# Another look at elliptic homogenization

Andrea Braides, Giuseppe Cosma Brusca, and Davide Donati  SISSA, via Bonomea
265, Trieste, Italy

###### Abstract.

We consider the limit of sequences of normalized $(s,2)$-Gagliardo seminorms
with an oscillating coefficient as $s\to 1$. In a seminal paper by Bourgain,
Brezis and Mironescu (subsequently extended by Ponce) it is proven that if the
coefficient is constant then this sequence $\Gamma$-converges to a multiple of
the Dirichlet integral. Here we prove that, if we denote by $\varepsilon$ the
scale of the oscillations and we assume that $1-s<\\!<\varepsilon^{2}$, this
sequence converges to the homogenized functional formally obtained by
separating the effects of $s$ and $\varepsilon$; that is, by the
homogenization as $\varepsilon\to 0$ of the Dirichlet integral with
oscillating coefficient obtained by formally letting $s\to 1$ first.

MSC codes: 49J45, 35B27, 35R11.

Keywords: $\Gamma$-convergence, non-local functionals, fractional Sobolev
spaces, homogenization.

##  1\. Introduction

In their seminal paper [2] Bourgain, Brezis and Mironescu have studied the
asymptotic behaviour of Gagliardo seminorms $[u]_{W^{s,p}(\Omega)}$ as $s\to
1$, and in particular, in the case $p=2$, of the seminorm
$[u]_{W^{s,2}(\Omega)}$ given by

$[u]_{W^{s,2}(\Omega)}:=\Bigl{(}\iint_{\Omega\times\Omega}\frac{|u(x)-u(y)|^{2}}{|x-y|^{d+2s}}\,dxdy\Bigr{)}^{\frac{1}{2}},$

where $\Omega$ a bounded open subset of $\mathbb{R}^{d}$, $s\in(0,1)$, and the
fractional Sobolev space $W^{s,2}(\Omega)$ is defined as

$W^{s,2}(\Omega):=\\{u\in L^{2}(\Omega):[u]_{W^{s,2}(\Omega)}<+\infty\\}.$

Their results, subsequently extended by Ponce [10], imply that the functionals

$F_{s}(u)=(1-s)\iint_{\Omega\times\Omega}\frac{|u(x)-u(y)|^{2}}{|x-y|^{d+2s}}\,dxdy$

$\Gamma$-converge, with respect to the $L^{2}$-convergence, to the (multiple
of the) Dirichlet integral

$\frac{\sigma_{d-1}}{2d}\int_{\Omega}|\nabla u|^{2}\,dx,$

where $\sigma_{d-1}$ is the ${\mathcal{H}}^{d-1}$-dimensional measure of
$S^{d-1}$, with domain the usual Sobolev space $W^{1,2}(\Omega)$. Since the
functionals are equi-coercive in $L^{2}(\Omega)$, by the properties of
convergence of minima of $\Gamma$-convergence, minimum problems involving the
Dirichlet integral can be approximated by problems involving Gagliardo
seminorms, upon possibly suitably defining boundary-value problems if
necessary.

This fractional-Sobolev space approximation can be extended to other problems
involving elliptic integrals, such as homogenization problems, the most
classical of which is the asymptotic analysis of (isotropic) oscillating
energies

$\int_{\Omega}a\Bigl{(}\frac{x}{\varepsilon}\Bigr{)}|\nabla u|^{2}\,dx,$ (1)

where $a$ is a $1$-periodic function with $0<\alpha\leq
a(y)\leq\beta<+\infty$. The $\Gamma$-limit as $\varepsilon\to 0$ of such
energies is the homogenized functional

$F_{\rm hom}(u)=\int_{\Omega}\langle A_{\rm hom}\nabla u,\nabla u\rangle\,dx,$

where $A_{\rm hom}$ is a symmetric $d\times d$ matrix characterized by the
homogenization formula

$\langle A_{\rm
hom}z,z\rangle=\min\Bigl{\\{}\int_{(0,1)^{d}}a(y)|z+\nabla\varphi(y)|^{2}dy:\varphi\
1\hbox{-periodic}\Bigr{\\}}$ (2)

(see e.g. [4]). Formally, by the result of Bourgain, Brezis and Mironescu, we
can define an approximation of the homogenized functional using the energies

$F_{\varepsilon,s}(u)=(1-s)\iint_{\Omega\times\Omega}a\left(\frac{x}{\varepsilon}\right)\frac{\lvert
u(x)-u(y)\rvert^{2}}{\lvert x-y\rvert^{d+2s}}dxdy,$ (3)

defined for $u\in W^{s,2}(\Omega)$ and $\varepsilon>0$. Indeed, upon supposing
for simplicity that $a$ be continuous, we easily see that for fixed
$\varepsilon>0$ the $\Gamma$-limit of $F_{\varepsilon,s}$ as $s\to 1^{-}$ is
indeed

$\frac{\sigma_{d-1}}{2d}\int_{\Omega}a\Bigl{(}\frac{x}{\varepsilon}\Bigr{)}|\nabla
u|^{2}\,dx,$

which then $\Gamma$-converge to $\frac{\sigma_{d-1}}{2d}F_{\rm hom}$ as
$\varepsilon\to 0$. Since all functionals are equi-coercive in $L^{2}$, we can
use a diagonal argument (see [6]) and deduce that there exist
$s=s_{\varepsilon}$ such that $s\to 1^{-}$ as $\varepsilon\to 0$ and the
$\Gamma$-limit of $F_{\varepsilon,s}$ is still $\frac{\sigma_{d-1}}{2d}F_{\rm
hom}$.

In this paper we investigate the scales $s=s_{\varepsilon}$ at which this
limit holds. If we let both $\varepsilon,$ and $1-s$ simultaneously tend to
$0$, a heuristic argument in order to argue what can be a critical scale is as
follows. Let $z$ be fixed and let $\varphi$ be as in the formula
characterizing $\langle A_{\rm hom},z,z\rangle$. We can use
$u_{\varepsilon}(x)=\langle z,x\rangle+\varepsilon\varphi(x/\varepsilon)$ as
test functions for the $\Gamma$-limit of $F_{\varepsilon,s}$. We may suppose
that $\varphi$ be twice continuous differentiable, so that, using the Taylor
development of $\varphi$ at $x/\varepsilon$, we have

$\displaystyle\hskip-56.9055pt(1-s)\iint_{\Omega\times\Omega}a\Bigl{(}\frac{x}{\varepsilon}\Bigr{)}\frac{|u_{\varepsilon}(x)-u_{\varepsilon}(y)|^{2}}{|x-y|^{d+2s}}\,dxdy$
$\displaystyle=$
$\displaystyle(1-s)\iint_{\Omega\times\Omega}a\Bigl{(}\frac{x}{\varepsilon}\Bigr{)}\frac{|\langle
z+\nabla\varphi(\frac{x}{\varepsilon}),x-y\rangle|^{2}+O(\frac{1}{\varepsilon^{2}})}{|x-y|^{d+2s}}\,dxdy$
$\displaystyle=$
$\displaystyle\frac{\sigma_{d-1}}{2d}\int_{\Omega}a\Bigl{(}\frac{x}{\varepsilon}\Bigr{)}\Big{|}z+\nabla\varphi\left(\frac{x}{\varepsilon}\right)\Big{|}^{2}dx+O\Big{(}\frac{1-s}{\varepsilon^{2}}\Big{)}.$

This argument suggests that when

$1-s<\\!<\varepsilon^{2}\qquad\qquad\hbox{({\em subcritical case})},$ (4)

we may construct recovery sequences optimizing the oscillations of $a$, and
actually is a proof of the upper bound in this case for the target function
$u(x)=\langle z,x\rangle$. We will prove that under assumption (4) a
separation of the scales $\varepsilon$ and $s$ occurs, and the limit is the
one computed above by letting $s\to 1^{-}$ first and then $\varepsilon\to 0$.
The main argument in the proof of the lower bound is obtained by a
discretization procedure (which as a byproduct also gives a different proof of
the results in [10]) based on the use of Kuhn’s decomposition [8] and an
integration argument on the set of orthogonal bases in $\mathbb{R}^{d}$
borrowed from a recent paper by Solci [11]. Thanks to (4), we can then reduce
the computation to studying the limit of oscillating energies (1) on
piecewise-affine interpolations at a scale much smaller than $\varepsilon$,
for which the known homogenization result can be applied. This reduction is
made possible by a lemma (Lemma 3), which allows to consider only points $x,y$
sufficiently close to each other in the computation of (3).

The cases other than subcritical are not dealt with here. They will require
different, more complex, techniques and will be treated in future work. In
order to give a hint of the necessity of different types of arguments in those
cases, we can consider a similar type of non-local functionals treated in a
general setting in [1], of the form

$F_{\varepsilon,\delta}(u)=\frac{1}{\delta^{d}}\iint_{\Omega\times\Omega}\varrho\Big{(}\frac{|x-y|}{\delta}\Bigr{)}a\left(\frac{x}{\varepsilon}\right)\frac{\lvert
u(x)-u(y)\rvert^{2}}{\lvert x-y\rvert^{2}}dxdy,$ (5)

where $\varrho$ is a suitably integrable positive kernel and $\delta>0$ plays
a similar role as $s$ above, in that it forces concentration as $\delta\to 0$.
Such functionals are also studied in [2] if the function $a$ is a constant,
and still approximate the Dirichlet integrand as $\delta\to 0$. The critical
case for such functionals is $\delta\sim\varepsilon$, and if
$\delta/\varepsilon\to\kappa\in(0,+\infty)$ the $\Gamma$-limit can be computed
using a non-local-to-local homogenization procedure giving a local homogenized
limit energy of the form

$F^{\kappa}_{\rm hom}(u)=\int_{\Omega}\langle A^{\kappa}_{\rm hom}\nabla
u,\nabla u\rangle\,dx,$

where $A^{\kappa}_{\rm hom}$ is now characterized by a non-local
homogenization formula

$\displaystyle\hskip-14.22636pt\langle A^{\kappa}_{\rm hom}z,z\rangle$
$\displaystyle=\min\Bigl{\\{}\int_{\mathbb{R}^{d}\times(0,1)^{d}}\frac{1}{\kappa^{d}}\varrho\Bigl{(}\frac{|x-y|}{\kappa}\Bigr{)}a(y)\frac{|\langle
z,y-x\rangle+\varphi(y)-\varphi(x)|^{2}}{|x-y|^{2}}dxdy:\varphi\
1\hbox{-periodic}\Bigr{\\}}.$

In order to prove this result, localization techniques for limits of non-local
functionals must be used (see [1] and also [5]). Comparing functionals
$F_{\varepsilon,s}$ with those in (5), a main additional difficulty is due to
the different nature of the dependence on $\delta$ and $s$, respectively, so
that the case $1-s\sim\varepsilon^{2}$ cannot be directly set as a
homogenization-concentration problem, contrary to the case
$\delta\sim\varepsilon$.

##  2\. Statement of the result and preliminaries

Let $a$ be a $1$-periodic continuous function such that

$0<\alpha\leq a(x)\leq\beta<+\infty$ (6)

for every $x\in\mathbb{R}^{d}$. For $\varepsilon,s\in(0,1)$ we introduce the
functional

$F_{\varepsilon,s}(u)=(1-s)\iint_{\Omega\times\Omega}a\left(\frac{x}{\varepsilon}\right)\frac{\lvert
u(x)-u(y)\rvert^{2}}{\lvert x-y\rvert^{d+2s}}dxdy$ (7)

for $u\in W^{s,2}(\Omega)$.

We assume that the $\varepsilon$ is a positive parameter and that $s$ is a
function of $\varepsilon$ valued in $(0,1)$ approaching $1$ as $\varepsilon\to
0$. Since $s=s(\varepsilon)$, as a shorthand, $F_{\varepsilon,s}$ will be
denoted by $F_{\varepsilon}$. We prove that if $1-s\ll\varepsilon^{2}$, then
the separation of scales described in the Introduction holds, as if we were
passing to the limit first letting $s\to 1$ and then letting $\varepsilon\to
0$.

######  Theorem 1.

Let $\Omega$ be a bounded open set with Lipschitz boundary,
$\varepsilon\in(0,1),\,s=s_{\varepsilon}$, and let $F_{\varepsilon}$ be the
functional defined in (7). If

$\lim_{\varepsilon\to 0}\frac{1-s}{\varepsilon^{2}}=0,$ (8)

then

$\Gamma\hbox{-}\lim_{\varepsilon\to
0}F_{\varepsilon}(u)=\frac{\sigma_{d-1}}{2d}\int_{\Omega}\langle A_{\rm
hom}\nabla u,\nabla u\rangle\,dx=:F_{\rm hom}(u)$ (9)

for every $u\in W^{1,2}(\Omega)$, where the $\Gamma$-limit is computed with
respect to the $L^{2}(\Omega)$ convergence, $A_{\rm hom}$ is given by (2), and
$\sigma_{d-1}:={\mathcal{H}}^{d-1}(S^{d-1})$ is the $(d-1)$-dimensional
Hausdorff measure of the unit sphere in $\mathbb{R}^{d}$.

We note that the hypothesis that $\Omega$ has a Lipschtiz boundary is
exploited only in proving, by means of a density argument, the
$\Gamma$-$\limsup$ inequality.

