# Emergence of Wulff-Crystals from atomistic systems on the FCC and HCP
lattices

Marco Cicalese  Zentrum Mathematik - M7, Technische Universität München,
Garching, Germany<EMAIL_ADDRESS>,  Leonard Kreutz  Applied Mathematics
Münster, WWU Münster, 48149 Münster, Germany<EMAIL_ADDRESS>and
Gian Paolo Leonardi  Department of Mathematics, University of Trento, 38050
Povo (TN), Italy<EMAIL_ADDRESS>
###### Abstract.

We consider a system of $N$ hard spheres sitting on the nodes of either the
$\mathrm{FCC}$ or $\mathrm{HCP}$ lattice and interacting via a sticky-disk
potential. As $N$ tends to infinity (continuum limit), assuming the
interaction energy does not exceed that of the ground-state by more than
$N^{2/3}$ (surface scaling), we obtain the variational coarse grained model by
$\Gamma$-convergence. More precisely, we prove that the continuum limit
energies are of perimeter type and we compute explicitly their Wulff shapes.
Our analysis shows that crystallization on $\mathrm{FCC}$ is preferred to that
on $\mathrm{HCP}$ for $N$ large enough. The method is based on integral
representation and concentration-compactness results that we prove for general
periodic lattices in any dimension.

###### Key words and phrases:

Wulff shape, isoperimetric inequality, crystallization, $\Gamma$-convergence,
lattice systems, concentration-compactness

##  1\. Introduction

A fundamental problem in crystallography is to understand why ensembles of
large number of atoms arrange themselves into crystals at low temperatures.
From the mathematical point of view, proving that equilibrium configurations
of certain phenomenological interaction energies exhibit these structures is
referred to as the crystallization problem [8].

At zero temperature the internal energy of a configuration of atoms is
expected to be solely governed by its geometric arrangement. Within the
framework of molecular mechanics [1, 24, 31], one identifies each ensemble of
atoms with its atomic positions
$X=\\{x_{1},\ldots,x_{N}\\}\subset\mathbb{R}^{3}$ and associates to it a
configurational energy of the form

$\displaystyle\mathcal{E}(X):=\frac{1}{2}\sum_{i\neq j}V(|x_{i}-x_{j}|)\,,$

where $V\colon\mathbb{R}\to\mathbb{R}\cup\\{+\infty\\}$ is an empirical pair
interaction potential (the factor $\frac{1}{2}$ accounts for double counting).
Such potentials are typically repulsive at short distances and attractive at
large distances. While clustering is favored by long range attraction, the
density of a cluster cannot get too large due to short-range repulsion.

Notably, even under simplifying assumptions on the interaction potentials, the
mathematical literature on rigorous crystallization results is scarce. In
fact, for finite $N$, only results in one and two space dimensions are
available. For example, if $V$ is of Lennard–Jones type, crystallization has
been proved only in one space dimension [26]. In higher space dimensions only
partial results are available. Most notably, in [18, 20, 39] it has been
proven that crystalline structures have optimal bulk energy scaling. In two
dimensions, only results for (some variants of) the sticky disc potential (see
Fig. 1)

$\displaystyle V(r):=\begin{cases}+\infty&\text{if }r<1\,,\\\ -1&\text{if
}r=1\,,\\\ 0&\text{otherwise}\end{cases}$ (1)

are available [16, 28, 30, 36].

$V(r)$$r$$+\infty$$1$$-1$ Figure 1. The sticky disc interaction potential $V$.

More recently, crystallization results have been proved for ionic compounds
[21, 22] and carbon structures in [35]. The potential given in (1) models the
atoms as hard spheres that interact exactly when two of them are tangent. In
$\mathbb{R}^{n}$ the kissing number $k(n)$ is the highest number of
$n$-dimensional spheres of radius $\frac{1}{2}$ which are tangent to a given
sphere of the same size. It is well known that $k(2)=6$ and $k(3)=12$, see
[37]. For a given configuration of non-overlapping equal balls centered at
$X=\\{x_{1},\ldots,x_{N}\\}\subset\mathbb{R}^{n}$,
$N\in\mathbb{N}\cup\\{+\infty\\}$ the coordination number of $x\in X$ is the
number of spheres centered at $y\in X\setminus\\{x\\}$ and tangent to the one
centered at $x$. In two dimensions there is a unique (up to a rigid motion)
configuration made of infinitely many particles such that all atoms have as
coordination number the kissing number. Such a set $X$ is the triangular
lattice with lattice spacing one. In three dimensions the problem is much more
intricate. In fact, there exist infinitely many configurations with constant
coordination number equal to $k(3)$. An infinite class of configurations can
be obtained by stacking in an appropriate way layers of triangular lattices. A
remarkable result by Hales [27] shows that all such structures solve Kepler’s
conjecture which is to say that they have the maximal packing density in
$\mathbb{R}^{3}$. Two notable cases of the aforementioned structures are the
face-centered cubic lattice $\mathcal{L}_{\mathrm{FCC}}$ and the hexagonal
closed-packed lattice $\mathcal{L}_{\mathrm{HCP}}$ (see (7)–(9) for their
precise definition) which are the most prevalent among the crystalline
arrangements in the periodic table of elements.

In this paper we want to investigate already crystallized configurations, i.e.
configurations $X\subset\mathcal{L}$ where
$\mathcal{L}=\mathcal{L}_{\mathrm{FCC}}$ or
$\mathcal{L}=\mathcal{L}_{\mathrm{HCP}}$. For such
$X=\\{x_{1},\ldots,x_{N}\\}\subset\mathcal{L}$, fixing the lattice spacing to
be $1$, we have

$\displaystyle\mathcal{E}(X)=\frac{1}{2}\sum_{i\neq
j}V(|x_{i}-x_{j}|)=-\sum_{i=1}^{N}\\#(\mathcal{N}(x_{i})\cap X)\,,\text{ where
}\mathcal{N}(x)=\\{y\in\mathcal{L}\colon|x-y|=1\\}\,.$

As described above the minimal energy per atom is $-k(3)=-12$. Further
information on $\mathcal{E}$ as $N$ grows can be obtained by referring it to
the minimal energy per atom and calculating the excess energy $E_{N}(X)$
defined below. More precisely, in Theorem 2.3, we carry out a rigorous
variational asymptotic expansion (see [10]) of $\mathcal{E}(X)$, by
considering

$\displaystyle
E_{N}(X)=N^{-2/3}\left(\mathcal{E}(X)+12N\right)=N^{-2/3}\sum_{i=1}^{N}(12-\\#(\mathcal{N}(x)\cap
X))\,$ (2)

and calculating its $\Gamma$-limit [9, 13] as $N$ tends to infinity. This
analysis has been done in two dimensions for configurations confined to the
triangular lattice [7] as well as without any confinement assumption [23].
Note that, the scaling factor $N^{-2/3}$ is used in order to keep the energy
bounded as the number of atoms grows. In fact, given a low energy
configuration of $N$ atoms, the number of those contributing to the energy
scales like $N^{2/3}$ for $N$ large. By associating to each configuration its
rescaled emprical measure

$\displaystyle\mu_{N}(X):=\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\frac{1}{N}\sum_{i=1}^{N}\delta_{N^{-1/3}x_{i}}\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\,,$

we show in Theorem 2.3 i) that the sequence of rescaled energies (2) is equi-
coercive with respect to the weak*-convergence of the associated empirical
measures. In Theorem 2.3 ii), iii) we exploit integral representation theorems
[2, 3, 5] to show that the limit energy is finite on the set of measures
$\mu=\sqrt{2}\mathcal{L}^{3}\llcorner V$, where $V\subset\mathbb{R}^{3}$ is a
set of finite perimeter, on which the energy takes the form

$\displaystyle
E_{\mathcal{L}}(\mu):=\int_{\partial^{*}V}\varphi_{\mathcal{L}}(\nu)\,\mathrm{d}\mathcal{H}^{2}\,.$
(3)

Here, $\partial^{*}V$ denotes the reduced boundary of the set $V$, $\nu(x)$
denotes its unit outer normal at the point $x\in\partial^{*}V$ and
$\varphi_{\mathcal{L}}$ is an anisotropic surface energy density depending on
the underlying lattice $\mathcal{L}$. In the case of multi-lattices, like the
HCP-lattice, this integral representation result has not yet been proven in
the literature. We defer to Section 5 for a proof of this result whose main
ingredient is the integral representation theorem in [3]. Furthermore, in the
same section we prove general compactness and concentration lemmata that
ensure the convergence of the rescaled empircal measures of minimizers of the
discrete problem (2) to the Wulff shape (up to a constant density factor) of
the associated limiting anisotropic perimeter energy (21). Such kind of result
was previously known only in two dimensions [7]. Its extension to higher
dimensions, see Lemma 5.16, requires more refined tools from geometric measure
theory that, to the best of our knowledge, are exploited in this setting here
for the first time. The main body of this work lies in the calculation of the
surface energy density
$\varphi_{\mathcal{L}}\colon\mathbb{R}^{3}\to[0,+\infty)$ both for the
$\mathrm{FCC}$ and the $\mathrm{HCP}$ lattices. Here, we take advantage of a
recently proved finite cell formula [11]. Finally, for both lattices, we solve
the associated isoperimetric problem [19]

$\displaystyle
m_{\mathcal{L}}:=\min\left\\{\int_{\partial^{*}V}\varphi_{\mathcal{L}}(\nu)\,\mathrm{d}\mathcal{H}^{2}\colon|V|=1\right\\}\,$
(4)

by calculating the (up to translation unique) set realizing the minimum in
(4), also known as the Wulff shape [40]. We show that
$m_{\mathrm{FCC}}<m_{\mathrm{HCP}}$ which also implies (since
$\Gamma$-convergence and coercivity implies the convergence of minimum values)
that, for large number of atoms, crystallization on the face-centered cubic
lattice is preferred to that on the hexagonal-closed packed lattice. We
finally mention [6] for some preliminary computations on the Wulff shape of
the FCC and HCP.

In contrast to the uniqueness of the Wulff crystal in the continuum setting,
minimizers to the discrete isoperimetric problem [29] are non-unique [17].
Over the last years there has been a remarkable interest in establishing
fluctuation estimates between different minimizers, i.e., estimating (several
notions of) distances between different minimizers. Maximal fluctuation
estimates between two minimizers have been first conjectured in [7] in the
case of the crystallization on the triangular lattice and have been later
proved in [38] and [15]. The same estimates have been proved in [33, 21] and
[14, 22] for the square and the honeycomb lattices, respectively. A general
approach linking the quantitative anistropic isoperimetric inequality to such
fluctuation estimates has been set up in [12] by two of the authors. In
dimensions larger than two these fluctuation estimates have been only
established for the cubic lattice in [32] and for $\mathbb{Z}^{d}$ in [34]. In
order to establish the aforementioned fluctuation estimates, however, an
understanding of the limiting macroscopic Wulff shape is essential. Since the
present work yields these shapes for the FCC and HCP lattices, it is our
opinion that it may be considered an indispensable first step to prove
fluctuation estimates also for such lattices.

The article is structured as follows. In Section 2, we introduce the necessary
mathematical preliminaries, the model, and the main results. In Section 3 we
prove Proposition 2.4 and 2.5, by calculating the surface energy density as
well as the Wulff crystal associated to both the $\mathrm{FCC}$ and the
$\mathrm{HCP}$ lattices. In Section 4 we prove the main $\Gamma$-convergence
Theorem 2.3. The latter is a consequence of a more general theory for discrete
perimeter energies on general periodic lattices developed in Section 5.

##  2\. Setting and Notation

Given a set of vectors $V\subset\mathbb{R}^{n}$ we denote by
$\mathrm{span}_{\mathbb{Z}}V$ the set of finite linear combinations of
elements of $V$ with coefficients in $\mathbb{Z}$. We denote by $\mathfrak{M}$
the collection of all Lebesgue measurable subsets of $\mathbb{R}^{n}$. Given
$A\in\mathfrak{M}$ we denote by $|V|$ its $n$-dimensional Lebesgue measure,
i.e., $|V|=\mathcal{L}^{n}(V)$, and $\mathcal{H}^{n}$ its $n$-dimensional
Hausdorff measure. Given a countable set $X$, we denote by $\\#X$ the
cardinality of $X$. Given $a,b\in\mathbb{R}^{n}$ we denote by $\langle
a,b\rangle$ their scalar product. We denote by $\mathbb{S}^{n-1}$ the set of
unitary vectors in $\mathbb{R}^{n}$. For any $\nu\in\mathbb{S}^{n-1}$ let
$\\{\nu_{1},\ldots,\nu_{n}=\nu\\}$ be an orthonormal basis of $\mathbb{R}^{n}$
, and let $Q^{\nu}:=\\{x\in\mathbb{R}^{n}\colon|\langle
x,\nu_{i}\rangle|<1/2,\,i=1,\ldots,n\\}$ be a unit cube centered at the origin
with faces parallel and orthogonal to $\nu$. For $T>0$ and
$x\in\mathbb{R}^{n}$ we set $Q_{T}^{\nu}(x)=x+TQ^{\nu}$ and we write
$Q_{T}^{\nu}=Q_{T}^{\nu}(0)$. For $r>0$ and $x\in\mathbb{R}^{n}$ we denote by
$B_{r}(x)$ the $n$-dimensional Euclidean ball of radius $r$ centered at $x$
(for $x=0$ we write $B_{r}$ in place of $B_{r}(0)$) and we set
$\omega_{n}=|B_{1}(x)|$. For $r>0$ and $A\subset\mathbb{R}^{n}$ we set
$(A)_{r}=B_{r}+A$. Given $A\subset\mathbb{R}^{n}$ open, we define the set of
positive Radon measures by $\mathcal{M}_{+}(A)$. We say that
$\\{\mu_{k}\\}_{k}\subset\mathcal{M}_{+}(A)$ converges to
$\mu\in\mathcal{M}_{+}(A)$ with respect to the weak star topology and we write
$\mu_{k}\overset{*}{\rightharpoonup}\mu$ if

$\displaystyle\lim_{k\to\infty}\int_{A}\varphi\,\mathrm{d}\mu_{k}=\int_{A}\varphi\,\mathrm{d}\mu\text{
for all }\varphi\in C_{c}(A)\,.$

we denote by $BV(A)$ the space of functions of bounded variation in $A$ and we
denote by $BV_{\mathrm{loc}}(A)=\\{u\colon u\in BV(K)\text{ for all
}K\subset\subset A,K\text{ open}\\}$. Given a function $u\in BV(A)$ we use the
notation of [4] for the jump set $J(u)$ and the measure theoretic normal
$\nu_{u}\colon J(u)\to\mathbb{S}^{n-1}$. For $V\subset A$, $V\in\mathfrak{M}$
we denote the relative perimeter of $V$ in $A$ by

$\displaystyle\mathrm{Per}(V,A)=\sup\left\\{\int_{V}\mathrm{div}\,v\,\mathrm{d}x\colon
v\in C_{c}^{\infty}(A;\mathbb{R}^{n}),\|v\|_{\infty}\leq 1\right\\}\,.$

For $\nu\in\mathbb{S}^{n-1}$ we set

$\displaystyle u_{\nu}(i):=\begin{cases}1&\text{if }\langle x,\nu\rangle\geq
0\,;\\\ 0&\text{otherwise.}\end{cases}$ (5)

In Section 2–Section 4 we set $n=3$.

### Definition of $\mathrm{HCP}$ and $\mathrm{FCC}$ lattices

In the following we define the _face-centered cubic lattice_ (short
$\mathrm{FCC}$-lattice) and the _hexagonal closed-packed lattice_ (short
$\mathrm{HCP}$-lattice). To this end, we introduce the verctors

$\displaystyle b_{1}:=\frac{1}{\sqrt{2}}\begin{pmatrix}1\\\ 1\\\
0\end{pmatrix}\,,\quad b_{2}:=\frac{1}{\sqrt{2}}\begin{pmatrix}1\\\ 0\\\
1\end{pmatrix}\,,\quad b_{3}:=\frac{1}{\sqrt{2}}\begin{pmatrix}0\\\ 1\\\
1\end{pmatrix}\,$ (6)

and

$\displaystyle e_{1}:=\begin{pmatrix}1\\\ 0\\\ 0\end{pmatrix}\,,\quad
e_{2}:=\frac{1}{2}\begin{pmatrix}1\\\ \sqrt{3}\\\ 0\end{pmatrix}\,,\quad
e_{3}:=\frac{2}{3}\sqrt{6}\begin{pmatrix}0\\\ 0\\\ 1\end{pmatrix}\,,\quad
v_{1}:=\frac{1}{3}\left(e_{1}+e_{2}\right)+\frac{1}{2}e_{3}\,.$ (7)

Figure 2. On the left: The $\mathrm{FCC}$-lattice. On the right: The
$\mathrm{HCP}$-lattice. Pairs of points at distance one are connected via the
dashed lines.

We define the $\mathrm{FCC}$-lattice as

$\displaystyle\mathcal{L}_{{\mathrm{FCC}}}:=\mathrm{span}_{\mathbb{Z}}\left\\{b_{1},b_{2},b_{3}\right\\}\,$
(8)

and the $\mathrm{HCP}$-lattice by

$\displaystyle\mathcal{L}_{{\mathrm{HCP}}}:=\mathrm{span}_{\mathbb{Z}}\left\\{e_{1},e_{2},e_{3}\right\\}\cup\left(\mathrm{span}_{\mathbb{Z}}\left\\{e_{1},e_{2},e_{3}\right\\}+v_{1}\right)\,.$
(9)

The two lattices are illustrated in Figure 2. We shall write $\mathcal{L}$ to
generically denote one of the two lattices defined above. We define the
_neighborhood_ of a point $x\in\mathcal{L}_{\mathrm{FCC}}$ as the set

$\displaystyle\mathcal{N}_{\mathrm{FCC}}(x):=\\{\pm b_{1},\pm b_{2},\pm
b_{3},\pm(b_{1}-b_{2}),\pm(b_{1}-b_{3}),\pm(b_{2}-b_{3})\\}+x\,.$ (10)

Similarly, for a point $x\in\mathcal{L}_{\mathrm{HCP}}$ we define its
_neighborhood_ as follows: if
$x\in\mathrm{span}_{\mathbb{Z}}\\{e_{1},e_{2},e_{3}\\}$ then

$\displaystyle\mathcal{N}_{\mathrm{HCP}}(x):=\\{\pm e_{1},\pm
e_{2},\pm(e_{1}-e_{2}),v_{1},v_{1}-e_{1},v_{1}-e_{2},v_{1}-e_{3},v_{1}-e_{1}-e_{3},v_{1}-e_{2}-e_{3}\\}+x\,,$
(11)

while if $x\in\mathrm{span}_{\mathbb{Z}}\\{e_{1},e_{2},e_{3}\\}+v_{1}$ then

$\displaystyle\mathcal{N}_{\mathrm{HCP}}(x):=\\{\pm e_{1},\pm
e_{2},\pm(e_{1}-e_{2}),-v_{1},e_{1}-v_{1},e_{2}-v_{1},e_{3}-v_{1},e_{1}+e_{3}-v_{1},e_{2}+e_{3}-v_{1}\\}+x\,.$
(12)

Note that $\mathcal{N}_{\mathrm{FCC}}(x)=\mathcal{N}_{\mathrm{FCC}}(0)+x$ for
all $x\in\mathcal{L}_{\mathrm{FCC}}$, while this is no more the case for
$x\in\mathcal{L}_{\mathrm{HCP}}$. Also for $\mathcal{N}$ we omit the subscript
if we do not need to distinguish between $\mathrm{FCC}$ and $\mathrm{HCP}$. It
is straightforward to check that for all $x,y\in\mathcal{L}$,

$\displaystyle x\in\mathcal{N}(y)\iff|x-y|=1\,.$

Given $\mathcal{L}$ we define the _Voronoi cell_ of $x\in\mathcal{L}$ (with
respect to $\mathcal{L}$) by

$\displaystyle\mathcal{V}_{\mathcal{L}}(x):=\\{y\in\mathbb{R}^{3}\colon|y-x|\leq|y-z|\text{
for all }z\in\mathcal{L}\\}\,.$ (13)

Accordingly, given $\varepsilon>0$ we write
$\mathcal{V}_{\varepsilon\mathcal{L}}(x)$ for the Voronoi cell centered at
$x\in\varepsilon\mathcal{L}$ with respect to the scaled lattice
$\varepsilon\mathcal{L}$. Given $X\subset\varepsilon\mathcal{L}$ we say that
$y\in\mathcal{N}_{\varepsilon}(x)$ if and only if
$\varepsilon^{-1}y\in\mathcal{N}(\varepsilon^{-1}x)$.

### Definition of the Energy

Given $X\subset\mathcal{L}$ and $A\subset\mathbb{R}^{3}$ we define the
configurational energy of $X$ localized to the set $A$ as

$\displaystyle E_{\mathcal{L}}(X,A):=\sum_{x\in X\cap
A}(12-\\#(\mathcal{N}(x)\cap X))\,.$

In the formula above we can interpret the set $X$ as the occupancy of the
crystal $\mathcal{L}$, i.e., the set of those nodes of $\mathcal{L}$ occupied
by atoms. The quantity $12-\\#(\mathcal{N_{\mathcal{L}}}(x)\cap X)$ is also
known as the valence of the point $x$ with respect to $X$, i.e., the number of
neighbours missing in $X$ in order to have a neighbourhood of maximal
cardinality (the number $12$ in the formula). Note that we can also rewrite
the energy as

$\displaystyle E_{\mathcal{L}}(X,A)=\frac{1}{2}\sum_{x\in\mathcal{L}\cap
A}\sum_{y\in\mathcal{L}}c(x,y)|\chi_{X}(y)-\chi_{X}(x)|\,,$

where

$\displaystyle c(x,y)=\begin{cases}1&\text{if }y\in\mathcal{N}(x)\,,\\\
0&\text{otherwise.}\end{cases}$ (14)

### Periodicity of the interaction coefficients

By definition

${\mathcal{L}}_{\mathrm{FCC}}={\mathcal{L}}_{\mathrm{FCC}}+b_{1}={\mathcal{L}}_{\mathrm{FCC}}+b_{2}={\mathcal{L}}_{\mathrm{FCC}}+b_{3}.$

As a consequence of that, for any $x,y\in\mathcal{L}_{\mathrm{FCC}}$ it holds
that

$\displaystyle\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}c(x+b_{1},y+b_{1})=c(x+b_{2},y+b_{2})=c(x+b_{3},y+b_{3})=c(x,y)\,.\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}$

According to the last two equalities, we say that the lattice
$\mathcal{L}_{\mathrm{FCC}}$ as well as the interaction coefficients of its
configurational energy are periodic with periodicity cell

$\displaystyle
T_{{\mathrm{FCC}}}=\\{\lambda_{1}b_{1}+\lambda_{2}b_{2}+\lambda_{3}b_{3}\colon\lambda_{i}\in[0,1)\\}\,,$
(15)

or simply that they are $T_{{\mathrm{FCC}}}$-periodic. Similarly, we observe
that ${\mathcal{L}}_{\mathrm{HCP}}$ and its interaction coefficients are
$T_{{\mathrm{HCP}}}$-periodic, where the periodicity cell is defined as

$\displaystyle
T_{{\mathrm{HCP}}}=\\{\lambda_{1}e_{1}+\lambda_{2}e_{2}+\lambda_{3}e_{3}\colon\lambda_{i}\in[0,1)\\}\,.$
(16)

### Surface scaling of the configurational energy

For $\varepsilon>0$ such that $\varepsilon^{3}\\#X\to 1$ as $\varepsilon\to 0$
we consider the following family of scaled energies

$\displaystyle G_{\mathcal{L},\varepsilon}(X):=\varepsilon^{2}\sum_{x\in
X}\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}(12-\\#(\mathcal{N}_{\varepsilon}(x)\cap
X))\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\,.$

Note that, modeling the points in $X\subset\varepsilon{\mathcal{L}}$ as hard-
spheres of $\varepsilon$ diameter, the quantity $\varepsilon^{3}\\#X$ is of
order one (according to the scaling assumption above) and proportional to the
volume of the union of the spheres in $X$. Hence, the scaling factor
$\varepsilon^{2}$ in the energy functional turns out to be a surface scaling.
We also define the rescaled empirical measures associated to the configuration
$X$ as

$\displaystyle\mu_{\varepsilon}:=\varepsilon^{3}\sum_{x\in X}\delta_{x}\,.$
(17)

Upon identifying $X\subset\varepsilon\mathcal{L}$ with its empirical measure
$\mu_{\varepsilon}$, we can regard these energies to be defined on
$\mathcal{M}_{+}(\mathbb{R}^{3})$ by setting

$\displaystyle
E_{\mathcal{L},\varepsilon}(\mu):=\begin{cases}G_{\mathcal{L},\varepsilon}(X)&\text{if
}\mu=\mu_{\varepsilon}\text{ given in }\eqref{def:empirical measures}\text{
for some }X\subset\varepsilon\mathcal{L}\,,\\\
+\infty&\text{otherwise.}\end{cases}$ (18)

### The coarse grained continuum energy

For $\mathcal{L}$ we define the _homogenized surface energy density_
$\varphi_{\mathcal{L}}\colon\mathbb{R}^{3}\to[0,+\infty]$ as the convex
positively homogeneous function of degree one such that for all
$\nu\in\mathbb{S}^{2}$ we have

$\displaystyle\varphi_{\mathcal{L}}(\nu):=\lim_{T\to+\infty}\frac{1}{T^{2}}\inf\left\\{E_{\mathcal{L}}(X,Q^{\nu}_{T}):X\subset\mathcal{L},\chi_{X}(i)=u_{\nu}(i)\text{
for }i\in\mathcal{L}\setminus Q^{\nu}_{T-3}\right\\}\,,$ (19)

where $u_{\nu}$ is given by (5).

