V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})^{\perp}$. For fixed $v\in
V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})^{\perp}$ the mapping
$\mathcal{J}(u+\cdot v):\mathbb{R}\to\mathbb{R}$,

$\displaystyle
t\mapsto\mathcal{J}(u+tv)=\mathcal{J}(u)+t\left[\mathcal{E}(u,v)-\int_{\Omega}f(x)v(x)\mathrm{d}x-\int_{\Omega^{c}}g(y)v(y)\mathrm{d}y\right]+\frac{t^{2}}{2}\mathcal{E}(v,v)$

is a polynomial of second order. For all $t\in\mathbb{R}$, $u+tv\in
V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})$ and since $u$ minimizes
$\mathcal{J}$ we get that $\mathcal{J}(u)\leq\mathcal{J}(u+tv)$ for all
$t\in\mathbb{R}$. Thus $\mathcal{J}(u+\cdot v):\mathbb{R}\to\mathbb{R}$ has a
critical point at $t=0$ which implies that

$\displaystyle 0=\lim_{t\to
0}\frac{\mathcal{J}(u+tv)-\mathcal{J}(u)}{t}=\lim_{t\to
0}\left[\mathcal{E}(u,v)-\int_{\Omega}f(x)v(x)\mathrm{d}x-\int_{\Omega^{c}}g(y)v(y)\mathrm{d}y+\frac{t}{2}\mathcal{E}(v,v)\right]$

equivalently

$\displaystyle\mathcal{E}(u,v)=\int_{\Omega}f(x)v(x)\mathrm{d}x+\int_{\Omega^{c}}g(y)v(y)\mathrm{d}y\,.$

This shows the equivalence between variational problem ($V^{\perp}$) and the
minimization problem ($M^{\perp}$). Meanwhile, if the compatibility condition
($C$) holds, then it is easy to observe that the relation in ($V^{\perp}$)
remains unchanged under additive constant and
$\mathcal{J}(v+c)=\mathcal{J}(v)$ for all $v\in
V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})$ and all $c\in\mathbb{R}.$
Accordingly, if $u\in V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})^{\perp}$
solves ($V^{\perp}$) then we have $\mathcal{J}(u+c)=\min\limits_{v\in
V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})}\mathcal{J}(v)$ which, by
similar arguments as above, is equivalent to ($V$). ∎

From 4.6 and 4.7 we deduce that, analogous to the case of the Laplace
operator, the complement condition $\operatorname{\mathcal{N}\\!\\!}u=0$ turns
out to be a natural condition in the variational context:

######  Corollary 4.8.

Let $f\in L^{2}(\Omega)$. Assume $u\in
V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})$ minimizes the functional
$v\mapsto\frac{1}{2}\operatorname{\mathcal{E}}(v,v)-\int_{\Omega}fv$ in the
space $V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})$. Then
$\operatorname{\mathcal{N}\\!\\!}u=0$ in $\Omega^{c}$.

A different version of this observation is given in [DLV21, Theorem 2.1].
[MPL19, Theorem 2.8] is similar in the translation invariant case. We are now
in position to state the existence and the uniqueness of a solution to
($V^{\perp}$) and hence to ($V$) up to additive constant. A direct application
of the Lax-Milgram lemma leads to the following observation.

######  Theorem 4.9.

We assume that $\Omega\subset\operatorname{\mathbb{R}}^{d}$ is open and
bounded. Let $\nu:\operatorname{\mathbb{R}}^{d}\to[0,\infty]$ be the density
of a symmetric Lévy measure with full support. We further assume that the
couple $(\nu,\Omega)$ belongs to one of the class
$\mathscr{A}_{i},\leavevmode\nobreak\ i=0,1,2,3$. Let $f\in
V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})^{\prime}$ and $g\in
T_{\nu}(\Omega^{c})^{\prime}$.

  1. (i)

There exists a unique solution $u\in
V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})^{\perp}$ to the problem
($V^{{}^{\prime}\perp}$) satisfying

$\displaystyle\|u\|_{V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})}\leq
C\left(\|f\|_{V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})^{\prime}}+\|g\|_{T_{\nu}(\Omega^{c})^{\prime}}\right)$

with a positive constant $C$, which depends only on $d,\Omega,\Lambda$ and
$\nu$.

  2. (ii)

Problem ($V^{\prime}$) is solvable if and only if $\langle f,1\rangle+\langle
g,1\rangle=0$. All solutions $w$ are of the form $w=u+c$ with $c\in\mathbb{R}$
and satisfy

$\displaystyle\|w-\hbox{$\fint_{\Omega}w$}\|_{V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})}\leq
C\left(\|f\|_{V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})^{\prime}}+\|g\|_{T_{\nu}(\Omega^{c})^{\prime}}\right)\,.$

###### Proof.

The existence and the uniqueness of solutions of ($V^{{}^{\prime}\perp}$)
follow from the Lax-Milgram lemma. The bilinear form
$\mathcal{E}(\cdot,\cdot)$ is continuous on
$V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})^{\perp}$. From the Poincaré
inequality (3.10) we conclude

$\displaystyle\|v\|_{L^{2}(\Omega)}^{2}\leq
C\mathcal{E}(v,v)\quad\quad\text{for all }\leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ v\in
V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})^{\perp}$

for some positive constant $C$. This implies coercivity of
$\mathcal{E}(\cdot,\cdot)$ on
$V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})^{\perp}$ and we obtain

$\displaystyle\mathcal{E}(v,v)\geq\big{(}1+C\big{)}^{-1}\|v\|^{2}_{V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})}.$
(4.7)

Note that, due to the continuity of the trace operator
$\operatorname{Tr}:V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})\to
T_{\nu}(\Omega^{c})$, the mapping $v\mapsto\langle f,v\rangle+\langle
g,v\rangle$ is linear and continuous on
$V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})^{\perp}$. The Lax-Milgram lemma
implies (i).

For $v\in V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})$ set
$v=\widetilde{v}+c^{\prime}$ with
$c^{\prime}=\mbox{$\fint_{\Omega}v\mathrm{d}x$}$ so that $\widetilde{v}\in
V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})^{\perp}$. In addition, every
constant function $w=c$ belongs to
$V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})$ for every
$c\in\operatorname{\mathbb{R}}^{d}$ because $\Omega$ is bounded. Hence,
$V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})=V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})^{\perp}\oplus\mathbb{R}$.
With this observation along with the identity
$\mathcal{E}(u+c,v+c^{\prime})=\mathcal{E}(u,v)$ for all
$c,c^{\prime}\in\mathbb{R}$ and the uniqueness of $u\in
V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})^{\perp}$ solving
($V^{{}^{\prime}\perp}$) it becomes easy to check that under the compatibility
condition $\langle f,1\rangle+\langle g,1\rangle=0$, all solutions of
($V^{\prime}$) are of the form $u+c$.

∎

######  Remark 4.10.

It worth to mention that, Theorem 4.9 (i) implies that the operator
$\Phi:V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})^{\prime}\times
T_{\nu}(\Omega^{c})^{\prime}\to
V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})^{\perp}$ mapping $(f,g)$ to the
unique solution $u\in V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})^{\perp}$
of the variational problem ($V^{{}^{\prime}\perp}$) is linear, one-to-one,
continuous with

$\displaystyle\|\Phi(f,g)\|_{V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})}\leq
C\|(f,g)\|_{V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})^{\prime}\times
T_{\nu}(\Omega^{c})^{\prime}}\,.$

Let us apply Theorem 4.9 in order to prove our main existence result.

######  Theorem 4.11.

Under the assumptions of Theorem 4.9 with $f\in L^{2}(\Omega)$ and $g\in
L^{2}(\Omega^{c},\widetilde{\nu}^{-1})$ the following holds true:

  1. (i)

There exists a unique solution $u\in
V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})^{\perp}$ to the problem
($V^{\perp}$) satisfying

$\displaystyle\|u\|_{V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})}\leq
C\left(\|f\|_{L^{2}(\Omega)}+\|g\|_{L^{2}(\Omega^{c},\widetilde{\nu}^{-1})}\right)$

with a positive constant $C$, which depends only on $d,\Omega,\Lambda$ and
$\nu$.

  2. (ii)

Problem ($V$) is solvable if and only if ($C$) holds true. All solutions $w$
are of the form $w=u+c$ with $c\in\mathbb{R}$ and satisfy

$\displaystyle\|w-\hbox{$\fint_{\Omega}w$}\|_{V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})}\leq
C\left(\|f\|_{L^{2}(\Omega)}+\|g\|_{L^{2}(\Omega^{c},\widetilde{\nu}^{-1})}\right)\,.$
(4.8)

###### Proof.

It suffices to show the continuity of the associated linear forms. For $v\in
V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})^{\perp}$

$\displaystyle\Big{|}\int_{\Omega}fv\mathrm{d}x\Big{|}\leq\|f\|_{L^{2}(\Omega)}\|v\|_{L^{2}(\Omega)}\leq\|f\|_{L^{2}(\Omega)}\|v\|_{V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})}\,.$
(4.9)

From $g\in L^{2}(\Omega^{c},\widetilde{\nu}^{-1})$ and the continuity of
$\operatorname{Tr}:V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})\hookrightarrow
T_{\nu}(\Omega^{c}),$ we obtain

$\displaystyle\Big{|}\int_{\Omega^{c}}g(x)v(x)\mathrm{d}x\Big{|}\leq\|g\|_{L^{2}(\Omega^{c},\widetilde{\nu}^{-1})}\|v\|_{L^{2}(\Omega^{c},\widetilde{\nu})}\leq
C\|g\|_{L^{2}(\Omega^{c},\widetilde{\nu}^{-1})}\|v\|_{V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})}.$
(4.10)

Application of Theorem 4.9 completes the proof. ∎

There is an alternative formulation of Theorem 4.11, which allows for more
general inhomogeneities $g$. Let us define a modified Neumann problem for the
operator $L$ associated to the data $f$ and $g$ as follows:

$\displaystyle Lu=f\quad\text{in}\leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \Omega\quad\quad\text{ and
}\quad\quad\mathcal{N}u=g\widetilde{\nu}\quad\text{ on
}\operatorname{\mathbb{R}}^{d}\setminus\Omega.$ ($N_{*}$)

######  Theorem 4.12.

Under the assumptions of Theorem 4.9 with $f\in L^{2}(\Omega)$ and $g\in
L^{2}(\Omega^{c},\widetilde{\nu})$, then the following holds true:

  1. (i)

There exists a unique weak solution $u_{*}\in
V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})^{\perp}$ to the problem
($N_{*}$), that is

$\displaystyle\mathcal{E}(u_{*},v)=\int_{\Omega}f(x)v(x)\mathrm{d}x+\int_{\Omega^{c}}g(y)v(y)\widetilde{\nu}(y)\mathrm{d}y\quad\text{
for all }v\in V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})^{\perp}\,.$
($V_{*}^{\perp}$)

satisfying

$\displaystyle\|u_{*}\|_{V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})}\leq
C\left(\|f\|_{L^{2}(\Omega)}+\|g\|_{L^{2}(\Omega^{c},\widetilde{\nu})}\right)$

with a positive constant $C$, which depends only on $d,\Omega,\Lambda$ and
$\nu$.

  2. (ii)

Problem ($V_{*}$) is solvable if and only if ($C_{*}$) holds true, where

$\displaystyle\mathcal{E}(u,v)$
$\displaystyle=\int_{\Omega}f(x)v(x)\mathrm{d}x+\int_{\Omega^{c}}g(y)v(y)\widetilde{\nu}(y)\mathrm{d}y\quad\text{
for all }v\in V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})\,,$ ($V_{*}$)
$\displaystyle\int_{\Omega}f(x)\mathrm{d}x+\int_{\Omega^{c}}g(y)\widetilde{\nu}(y)\mathrm{d}y=0\,.$
($C_{*}$)

All solutions $w_{*}$ are of the form $w=u_{*}+c$ with $c\in\mathbb{R}$ and
satisfy

$\displaystyle\|w-\hbox{$\fint_{\Omega}w$}\|_{V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})}\leq
C\left(\|f\|_{L^{2}(\Omega)}+\|g\|_{L^{2}(\Omega^{c},\widetilde{\nu})}\right)\,.$

The proof of Theorem 4.12 is analogous to the one of Theorem 4.11. Note that,
if $g\in L^{2}(\Omega^{c},\widetilde{\nu})$, then

$\displaystyle\Big{|}\int_{\Omega^{c}}g(x)v(x)\widetilde{\nu}(x)\mathrm{d}x\Big{|}\leq
C\|g\|_{L^{2}(\Omega^{c},\widetilde{\nu})}\|v\|_{V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})}\,.$

The last result in this section concerns the non-existence of weak solutions
when the Neumann data $g$ is not in the weighted trace space
$L^{2}(\Omega^{c},\widetilde{\nu}^{-1})$.