######  Remark 2.

Theorem 1 can be generalized to functionals modelled on $(s,p)$-Gagliardo
seminorms with $p>1$, of the form

$F_{\varepsilon,s}(u)=(1-s)\iint_{\Omega\times\Omega}a\left(\frac{x}{\varepsilon}\right)\frac{\lvert
u(x)-u(y)\rvert^{p}}{\lvert x-y\rvert^{d+ps}}dxdy$ (10)

for $u\in W^{s,p}(\Omega)$. The subcritical regime is then
$1-s<\\!<\varepsilon^{p}$, and the resulting $\Gamma$-limit is

$C_{d,p}\int_{\Omega}f_{\rm hom}(\nabla u)\,dx,$

defined on $W^{1,p}(\Omega)$, with $C_{d,p}$ the constant appearing in the
corresponding result in [2], and $f_{\rm hom}$ defined by

$f_{\rm
hom}(z)=\min\Bigl{\\{}\int_{(0,1)^{d}}a(y)|z+\nabla\varphi(y)|^{p}dy:\varphi\
1\hbox{-periodic}\Bigr{\\}}.$ (11)

This result can be obtained with few changes and some heavier notation from
the case $p=2$.

We preliminarily state a lemma that allows us to take into account only those
interactions due to pairs of sufficiently close points in $\Omega$.

######  Lemma 3.

Let $(u_{\varepsilon})_{\varepsilon}\subseteq L^{2}(\Omega)$ be bounded and
let $(r_{\varepsilon})_{\varepsilon}$ be a sequence of positive real numbers
such that

$\lim_{\varepsilon\to 0}\frac{1-s_{\varepsilon}}{r_{\varepsilon}^{2}}=0.$ (12)

Then

$\lim_{\varepsilon\to
0}\biggl{(}F_{\varepsilon}(u_{\varepsilon})-(1-s_{\varepsilon})\iint_{\Omega\times\Omega\cap\\{(x,y):|x-y|\leq
r_{\varepsilon}\\}}a\left(\frac{x}{\varepsilon}\right)\frac{|u_{\varepsilon}(x)-u_{\varepsilon}(y)|^{2}}{|x-y|^{d+2s_{\varepsilon}}}\,dxdy\biggr{)}=0.$

###### Proof.

By the convexity of $x\mapsto x^{2}$ and by (6), we have

$\displaystyle\hskip-56.9055pt(1-s_{\varepsilon})\iint_{\Omega\times\Omega\cap\\{(x,y):|x-y|>r_{\varepsilon}\\}}a\left(\frac{x}{\varepsilon}\right)\frac{|u_{\varepsilon}(x)-u_{\varepsilon}(y)|^{2}}{|x-y|^{d+2s_{\varepsilon}}}\,dxdy$
(13) $\displaystyle\leq$
$\displaystyle(1-s_{\varepsilon})\,4\beta\iint_{\Omega\times\Omega\cap\\{(x,y):|x-y|>r_{\varepsilon}\\}}\frac{|u_{\varepsilon}(x)|^{2}}{|x-y|^{d+2s_{\varepsilon}}}\,dxdy$
$\displaystyle=$
$\displaystyle(1-s_{\varepsilon})\,4\beta\int_{\mathbb{R}^{d}\setminus
B_{r_{\varepsilon}}(0)}\frac{1}{|\xi|^{d+2s_{\varepsilon}}}\,d\xi\int_{\Omega}|u_{\varepsilon}(x)|^{2}\,dx,$

where we also took advantage of the symmetry of the integrand to get the
inequality.

Since $(u_{\varepsilon})_{\varepsilon}$ is bounded in $L^{2}(\Omega)$, (13) is
estimated from above (up to a constant factor) by

$\displaystyle(1-s_{\varepsilon})\int_{\mathbb{R}^{d}\setminus
B_{r_{\varepsilon}}(0)}\frac{1}{|\xi|^{d+2s_{\varepsilon}}}\,d\xi$
$\displaystyle=$
$\displaystyle(1-s_{\varepsilon})\mathcal{H}^{d-1}(S^{d-1})\int_{r_{\varepsilon}}^{\infty}\rho^{-1-2s_{\varepsilon}}d\rho$
$\displaystyle=$
$\displaystyle\mathcal{H}^{d-1}(S^{d-1})(1-s_{\varepsilon})\frac{r_{\varepsilon}^{-2s_{\varepsilon}}}{2s_{\varepsilon}}$
$\displaystyle\leq$
$\displaystyle\mathcal{H}^{d-1}(S^{d-1})(1-s_{\varepsilon})\frac{r_{\varepsilon}^{-2}}{2s_{\varepsilon}},$

which tends to $0$ by (12), completing our proof. ∎

######  Remark 4.

We will apply the previous Lemma setting

$r_{\varepsilon}=\varepsilon\qquad\hbox{ or }\qquad
r_{\varepsilon}=\sqrt{\varepsilon\sqrt{1-s_{\varepsilon}}}$

in accordance with our convenience.

Note that with the second choice it holds

$\sqrt{1-s_{\varepsilon}}<\\!<r_{\varepsilon}<\\!<\varepsilon,$

so that, in particular,

$\lim_{\varepsilon\to 0}\frac{r_{\varepsilon}}{\varepsilon}=0.$ (14)

It is useful to observe that, in both circumstances, we have

$\lim_{\varepsilon\to 0}|\log
r_{\varepsilon}^{1-s_{\varepsilon}}|=\lim_{\varepsilon\to
0}(1-s_{\varepsilon})|\log r_{\varepsilon}|=0$

which implies

$\lim_{\varepsilon\to 0}r_{\varepsilon}^{1-s_{\varepsilon}}=1.$ (15)

In the proof of the liminf inequality we take advantage of a discretization
argument based on the construction of proper lattices, which can be
parameterized on orthogonal bases.

######  Remark 5 (Notation for the set of orthogonal bases).

Following the notation of [11] we define the set of orthonormal bases of
$\mathbb{R}^{d}$

$V:=\\{\overline{\nu}=(\nu_{1},...,\nu_{d}):\nu_{j}\in S^{d-1}\text{ such that
}\langle\nu_{i},\nu_{j}\rangle=0\text{ for }i\neq j\\}$

and observe that $V$ has Hausdorff dimension equal to $k_{d}:=d(d-1)/2$. For
every $n\in\\{1,...,d\\}$ and fixed $\nu\in S^{d-1}$ we define

$V_{n}^{\nu}:=\\{\overline{\nu}\in V\hbox{ such that }\nu_{n}=\nu\\},$

whose Hausdorff dimension is $k_{d}-(d-1)$. Note that we have

$\mathcal{H}^{k_{d}-(d-1)}(V_{n}^{\nu})=\frac{\mathcal{H}^{k_{d}}(V)}{\mathcal{H}^{d-1}(S^{d-1})},$
(16)

and that in general, the formula

$\int_{V}f(\overline{\nu})\,d\mathcal{H}^{k_{d}}(\overline{\nu})=\int_{S^{d-1}}\int_{V_{n}^{\nu}}f^{\nu}(\overline{\nu}_{n})\,d\mathcal{H}^{k_{d}-(d-1)}(\overline{\nu}_{n})\,d\mathcal{H}^{d-1}(\nu),$
(17)

holds with $\overline{\nu}_{n}:=(\nu_{1},...,\nu_{n-1},\nu_{n+1},...,\nu_{d})$
and $f^{\nu}(\overline{\nu}_{n}):=f(\overline{\nu})$ for every
$n\in\\{1,...,d\\}$ and $f$ non negative measurable function.

Given $\rho>0$ and $\overline{\nu}\in V$, we define
$\mathbb{Z}^{d}_{\rho\overline{\nu}}:=\\{z_{1}\rho\nu_{1}+z_{2}\rho\nu_{2}+...+z_{d}\rho\nu_{d}:(z_{1},...,z_{d})\in\mathbb{Z}^{d}\\}$
and $Q_{\rho\overline{\nu}}$ as the cube described by the orthogonal basis
$\\{\rho\nu_{1},...,\rho\nu_{d}\\}$.

######  Remark 6 (Kuhn’s decomposition).

A cube $Q_{\rho\overline{\nu}}$ defined as in the previous remark can be
further decomposed into $d!$ $d$-simplices of size $\rho^{d}/d!$ through
Kuhn’s decomposition in the following way: for every $\tau$ permutation of $d$
indices, we define $\Delta_{\rho\overline{\nu}}^{\tau}$ as the simplex
described by the vertices
$\rho\nu_{\tau(1)},\rho\nu_{\tau(1)}+\rho\nu_{\tau(2)},...,\rho\nu_{\tau(1)}+\rho\nu_{\tau(2)}+...+\rho\nu_{\tau(d)}$.
As shown in [8, Lemma 1], as $\tau$ vary among all the permutations, we get a
family of $d!$ simplices which constitutes the desired partition.

We denote the vertices of each simplex by

$\Delta_{\rho\overline{\nu}}^{\tau,0}=0,\quad\Delta_{\rho\overline{\nu}}^{\tau,j}:=\rho\nu_{\tau(1)}+\rho\nu_{\tau(2)}+...+\rho\nu_{\tau(j)}\quad\hbox{
for }j=1,...,d$

and observe that, with this choice, it holds

$\Delta_{\rho\overline{\nu}}^{\tau,j}-\Delta_{\rho\overline{\nu}}^{\tau,j-1}=\rho\nu_{\tau(j)}\qquad\hbox{
for }j=1,...,d.$ (18)

##  3\. Proof of the result

###  3.1. Liminf inequality

Throughout this section we write $s:=s_{\varepsilon}$, $r:=r_{\varepsilon}$
and $\sigma_{d-1}:=\mathcal{H}^{d-1}(S^{d-1})$ in order to simplify the
notation.

Let $u\in L^{2}(\Omega)$ and consider a sequence
$(u_{\varepsilon})_{\varepsilon}$ converging to $u$ in $L^{2}(\Omega)$.
Without loss of generality, we may assume that
$\sup_{\varepsilon}F_{\varepsilon}(u_{\varepsilon})<\infty$ which implies that
$u_{\varepsilon}\in W^{s_{,}2}(\Omega)$ for every $\varepsilon>0$.

In light of Lemma 3, we aim at proving that $u\in W^{1,2}(\Omega^{\prime})$
for every $\Omega^{\prime}$ open subset well contained in $\Omega$ and that

$\displaystyle\hskip-56.9055pt\liminf_{\varepsilon\to
0}(1-s)\iint_{\Omega\times\Omega\cap\\{(x,y):|x-y|\leq
r\\}}a\left(\frac{x}{\varepsilon}\right)\frac{|u_{\varepsilon}(x)-u_{\varepsilon}(y)|^{2}}{|x-y|^{d+2s}}\,dxdy$
$\displaystyle\geq$
$\displaystyle\frac{\sigma_{d-1}}{2d}\int_{\Omega^{\prime}}\langle A_{\rm
hom}\nabla u,\nabla u\rangle\,dx,$

where $r=\varepsilon^{\frac{1}{2}}(1-s)^{\frac{1}{4}}$. By the independence of
the left-hand side from $\Omega^{\prime}$ in particular this implies that
$u\in W^{1,2}(\Omega)$.

Applying Lemma 3, the change of variables $\eta:=y-x$ and $\xi:=\eta/r$, and
then the coarea formula, we get

$\displaystyle\hskip-42.67912ptF_{\varepsilon}(u_{\varepsilon})+o_{\varepsilon}(1)$
$\displaystyle=$
$\displaystyle(1-s)\iint_{\Omega\times\Omega\cap\\{(x,y):|x-y|\leq
r\\}}a\left(\frac{x}{\varepsilon}\right)\frac{|u_{\varepsilon}(x)-u_{\varepsilon}(y)|^{2}}{|x-y|^{d+2s}}\,dxdy$
$\displaystyle=$
$\displaystyle(1-s)\int_{B_{r}(0)}\frac{1}{|\eta|^{d-2(1-s)}}\int_{\\{x\in\Omega\,:\,x+\eta\in\Omega\\}}a\left(\frac{x}{\varepsilon}\right)\frac{|u_{\varepsilon}(x+\eta)-u_{\varepsilon}(x)|^{2}}{|\eta|^{2}}\,dxd\eta$
$\displaystyle=$
$\displaystyle(1-s)\int_{B_{1}(0)}\frac{r^{2(1-s)}}{|\xi|^{d-2(1-s)}}\int_{\\{x\in\Omega\,:\,x+r\xi\in\Omega\\}}a\left(\frac{x}{\varepsilon}\right)\frac{|u_{\varepsilon}(x+r\xi)-u_{\varepsilon}(x)|^{2}}{|r\xi|^{2}}\,dxd\xi$
$\displaystyle=$
$\displaystyle(1-s)\int_{0}^{1}\rho^{d-1}\int_{S^{d-1}}\frac{r^{2(1-s)}}{\rho^{d-2(1-s)}}$
$\displaystyle\qquad\qquad\qquad\int_{\\{x\in\Omega\,:\,x+r\rho\nu\in\Omega\\}}a\left(\frac{x}{\varepsilon}\right)\frac{|u_{\varepsilon}(x+r\rho\nu)-u_{\varepsilon}(x)|^{2}}{|r\rho|^{2}}dxd\mathcal{H}^{d-1}(\nu)d\rho.$

For $\overline{\nu}\in V$ (recall the notation of Remark 5) we set

$\mathcal{I}^{r}_{\rho\overline{\nu}}:=\\{k\in\mathbb{Z}_{\rho\overline{\nu}}^{d}:rk+rQ_{\rho\overline{\nu}}\subset\subset\Omega\\},$

and note that for every $\nu\in S^{d-1}$ it holds

$\bigcup_{k\in\mathcal{I}^{r}_{\rho\overline{\nu}}}rk+rQ_{\rho\overline{\nu}}\subseteq\\{x\in\Omega:x+r\rho\nu\in\Omega\\}$

for every $\overline{\nu}\in V$. Hence, we have

$\displaystyle\hskip-42.67912ptF_{\varepsilon}(u_{\varepsilon})+o_{\varepsilon}(1)$
$\displaystyle\hskip-42.67912pt\geq(1-s)\int_{0}^{1}\frac{r^{2(1-s)}}{\rho^{-1+2s}}\int_{S^{d-1}}\sum_{k\in\mathcal{I}^{r}_{\rho\overline{\nu}}}\int_{rk+rQ_{\rho\overline{\nu}}}{\hskip-8.5359pt}a\left(\frac{x}{\varepsilon}\right)\frac{|u_{\varepsilon}(x+r\rho\nu)-u_{\varepsilon}(x)|^{2}}{|r\rho|^{2}}dxd\mathcal{H}^{d-1}(\nu)d\rho.$
(19)

We shall exploit the uniform continuity of the function $a$ to factor it out
from the inner integral; for this reason, we introduce a modulus of continuity
$\omega:[0,+\infty)\rightarrow[0,+\infty)$; that is, an increasing continuous
function such that $\omega(0)=0$ and
$|a(x_{1})-a(x_{2})|\leq\omega(|x_{1}-x_{2}|)$ for every
$x_{1},x_{2}\in\mathbb{R}^{d}$.