In order to be able to apply [11, Proposition 2.6] and eventually obtain an
alternative representation of $\varphi_{\mathcal{L}}$ (up to a coordinate
transformation and reparametrization of the interaction coefficients), we
define for $u\colon\mathcal{L}\to\mathbb{R}$, $A\subset\mathbb{R}^{3}$ the
energy

$\displaystyle F_{\mathcal{L}}(u,A):=\frac{1}{2}\sum_{x\in\mathcal{L}\cap
A}\sum_{y\in\mathcal{L}}c(x,y)|u(y)-u(x)|\,.$

We are now in position to state [11, Proposition 2.6].

######  Proposition 2.1.

Let $c(x,y)$ be as in (14).Then

$\displaystyle\varphi_{\mathcal{L}}(\nu)=\frac{1}{|T_{\mathcal{L}}|}\inf\left\\{F_{\mathcal{L}}(u,T_{\mathcal{L}}):u\colon\mathcal{L}\to\mathbb{R},u(\cdot)-\langle\nu,\cdot\rangle\text{
is }T_{\mathcal{L}}\text{-periodic}\right\\}\,.$ (20)

With the definition of surface energy density at hand we can define the
coarse-grained continuum energy
$E_{\mathcal{L}}\colon\mathcal{M}_{+}(\mathbb{R}^{3})\to[0,+\infty]$ as

$\displaystyle
E_{\mathcal{L}}(\mu):=\begin{cases}\int_{\partial^{*}V}\varphi_{\mathcal{L}}(\nu)\,\mathrm{d}\mathcal{H}^{2}&\text{if
}\mu=\sqrt{2}\mathcal{L}^{3}\llcorner V,\chi_{V}\in
BV_{\mathrm{loc}}(\mathbb{R}^{3})\,,\\\ +\infty&\text{otherwise.}\end{cases}$
(21)

with $\varphi_{\mathcal{L}}$ given by (19). Here, $\partial^{*}V$ denotes the
reduced boundary of the set $V$, $\nu$ its outer normal and $\mathcal{H}^{2}$,
as noted at the beginning of this section, stands for the $2$-dimensional
Hausdorff measure in $\mathbb{R}^{3}$ (cf. [4], Chapters 2.8 and 3.5).

In what follows we say that $F_{\varepsilon}$ $\Gamma$-converges to $F$ if for
all sequences $\\{\varepsilon_{j}\\}_{j}$ converging to $0$ we have
$\Gamma$-$\lim_{j}F_{\varepsilon_{j}}=F$.

### The Wulff Crystal

In this section we calculate the Wulff crystals of the coarse grained
$\mathrm{FCC}$ and $\mathrm{HCP}$ lattices. To the best of our knowledge, this
is the first time that such a calculation has been carried out in a rigorous
analytical way. In what follows we introduce the notion of Wulff shape in the
general case of $\mathbb{R}^{n}$. While in the rest of this section we limit
ourselves to the case $n=3$, in Section 5 we consider general $n$.

Given $\varphi\colon\mathbb{R}^{n}\to[0,+\infty)$ convex, non-degenerate,
(i.e. there exist $0<c<C$ such that $c\leq\varphi(\nu)\leq C$ for all
$\nu\in\mathbb{S}^{n-1}$) positively homogeneous of degree one, we define the
Wulff set of $\varphi$ by

$\displaystyle
W_{\varphi}:=\\{\zeta\in\mathbb{R}^{n}\colon\langle\zeta,\nu\rangle\leq\varphi(\nu)\text{
for all }\nu\in\mathbb{S}^{n-1}\\}\,.$ (22)

Thanks to the anistropic isoperimetric inequality (cf. [19]), we have that
$W_{\varphi}$ is the unique (up to rigid motions) minimizer of

$\displaystyle\min\left\\{\int_{\partial^{*}A}\varphi(\nu)\,\mathrm{d}\mathcal{H}^{n-1}\colon|A|=|W_{\varphi}|\right\\}\,.$

Given $\lambda>0$ we set
$W_{\lambda}=\left(\frac{\lambda}{|W_{\varphi}|}\right)^{1/n}W_{\varphi}$ so
that $|W_{\lambda}|=\lambda$ and, by scaling, it solves the minimum problem
above among all sets $A\subset\mathbb{R}^{n}$ with $|A|=\lambda$. We recall
here that, defining $\varphi^{\circ}\colon\mathbb{R}^{n}\to[0,+\infty)$ by

$\displaystyle\varphi^{\circ}(\xi)=\sup_{\nu\in\mathbb{S}^{n-1}}\frac{\langle\nu,\xi\rangle}{\varphi(\nu)}\,,$

it holds that $W_{\varphi}=\\{\varphi^{\circ}\leq 1\\}$.

######  Definition 2.2.

Let $(X,\tau)$ be a topological space and let $F_{k}\colon X\to[0,+\infty]$.
For $x\in X$ we set

$\displaystyle\Gamma\text{-}\limsup_{k\to+\infty}F_{k}(x)=\inf\left\\{\limsup_{k\to+\infty}F_{k}(x_{k})\colon
x_{k}\overset{\tau}{\to}x\right\\}$

and

$\displaystyle\Gamma\text{-}\liminf_{k\to+\infty}F_{k}(x)=\inf\left\\{\liminf_{k\to+\infty}F_{k}(x_{k})\colon
x_{k}\overset{\tau}{\to}x\right\\}\,.$

If there exists $F\colon X\to[0,+\infty]$ such that

$\displaystyle
F(x)=\Gamma\text{-}\limsup_{k\to+\infty}F_{k}(x)=\Gamma\text{-}\liminf_{k\to+\infty}F_{k}(x)\,,$

we say that $F_{k}$ $\Gamma$-converges with respect to $\tau$ to $F$ and we
write

$\displaystyle F(x)=\Gamma\text{-}\lim_{k\to+\infty}F_{k}(x)\,.$

If we have $(F_{\varepsilon})_{\varepsilon>0}\colon X\to[0,+\infty]$ we say
that $F_{\varepsilon}$ $\Gamma$-converges with respect to $\tau$ to $F$ if
$F_{\varepsilon_{k}}$ $\Gamma$-converges with respect to $\tau$ to $F$ for all
$\varepsilon_{k}\to 0$.

The following variational coarse-graining result is proved in Section 4.

######  Theorem 2.3.

Let $\varepsilon\to 0$, and let $E_{\mathcal{L},\varepsilon}$ and
$E_{\mathcal{L}}$ be the energy functionals defined in (18) and (21),
respectively.

  * i)

(Compactness) Let
$\\{\mu_{\varepsilon}\\}_{\varepsilon}\subset\mathcal{M}_{+}(\mathbb{R}^{3})$
be such that

$\displaystyle\sup_{\varepsilon>0}E_{\mathcal{L},\varepsilon}(\mu_{\varepsilon})<+\infty\,.$

Then there exists $V\subset\mathbb{R}^{3}$ such that $\chi_{V}\in
BV_{\mathrm{loc}}(\mathbb{R}^{3})$, $\mu=\sqrt{2}\mathcal{L}^{3}\llcorner V$,
and a subsequence (not relabeled) such that
$\mu_{\varepsilon}\overset{*}{\rightharpoonup}\mu$. Furthermore, if
$\mu_{\varepsilon}$ is such that

$\displaystyle
E_{\mathcal{L},\varepsilon}(\mu_{\varepsilon})=\inf_{\nu\in\mathcal{M}_{+}(\mathbb{R}^{3}):\,|\nu|(\mathbb{R}^{3})=\varepsilon^{3}n_{\varepsilon}}E_{\mathcal{L},\varepsilon}(\mu)\,,$

with $\varepsilon^{3}n_{\varepsilon}\to\sqrt{2}v$, then
$\mu=\sqrt{2}\mathcal{L}^{3}\llcorner W_{\varphi_{\mathcal{L}}}^{v}$, where
$W_{\varphi_{\mathcal{L}}}^{v}=\lambda W_{\varphi_{\mathcal{L}}}$ (defined in
(22)) for $\lambda>0$ such that $|W_{\varphi_{\mathcal{L}}}^{v}|=v$.

  * ii)

(Liminf inequality) Let $\mu\in\mathcal{M}_{+}(\mathbb{R}^{3})$ be such that
$\mu_{\varepsilon}\overset{*}{\rightharpoonup}\mu$. Then

$\displaystyle E_{\mathcal{L}}(\mu)\leq\liminf_{\varepsilon\to
0}E_{\mathcal{L},\varepsilon}(\mu_{\varepsilon})\,.$

  * iii)

(Limsup inequality) Let $\mu\in\mathcal{M}_{+}(\mathbb{R}^{3})$. Then there
exists
$\\{\mu_{\varepsilon}\\}_{\varepsilon}\subset\mathcal{M}_{+}(\mathbb{R}^{3})$
such that $\mu_{\varepsilon}\overset{*}{\rightharpoonup}\mu$ and

$\displaystyle E_{\mathcal{L}}(\mu)\geq\liminf_{\varepsilon\to
0}E_{\mathcal{L},\varepsilon}(\mu_{\varepsilon})\,.$

Figure 3. The Wulff Crystal of the FCC-lattice on the left and HCP-lattice on
the right. Figure 4. The sublevel set $\\{\varphi_{\mathrm{FCC}}\leq 1\\}$ on
the left and the sublevel set $\\{\varphi_{\mathrm{HCP}}\leq 1\\}$ on the
right.

### Explicit formula of the surface energy densities

Taking advantage of the representation formula (20) stated in Proposition
(2.1), we provide the explicit formulas of the surface energy density
$\varphi_{{\mathcal{L}}_{\mathrm{FCC}}}$ and
$\varphi_{{\mathcal{L}}_{\mathrm{HCP}}}$. Their sublevel sets are depicted in
Figure 4. With the two explicit formulas at hand we can calculate the polar
functions of both densities, the associated Wulff shapes and the surface
energy per unit volume of both the $\mathrm{FCC}$ and $\mathrm{HCP}$ crystals.
In order not to overburden the reader with notation, we write
$\varphi_{\mathrm{FCC}}$ and $\varphi_{\mathrm{HCP}}$ for
$\varphi_{{\mathcal{L}}_{\mathrm{FCC}}}$ and
$\varphi_{{\mathcal{L}}_{\mathrm{HCP}}}$ as well as $W_{\mathrm{FCC}}$ and
$W_{\mathrm{HCP}}$ instead of $W_{\varphi_{\mathcal{L}_{\mathrm{FCC}}}}$ and
$W_{\varphi_{\mathcal{L}_{\mathrm{HCP}}}}$. $W_{\mathrm{FCC}}$ and
$W_{\mathrm{HCP}}$ are depicted in Figure 3.

######  Proposition 2.4.

The following formulas hold true.

$\displaystyle\varphi_{\mathrm{FCC}}(\nu)=|\nu_{1}+\nu_{2}|+|\nu_{1}+\nu_{3}|+|\nu_{2}+\nu_{3}|+|\nu_{1}-\nu_{2}|+|\nu_{1}-\nu_{3}|+|\nu_{2}-\nu_{3}|\,,$
(23)

and

$\displaystyle\varphi_{\mathrm{FCC}}^{\circ}(\zeta)=\max\left\\{\frac{1}{4}\|\zeta\|_{\infty},\frac{1}{6}\|\zeta\|_{1}\right\\}\,.$
(24)

In particular, $W_{\mathrm{FCC}}$ is a truncated octahedron and its surface
energy per unit volume is

$\displaystyle|W_{\mathrm{FCC}}|^{-2/3}\int_{\partial^{*}W_{\mathrm{FCC}}}\varphi_{\mathrm{FCC}}(\nu)\,\mathrm{d}\mathcal{H}^{2}=3\cdot
2^{2}\cdot 2^{2/3}\,.$ (25)

######  Proposition 2.5.

The following formulas hold true.

$\displaystyle\begin{split}\varphi_{\mathrm{HCP}}(\nu)&=\sqrt{2}\left(|\langle
e_{1},\nu\rangle|+|\langle e_{2},\nu\rangle|+|\langle
e_{1}-e_{2},\nu\rangle|\right)+\frac{1}{\sqrt{2}}|\langle e_{3},\nu\rangle|\\\
&\quad+\sqrt{2}\max\left\\{|\langle e_{1},\nu\rangle|,|\langle
e_{2},\nu\rangle|,|\langle e_{3},\nu\rangle|,|\langle
e_{1}-e_{2},\nu\rangle|\right\\}\,,\end{split}$ (26)

and

$\displaystyle\begin{split}\varphi^{\circ}_{\mathrm{HCP}}(\zeta)=\max\Big{\\{}&\frac{2}{7\sqrt{2}}\left(|\zeta_{1}|+\frac{1}{\sqrt{3}}|\zeta_{2}|+\frac{3}{2\sqrt{6}}|\zeta_{3}|\right),\frac{1}{2\sqrt{3}}|\zeta_{3}|,\\\
&\frac{2}{3\sqrt{6}}|\zeta_{2}|,\frac{4}{7\sqrt{6}}|\zeta_{2}|+\frac{3}{14\sqrt{3}}|\zeta_{3}|,\frac{1}{3\sqrt{2}}\left(|\zeta_{1}|+\frac{1}{\sqrt{3}}|\zeta_{2}|\right)\Big{\\}}\,.\end{split}$
(27)

In particular, $W_{\mathrm{HCP}}$ is a truncated elongated hexagonal bipyramid
and its surface energy per unit volume is

$\displaystyle|W_{\mathrm{HCP}}|^{-2/3}\int_{\partial^{*}W_{\mathrm{HCP}}}\varphi_{\mathrm{HCP}}(\nu)\,\mathrm{d}\mathcal{H}^{2}=3\cdot
2^{2/3}\cdot 65^{1/3}\,.$ (28)

##  3\. Proof of Proposition 2.4 and Proposition 2.5

In this section we prove Proposition 2.4 and Proposition 2.5. To this end, we
use Proposition 2.1 to note that $\varphi_{\mathcal{L}}$ is given by (20).

###### Proof of Proposition 2.4.

We divide the proof into several steps. First, we calculate
$\varphi_{{\mathrm{FCC}}}$. Then, we calculate
$\varphi^{\circ}_{\mathrm{FCC}}$. Lastly, we calculate (25). Recall (6).
Step 1.(Calculation of $\varphi_{\mathrm{FCC}}$)We make use of Proposition 2.1
in order to calculate $\varphi_{\mathrm{FCC}}$. First of all, owing to (15),
we note that

$\displaystyle|T_{{\mathrm{FCC}}}|=\frac{1}{3}\sqrt{6}\cdot\frac{1}{2}\sqrt{3}=\frac{1}{2}\sqrt{2}\,.$
(29)

Given $u\colon\mathcal{L}_{\mathrm{FCC}}\to\mathbb{R}$ such that
$u(\cdot)-\langle\nu,\cdot\rangle$ is $T_{{\mathrm{FCC}}}$-periodic we have
that $u(x+b_{i})=u(x)+\langle b_{i},\nu\rangle$ for all $i=1,2,3$. Therefore,
$u$ is an affine function of the form $u(x)=\langle
x,\nu\rangle+c\,,x\in\mathcal{L}_{\mathrm{FCC}}$ for some $c\in\mathbb{R}$.
Lastly, note that $\mathcal{L}_{\mathrm{FCC}}\cap T_{{\mathrm{FCC}}}=\\{0\\}$.
Using (20) and (29), we obtain

$\displaystyle\varphi_{\mathrm{FCC}}(\nu)=\frac{1}{2}\sqrt{2}\sum_{\xi\in\mathcal{N}_{{\mathrm{FCC}}}}|u(\xi)-u(0)|=\frac{1}{2}\sqrt{2}\sum_{\xi\in\mathcal{N}_{{\mathrm{FCC}}}}|\langle\xi,\nu\rangle|\,.$

Employing now (10), we obtain (23).
Step 2.(Calculation of $\varphi^{\circ}_{\mathrm{FCC}}$) Let $G$ be the
isometry group on $\mathbb{R}^{3}$ whose elements $g\in G$ are the linear
isometries $g\colon\mathbb{R}^{3}\to\mathbb{R}^{3}$ defined by
$g(\nu_{1},\nu_{2},\nu_{3})=(b_{1}\nu_{\pi_{1}},b_{2}\nu_{\pi_{2}},b_{3}\nu_{\pi_{3}})$
where $\pi$ is a permutation on $\\{1,2,3\\}$ and $b_{i}\in\\{-1,1\\}$. Since
$\varphi_{\mathrm{FCC}}(g(\nu))=\varphi_{\mathrm{FCC}}(\nu)$ for all $g\in G$,
$\nu\in\mathbb{R}^{3}$, we infer that

$\displaystyle\varphi^{\circ}_{\mathrm{FCC}}(\zeta):=\underset{\varphi_{\mathrm{FCC}}(\nu)\leq
1}{\max_{\nu\in\mathbb{R}^{3}}}\langle\zeta,\nu\rangle=\underset{\varphi_{\mathrm{FCC}}(g^{-1}(\nu))\leq
1}{\max_{\nu\in\mathbb{R}^{3}}}\langle\zeta,g^{-1}(\nu)\rangle=\underset{\varphi_{\mathrm{FCC}}(\nu)\leq
1}{\max_{\nu\in\mathbb{R}^{3}}}\langle
g(\zeta),\nu\rangle=\varphi_{\mathrm{FCC}}^{\circ}(g(\zeta))\,,$

also relying on the property $g^{T}=g^{-1}$. Therefore, we can assume that
$0\leq\zeta_{1}\leq\zeta_{2}\leq\zeta_{3}$. Thus, if we want to maximize
$\langle\zeta,\nu\rangle$ under the condition $\varphi_{\mathrm{FCC}}(\nu)\leq
1$, we can as well assume that $0\leq\nu_{1}\leq\nu_{2}\leq\nu_{3}$, so that
condition $\varphi_{\mathrm{FCC}}(\nu)\leq 1$ becomes equivalent to

$\displaystyle 4\nu_{3}+2\nu_{2}\leq 1\,.$

Therefore, noting that any linear function attains its maximum at the extreme
points of a convex set and consulting Figure 5, we obtain

$\displaystyle\underset{4\nu_{3}+2\nu_{2}\leq
1}{\max_{0\leq\nu_{1}\leq\nu_{2}\leq\nu_{3}}}\zeta_{1}\nu_{1}+\zeta_{2}\nu_{2}+\zeta_{3}\nu_{3}=\underset{4\nu_{3}+2\nu_{2}\leq
1}{\max_{0\leq\nu_{2}\leq\nu_{3}}}(\zeta_{1}+\zeta_{2})\nu_{2}+\zeta_{3}\nu_{3}$
$\displaystyle=\max\left\\{\frac{1}{4}\zeta_{3},\frac{1}{6}(\zeta_{1}+\zeta_{2}+\zeta_{3})\right\\}$
$\displaystyle=\max\left\\{\frac{1}{4}\|\zeta\|_{\infty},\frac{1}{6}\|\zeta\|_{1}\right\\}\,.$

$\nu_{3}$$\frac{1}{4}$$\nu_{2}$$\frac{1}{2}$ Figure 5. The set
$\\{0\leq\nu_{2}\leq\nu_{3}\\}\cap\\{4\nu_{3}+2\nu_{1}\leq 1\\}$ depicted in
gray.

