorem 2 in [SV14], which shows the
global continuity of weak solutions of an equation which includes the operator
of our interest.

######  Proposition 4.11.

Let $K$ be as in (1.8) (2.4), (4.35) and (4.36). Let $f\in L^{\infty}(B_{1})$
and $g\in C^{\alpha}(\mathbb{R}^{n})$ for some $\alpha\in(0,\min\\{2s,1\\})$.
Assume that

$|g(x)|\leqslant C|x|^{\alpha}\qquad{\mbox{ for all
}}x\in\mathbb{R}^{n}\setminus B_{1}.$

Let also $u\in J_{g}(B_{1})$ be a weak solution of

(4.39) $Au=f\quad{\mbox{ in }}B_{1}.$

Then, $u\in C(\mathbb{R}^{n})$.

###### Proof.

First of all, we exploit Proposition 4.10 with $g:=0$ to find a weak solution
$v$ of

$\begin{cases}Av=f&{\mbox{ in }}B_{1},\\\ v=0&{\mbox{ in
}}B_{1}^{c}.\end{cases}$

By Proposition 7.2 in [R16] (see also [MR3293447] for related results), we
have that $v\in C(\mathbb{R}^{n})$.

Let now $w:=u-v$. We see that $w$ is a weak solution of

$\begin{cases}Aw=0&{\mbox{ in }}B_{1},\\\ w=u=g&{\mbox{ in
}}B_{1}^{c}.\end{cases}$

We thus exploit Theorem 1.4 in [MR4135310] and find that $w\in
C(\mathbb{R}^{n})$. From these observations, we find that $u=v+w\in
C(\mathbb{R}^{n})$, as desired. ∎

With this, we can now prove that, in this setting, week solutions are also
viscosity solutions.

######  Proposition 4.12.

Let $K$ be as in (1.8) (2.4), (4.18), (4.35) and (4.36). Let $f$ be bounded
and continuous in $B_{1}$ and $g\in C^{\alpha}(\mathbb{R}^{n})$ for some
$\alpha\in(0,\min\\{2s,1\\})$. Assume that

(4.40) $|g(x)|\leqslant C|x|^{\alpha}\qquad{\mbox{ for all
}}x\in\mathbb{R}^{n}\setminus B_{1}.$

Let also $u\in J_{g}(B_{1})$ be a weak solution of

(4.41) $\begin{cases}Au=f&{\mbox{ in }}B_{1},\\\ u=g&{\mbox{ in
}}B_{1}^{c}.\end{cases}$

Then, $u$ is a viscosity solution of (4.41).

###### Proof.

By Proposition 4.11, we know that $u\in C(\mathbb{R}^{n})$.

Now, we take a point $x_{0}\in B_{1}$ and a function $\rho\in
C^{\infty}_{0}(B_{1},[0,1])$ and we consider an even mollifier
$\rho_{\varepsilon}:=\varepsilon^{-n}\rho(x/\varepsilon)$, for any
$\varepsilon\in(0,1)$. We set $u_{\varepsilon}:=u\ast\rho_{\varepsilon}$ and
$f_{\varepsilon}:=f\ast\rho_{\varepsilon}$ (where we identified $f$ with its
null extension outside $B_{1}$).

We claim that

(4.42) $Au_{\varepsilon}=f_{\varepsilon}$ in the weak sense in any ball
$B_{r}(x_{0})$ such that $B_{r}(x_{0})\Subset B_{1}$.

To prove this, we take a ball $B_{r}(x_{0})$ such that $B_{r}(x_{0})\Subset
B_{1}$ and a function $\varphi\in C^{\infty}_{0}(B_{\rho}(x_{0}))$. We observe
that

