#  $\Gamma$-Convergence for plane to wrinkles transition problem

Peter Bella111 Technische Universität Dortmund Fakultät für Mathematik, 44227
Dortmund, Germany. E-mail<EMAIL_ADDRESS>Roberta
Marziani222Dipartimento di Ingegneria e Scienze dell’Informazione e
Matematica, 67100 L’Aquila, Italy. E-mail<EMAIL_ADDRESS>
###### Abstract

We consider a variational problem modeling transition between flat and
wrinkled region in a thin elastic sheet, and identify the $\Gamma$-limit as
the sheet thickness goes to $0$, thus extending the previous work of the first
author [Bella, ARMA 2015]. The limiting problem is scalar and convex, but
constrained and posed for measures. For the $\Gamma-\liminf$ inequality we
first pass to quadratic variables so that the constraint becomes linear, and
then obtain the lower bound using Reshetnyak’s theorem. The construction of
the recovery sequence for the $\Gamma-\limsup$ inequality relies on
mollification of quadratic variables, and careful choice of multiple
construction parameters. Eventually for the limiting problem we show existence
of a minimizer and equipartition of the energy for each frequency.

##  1 Introduction

This paper is about fine analysis of minimizers of a nonconvex variational
problem which describes wrinkling of thin elastic sheets.

Motivated by some physical experiments with thin elastic sheets [22, 23, 24],
the first author, in his PhD thesis [8] (see also [5]), considered a specific
variational problem describing deformations of a thin elastic sheet
$\Omega_{h}\subset\mathbb{R}^{3}$ of thickness $h$ and cross section of
annular shape. The elastic energy, corresponding to a deformation
$v\colon\Omega_{h}\to\mathbb{R}^{3}$, consists of a membrane term, measuring
stretching and compression of the sheet, and of a bending term, which
penalizes curvature. As a proxy for the energy one can think of

$E_{h}(v)=\int_{\Omega_{h}}\mathrm{dist}^{2}(\nabla v,{\mathrm{SO}}(3))\,{\rm
d}x+h^{2}\int_{\Omega_{h}}|\nabla^{2}v|^{2}\,{\rm d}x.$ (1.1)

The membrane part is non-convex, possibly giving rise to oscillations. In
contrast, the latter bending part is convex and of higher-order, thus
regularizing the problem. Since the bending resistance is related to the sheet
thickness $h$, the magnitude of this contribution asymptotically vanishes in
the limit $h\searrow 0$.

The physics approach to tackle these problems consists of a specific choice of
an ansatz (guess) for the form (shape) of a minimizer. In other words, one
restricts the analysis to a class of competitors having specific
characteristics, and look for a minimizer of the energy within that class. On
the other hand, the rigorous analytical approach does not make any assumptions
on the form of a minimizer, i.e., the energy is minimized over all possible
deformations. The problem in (1.1) being non-convex, hence possibly possessing
many (local) minimizers or critical points, the first step is to understand
the minimal value of the energy, with possibly learning some clues by which
deformations is this minimal value, at least approximately, achieved.

Figure 1: Elastic annular membrane stretched in the radial direction. The blue
dotted curve represents the free boundary that separates the outer stretched
region from the inner wrinkled one.

Hence, we first try to identify the minimal value of the energy. Precisely, in
the present situation, the goal is to understand its dependence on the (small)
sheets thickness $h$. It turns out that the minimal value $\min_{v}E_{h}(v)$
consists of a leading zeroth-order term $\mathcal{E}_{0}$ (coming from the
stretching of the sheet) plus a linear correction in $h$, which corresponds to
the cost of _wrinkling_ of the sheet [8, 5]. More precisely, there exist two
constants $0<C_{0}<C_{1}<\infty$ such that for any thickness $0<h<1$ there
holds

$\mathcal{E}_{0}+C_{0}h\leq\min_{v}E_{h}(v)\leq\mathcal{E}_{0}+C_{1}h.$ (1.2)

The wrinkling serves as a mechanism to relieve compressive stresses, which are
caused by specific geometrical effects. An alternative to wrinkling would be
simply compression, which contributes to the membrane part at the order
$O(1)$. Hence, in the case of small thickness (present situation) compression
is much less energetically favorable ($O(1)$ vs $O(h^{2})$), and thus not
expected.

The identified linear scaling law (1.2) in $h$ for the minimal value of the
energy raised a lot of discussion among the physics community, having improved
their ansatz-based prediction by a factor of $\log h$ (i.e. $h$ in [5] vs
$h{(|\log h|+1)}$ in [22]). It turns out that this discrepancy is related to a
suboptimal choice of the ansatz close to the interface between the wrinkled
and the flat region. Moreover, the upper bound in (1.2) is achieved through a
complex construction involving branching effects, a pattern which was not
observed experimentally.