This is the desired formula (24) and concludes Step 2.
Step 3.(Calculation of (25)) Note that the set $W_{\varphi_{\mathrm{FCC}}}$ is
the intersection of a cube $\|\zeta\|_{\infty}\leq 4$ with an octahedron
$\|\zeta\|_{1}\leq 6$, see Fig. 3. Its boundary has $6$ square faces, where
$\nu=\pm(1,0,0)$ (resp. $\pm(0,1,0)$ or $\pm(0,0,1)$) and $8$ hexagonal faces,
where $\nu=\frac{1}{\sqrt{3}}(\pm 1\pm 1\pm 1)$. First, we consider the set
where $\nu=(1,0,0)$, the other cases where
$\varphi_{\mathrm{FCC}}^{\circ}(\zeta)=\frac{1}{4}\|\zeta\|_{\infty}=1$
contributing with the same value. The square is given by

$\displaystyle
S_{1}^{+}=\\{(4,\zeta_{2},\zeta_{3})\colon|\zeta_{2}|+|\zeta_{3}|\leq
2\\}=\left\\{\frac{1}{4}\|\zeta\|_{\infty}=\frac{1}{4}\zeta_{1}=1\right\\}\cap\left\\{\frac{1}{6}\|\zeta\|_{1}\leq
1\right\\}\,.$

Therefore, $\mathcal{H}^{2}(S_{1}^{+})=8$ and
$\varphi_{\mathrm{FCC}}((1,0,0))=4$. Similarly, we obtain the same measure and
value of $\varphi_{\mathrm{FCC}}$ for the other squares $S_{1}^{-}$,
$S_{2}^{\pm}$,$S_{3}^{\pm}$, where $\nu$ is (up to sign) one of the coordinate
unit vectors. Hence,

$\displaystyle\sum_{i=1}^{3}\int_{S_{i}^{+}}\varphi_{\mathrm{FCC}}(\nu)\,\mathrm{d}\mathcal{H}^{2}+\sum_{i=1}^{3}\int_{S_{i}^{-}}\varphi_{\mathrm{FCC}}(\nu)\,\mathrm{d}\mathcal{H}^{2}=6\cdot
8\cdot 4=3\cdot 2^{6}\,.$ (30)

Next, we consider the contribution of a hexagon. We consider the hexagon
contained in the set $\zeta_{i}\geq 0$ for all $i$. Here, we have
$\nu=\frac{1}{\sqrt{3}}(1,1,1)$ and $\varphi_{\mathrm{FCC}}(\nu)=2\sqrt{3}$.
The $6$ sides of the hexagon have all side-length $2\sqrt{2}$. To see this,
there are sides of the form $(4,2-t,t),t\in[0,2]$ or $(4-t,0,2+t),t\in[0,2]$
and their permutations (up to identifying $t$ with $2-t$ in the first case and
$4-t$ and $2+t$ in the second case). An equilateral hexagon $H$ of side-length
$2\sqrt{2}$ satisfies $\mathcal{H}^{2}(H)=12\sqrt{3}$. Labeling the hexagons
by $H_{i}$, $i=0,\ldots,7$, we obtain

$\displaystyle\sum_{i=0}^{7}\int_{H_{i}}\varphi_{\mathrm{FCC}}(\nu)\,\mathrm{d}\mathcal{H}^{2}=8\cdot\mathcal{H}^{2}(H_{i})\cdot\varphi_{\mathrm{FCC}}\left(\frac{1}{\sqrt{3}}(\pm
1,\pm 1,\pm 1)\right)=8\cdot 12\sqrt{3}\cdot 2\sqrt{3}=3^{2}\cdot 2^{6}\,.$
(31)

Using (30) and (31), we obtain

$\displaystyle\int_{\partial
W_{\mathrm{FCC}}}\varphi_{\mathrm{FCC}}(\nu)\,\mathrm{d}\mathcal{H}^{2}=3\cdot
2^{6}+3^{2}\cdot 2^{6}=3\cdot 2^{8}\,.$ (32)

Let $C:=\\{\zeta\in\mathbb{R}^{3}\colon\zeta_{i}\geq 0\text{ for all
}i=1,2,3\text{ and
}\frac{1}{4}\|\zeta\|_{\infty}\geq\frac{1}{6}\|\zeta\|_{1}\\}$ and
$C^{c}:=\\{\zeta\in\mathbb{R}^{3}\colon\zeta_{i}\geq 0\text{ for all
}i=1,2,3\text{ and
}\frac{1}{4}\|\zeta\|_{\infty}<\frac{1}{6}\|\zeta\|_{1}\\}$. We split the
calculation of the volume $W\cap\\{\zeta\in\mathbb{R}^{3}\colon\zeta_{i}\geq
0\text{ for all i}\\}$ into the set $C\cap W_{{\mathrm{FCC}}}$ and $C^{c}\cap
W_{\mathrm{FCC}}$. Noting that on this set
$|\nabla\varphi^{\circ}_{\mathrm{FCC}}(\zeta)|=\frac{1}{4}$
$\mathcal{L}^{3}$-a.e. on $C$, due to the coarea-formula, we have

$\displaystyle|\\{C\cap W_{\mathrm{FCC}}\\}|$ $\displaystyle=4\int_{C\cap
W_{\mathrm{FCC}}}|\nabla\varphi_{\mathrm{FCC}}^{\circ}(\zeta)|\,\mathrm{d}\zeta=4\int_{0}^{1}\mathcal{H}^{2}(C\cap\\{\varphi^{\circ}_{\mathrm{FCC}}(\zeta)=s\\})\,\mathrm{d}s$
$\displaystyle=\int_{0}^{1}4\cdot s^{2}\cdot 6\,\mathrm{d}s=8\,.$

Here we used that,
$C\cap\\{\varphi^{\circ}_{\mathrm{FCC}}(\zeta)=s\\}=s(S_{1}^{+}\cup
S_{2}^{+}\cup S_{3}^{+})\cap\\{\zeta_{i}\geq 0\\}$ and the scaling properties
of the $2$-dimensional Hausdorff-measure. On the other hand, using that
$|\nabla\varphi^{\circ}_{\mathrm{FCC}}(\zeta)|=\frac{\sqrt{3}}{6}$
$\mathcal{L}^{3}$-a.e. on $C^{c}$, we have

$\displaystyle|\\{C^{c}\cap W_{\mathrm{FCC}}\\}|$
$\displaystyle=2\sqrt{3}\int_{C^{c}\cap
W_{\mathrm{FCC}}}|\nabla\varphi^{\circ}_{\mathrm{FCC}}(\zeta)|\,\mathrm{d}\zeta=2\sqrt{3}\int_{0}^{1}\mathcal{H}^{2}(C^{c}\cap\\{\varphi^{\circ}_{\mathrm{FCC}}(\zeta)=s\\})\,\mathrm{d}s$
$\displaystyle=2\sqrt{3}\int_{0}^{1}s^{2}\cdot 12\sqrt{3}\,\mathrm{d}s=3\cdot
2^{3}\,.$

Taking into account also the sets $\\{\pm\zeta_{i}\geq 0\\}$, we obtain

$\displaystyle|W_{\mathrm{FCC}}|=8(8+3\cdot 2^{3})=2^{8}\,.$

Now, this together with $\eqref{eq:intWFCC}$ yields (25). ∎

###### Proof of Proposition 2.5.

We divide the proof into several steps. First, we calculate
$\varphi_{\mathrm{HCP}}$. Then, we calculate $\varphi^{\circ}_{\mathrm{HCP}}$.
Lastly, we calculate (28).
Step 1.(Calculation of $\varphi_{\mathrm{HCP}}$) We make use of Proposition
2.1 in order to calculate $\varphi_{\mathrm{HCP}}$. First of all, due to (16),
note that

$\displaystyle|T_{{\mathrm{HCP}}}|=\frac{2}{3}\sqrt{6}\cdot\frac{1}{2}\sqrt{3}=\sqrt{2}\,.$
(33)

Given $u\colon\mathcal{L}_{\mathrm{FCC}}\to\mathbb{R}$ such that
$u(\cdot)-\langle\nu,\cdot\rangle$ is $T_{\mathrm{HCP}}$-periodic we have that
$u(x+e_{i})=u(x)+\langle e_{i},\nu\rangle$ for all $i=,1,2,3$ and
$\mathcal{L}_{\mathrm{HCP}}\cap T_{{\mathrm{HCP}}}=\\{0,v_{1}\\}$. Hence,
there exist $c_{1},c_{2}\in\mathbb{R}$ such that

$\displaystyle u(x)=\begin{cases}\langle
x,\nu\rangle+c_{1}&x\in\mathrm{span}_{\mathbb{Z}}\left\\{e_{1},e_{2},e_{3}\right\\}\,;\\\
\langle x_{0},\nu\rangle+c_{2}&x=x_{0}+v_{1},\text{ with
}x_{0}\in\mathrm{span}_{\mathbb{Z}}\left\\{e_{1},e_{2},e_{3}\right\\}\,.\end{cases}$

Setting $c_{2}-c_{1}=t$, recalling (11) and (12), we therefore obtain

$\displaystyle F_{\mathcal{L}_{\mathrm{HCP}}}(u,T_{{\mathrm{HCP}}})$
$\displaystyle=2\left(|\langle e_{1},\nu\rangle|+|\langle
e_{2},\nu\rangle|+|\langle e_{1}-e_{2},\nu\rangle|\right)+|t|+|t-\langle
e_{1},\nu\rangle|+|t-\langle e_{2},\nu\rangle|$ $\displaystyle\quad+|t-\langle
e_{3},\nu\rangle|+|t-\langle e_{3}+e_{1},\nu\rangle|+|t-\langle
e_{3}+e_{2},\nu\rangle|\,.$

Employing Proposition 2.1 and (33), we have

$\displaystyle\varphi_{\mathrm{HCP}}(\nu)=\sqrt{2}\left(|\langle
e_{1},\nu\rangle|+|\langle e_{2},\nu\rangle|+|\langle
e_{1}-e_{2},\nu\rangle|\right)+\frac{1}{\sqrt{2}}\min_{t\in\mathbb{R}}g_{\nu}(t)\,,$
(34)

where

$\displaystyle g_{\nu}(t):=|t|+|t-\langle e_{1},\nu\rangle|+|t-\langle
e_{2},\nu\rangle|+|t-\langle e_{3},\nu\rangle|+|t-\langle
e_{3}+e_{1},\nu\rangle|+|t-\langle e_{3}+e_{2},\nu\rangle|\,.$

Next, we show that

$\displaystyle\min_{t\in\mathbb{R}}g_{\nu}(t)=|\langle
e_{3},\nu\rangle|+2\max\\{|\langle e_{1},\nu\rangle|,\,|\langle
e_{2},\nu\rangle|,\,|\langle e_{1}-e_{2},\nu\rangle|,\,|\langle
e_{3},\nu\rangle|\\}\,.$ (35)

Note that if (35) is shown, (26) is proven and Step 1 is concluded. In order
to prove (35), we first note that $g_{\nu}(t)$ is a piecewise affine function
such that $g_{\nu}(t)\to+\infty$ as $|t|\to+\infty$. Hence, it attains its
minimum at a point of non-differentiability. The function $g_{\nu}$ is not
differentiable for $t\in\\{0,\langle e_{1},\nu\rangle,\langle
e_{2},\nu\rangle,\langle e_{3},\nu\rangle,\langle
e_{3}+e_{1},\nu\rangle,\langle e_{3}+e_{2},\nu\rangle\\}$ and therefore

$\displaystyle\min_{t\in\mathbb{R}}g_{\nu}(t)=|\langle
e_{3},\nu\rangle|+\min\\{f_{k}(\nu)\colon k\in\\{0,\ldots,5\\}\\}\,,$

where

$\displaystyle f_{0}(\nu)$ $\displaystyle=|\langle e_{1},\nu\rangle|+|\langle
e_{2},\nu\rangle|+|\langle e_{3}+e_{1},\nu\rangle|+|\langle
e_{3}+e_{2},\nu\rangle|\,,$ $\displaystyle f_{1}(\nu)$ $\displaystyle=|\langle
e_{1},\nu\rangle|+|\langle e_{1}-e_{2},\nu\rangle|+|\langle
e_{3}-e_{1},\nu\rangle|+|\langle e_{3}+e_{2}-e_{1},\nu\rangle|\,,$
$\displaystyle f_{2}(\nu)$ $\displaystyle=|\langle e_{2},\nu\rangle|+|\langle
e_{1}-e_{2},\nu\rangle|+|\langle e_{3}-e_{2},\nu\rangle|+|\langle
e_{3}+e_{1}-e_{2},\nu\rangle|\,,$ $\displaystyle f_{3}(\nu)$
$\displaystyle=|\langle e_{1},\nu\rangle|+|\langle e_{2},\nu\rangle|+|\langle
e_{3}-e_{1},\nu\rangle|+|\langle e_{3}-e_{2},\nu\rangle|\,,$ $\displaystyle
f_{4}(\nu)$ $\displaystyle=|\langle e_{1},\nu\rangle|+|\langle
e_{1}-e_{2},\nu\rangle|+|\langle e_{3}+e_{1},\nu\rangle|+|\langle
e_{3}+e_{1}-e_{2},\nu\rangle|\,,$ $\displaystyle f_{5}(\nu)$
$\displaystyle=|\langle e_{2},\nu\rangle|+|\langle
e_{1}-e_{2},\nu\rangle|+|\langle e_{3}+e_{2},\nu\rangle|+|\langle
e_{3}+e_{2}-e_{1},\nu\rangle|\,.$

It is easy to see that

$\displaystyle\min_{t\in\mathbb{R}}g_{\nu}(t)=|\langle
e_{3},\nu\rangle|+\min\big{\\{}|\langle e_{1},R_{k}\nu\rangle|$
$\displaystyle+|\langle e_{2},R_{k}\nu\rangle|+|\langle
e_{3}+e_{1},R_{k}\nu\rangle|$ $\displaystyle+|\langle
e_{3}+e_{2},R_{k}\nu\rangle|\colon k\in\\{0,\ldots,5\\}\big{\\}}\,,$

where $R_{k}$ is the rotation of angle $k\pi/3$ around the $x_{3}$-axis.
Noting also that
$\min_{t\in\mathbb{R}}g_{\nu}(t)=\min_{t\in\mathbb{R}}g_{-\nu}(t)$, it is not
restrictive to assume that $\langle e_{1},\nu\rangle\geq 0,\langle
e_{2},\nu\rangle\geq 0,\langle e_{3},\nu\rangle\geq 0$. We only consider the
case, where $\langle e_{1},\nu\rangle\geq\langle e_{2},\nu\rangle\geq\langle
e_{3},\nu\rangle\geq 0$, the other being dealt with in a similar fashion. In
this case we have $f_{0}(\nu)\geq f_{3}(\nu)$, $f_{4}(\nu)\geq f_{5}(\nu)$,
and

$\displaystyle f_{1}(\nu)$ $\displaystyle=\langle e_{1},\nu\rangle+\langle
e_{1}-e_{2},\nu\rangle+\langle e_{1}-e_{3},\nu\rangle+|\langle
e_{3}+e_{2}-e_{1},\nu\rangle|$ $\displaystyle=2\langle
e_{1},\nu\rangle+\langle e_{1}-e_{2}-e_{3},\nu\rangle+|\langle
e_{3}+e_{2}-e_{1},\nu\rangle|\geq 2\langle e_{1},\nu\rangle\,;$ $\displaystyle
f_{2}(\nu)$ $\displaystyle=\langle e_{2},\nu\rangle+\langle
e_{1}-e_{2},\nu\rangle+\langle e_{2}-e_{3},\nu\rangle+\langle
e_{3}+e_{1}-e_{2},\nu\rangle=2\langle e_{1},\nu\rangle\,;$ $\displaystyle
f_{3}(\nu)$ $\displaystyle=\langle e_{1},\nu\rangle+\langle
e_{2},\nu\rangle+\langle e_{1}-e_{3},\nu\rangle+\langle
e_{2}-e_{3},\nu\rangle=2\langle e_{1},\nu\rangle+2\langle
e_{2},\nu\rangle-2\langle e_{3},\nu\rangle\geq 2\langle e_{1},\nu\rangle\,;$
$\displaystyle f_{5}(\nu)$ $\displaystyle=\langle e_{2},\nu\rangle+\langle
e_{1}-e_{2},\nu\rangle+\langle e_{3}+e_{2},\nu\rangle+|\langle
e_{3}+e_{2}-e_{1},\nu\rangle|$ $\displaystyle=2\langle
e_{1},\nu\rangle+\langle e_{3}+e_{2}-e_{1},\nu\rangle+|\langle
e_{3}+e_{2}-e_{1},\nu\rangle|\geq 2\langle e_{1},\nu\rangle\,.$

Hence, we see that (35) holds true. This together with (34) establishes (26)
and concludes Step 1.
Step 2.(Calculation of $\varphi^{\circ}_{\mathrm{HCP}}$) In order to calculate
$\varphi^{\circ}_{\mathrm{HCP}}$, we exploit the symmetries of
$\varphi^{\circ}_{\mathrm{HCP}}$. Note that

$\displaystyle\varphi_{\mathrm{HCP}}\left(-\begin{pmatrix}\nu_{1}\\\
\nu_{2}\\\
\nu_{3}\end{pmatrix}\right)=\varphi_{\mathrm{HCP}}\left(\begin{pmatrix}\nu_{1}\\\
\nu_{2}\\\
-\nu_{3}\end{pmatrix}\right)=\varphi_{\mathrm{HCP}}\left(\begin{pmatrix}-\nu_{1}\\\
\nu_{2}\\\
\nu_{3}\end{pmatrix}\right)=\varphi_{\mathrm{HCP}}\left(\begin{pmatrix}\nu_{1}\\\
\nu_{2}\\\ \nu_{3}\end{pmatrix}\right)\,.$ (36)

Given $\zeta\in\mathbb{R}^{3}$ we can find $R=T_{1}^{\alpha_{1}}\circ
T_{2}^{\alpha_{2}}\circ T_{3}^{\alpha_{3}}$, $\alpha_{i}\in\\{0,1\\}$ such
that $(R\zeta)_{i}\geq 0$ for all $i$. Thus,

$\displaystyle\begin{split}\varphi_{\mathrm{HCP}}^{\circ}(\zeta)&=\max_{\varphi_{\mathrm{HCP}}(\nu)\leq
1}\langle\nu,\zeta\rangle=\max_{\varphi_{\mathrm{HCP}}(\nu)\leq 1}\langle
R\nu,R\zeta\rangle\\\ &=\max_{\varphi_{\mathrm{HCP}}(R^{-1}\nu)\leq
1}\langle\nu,R\zeta\rangle=\max_{\varphi_{\mathrm{HCP}}(\nu)\leq
1}\langle\nu,R\zeta\rangle=\varphi_{\mathrm{HCP}}^{\circ}(R\zeta)\,.\end{split}$
(37)

It therefore suffices to calculate $\varphi_{\mathrm{HCP}}^{\circ}$ for
$\zeta\in\mathbb{R}^{3}$ such that $\zeta_{i}\geq 0$. This together with (36)
implies that if $\nu=(\nu_{1},\nu_{2},\nu_{3})$ is such that
$\varphi_{{\mathrm{HCP}}}(\nu)\leq 1$ and

$\displaystyle\langle\nu,\zeta\rangle=\max_{\varphi_{\mathrm{HCP}}(\nu)\leq
1}\langle\nu,\zeta\rangle\,,$

then $\nu_{i}\geq 0$ for all $i$. Additionally, a maximizer $\nu$ can be
chosen such that $\varphi_{\mathrm{HCP}}$ is not differentiable at $\nu$.
Therefore, there are the following cases to consider:

  * (a)

$\langle e_{1}-e_{2},\nu\rangle=0$;

  * (b)

$\langle e_{1}-e_{3},\nu\rangle=0,\langle e_{1}-e_{2},\nu\rangle\geq 0$;

  * (c)

$\langle e_{2}-e_{3},\nu\rangle=0,\langle e_{3}-e_{1},\nu\rangle\geq 0$;

  * (d)

$\langle e_{3},\nu\rangle=0$;

  * (e)

$\langle e_{1},\nu\rangle=0$.

Here, we point out that the points on the boundary $\nu_{2}=0$ are excluded as
possible maximum points by arguing in the following way: If there were a point
$\nu$ such that $\nu_{2}=0$, then $\varphi_{\mathrm{HCP}}(\nu)$ would either
be differentiable and thus $\nu$ would not be a maximum point or $\nu$ would
satisfy one of the cases (a)-(e).
_Maximum of case(a)._ Since $\langle e_{1}-e_{2},\nu\rangle=0$, we have
$\nu_{1}=\sqrt{3}\nu_{2}$. Hence, $\nu=(t,\frac{1}{\sqrt{3}}t,s)$ for some
$t,s\geq 0$. Now, using (26), we have

$\displaystyle\varphi_{\mathrm{HCP}}(\nu)=\sqrt{2}\left(2\nu_{1}+\frac{1}{3}\sqrt{6}\nu_{3}+\max\\{\nu_{1},\frac{2}{3}\sqrt{6}\nu_{3}\\}\right)\,.$

(a.1) $t\geq\frac{2}{3}\sqrt{6}s$: Since the maximum is attained for
$\varphi_{\mathrm{HCP}}(\nu)=1$, we have
$t=\frac{1}{3\sqrt{2}}-\frac{1}{9}\sqrt{6}s$. Now, $t\geq 0$ together with
$t\geq\frac{2}{3}\sqrt{6}s$ implies $0\leq s\leq\frac{3}{14\sqrt{3}}$. Noting
that

$\displaystyle\langle\nu,\zeta\rangle=t\left(\zeta_{1}+\frac{1}{\sqrt{3}}\zeta_{2}\right)+s\zeta_{3}=\left(\frac{1}{3\sqrt{2}}-\frac{1}{9}\sqrt{6}s\right)\left(\zeta_{1}+\frac{1}{\sqrt{3}}\zeta_{2}\right)+s\zeta_{3}\,,$

we obtain

$\displaystyle\max_{\nu\text{ sat.
(a.1)}}\langle\nu,\zeta\rangle=\max\left\\{\frac{2}{7\sqrt{2}}\left(\zeta_{1}+\frac{1}{\sqrt{3}}\zeta_{2}+\frac{3}{2\sqrt{6}}\zeta_{3}\right),\frac{1}{3\sqrt{2}}\left(\zeta_{1}+\frac{1}{\sqrt{3}}\zeta_{2}\right)\right\\}\,.$
(38)

(a.2) $t\leq\frac{2}{3}\sqrt{6}s$: Using $\varphi_{\mathrm{HCP}}(\nu)=1$, we
obtain $t=\frac{1}{2\sqrt{2}}-\frac{\sqrt{6}}{2}s$. Now, $t\geq 0$ together
with $t\leq\frac{2}{3}\sqrt{6}s$ implies $\frac{3}{14\sqrt{3}}\leq
s\leq\frac{1}{2\sqrt{3}}$. Noting that

$\displaystyle\langle\nu,\zeta\rangle=t\left(\zeta_{1}+\frac{1}{\sqrt{3}}\zeta_{2}\right)+s\zeta_{3}=\left(\frac{1}{3\sqrt{2}}-\frac{1}{9}\sqrt{6}s\right)\left(\zeta_{1}+\frac{1}{\sqrt{3}}\zeta_{2}\right)+s\zeta_{3}\,,$

we obtain

$\displaystyle\max_{\nu\text{ sat.
}(a.2)}\langle\nu,\zeta\rangle=\max\left\\{\frac{2}{7\sqrt{2}}\left(\zeta_{1}+\frac{1}{\sqrt{3}}\zeta_{2}+\frac{3}{2\sqrt{6}}\zeta_{3}\right),\frac{1}{2\sqrt{3}}\zeta_{3}\right\\}\,.$
(39)

_Maximum of case(b)._ Since $\langle e_{1}-e_{3},\nu\rangle=0$, we have
$\nu_{1}=\frac{2}{3}\sqrt{6}\nu_{3}$. Hence, $\nu=(t,s,\frac{3}{2\sqrt{6}}t)$
for some $t,s\geq 0$. Now using (26), we have

$\displaystyle\varphi_{\mathrm{HCP}}(\nu)=\frac{7}{2}\sqrt{2}\nu_{1}\,.$

Hence, since the maximum is attained for $\varphi_{\mathrm{HCP}}(\nu)=1$, we
have $\nu_{1}=\frac{2}{7\sqrt{2}}$. Additionally, since $\langle
e_{1}-e_{2},\nu\rangle\geq 0$, we have $\nu_{2}\leq\frac{2}{7\sqrt{6}}$, and
due to the form of $\nu$, we have $\nu_{3}=\frac{3}{14\sqrt{3}}$. This implies

$\displaystyle\max_{\nu\text{ sat.
(b)}}\langle\nu,\zeta\rangle=\frac{2}{7\sqrt{2}}\left(\zeta_{1}+\frac{1}{\sqrt{3}}\zeta_{2}+\frac{3}{2\sqrt{6}}\zeta_{3}\right)\,.$
(40)

_Maximum of case(c)._ Since $\langle e_{2}-e_{3},\nu\rangle=0$, we have
$\frac{1}{2}\nu_{1}+\frac{1}{2}\sqrt{3}\nu_{2}=\frac{2}{3}\sqrt{6}\nu_{3}$.
Now using (26), we have

$\displaystyle\varphi_{\mathrm{HCP}}(\nu)=\frac{7}{2}\sqrt{2}\langle
e_{3},\nu\rangle=\frac{14}{3}\sqrt{3}\nu_{3}\,.$

Hence, since the maximum is attained for $\varphi_{\mathrm{HCP}}(\nu)=1$, we
have $\nu_{3}=\frac{3}{14\sqrt{3}}$. Additionally, since $\langle
e_{3}-e_{1},\nu\rangle\geq 0$, we have $\nu_{1}\leq\frac{2}{7\sqrt{2}}$. Due
to the form of $\langle e_{2}-e_{3},\nu\rangle=0$, we have
$\nu_{2}=\frac{4}{7\sqrt{6}}-\frac{1}{\sqrt{3}}\nu_{1}$. Note that
$\nu_{2}\geq 0$ for all $0\leq\nu_{1}\leq\frac{2}{7\sqrt{2}}$. Therefore

$\displaystyle\langle\nu,\zeta\rangle=\nu_{1}\zeta_{1}+\left(\frac{4}{7\sqrt{6}}-\frac{1}{\sqrt{3}}\nu_{1}\right)\zeta_{2}+\frac{3}{14\sqrt{3}}\zeta_{3}\,.$

This implies

$\displaystyle\max_{\nu\text{ sat.
(c)}}\langle\nu,\zeta\rangle=\max\left\\{\frac{4}{7\sqrt{6}}\zeta_{2}+\frac{3}{14\sqrt{3}}\zeta_{3},\frac{2}{7\sqrt{2}}\left(\zeta_{1}+\frac{1}{\sqrt{3}}\zeta_{2}+\frac{3}{2\sqrt{6}}\zeta_{3}\right)\right\\}\,.$
(41)

_Maximum of case(d)._ We have $\nu_{3}=0$ and therefore

$\displaystyle\varphi_{\mathrm{HCP}}(\nu)=\sqrt{2}\left(\langle
e_{1},\nu\rangle+\langle e_{2},\nu\rangle+|\langle
e_{1}-e_{2},\nu\rangle|+\max\\{\langle e_{1},\nu\rangle,\langle
e_{2},\nu\rangle\\}\right)$

We distinguish to cases

  * (d.1)

$\langle e_{1}-e_{2},\nu\rangle\geq 0$;

  * (d.2)

$\langle e_{1}-e_{2},\nu\rangle\leq 0$.