(4.43)
$\begin{split}&\int_{\mathbb{R}^{n}}\left(\;\iint_{\mathbb{R}^{2n}}(u(x+z)-u(y+z))(\varphi(x)-\varphi(y))\rho_{\varepsilon}(z)K(x,y)\,dx\,dy\right)\,dz\\\
=\,&\int_{B_{\varepsilon}}\left(\quad\iint_{\mathbb{R}^{2n}\setminus(B_{r}^{c}(x_{0})\times
B^{c}_{r}(x_{0}))}(u(x+z)-u(y+z))(\varphi(x)-\varphi(y))\rho_{\varepsilon}(z)K(x,y)\,dx\,dy\right)\,dz\\\
\leqslant\,&\frac{\varepsilon^{-n}}{2}\int_{B_{\varepsilon}}\left(\quad\iint_{\mathbb{R}^{2n}\setminus(B_{1}^{c}\times
B^{c}_{1})}(u(x)-u(y))^{2}K(x,y)\,dx\,dy+\iint_{\mathbb{R}^{2n}\setminus(B_{1}^{c}\times
B^{c}_{1})}(\varphi(x)-\varphi(y))^{2}K(x,y)\,dx\,dy\right)\,dz\\\
<\,&+\infty,\end{split}$

thanks to (4.36).

Therefore, Tonelli’s Theorem gives us that the function

$(x,y,z)\in\mathbb{R}^{2n}\times\mathbb{R}^{n}\mapsto(u(x+z)-u(y+z))(\varphi(x)-\varphi(y))\rho_{\varepsilon}(z)K(x,y)\quad{\mbox{lies
in }}L^{1}(\mathbb{R}^{2n}\times\mathbb{R}^{n}).$

One can interchange the order of integration in (4.43), thanks to Fubini’s
Theorem, and exploit the definition of $u_{\varepsilon}$ to obtain

(4.44)
$\begin{split}\int_{\mathbb{R}^{n}}&\left(\;\iint_{\mathbb{R}^{2n}}(u(x+z)-u(y+z))(\varphi(x)-\varphi(y))\rho_{\varepsilon}(z)K(x,y)\,dx\,dy\right)\,dz\\\
&=\iint_{\mathbb{R}^{2n}}\left(\;\int_{\mathbb{R}^{n}}(u(x+z)-u(y+z))(\varphi(x)-\varphi(y))\rho_{\varepsilon}(z)K(x,y)\,dz\right)\,dx\,dy\\\
&=\iint_{\mathbb{R}^{2n}}(u_{\varepsilon}(x)-u_{\varepsilon}(y))(\varphi(x)-\varphi(y))K(x,y)\,dx\,dy.\end{split}$

Then, we can use Fubini’s Theorem once again to get

(4.45) $\begin{split}\int_{\mathbb{R}^{n}}&f_{\varepsilon}(x)\varphi(x)\,dx\\\
&=\int_{\mathbb{R}^{n}}\left(\;\int_{\mathbb{R}^{n}}f(x+z)\rho_{\varepsilon}(z)\varphi(x)\,dx\right)\,dz\\\
&=\int_{B_{\varepsilon}}\left(\;\int_{\mathbb{R}^{n}}f(\tilde{x})\varphi(\tilde{x}-z)\rho_{\varepsilon}(z)d\tilde{x}\right)\,dz\\\
&=\frac{1}{2}\int_{B_{\varepsilon}}\left(\;\iint_{\mathbb{R}^{2n}}(u(\tilde{x})-u(\tilde{y}))(\varphi(\tilde{x}-z)-\varphi(\tilde{y}-z))\rho_{\varepsilon}(z)K(\tilde{x},\tilde{y})d\tilde{x}d\tilde{y}\right)\,dz\\\
&=\frac{1}{2}\int_{\mathbb{R}^{n}}\left(\;\iint_{\mathbb{R}^{2n}}(u(x+z)-u(y+z))(\varphi(x)-\varphi(y))\rho_{\varepsilon}(z)K(x,y)\,dx\,dy\right)\,dz\\\
&=\frac{1}{2}\iint_{\mathbb{R}^{2n}}(u_{\varepsilon}(x)-u_{\varepsilon}(y))(\varphi(x)-\varphi(y))K(x,y)\,dx\,dy,\end{split}$

since the kernel $K$ is translation invariant and $u$ satisfies (4.41) in weak
sense. This shows (4.42).