######  Theorem 4.13 (Non-existence of weak solution).

Let $B_{1}=B_{1}(0)$ be the unit ball in $\operatorname{\mathbb{R}}^{d}$. Let
$\nu(h)=|h|^{-d-\alpha}$, $\alpha\in(0,2)$ so that
$\widetilde{\nu}(h)\asymp(1+|h|)^{-d-\alpha}$. Let $f=0$. There exists $g\in
L^{1}(B_{1}^{c})\setminus L^{2}(B_{1}^{c},\widetilde{\nu}^{-1})$ with
$\int_{B_{1}^{c}}g(y)\mathrm{d}y=0$, for which the Neumann problem $Lu=0$ on
$B_{1}$ and $\mathcal{N}u=g$ on $\operatorname{\mathbb{R}}^{d}\setminus B_{1}$
has no weak solution in $V_{\nu}(B_{1}|\operatorname{\mathbb{R}}^{d})$.

###### Proof.

We will construct a function $g$ of the form $g=g_{\gamma}\widetilde{\nu}$
where, given $x\in\operatorname{\mathbb{R}}^{d}$,

$\displaystyle
g_{\gamma}(x)=\frac{x_{1}}{|x|}(|x|-1)^{\gamma}\mathds{1}_{B^{c}_{1}(0)}(x)\,,$

with an appropriate choice of
$\gamma\in(-1,-\frac{\alpha+1}{2})\cup(\frac{\alpha}{2},\frac{\alpha+1}{2})$.
Note that for $x\in B^{c}_{1}(0)$ we have $\operatorname{dist}(x,\partial
B_{1}(0))=(|x|-1)$ and

$\displaystyle\int_{B_{1}(0)}\frac{\mathrm{d}y}{|x-y|^{d+\alpha}}\asymp(|x|-1)^{-\alpha}\land(|x|-1)^{-d-\alpha}.$

Claim 1: $g_{\gamma}\in V_{\nu}(B_{1}|\operatorname{\mathbb{R}}^{d})$ if and
only if $\gamma\in(\frac{\alpha-1}{2},\frac{\alpha}{2})$. Integration in polar
coordinates yields

$\displaystyle\|g_{\gamma}\|^{2}_{V_{\nu}(B_{1}|\operatorname{\mathbb{R}}^{d})}$
$\displaystyle=2\int_{B^{c}_{1}(0)}\frac{x_{1}^{2}}{|x|^{2}}(|x|-1)^{2\gamma}\int_{B_{1}(0)}|x-y|^{-d-\alpha}\mathrm{d}y\,\mathrm{d}x$
$\displaystyle\asymp\int_{B^{c}_{1}(0)}\frac{x_{1}^{2}}{|x|^{2}}(|x|-1)^{2\gamma-\alpha}(1\land(|x|-1)^{-d})\mathrm{d}x$
$\displaystyle\asymp\Big{(}\int_{0}^{1}r^{2\gamma-\alpha}\mathrm{d}r+\int_{1}^{\infty}r^{2\gamma-\alpha-1}\mathrm{d}r\Big{)}.$

Claim 2: $g_{\gamma+\beta}\in L^{1}(B_{1}^{c},\widetilde{\nu})$ if and only if
$\gamma+\beta\in(-1,\alpha)$. Indeed,

$\displaystyle\|g_{\gamma+\beta}\|_{L^{1}(B_{1}^{c},\widetilde{\nu})}$
$\displaystyle=\int_{B^{c}_{1}(0)}\frac{x_{1}^{2}}{|x|^{2}}(|x|-1)^{\gamma+\beta}(1+|x|)^{-d-\alpha}\mathrm{d}x$
$\displaystyle\asymp|\mathbb{S}^{d-1}|\Big{(}\int_{0}^{1}r^{\gamma+\beta}\mathrm{d}r+\int_{1}^{\infty}r^{\gamma+\beta-\alpha-1}\mathrm{d}r\Big{)}.$

Claim 3: Analogously, $g_{\gamma}\in L^{2}(B_{1}^{c},\widetilde{\nu})$ if and
only if $\gamma\in(-\frac{1}{2},\frac{\alpha}{2})$.
Claim 4: $g\in L^{1}(B_{1}^{c})\setminus
L^{2}(B_{1}^{c},\widetilde{\nu}^{-1})$ if and only if $g_{\gamma}\in
L^{1}(B_{1}^{c},\widetilde{\nu})\setminus L^{2}(B_{1}^{c},\widetilde{\nu})$ if
and only if $\gamma\in(-1,-\frac{1}{2}]\cup[\frac{\alpha}{2},\alpha)$ by Claim
2 and Claim 3.

From now on, we assume
$\gamma\in(1,-\frac{\alpha+1}{2})\cup(\frac{\alpha}{2},\frac{\alpha+1}{2})\subset(-1,-\frac{1}{2}]\cup[\frac{\alpha}{2},\alpha)$.
Claim 5: Since $g_{\gamma}(x)=-g_{\gamma}(-x)$ it follows that
$g=g_{\gamma}\widetilde{\nu}$ satisfies the compatibility condition

$\displaystyle\int_{B_{1}^{c}}g(y)\mathrm{d}y=\int_{B_{1}^{c}}g_{\gamma}(y)\widetilde{\nu}(y)\mathrm{d}y=0.$

We assume the Neumann problem $Lu=0$ on $B_{1}$ and $\mathcal{N}u=g$ on
$\operatorname{\mathbb{R}}^{d}\setminus B_{1}$ has a weak solution $u\in
V_{\nu}(B_{1}|\operatorname{\mathbb{R}}^{d})$, that is we have

$\displaystyle\mathcal{E}(u,v)=\int_{B_{1}^{c}}g_{\gamma}(y)v(y)\widetilde{\nu}(y)\mathrm{d}y\qquad\text{for
all}\quad v\in V_{\nu}(B_{1}|\operatorname{\mathbb{R}}^{d}).$

This implies that for the constant
$C=1+\|u\|_{V_{\nu}(B_{1}|\operatorname{\mathbb{R}}^{d})}>0$ we have

$\displaystyle\Big{|}\int_{B_{1}^{c}}g_{\gamma}(y)v(y)\widetilde{\nu}(y)\mathrm{d}y\Big{|}\leq
C\|v\|_{V_{\nu}(B_{1}|\operatorname{\mathbb{R}}^{d})}\quad\text{ for all}\quad
v\in V_{\nu}(B_{1}|\operatorname{\mathbb{R}}^{d}).$ (4.11)

Claim 1 allows us to take $v=g_{\beta}\in
V_{\nu}(B_{1}|\operatorname{\mathbb{R}}^{d})$ for
$\beta\in(\tfrac{\alpha-1}{2},\tfrac{\alpha}{2})$. Then we obtain

$\displaystyle\|g_{\gamma+\beta}\|_{L^{1}(B_{1}^{c},\widetilde{\nu})}\leq
c(d)\int_{B_{1}^{c}}g_{\gamma}(y)g_{\beta}(y)\widetilde{\nu}(y)\mathrm{d}y\leq
C\|g_{\beta}\|_{V_{\nu}(B_{1}|\operatorname{\mathbb{R}}^{d})}\,,$ (4.12)

where we have used (4.11). Finally, we consider two cases. If $\alpha\geq 1$,
then we choose $\gamma\in(\frac{\alpha}{2},\frac{\alpha+1}{2})$ and
$\beta=\alpha-\gamma$. If $\alpha\leq 1$, then we choose
$\gamma\in(-1,-\frac{\alpha+1}{2})$ and $\beta=-\gamma-1$. In both cases,
$\gamma\in(-1,-\frac{1}{2}]\cup[\frac{\alpha}{2},\alpha)$,
$\beta\in(\frac{\alpha-1}{2},\frac{\alpha}{2})$ and
$\gamma+\beta\in\\{-1,\alpha\\}$. This implies $g_{\gamma}\in
L^{1}(B_{1}^{c},\widetilde{\nu})\setminus L^{2}(B_{1}^{c},\widetilde{\nu})$
and $g_{\beta}\in V_{\nu}(B_{1}|\operatorname{\mathbb{R}}^{d})$ whereas
$\|g_{\gamma+\beta}\|_{L^{1}(B_{1}^{c},\widetilde{\nu})}=\infty$. This
contradicts (4.12) since
$\|g_{\beta}\|_{V_{\nu}(B_{1}|\operatorname{\mathbb{R}}^{d})}<\infty$.

∎

###  4.2. Neumann eigenvalues of $L$

Let $f\in L^{2}(\Omega)$, for $g=0$ it is worthwhile to see that the
variational problem ($V$) coincides with ($V_{*}$) and both correspond to the
variational(weak) formulation of the homogeneous Neumann problem $Lu=f$ in
$\Omega$ and $\mathcal{N}u=0$ on $\Omega^{c}$.

######  Definition 4.14 (Neumann eigenvalue of $L$).

A non-zero function $u\in V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})$ is
called a Neumann eigenfunction of the operator $L$ on $\Omega$ if there exists
a real number $\mu$, which is the eigenvalue associated to $u$, such that for
all $v\in V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})$

$\displaystyle\mathcal{E}(u,v)=\mu\int_{\Omega}u(x)v(x)\mathrm{d}x.$

One formally writes $Lu=\mu u$ in $\Omega$ and $\mathcal{N}u=0$ on
$\Omega^{c}$, which corresponds to the aforementioned weak formulation
provided that $u$ is sufficiently regular.

It is worth noticing that, if $u$ is a Neumann eigenfunction of $L$ with
associated eigenvalue $\mu$, then either $u\in
V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})^{\perp}$ when $\mu\neq 0$ or
else, $\mu=0$ and the constant functions $u=c,\leavevmode\nobreak\
c\in\mathbb{R}\setminus\\{0\\}$, are the related eigenfunctions.

######  Theorem 4.15.

Assume $\Omega\subset\operatorname{\mathbb{R}}^{d}$ is bounded and open and
$\nu:\operatorname{\mathbb{R}}^{d}\setminus\\{0\\}\to[0,\infty)$ is the
density of a symmetric Lévy measure with full support. Assume that the couple
$(\nu,\Omega)$ belongs to one of the classes
$\mathscr{A}_{i},\leavevmode\nobreak\ i=1,2,3$. Then there exists a sequence
$(\phi_{n})_{\in\mathbb{N}_{0}}$ in
$V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})$, which forms an orthonormal
basis of $L^{2}(\Omega)$, and an increasing sequence of real numbers
$0=\mu_{0}<\mu_{1}\leq\cdots\leq\mu_{n}\leq\cdots.$ such that
$\mu_{n}\to\infty$ as $n\to\infty$ and each $\phi_{n}$ is a Neumann
eigenfunction of $L$ with corresponding eigenvalue $\mu_{n}$. The number of
each eigenvalue is given by its geometric multiplicity.