We rewrite the right-hand side of (LABEL:liminf1) as

$\displaystyle\hskip-22.76228pt(1-s)\int_{0}^{1}\frac{r^{2(1-s)}}{\rho^{-1+2s}}\int_{S^{d-1}}\sum_{k\in\mathcal{I}^{r}_{\rho\overline{\nu}}}a\left(\frac{rk}{\varepsilon}\right)\int_{rk+rQ_{\rho\overline{\nu}}}\frac{|u_{\varepsilon}(x+r\rho\nu)-u_{\varepsilon}(x)|^{2}}{|r\rho|^{2}}dxd\mathcal{H}^{d-1}(\nu)d\rho$
$\displaystyle\hskip-31.2982pt+(1-s)\int_{0}^{1}\frac{r^{2(1-s)}}{\rho^{-1+2s}}$
$\displaystyle\int_{S^{d-1}}\sum_{k\in\mathcal{I}^{r}_{\rho\overline{\nu}}}\int_{rk+rQ_{\rho\overline{\nu}}}{\hskip-2.84544pt}\Bigl{(}a\left(\frac{x}{\varepsilon}\right)-a\Bigl{(}\frac{rk}{\varepsilon}\Bigr{)}\Bigr{)}\frac{|u_{\varepsilon}(x+r\rho\nu)-u_{\varepsilon}(x)|^{2}}{|r\rho|^{2}}dxd\mathcal{H}^{d-1}(\nu)d\rho,$

and we note that the last term is negligible as $\varepsilon\to 0$ as its
absolute value is estimated from above by

$\displaystyle\hskip-56.9055pt\omega\Bigl{(}\frac{r\sqrt{d}}{\varepsilon}\Bigr{)}(1-s)\int_{0}^{1}\frac{r^{2(1-s)}}{\rho^{-1+2s}}\int_{S^{d-1}}\int_{\\{x+r\rho\nu\in\Omega\\}}\frac{|u_{\varepsilon}(x+r\rho\nu)-u_{\varepsilon}(x)|^{2}}{|r\rho|^{2}}dxd\mathcal{H}^{d-1}(\nu)d\rho$
$\displaystyle=$
$\displaystyle\omega\Bigl{(}\frac{r\sqrt{d}}{\varepsilon}\Bigr{)}(1-s)\int_{B_{1}(0)}\frac{r^{2(1-s)}}{|\xi|^{d-2(1-s)}}\int_{\\{x+r\xi\in\Omega\\}}\frac{|u_{\varepsilon}(x+r\xi)-u_{\varepsilon}(x)|^{2}}{|r\xi|^{2}}\,dxd\xi$
$\displaystyle\leq$
$\displaystyle\omega\Bigl{(}\frac{r\sqrt{d}}{\varepsilon}\Bigr{)}\frac{1}{\alpha}F_{\varepsilon}(u_{\varepsilon}),$

which tends to $0$ since by the assumption
$\sup_{\varepsilon}F_{\varepsilon}(u_{\varepsilon})<+\infty$ and since
$r/\varepsilon\to 0$ by (14).

By this fact and by formulas (16) and (17), inequality (LABEL:liminf1) turns
into

$\displaystyle\hskip-22.76228ptF_{\varepsilon}(u_{\varepsilon})+o_{\varepsilon}(1)$
(20) $\displaystyle\geq$
$\displaystyle(1-s)\int_{0}^{1}\frac{r^{2(1-s)}}{\rho^{-1+2s}}\int_{S^{d-1}}\sum_{k\in\mathcal{I}^{r}_{\rho\overline{\nu}}}a\left(\frac{rk}{\varepsilon}\right)\int_{rk+rQ_{\rho\overline{\nu}}}\frac{|u_{\varepsilon}(x+r\rho\nu)-u_{\varepsilon}(x)|^{2}}{|r\rho|^{2}}dxd\mathcal{H}^{d-1}(\nu)d\rho$
$\displaystyle=$
$\displaystyle(1-s)\int_{0}^{1}\frac{r^{2(1-s)}}{\rho^{-1+2s}}\int_{S^{d-1}}\frac{1}{d}\sum_{n=1}^{d}\frac{\sigma_{d-1}}{\mathcal{H}^{k_{d}}(V)}$
$\displaystyle\qquad\int_{V_{n}^{\nu}}\sum_{k\in\mathcal{I}^{r}_{\rho\overline{\nu}}}a\left(\frac{rk}{\varepsilon}\right)\int_{rk+rQ_{\rho\overline{\nu}}}\frac{|u_{\varepsilon}(x+r\rho\nu)-u_{\varepsilon}(x)|^{2}}{|r\rho|^{2}}dxd\mathcal{H}^{k_{d}-(d-1)}(\overline{\nu}_{n})d\mathcal{H}^{d-1}(\nu)d\rho$
$\displaystyle=$
$\displaystyle\frac{1-s}{d}\frac{\sigma_{d-1}}{\mathcal{H}^{k_{d}}(V)}\int_{0}^{1}\frac{r^{2(1-s)}}{\rho^{-1+2s}}$
$\displaystyle\qquad\qquad\int_{V}\sum_{n=1}^{d}\sum_{k\in\mathcal{I}^{r}_{\rho\overline{\nu}}}a\left(\frac{rk}{\varepsilon}\right)\int_{rk+rQ_{\rho\overline{\nu}}}\frac{|u_{\varepsilon}(x+r\rho\nu_{n})-u_{\varepsilon}(x)|^{2}}{|r\rho|^{2}}dxd\mathcal{H}^{k_{d}}(\overline{\nu})d\rho.$

To produce a lower bound for (20), we take advantage of a discretization
argument based on piecewise-affine auxiliary functions.

Given sufficiently small $\varepsilon>0$, $\rho\in(0,1)$ and
$\overline{\nu}\in V$ we define the function
$u_{\varepsilon}^{\rho\overline{\nu}}$ in two steps:

$\bullet$ first, we assign values on the lattice
$\mathcal{I}^{r}_{\rho\overline{\nu}}$ setting

$u_{\varepsilon}^{\rho\overline{\nu}}(rk)=\displaystyle\frac{1}{|r\rho|^{d}}\int_{rk+rQ_{\rho\overline{\nu}}}u_{\varepsilon}\,dx\qquad\hbox{
for every }k\in\mathcal{I}^{r}_{\rho\overline{\nu}};$

$\bullet$ then, given $k$ in the coarser lattice
$\mathcal{I}^{2r}_{\rho\overline{\nu}}$, consider the cube
$rk+rQ_{\rho\overline{\nu}}$, $\tau$ a permutation of the indices
$\\{1,...,d\\}$, and $rk+r\Delta^{\tau}_{\rho\overline{\nu}}$ the
corresponding simplex in Kuhn’s decomposition, on such simplex we define
$u^{\rho\overline{\nu}}_{\varepsilon}$ being the affine interpolation of the
previously defined values
$u^{\rho\overline{\nu}}_{\varepsilon}(rk)=u^{\rho\overline{\nu}}_{\varepsilon}(rk+r\Delta^{\tau,0}_{\rho\overline{\nu}}),\,u^{\rho\overline{\nu}}_{\varepsilon}(rk+r\Delta^{\tau,1}_{\rho\overline{\nu}}),...,\,u^{\rho\overline{\nu}}_{\varepsilon}(rk+r\Delta^{\tau,d}_{\rho\overline{\nu}})$.

Note that $u_{\varepsilon}^{\rho\overline{\nu}}$ is well defined as a
continuous piecewise-affine function on
$\bigcup_{k\in\mathcal{I}^{2r}_{\rho\overline{\nu}}}rk+rQ_{\rho\overline{\nu}}$;
hence, on each simplex $rk+r\Delta_{\rho\overline{\nu}}^{\tau}$, its gradient
is constant and by (18) it holds

$\displaystyle\hskip-42.67912pt\int_{rk+r\Delta_{\rho\overline{\nu}}^{\tau}}|\nabla
u_{\varepsilon}^{\rho\overline{\nu}}|^{2}\,dx$ $\displaystyle=$
$\displaystyle\sum_{j=1}^{d}\int_{rk+r\Delta_{\rho\overline{\nu}}^{\tau}}\frac{|u_{\varepsilon}^{\rho\overline{\nu}}(rk+r\Delta_{\rho\overline{\nu}}^{\tau,j})-u_{\varepsilon}^{\rho\overline{\nu}}(rk+r\Delta_{\rho\overline{\nu}}^{\tau,j-1})|^{2}}{|r\rho|^{2}}\,dx$
$\displaystyle=$
$\displaystyle\frac{1}{|r\rho|^{2}}\sum_{j=1}^{d}\int_{rk+r\Delta_{\rho\overline{\nu}}^{\tau}}\biggl{|}\frac{1}{|r\rho|^{d}}\Bigl{(}\int_{rk+r\Delta_{\rho\overline{\nu}}^{\tau,j}+rQ_{\rho\overline{\nu}}}u_{\varepsilon}\,dy-\int_{rk+r\Delta_{\rho\overline{\nu}}^{\tau,j-1}+rQ_{\rho\overline{\nu}}}u_{\varepsilon}\,dy\Bigr{)}\biggr{|}^{2}dx$
$\displaystyle=$
$\displaystyle\frac{1}{|r\rho|^{2}}\sum_{j=1}^{d}\int_{rk+r\Delta_{\rho\overline{\nu}}^{\tau}}\biggl{|}\frac{1}{|r\rho|^{d}}\Bigl{(}\int_{rk+r\Delta_{\rho\overline{\nu}}^{\tau,j-1}+rQ_{\rho\overline{\nu}}}u_{\varepsilon}(y+r\rho\nu_{\tau(j)})-u_{\varepsilon}(y)\,dy\Bigr{)}\biggr{|}^{2}dx$
$\displaystyle\leq$
$\displaystyle\frac{1}{|r\rho|^{2}}\sum_{j=1}^{d}\int_{rk+r\Delta_{\rho\overline{\nu}}^{\tau}}\frac{1}{|r\rho|^{d}}\int_{rk+r\Delta_{\rho\overline{\nu}}^{\tau,j-1}+rQ_{\rho\overline{\nu}}}|u_{\varepsilon}(y+r\rho\nu_{\tau(j)})-u_{\varepsilon}(y)|^{2}dy\,dx$
$\displaystyle=$
$\displaystyle\frac{1}{|r\rho|^{d}}\sum_{j=1}^{d}\int_{rk+r\Delta_{\rho\overline{\nu}}^{\tau}}\int_{rk+r\Delta_{\rho\overline{\nu}}^{\tau,j-1}+rQ_{\rho\overline{\nu}}}\frac{|u_{\varepsilon}(y+r\rho\nu_{\tau(j)})-u_{\varepsilon}(y)|^{2}}{|r\rho|^{2}}\,dy\,dx$
$\displaystyle=$
$\displaystyle\frac{1}{d!}\sum_{j=1}^{d}\int_{rk+r\Delta_{\rho\overline{\nu}}^{\tau,j-1}+rQ_{\rho\overline{\nu}}}\frac{|u_{\varepsilon}(y+r\rho\nu_{\tau(j)})-u_{\varepsilon}(y)|^{2}}{|r\rho|^{2}}\,dy,$

where we used Jensen’s inequality and the fact that
$|rk+r\Delta_{\rho\overline{\nu}}^{\tau}|=|r\rho|^{d}/d!$.