Now, given $B_{r}(x_{0})\Subset B_{1}$, we show that

(4.46) ${\mbox{the map }}B_{r}(x_{0})\ni
x\mapsto\int_{\mathbb{R}^{n}}(u_{\varepsilon}(x)-u_{\varepsilon}(y))\,K(x,y)\,dy{\mbox{
is continuous.}}$

For this, we let $x_{k}$ be a sequence converging to a given point $x\in
B_{r}(x_{0})$ and we define

$\zeta_{k}(z):=(2u_{\varepsilon}(x_{k})-u_{\varepsilon}(x_{k}+z)-u_{\varepsilon}(x_{k}-z))\,K(0,z).$

Since $u_{\varepsilon}$ is smooth and its growth at infinity is controlled via
(4.40), we know that

$|2u_{\varepsilon}(x_{k})-u_{\varepsilon}(x_{k}+z)-u_{\varepsilon}(x_{k}-z)|\leqslant
C_{\varepsilon}\min\\{|z|^{2},|z|^{\alpha}\\},$

for some $C_{\varepsilon}>0$. For this reason and (1.8),

$|\zeta_{k}(z)|\leqslant\frac{C_{\varepsilon}\min\\{|z|^{2},|z|^{\alpha}\\}}{|z|^{n+2s}},$

up to renaming $C_{\varepsilon}$ and therefore we are in the position of
applying the Dominated Convergence Theorem and conclude that

$\lim_{k\to+\infty}\int_{\mathbb{R}^{n}}(2u_{\varepsilon}(x_{k})-u_{\varepsilon}(x_{k}+z)-u_{\varepsilon}(x_{k}-z))\,K(0,z)\,dz=\int_{\mathbb{R}^{n}}(2u_{\varepsilon}(x)-u_{\varepsilon}(x+z)-u_{\varepsilon}(x-z))\,K(0,z)\,dz.$

In view of (2.4) and (4.36), this proves (4.46).

We also observe that

(4.47) $Au_{\varepsilon}=f_{\varepsilon}$ pointwise in any ball $B_{r}(x_{0})$
such that $B_{r}(x_{0})\Subset B_{1}$.

Indeed, by (2.4), (4.42) and (4.46), if $x\in B_{r}(x_{0})$ and $\varphi\in
C^{\infty}_{0}(B_{r}(x_{0}))$,

$\displaystyle\int_{\mathbb{R}^{n}}f_{\varepsilon}(x)\varphi(x)\,dx=\iint_{\mathbb{R}^{2n}}(u_{\varepsilon}(x)-u_{\varepsilon}(y))\varphi(x)K(x,y)\,dx\,dy.$

Since $\varphi$ is arbitrary, we arrive at

$\displaystyle
f_{\varepsilon}(x)=\int_{\mathbb{R}^{n}}(u_{\varepsilon}(x)-u_{\varepsilon}(y))K(x,y)\,dy,$

from which we obtain (4.47).

We also have that

(4.48) $Au_{\varepsilon}=f_{\varepsilon}$ in the viscosity sense in any ball
$B_{r}(x_{0})$ such that $B_{r}(x_{0})\Subset B_{1}$.

For this, we take a smooth function $\psi$ touching, say from below, the
function $u_{\varepsilon}$ at some point $p\in B_{r}(x_{0})$. Since the kernel
$K$ is positive (thanks to (1.8)) and recalling (4.47), we have that

$\displaystyle
f_{\varepsilon}(p)=\int_{\mathbb{R}^{n}}(u_{\varepsilon}(p)-u_{\varepsilon}(y))K(p,y)\,dy=\int_{\mathbb{R}^{n}}(\psi(p)-u_{\varepsilon}(y))K(p,y)\,dy$
$\displaystyle\qquad\qquad\leqslant\int_{\mathbb{R}^{n}}(\psi(p)-\psi(y))K(p,y)\,dy=A\psi(p).$

This and a similar computation when $\psi$ touches from above give (4.48).