###### Proof.

For $f_{1},f_{2}\in L^{2}(\Omega)$ let us denote
$u_{f_{k}}=\Phi_{0}(f_{k})=\Phi(f_{k},0)\in
V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})^{\perp},\leavevmode\nobreak\
k=1,2$ the unique solution of ($V^{\perp}$) with Neumann data $f=f_{k}$ and
$g=0$. Precisely,

$\displaystyle\mathcal{E}(\Phi_{0}(f_{k}),v)=\int_{\Omega}f_{k}(x)v(x)\mathrm{d}x\quad\mbox{for
all}\leavevmode\nobreak\ \leavevmode\nobreak\ v\in
V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})^{\perp}\,.$ (4.13)

Testing (4.13) against $v=\Phi_{0}(f_{2})$ and $v=\Phi_{0}(f_{1})$
successively when $k=1$ and $k=2$ yields

$\displaystyle\big{(}f_{1},\Phi_{0}(f_{2})\big{)}_{L^{2}(\Omega)}=\mathcal{E}(\Phi_{0}(f_{1}),\Phi_{0}(f_{2}))=\mathcal{E}(\Phi_{0}(f_{2}),\Phi_{0}(f_{1}))=\big{(}f_{2},\Phi_{0}(f_{1})\big{)}_{L^{2}(\Omega)}.$

Therefore, the operator
$R_{\Omega}\circ\Phi_{0}:L^{2}(\Omega)\xrightarrow[]{\Phi_{0}}V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})^{\perp}\xrightarrow[]{R_{\Omega}}L^{2}(\Omega)^{\perp}$
is compact (by Theorem 3.10) and symmetric hence self-adjoint. It is a fact
from the spectral theory of compact self-adjoint operators that
$L^{2}(\Omega)^{\perp}$ has an orthonormal basis $(e_{n})_{n}$ whose elements
are eigenfunctions of $R_{\Omega}\circ\Phi_{0}$ and the sequence of the
corresponding eigenvalues are non-negative real numbers $(r_{n})_{n}$ which we
assume ordered in the decreasing order, $r_{1}\geq r_{2}\geq\cdots\geq
r_{n}\geq\cdots 0$ such that $r_{n}\to 0$ as $n\to\infty$. Precisely, for each
$n\geq 1$, $R_{\Omega}\circ\Phi_{0}(e_{n})=r_{n}e_{n}$ or simply write
$\Phi_{0}(e_{n})=r_{n}e_{n}$ a.e in $\Omega$. Combining the latter relation
with definition of $\Phi_{0}(e_{n})$ we get

$\displaystyle\mathcal{E}(\Phi_{0}(e_{n}),v)$
$\displaystyle=\int_{\Omega}e_{n}(x)v(x)\mathrm{d}x=r^{-1}_{n}\int_{\Omega}\Phi_{0}(e_{n})(x)v(x)\mathrm{d}x\,\quad\mbox{for
all}\leavevmode\nobreak\ \leavevmode\nobreak\ v\in
V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})^{\perp}\,.$

Equivalently, setting $\mu_{n}=r^{-1}_{n}$ and
$\phi_{n}=\Phi_{0}(e_{n})/\|\Phi_{0}(e_{n})\|_{L^{2}(\Omega)}=r_{n}^{-1}\Phi_{0}(e_{n})$
which is clearly an element of $\in
V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})^{\perp}$ yields

$\displaystyle\mathcal{E}(\phi_{n},v)$
$\displaystyle=\mu_{n}\int_{\Omega}\phi_{n}(x)v(x)\mathrm{d}x\,\quad\mbox{for
all}\leavevmode\nobreak\ \leavevmode\nobreak\ v\in
V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})^{\perp}\,.$

Hereby, along with $\mu_{0}=0$ and $\phi_{0}=|\Omega|^{-1}$ provides the
sequences sought for. Now if we assume $\mu_{1}=0$ then we have $\phi_{1}\in
V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})^{\perp}$ and
$\mathcal{E}(\phi_{1},v)=0$ for all $v\in
V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})^{\perp}$ in particular
$\mathcal{E}(\phi_{1},\phi_{1})=0$ i.e $\phi_{1}$ is a constant function in
$V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})^{\perp}$ necessarily
$\phi_{1}=0$ since $u_{1}$ has zero mean over $\Omega$. We have therefore
reached a contradiction as $\phi_{1}$ is supposed to be an eigenfunction i.e
$\phi_{1}\neq 0$. Thus, $\mu_{1}>0$ and the proof is complete.

∎

###  4.3. Robin boundary condition

In this section we treat a Robin-type problem with respect to the nonlocal
operator $L$ on $\Omega$. In the classical setting for the Laplace operator,
the Robin boundary problem – also known as Fourier boundary problem or third
boundary problem – is a combination of the Dirichlet and Neumann boundary
problem in the form333According to the over 20 years survey work [GA98], there
is no historical evidence why the problem (4.14) is termed after Robin’s name.
The survey [GA98, p.69] also points out that the first mathematical appearance
of the problem (4.14) goes back at least to the works on cooling law by
Fourier(1822) and/or Newton (1701, but mathematical contribution by Newton is
uncertain).

$\displaystyle-\Delta u=f\text{ in }\Omega\quad\text{ and }\quad\frac{\partial
u}{\partial n}+\beta u=g\text{ on }\partial\Omega.$ (4.14)

Here $f\in L^{2}(\Omega)$ and the measurable functions
$\beta,g:\partial\Omega\to\mathbb{R}$ are given. Analogously, in the nonlocal
set up, we assume that $\beta,g:\Omega^{c}\to\mathbb{R}$ are measurable
functions. The Robin problem consists in finding a measurable function
$u:\operatorname{\mathbb{R}}^{d}\to\mathbb{R}$ such that

$\displaystyle Lu=f\text{ in }\Omega\quad\text{ and }\quad\mathcal{N}u+\beta
u=g\text{ on }\Omega^{c}.$ (4.15)

Note that, for $\beta=0$ one recovers the inhomogeneous Neumann problem.
Informally, for $\beta\to\infty$ it leads to the homogeneous Dirichlet
problem. Define the quadratic form

$\displaystyle
Q_{\beta}(u,v)=\mathcal{E}(u,v)+\int_{\Omega^{c}}u(y)v(y)\beta(y)\mathrm{d}y.$

A function $u\in V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})$ is called a
weak solution of the Robin problem (4.15) if

$\displaystyle
Q_{\beta}(u,v)=\int_{\Omega}f(x)v(x)\mathrm{d}x+\int_{\Omega^{c}}g(y)v(y)\mathrm{d}y\text{
for all }v\in V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d}).$ (4.16)

######  Theorem 4.16.

Let $\nu$ and $\Omega$ be as in Theorem 4.15. Assume that
$\beta\widetilde{\nu}^{-1}:\Omega^{c}\to[0,\infty)$ is essentially bounded and
$\beta$ is non-trivial that is, $|\Omega^{c}\cap\\{\beta>0\\}|>0$. Let $f\in
L^{2}(\Omega)$ and $g\in L^{2}(\Omega^{c},\widetilde{\nu}^{-1})$. There exists
a unique function $u\in V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})$
solution to (4.16) satisfying

$\displaystyle\|u\|_{V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})}\leq
C\left(\|f\|_{L^{2}(\Omega)}+\|g\|_{L^{2}(\Omega^{c},\widetilde{\nu}^{-1})}\right)\,,$
(4.17)

where $C:=C(d,\Omega,\Lambda,\nu,\beta)>0$ can be chosen independently of $u$,
$f$ and $g$.

######  Remark 4.17.

The operator $\Psi:L^{2}(\Omega)\times
L^{2}(\Omega^{c},\widetilde{\nu}^{-1})\to
V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})$ mapping the data $(f,g)$ to the
unique solution $u\in V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})$ of the
variational problem (4.16) is linear, one-to-one, and continuous. Moreover,
with $C$ as above,

$\displaystyle\|\Psi(f,g)\|_{V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})}\leq
C\|(f,g)\|_{L^{2}(\Omega)\times L^{2}(\Omega^{c},\widetilde{\nu}^{-1})}.$

###### Proof.

First of all, we claim that the form $Q_{\beta}(\cdot,\cdot)$ is coercive on
$V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})$. Assume it is not true. Then
for each $n\geq 1$ there exists $u_{n}\in
V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})$ with
$\|u_{n}\|_{V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})}=1$ such that

$\displaystyle\mathcal{E}(u_{n},u_{n})+\int_{\Omega^{c}}|u_{n}(y)|^{p}\beta(y)\mathrm{d}y=Q_{\beta}(u_{n},u_{n})<\frac{1}{2^{n}}\,.$

In virtue of our compactness result, Theorem 3.10, $(u_{n})_{n}$ converges up
to a subsequence in $L^{2}(\Omega)$ to some $u\in
V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})$. We deduce
$\|u\|_{L^{2}(\Omega)}=1$, since
$\mathcal{E}(u_{n},u_{n})\xrightarrow[]{n\to\infty}0$ and for all $n\geq 1$,
$\|u_{n}\|_{V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})}=1$. From
$\mathcal{E}(u_{n},u_{n})\xrightarrow{n\to\infty}0$ and
$\|u_{n}-u\|_{L^{p}(\Omega)}\xrightarrow{n\to\infty}0$ we obtain that $u_{n}$
converges to $u$ in $V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})$ with
$\mathcal{E}(u,u)=0$. Thus $u$ is constant almost everywhere in
$\operatorname{\mathbb{R}}^{d}$. On the other hand, since $\beta$ is bounded
and the embedding
$V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})\hookrightarrow
L^{2}(\Omega^{c},\widetilde{\nu})$, see 2.17, is continuous, we have

$\displaystyle\int_{\Omega^{c}}u^{2}(y)\beta(y)\mathrm{d}y$ $\displaystyle\leq
2\int_{\Omega^{c}}u_{n}^{2}(y)\beta(y)\mathrm{d}y+2\|\beta\widetilde{\nu}^{-1}\|_{L^{\infty}(\Omega^{c})}\int_{\Omega^{c}}(u_{n}(y)-u(y))^{2}\widetilde{\nu}(y)\mathrm{d}y$
$\displaystyle\leq
2Q_{\beta}(u_{n},u_{n})+C\|u_{n}-u\|^{2}_{V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})}\xrightarrow[]{n\to\infty}0\,.$

From this, we conclude $u=0$ since we know that $u$ is a constant function and
$\beta>0$ almost everywhere on a set of positive measure $U\subset\Omega^{c}$
on which $u$ vanishes. This contradicts $\|u\|_{L^{2}(\Omega)}=1$ and hence
our initial assumption was wrong. Therefore there exists a constant
$C=C(d,\Omega,\nu,\beta)>0$ such that

$\displaystyle Q_{\beta}(u,u)\geq
C\|u\|^{2}_{V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})}\quad\text{for all
}\leavevmode\nobreak\ u\in V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d}).$
(4.18)

The remaining requirements for the application of the Lax-Milgram lemma can be
checked easily. Existence of a unique solution to (4.16) follows. The estimate
(4.17) is a direct consequence of (4.18). ∎

###  4.4. Dirichlet-to-Neumann map

In this section we define the Dirichlet-to-Neumann map related to the nonlocal
Lévy operator $L$ under consideration. Afterwards we prove that its spectrum
is strongly connected to the Robin eigenvalues of the operator $L$. This was
originally introduced in [FG20]. We refer the interested reader to the
expositions [AM12, BtE15] where the Dirichlet-to-Neumann map is treated in the
local setting for the Laplacian. We point out that an attempt to define the
Dirichlet-to-Neumann map is provided in [Von21]. For the case for the
fractional Laplacian a different Dirichlet-to-Neumann map to ours is derived
in [GSU16], see also the variant for fractional regional operators in [War15,
War18]. Let us first review the nonlocal Dirichlet problem. In the spirit of
[FKV15] one can easily prove the following

######  Theorem 4.18.

Let $\Omega\subset\operatorname{\mathbb{R}}^{d}$ be open and bounded. Given
$f\in L^{2}(\Omega)$ and $g\in T_{\nu}(\Omega^{c})$, there exists a unique
function $u\in V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})$ with $u=g$ a.e.
on $\Omega^{c}$ and

$\displaystyle\mathcal{E}(u,v)=\int_{\Omega}f(x)v(x)\mathrm{d}x\qquad\text{for
all }\leavevmode\nobreak\ \leavevmode\nobreak\ v\in
V_{\nu,0}(\Omega|\operatorname{\mathbb{R}}^{d}).$ (4.19)

In fact, $u$ is the weak solution to the nonlocal Dirichlet problem $Lu=f$ in
$\Omega$ and $u=g$ on $\Omega^{c}$. Moreover, there exists
$C=C(\Omega,d,\Lambda,\nu)>0$ independent of $f$ and $g$,

$\displaystyle\|u\|_{V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})}\leq
C(\|f\|_{L^{2}(\Omega)}+\|g\|_{T_{\nu}(\Omega^{c})}).$ (4.20)

The result follows from the Lax-Milgram lemma because the linear form
$v\mapsto\int_{\Omega}fv$ is continuous on
$V_{\nu,0}(\Omega|\operatorname{\mathbb{R}}^{d})$ and the bilinear form
$\mathcal{E}(\cdot,\cdot)$ bounded and coercive on
$V_{\nu,0}(\Omega|\operatorname{\mathbb{R}}^{d})$ (see Theorem 3.13). It is
noteworthy to recall that under the non-integrability condition (I) and the
Lévy integrability condition (L),
$V_{\nu,0}(\Omega|\operatorname{\mathbb{R}}^{d})$ is compactly embedded in
$L^{2}(\Omega)$. With this at hand, analogously to Theorem 4.15 there exist a
family $(\psi_{n})_{n}$ elements of
$V_{\nu,0}(\Omega|\operatorname{\mathbb{R}}^{d})$, orthonormal basis of
$L^{2}(\Omega)$ and an increasing sequence of real number
$0<\lambda_{1}\leq\cdots\leq\lambda_{n}\leq\cdots.$ such that
$\lambda_{n}\to\infty$ as $n\to\infty$ and each $\psi_{n}$ is a Dirichlet
eigenfunction of $L$ whose corresponding eigenvalue is $\lambda_{n}$ namely