Summing over all permutations $\tau$ and
$k\in\mathcal{I}^{2r}_{\rho\overline{\nu}}$ and reinstating the coefficient
$a$, we get

$\displaystyle\hskip-51.21504pt\sum_{k\in\mathcal{I}^{2r}_{{\rho\overline{\nu}}}}a\left(\frac{rk}{\varepsilon}\right)\int_{rk+rQ_{\rho\overline{\nu}}}|\nabla
u_{\varepsilon}^{\rho\overline{\nu}}|^{2}\,dx$ (21) $\displaystyle=$
$\displaystyle\sum_{k\in\mathcal{I}^{2r}_{{\rho\overline{\nu}}}}a\left(\frac{rk}{\varepsilon}\right)\sum_{\tau}\int_{rk+r\Delta_{\rho\overline{\nu}}^{\tau}}|\nabla
u_{\varepsilon}^{\rho\overline{\nu}}|^{2}\,dx$ $\displaystyle\leq$
$\displaystyle\sum_{k\in\mathcal{I}^{2r}_{{\rho\overline{\nu}}}}a\left(\frac{rk}{\varepsilon}\right)\sum_{\tau}\frac{1}{d!}\sum_{j=1}^{d}\int_{rk+r\Delta_{\rho\overline{\nu}}^{\tau,j-1}+rQ_{\rho\overline{\nu}}}\frac{|u_{\varepsilon}(y+r\rho\nu_{\tau(j)})-u_{\varepsilon}(y)|^{2}}{|r\rho|^{2}}\,dy$
$\displaystyle=$
$\displaystyle\frac{1}{d!}\sum_{n=1}^{d}\sum_{\\{\tau,\,j\,|\,\tau(j)=n\\}}\sum_{k\in\mathcal{I}^{2r}_{{\rho\overline{\nu}}}}a\left(\frac{rk}{\varepsilon}\right)\int_{rk+r\Delta_{\rho\overline{\nu}}^{\tau,j-1}+rQ_{\rho\overline{\nu}}}\frac{|u_{\varepsilon}(y+r\rho\nu_{n})-u_{\varepsilon}(y)|^{2}}{|r\rho|^{2}}\,dy.$

Keeping $n,j,\tau$ fixed, we put $h:=k+\Delta_{\rho\overline{\nu}}^{\tau,j-1}$
and note that $k\in\mathcal{I}^{2r}_{\rho\overline{\nu}}$ implies
$h\in\mathcal{I}^{r}_{\rho\overline{\nu}}$ so that (21) is less than or equal
to

$\displaystyle\hskip-71.13188pt\frac{1}{d!}\sum_{n=1}^{d}\sum_{\\{\tau,\,j\,|\,\tau(j)=n\\}}\sum_{h\in\mathcal{I}^{r}_{{\rho\overline{\nu}}}}a\left(\frac{rh}{\varepsilon}\right)\int_{rh+rQ_{\rho\overline{\nu}}}\frac{|u_{\varepsilon}(y+r\rho\nu_{n})-u_{\varepsilon}(y)|^{2}}{|r\rho|^{2}}\,dy+E(\overline{\nu})$
$\displaystyle=$
$\displaystyle\sum_{n=1}^{d}\sum_{h\in\mathcal{I}^{r}_{{\rho\overline{\nu}}}}a\left(\frac{rh}{\varepsilon}\right)\int_{rh+rQ_{\rho\overline{\nu}}}\frac{|u_{\varepsilon}(y+r\rho\nu_{n})-u_{\varepsilon}(y)|^{2}}{|r\rho|^{2}}\,dy+E(\overline{\nu}),$

where we used that the cardinality of the set $\\{\tau,j\,|\,\tau(j)=n\\}$ is
$d!$ for every $n=1,...,d$, and we set

$E(\overline{\nu}):=\sum_{n=1}^{d}\sum_{h\in\mathcal{I}^{r}_{{\rho\overline{\nu}}}}\biggl{(}a\biggl{(}\frac{rh-r\Delta_{\rho\overline{\nu}}^{\tau,j-1}}{\varepsilon}\biggr{)}-a\left(\frac{rh}{\varepsilon}\right)\biggr{)}\int_{rh+rQ_{\rho\overline{\nu}}}\frac{|u_{\varepsilon}(y+r\rho\nu_{n})-u_{\varepsilon}(y)|^{2}}{|r\rho|^{2}}\,dy.$

Summarizing, we have proved that for every $\varepsilon,\rho,\overline{\nu}$,
it holds

$\displaystyle\hskip-56.9055pt\sum_{n=1}^{d}\sum_{k\in\mathcal{I}^{r}_{{\rho\overline{\nu}}}}a\left(\frac{rk}{\varepsilon}\right)\int_{rk+rQ_{\rho\overline{\nu}}}\hskip-8.5359pt\frac{|u_{\varepsilon}(y+r\rho\nu_{n})-u_{\varepsilon}(y)|^{2}}{|r\rho|^{2}}\,dy$
$\displaystyle\geq\sum_{k\in\mathcal{I}^{2r}_{{\rho\overline{\nu}}}}a\left(\frac{rk}{\varepsilon}\right)\int_{rk+rQ_{\rho\overline{\nu}}}|\nabla
u_{\varepsilon}^{\rho\overline{\nu}}|^{2}\,dx-\,\,E(\overline{\nu}),$ (22)

and therefore, having achieved an estimate in terms of the inner intergal in
(20), we get

$\displaystyle\hskip-34.14322ptF_{\varepsilon}(u_{\varepsilon})+o_{\varepsilon}(1)$
(23) $\displaystyle\geq$
$\displaystyle\frac{1-s}{d}\frac{\sigma_{d-1}}{\mathcal{H}^{k_{d}}(V)}\int_{0}^{1}\frac{r^{2(1-s)}}{\rho^{-1+2s}}$
$\displaystyle\quad\quad\int_{V}\sum_{n=1}^{d}\sum_{k\in\mathcal{I}^{r}_{\rho\overline{\nu}}}a\left(\frac{rk}{\varepsilon}\right)\int_{rk+rQ_{\rho\overline{\nu}}}\frac{|u_{\varepsilon}(x+r\rho\nu_{n})-u_{\varepsilon}(x)|^{2}}{|r\rho|^{2}}dxd\mathcal{H}^{k_{d}}(\overline{\nu})d\rho$
$\displaystyle\geq$
$\displaystyle\frac{1-s}{d}\frac{\sigma_{d-1}}{\mathcal{H}^{k_{d}}(V)}\int_{0}^{1}\frac{r^{2(1-s)}}{\rho^{-1+2s}}\int_{V}\sum_{k\in\mathcal{I}^{2r}_{{\rho\overline{\nu}}}}a\left(\frac{rk}{\varepsilon}\right)\int_{rk+rQ_{\rho\overline{\nu}}}|\nabla
u_{\varepsilon}^{\rho\overline{\nu}}|^{2}\,dxd\mathcal{H}^{k_{d}}(\overline{\nu})d\rho$
(25)
$\displaystyle\qquad\qquad\qquad\qquad\qquad-\,\frac{1-s}{d}\frac{\sigma_{d-1}}{\mathcal{H}^{k_{d}}(V)}\int_{0}^{1}\frac{r^{2(1-s)}}{\rho^{-1+2s}}\int_{V}E(\overline{\nu})d\mathcal{H}^{k_{d}}(\overline{\nu})d\rho.$

Note that (25) vanishes as $\varepsilon\to 0$; indeed, considering the modulus
of continuity, the bound from above (6) and the bound from above involving
(23), we have

$\displaystyle\hskip-28.45274pt\frac{1-s}{d}\frac{\sigma_{d-1}}{\mathcal{H}^{k_{d}}(V)}\int_{0}^{1}\frac{r^{2(1-s)}}{\rho^{-1+2s}}\int_{V}E(\overline{\nu})\,d\mathcal{H}^{k_{d}}(\overline{\nu})d\rho$
$\displaystyle\leq$
$\displaystyle\omega\Bigl{(}\frac{r\sqrt{d}}{\varepsilon}\Bigr{)}\frac{1-s}{d}\frac{\sigma_{d-1}}{\mathcal{H}^{k_{d}}(V)}\int_{0}^{1}\frac{r^{2(1-s)}}{\rho^{-1+2s}}$
$\displaystyle\qquad\qquad\qquad\qquad\int_{V}\sum_{n=1}^{d}\sum_{h\in\mathcal{I}^{r}_{{\rho\overline{\nu}}}}\int_{rh+rQ_{\rho\overline{\nu}}}\frac{|u_{\varepsilon}(y+r\rho\nu_{n})-u_{\varepsilon}(y)|^{2}}{|r\rho|^{2}}\,dyd\mathcal{H}^{k_{d}}(\overline{\nu})d\rho$
$\displaystyle\leq$
$\displaystyle\omega\Bigl{(}\frac{r\sqrt{d}}{\varepsilon}\Bigr{)}\frac{1-s}{d}\frac{\sigma_{d-1}}{\mathcal{H}^{k_{d}}(V)}\frac{1}{\alpha}\int_{0}^{1}\frac{r^{2(1-s)}}{\rho^{-1+2s}}$
$\displaystyle\qquad\quad\qquad\int_{V}\sum_{n=1}^{d}\sum_{h\in\mathcal{I}^{r}_{{\rho\overline{\nu}}}}a\left(\frac{rh}{\varepsilon}\right)\int_{rh+rQ_{\rho\overline{\nu}}}\frac{|u_{\varepsilon}(y+r\rho\nu_{n})-u_{\varepsilon}(y)|^{2}}{|r\rho|^{2}}\,dyd\mathcal{H}^{k_{d}}(\overline{\nu})d\rho$
$\displaystyle\leq$
$\displaystyle\omega\Bigl{(}\frac{r\sqrt{d}}{\varepsilon}\Bigr{)}\frac{1}{\alpha}F_{\varepsilon}(u_{\varepsilon})+o_{\varepsilon}(1),$

which tends to $0$ since
$\sup_{\varepsilon}F_{\varepsilon}(u_{\varepsilon})<+\infty$ and since
$r/\varepsilon\to 0$ by (14).

As for (25), we can reinsert the coefficient $a$ in the integral; in
particular, taking into account the modulus of continuity and applying (6),
(3.1), and once more the inequality leading to (23), we have

$\displaystyle\hskip-28.45274pt\frac{1-s}{d}\frac{\sigma_{d-1}}{\mathcal{H}^{k_{d}}(V)}\biggl{|}\int_{0}^{1}\frac{r^{2(1-s)}}{\rho^{-1+2s}}\int_{V}\sum_{k\in\mathcal{I}^{2r}_{{\rho\overline{\nu}}}}\int_{rk+rQ_{\rho\overline{\nu}}}\biggl{(}a\biggl{(}\frac{rk}{\varepsilon}\biggr{)}-a\biggl{(}\frac{x}{\varepsilon}\biggr{)}\biggr{)}|\nabla
u_{\varepsilon}^{\rho\overline{\nu}}|^{2}\,dxd\mathcal{H}^{k_{d}}(\overline{\nu})d\rho\biggr{|}$
$\displaystyle\leq$
$\displaystyle\omega\Bigl{(}\frac{r\sqrt{d}}{\varepsilon}\Bigr{)}\frac{1-s}{d}\frac{\sigma_{d-1}}{\mathcal{H}^{k_{d}}(V)}\int_{0}^{1}\frac{r^{2(1-s)}}{\rho^{-1+2s}}\int_{V}\sum_{k\in\mathcal{I}^{2r}_{{\rho\overline{\nu}}}}\int_{rk+rQ_{\rho\overline{\nu}}}|\nabla
u_{\varepsilon}^{\rho\overline{\nu}}|^{2}\,dxd\mathcal{H}^{k_{d}}(\overline{\nu})d\rho$
$\displaystyle\leq$
$\displaystyle\omega\Bigl{(}\frac{r\sqrt{d}}{\varepsilon}\Bigr{)}\frac{1-s}{d}\frac{\sigma_{d-1}}{\mathcal{H}^{k_{d}}(V)}\frac{1}{\alpha}\int_{0}^{1}\frac{r^{2(1-s)}}{\rho^{-1+2s}}$
$\displaystyle\qquad\int_{V}\sum_{n=1}^{d}\sum_{k\in\mathcal{I}^{r}_{{\rho\overline{\nu}}}}a\left(\frac{rk}{\varepsilon}\right)\int_{rk+rQ_{\rho\overline{\nu}}}\hskip-8.5359pt\frac{|u(y+r\rho\nu_{n})-u(y)|^{2}}{|r\rho|^{2}}\,dy+E(\overline{\nu})d\mathcal{H}^{k_{d}}(\overline{\nu})d\rho$
$\displaystyle\leq$
$\displaystyle\omega\Bigl{(}\frac{r\sqrt{d}}{\varepsilon}\Bigr{)}\frac{1}{\alpha}F_{\varepsilon}(u_{\varepsilon})+o_{\varepsilon}(1),$

and we conclude as above.