_Maximum of case(d.1)._ In the case $\langle e_{1}-e_{2},\nu\rangle\geq 0$ we
have $\varphi_{\mathrm{HCP}}(\nu)=3\sqrt{2}\nu_{1}$ and therefore, since
$\varphi_{\mathrm{HCP}}(\nu)=1$, $\nu_{1}=\frac{1}{3\sqrt{2}}$. The inequality
$\langle e_{1}-e_{2},\nu\rangle\geq 0$ implies that
$0\leq\nu_{2}\leq\frac{1}{\sqrt{3}}\nu_{1}=\frac{1}{3\sqrt{6}}$. Hence,

$\displaystyle\max_{\nu\text{ sat.
(d.1)}}\langle\nu,\zeta\rangle=\frac{1}{3\sqrt{2}}\left(\zeta_{1}+\frac{1}{\sqrt{3}}\zeta_{2}\right)\,.$
(42)

_Maximum of case(d.2)._ In the case $\langle e_{1}-e_{2},\nu\rangle\leq 0$ we
have

$\displaystyle\varphi_{\mathrm{HCP}}(\nu)=3\sqrt{2}\langle
e_{2},\nu\rangle=\sqrt{2}\left(\frac{3}{2}\nu_{1}+\frac{3}{2}\sqrt{3}\nu_{2}\right)\,.$

This, together with $\varphi_{\mathrm{HCP}}(\nu)=1$, implies,
$\nu_{1}=\frac{2}{3\sqrt{2}}-\sqrt{3}\nu_{2}$ and therefore
$\nu_{2}\leq\frac{2}{3\sqrt{6}}$. Additionally, since $\langle
e_{1}-e_{2}\rangle\leq 0$, we have $\frac{1}{3\sqrt{6}}\leq\nu_{2}$.
Therefore,

$\displaystyle\langle\nu,\zeta\rangle=\nu_{1}\zeta_{1}+\nu_{2}\zeta_{2}=\left(\frac{3}{2\sqrt{2}}-\sqrt{3}\nu_{2}\right)\zeta_{1}+\nu_{2}\zeta_{2}\,.$

This implies

$\displaystyle\max_{\nu\text{ sat.
(d.2)}}\langle\nu,\zeta\rangle=\max\left\\{\frac{2}{3\sqrt{6}}\zeta_{2},\frac{1}{3\sqrt{2}}\left(\zeta_{1}+\frac{1}{\sqrt{3}}\zeta_{2}\right)\right\\}\,.$
(43)

_Maximum of case(e)._ In the case $\nu_{1}=0$ we have

$\displaystyle\varphi_{\mathrm{HCP}}(\nu)=\sqrt{2}\left(\sqrt{3}\nu_{2}+\frac{1}{3}\sqrt{6}\nu_{3}+\max\left\\{\frac{\sqrt{3}}{2}\nu_{2},\frac{2}{3}\sqrt{6}\nu_{3}\right\\}\right)\,.$

We distinguish between two cases:

  * (e.1)

$\langle e_{2},\nu\rangle\geq\langle e_{3},\nu\rangle$;

  * (e.2)

$\langle e_{2},\nu\rangle\leq\langle e_{3},\nu\rangle$.

_Maximum of case(e.1)._ In this case, we have

$\displaystyle\varphi_{\mathrm{HCP}}(\nu)=\sqrt{2}\left(\frac{3}{2}\sqrt{3}\nu_{2}+\frac{1}{3}\sqrt{6}\nu_{3}\right)\,.$

Therefore, since $\varphi_{\mathrm{HCP}}(\nu)=1$, we have
$\nu_{2}=\frac{2}{3\sqrt{6}}-\frac{2}{9}\sqrt{2}\nu_{3}$. Hence,
$\nu_{3}\leq\frac{3}{2\sqrt{3}}$. Additionally, since $\langle
e_{2}-e_{3},\nu\rangle\geq 0$, we have $\nu_{3}\leq\frac{3}{14\sqrt{3}}$.
Therefore,

$\displaystyle\langle\nu,\zeta\rangle=\nu_{2}\zeta_{2}+\nu_{3}\zeta_{3}=\left(\frac{2}{3\sqrt{6}}-\frac{2}{9}\sqrt{2}\nu_{3}\right)\zeta_{2}+\nu_{3}\zeta_{3}\,.$

Hence,

$\displaystyle\max_{\nu\text{ sat.
(e.1)}}\langle\nu,\zeta\rangle=\max\left\\{\frac{2}{3\sqrt{6}}\zeta_{2},\frac{4}{7\sqrt{6}}\zeta_{2}+\frac{3}{14\sqrt{3}}\zeta_{3}\right\\}\,.$
(44)

_Maximum of case(e.2)._ In this case, we have

$\displaystyle\varphi_{\mathrm{HCP}}(\nu)=\sqrt{2}\left(\sqrt{3}\nu_{2}+\sqrt{6}\nu_{3}\right)\,.$

Therefore, since $\varphi_{\mathrm{HCP}}(\nu)=1$, we have
$\nu_{2}=\frac{1}{\sqrt{6}}-\sqrt{2}\nu_{3}$. Hence,
$\nu_{3}\leq\frac{1}{2\sqrt{3}}$. Additionally, since $\langle
e_{2}-e_{3},\nu\rangle\leq 0$, we have $\nu_{3}\geq\frac{3}{14\sqrt{3}}$.
Therefore,

$\displaystyle\langle\nu,\zeta\rangle=\nu_{2}\zeta_{2}+\nu_{3}\zeta_{3}=\left(\frac{1}{3\sqrt{6}}-\sqrt{2}\nu_{3}\right)\zeta_{2}+\nu_{3}\zeta_{3}\,.$

Hence,

$\displaystyle\max_{\nu\text{ sat.
(e.1)}}\langle\nu,\zeta\rangle=\max\left\\{\frac{1}{2\sqrt{3}}\zeta_{3},\frac{4}{7\sqrt{6}}\zeta_{2}+\frac{3}{14\sqrt{3}}\zeta_{3}\right\\}\,.$
(45)

Exploiting (38)–(45), and (37), we obtain (27). This concludes Step 2.
Step 3.(Calculation of (28)) In order to calculate (28), we split the
calculation of
$\partial^{*}W_{\mathrm{HCP}}=\\{\varphi^{\circ}_{\mathrm{HCP}}(\zeta)=1\\}$
into different sets, where the maximum of $\varphi_{\mathrm{HCP}}^{\circ}$ is
attained. We consider the following cases

  * (a)

$A_{a}:=\\{\zeta\in\mathbb{R}^{3}\colon\varphi_{\mathrm{HCP}}^{\circ}(\zeta)=\frac{1}{2\sqrt{3}}|\zeta_{3}|=1\\}$;

  * (b)

$A_{b}:=\\{\zeta\in\mathbb{R}^{3}\colon\varphi_{\mathrm{HCP}}^{\circ}(\zeta)=\frac{2}{3\sqrt{6}}|\zeta_{2}|=1\\}$;

  * (c)

$A_{c}:=\\{\zeta\in\mathbb{R}^{3}\colon\varphi_{\mathrm{HCP}}^{\circ}(\zeta)=\frac{4}{7\sqrt{6}}|\zeta_{2}|+\frac{3}{14\sqrt{3}}|\zeta_{3}|=1\\}$;

  * (d)

$A_{d}:=\\{\zeta\in\mathbb{R}^{3}\colon\varphi_{\mathrm{HCP}}^{\circ}(\zeta)=\frac{1}{3\sqrt{2}}(|\zeta_{1}|+\frac{1}{\sqrt{3}}|\zeta_{2}|)=1\\}$;

  * (e)

$A_{e}:=\\{\zeta\in\mathbb{R}^{3}\colon\varphi_{\mathrm{HCP}}^{\circ}(\zeta)=\frac{2}{7\sqrt{2}}(|\zeta_{1}|+\frac{1}{\sqrt{3}}|\zeta_{2}|+\frac{3}{2\sqrt{6}}|\zeta_{3}|)=1\\}$.

In each of the cases, one can determine the area, shape and normal of the set,
by invoking the condition that the maximum is attained
$\varphi^{\circ}_{\mathrm{HCP}}$ for the respective function and therefore all
the other functions $f$ in the definition of $\varphi^{\circ}_{\mathrm{HCP}}$
satisfy $f\leq 1$. In the following, we only collect the results, since the
calculations are elementary (but very long).

_Calculations for case(a)._ In this case, we see that $\nu=(0,0,\pm 1)$
$\mathcal{H}^{2}$-a.e., since this set is contained in the level set of the
function $|\zeta_{3}|=c$ for some $c>0$. Additionally, we see that the set is
a union of two hexagons of side length $2\sqrt{2}$. Therefore, for each of the
two hexagons $H_{i}$ we have $\mathcal{H}^{2}(H_{i})=12\sqrt{3}$. Furthermore,
$\varphi_{\mathrm{HCP}}(\nu)=2\sqrt{3}$. Hence

$\displaystyle\int_{A_{a}}\varphi_{\mathrm{HCP}}(\nu)\,\mathrm{d}\mathcal{H}^{2}=2\cdot
12\sqrt{3}\cdot 2\sqrt{3}=2^{4}\cdot 3^{2}\,.$ (46)

_Calculations for case(b)._ In this case, we see that $\nu=(0,\pm 1,0)$
$\mathcal{H}^{2}$-a.e., since this set is contained in the level set of the
function $|\zeta_{2}|=c$ for some $c>0$. Additionally, we see that the set is
a union of two rectangles with side lengths $3\sqrt{2}$ and
$\frac{4}{3}\sqrt{3}$. Therefore, for each of the two rectangles $S_{i}$ we
have $\mathcal{H}^{2}(S_{i})=4\sqrt{6}$. Furthermore,
$\varphi_{\mathrm{HCP}}(\nu)=\frac{3}{2}\sqrt{6}$. Hence

$\displaystyle\int_{A_{b}}\varphi_{\mathrm{HCP}}(\nu)\,\mathrm{d}\mathcal{H}^{2}=2\cdot
4\sqrt{6}\cdot\frac{3}{2}\sqrt{6}=2^{3}\cdot 3^{2}\,.$ (47)

_Calculations for case(c)._ In this case, we see that $\nu=(3/41)^{1/2}(0,\pm
8/\sqrt{6},\pm\sqrt{3})$ $\mathcal{H}^{2}$-a.e., since this set is contained
in the level set of the function
$\frac{4}{7\sqrt{6}}|\zeta_{2}|+\frac{3}{14\sqrt{3}}|\zeta_{3}|=c$ for some
$c>0$. Additionally, we see that the set is a union of four trapezoids with
height $(41/6)^{1/2}$ and two parallel sides o lengths $3\sqrt{2}$ and
$2\sqrt{2}$. Therefore, for each of the four trapezoids $T_{i}$ we have
$\mathcal{H}^{2}(T_{i})=\frac{5}{2}(\frac{41}{3})^{1/2}$. Furthermore,
$\varphi_{\mathrm{HCP}}(\nu)=14(\frac{3}{41})^{1/2}$. Hence

$\displaystyle\int_{A_{c}}\varphi_{\mathrm{HCP}}(\nu)\,\mathrm{d}\mathcal{H}^{2}=4\cdot\frac{5}{2}\left(\frac{41}{3}\right)^{1/2}\cdot
14\left(\frac{3}{41}\right)^{1/2}=2^{2}\cdot 5\cdot 7\,.$ (48)

_Calculations for case(d)._ In this case, we see that
$\nu=\frac{1}{2}(\pm\sqrt{3},\pm 1,0)$ $\mathcal{H}^{2}$-a.e., since this set
is contained in the level set of the function
$|\zeta_{1}|+\frac{1}{\sqrt{3}}|\zeta_{2}|=c$ for some $c>0$. Additionally, we
see that the set is a union of four rectangles with side length $3\sqrt{2}$
and $\frac{4}{3}\sqrt{3}$. Therefore, for each of the four rectangles $R_{i}$
we have $\mathcal{H}^{2}(R_{i})=4\sqrt{6}$. Furthermore,
$\varphi_{\mathrm{HCP}}(\nu)=\frac{3}{2}\sqrt{6}$. Hence

$\displaystyle\int_{A_{d}}\varphi_{\mathrm{HCP}}(\nu)\,\mathrm{d}\mathcal{H}^{2}=4\cdot
4\sqrt{6}\cdot\frac{3}{2}\sqrt{6}=2^{4}\cdot 3^{2}\,.$ (49)

_Calculations for case(e)._ In this case, we see that $\nu=2(6/41)^{1/2}(\pm
1,\pm\frac{1}{\sqrt{3}},\frac{3}{2\sqrt{6}})$ $\mathcal{H}^{2}$-a.e., since
this set is contained in the level set of the function
$|\zeta_{1}|+\frac{1}{\sqrt{3}}|\zeta_{2}|+\frac{3}{2\sqrt{6}}|\zeta_{3}|=c$
for some $c>0$. Additionally, we see that the set is a union of eight
trapezoids with height $(41/6)^{1/2}$ and two parallel sides o lengths
$3\sqrt{2}$ and $2\sqrt{2}$. Therefore, for each of the eight trapezoids
$Z_{i}$ we have $\mathcal{H}^{2}(Z_{i})=\frac{5}{2}(\frac{41}{3})^{1/2}$.
Furthermore, $\varphi_{\mathrm{HCP}}(\nu)=14(\frac{3}{41})^{1/2}$. Hence

$\displaystyle\int_{A_{e}}\varphi_{\mathrm{HCP}}(\nu)\,\mathrm{d}\mathcal{H}^{2}=8\cdot\frac{5}{2}\left(\frac{41}{3}\right)^{1/2}\cdot
14\left(\frac{3}{41}\right)^{1/2}=2^{3}\cdot 5\cdot 7\,.$ (50)

Taking into account (46)–(50), we obtain

$\displaystyle\int_{\partial^{*}W_{\mathrm{HCP}}}\varphi_{\mathrm{HCP}}(\nu)\,\mathrm{d}\mathcal{H}^{2}=2^{5}\cdot
3^{2}+2^{3}\cdot 3^{2}+2^{2}\cdot 5\cdot 7+2^{4}\cdot 3^{2}+2^{3}\cdot 5\cdot
7=780\,.$ (51)

Next, we need to calculate $|W_{\mathrm{HCP}}|$, since
$W_{\mathrm{HCP}}=\\{\varphi_{\mathrm{HCP}}^{\circ}\leq 1\\}\cap(C_{a}\cup
C_{b}\cup C_{c}\cup C_{d}\cup C_{e})$, where

$\displaystyle
C_{a}=\\{\zeta\in\mathbb{R}^{3}\colon\varphi^{\mathrm{\circ}}_{\mathrm{HCP}}(\zeta)=\frac{1}{2\sqrt{3}}|\zeta_{3}|\\}\,,$
$\displaystyle
C_{b}:=\\{\zeta\in\mathbb{R}^{3}\colon\varphi_{\mathrm{HCP}}^{\circ}(\zeta)=\frac{2}{3\sqrt{6}}|\zeta_{2}|\\}\,,$
$\displaystyle
C_{c}:=\\{\zeta\in\mathbb{R}^{3}\colon\varphi_{\mathrm{HCP}}^{\circ}(\zeta)=\frac{4}{7\sqrt{6}}|\zeta_{2}|+\frac{3}{14\sqrt{3}}|\zeta_{3}|\\}\,,$
$\displaystyle
C_{d}:=\\{\zeta\in\mathbb{R}^{3}\colon\varphi_{\mathrm{HCP}}^{\circ}(\zeta)=\frac{1}{3\sqrt{2}}(|\zeta_{1}|+\frac{1}{\sqrt{3}}|\zeta_{2}|)\\}\,,$
$\displaystyle
C_{e}:=\\{\zeta\in\mathbb{R}^{3}\colon\varphi_{\mathrm{HCP}}^{\circ}(\zeta)=\frac{2}{7\sqrt{2}}(|\zeta_{1}|+\frac{1}{\sqrt{3}}|\zeta_{2}|+\frac{3}{2\sqrt{6}}|\zeta_{3}|)\\}\,.$

Note that
$\mathcal{H}^{2}(C_{\alpha}\cap\\{\varphi_{\mathrm{HCP}}^{\circ}(\zeta)=s\\})=s^{2}\mathcal{H}^{2}(A_{\alpha})$
for all $\alpha\in\\{a,b,c,d,e\\}$. In the set $C_{a}$ we have that
$|\nabla\varphi_{\mathrm{HCP}}^{\circ}(\zeta)|=\frac{1}{2\sqrt{3}}$
$\mathcal{L}^{3}$-a.e.. Due to the coarea formula, we have

$\displaystyle\begin{split}|C_{a}\cap
W_{\mathrm{HCP}}|&=2\sqrt{3}\int_{C_{a}\cap
W_{\mathrm{HCP}}}|\nabla\varphi_{\mathrm{HCP}}^{\circ}(\zeta)|\,\mathrm{d}\zeta\\\
&=2\sqrt{3}\int_{0}^{1}\mathcal{H}^{2}(C_{a}\cap\\{\varphi_{\mathrm{HCP}}^{\circ}(\zeta)=s\\})\,\mathrm{d}s=\frac{2}{3}\sqrt{3}\mathcal{H}^{2}(A_{a})=2^{4}\cdot
3\,.\end{split}$ (52)

In the set $C_{b}$, we have that
$|\nabla\varphi_{\mathrm{HCP}}^{\circ}(\zeta)|=\frac{2}{3\sqrt{6}}$
$\mathcal{L}^{3}$-a.e.. Due to the coarea formula, we have

$\displaystyle\begin{split}|C_{b}\cap
W_{{\mathrm{HCP}}}|&=\frac{3}{2}\sqrt{6}\int_{C_{b}\cap
W_{\mathrm{HCP}}}|\nabla\varphi_{\mathrm{HCP}}^{\circ}(\zeta)|\,\mathrm{d}\zeta\\\
&=\frac{3}{2}\sqrt{6}\int_{0}^{1}\mathcal{H}^{2}(C_{b}\cap\\{\varphi_{\mathrm{HCP}}^{\circ}(\zeta)=s\\})\,\mathrm{d}s=\frac{1}{2}\sqrt{6}\mathcal{H}^{2}(A_{b})=2^{3}\cdot
3\,.\end{split}$ (53)

In the set $C_{c}$, we have that
$|\nabla\varphi_{\mathrm{HCP}}^{\circ}(\zeta)|=\frac{1}{14}(41/3)^{1/2}$
$\mathcal{L}^{3}$-a.e.. Due to the coarea formula, we have

$\displaystyle\begin{split}|C_{c}\cap
W_{\mathrm{HCP}}|&=14\left(\frac{3}{41}\right)^{1/2}\int_{C_{c}\cap
W_{\mathrm{HCP}}}|\nabla\varphi_{\mathrm{HCP}}^{\circ}(\zeta)|\,\mathrm{d}\zeta\\\
&=14\left(\frac{3}{41}\right)^{1/2}\int_{0}^{1}\mathcal{H}^{2}(C_{c}\cap\\{\varphi_{\mathrm{HCP}}^{\circ}(\zeta)=s\\})\,\mathrm{d}s\\\
&=\frac{1}{3}14\left(\frac{3}{41}\right)^{1/2}\mathcal{H}^{2}(A_{c})=\frac{2^{2}\cdot
5\cdot 7}{3}\,.\end{split}$ (54)

In the set $C_{d}$, we have that
$|\nabla\varphi_{\mathrm{HCP}}^{\circ}(\zeta)|=\frac{2}{3\sqrt{6}}$
$\mathcal{L}^{3}$-a.e.. Due to the coarea formula, we have

$\displaystyle\begin{split}|C_{d}\cap
W_{\mathrm{HCP}}|&=\frac{3}{2}\sqrt{6}\int_{C_{d}\cap
W_{\mathrm{HCP}}}|\nabla\varphi_{\mathrm{HCP}}^{\circ}(\zeta)|\,\mathrm{d}\zeta\\\
&=\frac{3}{2}\sqrt{6}\int_{0}^{1}\mathcal{H}^{2}(C_{d}\cap\\{\varphi_{\mathrm{HCP}}^{\circ}(\zeta)=s\\})\,\mathrm{d}s\\\
&=\frac{1}{2}\sqrt{6}\mathcal{H}^{2}(A_{d})=2^{4}\cdot 3\,.\end{split}$ (55)

In the set $C_{e}$, we have that
$|\nabla\varphi_{\mathrm{HCP}}^{\circ}(\zeta)|=\frac{1}{14}(41/3)^{1/2}$
$\mathcal{L}^{3}$-a.e.. Due to the coarea formula, we have

$\displaystyle\begin{split}|C_{e}\cap
W_{\mathrm{HCP}}|&=14\left(\frac{3}{41}\right)^{1/2}\int_{C_{e}\cap
W_{\mathrm{HCP}}}|\nabla\varphi_{\mathrm{HCP}}^{\circ}(\zeta)|\,\mathrm{d}\zeta\\\
&=14\left(\frac{3}{41}\right)^{1/2}\int_{0}^{1}\mathcal{H}^{2}(C_{e}\cap\\{\varphi_{\mathrm{HCP}}^{\circ}(\zeta)=s\\})\,\mathrm{d}s\\\
&=\frac{1}{3}14\left(\frac{3}{41}\right)^{1/2}\mathcal{H}^{2}(A_{e})=\frac{2^{3}\cdot
5\cdot 7}{3}\,.\end{split}$ (56)

Using (52)–(56), we obtain $|W_{\mathrm{HCP}}|=260$. This together with (51)
yields (28). ∎

##  4\. $\Gamma$-convergence analysis on the $\mathrm{FCC}$ and $\mathrm{HCP}$
lattices

In this section we prove Theorem 2.3. In order to prove the compactness
statement, we provide some preliminary lemmata about the shape of the Voronoi
cells of the $\mathrm{FCC}$-lattice as well as the $\mathrm{HCP}$-lattice (see
Figure 6). In what follows we use the notation
$\mathcal{N}_{\mathrm{FCC}}=\mathcal{N}_{{\mathcal{L}_{\mathrm{FCC}}}}(0)$ and
$\mathcal{N}_{\mathrm{HCP}}=\mathcal{N}_{{\mathcal{L}_{\mathrm{HCP}}}}(0)$.

Figure 6. Left: the Voronoi cell $V_{\mathrm{FCC}}$ of the ${\mathrm{FCC}}$
lattice. Right: the Voronoi cell $V_{\mathrm{HCP}}$ of the ${\mathrm{HCP}}$
lattice.