$\displaystyle\mathcal{E}(\psi_{n},v)=\lambda_{n}\int_{\Omega}\psi_{n}(x)v(x)\mathrm{d}x\quad\text{for
all }\leavevmode\nobreak\ \leavevmode\nobreak\ v\in
V_{\nu,0}(\Omega|\operatorname{\mathbb{R}}^{d}).$

Note that the constants $\mu_{1}>0$ and $\lambda_{1}>0$ respectively satisfy
the Poincaré inequalities

$\displaystyle\mathcal{E}(u,u)$
$\displaystyle\geq\mu_{1}\|u\|^{2}_{L^{2}(\Omega)},\quad\text{for
all}\leavevmode\nobreak\ u\in
V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})^{\perp},$
$\displaystyle\mathcal{E}(u,u)$
$\displaystyle\geq\lambda_{1}\|u\|^{2}_{L^{2}(\Omega)},\quad\text{for
all}\leavevmode\nobreak\ u\in
V_{\nu,0}(\Omega|\operatorname{\mathbb{R}}^{d}).$

Before we formally define the Dirichlet-to-Neumann map, some prerequisites are
required. Let $f\in L^{2}(\Omega)$ and $g\in T_{\nu}(\Omega^{c})$. Assume,
$\lambda<\lambda_{1}$, then the bilinear form
$\mathcal{E}_{-\lambda}(u,u)=\mathcal{E}(u,u)-\lambda\|u\|^{2}_{L^{2}(\Omega)}$
is coercive on $V_{\nu,0}(\Omega|\operatorname{\mathbb{R}}^{d})$. Thus there
exists a function $u\in V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})$ unique
weak solution to the Dirichlet problem $Lu-\lambda u=f$ in $\Omega$ and $u=g$
on $\Omega^{c}$. Explicitly, $u=g$ on $\Omega^{c}$ and

$\displaystyle\mathcal{E}(u,v)-\lambda\int_{\Omega}u(x)v(x)\mathrm{d}x=\int_{\Omega}f(x)v(x)\mathrm{d}x\quad\text{for
all }\leavevmode\nobreak\ \leavevmode\nobreak\ v\in
V_{\nu,0}(\Omega|\operatorname{\mathbb{R}}^{d}).$ (4.21)

Moreover, the estimate (4.20) (with the estimating constant depending on
$\lambda$) remains true. More generally, by the mean of Fredholm alternative
and the closed graph theorem, the preceding facts (4.21) and (4.20)
respectively remain true for the operator $L-\lambda$, whenever
$\lambda\in\mathbb{R}\setminus\\{\lambda_{n}:n\geq 1\\}$.

From now on we suppose $f=0$ and
$\lambda\in\mathbb{R}\setminus\\{\lambda_{n}:n\geq 1\\}$ and label the
solution of (4.21) by $u=u_{g}$. Then the mapping $g\mapsto u_{g}$ is linear
and continuous from $T_{\nu}(\Omega^{c})$ to
$V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})$ since by (4.20) we have

$\displaystyle\|u_{g}\|_{V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})}\leq
C\|g\|_{T_{\nu}(\Omega^{c})}.$

Given $v\in T_{\nu}(\Omega^{c})$, put $\widetilde{v}=\operatorname{ext}(v)\in
V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})$ as an extension of $v$. Let
$\langle\cdot,\cdot\rangle$ be the dual pairing between $T_{\nu}(\Omega^{c})$
and $T_{\nu}(\Omega^{c})^{\prime}$.

######  Definition 4.19.

Let $\lambda\in\mathbb{R}\setminus\\{\lambda_{n}:n\geq 1\\}$. We call the
mapping $\mathscr{D}_{\lambda}:T_{\nu}(\Omega^{c})\to
T_{\nu}(\Omega^{c})^{\prime}$ with
$g\mapsto\mathscr{D}_{\lambda}g=\mathcal{E}_{-\lambda}(u_{g},\widetilde{\cdot})$
such that
$\langle\mathscr{D}_{\lambda}g,v\rangle=\mathcal{E}_{-\lambda}(u_{g},\widetilde{v})$,
the Dirichlet-to-Neumann map with respect to the operator $L-\lambda$.

######  Theorem 4.20.

The Dirichlet-to-Neumann operator
$\mathscr{D}_{\lambda}:T_{\nu}(\Omega^{c})\to T_{\nu}(\Omega^{c})^{\prime}$
with
$g\mapsto\mathscr{D}_{\lambda}g=\mathcal{E}_{-\lambda}(u_{g},\widetilde{\cdot})$
is well defined, linearly bounded and self-adjoint. Moreover, if we take
$c=\min(1,-\lambda)$ then for all $g\in T_{\nu}(\Omega^{c})$ we have

$\displaystyle\langle\mathscr{D}_{\lambda}g,g\rangle\geq
c\|u_{g}\|^{2}_{V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})}.$

In particular if $c>0$ it follows that

$\displaystyle\langle\mathscr{D}_{\lambda}g,g\rangle\geq
c\|g\|^{2}_{T_{\nu}(\Omega^{c})}.$

###### Proof.

Consider $v^{\prime}\in V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})$ another
extension of $v$ then $v^{\prime}-\widetilde{v}\in
V_{\nu,0}(\Omega|\operatorname{\mathbb{R}}^{d})$ and by definition of $u_{g}$
we have

$\displaystyle\mathcal{E}_{-\lambda}(u_{g},v^{\prime}-\widetilde{v})=0\quad\text{that
is}\quad\mathcal{E}_{-\lambda}(u_{g},\widetilde{v})=\mathcal{E}_{-\lambda}(u_{g},v^{\prime}).$

Therefore the mapping $v\mapsto\mathcal{E}_{-\lambda}(u_{g},\widetilde{v})$ is
well defined, linear and bounded on $T_{\nu}(\Omega^{c})$. Indeed,

$\displaystyle|\mathcal{E}_{-\lambda}(u_{g},\widetilde{v})|\leq(|\lambda|+1)\|u_{g}\|_{V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})}\|\widetilde{v}\|_{V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})}.$

Since the extension $\widetilde{v}$ of $v$ is arbitrarily chosen, upon the
estimate (4.21) we obtain

$\displaystyle|\mathcal{E}_{-\lambda}(u_{g},\widetilde{v})|\leq
C\|g\|_{T_{\nu}(\Omega^{c})}\|v\|_{T_{\nu}(\Omega^{c})}.$

This shows that, $\mathcal{E}_{-\lambda}(u_{g},\widetilde{\cdot})$ belongs
$T_{\nu}(\Omega^{c})^{\prime}$. Subsequently it also follows from this
estimate that the mapping $\mathscr{D}_{\lambda}:T_{\nu}(\Omega^{c})\to
T_{\nu}(\Omega^{c})^{\prime}$ with
$g\mapsto\mathscr{D}_{\lambda}g=\mathcal{E}_{-\lambda}(u_{g},\widetilde{\cdot})$
is linear and bounded. Now let $g,h\in T_{\nu}(\Omega^{c})$ specializing the
definition of $\mathscr{D}_{\lambda}$ with $\widetilde{g}=u_{g}$ and
$\widetilde{h}=u_{h}$ the self-adjointness is obtained as follows

$\displaystyle\langle\mathscr{D}_{\lambda}g,h\rangle=\mathcal{E}_{-\lambda}(u_{g},u_{h})=\mathcal{E}_{-\lambda}(u_{h},u_{g})=\langle\mathscr{D}_{\lambda}h,g\rangle.$

The choice $c=\min(1,-\lambda)$ leads to
$\langle\mathscr{D}_{\lambda}g,g\rangle=\mathcal{E}_{-\lambda}(u_{g},u_{g})\geq
c\|u_{g}\|^{2}_{V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})}$. ∎

######  Remark 4.21.

The above definition is motivated by the following observation. Assume $u_{g}$
is as before and $\varphi\in C_{c}^{\infty}(\operatorname{\mathbb{R}}^{d})$.
The Gauss-Green formula (A.5) gives

$\displaystyle\langle\mathscr{D}_{\lambda}g,\varphi\rangle$
$\displaystyle=\mathcal{E}_{-\lambda}(u_{g},\varphi)=\int_{\Omega^{c}}\mathcal{N}u_{g}(y)\varphi(y)\mathrm{d}y.$
(4.22)

From the second equality we can identify
$\mathscr{D}_{\lambda}g=\mathcal{N}u_{g}\in
L^{2}(\Omega^{c},\widetilde{\nu}^{-1})\subset T_{\nu}(\Omega^{c})^{\prime}$.
Hence $\mathscr{D}_{\lambda}:g\mapsto\mathcal{N}u_{g}$, which agrees with
conceptual idea behind the Dirichlet-to-Neumann map in the classical case.

######  Theorem 4.22.

Let the assumptions of Theorem 4.16 be in force. Denote by $L_{\beta}$ the
operator $L$ subject to the Robin boundary condition $\mathcal{N}u+\beta u=0$.
Then the point spectrum $\sigma_{p}(L_{\beta})=(\gamma_{n}(\beta))_{n}$ of
$L_{\beta}$ is infinitely countable say
$0<\gamma_{1}(\beta)\leq\gamma_{2}(\beta)\leq\cdots\leq\gamma_{n}(\beta)\leq\cdots$,
with $\gamma_{n}(\beta)\to\infty$ as $n\to\infty$, and the corresponding
eigenfunctions belong to $V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})$ and
form an orthonormal basis of $L^{2}(\Omega)$.

###### Proof.

It suffices to proceed as in the proof of Theorem 4.15, see also [FG20,
Theorem 4.36]. ∎

Next, we see the relation between the spectrum of the operator $L$ subject to
Robin boundary condition and that of Dirichlet-to-Neumann operator.

######  Theorem 4.23.

Let $\lambda\in\mathbb{R}\setminus\\{\lambda_{n}:n\geq 1\\}$ and
$\beta:\Omega^{c}\to\operatorname{\mathbb{R}}$ be measurable. Consider the
Dirichlet-to-Neumann map $\mathscr{D}_{\lambda}:T_{\nu}(\Omega^{c})\to
T_{\nu}(\Omega^{c})^{\prime},\,\,\mathscr{D}_{\lambda}g=\mathcal{N}u_{g}$.
Then, $0\in\sigma_{p}(\mathscr{D}_{\lambda}+\beta)$ if and only if
$\lambda\in\sigma_{p}(L_{\beta})$. In addition,
$\dim\ker(L_{\beta}-\lambda)=\dim\ker(\mathscr{D}_{\lambda}+\beta)$.