We finally obtain

$\displaystyle\hskip-48.36958ptF_{\varepsilon}(u_{\varepsilon})+o_{\varepsilon}(1)$
(26) $\displaystyle\geq$
$\displaystyle\frac{1-s}{d}\frac{\sigma_{d-1}}{\mathcal{H}^{k_{d}}(V)}\int_{0}^{1}\frac{r^{2(1-s)}}{\rho^{-1+2s}}\int_{V}\sum_{k\in\mathcal{I}^{2r}_{{\rho\overline{\nu}}}}\int_{rk+rQ_{\rho\overline{\nu}}}a\left(\frac{x}{\varepsilon}\right)|\nabla
u_{\varepsilon}^{\rho\overline{\nu}}|^{2}\,dxd\mathcal{H}^{k_{d}}(\overline{\nu})d\rho$
$\displaystyle\geq$
$\displaystyle\frac{1-s}{d}\frac{\sigma_{d-1}}{\mathcal{H}^{k_{d}}(V)}\int_{0}^{1}\frac{r^{2(1-s)}}{\rho^{-1+2s}}\int_{V}\int_{\Omega^{\prime}}a\left(\frac{x}{\varepsilon}\right)|\nabla
u_{\varepsilon}^{\rho\overline{\nu}}|^{2}\,dxd\mathcal{H}^{k_{d}}(\overline{\nu})d\rho,$

where $\Omega^{\prime}$ is any open subset well contained in $\Omega$ that we
may assume to be well contained in
$\displaystyle\bigcup_{k\in\mathcal{I}^{2r}_{\rho\overline{\nu}}}rk+rQ_{\rho\overline{\nu}}$
for $\varepsilon$ sufficiently small.

In the final part of the proof, we average the auxiliary functions obtained by
discretization weighting them with the appropriate kernel.

For every $\varepsilon>0$ sufficiently small we define the probability measure
$\mu_{\varepsilon}$ on $[0,1]\times V$ by

$d\mu_{\varepsilon}(\rho,\overline{\nu}):=2(1-s)\rho^{1-2s}\mathcal{L}^{1}\otimes\mathcal{H}^{k_{d}}[\mathcal{H}^{k_{d}}(V)]^{-1};$

then we rewrite (26) as

$\displaystyle\hskip-85.35826pt\frac{\sigma_{d-1}}{2d}\int_{[0,1]\times
V}r^{2(1-s)}\int_{\Omega^{\prime}}a\left(\frac{x}{\varepsilon}\right)|\nabla
u_{\varepsilon}^{\rho\overline{\nu}}|^{2}\,dx\,d\mu_{\varepsilon}(\rho,\overline{\nu})$
(27) $\displaystyle=$
$\displaystyle\frac{\sigma_{d-1}}{2d}\int_{\Omega^{\prime}}r^{2(1-s)}a\left(\frac{x}{\varepsilon}\right)\int_{[0,1]\times
V}|\nabla
u_{\varepsilon}^{\rho\overline{\nu}}|^{2}\,d\mu_{\varepsilon}(\rho,\overline{\nu})\,dx$
$\displaystyle\geq$
$\displaystyle\frac{\sigma_{d-1}}{2d}\int_{\Omega^{\prime}}r^{2(1-s)}a\left(\frac{x}{\varepsilon}\right)\Bigl{|}\int_{[0,1]\times
V}\nabla
u_{\varepsilon}^{\rho\overline{\nu}}\,d\mu_{\varepsilon}(\rho,\overline{\nu})\Bigr{|}^{2}\,dx$

by Jensen’s inequality.

We set

$\overline{u}_{\varepsilon}(x):=\int_{[0,1]\times
V}u^{\rho\overline{\nu}}_{\varepsilon}(x)\,d\mu_{\varepsilon}(\rho,\overline{\nu}),$

and note that, since $u_{\varepsilon}^{\rho\overline{\nu}}$ is continuous on
$\Omega^{\prime}$, this function belongs to $W^{1,2}(\Omega^{\prime})$ with

$\nabla\overline{u}_{\varepsilon}(x)=\int_{[0,1]\times V}\nabla
u^{\rho\overline{\nu}}_{\varepsilon}(x)\,d\mu_{\varepsilon}(\rho,\overline{\nu}).$

As a consequence, (27) reduces to

$\frac{\sigma_{d-1}}{2d}\int_{\Omega^{\prime}}r^{2(1-s)}a\Bigl{(}\frac{x}{\varepsilon}\Bigr{)}|\nabla\overline{u}_{\varepsilon}(x)|^{2}dx$

and our final goal is to show that

$\liminf_{\varepsilon\to
0}\frac{\sigma_{d-1}}{2d}\int_{\Omega^{\prime}}a\Bigl{(}\frac{x}{\varepsilon}\Bigr{)}|\nabla\overline{u}_{\varepsilon}(x)|^{2}dx\,\,\,\geq\,\,\,\frac{\sigma_{d-1}}{2d}\int_{\Omega^{\prime}}\langle
A_{\rm hom}\nabla u,\nabla u\rangle\,dx,$

where we used that $r^{2(1-s)}\to 1$, as already stated in (15).

To check this it is sufficient to prove that $\overline{u}_{\varepsilon}\to u$
in $L^{2}(\Omega^{\prime})$. Indeed, this implies that $u$ is the
$L^{2}(\Omega^{\prime})$-limit of a bounded sequence in
$W^{1,2}(\Omega^{\prime})$, so that $u\in\ W^{1,2}(\Omega^{\prime})$ and the
conclusion follows by the liminf inequality provided by the Homogenization
Theorem (see [4, Theorem 14.7]).

First we prove the convergence under the additional assumption that
$(u_{\varepsilon})_{\varepsilon}$ is bounded in $L^{\infty}(\Omega^{\prime})$.
If this is the case, it suffices to prove that
$\overline{u}_{\varepsilon}-u_{\varepsilon}\to 0$ in $L^{1}(\Omega^{\prime})$,
indeed, since $(\overline{u}_{\varepsilon})_{\varepsilon}$ is bounded by
construction in $L^{\infty}(\Omega^{\prime})$, we conclude by interpolation.

As
$\Omega^{\prime}\subseteq\displaystyle\bigcup_{k\in\mathcal{I}^{2r}_{\rho\overline{\nu}}}rk+rQ_{\rho\overline{\nu}}$,
we have

$\displaystyle\|\overline{u}_{\varepsilon}-u_{\varepsilon}\|_{L^{1}(\Omega^{\prime})}$
$\displaystyle=$
$\displaystyle\int_{\Omega^{\prime}}\biggl{|}u_{\varepsilon}(x)-\int_{[0,1]\times
V}u_{\varepsilon}^{\rho\overline{\nu}}(x)d\mu_{\varepsilon}(\rho,\overline{\nu})\biggr{|}\,dx$
(28) $\displaystyle\leq$ $\displaystyle\int_{[0,1]\times
V}\int_{\Omega^{\prime}}|u_{\varepsilon}(x)-u_{\varepsilon}^{\rho\overline{\nu}}(x)|\,dx\,d\mu_{\varepsilon}(\rho,\overline{\nu})$
$\displaystyle\leq$ $\displaystyle\int_{[0,1]\times
V}\sum_{k\in\mathcal{I}^{2r}_{\rho\overline{\nu}}}\int_{rk+rQ_{\rho\overline{\nu}}}|u_{\varepsilon}(x)-u_{\varepsilon}^{\rho\overline{\nu}}(x)|\,dx\,d\mu_{\varepsilon}(\rho,\overline{\nu})$
$\displaystyle\leq$ $\displaystyle\int_{[0,1]\times
V}\sum_{k\in\mathcal{I}^{2r}_{\rho\overline{\nu}}}\int_{rk+rQ_{\rho\overline{\nu}}}|u_{\varepsilon}(x)-u_{\varepsilon}^{\rho\overline{\nu}}(rk)|\,dx\,d\mu_{\varepsilon}(\rho,\overline{\nu})$
$\displaystyle+$ $\displaystyle\int_{[0,1]\times
V}\sum_{k\in\mathcal{I}^{2r}_{\rho\overline{\nu}}}\int_{rk+rQ_{\rho\overline{\nu}}}|u^{\rho\overline{\nu}}_{\varepsilon}(x)-u_{\varepsilon}^{\rho\overline{\nu}}(rk)|\,dx\,d\mu_{\varepsilon}(\rho,\overline{\nu}).$
(29)

We first prove that (28) tends to $0$ as $\varepsilon$ vanishes. Applying
Hölder’s inequality and a rescaled version of Poincaré-Wirtinger’s inequality
(see [9, Theorem 6.33]) we get

$\displaystyle\hskip-71.13188pt\sum_{k\in\mathcal{I}^{2r}_{\rho\overline{\nu}}}\int_{rk+rQ_{\rho\overline{\nu}}}|u_{\varepsilon}(x)-u_{\varepsilon}^{\rho\overline{\nu}}(rk)|\,dx$
$\displaystyle\leq$
$\displaystyle|r\rho|^{\frac{d}{2}}\sum_{k\in\mathcal{I}^{2r}_{\rho\overline{\nu}}}\Bigl{(}\int_{rk+rQ_{\rho\overline{\nu}}}|u_{\varepsilon}(x)-u_{\varepsilon}^{\rho\overline{\nu}}(rk)|^{2}\,dx\Bigr{)}^{\frac{1}{2}}$
$\displaystyle\leq$ $\displaystyle
P|r\rho|^{\frac{d}{2}+s}\sum_{k\in\mathcal{I}^{2r}_{\rho\overline{\nu}}}\Bigl{(}\iint_{(rk+rQ_{\rho\overline{\nu}})\times(rk+rQ_{\rho\overline{\nu}})}\frac{|u_{\varepsilon}(x)-u_{\varepsilon}(y)|^{2}}{|x-y|^{d+2s}}\,dxdy\Bigr{)}^{\frac{1}{2}},$

where $P$ is the Poincarè-Wirtinger constant for the $d$-dimensional unit
cube. Then we make use of the concavity of $x\mapsto x^{\frac{1}{2}}$ so that

$\displaystyle\hskip-51.21504ptP|r\rho|^{\frac{d}{2}+s}\\#\mathcal{I}^{2r}_{\rho\overline{\nu}}\sum_{k\in\mathcal{I}^{2r}_{\rho\overline{\nu}}}\frac{1}{\\#\mathcal{I}^{2r}_{\rho\overline{\nu}}}\Bigl{(}\iint_{(rk+rQ_{\rho\overline{\nu}})\times(rk+rQ_{\rho\overline{\nu}})}\frac{|u_{\varepsilon}(x)-u_{\varepsilon}(y)|^{2}}{|x-y|^{d+2s}}\,dxdy\Bigr{)}^{\frac{1}{2}}$
$\displaystyle\leq$ $\displaystyle
P|r\rho|^{\frac{d}{2}+s}\\#\mathcal{I}^{2r}_{\rho\overline{\nu}}\Bigl{(}\sum_{k\in\mathcal{I}^{2r}_{\rho\overline{\nu}}}\frac{1}{\\#\mathcal{I}^{2r}_{\rho\overline{\nu}}}\iint_{(rk+rQ_{\rho\overline{\nu}})\times(rk+rQ_{\rho\overline{\nu}})}\frac{|u_{\varepsilon}(x)-u_{\varepsilon}(y)|^{2}}{|x-y|^{d+2s}}\,dxdy\Bigr{)}^{\frac{1}{2}}$
$\displaystyle\leq$ $\displaystyle
P|r\rho|^{\frac{d}{2}+s}(\\#\mathcal{I}^{2r}_{\rho\overline{\nu}})^{\frac{1}{2}}\Bigl{(}\iint_{\Omega\times\Omega}\frac{|u_{\varepsilon}(x)-u_{\varepsilon}(y)|^{2}}{|x-y|^{d+2s}}\,dxdy\Bigr{)}^{\frac{1}{2}}$
$\displaystyle\leq$ $\displaystyle
P|r\rho|^{\frac{d}{2}+s}|2r\rho|^{-\frac{d}{2}}|\Omega|^{\frac{1}{2}}[u_{\varepsilon}]_{W^{s,2}(\Omega)}$
$\displaystyle=$
$\displaystyle|r\rho|^{s}\,2^{-\frac{d}{2}}|\Omega|^{\frac{1}{2}}[u_{\varepsilon}]_{W^{s,2}(\Omega)}.$

After integration we obtain that (28) is bounded above by

$r^{s}\,2^{1-\frac{d}{2}}|\Omega|^{\frac{1}{2}}(1-s)[u_{\varepsilon}]_{W^{s,2}(\Omega)}\int_{0}^{1}\rho^{1-s}d\rho\,\,\,=\,\,\,r^{s}\,2^{1-\frac{d}{2}}|\Omega|^{\frac{1}{2}}(1-s)[u_{\varepsilon}]_{W^{s,2}(\Omega)}\frac{1}{2-s}.$

Then we use that $[u_{\varepsilon}]_{W^{s,2}(\Omega)}(1-s)\leq
F_{\varepsilon}(u_{\varepsilon})/\alpha$, which is uniformly bounded in
$\varepsilon$ by assumption, and that $r^{s}\to 0$ to deduce that the above
term vanishes.