######  Lemma 4.1.

(Voronoi cell in the $\mathrm{FCC}$-lattice) Let us take
$x\in\mathcal{L}_{\mathrm{FCC}}$. Then

$\displaystyle{\mathcal{V}}_{\mathcal{L}_{\mathrm{FCC}}}(x)=x+V_{\mathrm{FCC}}\,,\quad\text{
where
}V_{\mathrm{FCC}}:=\left\\{y\in\mathbb{R}^{3}\colon\max_{b\in\mathcal{N}_{\mathrm{FCC}}}\langle
b,y\rangle\leq\frac{1}{2}\right\\}\,.$ (57)

Given $b_{0}\in\mathcal{N}_{\mathrm{FCC}}$ the face

$\displaystyle
S_{b_{0}}:=\left\\{y\in\mathbb{R}^{3}\colon\max_{b\in\mathcal{N}_{\mathrm{FCC}}}\langle
b,y\rangle=\langle b_{0},y\rangle=\frac{1}{2}\right\\}$ (58)

is a rhombus with $\mathcal{H}^{2}(S_{b_{0}})=\frac{1}{4}\sqrt{2}$. Moreover,
for each $b_{0}\in\mathcal{N}_{\mathrm{FCC}}$ the face $S_{b_{0}}$ of
${\mathcal{V}}_{\mathrm{FCC}}(0)$ is shared with the Voronoi cell
${\mathcal{V}}_{\mathrm{FCC}}(b_{0})$. Lastly, we have
$|{\mathcal{V}}_{\mathrm{FCC}}(x)|=\frac{1}{2}\sqrt{2}$ for all
$x\in\mathcal{L}_{\mathrm{FCC}}$.

######  Lemma 4.2.

(Voronoi cell in the $\mathrm{HCP}$-lattice) Let us take
$x\in\mathcal{L}_{\mathrm{HCP}}$. Then

$\displaystyle{\mathcal{V}}_{\mathcal{L}_{\mathrm{HCP}}}(x)=\begin{cases}x+V_{\mathrm{HCP}}&\text{if
}x\in\mathrm{span}_{\mathbb{Z}}\\{e_{1},e_{2},e_{3}\\}\,,\\\
x-V_{\mathrm{HCP}}&\text{if
}x\in(v_{1}+\mathrm{span}_{\mathbb{Z}}\\{e_{1},e_{2},e_{3}\\})\,,\end{cases}$
(59)

where

$\displaystyle
V_{\mathrm{HCP}}:=\left\\{y\in\mathbb{R}^{3}\colon\max_{b\in\mathcal{N}_{\mathrm{HCP}}}\langle
b,y\rangle\leq\frac{1}{2}\right\\}\,.$

For $b_{0}\in\mathcal{N}_{\mathrm{HCP}}$ we set

$\displaystyle
S_{b_{0}}:=\left\\{y\in\mathbb{R}^{3}\colon\max_{b\in\mathcal{N}_{\mathrm{HCP}}}\langle
b,y\rangle=\langle b_{0},y\rangle=\frac{1}{2}\right\\}\,.$ (60)

If $b_{0}\in\\{\pm e_{1},\pm e_{2},\pm(e_{1}-e_{2})\\}$ the face $S_{b_{0}}$
is a trapezoid of area $\frac{1}{4}\sqrt{2}$. If
$b_{0}\in\\{v_{1},v_{1}-e_{1},v_{1}-e_{2},v_{1}-e_{3},v_{1}-e_{1}-e_{3},v_{1}-e_{2}-e_{3}\\}$
the face $S_{b_{0}}$ is a rhombus of area $\frac{1}{8}\sqrt{6}$. Moreover, for
each $b_{0}\in\mathcal{N}_{\mathrm{HCP}}$ the face $S_{b_{0}}$ is shared with
the Voronoi cell $\mathcal{V}_{\mathcal{L}_{\mathrm{HCP}}}(b_{0})$. Lastly, we
have $|\mathcal{V}_{\mathcal{L}_{\mathrm{HCP}}}(x)|=\frac{1}{2}\sqrt{2}$ for
all $x\in\mathcal{L}_{\mathrm{HCP}}$.

###### Proof of Lemma 4.1.

We split the proof of the lemma into four steps. First, we prove (57). In the
second step, we show that each face is a rhombus and calculate its area.
Lastly, we show that each neighboring Voronoi cell $V_{\mathrm{FCC}}(b)$,
$b\in\mathcal{N}_{\mathrm{FCC}}$ shares one face with the Voronoi cell
$V_{\mathrm{FCC}}(0)$.
Step 1.(Proof of (57)) To check (57), since $\mathcal{L}_{\mathrm{FCC}}$ is a
Bravais-lattice (see (8)), it suffices to consider the case $x=0$. Let
$\mathcal{V}_{{\mathcal{L}_{\mathrm{FCC}}}}(0)$ denote the Voronoi cell of
$\mathcal{L}_{\mathrm{FCC}}$ at $x=0$ defined according to (13).
Step 1.1.($\mathcal{V}_{{\mathcal{L}_{\mathrm{FCC}}}}(0)\subset
V_{\mathrm{FCC}}$) Let $y\in\mathcal{V}_{{\mathcal{L}_{\mathrm{FCC}}}}(0)$. By
the very definition of Voronoi cell we have that for all
$b\in{\mathcal{N}_{\mathrm{FCC}}}$ it holds $|y|\leq|y-b|\,.$ Noting that
$|b|=1$ for all
$b\in\mathcal{N}_{\mathrm{FCC}}\subset\mathcal{L}_{\mathrm{FCC}}$, we have

$\displaystyle|y|\leq|y-b|\iff|y|^{2}\leq|y-b|^{2}=|y|^{2}-2\langle
b,y\rangle+|b|^{2}\iff\langle b,y\rangle\leq\frac{1}{2}\,,$

that is the inclusion $\mathcal{V}_{\mathcal{L}_{\mathrm{FCC}}}(0)\subset
V_{\mathrm{FCC}}$.
Step
1.2.($V_{\mathrm{FCC}}\subset\mathcal{V}_{\mathcal{L}_{\mathrm{FCC}}}(0)$) We
show that for $y\in V_{\mathrm{FCC}}$ we have $|y|\leq|y-z|$ for all
$z\in\mathcal{L}_{\mathrm{FCC}}$. This is equivalent to

$\displaystyle y\in V_{\mathrm{FCC}}\implies\langle
y,z\rangle\leq\tfrac{1}{2}|z|^{2}\quad\text{ for all
}z\in\mathcal{L}_{\mathrm{FCC}}\,.$ (61)

We first observe that if $z\in\mathcal{N}_{\mathrm{FCC}}$, (61) is trivial
since $|z|=1$. Next, we prove (61) for all
$z\in\mathcal{L}_{\mathrm{FCC}}\setminus\mathcal{N}_{\mathrm{FCC}}$. We
distinguish two cases:

  * (a)

$z=\lambda_{1}b_{j}+\lambda_{2}b_{k}$ $\lambda_{1},\lambda_{2}\in\mathbb{Z}$,
$j,k\in\\{1,2,3\\},j\neq k\,$;

  * (b)

$z=\lambda_{1}b_{1}+\lambda_{2}b_{2}+\lambda_{3}b_{3}$
$\lambda_{1},\lambda_{2},\lambda_{3}\in\mathbb{Z}\,$.

_Proof in case(a)._ We only show the statement for
$z=\lambda_{1}b_{1}+\lambda_{2}b_{2}$ for
$\lambda_{1},\lambda_{2}\in\mathbb{Z}$, the cases with any other combination
of two vectors being analogous. If $\lambda_{1}\lambda_{2}\geq 0$, since
$\langle b_{1},b_{2}\rangle\geq 0$, we have

$\displaystyle\langle y,z\rangle$ $\displaystyle=\langle
y,\lambda_{1}b_{1}+\lambda_{2}b_{2}\rangle\leq\frac{1}{2}|\lambda_{1}b_{1}|^{2}+\frac{1}{2}|\lambda_{2}b_{2}|^{2}$
$\displaystyle=\frac{1}{2}|\lambda_{1}b_{1}+\lambda_{2}b_{2}|^{2}-\lambda_{1}\lambda_{2}\langle
b_{1},b_{2}\rangle\leq\frac{1}{2}|\lambda_{1}b_{1}+\lambda_{2}b_{2}|^{2}=\frac{1}{2}|z|^{2}\,.$

On the other hand, if $\lambda_{1}\lambda_{2}\leq 0$ and without loss of
generality $|\lambda_{1}|\leq|\lambda_{2}|$, noting that
$b_{1}-b_{2}\in\mathcal{N}_{\mathrm{FCC}}$, we have

$\displaystyle\langle y,z\rangle$ $\displaystyle=\langle
y,\lambda_{1}b_{1}+\lambda_{2}b_{2}\rangle=\langle
y,(\lambda_{2}+\lambda_{1})b_{2}+\lambda_{1}(b_{1}-b_{2})\rangle\leq\frac{1}{2}|(\lambda_{2}+\lambda_{1})b_{2}|^{2}+\frac{1}{2}|\lambda_{1}(b_{1}-b_{2})|^{2}$
$\displaystyle=\frac{1}{2}|\lambda_{1}b_{1}+\lambda_{2}b_{2}|^{2}-\lambda_{1}(\lambda_{2}+\lambda_{1})\langle(b_{1}-b_{2}),b_{2}\rangle\leq\frac{1}{2}|\lambda_{1}b_{1}+\lambda_{2}b_{2}|^{2}\,.$

Here, the last inequality follows, since $|b_{1}|=|b_{2}|$ and therefore
$\lambda_{1}(\lambda_{2}+\lambda_{1})\langle(b_{1}-b_{2}),b_{2}\rangle\geq 0$.
This concludes case (a).
_Proof in case(b)._ We now show that (57) holds true in the case of
$b=\lambda_{1}b_{1}+\lambda_{2}b_{2}+\lambda_{3}b_{3}$ with
$\lambda_{i}\in\mathbb{Z}$. We restrict to the case $\lambda_{1}\geq
0,\lambda_{2}\geq 0$ and $\lambda_{3}\leq 0$, since if all $\lambda_{i}$ are
of the same sign, (57) can be deduced from the fact that it holds true for
$b\in\mathcal{N}_{\mathrm{FCC}}$ and the fact that $\langle
b_{j},b_{k}\rangle\geq 0$. Without loss of generality, we assume
$|\lambda_{2}|\leq|\lambda_{3}|$. Hence, observing that
$b_{2}-b_{3}\in\mathcal{N}_{\mathrm{FCC}}$, we have

$\displaystyle\langle y,z\rangle$ $\displaystyle=\langle
y,\lambda_{1}b_{1}+\lambda_{2}b_{2}+\lambda_{3}b_{3}\rangle=\langle
y,\lambda_{1}b_{1}+(\lambda_{3}+\lambda_{2})b_{3}+\lambda_{2}(b_{2}-b_{3})\rangle$
$\displaystyle\leq\frac{1}{2}|\lambda_{1}b_{1}+(\lambda_{3}+\lambda_{2})b_{3}|^{2}+\frac{1}{2}|\lambda_{2}(b_{2}-b_{3})|^{2}$
$\displaystyle=\frac{1}{2}|\lambda_{1}b_{1}+\lambda_{2}b_{2}+\lambda_{3}b_{3}|^{2}-\langle(\lambda_{1}b_{1}+(\lambda_{3}+\lambda_{2})b_{3}),\lambda_{2}(b_{2}-b_{3})\rangle$
$\displaystyle=\frac{1}{2}|\lambda_{1}b_{1}+\lambda_{2}b_{2}+\lambda_{3}b_{3}|^{2}-(\lambda_{3}+\lambda_{2})\lambda_{2}\langle
b_{2}-b_{3},b_{3}\rangle\leq\frac{1}{2}|\lambda_{1}b_{1}+\lambda_{2}b_{2}+\lambda_{3}b_{3}|^{2}\,.$

Here, the last inequality follows from $|b_{2}|=|b_{3}|$ and
$\lambda_{3}+\lambda_{2}\leq 0$ whereas the equality in the last line is due
to $\langle b_{1},b_{2}\rangle=\langle b_{1},b_{3}\rangle=\langle
b_{2},b_{3}\rangle$. This concludes case (b) and with that Step 1.2.
Step 2.(The faces of the Voronoi cell) To show that each face of the Voronoi
cell $V_{\mathrm{FCC}}$ is a rhombus with area $\frac{1}{4}\sqrt{2}$ we first
exploit its symmetries. Let $i\in\\{1,2,3\\}$ and let
$T_{i}\colon\mathbb{R}^{3}\to\mathbb{R}^{3}$ be the linear mapping that flips
the $i$-th entry, i.e.

$\displaystyle(T_{i}x)_{j}=\begin{cases}-x_{i}&\text{if }i=j\,,\\\
x_{j}&\text{otherwise.}\end{cases}$

We observe that

$\displaystyle T_{i}\mathcal{N}_{\mathrm{FCC}}=\\{T_{i}b\colon
b\in\mathcal{N}_{\mathrm{FCC}}\\}=\mathcal{N}_{\mathrm{FCC}}\,,\text{ for all
}i\in\\{1,2,3\\}.$

Moreover, given a permutation $\pi\in S_{3}$ we have that

$\displaystyle\pi\mathcal{N}_{\mathrm{FCC}}=\\{\pi b\colon
b\in\mathcal{N}_{\mathrm{FCC}}\\}=\mathcal{N}_{\mathrm{FCC}}\,.$

It therefore suffices to restrict only to the case in which the vector $b_{0}$
agrees with the vector $b_{1}\in\mathcal{N}_{\mathrm{FCC}}$. We claim that
this face has corners given by

$\displaystyle
c_{1}=\left(\frac{1}{2}\sqrt{2},0,0\right)\,,c_{2}=\left(0,\frac{1}{2}\sqrt{2},0\right)\,,c_{3}=\frac{1}{4}\left(\sqrt{2},\sqrt{2},\sqrt{2}\right)\,,c_{4}=\frac{1}{4}\left(\sqrt{2},\sqrt{2},-\sqrt{2}\right)\,.$
(62)

Note that, if this were true then it is easy to see that $S_{b_{0}}$ is a
rhombus and $\mathcal{H}^{2}(S_{b_{0}})=\frac{1}{4}\sqrt{2}$. It remains to
prove (62). Let us denote by $y$ a corner of $S_{b_{0}}$. We can assume that
$y_{1},y_{2}\geq 0$. Were this not the case, then there could be
$b^{\prime}\in\mathcal{N}_{\mathrm{FCC}}$ such that $\langle
b^{\prime},y\rangle>\langle b,y\rangle$, thus contradicting the definition of
$S_{b_{0}}$ in (58). If $y_{1}=0$ (or $y_{2}=0$), then
$y_{2}=\frac{1}{2}\sqrt{2}$ (resp. $y_{1}=\frac{1}{2}\sqrt{2}$) and since,
$\langle b^{\prime},y\rangle\leq\frac{1}{2}$ for all
$b^{\prime}\in\mathcal{N}_{\mathrm{FCC}}$ we have $y_{3}=0$. Hence, we find
the two corners with coordinates $(\frac{1}{2}\sqrt{2},0,0)$ and
$(0,\frac{1}{2}\sqrt{2},0)$. Now, if $y_{1}>0$ and $y_{2}>0$, then assuming
that $y_{3}\geq 0$ we have that the corner is equal to $\langle
b_{1},y\rangle=\langle b_{2},y\rangle=\langle
b_{3},y\rangle=\frac{1}{2}\sqrt{2}$. Thus, necessarily
$y_{1}=y_{2}=y_{3}=\frac{1}{4}\sqrt{2}$. If instead $y_{3}<0$, then the corner
is equal to $\langle b_{1},y\rangle=\langle b_{2},y\rangle=\langle
b_{1}-b_{3},y\rangle=\frac{1}{2}\sqrt{2}$ which implies
$y_{1}=y_{2}=-y_{3}=\frac{1}{4}\sqrt{2}$. Hence (62) holds true and this
concludes Step 2.
Step 3.(Neighbors share faces) We want to show that for each
$b_{0}\in\mathcal{N}_{\mathrm{FCC}}$ we have that the face $S_{b_{0}}$ of
$V_{\mathrm{FCC}}(0)$ is shared with the Voronoi cell
$V_{\mathrm{FCC}}(b_{0})$. By the symmetries shown in Step 2 it suffices to
prove this statement only for $b_{0}=b_{1}$. Using (62) we see that the
corners of the face $S_{b_{0}}$ of the Voronoi cell $V_{\mathrm{FCC}}(0)$
coincide with the corners of the face $S_{-b_{0}}+b_{0}$ of the Voronoi cell
$V_{\mathrm{FCC}}(b_{0})$.
Step 4.(Volume of the Voronoi cell) In order to calculate the volume of the
Voronoi cell we note that $\mathcal{L}_{\mathrm{FCC}}$ is a Bravais-lattice
with spanning vectors $b_{1},b_{2},b_{3}$. Since, the Voronoi cells of all the
points are the same, it suffices to calculate the fraction of points per unit
volume. This, then gives also the volume per point. Since, the Voronoi cells
are space filling the volume per point is equal to the volume of each Voronoi
cell. Due to (15) we have that

$\displaystyle|T_{\mathrm{FCC}}|=\frac{1}{2}\sqrt{2}\,.$

Furthermore, we have that

$\displaystyle\bigcup_{x\in\mathcal{L}_{\mathrm{FCC}}}(x+T_{{\mathrm{FCC}}})=\mathbb{R}^{3}\,,\text{
and }\mathcal{L}_{\mathrm{FCC}}\cap T_{\mathrm{FCC}}=\\{0\\}\,.$

Hence, each points of the lattice occupies a volume
$|T_{\mathrm{FCC}}|=\frac{1}{2}\sqrt{2}$ and the volume of the Voronoi cell
must be the same. This concludes Step 3 and thus the proof of the lemma. ∎

###### Proof of Lemma 4.2.

We split the proof of the lemma into four steps. First, we prove (59). In the
second step, we show that $6$ of the faces are rhombi, the $6$ other faces are
trapezoids, and we calculate the area of each face. Lastly, given
$x\in\mathcal{L}_{\mathrm{HCP}}$, we show that each neighboring Voronoi cell
$\mathcal{V}_{\mathcal{L}_{\mathrm{HCP}}}(y),y\in\mathcal{N}_{\mathrm{HCP}}(x)$
shares a face with the Voronoi cell $V_{\mathrm{HCP}}(x)$.
Step 1.(Shape of the Voronoi cell) The purpose of this step is to prove (59).
Here, we only show this equality in the case that $x=0$, the case $x\neq 0$
being treated in a similar fashion. Step
1.1.($\mathcal{V}_{\mathcal{L}_{\mathrm{HCP}}}(0)\subset V_{\mathrm{HCP}}$)
Given $y\in\mathcal{V}_{\mathcal{L}_{\mathrm{HCP}}}(0)$ we have that
$|y|\leq|y-b|\,.$ Now, noting that $|b|=1$ for all
$b\in\mathcal{N}_{\mathrm{HCP}}\subset\mathcal{L}_{\mathrm{HCP}}$, we have

$\displaystyle|y|^{2}\leq|y-b|^{2}=|y|^{2}-2\langle
y,b\rangle+|b|^{2}\iff\langle b,y\rangle\leq\frac{1}{2}\,.$

This concludes Step 1.1.
Step
1.2.($V_{\mathrm{HCP}}\subset\mathcal{V}_{\mathcal{L}_{\mathrm{HCP}}}(0)$) We
show that for $y\in V_{\mathrm{HCP}}$ we have $|y|\leq|y-z|$, for all
$z\in\mathcal{L}_{\mathrm{HCP}}$. This is equivalent to

$\displaystyle y\in V_{\mathrm{HCP}}\implies\langle
y,z\rangle\leq\tfrac{1}{2}|z|^{2}\quad\text{ for all
}z\in\mathcal{L}_{\mathrm{HCP}}\,.$ (63)

Since, $|b|=1$ for all $b\in\mathcal{N}_{\mathrm{HCP}}$ (63) is true for all
$b\in\mathcal{N}_{\mathrm{HCP}}$. Next, we prove (63) for all
$z\in\mathcal{L}_{\mathrm{HCP}}\setminus\mathcal{N}_{\mathrm{HCP}}$. We
distinguish several cases:

  * (a)

$z=\lambda_{1}e_{1}+\lambda_{2}e_{2},\lambda_{1},\lambda_{2}\in\mathbb{Z}\,;$

  * (b)

$z=\lambda_{1}e_{1}+\lambda_{2}e_{2}+\lambda_{3}e_{3},\lambda_{1},\lambda_{2},\lambda_{3}\in\mathbb{Z}\,;$

  * (c)

$z=v_{1}+\lambda_{1}e_{1}+\lambda_{2}e_{2},\lambda_{1},\lambda_{2}\in\mathbb{Z}\,;$

  * (d)

$z=v_{1}+\lambda_{1}e_{1}+\lambda_{2}e_{2}+\lambda_{3}e_{3},\lambda_{1},\lambda_{2},\lambda_{3}\in\mathbb{Z}\,;$

_Proof in case(a)._ If $\lambda_{1},\lambda_{2}\geq 0$, using that $\langle
e_{1},e_{2}\rangle\geq 0$, we have

$\displaystyle\langle y,z\rangle$ $\displaystyle=\langle
y,\lambda_{1}e_{1}+\lambda_{2}e_{2}\rangle\leq\frac{1}{2}|\lambda_{1}e_{1}|^{2}+\frac{1}{2}|\lambda_{2}e_{2}|^{2}$
$\displaystyle=\frac{1}{2}|\lambda_{1}e_{1}+\lambda_{2}e_{2}|^{2}-\lambda_{1}\lambda_{2}\langle
e_{1},e_{2}\rangle\leq\frac{1}{2}|\lambda_{1}e_{1}+\lambda_{2}e_{2}|^{2}=\frac{1}{2}|z|^{2}\,.$

On the other hand, if $\lambda_{1}\lambda_{2}\leq 0$ and without loss of
generality $\lambda_{1}\geq|\lambda_{2}|\geq 0$, noting that
$e_{2}-e_{1}\in\mathcal{N}_{\mathrm{HCP}}$, we have

$\displaystyle\langle y,z\rangle$ $\displaystyle=\langle
y,\lambda_{1}e_{1}+\lambda_{2}e_{2}\rangle=\langle
y,\lambda_{2}(e_{2}-e_{1})+(\lambda_{1}+\lambda_{2})e_{1}\rangle\leq\frac{1}{2}|\lambda_{2}(e_{2}-e_{1})|^{2}+\frac{1}{2}|(\lambda_{2}+\lambda_{1})e_{1}|^{2}$
$\displaystyle=\frac{1}{2}|\lambda_{1}e_{1}+\lambda_{2}e_{2}|^{2}-\lambda_{2}(\lambda_{1}+\lambda_{2})\langle
e_{2}-e_{1},e_{1}\rangle\leq\frac{1}{2}|\lambda_{1}e_{1}+\lambda_{2}e_{2}|^{2}=\frac{1}{2}|z|^{2}\,.$

Here, the last inequality follows, since $\lambda_{2}\leq
0\leq\lambda_{1}+\lambda_{2}$ and $\langle e_{2}-e_{1},e_{1}\rangle\leq 0$.
This concludes case (a).
_Proof in case(b)._ We first show that $\langle
y,e_{3}\rangle\leq\frac{1}{2}|e_{3}|^{2}$. Using that
$v_{1},v_{1}-e_{1},v_{1}-e_{2}\in\mathcal{N}_{\mathrm{HCP}}$, that
$3v_{1}-e_{1}-e_{2}=\frac{3}{2}e_{3}$ we have