###### Proof.

Let $u\in\ker(L_{\beta}-\lambda)$ then for all $v\in
V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})$,

$\displaystyle
Q_{\beta}(u,v)=\lambda\int_{\Omega}u(x)v(x)\mathrm{d}x\quad\text{equivalently}\quad\mathcal{E}_{-\lambda}(u,v)=-\int_{\Omega^{c}}u(y)v(y)\beta(y)\mathrm{d}y.$

Set $g=\operatorname{Tr}(u)=u|_{\Omega^{c}}$, with the aid of (4.22) the above
relation reduces to

$\displaystyle\int_{\Omega^{c}}\mathcal{N}u_{g}(y)v(y)\mathrm{d}y=-\int_{\Omega^{c}}g(y)v(y)\beta(y)\mathrm{d}y.$

Thus $g\in\ker(\mathscr{D}_{\lambda}+\beta)$. We have shown that the mapping
$T:\ker(L_{\beta}-\lambda)\to\ker(\mathscr{D}_{\lambda}+\beta)$ with
$u\mapsto\operatorname{Tr}(u)$ is well defined and onto. Both assertions will
follow once we show that $T$ defines a bijection, in other words we only have
to show that $T$ is one-to-one. For $u\in\ker(L_{\beta}-\lambda)$ if
$\operatorname{Tr}(u)=0$ then from the first relation above, we have
$\mathcal{E}(u,v)=\lambda\int_{\Omega}u(x)v(x)\mathrm{d}x$ for all $v\in
V_{\nu,0}(\Omega|\operatorname{\mathbb{R}}^{d})$. Necessarily, $u=0$ otherwise
$\lambda$ is a Dirichlet eigenvalue which is not the case by assumption. ∎

##  5\. Transition from nonlocal to local

The main purpose of this section is to prove the convergence of a sequence of
nonlocal Neumann problems to a local Neumann problem, i.e., the corresponding
solutions converge. The main result of this section is Theorem 5.4. We
consider the following set-up: Let $(\nu_{\alpha})_{\alpha\in(0,2)}$ be a
family of Lévy radial functions approximating the Dirac measure at the origin,
i.e., for every $\alpha,\delta>0$

$\displaystyle\begin{split}\nu_{\alpha}\geq 0\,\,\text{ is
radial},\quad\int_{\mathbb{R}^{d}}(1\land|h|^{2})\nu_{\alpha}(h)\mathrm{d}h=d,\quad\lim_{\alpha\to
2}\int_{|h|>\delta}\nu_{\alpha}(h)\mathrm{d}h=0\,.\end{split}$ (5.1)

Note that there is no restriction on the support of $\nu_{\alpha}$. The above
definition of $(\nu_{\alpha})_{0<\alpha<2}$ generalizes the spectrum of
possible approximation sequences in [FKV20, DTZ22]. Note that, convergence of
nonlocal variational structures including finite dimensional Galerkin methods
have already been considered in [MD15] and [BMCP15] for homogeneous nonlocal
problems of vanishing horizon-type.

We denote $L_{\alpha}$ and $\mathcal{N}_{\alpha}$ to be the nonlocal operators
associated with $\nu_{\alpha}$, i.e.,

$\displaystyle L_{\alpha}u(x)$
$\displaystyle=2\operatorname{\operatorname{p.\\!v.}}\int_{\operatorname{\mathbb{R}}^{d}}(u(x)-u(y))\nu_{\alpha}(x-y)\,\mathrm{d}y,$
$\displaystyle\mathcal{N}_{\alpha}u(x)$
$\displaystyle=2\int_{\Omega}(u(x)-u(y))\nu_{\alpha}(x-y)\,\mathrm{d}y.$

The associated energy forms are defined by

$\displaystyle\mathcal{E}^{\alpha}_{\Omega}(u,v)$
$\displaystyle=\iint\limits_{\Omega\Omega}\big{(}u(y)-u(x)\big{)}\big{(}v(y)-v(x)\big{)}\nu_{\alpha}(x-y)\mathrm{d}x\,\mathrm{d}y\,,$
$\displaystyle\mathcal{E}^{\alpha}(u,v)$
$\displaystyle=\iint\limits_{(\Omega^{c}\times\Omega^{c})^{c}}\big{(}u(y)-u(x)\big{)}\big{(}v(y)-v(x)\nu_{\alpha}(x-y)\mathrm{d}x\,\mathrm{d}y.$

Let us mention two prototypical examples of interest here. For more concrete
examples we refer the reader to [FKV20, FG20].

######  Example 5.1.

Define $\nu_{\alpha}(h)=a_{d,\alpha}|h|^{-d-\alpha}$ with
$a_{d,\alpha}=\tfrac{d\alpha(2-\alpha)}{2|\mathbb{S}^{d-1}|}.$ Indeed, passing
through polar coordinates yields

$\displaystyle\int_{\mathbb{R}^{d}}(1\land|h|^{2})|h|^{-d-\alpha}\,\mathrm{d}h=|\mathbb{S}^{d-1}|\Big{(}\int_{0}^{1}r^{1-\alpha}\,\mathrm{d}r+\int_{1}^{\infty}r^{-1-\alpha}\,\mathrm{d}r\Big{)}=\frac{2d|\mathbb{S}^{d-1}|}{d\alpha(2-\alpha)}=da^{-1}_{d,\alpha}.$

For $\delta>0,$ a similar computation gives

$\displaystyle
a_{d,\alpha}\int\limits_{|h|\geq\delta}(1\land|h|^{2})|h|^{-d-\alpha}\,\mathrm{d}h\leq\frac{d\alpha(2-\alpha)}{2d}\int_{\delta}^{\infty}r^{-1-\alpha}\,\mathrm{d}r=\frac{d}{2}(2-\alpha)\delta^{-\alpha}\xrightarrow{\alpha\to
2}0.$

The choice of $\nu_{\alpha}(h)=a_{d,\alpha}|h|^{-d-\alpha}$ gives rise to a
multiple of fractional Laplace operator, i.e.,
$L_{\alpha}=\frac{a_{d,\alpha}}{C_{d,\alpha}}(-\Delta)^{\alpha/2}$, where we
recall that $C_{d,\alpha}$ is the normalizing constant of
$(-\Delta)^{\alpha/2}$. Note however that
$\frac{a_{d,\alpha}}{C_{d,\alpha}}\to 1$ as $\alpha\to 2$ see [AAS67, NPV12,
FG20].

######  Example 5.2.

Let $\nu\in L^{1}(\operatorname{\mathbb{R}}^{d},1\land|h|^{2})$ be any radial
Lévy density that is normalized, i.e.,

$\int_{\operatorname{\mathbb{R}}^{d}}1\land|h|^{2}\,\nu(h)\mathrm{d}h=d.$

Let a family $(\nu_{\alpha})_{\alpha\in(0,2)}$ be defined by
$\nu_{\alpha}=\nu^{2-\alpha}$ where $\nu^{\varepsilon}$ is a rescaled version
of $\nu$ in the following sense:

$\displaystyle\begin{split}\nu^{\varepsilon}(h)=\begin{cases}\varepsilon^{-d-2}\nu\big{(}h/\varepsilon\big{)}&\text{if}\leavevmode\nobreak\
\leavevmode\nobreak\ |h|\leq\varepsilon\\\
\varepsilon^{-d}|h|^{-2}\nu\big{(}h/\varepsilon\big{)}&\text{if}\leavevmode\nobreak\
\leavevmode\nobreak\ \varepsilon<|h|\leq 1\\\
\varepsilon^{-d}\nu\big{(}h/\varepsilon\big{)}&\text{if}\leavevmode\nobreak\
\leavevmode\nobreak\ |h|>1.\end{cases}\end{split}$

Then, as shown in [Fog23b, Proposition 2.2], $(\nu_{\alpha})_{\alpha\in(0,2)}$
satisfies (5.1). Note that, as a possible simple example, one could consider
$\nu(h)=c\mathds{1}_{B_{1}(0)}(h)$, so that $(\nu_{\alpha})$ would correspond
to what is known as vanishing horizon in peridynamics, see [DY21, DTZ22].

The next result implies the convergence of the nonlocal normal derivative to
the local one. We point out that a similar convergence has been recently
established in [HK23].

######  Lemma 5.3.

Assume $\Omega\subset\operatorname{\mathbb{R}}^{d}$ is an open bounded set
with Lipschitz boundary. Let $\varphi\in
C^{2}_{b}(\operatorname{\mathbb{R}}^{d})$ and $v\in
V_{\nu_{\alpha}}(\Omega|\operatorname{\mathbb{R}}^{d})$. The following
assertions hold true.

  1. $(i)$

There is a constant $C>0$ independent of $\alpha$ such that

$\displaystyle\sup_{\alpha\in(0,2)}\Big{|}\int_{\Omega^{c}}\mathcal{N}_{\alpha}\varphi(y)v(y)\mathrm{d}y\Big{|}\leq
C\|\varphi\|_{C^{2}_{b}(\operatorname{\mathbb{R}}^{d})}\|v\|_{V_{\nu_{\alpha}}(\Omega|\operatorname{\mathbb{R}}^{d})}\,.$

  2. $(ii)$

Assume $v\in H^{1}(\operatorname{\mathbb{R}}^{d})$ then

$\displaystyle\lim_{\alpha\to
2}\int_{\Omega^{c}}\mathcal{N}_{\alpha}\varphi(y)v(y)\mathrm{d}y=\int_{\partial\Omega}\frac{\partial\varphi}{\partial
n}(x)v(x)\mathrm{d}\sigma(x)\,.$

###### Proof.

In view of the estimates (A.1) and (A.4) respectively, we have

$\displaystyle|L_{\alpha}\varphi|\leq
4d\|\varphi\|_{C^{2}_{b}(\operatorname{\mathbb{R}}^{d})}\quad\text{and}\quad\mathcal{E}^{\alpha}(\varphi,\varphi)\leq
4d|\Omega|\|\varphi\|^{2}_{C^{1}_{b}(\operatorname{\mathbb{R}}^{d})}\quad\text{for
all }\leavevmode\nobreak\ \leavevmode\nobreak\ \alpha\in(0,2)\,.$

By the continuity of the linear mapping
$v\mapsto\mathcal{E}^{\alpha}(\varphi,v)-\int_{\Omega}L_{\alpha}\varphi(x)v(x)\mathrm{d}x$,
the Gauss-Green formula (A.5) is applicable for $\varphi\in
C_{b}^{2}(\operatorname{\mathbb{R}}^{d})$ and $v\in
V_{\nu_{\alpha}}(\Omega|\operatorname{\mathbb{R}}^{d})$. Therefore, with the
help of the above estimates we get $(i)$ as follows

$\displaystyle\Big{|}\int_{\Omega^{c}}\mathcal{N}_{\alpha}\varphi(y)v(y)\mathrm{d}y\Big{|}$
$\displaystyle=\Big{|}\mathcal{E}^{\alpha}(\varphi,v)-\int_{\Omega}L_{\alpha}\varphi(x)v(x)\mathrm{d}x\Big{|}$
$\displaystyle\leq\mathcal{E}^{\alpha}(\varphi,\varphi)^{1/2}\mathcal{E}^{\alpha}(v,v)^{1/2}+\|L_{\alpha}\varphi\|_{L^{2}(\Omega)}\|v\|_{L^{2}(\Omega)}$
$\displaystyle\leq
C\|\varphi\|_{C^{2}_{b}(\operatorname{\mathbb{R}}^{d})}\|v\|_{V_{\nu_{\alpha}}(\Omega|\operatorname{\mathbb{R}}^{d})}\,.$

Noting that $L_{\alpha}\varphi(x)\xrightarrow[]{\alpha\to 2}-\Delta\varphi(x)$
for all $x\in\operatorname{\mathbb{R}}^{d}$ (see [Fog23b, Proposition 2.4])
and that
$|L_{\alpha}\varphi|\leq\frac{4}{d}\|\varphi\|_{C^{2}_{b}(\operatorname{\mathbb{R}}^{d})}$,
the Lebesgue dominated convergence theorem yields

$\displaystyle\int_{\Omega}L_{\alpha}\varphi(x)v(x)\mathrm{d}x\xrightarrow[]{\alpha\to
2}\int_{\Omega}-\Delta\varphi(x)v(x)\mathrm{d}x\,.$

On the other hand, according to [Fog23b] and [FKV20, Theorem 3.4], we have
that

$\displaystyle\begin{split}&\iint\limits_{\Omega\Omega}(v(x)-v(y))^{2}\nu_{\alpha}(x-y)\mathrm{d}x\mathrm{d}y\xrightarrow[]{\alpha\to
2}\int_{\Omega}|\nabla v(x)|^{2}\mathrm{d}x\\\
&\iint\limits_{\Omega\Omega^{c}}(v(x)-v(y))^{2}\nu_{\alpha}(x-y)\mathrm{d}x\mathrm{d}y\xrightarrow{\alpha\to
2}0.\end{split}$ (5.2)

So that, $\mathcal{E}^{\alpha}(v,v)\xrightarrow[]{\alpha\to
2}\int_{\Omega}|\nabla v(x)|^{2}\mathrm{d}x$. Thus we also have

$\displaystyle\mathcal{E}^{\alpha}(\varphi,v)\xrightarrow[]{\alpha\to
2}\int_{\Omega}\nabla\varphi(x)\cdot\nabla v(x)\mathrm{d}x.$

Finally from the foregoing and the local Gauss-Green formula we obtain $(ii)$
as follows

$\displaystyle\lim_{\alpha\to
2}\int_{\Omega^{c}}\mathcal{N}_{\alpha}\varphi(y)v(y)\mathrm{d}y$
$\displaystyle=\lim_{\alpha\to
2}\mathcal{E}^{\alpha}(\varphi,v)-\lim_{\alpha\to
2}\int_{\Omega}L_{\alpha}\varphi(x)v(x)\mathrm{d}x$
$\displaystyle=\int_{\Omega}\nabla\varphi(x)\cdot\nabla
v(x)\mathrm{d}x-\int_{\Omega}\Delta\varphi(x)v(x)\mathrm{d}x$
$\displaystyle=\int_{\partial\Omega}\frac{\partial\varphi}{\partial
n}(x)v(x)\mathrm{d}\sigma(x)\,.$