We also remark that $u_{\varepsilon}$ and $f_{\varepsilon}$ converge uniformly
to $u$ and $f$, respectively, in any ball $B_{r}(x_{0})\Subset B_{1}$, due to
Theorem 9.8 in [MR3381284]. In addition, by (4.40), we see that, for every
$y\in\mathbb{R}^{n}\setminus B_{3r}(x_{0})$,

$|u_{\varepsilon}(y)|\leqslant\int_{B_{\varepsilon}}|u(y-z)|\rho_{\varepsilon}(z)\,dz\leqslant
C\int_{B_{\varepsilon}}|y-z|^{\alpha}\rho_{\varepsilon}(z)\,dz\leqslant
C|y|^{\alpha},$

up to renaming $C>0$. As a consequence of this and (1.8), we have that, for
every $y\in\mathbb{R}^{n}\setminus B_{3r}(x_{0})$,

$|u(y)-u_{\varepsilon}(y)|\,\sup_{x\in B_{r}(x_{0})}K(x,y)\leqslant
C|y|^{\alpha}\sup_{x\in
B_{r}(x_{0})}\frac{1}{|x-y|^{n+2s}}\leqslant\frac{C}{|y|^{n+2s-\alpha}},$

up to relabeling $C>0$. Since $\alpha<\min\\{2s,1\\}$, this function is in
$L^{1}(\mathbb{R}^{n}\setminus B_{3r}(x_{0}))$, and therefore we exploit the
Dominated Convergence Theorem to obtain that

$\lim_{\varepsilon\searrow 0}\int_{\mathbb{R}^{n}\setminus
B_{3r}(x_{0})}|u(y)-u_{\varepsilon}(y)|\,\sup_{x\in B_{1}}K(x,y)\,dy=0.$

Consequently, condition (4.23) is satisfied, and therefore we can apply Lemma
4.7, thus obtaining that $Au=f$ in the viscosity sense, as desired. ∎

With this preliminary work, we can now address the existence of solutions for
a Dirichlet problem in a generalized setting.

######  Theorem 4.13.

Let $K$ be as in (1.8) (2.4), (4.18), (4.35) and (4.36). Let $f$ be bounded
and continuous in $B_{1}$ and $g\in C^{\alpha}(\mathbb{R}^{n})$ for some
$\alpha\in(0,\min\\{2s,1\\})$. Assume that

$|g(x)|\leqslant C|x|^{\alpha}\qquad{\mbox{ for all
}}x\in\mathbb{R}^{n}\setminus B_{1}.$

Then, there exists a function $u\in\mathcal{V}_{K}$ such that

(4.49) $\begin{cases}Au\stackrel{{\scriptstyle m}}{{=}}f&{\mbox{ in
}}B_{1},\\\ u=g&{\mbox{ in }}B_{1}^{c}.\end{cases}$

Also, the solution to (4.49) is not unique, since the space of solutions of
(4.49) has dimension $N_{m}$, with

$N_{m}:=\sum_{j=0}^{m-1}\binom{j+n-1}{n-1}.$

###### Proof.

Firstly, we prove the existence of solutions for (4.49). For this goal, we set

$u_{1}:=\chi_{B_{4}^{c}}\,g\qquad{\mbox{and
}}\qquad\tilde{g}:=\chi_{B_{4}\setminus B_{1}}\,g.$

Since $u_{1}$ is identically zero in $B_{4}$ and
$K\in\mathcal{K}_{m,\vartheta}^{+}$, we can write
$Au_{1}\stackrel{{\scriptstyle m}}{{=}}f_{u_{1}}$ in $B_{1}$ in both pointwise
and viscosity sense, for some function $f_{u_{1}}$, due to Remark 2.6 and
Lemma 4.6.

We now define $\tilde{f}:=f-f_{u_{1}}$ and consider the Dirichlet problem
given by

(4.50) $\begin{cases}A\tilde{u}=\tilde{f}&{\mbox{ in }}B_{1},\\\
\tilde{u}=\tilde{g}&{\mbox{ in }}B_{1}^{c}.\end{cases}$

By Proposition 4.10, we find that (4.50) has a unique weak solution
$\tilde{u}$. Moreover, thanks to Proposition 4.12, we get that $\tilde{u}$ is
a viscosity solution of (4.50).