We treat (29) with a similar argument. By the Cauchy-Schwarz inequality we
have

$\displaystyle\hskip-85.35826pt\int_{[0,1]\times
V}\sum_{k\in\mathcal{I}^{2r}_{\rho\overline{\nu}}}\int_{rk+rQ_{\rho\overline{\nu}}}|u^{\rho\overline{\nu}}_{\varepsilon}(x)-u_{\varepsilon}^{\rho\overline{\nu}}(rk)|\,dx\,d\mu_{\varepsilon}(\rho,\overline{\nu})$
$\displaystyle\leq$ $\displaystyle\int_{[0,1]\times
V}\sum_{k\in\mathcal{I}^{2r}_{\rho\overline{\nu}}}\int_{rk+rQ_{\rho\overline{\nu}}}|\nabla
u^{\rho\overline{\nu}}_{\varepsilon}(x)||x-rk|\,dx\,d\mu_{\varepsilon}(\rho,\overline{\nu})$
$\displaystyle\leq$ $\displaystyle\sqrt{d}\int_{[0,1]\times
V}|r\rho|\sum_{k\in\mathcal{I}^{2r}_{\rho\overline{\nu}}}\int_{rk+rQ_{\rho\overline{\nu}}}|\nabla
u^{\rho\overline{\nu}}_{\varepsilon}(x)|\,dx\,d\mu_{\varepsilon}(\rho,\overline{\nu})$

and by Hölder’s inequality and the concavity of $x\mapsto x^{\frac{1}{2}}$,
this is bounded above by

$\displaystyle\hskip-34.14322pt\hskip-17.07182pt\sqrt{d}\int_{[0,1]\times
V}|r\rho|^{\frac{d}{2}+1}\sum_{k\in\mathcal{I}^{2r}_{\rho\overline{\nu}}}\biggl{(}\int_{rk+rQ_{\rho\overline{\nu}}}|\nabla
u^{\rho\overline{\nu}}_{\varepsilon}(x)|^{2}\,dx\biggr{)}^{\frac{1}{2}}d\mu_{\varepsilon}(\rho,\overline{\nu})$
$\displaystyle=$ $\displaystyle\sqrt{d}\int_{[0,1]\times
V}|r\rho|^{\frac{d}{2}+1}\\#\mathcal{I}^{2r}_{\rho\overline{\nu}}\sum_{k\in\mathcal{I}^{2r}_{\rho\overline{\nu}}}\frac{1}{\\#\mathcal{I}^{2r}_{\rho\overline{\nu}}}\biggl{(}\int_{rk+rQ_{\rho\overline{\nu}}}|\nabla
u^{\rho\overline{\nu}}_{\varepsilon}(x)|^{2}\,dx\biggr{)}^{\frac{1}{2}}d\mu_{\varepsilon}(\rho,\overline{\nu})$
$\displaystyle\leq$ $\displaystyle\sqrt{d}\int_{[0,1]\times
V}|r\rho|^{\frac{d}{2}+1}(\\#\mathcal{I}^{2r}_{\rho\overline{\nu}})^{\frac{1}{2}}\biggl{(}\sum_{k\in\mathcal{I}^{2r}_{\rho\overline{\nu}}}\int_{rk+rQ_{\rho\overline{\nu}}}|\nabla
u^{\rho\overline{\nu}}_{\varepsilon}(x)|^{2}\,dx\biggr{)}^{\frac{1}{2}}d\mu_{\varepsilon}(\rho,\overline{\nu})$
$\displaystyle\leq$ $\displaystyle r\sqrt{d\,2^{-d}|\Omega|}\int_{[0,1]\times
V}\biggl{(}\sum_{k\in\mathcal{I}^{2r}_{\rho\overline{\nu}}}\int_{rk+rQ_{\rho\overline{\nu}}}|\nabla
u^{\rho\overline{\nu}}_{\varepsilon}(x)|^{2}\,dx\biggr{)}^{\frac{1}{2}}d\mu_{\varepsilon}(\rho,\overline{\nu}),$

which by Jensen’s inequality is less than or equal to

$r\sqrt{d\,2^{-d}|\Omega|}\Bigl{(}\int_{[0,1]\times
V}\sum_{k\in\mathcal{I}^{2r}_{\rho\overline{\nu}}}\int_{rk+rQ_{\rho\overline{\nu}}}|\nabla
u_{\varepsilon}^{\rho\overline{\nu}}(x)|^{2}dx\,d\mu_{\varepsilon}(\rho,\overline{\nu})\Bigr{)}^{\frac{1}{2}}.$
(30)

Now we write

$\displaystyle\hskip-28.45274pt\int_{[0,1]\times
V}\sum_{k\in\mathcal{I}^{2r}_{\rho\overline{\nu}}}\int_{rk+rQ_{\rho\overline{\nu}}}|\nabla
u_{\varepsilon}^{\rho\overline{\nu}}(x)|^{2}dx\,d\mu_{\varepsilon}(\rho,\overline{\nu})$
$\displaystyle=$
$\displaystyle\frac{2(1-s)}{\mathcal{H}^{k_{d}}(V)}\int_{0}^{1}\frac{1}{\rho^{-1+2s}}\int_{V}\sum_{k\in\mathcal{I}^{2r}_{{\rho\overline{\nu}}}}\int_{rk+rQ_{\rho\overline{\nu}}}|\nabla
u_{\varepsilon}^{\rho\overline{\nu}}|^{2}\,dxd\mathcal{H}^{k_{d}}(\overline{\nu})d\rho$
$\displaystyle\leq$
$\displaystyle\hskip-8.5359pt\Bigl{(}\frac{2d}{\alpha\sigma_{d-1}}\Bigr{)}\frac{1-s}{d}\frac{\sigma_{d-1}}{\mathcal{H}^{k_{d}}(V)}\int_{0}^{1}\frac{r^{2(1-s)}}{\rho^{-1+2s}}\int_{V}\sum_{k\in\mathcal{I}^{2r}_{{\rho\overline{\nu}}}}a\left(\frac{rk}{\varepsilon}\right)\int_{rk+rQ_{\rho\overline{\nu}}}|\nabla
u_{\varepsilon}^{\rho\overline{\nu}}|^{2}\,dxd\mathcal{H}^{k_{d}}(\overline{\nu})d\rho$

and we note that this last term equals (25) (up to a constant factor),
therefore, as a byproduct of the previous computations, it is bounded above by

$\frac{2d}{\alpha\sigma_{d-1}}F_{\varepsilon}(u_{\varepsilon})+o_{\varepsilon}(1),$

which is uniformly bounded for small $\varepsilon$. We conclude that (30)
vanishes since $r\to 0$.

Finally, we remove the boundedness condition by a truncation argument. Let
$u^{M}:=(u\land M)\lor-M$, and note that $u_{\varepsilon}\to u$ in
$L^{2}(\Omega)$ implies $u_{\varepsilon}^{M}\to u^{M}$ in $L^{2}(\Omega)$.
Since $F_{\varepsilon}(u_{\varepsilon})\geq
F_{\varepsilon}(u_{\varepsilon}^{M})$, we have

$\liminf_{\varepsilon\to 0}F_{\varepsilon}(u_{\varepsilon})\ \geq\
\liminf_{\varepsilon\to 0}F_{\varepsilon}(u^{M}_{\varepsilon})\ \geq\
\frac{\sigma_{d-1}}{2d}\int_{\Omega}\langle A_{\rm hom}\nabla u^{M},\nabla
u^{M}\rangle\,dx,$

and then

$\sup_{M}\|\nabla u^{M}\|_{L^{2}(\Omega)}\ \leq\
\frac{2d}{\alpha\sigma_{d-1}}\sup_{\varepsilon}F_{\varepsilon}(u_{\varepsilon})\
<\ \ +\infty,$

so that $u\in W^{1,2}(\Omega)$ with $\nabla u^{M}\to\nabla u$. The proof is
concluded once we let $M\to+\infty$.

###  3.2. Limsup inequality

To prove the $\Gamma$-$\limsup$ inequality, we first study the case where $u$
is piecewise-affine. We then recover the inequality on the whole space
$W^{1,2}(\Omega)$ by means of a density argument. For the sake of clarity, we
first deal with the case where $u$ is affine.

Step $1$. We first suppose that $u(x)=\langle z,x\rangle$, for some
$z\in\mathbb{R}^{d}$. We recall that by the classical homogenization formula
for $A_{\rm hom}$ (see [4, Theorem 14.7]) and by density, it holds

$\langle A_{\rm
hom}z,z\rangle=\inf\Bigl{\\{}\int_{(0,1)^{d}}a(y)\lvert\nabla{\varphi}(y)+z\rvert^{2}dy\,:\,\varphi\in
C^{\infty}(\mathbb{R}^{d}),\,\,1\text{-periodic}\Bigr{\\}}.$ (31)

Fix $\delta>0$ and let $\varphi\in C^{\infty}(\mathbb{R}^{d})$ be a
$1$-periodic function such that

$\int_{(0,1)^{d}}a(y)\lvert\nabla\varphi(y)+z\rvert^{2}dy<\langle A_{\rm
hom}z,z\rangle+\delta;$ (32)

we claim that $u_{\varepsilon}(x)=\langle
z,x\rangle+\varepsilon\varphi\left(\frac{x}{\varepsilon}\right)$ is an
approximate recovery sequence for $u$, i.e., it is such that

$\limsup_{\varepsilon\to
0}F_{\varepsilon}(u_{\varepsilon})\leq\frac{\sigma_{d-1}}{2d}\int_{\Omega}\langle
A_{\rm hom}z,z\rangle dx+\delta.$ (33)

This, combined with a diagonal argument and the arbitrariness of $\delta>0$,
will lead to the limsup inequality.

In order to prove (33), we take advantage of Lemma 3 with $r_{\varepsilon}=1$,
and consider the first-order Taylor expansion of $\varphi$ to get

$\displaystyle\hskip-28.45274ptF_{\varepsilon}(u_{\varepsilon})+o_{\varepsilon}(1)$
$\displaystyle=$
$\displaystyle(1-s)\iint_{\Omega\times\Omega}a\left(\frac{x}{\varepsilon}\right)\frac{\lvert\langle
z,x-y\rangle+\varepsilon\varphi\left(\frac{x}{\varepsilon}\right)-\varepsilon\varphi\left(\frac{y}{\varepsilon}\right)\rvert^{2}}{\lvert
x-y\rvert^{d+2s}}dxdy$ $\displaystyle=$
$\displaystyle(1-s)\iint_{\Omega\times\Omega\cap\\{\lvert x-y\rvert\leq
1\\}}a\left(\frac{x}{\varepsilon}\right)\frac{|\langle
z+\nabla{\varphi}\left(\frac{x}{\varepsilon}\right),x-y\rangle+\frac{1}{\varepsilon}R(|x-y|)\rvert^{2}}{\lvert
x-y\rvert^{d+2s}}dxdy,$

where the function $R(|x-y|)$ is the remainder in Lagrange form of $\varphi$
and it is such that

$R(|x-y|)\leq d^{2}||\nabla^{2}\varphi||_{\rm\infty}|x-y|^{2}.$ (34)

By the inequality $|a+b|^{2}\leq(1+\eta)a^{2}+(1+1/\eta)b^{2}$, with $\eta>0$,
and by the change of variables $\xi:=y-x$, we deduce

$\displaystyle F_{\varepsilon}(u_{\varepsilon})+o_{\varepsilon}(1)$
$\displaystyle\leq$
$\displaystyle(1+\eta)(1-s)\int_{\Omega}\int_{B_{1}(0)}a\left(\frac{x}{\varepsilon}\right)\frac{|\langle
z+\nabla\varphi\left(\frac{x}{\varepsilon}\right),\xi\rangle|^{2}}{|\xi|^{d+2s}}d\xi
dx$ (36)
$\displaystyle+\left(1+\frac{1}{\eta}\right)\frac{1-s}{\varepsilon^{2}}\int_{\Omega}\int_{B_{1}(0)}a\left(\frac{x}{\varepsilon}\right)\frac{R(|\xi|)^{2}}{|\xi|^{d+2s}}d\xi
dx.$

For a fixed $\eta>0$, the term in (36) vanishes as $\varepsilon$ tends to
zero. Indeed, by assumption, the term $\frac{1-s}{\varepsilon^{2}}$ vanishes
as $\varepsilon\to 0$, and using (6) we have that

$\displaystyle\int_{\Omega}\int_{B_{1}(0)}a\left(\frac{x}{\varepsilon}\right)\frac{R(|\xi|)^{2}}{|\xi|^{d+2s}}d\xi
dx$ $\displaystyle\leq$ $\displaystyle
d^{4}\beta|\Omega|||\nabla^{2}\varphi||_{\rm\infty}\int_{B_{1}(0)}\frac{1}{|\xi|^{d+2s-4}}d\xi$
(37) $\displaystyle=$ $\displaystyle
d^{4}\sigma_{d-1}\beta|\Omega|||\nabla^{2}\varphi||_{\rm\infty}\int_{0}^{1}\rho^{3-2s}d\rho<+\infty.$

As for the term in (36), note that by the symmetry of the integrand, it holds

$\displaystyle\int_{B_{1}(0)}\frac{|\langle
z+\nabla{\varphi}\left(\frac{x}{\varepsilon}\right),\xi\rangle|^{2}}{\lvert\xi\rvert^{d+2s}}d\xi$
$\displaystyle=$ $\displaystyle\int_{0}^{1}\int_{\partial
B_{\rho}(0)}\frac{\left|\langle
z+\nabla\varphi\left(\frac{x}{\varepsilon}\right),\nu\rangle\right|^{2}}{\rho^{d+2s}}d\mathcal{H}^{d-1}(\nu)d\rho$
(38) $\displaystyle=$
$\displaystyle\int_{0}^{1}\frac{1}{\rho^{d+2(s-1)}}\frac{1}{d}\int_{\partial
B_{\rho}(0)}\left|z+\nabla\varphi\left(\frac{x}{\varepsilon}\right)\right|^{2}d\mathcal{H}^{d-1}(\nu)d\rho$
$\displaystyle=$
$\displaystyle\frac{\sigma_{d-1}}{d}\left|z+\nabla\varphi\left(\frac{x}{\varepsilon}\right)\right|^{2}\int_{0}^{1}\rho^{1-2s}d\rho$
$\displaystyle=$
$\displaystyle\frac{\sigma_{d-1}}{2d}\left|z+\nabla\varphi\left(\frac{x}{\varepsilon}\right)\right|^{2}\frac{1}{(1-s)}.$

Recalling that $\varphi$ and $a$ are $1$-periodic, we may substitute (38) in
(36) to get

$\displaystyle\hskip-85.35826pt(1+\eta)(1-s)\int_{\Omega}\int_{B_{1}(0)}a\left(\frac{x}{\varepsilon}\right)\frac{|\langle
z+\nabla\varphi\left(\frac{x}{\varepsilon}\right),\xi\rangle|^{2}}{|\xi|^{d+2s}}d\xi
dx$ (39) $\displaystyle=$
$\displaystyle(1+\eta)\frac{\sigma_{d-1}}{2d}\int_{\Omega}a\left(\frac{x}{\varepsilon}\right)\left\lvert\nabla{\varphi}\left(\frac{x}{\varepsilon}\right)+z\right\rvert^{2}dx$
$\displaystyle\leq$
$\displaystyle(1+\eta)\frac{\sigma_{d-1}}{2d}\frac{|\Omega|}{\varepsilon^{d}}\int_{(0,\varepsilon)^{d}}a\left(\frac{x}{\varepsilon}\right)\left|\nabla\varphi\left(\frac{x}{\varepsilon}\right)+z\right|^{2}dx$
$\displaystyle=$
$\displaystyle(1+\eta)\frac{\sigma_{d-1}}{2d}\lvert\Omega\rvert\int_{(0,1)^{d}}a(y)\left\lvert\nabla{\varphi}\left(y\right)+z\right\rvert^{2}dy,$

which, in view of (32), implies

$\limsup_{\varepsilon\to
0}F_{\varepsilon}(u_{\varepsilon})\leq(1+\eta)\left(\frac{\sigma_{d-1}}{2d}\int_{\Omega}\langle
A_{\rm hom}z,z\rangle dx+\delta\right),$

and we conclude by letting $\eta$ tend to $0$.