∎

######  Theorem 5.4 (Convergence of weak solution).

Let $\Omega\subset\operatorname{\mathbb{R}}^{d}$ be an open bounded and
connected domain with Lipschitz boundary. Let $(f_{\alpha})_{\alpha}$ be
functions converging in the weak sense to another function $f$ in
$L^{2}(\Omega)$ and let $g_{\alpha}=\mathcal{N}_{\alpha}\varphi$ and
$g=\frac{\partial\varphi}{\partial n}$ for some $\varphi\in
C^{2}_{b}(\operatorname{\mathbb{R}}^{d})$. Assume $u_{\alpha}\in
V_{\nu_{\alpha}}(\Omega|\operatorname{\mathbb{R}}^{d})^{\perp}$ is a weak
solution to $L_{\alpha}u=f_{\alpha}$ on $\Omega$ and
$\mathcal{N}_{\alpha}u=g_{\alpha}$ on $\Omega^{c}$ that is,

$\displaystyle\mathcal{E}^{\alpha}(u_{\alpha},v)=\int_{\Omega}f_{\alpha}(x)v(x)+\int_{\Omega^{c}}g_{\alpha}(x)v(x)\quad\text{for
all }\leavevmode\nobreak\ \leavevmode\nobreak\ v\in
V_{\nu_{\alpha}}(\Omega|\operatorname{\mathbb{R}}^{d})^{\perp}\,.$

Let $u\in H^{1}(\Omega)^{\perp}$ be the unique weak solution in to the Neumann
problem $-\Delta u=f$ in $\Omega$ and $\frac{\partial u}{\partial n}=g$ on
$\partial\Omega$ i.e.

$\displaystyle\int_{\Omega}\nabla u(x)\cdot\nabla
v(x)\mathrm{d}x=\int_{\Omega}f(x)v(x)\mathrm{d}x+\int_{\partial\Omega}g(x)v(x)\mathrm{d}\sigma(x)\qquad\text{for
all $u\in H^{1}(\Omega)^{\perp}$}.$

Then $(u_{\alpha})_{\alpha}$ strongly converges to $u$ in $L^{2}(\Omega)$,
i.e., $\|u_{\alpha}-u\|_{L^{2}(\Omega)}\xrightarrow[]{\alpha\to 2}0$.
Moreover, the following weak convergence of the energy forms holds true

$\displaystyle\mathcal{E}^{\alpha}(u_{\alpha},v)\xrightarrow[]{\alpha\to
2}\int_{\Omega}\nabla u(x)\cdot\nabla v(x)\mathrm{d}x\qquad\text{for all $v\in
H^{1}(\operatorname{\mathbb{R}}^{d})$}.$ (5.3)

######  Remark 5.5.

  1. (i)

In case of the homogeneous problem, i.e., for $\varphi=0$ and $f_{\alpha}=f$,
the corresponding result is a direct consequence of the Mosco-convergence of
$(\mathcal{E}^{\alpha}(\cdot,\cdot),V_{\nu_{\alpha}}(\Omega|\operatorname{\mathbb{R}}^{d}))_{\alpha}$
to the gradient form $\int_{\Omega}|\nabla u(x)|^{2}\mathrm{d}x$ with domain
$H^{1}(\Omega)$, see [FKV20].

  2. (ii)

The convergence in result of Theorem 5.4 remains true if one replaces the
Neumann condition with the Dirichlet condition, see [FG20].

  3. (iii)

Examples of the type of 5.2 have been considered in relation to models in
peridynamics, see [DTZ22, Section 4.2] and [BMCP15] for a natural nonlinear
setting.

  4. (iv)

The assertion of the theorem remains true under the weaker assumption that
$(g_{\alpha},\psi)_{L^{2}(\Omega^{c})}$ convergences to
$(g,\psi)_{L^{2}(\partial\Omega)}$ for all $\psi\in
H^{1}(\operatorname{\mathbb{R}}^{d})$.

  5. (v)

It is desirable to study Theorem 5.4 under more general assumptions, e.g.,
under a weaker assumption than $\varphi\in
C^{2}_{b}(\operatorname{\mathbb{R}}^{d})$. A sufficient condition to be
expected is $g\in H^{1/2}(\partial\Omega)$.

###### Proof.

A compactness argument as in [Pon04, Corollary 2.1], see [FG20, Chapter 5],
shows that for certain $\alpha_{0}\in(0,2)$ there exists a constant positive
$C>0$ depending only on $\alpha_{0},\Omega$ and $d$ such that for all $v\in
L^{2}(\Omega)^{\perp}$ and all $\alpha\in(\alpha_{0},2)$

$\displaystyle\|v\|^{2}_{V_{\nu_{\alpha}}(\Omega|\operatorname{\mathbb{R}}^{d})}\leq
C\mathcal{E}^{\alpha}(v,v).$ (5.4)

In view of the weak convergence, we can assume without loss generality that
$\sup_{\alpha\in(0,2)}\|f_{\alpha}\|_{L^{2}(\Omega)}<\infty$. This together
with the definition of $u_{\alpha}$ along with 5.3 $(i)$ yields

$\displaystyle\mathcal{E}^{\alpha}(u_{\alpha},u_{\alpha})$
$\displaystyle=\int_{\Omega}f_{\alpha}(x)u_{\alpha}(x)\mathrm{d}x+\int_{\Omega^{c}}g_{\alpha}(y)u_{\alpha}(y)\mathrm{d}y$
$\displaystyle\leq\|u_{\alpha}\|_{V_{\nu_{\alpha}}(\Omega|\operatorname{\mathbb{R}}^{d})}(\|f_{\alpha}\|_{L^{2}(\Omega)}+\|\varphi\|_{C_{b}^{2}(\operatorname{\mathbb{R}}^{d})})$
$\displaystyle\leq
C\|u_{\alpha}\|_{V_{\nu_{\alpha}}(\Omega|\operatorname{\mathbb{R}}^{d})}.$

Combining this with (5.4), then for a generic constant $C>0$ independent of
$\alpha$ we have the following uniform boundedness

$\displaystyle\|u_{\alpha}\|_{H_{\nu_{\alpha}}(\Omega)}\leq\|u_{\alpha}\|_{V_{\nu_{\alpha}}(\Omega|\operatorname{\mathbb{R}}^{d})}\leq
C\qquad\text{for all }\alpha\in(\alpha_{0},2)\,.$ (5.5)

Recall that, see [Fog23b, FG20],
$\|u\|_{H_{\nu_{\alpha}}(\Omega)}\xrightarrow[]{\alpha\to
2}\|u\|_{H^{1}(\Omega)}$ for all $u\in H^{1}(\Omega)$. Whence from [KS03,
Lemma 2.2] there exists $u^{\prime}\in H^{1}(\Omega)$ and a subsequence
$\alpha_{n}\xrightarrow[]{n\to\infty}2$ such that,

$\displaystyle\lim_{n\to\infty}\big{(}u_{\alpha_{n}},v\big{)}_{H_{\nu_{\alpha_{n}}}(\Omega)}=\big{(}u^{\prime},v\big{)}_{H^{1}(\Omega)},$

where, we recall that

$\displaystyle\big{(}w,\,v\big{)}_{H_{\nu_{\alpha}}(\Omega)}$
$\displaystyle=\int_{\Omega}w(x)v(x)\mathrm{d}x+\iint\limits_{\Omega\Omega}(w(x)-w(y))(v(x)-w(y))\nu_{\alpha}(x-y)\mathrm{d}x\mathrm{d}y$
$\displaystyle\big{(}w,v\big{)}_{H^{1}(\Omega)}$
$\displaystyle=\int_{\Omega}w(x)v(x)\mathrm{d}x+\int_{\Omega}\nabla
w(x)\cdot\nabla v(x)\mathrm{d}x\,.$

By virtue of the asymptotic compactness, see [BBM01, FG20, Pon04]; see also
[AAS67, Section 4], there exists a further subsequence that we still denote by
$(\alpha_{n})_{n}$ and a function $u^{\prime\prime}\in H^{1}(\Omega)$ such
that
$\|u_{\alpha_{n}}-u^{\prime\prime}\|_{L^{2}(\Omega)}\xrightarrow[]{n\to\infty}0$.
It is not difficult to show that $u^{\prime}=u^{\prime\prime}$ almost
everywhere in $\Omega$, $u\in H^{1}(\Omega)^{\perp}$ where we let
$u=u^{\prime}$, and that for all $v\in H^{1}(\Omega)$

$\displaystyle\iint\limits_{\Omega\Omega}(u_{\alpha_{n}}(x)-u_{\alpha_{n}}(y))(v(x)-v(y))\nu_{\alpha_{n}}(x-y)\mathrm{d}x\mathrm{d}y\xrightarrow[]{n\to\infty}\int_{\Omega}\nabla
u(x)\cdot\nabla v(x)\mathrm{d}x\,.$ (5.6)

It remains to show that $u$ is the weak solution of the corresponding local
Neumann problem. To this end, we fix $v\in H^{1}(\Omega)^{\perp}$, given that
$\Omega$ has a Lipschitz boundary we let $\overline{v}\in
H^{1}(\operatorname{\mathbb{R}}^{d})$ be an extension of $v$. The uniform
boundedness in (5.5) and the convergence in (5.2) yield

$\displaystyle\iint\limits_{\Omega\Omega^{c}}\big{|}(u_{\alpha_{n}}(x)-u_{\alpha_{n}}(y))(\overline{v}(x)-\overline{v}(y))\big{|}$
$\displaystyle\nu_{\alpha_{n}}(x-y)\mathrm{d}x\mathrm{d}y$ $\displaystyle\leq
C\iint\limits_{\Omega\Omega^{c}}(\overline{v}(x)-\overline{v}(y))^{2}\nu_{\alpha_{n}}(x-y)\mathrm{d}x\mathrm{d}y\xrightarrow[]{n\to\infty}0\,.$

This combined with (5.6) gives

$\displaystyle\mathcal{E}^{\alpha_{n}}(u_{\alpha_{n}},\overline{v})\xrightarrow[]{n\to\infty}\int_{\Omega}\nabla
u(x)\cdot\nabla v(x)\mathrm{d}x\,.$

In particular, since $\overline{v}\in H^{1}(\operatorname{\mathbb{R}}^{d})$
can be arbitrarily chosen, we have the weak convergence

$\displaystyle\mathcal{E}^{\alpha_{n}}(u_{\alpha_{n}},v)\xrightarrow[]{n\to\infty}\int_{\Omega}\nabla
u(x)\cdot\nabla v(x)\mathrm{d}x\,\qquad\text{for all $v\in
H^{1}(\operatorname{\mathbb{R}}^{d})$}.$

We know that $\overline{v}\in
V_{\nu_{\alpha}}(\Omega|\operatorname{\mathbb{R}}^{d})^{\perp}$ for all
$\alpha\in(0,2)$, thus by definition of $u_{\alpha_{n}}$ it follows that,

$\displaystyle\mathcal{E}^{\alpha_{n}}(u_{\alpha_{n}},\overline{v})=\int_{\Omega}f_{\alpha_{n}}(x)v(x)\mathrm{d}x+\int_{\Omega^{c}}g_{\alpha_{n}}(y)\overline{v}(y)\mathrm{d}y\,.$

By 5.3 $(ii)$ and the fact that $f_{\alpha_{n}}\rightharpoonup f$ weakly in
$L^{2}(\Omega)$, letting $n\to\infty$ we obtain

$\displaystyle\int_{\Omega}\nabla u(x)\cdot\nabla
v(x)\mathrm{d}x\,=\int_{\Omega}f(x)v(x)\mathrm{d}x+\int_{\partial\Omega}g(x)v(x)\mathrm{d}\sigma(x)\,.$

By virtue of the uniqueness of the limit $u\in H^{1}(\Omega)^{\perp}$, the
same reasoning can be applied to any other subsequence $(\alpha_{n})_{n}$ with
$\alpha_{n}\xrightarrow[]{n\to\infty}2$ and hence the claimed convergences
hold true for the whole sequence as desired. ∎

## Appendix A

In the following appendices we explain basic properties of translation-
invariant nonlocal operators $L$ driven by the density of a Lévy measure
$\nu:\operatorname{\mathbb{R}}^{d}\setminus\\{0\\}\to[0,\infty)$ satisfying
condition (L). Throughout this section we assume $k(x,y)=\nu(x-y)$ for all
$x\neq y$. The main goals include a definition of $Lu$ as a distribution in
Proposition A.2 and the Gauss-Green formula for nonlocal operators in A.5.