Furthermore, by Remark 4.4 and Corollary 4.8 we obtain that

$\begin{cases}A\tilde{u}\stackrel{{\scriptstyle m}}{{=}}\tilde{f}&{\mbox{ in
}}B_{1},\\\ \tilde{u}=\tilde{g}&{\mbox{ in }}B_{1}^{c}.\end{cases}$

Now, we set $u:=u_{1}+\tilde{u}$ and we get that
$Au=Au_{1}+A\tilde{u}\stackrel{{\scriptstyle m}}{{=}}f_{u_{1}}+\tilde{f}=f$ in
$B_{1}$. Moreover, we have that $u=u_{1}+\tilde{g}=g$ in $B_{1}^{c}$. These
observations give that is $u$ is solution of (4.49). This proves the existence
of solution for (4.49).

Now, we focus on the second part of the proof. Namely we establish that
solutions of (4.49) are not unique and we determine the dimension of the
corresponding linear space. For this, we notice that, exploiting Propositions
4.10 and 4.12, one can find a unique solution $\tilde{u}_{P}$ of the problem

(4.51) $\begin{cases}A\tilde{u}_{P}=P&{\mbox{ in }}B_{1},\\\
\tilde{u}_{P}=0&{\mbox{ in }}B_{1}^{c}.\end{cases}$

Furthermore, $A\tilde{u}_{P}\stackrel{{\scriptstyle 0}}{{=}}P$ in $B_{1}$, due
to Corollary 4.8. Using Remark 4.4, we obtain that
$A\tilde{u}_{P}\stackrel{{\scriptstyle m}}{{=}}P$ in $B_{1}$. Moreover, from
Remark 4.5, we obtain that $\tilde{u}_{P}$ is a solution of

(4.52) $\begin{cases}A\tilde{u}_{P}\stackrel{{\scriptstyle m}}{{=}}0&{\mbox{
in }}B_{1},\\\ \tilde{u}_{P}=0&{\mbox{ in }}B_{1}^{c}.\end{cases}$

This yields that if $u$ is a solution of (4.49), then $u+\tilde{u}_{P}$ is
also a solution of (4.49).

Viceversa, if $u$ and $v$ are two solutions of (4.49), then $w:=u-v$ is a
solution of

$\begin{cases}Aw\stackrel{{\scriptstyle m}}{{=}}0&{\mbox{ in }}B_{1},\\\
w=0&{\mbox{ in }}B_{1}^{c}.\end{cases}$

Here we can apply Lemma 4.9 with $j:=0$ thus obtaining that
$Aw\stackrel{{\scriptstyle 0}}{{=}}P$ in $B_{1}$, where $P$ is a polynomial of
$\deg P\leqslant m-1$. We use Corollary 4.8 one more time to find that

(4.53) $\begin{cases}Aw=P&{\mbox{ in }}B_{1},\\\ w=0&{\mbox{ in
}}B_{1}^{c}.\end{cases}$

Therefore, the uniqueness of the solution of (4.53), confronted with (4.51),
gives us that $w=\tilde{u}_{P}$, and thus $v=u+\tilde{u}_{P}$.

This reasoning gives that the space of solutions of (4.49) is isomorphic to
the space of polynomials with degree less than or equal to $m-1$, which has
dimension $N_{m}$, given by (3.3). ∎

## Acknowledgements

The first and third authors are members of INdAM and AustMS. The first author
has been supported by the Australian Research Council DECRA DE180100957 “PDEs,
free boundaries and applications”. The second author is supported by the
fellowship INDAM-DP-COFUND-2015 “INdAM Doctoral Programme in Mathematics
and/or Applications Cofunded by Marie Sklodowska-Curie Actions”, Grant 713485.
The third author has been supported by the Australian Laureate Fellowship
FL190100081 “Minimal surfaces, free boundaries and partial differential
equations”. Part of this work has been completed during a very pleasant visit
of the second author to the University of Western Australia, that we thank for
the warm hospitality.

## References