To shed light on this discrepancy, the first author considered a variational
problem modelling the transition region [9] with the aim of better
understanding the behavior of the minimizer in that region. Working at the
level of the energy, this means to consider the quantity
$\frac{\min_{u}E_{h}(u)-\mathcal{E}_{0}}{h}$, which not only is bounded away
from $0$ and $\infty$ (see (1.2)), but as $h\searrow 0$ it actually converges
to some value $\sigma$ (as proven in [9]). Even though the value $\sigma$ is
characterized as a limit of minima of simpler scalar and convex variational
problems, it _does not_ provide any information on the form of sequence of
minimizers.

In that respect, the goal of this paper is to overcome this shortcoming by
showing $\Gamma$-convergence of the functionals
$\frac{E_{h}-\mathcal{E}_{0}}{h}$ as $h\searrow 0$. As usual, a consequence of
$\Gamma$-convergence is convergence of minimizers $u_{h}$ of $E_{h}$ to a
minimizer of the limiting problem, hence providing information on $u_{h}$, at
least for $0<h\ll 1$. Denoting by $\mathcal{F}_{\infty}$ the $\Gamma$-limit
functional (see (2.9) below), it turns out that as expected from [9],
$\mathcal{F}_{\infty}$ is scalar and convex, thus possibly much easier to
analyze than the original $E_{h}$. Nevertheless, the study of minimizers of
$\mathcal{F}_{\infty}$ is still far from trivial and we postpone it to a
future work – except for some preliminary results collected in Section 6.

There are many areas of material science, most of them falling within a class
of energy-driven pattern formation [29], where the idea to study energy
scaling laws for variational problems turned out to be very fruitful. The
common features of these problems is the presence of a nonconvex term in the
energy, which is regularized by a higher-order term with a small prefactor.
This small parameter (for now denoted $\varepsilon$) has different meanings:
thickness in the case of elastic films, inverse Ginzburg-Landau parameter in
the theory of superconductors, strength of the interfacial energy for models
of shape-memory alloys or micromagnetics, to name just few. As
$\varepsilon\searrow 0$, the oscillations caused by the nonconvexity are less
penalized, giving the energy more freedom to form patterns/microstructure.

The first paper in this direction, in the context of shape memory alloys, is a
seminal work of Kohn and Müller [28], where they studied a toy problem to
model the interface in the austenite-martensite phase transformation. They
showed that the energy minimum scales like $\epsilon^{2/3}$, which was in
contrast with the scaling $\varepsilon^{1/2}$, widely accepted in the physics
community. More precisely, the physics arguments were based on an ansatz of
“one-dimensional” structure of minimizers, whereas Kohn and Müller used a
branching construction to achieve lower energy. While they did not show the
form of minimizers, they provided localized (in one direction) estimates on
the energy distribution for the minimizer – thus providing hints on scales
used for branching. Subsequently, Conti [20] used an intricate upper bound
construction to show localized energy bounds (in both directions), which in
particular implies asymptotical self-similarity of the minimizer close to the
interface. The analysis of the toy model was later generalized in several
directions, for example analysis based on energy scalings laws for the cubic-
to-tetragonal phase transformation - e.g. rigidity of the microstructure [17,
18] or study of the energy barrier for the nucleation in the bulk [27] and at
the boundary [3]. In that respect it is worth to also mention recent works of
Rüland and Tribuzio [44, 43], where a novel use of Fourier Analysis allows to
obtain sharp lower bounds on the energy on a more advanced model.

The work of Kohn and Müller initiated many developments in other areas of
material science to study pattern formation driven by the energy minization,
for example in micromagnetics [37, 16, 41], island growth on epitaxially
strained films [4], diblock copolymers [19], optimal design [33, 32], or
superconductors [46]. Picking one of them as an example, the Ginzburg-Landau
model describes behavior of superconductors in different regimes of the
applied magnetics field. While for extreme values of the magnetic field (very
small or very large) there is only one (normal or superconducting) phase, for
intermediate values of the field the mixed states consisting of many vortices
are observed. There the leading order energy characterizes the number of
vortices, and the next order in the energy describes interaction between them
(see [46] for a survey, [42] for analysis in three spatial dimensions, and
[39] for a similar work in the context of $2d$ Coulomb gases).

The models for wrinkling of thin elastic films have similar feature, with the
leading order term in the energy expansion encoding the wrinkled regions while
the next term in the energy expansion being related to the form (e.g.
lengthscale) of wrinkling. The relevant physical object being a two
dimensional (thickened) surface in $\mathbb{R}^{3}$, the local energy expense
of a deformation $u$ is encoded using two principal values of a $3\times 2$
matrix $\nabla u$ – heuristically, singular value greater or smaller than $1$
corresponds to a tension or a compression, respectively. Wrinkling being an
energetically efficient alternative to a compression, we expect it to appear
in the case of (at least) one singular value being less than one.