$\displaystyle\langle y,e_{3}\rangle$ $\displaystyle=\frac{2}{3}\langle
y,v_{1}+v_{1}-e_{1}+v_{1}-e_{2}\rangle$
$\displaystyle\leq\frac{1}{3}\left(\left|\frac{1}{3}(e_{1}+e_{2})\right|^{2}+\left|\frac{1}{3}(e_{2}-2e_{1})\right|^{2}+\left|\frac{1}{3}(e_{1}-2e_{2})\right|^{2}\right)+\left|\frac{1}{2}e_{3}\right|^{2}\leq\frac{1}{2}|e_{3}|^{2}\,.$

Here, the last inequality follows by calculating the norms of
$e_{1}+e_{2},e_{1}-2e_{2},e_{2}-2e_{1}$ and $e_{3}$ by using (7). Note that
now, the case of $z=\lambda_{1}e_{1}+\lambda_{2}e_{2}+\lambda_{3}e_{3}$
follows from case (a) using that $\langle e_{3},e_{1}\rangle=\langle
e_{3},e_{2}\rangle=0$.
_Proof of case(c)._ Let $z=v_{1}+\lambda_{1}e_{1}+\lambda_{2}e_{2}$. If
$\lambda_{1},\lambda_{2}\geq 0$ we have

$\displaystyle\langle y,z\rangle$ $\displaystyle=\langle
y,v_{1}+\lambda_{1}e_{1}+\lambda_{2}e_{2},\rangle\leq\frac{1}{2}|v_{1}|^{2}+\frac{1}{2}|\lambda_{1}e_{1}|^{2}+\frac{1}{2}|\lambda_{2}e_{2}|^{2}\leq\frac{1}{2}|v_{1}|^{2}+\frac{1}{2}|\lambda_{1}e_{1}+\lambda_{2}e_{2}|^{2}$
$\displaystyle=\frac{1}{2}|v_{1}+\lambda_{1}e_{1}+\lambda_{2}e_{2}|^{2}-\langle
v_{1},\lambda_{1}e_{1}+\lambda_{2}e_{2}\rangle\leq\frac{1}{2}|v_{1}+\lambda_{1}e_{1}+\lambda_{2}e_{2}|^{2}\,.$

The second inequality uses that $\langle e_{1},e_{2}\rangle\geq 0$ and the
last inequality uses that $\langle v_{1},e_{1}\rangle,\langle
v_{1},e_{2}\rangle\geq 0$. Now assume that $\lambda_{1}\geq 0,\lambda_{2}<0$.
Then, since $\langle(v_{1}-e_{2}),e_{1}\rangle=0$ and $\langle
v_{1}-e_{2},e_{2}\rangle\leq 0$, again exploiting that
$v_{1}-e_{2}\in\mathcal{N}_{\mathrm{HCP}}$ it holds that

$\displaystyle\langle y,z\rangle$ $\displaystyle=\langle
y,(v_{1}-e_{2})+\lambda_{1}e_{1}+(\lambda_{2}+1)e_{2}\rangle\leq\frac{1}{2}|v_{1}-e_{2}|^{2}+\frac{1}{2}|\lambda_{1}e_{1}+(\lambda_{2}+1)e_{2}|^{2}$
$\displaystyle=\frac{1}{2}|v_{1}+\lambda_{1}e_{1}+\lambda_{2}e_{2}|^{2}-\langle
v_{1}-e_{2},\lambda_{1}e_{1}+(\lambda_{2}+1)e_{2}\rangle\leq\frac{1}{2}|v_{1}+\lambda_{1}e_{1}+\lambda_{2}e_{2}|^{2}=\frac{1}{2}|z|^{2}\,.$

The case where $\lambda_{1}<0,\lambda_{2}\geq 0$ (resp.
$\lambda_{1},\lambda_{2}<0$) is being treated in a similar fashion by
replacing $v_{1}-e_{2}$ with $v_{1}-e_{1}$ (resp. $v_{1}-e_{1}-e_{2}$).
_Proof of case(d)._ Here, we only treat the case of
$z=v_{1}+\lambda_{1}e_{1}+\lambda_{2}e_{2}+\lambda_{3}e_{3}$, $\lambda_{3}\geq
0$. Since $\langle v_{1}+\lambda_{1}e_{1}+\lambda_{2}e_{2},e_{3}\rangle\geq
0$, we have

$\displaystyle\langle y,z\rangle$ $\displaystyle=\langle
y,v_{1}+\lambda_{1}e_{1}+\lambda_{2}e_{2}+\lambda_{3}e_{3}\rangle\leq\frac{1}{2}|v_{1}+\lambda_{1}e_{1}+\lambda_{2}e_{2}|^{2}+\frac{1}{2}|\lambda_{3}e_{3}|^{2}$
$\displaystyle=\frac{1}{2}|v_{1}+\lambda_{1}e_{1}+\lambda_{2}e_{2}+\lambda_{3}e_{3}|^{2}-\langle
v_{1}+\lambda_{1}e_{1}+\lambda_{2}e_{2},e_{3}\rangle\leq\frac{1}{2}|v_{1}+\lambda_{1}e_{1}+\lambda_{2}e_{2}+\lambda_{3}e_{3}|^{2}=\frac{1}{2}|z|^{2}\,.$

The case of $\lambda_{3}<0$ follows by replacing $v_{1}$ with $v_{1}-e_{3}$ in
the last two cases (c) and (d). This concludes Step 1.2 and, together with
Step 1.1, shows (59).
Step 2.(The faces of the Voronoi cell) In order to calculate the faces of
$V_{\mathrm{HCP}}$ we use (59) and exploit its symmetries. We note that if
$R\in SO(3)$ is any rotation of integer multiples of $2\pi/3$ around the
$x_{3}$-axis we have that

$\displaystyle R\mathcal{N}_{\mathrm{HCP}}=\\{Rb\colon
b\in\mathcal{N}_{\mathrm{HCP}}\\}=\mathcal{N}_{\mathrm{HCP}}\,.$ (64)

Moreover, if $T_{3}\colon\mathbb{R}^{3}\to\mathbb{R}^{3}$ is the reflection
with respect to the $(x_{1},x_{2})$-plane, i.e.

$\displaystyle(T_{3}x)_{j}:=\begin{cases}x_{j}&j=1,2\,,\\\
-x_{3}&j=3\,,\end{cases}$ (65)

we have that

$\displaystyle T_{3}\mathcal{N}_{\mathrm{HCP}}=\\{T_{3}b\colon
b\in\mathcal{N}_{\mathrm{HCP}}\\}=\mathcal{N}_{\mathrm{HCP}}\,.$ (66)

Exploiting (64) and (66), it suffices to find the corners of $S_{b_{0}}$ in
(60) for

$\displaystyle{\rm(a)}\quad b_{0}=e_{1}\,,\quad{\rm(b)}\quad
b_{0}=-e_{1}\,,\quad{\rm(c)}\quad b_{0}=v_{1}\,.$

_Corners in case(a)._ We claim that in the case of $b_{0}=e_{1}$ that the
corners of $S_{b_{0}}$ are given by the points

$\displaystyle\begin{split}&c_{1}=\left(\frac{1}{2},\frac{1}{6}\sqrt{3},\frac{1}{12}\sqrt{6}\right)\,,c_{2}=\left(\frac{1}{2},\frac{1}{6}\sqrt{3},-\frac{1}{12}\sqrt{6}\right)\,,\\\
&c_{3}=\left(\frac{1}{2},-\frac{1}{6}\sqrt{3},\frac{1}{6}\sqrt{6}\right)\,,c_{4}=\left(\frac{1}{2},-\frac{1}{6}\sqrt{3},-\frac{1}{6}\sqrt{6}\right)\,.\end{split}$
(67)

In particular, the face $S_{b_{0}}$ is a trapezoid with two bases of length
$\frac{1}{6}\sqrt{6}$, $\frac{1}{3}\sqrt{6}$ and height $\frac{1}{3}\sqrt{3}$.
Hence, $\mathcal{H}^{2}(S_{b_{0}})=\frac{1}{4}\sqrt{2}$. It remains to prove
(67). Let $y\in S_{b_{0}}$ be a corner. Due to (66), we can assume that
$y_{3}\geq 0$, since the other corners are just found by applying the mapping
$T_{3}$ (see (65)) to the corners with positive coordinates. By the definition
of $S_{b_{0}}$ we have that $\langle y,e_{1}\rangle\geq\langle
y,e_{1}-e_{2}\rangle$ which is equivalent to $\langle y,e_{2}\rangle\geq 0$.
Now, if $\langle y,e_{2}\rangle>0$, then $y$ is given by $\langle
y,e_{1}\rangle=\langle y,e_{2}\rangle=\langle y,v_{1}\rangle=\frac{1}{2}$.
This linear system has a unique solution given by
$c_{1}=\left(\frac{1}{2},\frac{1}{6}\sqrt{3},\frac{1}{12}\sqrt{6}\right)$. On
the other hand, if $\langle y,e_{2}\rangle=0$, then $y$ is given by $\langle
y,e_{2}\rangle=0,\langle y,e_{1}\rangle=\langle y,v_{1}\rangle=\frac{1}{2}$.
The unique solution of this linear system is given by
$c_{3}=\left(\frac{1}{2},-\frac{1}{6}\sqrt{3},\frac{1}{6}\sqrt{6}\right)$.
This shows (67) and concludes case (a).
_Corners in case(b)._ We claim that in the case of $b_{0}=-e_{1}$ that the
corners of $S_{b_{0}}$ are given by the points

$\displaystyle\begin{split}&c_{1}=\left(-\frac{1}{2},\frac{1}{6}\sqrt{3},\frac{1}{12}\sqrt{6}\right)\,,c_{2}=\left(-\frac{1}{2},\frac{1}{6}\sqrt{3},-\frac{1}{12}\sqrt{6}\right)\,,\\\
&c_{3}=\left(-\frac{1}{2},-\frac{1}{6}\sqrt{3},\frac{1}{6}\sqrt{6}\right)\,,c_{4}=\left(-\frac{1}{2},-\frac{1}{6}\sqrt{3},-\frac{1}{6}\sqrt{6}\right)\,.\end{split}$
(68)

In particular, the face $S_{b_{0}}$ is a trapezoid with two bases of length
$\frac{1}{6}\sqrt{6}$, $\frac{1}{3}\sqrt{6}$ and height $\frac{1}{3}\sqrt{3}$.
Hence, $\mathcal{H}^{2}(S_{b_{0}})=\frac{1}{4}\sqrt{2}$. It remains to prove
(68). Let $y\in S_{b_{0}}$ be a corner. Due to (66), as in case $\rm{(a)}$, we
can assume that $y_{3}\geq 0$. By the definition of $S_{b_{0}}$ we have that
$\langle y,-e_{1}\rangle\geq\langle y,e_{2}-e_{1}\rangle$ which is equivalent
to $\langle y,e_{2}\rangle\leq 0$. Now, if $\langle y,e_{2}\rangle=0$, then
$y$ is given by $\langle y,e_{2}\rangle=0,\langle y,v_{1}-e_{1}\rangle=\langle
y,-e_{1}\rangle=\frac{1}{2}$. We see that the unique solution of this linear
system is given by
$c_{1}=\left(-\frac{1}{2},\frac{1}{6}\sqrt{3},\frac{1}{12}\sqrt{6}\right)$. On
the other hand, if $\langle y,e_{2}\rangle<0$, then $y$ is given by $\langle
y,v_{1}\rangle=0,\langle y,-e_{1}\rangle=\langle y,-e_{2}\rangle=\frac{1}{2}$.
The unique solution is now given by
$c_{3}=\left(-\frac{1}{2},-\frac{1}{6}\sqrt{3},\frac{1}{6}\sqrt{6}\right)$.
This shows (68) and concludes case (b).
_Corners in case(c)._ We claim that in the case of $b_{0}=v_{1}$ that the
corners of $S_{b_{0}}$ are given by the points

$\displaystyle\begin{split}&c_{1}=\left(\frac{1}{2},\frac{1}{6}\sqrt{3},\frac{1}{12}\sqrt{6}\right)\,,c_{2}=\left(0,0,\frac{1}{4}\sqrt{6}\right)\,,\\\
&c_{3}=\left(0,\frac{1}{3}\sqrt{3},\frac{1}{6}\sqrt{6}\right)\,,c_{4}=\left(\frac{1}{2},-\frac{1}{6}\sqrt{3},\frac{1}{6}\sqrt{6}\right)\,.\end{split}$
(69)

In particular, the face $S_{b_{0}}$ is a rhombus. Hence,
$\mathcal{H}^{2}(S_{b_{0}})=\frac{1}{8}\sqrt{6}$. It remains to prove (67).
Let $y\in S_{b_{0}}$ be a corner. By the definition of $S_{b_{0}}$ we have
that $\langle y,v_{1}\rangle\geq\langle y,v_{1}-e_{1}\rangle,\langle
y,v_{1}-e_{2}\rangle$ which is equivalent to $\langle y,e_{1}\rangle,\langle
y,e_{2}\rangle\geq 0$. Now if, $\langle y,e_{2}\rangle>0$ then the corner
solves the linear system $\langle y,e_{1}\rangle=\langle
y,e_{2}\rangle=\langle y,v_{1}\rangle=\frac{1}{2}$. Its unique solution is
$c_{1}=(\frac{1}{2},\frac{1}{6}\sqrt{3},\frac{1}{12}\sqrt{6})$. On the other
hand if $\langle y,e_{2}\rangle=0$, then the corners are given by those $y$
such that $\langle y,e_{2}\rangle=0,\langle y,e_{1}\rangle=\langle
y,v_{1}\rangle=\frac{1}{2}$ or $\langle y,e_{1}\rangle=\langle
y,e_{2}\rangle=0,\langle y,v_{1}\rangle=\frac{1}{2}$. These points have
coordinates $c_{2}=(\frac{1}{2},-\frac{1}{6}\sqrt{3},\frac{1}{6}\sqrt{6})$ and
$c_{3}=(0,0,\frac{1}{4}\sqrt{6})$. Finally, if $\langle y,e_{1}\rangle=0$ and
$\langle y,e_{2}\rangle>0$, then $y$ is obtained by solving $\langle
y,e_{1}\rangle=0,\langle y,e_{2}\rangle=\langle y,v_{1}\rangle=\frac{1}{2}$.
Hence it has coordinates $c_{4}=(0,\frac{\sqrt{3}}{3},\frac{\sqrt{6}}{6})$.
This proves (69) and concludes Step 2.
Step 3.(Neighbors share faces) We want to show that for each
$b_{0}\in\mathcal{N}_{\mathrm{HCP}}$ we have that the face $S_{b_{0}}$ of
$\mathcal{V}_{\mathcal{L}_{\mathrm{HCP}}}(0)$ is shared with the Voronoi cell
$\mathcal{V}_{\mathcal{L}_{\mathrm{HCP}}}(b_{0})$. By Step 1 we have that
$\mathcal{V}_{\mathcal{L}_{\mathrm{HCP}}}(0)=V_{\mathrm{HCP}}$ and
$\mathcal{V}_{\mathcal{L}_{\mathrm{HCP}}}(b_{0})=b_{0}-V_{\mathrm{HCP}}$.
Hence, they share the side $\langle y,b_{0}\rangle=\frac{1}{2}=\langle
b_{0}-y,b_{0}\rangle$.

Step 4.(Volume of the Voronoi cell) In order to calculate the volume of the
Voronoi cell we note that $\mathcal{L}_{\mathrm{HCP}}$ is periodic with
respect to the vectors $e_{1},e_{2},e_{3}$. Since the Voronoi cells of all the
points occupy the same volume, it suffices to calculate the fraction of points
per unit volume. The inverse of this number is the volume per point. Since the
Voronoi cells are space filling the volume per point is equal to the volume of
each Voronoi cell. Due to (16) we have that

$\displaystyle|T_{\mathrm{HCP}}|=\sqrt{2}\,.$

Furthermore, we have that

$\displaystyle\bigcup_{x\in\mathrm{span}_{\mathbb{Z}}\\{e_{1},e_{2},e_{3}\\}}(x+T_{{\mathrm{HCP}}})=\mathbb{R}^{3}\,,\text{
and }\mathcal{L}_{\mathrm{HCP}}\cap T_{{\mathrm{HCP}}}=\\{0,v_{1}\\}\,.$

Hence, the volume per point is
$\frac{1}{2}|T_{{\mathrm{HCP}}}|=\frac{1}{2}\sqrt{2}$ and it agrees with the
volume of the Voronoi cell. This concludes Step 4 and thus the proof of the
lemma. ∎

We are now in the position to prove Theorem 2.3.

###### Proof of Theorem 2.3.

All the statements are consequences of Proposition 5.10, Lemma 5.11, Theorem
5.14 and Lemma 5.16 once we show that $\mathcal{L}_{\mathrm{FCC}}$ and
$\mathcal{L}_{\mathrm{HCP}}$ are periodic admissible sets (according to
Definition 5.1 and Definition 5.8) and we observe that, due to Lemma 4.1 and
Lemma 4.2, $\mathcal{N}_{\mathrm{FCC}}(x)=\mathcal{NN}(x)$ (in the sense of
Definition 5.2) as well as $\mathcal{N}_{\mathrm{HCP}}(x)=\mathcal{NN}(x)$ in
the respective cases. We first show that both lattices are admissible sets.
Let us first observe that

$\displaystyle(T_{\mathrm{FCC}}+x)\cap\mathcal{L}_{\mathrm{FCC}}\neq\emptyset\quad\text{and}\quad(T_{\mathrm{HCP}}+x)\cap\mathcal{L}_{\mathrm{HCP}}\neq\emptyset\text{
for all }x\in\mathbb{R}^{3}\,.$

Therefore, (L1) is satisfied for both $\mathcal{L}_{\mathrm{FCC}}$ and
$\mathcal{L}_{\mathrm{HCP}}$ with

$\displaystyle
R:=\max\\{\mathrm{diam}(T_{\mathrm{FCC}}),\mathrm{diam}(T_{\mathrm{FCC}})\\}\leq\max\left\\{\sum_{k=1}^{3}|e_{k}|,\sum_{k=1}^{3}|b_{k}|\right\\}<+\infty\,,$

where we recalled Definition 15 and Definition 16. On the other hand, (L2) is
satisfied with $r=1$, see the discussion at the definition of the FCC and HCP
lattice in Section 2. Concerning periodicity: We observe that for all
$z=(z_{1},z_{2},z_{3})\in\mathbb{Z}^{3}$ we have

$\displaystyle\mathcal{L}_{\mathrm{FCC}}=\mathcal{L}_{\mathrm{FCC}}+\sum_{k=1}^{3}z_{k}b_{k}\,,\text{
and
}\mathcal{L}_{\mathrm{HCP}}=\mathcal{L}_{\mathrm{HCP}}+\sum_{k=1}^{3}z_{k}e_{k}$

and thus both $\mathcal{L}_{\mathrm{FCC}}$ and $\mathcal{L}_{\mathrm{HCP}}$
are periodic according to Definition 5.8. The statement follows by Theorem
5.14 with $c_{nn}(x)=1$. ∎

##  5\. General periodic lattices

This section deals with integral representation and concentrated-compactness
properties of energies defined on general periodic lattices.

######  Definition 5.1.

Let $\Sigma\subset\mathbb{R}^{n}$ be a countable set of points in
$\mathbb{R}^{n}$. We call $\Sigma$ an _admissible_ set of points if the
following two conditions hold:

  * (L1)

There exists $R>0$ such that $\inf_{x\in\mathbb{R}^{n}}\\#(\Sigma\cap
B_{R}(x))\geq 1$;

  * (L2)

There exists $r>0$ such that $\mathrm{dist}(x,\Sigma\setminus\\{x\\})\geq r$
for all $x\in\Sigma$.

######  Definition 5.2.

We define the Voronoi cell of $x\in\Sigma$ as

$\displaystyle\mathcal{V}(x):=\\{z\in\mathbb{R}^{n}\colon|x-z|\leq|y-z|\text{
for all }y\in\Sigma\\}\,.$ (70)

The set of nearest neighbors of $\Sigma$ is defined by

$\displaystyle\mathcal{NN}(\Sigma):=\\{(x,y)\in\Sigma\times\Sigma\colon\mathcal{H}^{n-1}(\mathcal{V}(x)\cap\mathcal{V}(y))>0\\}\,,$

We set $\mathcal{NN}(x)=\\{y\in\Sigma\colon(x,y)\in\mathcal{NN}(\Sigma)\\}$.
Given $\varepsilon>0$ we denote by $\varepsilon\Sigma:=\\{\varepsilon x\colon
x\in\Sigma\\}$ and for $x\in\varepsilon\Sigma$ we set
$\mathcal{V}_{\varepsilon}(x)=\varepsilon\mathcal{V}(\varepsilon^{-1}x)$ the
Voronoi cell of $x\in\varepsilon\Sigma$, and
$\mathcal{NN}_{\varepsilon}(x)=\\{y\in\varepsilon\Sigma\colon\varepsilon^{-1}(x,y)\in\mathcal{NN}(\Sigma)\\}$
the set of nearest neighbors of $x$ in $\varepsilon\Sigma$.

We now define for $u\colon\Sigma\to\\{0,1\\}$ the two energy functionals given
by

$\displaystyle F_{\varepsilon}(u,A):=\underset{\varepsilon x\in
A}{\sum_{(x,y)\in\mathcal{NN}(\Sigma)}}\varepsilon^{n-1}c_{nn}(x-y)|u(\varepsilon
x)-u(\varepsilon y)|\,$ (71)

and

$\displaystyle\hat{F}_{\varepsilon}(u,A):=\underset{\varepsilon x,\varepsilon
y\in
A}{\sum_{(x,y)\in\mathcal{NN}(\Sigma)}}\varepsilon^{n-1}c_{nn}(x-y)|u(\varepsilon
x)-u(\varepsilon y)|\,,$ (72)

where $c_{nn}\colon\mathbb{R}^{n}\to[0,+\infty]$ satisfies

$\displaystyle C^{-1}\leq c_{nn}(x)\leq C\,,\quad\text{for all
}x\in\mathbb{R}^{n}\,.$ (73)

When $A=\mathbb{R}^{n}$ we omit the dependence on it and write
$F_{\varepsilon}(u)=F_{\varepsilon}(u,\mathbb{R}^{n})$ and
$\hat{F}_{\varepsilon}(u)=\hat{F}_{\varepsilon}(u,\mathbb{R}^{n})$.

######  Remark 5.3.

(Difference between $F_{\varepsilon}$ and $\hat{F}_{\varepsilon}$) We want to
point out the difference between $F_{\varepsilon}$ and
$\hat{F}_{\varepsilon}$: In the formula defining $F_{\varepsilon}$ the sum is
taken over all $(x,y)\in\mathcal{NN}(\Sigma)$ such that $\varepsilon x\in A$.
Instead in the case of $\hat{F}_{\varepsilon}$ the sum takes only those
$(x,y)\in\mathcal{NN}(\Sigma)$ such that both $\varepsilon x\in A$ and
$\varepsilon y\in A$. The functional $F_{\varepsilon}(u,\cdot)$ is an additive
set function on disjoint sets, i.e., given $A,B\subset\mathbb{R}^{n}$ such
that $A\cap B=\emptyset$, we have

$\displaystyle F_{\varepsilon}(u,A\cup
B)=F_{\varepsilon}(u,A)+F_{\varepsilon}(u,B)\,,$

whereas $\hat{F}_{\varepsilon}(u,\cdot)$ is only super-additive on disjoint
sets. Our $\Gamma$-convergence result will be stated for the functional
$F_{\varepsilon}$. The reason for us to introduce $\hat{F}_{\varepsilon}$ is
that our proof will use the integral representation result proven in [3], see
Theorem 5.7. However, we will show later on that the $\Gamma$-convergence of
$F_{\varepsilon}$ is equivalent to that of $\hat{F}_{\varepsilon}$.