Step $2$. We now suppose that the function $u$ be piecewise-affine in
$\Omega$. More precisely, we suppose that there exists a finite family of
$d$-simplices $(\Delta)_{i\in I}$ covering $\Omega$, vectors $(z_{i})_{i\in
I}\subseteq\mathbb{R}^{d}$, and constants $(m_{i})_{i\in
I}\subseteq\mathbb{R}$ such that $u(x)=\langle x,z_{i}\rangle+m_{i}$ for every
$x\in\Delta_{i}\cap\Omega$ and for every $i\in I$. To simplify the notation,
from now on we will always write $\Delta_{i}$ in place of
$\Delta_{i}\cap\Omega$.

Similarly to what we have done in the previous case, for a fixed $\delta>0$
and for every $i\in I$, we consider $\varphi_{i}\in
C^{\infty}_{c}(\mathbb{R}^{d})$ a $1$-periodic function satisfying

$\int_{(0,1)^{d}}a(y)\lvert\nabla\varphi_{i}(y)+z_{i}\rvert^{2}dy<\langle
A_{\rm hom}z_{i},z_{i}\rangle+\frac{\delta}{\\#I}.$ (40)

Having in mind the construction of the recovery sequence performed in the
previous case, we aim at constructing an approximate recovery sequence for $u$
by perturbing, in a suitable way, $u$ with the functions $\varphi_{i}$. To
this end, for every $\Delta_{i}$, we consider the subset defined by

$\Delta_{i}^{\varepsilon}:=\left\\{x\in\Omega:\,\,\text{dist}(x,\Delta_{i}^{c})>\varepsilon\right\\}.$

For fixed $\varepsilon>0$ and $i\in I$, we consider a positive function
$\psi^{i}_{\varepsilon}\in C^{\infty}_{c}(\mathbb{R}^{d})$, satisfying the
following conditions

$\displaystyle\ \psi^{i}_{\varepsilon}(x)=0\qquad\qquad\text{for
}x\notin\Delta_{i}^{\varepsilon}$ $\displaystyle\
\psi^{i}_{\varepsilon}(x)=1\qquad\qquad\text{for
}x\in\Delta_{i}^{2\varepsilon}$ $\displaystyle\
0\leq\psi_{\varepsilon}^{i}(x)\leq 1\qquad\,\text{for }x\in\mathbb{R}^{d}$
(41a) $\displaystyle\
\|\nabla\psi^{i}_{\varepsilon}\|_{\infty}\leq\frac{2}{\varepsilon}$ (41b)
$\displaystyle\
\|\nabla^{2}\,\psi^{\textit{i}}_{\varepsilon}\|_{\infty}\leq\frac{4}{\varepsilon^{2}}$
$\displaystyle\,,$ (41c)

where $\nabla^{2}\psi_{\varepsilon}^{i}$ is the Hessian matrix of
$\psi_{\varepsilon}^{i}$.

We claim that an approximate recovery sequence for $u$ is given by

$u_{\varepsilon}(x)=\sum_{i\in I}\left(\langle
z_{i},x\rangle+\varepsilon\varphi_{i}\left(\frac{x}{\varepsilon}\right)\psi_{\varepsilon}^{i}(x)+m_{i}\right)\chi_{\Delta_{i}}(x),$

where $\chi_{\Delta_{i}}$ denotes the characteristic function of $\Delta_{i}$.
Note that $u_{\varepsilon}$ is a continuous function, since it coincides with
$u$ on the boundary of each $\Delta_{i}$.

Let $A:={\Omega\times\Omega\cap\\{\lvert x-y\rvert\leq\varepsilon\\}}$. Thanks
to Lemma 3 and Remark 4, it is enough to show that

$\limsup_{\varepsilon\to
0}(1-s)\iint_{A}a\left(\frac{x}{\varepsilon}\right)\frac{\lvert
u_{\varepsilon}(x)-u_{\varepsilon}(y)\rvert^{2}}{\lvert
x-y\rvert^{d+2s}}dxdy\leq\frac{\sigma_{d-1}}{2d}\int_{\Omega}\langle A_{\rm
hom}\nabla{u},\nabla{u}\rangle dx+\delta.$

To make the computations more manageable, we consider a suitable partition of
$A$. For $i\in I$, we set

$B_{i}=\Delta_{i}^{2\varepsilon}\times\Delta_{i}^{2\varepsilon}\cap\\{\lvert
x-y\rvert\leq\varepsilon\\},$ $C_{i}=\Delta_{i}\times\Delta_{i}\cap\\{\lvert
x-y\rvert\leq\varepsilon\text{ and }x\text{ or
}y\in\Delta_{i}\setminus{\Delta_{i}^{2\varepsilon}}\\},$

and for $j\in I,\,j\neq i$, we set

$D_{ij}=\Delta_{i}\times\Delta_{j}\cap\\{\lvert x-y\rvert\leq\varepsilon\\},$

so that $A=\bigcup_{i\in I}(B_{i}\cup C_{i})\cup\bigcup_{i\neq j}D_{ij}$.

Taking advantage of this decomposition and of the subadditivity of the
$\limsup$, we have

$\displaystyle\hskip-56.9055pt\limsup_{\varepsilon\to
0}\,(1-s)\iint_{A}a\left(\frac{x}{\varepsilon}\right)\frac{\lvert
u_{\varepsilon}(x)-u_{\varepsilon}(y)\rvert^{2}}{\lvert x-y\rvert^{d+2s}}dxdy$
$\displaystyle\leq\limsup_{\varepsilon\to 0}\sum_{i\in
I}(1-s)\iint_{B_{i}}a\left(\frac{x}{\varepsilon}\right)\frac{\lvert
u_{\varepsilon}(x)-u_{\varepsilon}(y)\rvert^{2}}{\lvert x-y\rvert^{d+2s}}dxdy$
(42) $\displaystyle+\sum_{i\in I}\limsup_{\varepsilon\to
0}\,(1-s)\iint_{C_{i}}a\left(\frac{x}{\varepsilon}\right)\frac{\lvert
u_{\varepsilon}(x)-u_{\varepsilon}(y)\rvert^{2}}{\lvert x-y\rvert^{d+2s}}dxdy$
(43) $\displaystyle+\sum_{i\in I}\sum_{j\in I,\,j\neq
i}\limsup_{\varepsilon\to
0}\,(1-s)\iint_{D_{ij}}a\left(\frac{x}{\varepsilon}\right)\frac{\lvert
u_{\varepsilon}(x)-u_{\varepsilon}(y)\rvert^{2}}{\lvert
x-y\rvert^{d+2s}}dxdy.$ (44)

We begin by estimating (42). Since for fixed $i\in I$ we have
$\psi_{\varepsilon}^{i}=1$ on $B_{i}$, by the first-order Taylor expansion of
$\varphi_{i}$, for a fixed $\eta>0$ we get

$\displaystyle\hskip-71.13188pt\sum_{i\in
I}(1-s)\iint_{B_{i}}a\left(\frac{x}{\varepsilon}\right)\frac{|\langle
z_{i}+\nabla{\varphi}_{i}\left(\frac{x}{\varepsilon}\right),x-y\rangle+\frac{1}{\varepsilon}R_{i}(|x-y|)\rvert^{2}}{\lvert
x-y\rvert^{d+2s}}dxdy$ (45) $\displaystyle\leq$ $\displaystyle\sum_{i\in
I}(1+\eta)(1-s)\int_{\Delta_{i}^{2\varepsilon}}\int_{B_{\varepsilon}(0)}a\left(\frac{x}{\varepsilon}\right)\frac{|\langle
z_{i}+\nabla\varphi_{i}\left(\frac{x}{\varepsilon}\right),\xi\rangle|^{2}}{|\xi|^{d+2s}}d\xi
dx$ $\displaystyle+$ $\displaystyle\sum_{i\in
I}\left(1+\frac{1}{\eta}\right)\frac{1-s}{\varepsilon^{2}}\int_{\Delta_{i}^{2\varepsilon}}\int_{B_{\varepsilon}(0)}a\left(\frac{x}{\varepsilon}\right)\frac{R_{i}(|\xi|)^{2}}{|\xi|^{d+2s}}d\xi
dx,$ (46)

where $R_{i}$ is the remainder in Lagrange form of the first-order Taylor
expansion of $\varphi_{i}$, which satisfies

$R_{i}(|x-y|)\leq d^{2}\max_{i\in
I}||\nabla^{2}\,\varphi_{i}||_{\infty}|x-y|^{2}=:M|x-y|^{2}.$ (47)

Thus, we may estimate (46) with

$\beta M^{2}\sum_{i\in
I}\left(1+\frac{1}{\eta}\right)\frac{1-s}{\varepsilon^{2}}\int_{\Delta_{i}}\int_{B_{1}(0)}\frac{|\xi|^{4}}{|\xi|^{d+2s}}d\xi
dx,$

which vanishes for $\varepsilon\to 0$ in view of the fact that
$\frac{1-s}{\varepsilon^{2}}$ goes to zero for $\varepsilon\to 0$ and of (37)
(with $\Omega$ replaced by $\Delta_{i}$).

As for (45), considering larger domains and taking advantage of the
computations leading to (39), we see that

$\displaystyle\displaystyle\sum_{i\in
I}(1+\eta)(1-s)\int_{\Delta_{i}^{2\varepsilon}}\int_{B_{\varepsilon}(0)}a\left(\frac{x}{\varepsilon}\right)\frac{|\langle
z_{i}+\nabla\varphi_{i}\left(\frac{x}{\varepsilon}\right),\xi\rangle|^{2}}{|\xi|^{d+2s}}d\xi
dx$ $\displaystyle\leq$ $\displaystyle\sum_{i\in
I}(1+\eta)(1-s)\int_{\Delta_{i}}\int_{B_{1}(0)}a\left(\frac{x}{\varepsilon}\right)\frac{|\langle
z_{i}+\nabla\varphi_{i}\left(\frac{x}{\varepsilon}\right),\xi\rangle|^{2}}{|\xi|^{d+2s}}d\xi
dx;$ $\displaystyle\leq$
$\displaystyle(1+\eta)\frac{\sigma_{d-1}}{2d}\sum_{i\in
I}\lvert\Delta_{i}\rvert\int_{(0,1)^{d}}a(y)\left\lvert\nabla{\varphi_{i}}\left(y\right)+z_{i}\right\rvert^{2}dy.$

Recalling (40), we finally conclude

$\displaystyle\limsup_{\varepsilon\to
0}\sum_{i=1}^{n}(1+\eta)(1-s)\iint_{B_{i}}a\left(\frac{x}{\varepsilon}\right)\frac{\lvert
u_{\varepsilon}(x)-u_{\varepsilon}(y)\rvert^{2}}{\lvert x-y\rvert^{d+2s}}$
$\displaystyle\leq$ $\displaystyle\limsup_{\varepsilon\to
0}\,(1+\eta)\frac{\sigma_{d-1}}{2d}\sum_{i\in
I}\Bigl{(}\int_{\Delta_{i}}\langle A_{\rm
hom}z_{i},z_{i}\rangle+\frac{\delta}{\\#I}\Bigr{)}$ $\displaystyle\leq$
$\displaystyle(1+\eta)\frac{\sigma_{d-1}}{2d}\int_{\Omega}\langle A_{\rm
hom}\nabla{u},\nabla{u}\rangle dx+\delta,$

so that letting $\eta\to 0$, (42) is estimated.