###  A.1. Basics on the operator $L$

Given $k\in\mathbb{N}$, denote $C_{b}^{k}(\operatorname{\mathbb{R}}^{d})$ as
the space of bounded functions of class $C^{k}$ whose derivatives up to order
$k$ are bounded. Recall that for a sufficiently smooth function
$v:\operatorname{\mathbb{R}}^{d}\to\mathbb{R}$, the operator $L$ is defined by

$\displaystyle Lv(x)$
$\displaystyle=\operatorname{\operatorname{p.\\!v.}}\int\limits_{\operatorname{\mathbb{R}}^{d}}(v(x)-v(y))\nu(x\\!-\\!y)\mathrm{d}y=\lim_{\varepsilon\to
0^{+}}L_{\varepsilon}v(x)$

where

$\displaystyle L_{\varepsilon}v(x)$
$\displaystyle=\int\limits_{\operatorname{\mathbb{R}}^{d}\setminus
B_{\varepsilon}(x)}\\!\\!(v(x)-v(y))\nu(x\\!-\\!y)\mathrm{d}y\qquad\hbox{$(x\in\operatorname{\mathbb{R}}^{d};\varepsilon>0)$}.$

Here are some basic properties of the operator $L$.

######  Proposition A.1.

Let $u\in C^{2}_{b}(\operatorname{\mathbb{R}}^{d})$. Then the following
properties are satisfied.

  1. (i)

The map $x\mapsto Lu(x)$ is bounded and uniformly continuous. Moreover,

$\displaystyle
Lu(x)=-\frac{1}{2}\int_{\operatorname{\mathbb{R}}^{d}}(u(x+h)+u(x-h)-2u(x))\nu(h)\,\mathrm{d}h.$

  2. (ii)

For each $\varepsilon>0$, the map $x\mapsto L_{\varepsilon}u(x)$ is uniformly
continuous.

  3. (iii)

The family $(L_{\varepsilon}u(x))_{\varepsilon}$ is uniformly bounded and
uniformly converges to $Lu,$ i.e.

$\displaystyle\|L_{\varepsilon}u-Lu\|_{L^{\infty}(\operatorname{\mathbb{R}}^{d})}\xrightarrow[]{\varepsilon\to
0}0.$

###### Proof.

Let $u\in C^{2}_{b}(\operatorname{\mathbb{R}}^{d})$. A simple change of
variables implies

$\displaystyle
L_{\varepsilon}u(x)=-\frac{1}{2}\int\limits_{\operatorname{\mathbb{R}}^{d}\setminus
B_{\varepsilon}(0)}(u(x+h)+u(x-h)-2u(x))\nu(h)\,\mathrm{d}h.$

An application of the fundamental theorem of calculus yields

$\displaystyle(u(x+h)+u(x-h)-2u(x))$ $\displaystyle=\int_{0}^{1}\big{[}\nabla
u(x+th)-\nabla u(x-th)\big{]}\cdot h\,\mathrm{d}t$
$\displaystyle=\int_{0}^{1}\int_{0}^{1}2t\big{[}D^{2}u(x-th+2sth)\cdot
h\big{]}\cdot h\,\mathrm{d}s\mathrm{d}t\,.$

Since $u$ and its Hessian $D^{2}u$ are bounded functions, we deduce

$\displaystyle\left|u(x+h)+u(x-h)-2u(x)\right|\leq
2\|u\|_{C^{2}_{b}(\operatorname{\mathbb{R}}^{d})}(1\land|h|^{2}),\quad\quad
x,h\in\operatorname{\mathbb{R}}^{d}.$ (A.1)

The integrability of the function $h\mapsto(1\land|h|^{2})\nu(h)$ entails the
boundedness of $x\mapsto Lu(x)$ and the uniform boundedness of $x\mapsto
L_{\varepsilon}u(x)$. It also allows us to get rid of the principal value
formulation. Furthermore, we can prove the uniform convergence of
$(L_{\varepsilon}u)_{\varepsilon}$ to $Lu$ by

$\displaystyle\|L_{\varepsilon}u-Lu\|_{L^{\infty}(\operatorname{\mathbb{R}}^{d})}\leq
2\|u\|_{C^{2}_{b}(\operatorname{\mathbb{R}}^{d})}\int_{B_{\varepsilon}(0)}(1\land|h|^{2})\nu(h)\mathrm{d}h\xrightarrow[]{\varepsilon\to
0}0.$

In order to prove the uniform continuity, we fix
$x,z\in\operatorname{\mathbb{R}}^{d}$ close enough, say $|x-z|\leq\delta$ with
$0<\delta<1$. Then for every h $\in\operatorname{\mathbb{R}}^{d}$, $h\neq 0$,

$\displaystyle 2|u(x)-u(z)|+|u(x+h)-u(z+h)|+|u(x-h)-u(z-h)|\leq
4\delta\|u\|_{C^{2}_{b}(\operatorname{\mathbb{R}}^{d})}.$

This combined with (A.1) yields the uniform continuity via the integrability
of $h\mapsto(1\land|h|^{2})\nu(h)$ as follows,

$\displaystyle\|Lu(x)-Lu(z)\|_{L^{\infty}(\operatorname{\mathbb{R}}^{d})}\leq
2\|u\|_{C^{2}_{b}(\operatorname{\mathbb{R}}^{d})}\int_{\operatorname{\mathbb{R}}^{d}}(\delta\land|h|^{2})\nu(h)\mathrm{d}h\xrightarrow[]{\delta\to
0}0.$

The uniform continuity of $x\mapsto L_{\varepsilon}u(x)$ follows analogously.
∎

In order for $Lu(x)$ to be defined, $u$ needs to possess two properties: some
regularity in the neighborhood of the point $x$ and some weighted
integrability for $|x|\to\infty$. As shown above, being $C^{2}$ in the
neighborhood of $x$ is more than sufficient as is boundedness for
$|x|\to\infty$. Let us investigate some mild condition on $u$ as
$|x|\to\infty$ that still allows a suitable definition of $Lu$. In order to do
so, we additionally assume that $\nu$ is unimodal.

######  Proposition A.2.

Let $\nu$ be a unimodal Lévy measure. Define a weight $\widehat{\nu}$ on
$\operatorname{\mathbb{R}}^{d}$ by $\widehat{\nu}(x)=\nu(\frac{1}{2}(1+|x|))$.

  1. (i)

For $u\in C^{2}(\operatorname{\mathbb{R}}^{d})\cap
L^{1}(\operatorname{\mathbb{R}}^{d},\widehat{\nu})$, the expression $Lu(x)$
exists for every $x\in\operatorname{\mathbb{R}}^{d}$.

  2. (ii)

Assume that $\nu$ has full support. For $u\in
L^{1}(\operatorname{\mathbb{R}}^{d},\widehat{\nu})$ the expression $Lu$ is
defined in the distributional sense via the mapping $\varphi\mapsto\langle
Lu,\varphi\rangle=(u,L\varphi)_{L^{2}(\operatorname{\mathbb{R}}^{d})}$.

  3. (iii)

Assume that $\nu$ satisfies the scaling condition (2.5). Let
$\Omega\subset\operatorname{\mathbb{R}}^{d}$ be open and bounded and $u\in
V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})$. Then $Lu$ is defined in the
distributional sense.

######  Remark A.3.

Note that $\widehat{\nu}\in L^{1}(\operatorname{\mathbb{R}}^{d})\cap
L^{\infty}(\operatorname{\mathbb{R}}^{d})$ and that
$L^{1}(\operatorname{\mathbb{R}}^{d},\widehat{\nu})$ contains
$L^{\infty}(\operatorname{\mathbb{R}}^{d})$.

######  Example A.4.

If $\nu(h)=|h|^{-d-\alpha}$ for some $\alpha\in(0,2)$, then
$\widehat{\nu}(h)\asymp(1+|h|)^{-d-\alpha}$.

###### Proof.

For the proof of (i) we decompose the integral in the definition of $Lu(x)$
into the two domains $\\{|y|\leq 2|x|+1\\}$ and $\\{|y|>2|x|+1\\}$. In the
first domain we employ the Taylor formula as in the proof of A.1. For $y$ in
the second domain we observe

$\displaystyle|x-y|\geq|y|-|x|\geq\frac{|y|}{2}+\frac{|y|}{2}-|x|\geq\frac{|y|+1}{2}\,.$

Thus, for $y$ from the second domain, by property 2.4 we conclude
$\nu(x\\!-\\!y)\leq c\widehat{\nu}(y)$ and thus

$\displaystyle\int\limits_{\\{|y|>2|x|+1\\}}|u(x)|\nu(x\\!-\\!y)\mathrm{d}y+\int\limits_{\\{|y|>2|x|+1\\}}|u(y)|\nu(x\\!-\\!y)\mathrm{d}y\leq
K_{\nu}|u(x)|+\|u\|_{L^{1}(\operatorname{\mathbb{R}}^{d},\widehat{\nu})}.$

For the proof of (ii), let $\varphi\in
C_{c}^{\infty}(\operatorname{\mathbb{R}}^{d})$ be supported in $B_{R}(0)$ for
some $R\geq 1$. We claim

$\displaystyle|L\varphi(x)|\leq
C\|\varphi\|_{C^{2}_{b}(\operatorname{\mathbb{R}}^{d})}\widehat{\nu}(x)\qquad\text{for
all}\leavevmode\nobreak\ x\in\operatorname{\mathbb{R}}^{d}.$ (A.2)

with some constant $C=C(R,d,\nu)$ depending only on $R,d$ and $\nu$. Indeed,
suppose $|x|\geq 4R$, so that $\varphi(x)=0$. Since
$|x-y|\geq\frac{|x|}{2}+R\geq\frac{1}{2}(1+|x|)$ for $y\in B_{R}(0)$, the
property 2.4 implies $\nu(x\\!-\\!y)\leq c\widehat{\nu}(x)$. Accordingly,

$\displaystyle|L\varphi(x)|\leq\int_{B_{R}(0)}|\varphi(y)|\nu(x\\!-\\!y)\mathrm{d}y\leq
c|B_{R}(0)|\|\varphi\|_{C^{2}(\operatorname{\mathbb{R}}^{d})}\widehat{\nu}(x).$

Whereas, if $|x|\leq 4R$ the proof of (A.2) is complete using (A.1) as
follows. Since $\frac{1}{2}(1+|x|)\leq 4R$ we have $\widehat{\nu}(x)\geq
c_{1}$ for an appropriate constant $c_{1}>0$ depending on $R$ and $\nu$. Thus
we conclude

$\displaystyle|L\varphi(x)|\leq
4\Theta\|\varphi\|_{C^{2}_{b}(\operatorname{\mathbb{R}}^{d})}\leq
c_{1}^{-1}4\Theta\|\varphi\|_{C^{2}_{b}(\operatorname{\mathbb{R}}^{d})}\widehat{\nu}(x)$

with
$\Theta=\int_{\operatorname{\mathbb{R}}^{d}}(1\land|h|^{2})\nu(h)\mathrm{d}h$.
Note that in case of the fractional Laplace operator the estimate (A.2) is
analogous to [FW12, Lemma 2.1]. Finally, (A.2) yields

$\displaystyle|(u,L\varphi)_{L^{2}(\operatorname{\mathbb{R}}^{d})}|\leq
C\|\varphi\|_{C^{2}_{b}(\operatorname{\mathbb{R}}^{d})}\int_{\operatorname{\mathbb{R}}^{d}}|u(x)|\widehat{\nu}(x)\mathrm{d}x.$

This shows that $Lu$ is a distribution when $u\in
L^{1}(\operatorname{\mathbb{R}}^{d},\widehat{\nu})$. With regard to (iii) let
$\Omega\subset\operatorname{\mathbb{R}}^{d}$ be open and bounded. We show that
the embedding $V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})\hookrightarrow
L^{1}(\operatorname{\mathbb{R}}^{d},\widehat{\nu})$ is continuous under the
additional scaling assumption (2.5). Indeed, for $u\in
V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})$ we assume $\Omega\subset
B_{R}(0)$ for some $R\geq 1$. Then $|x-y|\leq R(1+|x|)$ for all
$x\in\operatorname{\mathbb{R}}^{d}$ and all $y\in\Omega$ so that by (2.5) and
2.4 we deduce $\widehat{\nu}(x)\leq C\nu(R(1+|x|))\leq cC\nu(x\\!-\\!y)$. Here
$c,C>0$ are constants independent of $x$ and $y$. Proceeding as in 2.17, one
arrives at the estimate