Given $X\subset\varepsilon\Sigma$ we write with a slight abuse of notation

$\displaystyle F_{\varepsilon}(X,A)=F_{\varepsilon}(\chi_{X},A)\,.$

Note that, due to (73), we have

$\displaystyle C^{-1}\sum_{x\in
X}\varepsilon^{n-1}\\#(\mathcal{NN}_{\varepsilon}(x)\setminus X)\leq
F_{\varepsilon}(X,A)\leq C\sum_{x\in
X}\varepsilon^{n-1}\\#(\mathcal{NN}_{\varepsilon}(x)\setminus X)\,.$

Hypothesis (73) corresponds to [3, Hypothesis 1] in the case that, according
to the notation in [3, Equation (5.23)],
$c_{nn}^{\varepsilon}(x,y)=c_{nn}(x-y)$ and $c_{lr}^{\varepsilon}(x,y)=0$. It
is worth observing that in [3] a more general class of functionals was
investigated, namely those for which also certain long-range interactions
between points in $\Sigma$ contribute to the energy, i.e., $c_{lr}(x,y)\neq
0$. For the sake of exposition and simplicity, here we consider the case
$c_{lr}=0$, that is the energy accounts only for the nearest neighbor
interactions. However, with some more involved multi-scale constructions, all
the statements below extend to the more general case where also long-range
interactions are considered.

######  Definition 5.4.

Given $X\subset\varepsilon\mathcal{L}$ we define the rescaled empirical
measures associated to $X$ as

$\displaystyle\mu_{\varepsilon}=\varepsilon^{n}\sum_{x\in X}\delta_{x}\,.$
(74)

Furthermore, recalling (70), we define

$\displaystyle V_{\varepsilon}(X):=\bigcup_{x\in
X}{\mathcal{V}}_{\varepsilon}(x)\,.$ (75)

Henceforth, we drop the dependence on $X$ and simply write $V_{\varepsilon}$.
Given $A\subset\mathbb{R}^{n}$ open with $\partial A\in\mathrm{Lip}$, with
slight abuse of notation we define
$F_{\varepsilon}\colon\mathcal{M}_{+}(A)\to[0,+\infty]$ (similarly
$\hat{F}_{\varepsilon}\colon\mathcal{M}_{+}(A)\to[0,+\infty]$) by

$\displaystyle
F_{\varepsilon}(\mu,A)=\begin{cases}F_{\varepsilon}(X,A)&\mu\text{ is given by
}\eqref{def:empiricalmeasures}\text{ for some
}X\subset\varepsilon\mathcal{L}\,;\\\ +\infty&\text{otherwise.}\end{cases}$

Additionally, we define $F_{\varepsilon}\colon
L^{1}_{\mathrm{loc}}(A)\to[0,+\infty]$ (similarly $\hat{F}_{\varepsilon}\colon
L^{1}_{\mathrm{loc}}(A)\to[0,+\infty]$) by

$\displaystyle
F_{\varepsilon}(u,A)=\begin{cases}F_{\varepsilon}(X,A)&u=\chi_{V_{\varepsilon}}\text{
and }V_{\varepsilon}\text{ is given by
}\eqref{def:piecewiseconstantinterpolant}\text{ for some
}X\subset\varepsilon\mathcal{L}\,;\\\ +\infty&\text{otherwise.}\end{cases}$

It is necessary for us to introduce two different domains of definition for
the extended functional $F_{\varepsilon}$, since we want to make use of [3,
Theorem 5.5]. As it will turn out the two types of extension are equivalent,
cf. Lemma 5.11 and Corollary 5.12.

Let $(\Omega,\mathbb{P},\mathcal{F})$ be a probability space. Hereafter we
recall some definitions from [3] (Definition 5.1 and Definition 5.4):

######  Definition 5.5.

We say that a family $(\tau_{z})_{z\in\mathbb{Z}^{n}}$,
$\tau_{z}\colon\Omega\to\Omega$, is an _additive group action_ on $\Omega$ if

$\displaystyle\tau_{z_{1}+z_{2}}=\tau_{z_{2}}\circ\tau_{z_{1}}\quad\text{for
all }z_{1},z_{2}\in\mathbb{Z}^{n}\,.$

Such an additive group action is called _measure preserving_ if

$\displaystyle\mathbb{P}(\tau_{z}B)=\mathbb{P}(B)\quad\text{for all
}B\in\mathcal{F},z\in\mathbb{Z}^{n}\,.$

If in addition, for all $B\in\mathcal{F}$ we have

$\displaystyle(\tau_{z}(B)=B\quad\text{for all
}z\in\mathbb{Z}^{n})\implies\mathbb{P}(B)\in\\{0,1\\}\,,$

then $(\tau_{z})_{z\in\mathbb{Z}^{n}}$ is called _ergodic_.

######  Definition 5.6.

A random variable
$\mathcal{L}\colon\Omega\to(\mathbb{R}^{n})^{\mathbb{Z}^{n}}$,
$\omega\mapsto\mathcal{L}(\omega)=\\{\mathcal{L}(\omega)(i)\\}_{i\in\mathbb{Z}^{n}}$is
called a stochastic lattice. We say that $\mathcal{L}$ is admissible if
$\mathcal{L}(\omega)$ is admissible in the sense of Definition 5.1 and the
constants $r,R$ can be chosen independent of $\omega$ $\mathbb{P}$-almost
surely. The stochastic lattice $\mathcal{L}$ is said to be _stationary_ if
there exists a measure preserving group action
$(\tau_{z})_{z\in\mathbb{Z}^{n}}$ on $\Omega$ such that, for
$\mathbb{P}$-almost every $\omega\in\Omega$,
$\mathcal{L}(\tau_{z}\omega)=\mathcal{L}(\omega)+z$. If in addition
$(\tau_{z})_{z\in\mathbb{Z}^{n}}$ is ergodic, then $\mathcal{L}$ is called
ergodic, too.

We now state a simplified version of [3, Theorem 5.5] which is enough for our
purposes.

######  Theorem 5.7.

(Stochastic homogenization of spin systems) Let $\mathcal{L}$ be a stationary
and ergodic stochastic lattice and let $\hat{F}_{\varepsilon}$ be defined by
(72). Let $A\subset\mathbb{R}^{n}$ be open and bounded with $\partial
A\in\mathrm{Lip}$. For $\mathbb{P}$-almost every $\omega$ the functionals
$F_{\varepsilon}(\omega)$ $\Gamma$-converge with respect to the strong
$L^{1}(A)$-topology to the functional $F_{\mathrm{hom}}\colon
L^{1}(A)\to[0,+\infty]$ defined by

$\displaystyle F_{\mathrm{hom}}(u,A):=\begin{cases}\int_{J(u)\cap
A}\varphi_{\mathrm{hom}}(\nu_{u})\,\mathrm{d}\mathcal{H}^{n-1}&\text{if }u\in
BV(A;\\{0,1\\}),\\\ +\infty&\text{otherwise.}\end{cases}$

The function $\varphi_{\mathrm{hom}}\colon\mathbb{R}^{n}\to[0,+\infty]$ is
given by

$\displaystyle\varphi_{\mathrm{hom}}(\nu)=\lim_{T\to+\infty}\frac{1}{T^{n-1}}\int_{\Omega}\inf\big{\\{}$
$\displaystyle F(u,Q_{T}^{\nu})\colon u\colon\mathcal{L}(\omega)\to\\{0,1\\},$
$\displaystyle u(i)=u_{\nu}(i)\text{ for }i\in\mathcal{L}(\omega)\setminus
Q^{\nu}_{T-l_{T}}\big{\\}}\,\mathrm{d}\mathbb{P}(\omega)\,,$

where $l_{T}\to+\infty$ and $l_{T}/T\to 0$ as $T\to+\infty$.

######  Definition 5.8.

Let $\mathcal{L}\subset\mathbb{R}^{n}$ be an admissible set of points. We say
that $\mathcal{L}$ is periodic if there exists a basis
$\\{e_{1},\ldots,e_{n}\\}\subset\mathbb{R}^{n}$ such that

$\displaystyle\mathcal{L}+e_{k}=\mathcal{L}\text{ for all }k=1,\ldots,n\,.$

We denote by

$\displaystyle Q:=\left\\{\sum_{k=1}^{n}\lambda_{k}e_{k}\colon
0\leq\lambda_{k}<1\right\\}$

the periodicity cell of $\mathcal{L}$. We set

$\displaystyle\rho:=\frac{\\#(\mathcal{L}\cap Q)}{|Q|}\,.$ (76)

In the following we assume, up to a change of coordinates that
$\\{e_{1},\ldots,e_{n}\\}$ is the standard orthonormal basis of
$\mathbb{R}^{n}$.

We collect the following general properties of periodic admissible set of
points.

######  Lemma 5.9.

(Properties of periodic admissible sets) Let $\mathcal{L}$ be a periodic
admissible set of points. The following holds true:

  * (i)

$B_{r/2}(x)\subset\mathcal{V}(x)\subset B_{R}(x)$ for all $x\in\mathcal{L}$;

  * (ii)

$\mathcal{NN}(x)\subset B_{2R}(x)$ for all $x\in\mathcal{L}$;

  * (iii)

There exists $C=C(n,r,R)\in(0,+\infty)$ such that
$\displaystyle\sup_{x\in\mathcal{L}}\\#\mathcal{NN}(x)\leq C$. In particular,
$\partial\mathcal{V}(x)$ is made out of finitely many $(n-1)$-dimensional
polyhedral faces;

  * (iv)

There exists $C=C_{\mathcal{L}}>0$ such that for all
$X\subset\varepsilon\mathcal{L}$ and $A\subset\mathbb{R}^{n}$ there holds

$\displaystyle F_{\varepsilon}(X,A)\leq
C\,\mathrm{Per}(V_{\varepsilon},(A)_{R\varepsilon})\text{ and
}\,\mathrm{Per}(V_{\varepsilon},A)\leq
CF_{\varepsilon}(X,(A)_{R\varepsilon})\,.$

###### Proof.

Apart from (iv) all of these facts are classical. We collect their proof here
for completeness.
proof of (i), (ii): Let $x\in\mathcal{L}$. The inclusion
$B_{r/2}(x)\subset\mathcal{V}(x)$ follows from (L2) since for all $y\in
B_{r/2}(x)$ and $z\in\mathcal{L}\setminus\\{x\\}$

$\displaystyle|z-y|\geq|x-z|-|x-y|\geq r-r/2=r/2\geq|x-y|\,.$

As for the inclusion $\mathcal{V}(x)\subset B_{R}(x)$ assume that there exists
$y\in\mathcal{V}(x)\setminus B_{R}(x)$. We have for all
$z\in\mathcal{L}\setminus\\{x\\}$

$\displaystyle|y-z|\geq|y-x|\geq R\,.$

This implies that $B_{R}(y)\cap\mathcal{L}=\emptyset$ contradicting (L1).
Finally $\mathcal{NN}(x)\subset B_{2R}(x)$ since for $y\in\mathcal{NN}(x)$ we
have that $\mathcal{V}(x)\cap\mathcal{V}(y)\neq\emptyset$ which implies
$B_{R}(x)\cap B_{R}(y)\neq\emptyset$.

proof of (iii): Due to (i) and (ii) we have that $B_{r/2}(y)\cap
B_{r/2}(z)=\emptyset$ $y,z\in\mathcal{NN}(x)$, $y\neq z$ and
$B_{r/2}(y)\subset B_{2R+r}(x)$ for all $y\in\mathcal{NN}(x)$. Therefore

$\displaystyle\omega_{n}\left(\frac{r}{2}\right)^{n}\\#\mathcal{NN}(x)=\bigcup_{y\in\mathcal{NN}(x)}|B_{r/2}(y)|\leq|B_{2R+r}(x)|\leq\omega_{n}(2R+r)^{n}$

and thus the claim follows with $C=(2+4R/r)^{n}$.
proof of (iv): The desired inequalities follow from the following observation:
Given $x\in X$, we have that

$\displaystyle\mathcal{H}^{n-1}(\mathcal{V}_{\varepsilon}(x)\cap\partial
V_{\varepsilon})>0\iff\mathcal{NN}_{\varepsilon}(x)\setminus
X\neq\emptyset\,.$ (77)

Additionally, we note that, for $x\in X$ such that
$\mathcal{NN}_{\varepsilon}(x)\setminus X\neq\emptyset$, there exists $C>0$
such that

$\displaystyle
C^{-1}\varepsilon^{n-1}\leq\mathcal{H}^{n-1}(\mathcal{V}_{\varepsilon}(x)\cap\partial
V_{\varepsilon})\leq C\varepsilon^{n-1}\,.$ (78)

Now, summing over all $x\in X\cap A$ and noting that each Voronoi cell
intersects only a finite number of other Voronoi cells, using (77),(78),(i),
and (iii), we obtain

$\displaystyle F_{\varepsilon}(X,A)=\sum_{x\in X\cap
A}\varepsilon^{n-1}\\#(\mathcal{NN}_{\varepsilon}(x)\setminus X)$
$\displaystyle\leq C\sum_{x\in X\cap
A}\mathcal{H}^{n-1}(\mathcal{V}_{\varepsilon}(x)\cap\partial V_{\varepsilon})$
$\displaystyle\leq C\mathcal{H}^{n-1}(\partial
V_{\varepsilon}\cap(A)_{R\varepsilon})=C\,\mathrm{Per}(V_{\varepsilon},(A)_{R\varepsilon})\,.$

This yields the first inequality in (iv). On the other hand, owing to (i) we
have $\partial V_{\varepsilon}\cap A\subset\bigcup_{x\in
X\cap(A)_{R\varepsilon}}(\mathcal{V}_{\varepsilon}(x)\cap\partial
V_{\varepsilon})$, and thus, by (77) and (78),

$\displaystyle\mathrm{Per}(V_{\varepsilon},A)=\mathcal{H}^{n-1}(\partial
V_{\varepsilon}\cap A)$ $\displaystyle\leq\sum_{x\in
X\cap(A)_{R\varepsilon}}\mathcal{H}^{n-1}(\mathcal{V}_{\varepsilon}(x)\cap\partial
V_{\varepsilon})$ $\displaystyle\leq C\varepsilon^{n-1}\sum_{x\in
X\cap(A)_{R\varepsilon}}\\#(\mathcal{NN}_{\varepsilon}(x)\setminus
X)=CF_{\varepsilon}(X,(A)_{R\varepsilon})\,.$

This shows the second inequality in (iv) and concludes the proof. ∎

######  Proposition 5.10.

(Compactness of the piecewise-constant interpolants) Let $\mathcal{L}$ be an
admissible periodic set of points and $F_{\varepsilon}$ defined in (71) with
$\mathcal{L}$ in place of $\Sigma$. Let $A\subset\mathbb{R}^{n}$ be open and
let $\\{X_{\varepsilon}\\}_{\varepsilon}\subset\varepsilon\mathcal{L}$ be such
that

$\displaystyle\sup_{\varepsilon>0}F_{\varepsilon}(X_{\varepsilon},A)<+\infty\,.$

Then there exists a set of finite perimeter $V\subset A$ and a subsequence
(not relabeled) such that $\chi_{V_{\varepsilon}}\to\chi_{V}$ with respect to
the strong $L^{1}_{\mathrm{loc}}(A)$-topology.

###### Proof.

Let $X_{\varepsilon}$ be as above and let $A^{\prime}\subset\subset A$ such
that $(A^{\prime})_{R\varepsilon}\subset A$, and $\partial
A^{\prime}\in\mathrm{Lip}$. We observe, due to the second inequality of Lemma
5.9(iv),

$\displaystyle\mathrm{Per}(V_{\varepsilon},A^{\prime})\leq
CF_{\varepsilon}(X_{\varepsilon},(A^{\prime})_{R\varepsilon})\leq
CF_{\varepsilon}(X_{\varepsilon},A)\leq C\,.$

Therefore

$\displaystyle\|\chi_{V_{\varepsilon}}\|_{L^{1}(A^{\prime})}+|D\chi_{V_{\varepsilon}}|(A^{\prime})\leq
C(|A^{\prime}|+\mathrm{Per}(V_{\varepsilon},A^{\prime}))\leq C\,.$

We use [4, Theorem 3.39] to deduce that there exists a subsequence (depending
on $A^{\prime}$) and a set of finite perimeter $V$ such that
$\chi_{V_{\varepsilon}}\to\chi_{V}$ in $L^{1}(A^{\prime})$. By a diagonal
argument on a sequence $A^{\prime}_{k}\uparrow A$ as $k\to+\infty$, we obtain
the claim. ∎

######  Lemma 5.11.

(Equivalence of convergences) Let $\mathcal{L}$ be an admissible periodic set
of points and $F_{\varepsilon}$ defined in (71) with $\mathcal{L}$ in place of
$\Sigma$. Let $A\subset\mathbb{R}^{n}$ be open and let $V\subset A$ be a set
of finite perimeter and let
$\\{X_{\varepsilon}\\}_{\varepsilon}\subset\varepsilon\mathcal{L}$ for each
$\varepsilon>0$ be such that

$\displaystyle\sup_{\varepsilon>0}F_{\varepsilon}(X_{\varepsilon},A)<+\infty\,.$
(79)

Then, setting $\mu_{\varepsilon}$ and $V_{\varepsilon}$ as in (74) and (75),
the following are equivalent:

  * (i)

$\mu_{\varepsilon}\overset{*}{\rightharpoonup}\mu$ with respect to the weak
star topology of measures and $\mu=\rho\mathcal{L}^{n}\llcorner V$.

  * (ii)

$\chi_{V_{\varepsilon}}\to\chi_{V}$ with respect to the strong
$L^{1}_{\mathrm{loc}}(A)$-topology.

###### Proof.

We proceed in two steps. First, we construct a sequence of auxiliary measure
$\nu_{\varepsilon}$ and show that its weak*-convergence is equivalent to the
weak*-convergence of the sequence of measures $\mu_{\varepsilon}$. Then, for
this sequence of measures we show that its weak*-convergence is equivalent to
(ii).
Step 1.(Construction of the auxiliary measure) Let
$\\{X_{\varepsilon}\\}_{\varepsilon}$ be as in the assumptions of the lemma
and let $v\in C_{c}(A)$ such that $\mathrm{supp}\,v\subset\subset A$. We
assume that $\varepsilon>0$ is small enough such that for all
$k\in\varepsilon\mathbb{Z}^{n}$ there holds

$\displaystyle
Q_{(3+R)\varepsilon}(k)\cap\mathrm{supp}\,v\neq\emptyset\implies
Q_{(3+R)\varepsilon}(k)\subset\subset A\,.$ (80)

Fix $R_{0}>0$ such that $\mathrm{supp}\,v\subset B_{R_{0}}$. Since $v$ is
uniformly continuous, it admits a modulus of continuity
$\omega=\omega_{v}\colon[0,+\infty)\to[0,+\infty)$, i.e., an increasing
function such that $\omega(0)=0$ and

$\displaystyle|v(x)-v(y)|\leq\omega(|x-y|)\text{ for all
}x,y\in\mathbb{R}^{n}\,.$ (81)

We set

$\displaystyle\begin{split}&\mathcal{I}_{\varepsilon}^{\mathrm{full}}:=\\{k\in\varepsilon\mathbb{Z}^{d}\colon
Q_{(3+R)\varepsilon}(k)\cap
X=Q_{(3+R)\varepsilon}(k)\cap\varepsilon\mathcal{L}\\}\,,\\\
&\mathcal{I}_{\varepsilon}^{\mathrm{empty}}:=\\{k\in\varepsilon\mathbb{Z}^{d}\colon
Q_{(3+R)\varepsilon}(k)\cap X=\emptyset\\}\,,\end{split}$

and
$\mathcal{I}_{\varepsilon}^{\mathrm{bad}}:=\\{k\in\varepsilon\mathbb{Z}^{n}\colon
Q_{(3+R)\varepsilon}(k)\cap\mathrm{supp}\,v\neq\emptyset\\}\setminus(\mathcal{I}_{\varepsilon}^{\mathrm{full}}\cup\mathcal{I}_{\varepsilon}^{\mathrm{empty}})$.
We now set

$\displaystyle\nu_{\varepsilon}:=\sum_{k\in\mathcal{I}_{\varepsilon}^{\mathrm{full}}}\varepsilon^{n}\\#(\mathcal{L}\cap
Q)\delta_{k}\,.$

Our goal is to show that

$\displaystyle\mu_{\varepsilon}\overset{*}{\rightharpoonup}\mu\quad\text{in
the sense of
measures}\quad\iff\quad\nu_{\varepsilon}\overset{*}{\rightharpoonup}\mu\quad\text{in
the sense of measures.}$ (82)

First of all we claim that

$\displaystyle\\#\mathcal{I}_{\varepsilon}^{\mathrm{bad}}\leq
C\varepsilon^{1-n}\,.$ (83)

In fact, due to its definition and (80), we have that for all
$k\in\mathcal{I}_{\varepsilon}^{\mathrm{bad}}$ there exists $x\in X\cap
Q_{(3+R)\varepsilon}(k)\subset A$ such that $\\#(\mathcal{NN}(x)\setminus
X)\neq\emptyset$. Therefore, since for $k\in\varepsilon\mathbb{Z}^{n}$ fixed
$Q_{(3+R)\varepsilon}(k)\cap Q_{(3+R)\varepsilon}(j)\neq\emptyset$ for only
finitely many $j$ independent of $\varepsilon$, we have

$\displaystyle\varepsilon^{n-1}\\#\mathcal{I}_{\varepsilon}^{\mathrm{bad}}\leq\sum_{k\in\mathcal{I}_{\varepsilon}^{\mathrm{bad}}}\sum_{x\in
Q_{(3+R)\varepsilon}(k)\cap X}\varepsilon^{n-1}\\#(\mathcal{NN}(x)\setminus
X)\leq CF_{\varepsilon}(X_{\varepsilon},A)\,.$

Using (79) yields (83). Let now
$k\in\mathcal{I}_{\varepsilon}^{\mathrm{full}}$. Then

$\displaystyle\int_{Q_{\varepsilon}(k)}v(x)\,\mathrm{d}\nu_{\varepsilon}(x)=v(k)\varepsilon^{n}\\#(\mathcal{L}\cap
Q)$ $\displaystyle=\sum_{x\in
Q_{\varepsilon}(k)\cap\varepsilon\mathcal{L}}\varepsilon^{n}v(k)$
$\displaystyle=\sum_{x\in Q_{\varepsilon}(k)\cap
X}\varepsilon^{n}v(x)+\sum_{x\in
Q_{\varepsilon}(k)\cap\varepsilon\mathcal{L}}\varepsilon^{n}(v(k)-v(x))$
$\displaystyle=\int_{Q_{\varepsilon}(k)}v(x)\,\mathrm{d}\mu_{\varepsilon}(x)+\sum_{x\in
Q_{\varepsilon}(k)\cap\varepsilon\mathcal{L}}\varepsilon^{n}(v(k)-v(x))\,.$

Thus,

$\displaystyle\left|\int_{Q_{\varepsilon}(k)}v(x)\,\mathrm{d}(\nu_{\varepsilon}-\mu_{\varepsilon})(x)\right|\leq\sum_{x\in
Q_{\varepsilon}(k)\cap\varepsilon\mathcal{L}}\varepsilon^{n}|v(k)-v(x)|\leq\omega(\varepsilon\sqrt{n})|Q_{\varepsilon}(k)|$

and, recalling that $\mathrm{supp}\,v\subset B_{R_{0}}$ and (81), we have for
$\varepsilon>0$ small enough

$\displaystyle\sum_{k\in\mathcal{I}_{\varepsilon}^{\mathrm{full}}}\left|\int_{Q_{\varepsilon}(k)}v(x)\,\mathrm{d}(\nu_{\varepsilon}-\mu_{\varepsilon})(x)\right|\leq\omega(\varepsilon\sqrt{n})|B_{2R_{0}}|\,.$
(84)