We now show that (43) vanishes for $\varepsilon\to 0$. As $\\#I$ is finite and
independent on $\varepsilon$, it suffices to take into account a single term
in the sum. By the convexity of $x\mapsto x^{2}$, we have

$\displaystyle(1-s)\iint_{C_{i}}a\left(\frac{x}{\varepsilon}\right)\frac{\lvert
u_{\varepsilon}(x)-u_{\varepsilon}(y)\rvert^{2}}{\lvert x-y\rvert^{d+2s}}dxdy$
(50) $\displaystyle\leq$ $\displaystyle 3\beta(1-s)\iint_{C_{i}}\frac{\lvert
z_{i}\rvert^{2}\lvert x-y\rvert^{2}}{\lvert x-y\rvert^{d+2s}}dxdy$
$\displaystyle+3\beta(1-s)\iint_{C_{i}}\frac{\left\lvert\varepsilon\psi^{i}_{\varepsilon}(y)\left(\varphi_{i}\left(\frac{x}{\varepsilon}\right)-\varphi_{i}\left(\frac{y}{\varepsilon}\right)\right)\right\rvert^{2}}{\lvert
x-y\rvert^{d+2s}}dxdy$
$\displaystyle+3\beta(1-s)\iint_{C_{i}}\frac{\left\lvert\varepsilon\varphi_{i}\left(\frac{x}{\varepsilon}\right)(\psi^{i}_{\varepsilon}(x)-\psi^{i}_{\varepsilon}(y))\right\rvert^{2}}{\lvert
x-y\rvert^{d+2s}}dxdy.$

We start with (50). Via a change of variables, we see that

$\displaystyle(1-s)\iint_{C_{i}}\frac{\lvert z_{i}\rvert^{2}\lvert
x-y\rvert^{2}}{\lvert x-y\rvert^{d+2s}}dxdy$ $\displaystyle\leq$
$\displaystyle(1-s)\lvert
z_{i}\rvert^{2}\int_{\Delta_{i}\setminus{\Delta_{i}^{2\varepsilon}}}\int_{B_{1}(0)}\frac{\lvert\xi\rvert^{2}}{\lvert\xi\rvert^{d+2s}}d\xi
dx$ $\displaystyle\leq$ $\displaystyle(1-s)\rvert\lvert
z_{i}\rvert^{2}\lvert\Delta_{i}\setminus{\Delta_{i}^{2\varepsilon}}\rvert\sigma_{d-1}\int_{B_{1}(0)}\frac{\lvert\xi\rvert^{2}}{\lvert\xi\rvert^{d+2s}}d\xi$
$\displaystyle=$ $\displaystyle\frac{\sigma_{d-1}}{2}\lvert
z_{i}\rvert^{2}\lvert\Delta_{i}\setminus{\Delta_{i}^{2\varepsilon}}\rvert,$

which converges to zero as $\varepsilon\to 0$.

As for (50), using the first-order Taylor expansions of $\varphi_{i}$, (41a),
(41b), and recalling (47), one gets

$\displaystyle(1-s)\iint_{C_{i}}\frac{\left\lvert\varepsilon\psi^{i}_{\varepsilon}(y)\left(\varphi_{i}\left(\frac{x}{\varepsilon}\right)-\varphi_{i}\left(\frac{y}{\varepsilon}\right)\right)\right\rvert^{2}}{\lvert
x-y\rvert^{d+2s}}dxdy$ $\displaystyle\leq$
$\displaystyle(1-s)\int_{\Delta_{i}\setminus{\Delta_{i}^{2\varepsilon}}}\int_{B_{\varepsilon}(0)}\frac{\left\lvert\langle\nabla{\varphi_{i}}\left(\frac{x}{\varepsilon}\right),\xi\rangle+\frac{1}{\varepsilon}R_{i}(|\xi|)\right\rvert^{2}}{\lvert\xi\rvert^{d+2s}}d\xi
dx$ $\displaystyle\leq$ $\displaystyle
2(1-s)\Bigl{(}\int_{\Delta_{i}\setminus{\Delta_{i}^{2\varepsilon}}}\int_{B_{\varepsilon}(0)}\frac{\left\lvert\langle\nabla{\varphi_{i}}\left(\frac{x}{\varepsilon}\right),\xi\rangle\right\rvert^{2}}{\lvert\xi\rvert^{d+2s}}d\xi
dx+\frac{M^{2}}{\varepsilon^{2}}\int_{\Delta_{i}\setminus{\Delta_{i}^{2\varepsilon}}}\int_{B_{\varepsilon}(0)}\frac{\lvert\xi\rvert^{4}}{\lvert\xi\rvert^{d+2s}}d\xi
dx\Bigr{)}.$

The first term can be estimated with

$\displaystyle\hskip-56.9055pt(1-s)\int_{\Delta_{i}\setminus{\Delta_{i}^{2\varepsilon}}}\int_{B_{\varepsilon}(0)}\frac{\lvert\langle\nabla{\varphi_{i}}\left(\frac{x}{\varepsilon}\right),\xi\rangle\rvert^{2}}{\lvert\xi\rvert^{d+2s}}d\xi
dx$ $\displaystyle\leq$
$\displaystyle(1-s)\lvert\lvert\nabla{\varphi_{i}}\rvert\rvert_{\infty}^{2}\int_{\Delta_{i}\setminus{\Delta_{i}^{2\varepsilon}}}\int_{B_{\varepsilon}(0)}\frac{\lvert\xi\rvert^{2}}{\lvert\xi\rvert^{d+2s}}d\xi
dx$ $\displaystyle=$
$\displaystyle(1-s)\lvert\lvert\nabla{\varphi_{i}}\rvert\rvert_{\infty}^{2}\lvert\Delta_{i}\setminus\Delta_{i}^{2\varepsilon}\rvert\int_{B_{1}(0)}\frac{|\xi|^{2}}{|\xi|^{d+2s}}$
$\displaystyle\leq$
$\displaystyle\frac{\sigma_{d-1}}{2}\lvert\lvert\nabla{\varphi_{i}}\rvert\rvert_{\infty}^{2}|\Delta_{i}\setminus\Delta_{i}^{2\varepsilon}|\,,$

while for the second term we have

$\displaystyle\frac{1-s}{\varepsilon^{2}}\int_{\Delta_{i}\setminus{\Delta_{i}^{2\varepsilon}}}\int_{B_{\varepsilon}(0)}\frac{\lvert\xi\rvert^{4}}{\lvert\xi\rvert^{d+2s}}d\xi
dx\leq\frac{(1-s)}{\varepsilon^{2}}|\Delta_{i}\setminus\Delta_{i}^{2\varepsilon}|\frac{\sigma_{d-1}}{4-2s},$
(51)

and therefore both terms vanish as $\varepsilon$ tends to $0$.

In order to estimate (50), we denote by $\tilde{R}_{i}$ the Lagrange
remainders of the first-order Taylor expansion of $\psi_{\varepsilon}^{i}$ and
we note that by (41c) it follows that

$\tilde{R}_{i}(|x-y|)\leq\frac{4d^{2}}{\varepsilon^{2}}|x-y|^{2}.$ (52)

Using the first-order Taylor expansion of $\psi^{i}_{\varepsilon}$, we
conclude that

$\displaystyle(1-s)\iint_{C_{i}}\frac{\left\lvert\varepsilon\varphi_{i}\left(\frac{x}{\varepsilon}\right)(\psi^{i}_{\varepsilon}(x)-\psi^{i}_{\varepsilon}(y))\right\rvert^{2}}{\lvert
x-y\rvert^{d+2s}}dxdy$
$\displaystyle\leq(1-s)\lvert\lvert\varphi_{i}\rvert\rvert_{\infty}^{2}\int_{\Delta_{i}\setminus{\Delta_{i}^{2\varepsilon}}}\int_{B_{\varepsilon}(0)}\frac{\lvert\varepsilon\langle\nabla{\psi^{i}_{\varepsilon}(x)},\xi\rangle+\varepsilon\tilde{R}(|x-y|)\rvert^{2}}{\lvert\xi\rvert^{d+2s}}d\xi
dx$ $\displaystyle\leq
2(1-s)\lvert\lvert\varphi_{i}\rvert\rvert_{\infty}^{2}\Bigl{(}\int_{\Delta_{i}\setminus{\Delta_{i}^{2\varepsilon}}}\int_{B_{\varepsilon}(0)}\frac{\lvert\varepsilon\langle\nabla{\psi^{i}_{\varepsilon}(x)},\xi\rangle\rvert^{2}}{\lvert\xi\rvert^{d+2s}}d\xi
dx$ (53)
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\qquad+\int_{\Delta_{i}\setminus{\Delta_{i}^{2\varepsilon}}}\int_{B_{\varepsilon}(0)}\frac{16d^{4}}{\varepsilon^{2}}\frac{\lvert\xi\rvert^{4}}{\lvert\xi\rvert^{d+2s}}d\xi
dx\biggr{)}.$ (54)

In view of (51), we infer that the term in (54) converges to zero.

As for (53), we may use (41b) to obtain that

$\displaystyle(1-s)\int_{\Delta_{i}\setminus{\Delta_{i}^{2\varepsilon}}}\int_{B_{\varepsilon}(0)}\frac{\lvert\varepsilon\langle\nabla{\psi^{i}_{\varepsilon}(x)},\xi\rangle\rvert^{2}}{\lvert\xi\rvert^{d+2s}}d\xi
dx$ $\displaystyle\leq
2(1-s)|\Delta_{i}\setminus{\Delta_{i}^{2\varepsilon}}|\int_{B_{1}(0)}\frac{|\xi|^{2}}{|\xi|^{d+2s}}d\xi=\sigma_{d-1}|\Delta_{i}\setminus{\Delta_{i}^{2\varepsilon}}|,$

so that we have finally shown that (43) tends to $0$ for $\varepsilon\to 0$.

We conclude by proving that (44) vanishes for $\varepsilon\to 0$. Also in this
case, since $\\#I$ is finite and independent of $\varepsilon$, it suffices to
take into account just one summand.

Note that if $(x,y)\in D_{ij}$, then
$x\in\Delta_{i}\setminus{\Delta_{i}^{\varepsilon}}$ and
$y\in\Delta_{j}\setminus\Delta_{j}^{\varepsilon}$, so that
$\psi^{i}_{\varepsilon}(x)=\psi^{j}_{\varepsilon}(y)=0.$

Thus, taking advantage of the Lipschitz continuity of $u$, we get

$\displaystyle(1-s)\iint_{D_{ij}}a\left(\frac{x}{\varepsilon}\right)\frac{\lvert
u_{\varepsilon}(x)-u_{\varepsilon}(y)\rvert^{2}}{\lvert x-y\rvert^{d+2s}}$
$\displaystyle\leq$
$\displaystyle\beta(1-s)\int_{\Delta_{i}\setminus{\Delta_{i}^{\varepsilon}}}\int_{B_{\varepsilon}(0)}\frac{\lvert
u(x+\xi)-u(x)\rvert^{2}}{\lvert\xi\rvert^{d+2s}}d\xi dx$ $\displaystyle\leq$
$\displaystyle\beta\lvert\lvert\nabla{u}\rvert\rvert_{\infty}(1-s)\int_{\Delta_{i}\setminus{\Delta_{i}^{\varepsilon}}}\int_{B_{1}(0)}\frac{\lvert\xi\rvert^{2}}{\lvert\xi\rvert^{d+2s}}d\xi
dx$ $\displaystyle=$
$\displaystyle\beta\lvert\lvert\nabla{u}\rvert\rvert_{\infty}(1-s)|\Delta_{i}\setminus{\Delta_{i}^{\varepsilon}}|\sigma_{d-1}\frac{1}{2-2s}$
$\displaystyle\leq$
$\displaystyle\beta\lvert\lvert\nabla{u}\rvert\rvert_{\infty}\frac{\sigma_{d-1}}{2}|\Delta_{i}\setminus{\Delta_{i}^{\varepsilon}}|$

and this last term vanishes for $\varepsilon\to 0$ deducing that (44) tends to
$0$.

Step $3$. The functional $F_{\rm\hom}$ defined in (9), which has been proven
to be the desired $\Gamma$-limit on piecewise-affine functions, is continuous
in the strong topology of $W^{1,2}(\Omega)$. Since $\Omega$ is an open bounded
set with Lipschitz boundary, every function $u\in W^{1,2}(\Omega)$ may be
extended to $\overline{u}\in W^{1,2}_{0}(B_{R})$ via the standard extension
operator and a cut-off function. It is a known fact that a dense class in
$W^{1,2}_{0}(B_{R})$ is the class of piecewise-affine functions, namely,
functions which are piecewise-affine on a finite family of simplices covering
$\overline{B}_{R}$ (see [7, Proposition 2.1]). Therefore, the class of
piecewise-affine functions discussed in Step $2$ is dense in
$W^{1,2}_{0}(B_{R})$ and a fortiori in $W^{1,2}(\Omega)$. Having proved that
the $\Gamma$-$\limsup$ inequality holds on a dense subset of
$W^{1,2}(\Omega)$, we conclude that it holds for any $u\in W^{1,2}(\Omega)$
(see for instance [3, Remark 2.8]).

Acknowledgements. This paper is based on work supported by the National
Research Project (PRIN 2017BTM7SN) “Variational Methods for Stationary and
Evolution Problems with Singularities and Interfaces”, funded by the Italian
Ministry of University and Research. Andrea Braides and Davide Donati are
members of GNAMPA, INdAM.

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