$\displaystyle\int_{\operatorname{\mathbb{R}}^{d}}|u(x)|\widehat{\nu}(x)\mathrm{d}x\leq
C\|u\|_{V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})}.$

Therefore, regarding the preceding arguments $Lu$ is also a distribution
whenever $u\in V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})$. ∎

###  A.2. Gauss-Green type formula

Having at hand a nonlocal analog of the normal derivative as in 4.1, it makes
sense to study a formula that resembles the classical Gauss-Green formula.
Such formulas have been established in several contexts. See [DGLZ13] for
numerous identities of a nonlocal vector calculus in the case of bounded
kernels and [DROV17] for the case of the fractional Laplace operator. Recall
the classical Gauss-Green formula (see [Neč67, Chap 3], [Tri92, Appendix A.3]
or [BF13, Theorem III.1.8]) says for all $u\in
H^{2}(\Omega)\leavevmode\nobreak\ \hbox{and}\leavevmode\nobreak\ v\in
H^{1}(\Omega),$

$\displaystyle\int_{\Omega}(-\Delta)u(x)v(x)\,\mathrm{d}x=\int_{\Omega}\nabla
u(x)\cdot\nabla
v(x)\,\mathrm{d}x-\int_{\partial\Omega}\gamma_{1}u(x)\gamma_{0}v(x)\,\mathrm{d}\sigma(x).$
(A.3)

A reasonable explanation to this terminology is given in the 5.3. For a
function $u\in C_{b}^{1}(\operatorname{\mathbb{R}}^{d})$ we know

$\displaystyle|u(x)-u(y)|\leq
2\|u\|_{C_{b}^{1}(\operatorname{\mathbb{R}}^{d})}(1\land|x-y)|)\qquad\qquad(x,y\in\operatorname{\mathbb{R}}^{d})\,,$
(A.4)

which implies

$\displaystyle\iint\limits_{(\Omega^{c}\times\Omega^{c})^{c}}\big{(}u(x)-u(y)\big{)}^{2}\,\nu(x\\!-\\!y)\mathrm{d}x\,\mathrm{d}y\leq
8\|u\|^{2}_{C_{b}^{1}(\operatorname{\mathbb{R}}^{d})}\iint\limits_{\Omega\operatorname{\mathbb{R}}^{d}}(1\land|x-y|^{2})\,\nu(x\\!-\\!y)\mathrm{d}x\,\mathrm{d}y<\infty.$

######  Proposition A.5 (Gauss-Green type formula).

Let $\Omega$ be open and bounded. For $u\in
C_{b}^{2}(\operatorname{\mathbb{R}}^{d})$ and $v\in
C_{b}^{1}(\operatorname{\mathbb{R}}^{d})$

$\displaystyle\int_{\Omega}[Lu(x)]v(x)\mathrm{d}x=\mathcal{E}(u,v)-\int_{\Omega^{c}}\mathcal{N}u(y)v(y)\mathrm{d}y.$
(A.5)

In particular, by choosing $v=1$ one deduces

$\displaystyle\int_{\Omega}Lu(x)\mathrm{d}x=-\int_{\Omega^{c}}\mathcal{N}u(y)\mathrm{d}y.$
(A.6)

###### Proof.

Let $u\in C_{b}^{2}(\operatorname{\mathbb{R}}^{d})$ and $v\in
C_{b}^{1}(\operatorname{\mathbb{R}}^{d})$. With the aid of A.1 we can write

$\displaystyle\int_{\Omega}[Lu(x)]v(x)\mathrm{d}x=\lim_{\varepsilon\to
0}\int_{\Omega}v(x)\mathrm{d}x\int\limits_{\operatorname{\mathbb{R}}^{d}\setminus
B_{\varepsilon}(x)}((u(x)-u(y))\nu(x\\!-\\!y)\,\mathrm{d}y$
$\displaystyle=\lim_{\varepsilon\to
0}\int\limits_{\Omega}\int\limits_{\Omega\setminus
B_{\varepsilon}(x)}(u(x)-u(y))v(x)\nu(x\\!-\\!y)\,\mathrm{d}y\mathrm{d}x+\int\limits_{\Omega}\int\limits_{\Omega^{c}}(u(x)-u(y))v(x)\nu(x\\!-\\!y)\,\mathrm{d}y\mathrm{d}x.$

On one side, by a symmetry argument we have

$\displaystyle\lim_{\varepsilon\to 0}$
$\displaystyle\int\limits_{\Omega}\int\limits_{\Omega\setminus
B_{\varepsilon}(x)}(u(x)-u(y))v(x)\nu(x\\!-\\!y)\,\mathrm{d}y\mathrm{d}x=\lim_{\varepsilon\to
0}\iint\limits_{\Omega\times\Omega\cap\\{|x-y|>\varepsilon\\}}(u(x)-u(y))v(x)\nu(x\\!-\\!y)\,\mathrm{d}y\mathrm{d}x$
$\displaystyle=\lim_{\varepsilon\to
0}\frac{1}{2}\iint\limits_{\Omega\times\Omega\cap\\{|x-y|>\varepsilon\\}}(u(x)-u(y))(v(x)-v(y))\nu(x\\!-\\!y)\,\mathrm{d}y\mathrm{d}x$
$\displaystyle=\frac{1}{2}\iint\limits_{\Omega\Omega}(u(x)-u(y))(v(x)-v(y))\nu(x\\!-\\!y)\,\mathrm{d}y\mathrm{d}x$

where one gets rid of the principal value using the estimate (A.4) applied to
$u$ and $v$. On the other side, with the help of Fubini’s theorem we have

$\displaystyle\iint\limits_{\Omega\Omega^{c}}(u(x)-u(y))v(x)\nu(x\\!-\\!y)\,\mathrm{d}y\mathrm{d}x$
$\displaystyle=\iint\limits_{\Omega\Omega^{c}}(u(x)-u(y))(v(x)-v(y))\nu(x\\!-\\!y)\,\mathrm{d}y\mathrm{d}x+\int\limits_{\Omega^{c}}v(y)\mathrm{d}y\int\limits_{\Omega}(u(x)-u(y))\nu(x\\!-\\!y)\,\mathrm{d}x$
$\displaystyle=\frac{1}{2}\iint\limits_{\Omega\Omega^{c}}(u(x)-u(y))(v(x)-v(y))\nu(x\\!-\\!y)\,\mathrm{d}y\mathrm{d}x+\frac{1}{2}\iint\limits_{\Omega^{c}\Omega}(u(x)-u(y))(v(x)-v(y))\nu(x\\!-\\!y)\,\mathrm{d}y\mathrm{d}x$
$\displaystyle\qquad\qquad-\int\limits_{\Omega^{c}}\mathcal{N}u(y)v(y)\mathrm{d}y\,.$

Altogether inserted in the initial relation provide the desired relation. ∎

As a direct consequence of A.5 we have the following.

######  Corollary A.6 (Second Gauss-Green identity).

For all $u,v\in C_{b}^{2}(\operatorname{\mathbb{R}}^{d})$ we have

$\displaystyle\int_{\Omega}v(x)Lu(x)-u(x)Lv(x)\,\mathrm{d}x=\int_{\Omega^{c}}u(y)\mathcal{N}v(y)-v(y)\mathcal{N}u(y)\,\mathrm{d}y.$
(A.7)

We now look at certain aspect of the dual of the trace space
$T_{\nu}(\Omega^{c})$ in relation with the nonlocal normal derivative operator
$\mathcal{N}$.

######  Theorem A.7.

For any linear continuous form $\ell:T_{\nu}(\Omega^{c})\to\mathbb{R}$ there
exists a function $w\in V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})$ such
that for every $v\in C^{\infty}_{c}(\overline{\Omega}^{c})$

$\displaystyle\ell(v)=\int_{\Omega^{c}}\mathcal{N}w(y)v(y)\mathrm{d}y.$

In particular, given a measurable function $g:\Omega^{c}\to\mathbb{R}$ if the
linear mapping $\ell_{g}:v\mapsto\int_{\Omega^{c}}g(y)v(y)\mathrm{d}y$ is
continuous on $T_{\nu}(\Omega^{c})$ then, there exists $w\in
V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})$ such that $g=\mathcal{N}w$
almost everywhere on $\Omega^{c}$.

###### Proof.

Let $\ell\in T_{\nu}(\Omega^{c})^{\prime}$ then because of the continuity of
the trace operator $\operatorname{Tr}$, the linear form
$\ell\circ\operatorname{Tr}$ is also continuous on
$V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})$. By Riesz’s representation
theorem there exists $w\in V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})$ such
that
$\ell\circ\operatorname{Tr}(v)=\left(v,w\right)_{(V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})}$
for each $v\in V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})$. In particular,
for $v\in C^{\infty}_{c}(\overline{\Omega}^{c})$ identified with its zero
extension on $\Omega$ so that $\operatorname{Tr}(v)=v$, we remain with

$\displaystyle\ell(v)$
$\displaystyle=\int_{\Omega}w(x)v(x)\mathrm{d}x+\frac{1}{2}\iint\limits_{(\Omega^{c}\times\Omega^{c})^{c}}(w(x)-w(y))(v(x)-v(y))\nu(x\\!-\\!y)\,\mathrm{d}x\,\mathrm{d}y$
$\displaystyle=\int_{\Omega^{c}}v(y)\mathrm{d}y\int_{\Omega}(w(y)-w(x))\nu(x\\!-\\!y)\,\mathrm{d}x=\int_{\Omega^{c}}\mathcal{N}w(y)v(y)\mathrm{d}y.$

Furthermore, if $g:\Omega^{c}\to\mathbb{R}$ is such that $\ell_{g}$ is
continuous on $T_{\nu}(\Omega^{c})$ then by the above computation, it follows
that $g=\mathcal{N}w$ almost everywhere on $\Omega^{c}$ since

$\displaystyle\int_{\Omega^{c}}g(y)v(y)\mathrm{d}y=\int_{\Omega^{c}}\mathcal{N}w(y)v(y)\mathrm{d}y\qquad\hbox{for
all }\leavevmode\nobreak\ \leavevmode\nobreak\ v\in
C_{c}^{\infty}(\overline{\Omega}^{c}).$

∎

######  Remark A.8.

The second statement of Theorem A.7 particularly suggests that the space of
all measurable functions $g:\Omega^{c}\to\mathbb{R}$ for which the linear form
$v\mapsto\int_{\Omega^{c}}g(y)v(y)\mathrm{d}y$ is continuous on
$T_{\nu}(\Omega^{c})$ is contained in
$\mathcal{N}(V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d}))$ (the range of
$\mathcal{N}$).

######  Remark A.9.

The nonlocal normal derivative $\mathcal{N}u$ of a function measurable
$u:\operatorname{\mathbb{R}}^{d}\to\operatorname{\mathbb{R}}$ can be thought
of as the restriction of the regional operator on $\Omega$ associated with
$k(x,y)=\nu(x-y)$ on $\operatorname{\mathbb{R}}^{d}\setminus\Omega$. It might
be interesting to know some situations where the pointwise definition
$\mathcal{N}u(x)$ makes sense at least almost everywhere. It is
straightforward to verify the following: (i) if $u\in L^{\infty}(\Omega)$ then
$\mathcal{N}u(x)$ exists for almost every
$x\in\operatorname{\mathbb{R}}^{d}\setminus\overline{\Omega}$, (ii) if $u\in
V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})$ then $\mathcal{N}u\in
L^{2}_{\operatorname{loc}}(\operatorname{\mathbb{R}}^{d}\setminus\overline{\Omega})$,
(iii) more generally, if $u\in V_{\nu}(\Omega|\operatorname{\mathbb{R}}^{d})$
then $\mathcal{N}u\in
L^{2}(\operatorname{\mathbb{R}}^{d}\setminus\Omega,w^{-1}(x)\mathrm{d}x)$
where $w(x)=\int_{\Omega}\nu(x-y)\mathrm{d}y$,
$x\in\operatorname{\mathbb{R}}^{d}\setminus\Omega$.

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