Noting that both $|\mu_{\varepsilon}|(Q_{\varepsilon}(k))$ and
$|\nu_{\varepsilon}|(Q_{\varepsilon}(k))$ are bounded above by
$\varepsilon^{n}\\#(\mathcal{L}\cap Q)\leq C\varepsilon^{n}$ for all
$k\in\varepsilon\mathbb{Z}^{n}$, using (83), we observe

$\displaystyle\sum_{k\in\mathcal{I}_{\varepsilon}^{\mathrm{bad}}}\left|\int_{Q_{\varepsilon}(k)}v(x)\,\mathrm{d}(\nu_{\varepsilon}-\mu_{\varepsilon})(x)\right|\leq
2\|v\|_{\infty}\\#\mathcal{I}_{\varepsilon}^{\mathrm{bad}}\varepsilon^{n}\\#(\mathcal{L}\cap
Q)\leq C\varepsilon\|v\|_{\infty}\,.$ (85)

Therefore, noting that
$\mu_{\varepsilon}\lfloor_{Q_{\varepsilon}(k)}=\nu_{\varepsilon}\lfloor_{Q_{\varepsilon}(k)}$
for $k\in\mathcal{I}_{\varepsilon}^{\mathrm{empty}}$, using (84) and (85), we
obtain

$\displaystyle\begin{split}\left|\int_{\mathbb{R}^{n}}v(x)\,\mathrm{d}(\nu_{\varepsilon}-\mu_{\varepsilon})(x)\right|&\leq\omega(\varepsilon\sqrt{n})|B_{2R}|+C\varepsilon\|v\|_{\infty}\,.\end{split}$

This shows (82).
Step 2.(Equivalence of convergence) We now prove that

$\displaystyle\chi_{V_{\varepsilon}}\to\chi_{V}\quad\text{in
}L^{1}_{\mathrm{loc}}(A)\quad\iff\quad\nu_{\varepsilon}\overset{*}{\rightharpoonup}\mu\quad\text{in
the sense of measures.}$

First of all, recalling $\rho$ defined in (76), we note

$\displaystyle\begin{split}\int_{\mathbb{R}^{n}}v\,\mathrm{d}\nu_{\varepsilon}&=\sum_{k\in\mathcal{I}_{\varepsilon}^{\mathrm{full}}}\varepsilon^{n}\\#(\mathcal{L}\cap
Q)v(k)=\frac{\\#(\mathcal{L}\cap
Q)}{|Q|}\sum_{k\in\mathcal{I}_{\varepsilon}^{\mathrm{full}}}|Q_{\varepsilon}(k)|v(k)\\\
&=\rho\sum_{k\in\mathcal{I}_{\varepsilon}^{\mathrm{full}}}\int_{Q_{\varepsilon}(k)}v(y)\,\mathrm{d}y+\rho\sum_{k\in\mathcal{I}_{\varepsilon}^{\mathrm{full}}}\int_{Q_{\varepsilon}(k)}(v(k)-v(y))\,\mathrm{d}y\,.\end{split}$
(86)

Now, due to (81), we have for $\varepsilon>0$ small enough

$\displaystyle\left|\sum_{k\in\mathcal{I}_{\varepsilon}^{\mathrm{full}}}\int_{Q_{\varepsilon}(k)}(v(k)-v(y))\,\mathrm{d}y\right|\leq\sum_{k\in\mathcal{I}_{\varepsilon}^{\mathrm{full}}}\int_{Q_{\varepsilon}(k)}|v(k)-v(y)|\,\mathrm{d}y\leq
C\omega(\varepsilon\sqrt{n})|B_{2R_{0}}|\,.$ (87)

Note that, by Lemma 5.9(i), we have

$\displaystyle\rho\sum_{k\in\mathcal{I}_{\varepsilon}^{\mathrm{full}}}\int_{Q_{\varepsilon}(k)}v(y)\,\mathrm{d}y=\rho\sum_{k\in\mathcal{I}_{\varepsilon}^{\mathrm{full}}}\int_{Q_{\varepsilon}(k)\cap
V_{\varepsilon}}v(y)\,\mathrm{d}y$ (88)

and also $V_{\varepsilon}\cap Q_{\varepsilon}(k)=\emptyset$ for
$k\in\mathcal{I}_{\varepsilon}^{\mathrm{empty}}$. Note that by (83) we have

$\displaystyle\sum_{k\in\mathcal{I}_{\varepsilon}^{\mathrm{bad}}}\int_{Q_{\varepsilon}(k)}\left|v(y)\right|\,\mathrm{d}y\leq\varepsilon^{n}\\#\mathcal{I}_{\varepsilon}^{\mathrm{bad}}\|v\|_{\infty}\leq
C\varepsilon\|v\|_{\infty}\,.$ (89)

Due to (86)–(89), we obtain that

$\displaystyle\rho\chi_{V_{\varepsilon}}\overset{*}{\rightharpoonup}\mu\quad\text{in
the sense of
measures}\quad\iff\quad\nu_{\varepsilon}\overset{*}{\rightharpoonup}\mu\quad\text{in
the sense of measures.}$

Now clearly (ii) implies (i), since the $L^{1}_{\mathrm{loc}}(A)$ convergence
of the characteristic functions implies their weak* convergence as measures.
As for the implication (i) to (ii) we proceed as follows. Let $V\subset A$ be
a set of finite perimeter and assume that
$\mu_{\varepsilon}\overset{*}{\rightharpoonup}\mu$ and
$\mu=\rho\mathcal{L}^{n}\llcorner V$. By Step 1 this is equivalent to
$\nu_{\varepsilon}\overset{*}{\rightharpoonup}\mu$, hence to
$\rho\chi_{V_{\varepsilon}}\overset{*}{\rightharpoonup}\rho\chi_{V}$ in the
sense of measures (which is to say that
$\chi_{V_{\varepsilon}}\overset{*}{\rightharpoonup}\chi_{V}$) Take now an
arbitrary subsequence (not relabeled) of
$\\{X_{\varepsilon}\\}_{\varepsilon}$. We show that there exists a further
subsequence (again not relabeled) such that
$\chi_{V_{\varepsilon}}\to\chi_{V}$ with respect to the strong
$L^{1}_{\mathrm{loc}}(A)$-topology. Since $L^{1}_{\mathrm{loc}}(A)$-topology
satisfies the Urysohn property this implies the claim. By the compactness
statement in Proposition 5.10 we have that there exists a set of finite
perimeter $V^{\prime}\subset\mathbb{R}^{n}$ and a further subsequence
$\\{X_{\varepsilon_{k}}\\}_{k}\subset\\{X_{\varepsilon}\\}_{\varepsilon}$ such
that $\chi_{V_{\varepsilon}}\to\chi_{V^{\prime}}$ with respect to the strong
$L^{1}_{\mathrm{loc}}(A)$-topology. Since this implies their weak* convergence
as measures and we already know that the whole sequence converges to
$\chi_{V}$ we deduce $V=V^{\prime}$ which implies the claim and concludes the
proof of the lemma. ∎

######  Corollary 5.12.

(Equivalence of $\Gamma$-convergence) Let $\mathcal{L}$ be an admissible
periodic set of points and let $F_{\varepsilon}$ be defined in (71) with
$\mathcal{L}$ in place of $\Sigma$. Let $A\subset\mathbb{R}^{n}$ be a bounded
open set such that $\partial A\in\mathrm{Lip}$. Then the following statements
are equivalent:

  * (i)

$F_{\varepsilon}\colon\mathcal{M}_{+}(A)\to[0,+\infty]$ $\Gamma$-converges
with respect to the weak* convergence of measures to the functional
$F_{\mathrm{hom}}\colon\mathcal{M}_{+}(A)\to[0,+\infty]$ defined as

$\displaystyle F_{\mathrm{hom}}(\mu,A):=\begin{cases}\int_{\partial^{*}V\cap
A}\varphi_{\mathrm{hom}}(\nu)\,\mathrm{d}\mathcal{H}^{n-1}&\text{if
}\mu=\rho\mathcal{L}^{n}\llcorner V\,;\\\
+\infty&\text{otherwise.}\end{cases}$

  * (ii)

$F_{\varepsilon}\colon L^{1}_{\mathrm{loc}}(A)\to[0,+\infty]$
$\Gamma$-converges with respect to strong $L^{1}_{\mathrm{loc}}(A)$-topology
to the functional $F_{\mathrm{hom}}\colon
L^{1}_{\mathrm{loc}}(A)\to[0,+\infty]$ defined as

$\displaystyle F_{\mathrm{hom}}(u,A):=\begin{cases}\int_{\partial^{*}V\cap
A}\varphi_{\mathrm{hom}}(\nu)\,\mathrm{d}\mathcal{H}^{n-1}&\text{if
}u=\chi_{V}\text{ and }\chi_{V}\in BV(A)\,;\\\
+\infty&\text{otherwise.}\end{cases}$

###### Proof.

The statement follows directly from Lemma 5.11. ∎

######  Remark 5.13.

The analogous statements are true for $\hat{F}_{\varepsilon}$ as well.

######  Theorem 5.14.

($\Gamma$-convergence for periodic admissible lattices) Let $\mathcal{L}$ be
an admissible periodic set of points and let $F_{\varepsilon}$ be defined by
(71) with $\mathcal{L}$ in place of $\Sigma$. Let $A\subset\mathbb{R}^{n}$ be
bounded open set with $\partial A\in\mathrm{Lip}$ or $A=\mathbb{R}^{n}$. The
functionals $F_{\varepsilon}$ $\Gamma$-converge with respect to weak*
convergence of measures to the functional
$F_{\mathrm{hom}}\colon\mathcal{M}_{+}(A)\to[0,+\infty]$ defined by

$\displaystyle F_{\mathrm{hom}}(\mu,A):=\begin{cases}\int_{\partial^{*}V\cap
A}\varphi_{\mathrm{hom}}(\nu)\,\mathrm{d}\mathcal{H}^{n-1}&\text{if
}\mu=\rho\mathcal{L}^{n}\llcorner V\,;\\\
+\infty&\text{otherwise.}\end{cases}$

The function $\varphi_{\mathrm{hom}}\colon\mathbb{R}^{n}\to[0,+\infty]$ is
given by

$\displaystyle\varphi_{\mathrm{hom}}(\nu)=\lim_{T\to+\infty}\frac{1}{T^{n-1}}\inf\left\\{F(X,Q_{T}^{\nu})\colon
X\subset\mathcal{L},\chi_{X}(i)=u_{\nu}(i)\text{ for }i\in\mathcal{L}\setminus
Q^{\nu}_{T-l_{T}}\right\\}\,,$

where $l_{T}\to+\infty$ and $l_{T}/T\to 0$ as $T\to+\infty$.

###### Proof.

Step 1.(Probabilistic setup) We exploit the integral representation result
(Theorem 5.7) to obtain the specific form of the $\Gamma$-limit. We fix
$(\Omega,\mathcal{F},\mathbb{P})=(\\{0\\},\\{\\{0\\},\emptyset\\},\delta_{0})$
to be a probability space and a trivial additive and ergodic group action (see
Definition 5.1 in [3]) $\tau_{z}\colon\\{0\\}\to\\{0\\},z\in\mathbb{Z}^{3}$
given by $\tau_{z}(0)=0$. With respect to this group action
$\mathcal{L}(0)=\mathcal{L}$ is an admissible stationary and ergodic
stochastic lattice according to Definition 5.6. In fact, since $\mathcal{L}$
is periodic according to Definition 5.8, for all
$z=(z_{1},z_{2},z_{3})\in\mathbb{Z}^{3}$ we have

$\displaystyle\mathcal{L}(0)=\mathcal{L}\,,\quad\mathcal{L}(\tau_{z}(0))=\mathcal{L}(0)=\mathcal{L}(0)+\sum_{k=1}^{n}z_{k}e_{k}\,.$

Therefore, all conditions of Theorem 5.7 are satisfied. This shows that for
$\hat{F}_{\varepsilon}\colon L^{1}_{\mathrm{loc}}(A)\to[0,+\infty]$ we have

$\displaystyle\Gamma\text{-}\lim_{\varepsilon\to
0}\hat{F}_{\varepsilon}(\chi_{V},A)=F_{\mathrm{hom}}(\chi_{V},A)$

for all $A\subset\mathbb{R}^{n},\partial A\in\mathrm{Lip},\chi_{V}\in
BV(A,\\{0,1\\})$. Note that, by Corollary 5.12, this is equivalent to saying
that for $\hat{F}_{\varepsilon}\colon\mathcal{M}_{+}(A)\to[0,+\infty]$ we have

$\displaystyle\Gamma\text{-}\lim_{\varepsilon\to
0}\hat{F}_{\varepsilon}(\mu,A)=F_{\mathrm{hom}}(\mu,A)\,.$

This concludes Step 1.
Step 2.($\Gamma$-convergence of $F_{\varepsilon}$) We use the
$\Gamma$-convergence of $\hat{F}_{\varepsilon}$ obtained in Step 1 in order to
prove the $\Gamma$-convergence of $F_{\varepsilon}$. Let us first prove the
result for $A\subset\mathbb{R}^{n}$ open and bounded with $\partial
A\in\mathrm{Lip}$. We note that

$\displaystyle\hat{F}_{\varepsilon}(X,A)\leq F_{\varepsilon}(X,A)\,$

and therefore for all $\mu_{\varepsilon}\overset{*}{\rightharpoonup}\mu$, we
have

$\displaystyle\liminf_{\varepsilon\to
0}F_{\varepsilon}(\mu_{\varepsilon},A)\geq\liminf_{\varepsilon\to
0}\hat{F}_{\varepsilon}(\mu_{\varepsilon},A)\geq F_{\mathrm{hom}}(\mu,A)\,.$

Therefore $\Gamma\text{-}\liminf_{\varepsilon\to 0}F_{\varepsilon}(\mu,A)\geq
F_{\mathrm{hom}}(\mu,A)\,.$ Next, we show the Limsup-inequality. Let $V\subset
A$ be such that

$\displaystyle F_{\mathrm{hom}}(\rho\chi_{V},A)<+\infty\,$ (90)

since otherwise there is nothing to prove. Using Lemma 5.9(iv), we obtain that
$\mathrm{Per}(V,A)<+\infty$, i.e., $V$ is a set of finite perimeter. In
particular, by [25, Theorem 1.3] (noting that $\partial A\in\mathrm{Lip}$
implies that $A$ is an extension domain), there exists an extension
$V_{\mathrm{e}}\subset\mathbb{R}^{n}$ such that

  * (i)

$V_{\mathrm{e}}\cap A=V\cap A$ up to a set of zero $\mathcal{L}^{n}$-measure;

  * (ii)

$\mathrm{Per}(V_{\mathrm{e}},\mathbb{R}^{n})\leq C\mathrm{Per}(V,A)\,;$

  * (iii)

$\mathcal{H}^{d-1}(\partial^{*}V_{\mathrm{e}}\cap\partial A)=0\,.$

Now (ii) together with (90) implies, again by Lemma 5.9(iv), that

$\displaystyle
F_{\mathrm{hom}}(\rho\chi_{V_{\mathrm{e}}},\mathbb{R}^{n})=\int_{\partial^{*}V_{\mathrm{e}}}\varphi_{\mathrm{hom}}(\nu)\,\mathrm{d}\mathcal{H}^{n-1}<+\infty\,.$

Fix $\delta>0$ and let
$\mu_{\varepsilon}^{\delta}\to\rho\chi_{V_{\mathrm{e}}}$ weakly* in
$\mathcal{M}_{+}((A)_{\delta})$ be such that

$\displaystyle\limsup_{\varepsilon\to
0}\hat{F}_{\varepsilon}(\mu_{\varepsilon}^{\delta},(A)_{\delta})\leq
F_{\mathrm{hom}}(\rho\chi_{V_{\mathrm{e}}},(A)_{\delta})\,.$

Note that, due to Lemma 5.9(ii), for every $\varepsilon<\delta/(2R)$ we have
that

$\displaystyle
F_{\varepsilon}(X,A)\leq\hat{F}_{\varepsilon}(X,(A)_{\delta})\,.$

Therefore, recalling Definition 2.2, we obtain,

$\displaystyle\Gamma\text{-}\limsup_{\varepsilon\to
0}F_{\varepsilon}(\rho\chi_{V},A)\leq\limsup_{\varepsilon\to
0}F_{\varepsilon}(\mu_{\varepsilon}^{\delta},A)\leq\limsup_{\varepsilon\to
0}\hat{F}_{\varepsilon}(\mu_{\varepsilon}^{\delta},(A)_{\delta})\leq
F_{\mathrm{hom}}(\rho\chi_{V_{\mathrm{e}}},(A)_{\delta})\,.$

Sending $\delta\to 0$ we obtain

$\displaystyle\Gamma\text{-}\limsup_{\varepsilon\to
0}F_{\varepsilon}(\rho\chi_{V},A)\leq
F_{\mathrm{hom}}(\rho\chi_{V_{\mathrm{e}}},\overline{A})=F_{\mathrm{hom}}(\rho\chi_{V_{\mathrm{e}}},A)=F_{\mathrm{hom}}(\rho\chi_{V},A)\,,$

where the last equality follows by properties (i) and (iii) of
$V_{\mathrm{e}}$. This shows the desired integral representation for all
$A\subset\mathbb{R}^{n}$ such that $\partial A\in\mathrm{Lip}$.
Step 3.(Integral representation on unbounded sets) It remains to prove the
integral representation of the $\Gamma$-limit for $\mathbb{R}^{n}$. The
Liminf-inequality follows by monotonicity since for all $R>0$ and
$X\subset\varepsilon\mathcal{L}$ we have

$\displaystyle F_{\varepsilon}(X,B_{R})\leq F_{\varepsilon}(X)$

and therefore, given $\mu_{\varepsilon}\to\rho\chi_{V}$ weakly* in
$\mathcal{M}_{+}(\mathbb{R}^{n})$, we have

$\displaystyle\liminf_{\varepsilon\to
0}F_{\varepsilon}(\mu_{\varepsilon})\geq\liminf_{\varepsilon\to
0}F_{\varepsilon}(\mu_{\varepsilon},B_{R})\geq
F_{\mathrm{hom}}(\rho\chi_{V},B_{R})\,.$

The claim follows by taking the supremum over $R>0$. We now turn our attention
to the Limsup inequality. We can assume without loss of generality that
$V\subset\mathbb{R}^{n}$ is a set of finite perimeter and

$\displaystyle
C\mathrm{Per}(V,\mathbb{R}^{n})\leq\int_{\partial^{*}V}\varphi(\nu)\,\mathrm{d}\mathcal{H}^{n-1}=F_{\mathrm{hom}}(\rho\chi_{V},\mathbb{R}^{n})<+\infty$
(91)

since otherwise there is nothing to prove. By the isoperimetric inequality
there exists $C>0$ such that

$\displaystyle\min\\{|V|,|\mathbb{R}^{n}\setminus V|\\}^{\frac{n-1}{n}}\leq
C\mathrm{Per}(V,\mathbb{R}^{n})\,.$

Without loss of generality we assume that $|V|<+\infty$. By the Fleming-Rishel
formula we can find $\\{R_{k}\\}_{k}$ such that $R_{k}\to+\infty$ and

$\displaystyle{\rm(i)}\quad\mathcal{H}^{n-1}(V\cap\partial
B_{R_{k}})\leq\frac{1}{k}\quad\text{and}\quad{\rm(ii)}\quad|V\cap
B_{R_{k}}^{c}|\leq\frac{1}{k}\,.$ (92)

We define $V_{k}=V\cap B_{R_{k}}$. Then, thanks to (92)(ii),
$\chi_{V_{k}}\to\chi_{V}$ in $L^{1}(\mathbb{R}^{n})$ and thus also
$\rho\chi_{V_{k}}\to\rho\chi_{V}$ weakly* in
$\mathcal{M}_{+}(\mathbb{R}^{n})$. Furthermore, (92)(i) implies that

$\displaystyle\lim_{k\to+\infty}F_{\mathrm{hom}}(\rho\chi_{V_{k}})=F_{\mathrm{hom}}(\rho\chi_{V})\,.$

It therefore suffices to construct the recovery sequence for
$\rho\chi_{V_{k}}$. Let $S_{k}=R_{k}+2R$ and let
$\mu_{\varepsilon}\to\rho_{V_{k}}$ in $L^{1}(B_{S})$ such that

$\displaystyle\limsup_{\varepsilon\to
0}F_{\varepsilon}(\mu_{\varepsilon},B_{S_{k}})\leq
F_{\mathrm{hom}}(\rho\chi_{V},B_{S_{k}})\,.$ (93)

We modify $\mu_{\varepsilon}$ such that

$\displaystyle F_{\varepsilon}(\mu_{\varepsilon})\leq
F_{\varepsilon}(\mu_{\varepsilon},B_{S_{k}})+o(1)\,,$ (94)

and this, by (93), proves the statement. By Lemma 5.11, there exists
$\eta_{\varepsilon}\to 0$ such that $|V_{\varepsilon}\cap
A_{R_{k},S}|\leq\eta_{\varepsilon}$ with $V_{\varepsilon}$ defined in (75).
Now, let us take
$k\in\\{\lceil\frac{R_{k}}{\varepsilon}+R\rceil,\ldots,\lfloor\frac{S}{\varepsilon}-3R\rfloor\\}=:\mathcal{K}_{\varepsilon}$.
Noting that $\\#\mathcal{K}_{\varepsilon}\approx\varepsilon^{-1}$, Lemma
5.9(i) implies that there exists $k_{\varepsilon}\in\mathcal{K}_{\varepsilon}$
such that

$\displaystyle\omega_{n}\left(\frac{r}{2}\right)^{n}\varepsilon^{n-1}\\#(X_{\varepsilon}\cap
A_{k_{\varepsilon}\varepsilon,(k_{\varepsilon}+3R)\varepsilon})$
$\displaystyle\leq\omega_{n}\left(\frac{r}{2}\right)^{n}\varepsilon^{n}\sum_{k\in\mathcal{K}_{\varepsilon}}\\#(X_{\varepsilon}\cap
A_{k\varepsilon,(k+3R)\varepsilon})$ $\displaystyle\leq C|V_{\varepsilon}\cap
A_{R_{k},S}|\leq C\eta_{\varepsilon}\,.$

Here, we have used that for $k\in\mathcal{K}_{\varepsilon}$ fixed

$\displaystyle A_{k\varepsilon,(k+3R)\varepsilon}\cap
A_{j\varepsilon,(j+3R)\varepsilon}\neq\emptyset$

for finitely many indices $j$ independent of $\varepsilon$ (clearly for
$j>k+3R$ or $j<k-3R$ the intersection is empty). We can therefore define
$\hat{X}_{\varepsilon}=\mathcal{L}\cap B_{k_{\varepsilon}}$ and employ Lemma
5.9(ii) to obtain

$\displaystyle
F_{\varepsilon}(\hat{X}_{\varepsilon})=F_{\varepsilon}(\hat{X}_{\varepsilon},B_{k_{\varepsilon}})$
$\displaystyle\leq
F_{\varepsilon}(X_{\varepsilon},B_{k_{\varepsilon}})+\overline{c}\max_{x\in\mathcal{L}}\\#\mathcal{NN}(x)\varepsilon^{n-1}\\#(X_{\varepsilon}\cap
A_{k_{\varepsilon}\varepsilon,(k_{\varepsilon}+3R)\varepsilon})$
$\displaystyle\leq
F_{\varepsilon}(X_{\varepsilon},B_{S})+C\eta_{\varepsilon}\,,$

which proves (94) and with it the claim. ∎

In the following we show that for minimizing sequences we can improve
Proposition 5.10 to obtain strong $L^{1}(\mathbb{R}^{n})$ compactness. This
implies that the rescaled empirical measures of such sequences converge to a