A compressed elastic sheet can feel the compression in one (“tensile
wrinkling”) or both directions (“compressive wrinkling”). A class of problems
falling into the latter category for which the energy scaling laws were
identified include for instance blistering/delamination problem (with [30, 2,
14, 38] or without [26, 13, 12] substrate effects), crumpling of elastic
sheets [21, 50], or analysis of conical singularities in elastic sheets [36,
35]. A common feature of this problem is degeneracy of the relaxed energy: the
minimum of the relaxed energy $\mathcal{E}_{0}$ equals zero, and more
importantly it is achieved by many different minimizers, making the analysis
of the next order expansion of the energy often difficult.

In contrast, tensile wrinkling problems usually have relaxed problem with
unique minimizer, making the analysis of the next order term (which describes
wrinkling) more accessible. The need for compression usually comes from the
prescribed boundary conditions (as for example in the raft problem [15, 25],
twisted ribbon [31], hanging drapes [49, 6], or compressed cylinder [47]),
through prescribed incompatible strain [10, 34] or curvature effects [7, 11,
48].

The model we consider here is a mixture of the first and the second case,
i.e., it is driven both by the boundary conditions as well as prescribed
nontrivial metric (prestrain). The latter should mimic the need to “waste the
length” in one direction, this need coming from geometric effects in our
original motivation [5]. More precisely, in [5] an elastic annulus is
stretched radially with stronger inner loads, forcing some of the concentric
circles of material to move closer to the center. Pushing some circles into
less space naturally force compression or wrinkling out of plane, while the
circles towards outer boundary stay planar (and are actually stretched in the
azimuthal direction). As we will see, it is crucial that the amount of
arclength grows linearly in the distance from the free boundary (between the
wrinkled and planar region) and not slower (e.g. quadratically) – the latter
case is expected to be quite boring with the minimizer using only _one_
frequency. In contrast, the present problem requires infinitely many
frequencies, in particular near the transition higher and higher frequencies
are needed.

The rest of this section will provide an overview of our results and
organization of the paper. As in [9], we consider a specific thickness
dependent energy $E_{h}$ (see (2.1) for its precise definition), a model
problem describing transition between planar and wrinkled region in thin
elastic sheet, and are interested to understand structure of minimizers of the
energies as $h\searrow 0$. We consider a thin elastic sheet of thickness $h$
and cross section of rectangular shape
$[-1,1]\times[-1,1]\subset\mathbb{R}^{2}$, which represents a piece of the
elastic annulus depicted above in Figure 1 by a green region, and assume the
sheet is i) stretched in the $x$-direction, and ii) stretched/compressed in
the $y$-direction proportional to $x$ (i.e., it is unstrained for $x=0$,
stretched in the $y$-direction in the left half and compressed in the right
half of the domain). The streching/compression in the $y$-direction is modeled
via prescribed metric together with periodic boundary conditions at the top
and bottom boundary.

To relax the compression in the region $\\{x>0\\}$ we expect the sheet to
wrinkle, with the lengthscale of wrinkles of order $h^{1/2}$ [5]. In order to
analyze the limit of $\frac{E_{h}-\mathcal{E}_{0}}{h}$ as $h\searrow 0$, we
rescale the $y$-variable by $h^{-1/2}$ so that the wrinkles lengthscales stay
of order $1$, and the out-of-plane displacement has chance to converge to some
limiting shape. Though the rescaling cause changes of the domain to
$[-1,1]\times[-L,L]$ for $L:=h^{-1/2}\to\infty$, in particular the
$\Gamma$-limit of the functionals is not clear due to changing domain, we pass
to the Fourier space to avoid these complications. More precisely, we rewrite
the energy using Fourier expansion in $y$, with $L$ appearing through the
summation set $\frac{\pi\mathbb{Z}}{L}$. Heuristically, as $L\to 0$ the
Fourier sum will turn into an integral, hence there is a hope for a limiting
functional to make sense.

This strategy was successfully pursued by the first author in [9], by
observing i) the out-of-plane displacement $u$ being the only relevant
quantity to be monitored in this limit, and ii) for fixed (large) $L=h^{-1/2}$
the minimum of the excess energy $\frac{E_{h}-\mathcal{E}_{0}}{h}$ is well
approximated by minimum of a _scalar, convex, and constrained_ variational
problem for $u$ of the form

$\mathcal{S}_{L}(u):=\int_{0}^{1}\fint_{-L}^{L}u_{,x}^{2}+u_{,yy}^{2}\,{\rm
d}x\,{\rm d}y\quad\textrm{ subject to
}\quad\fint_{-L}^{L}u_{,y}^{2}(x,y)\,{\rm d}y=2x+o(1)\quad\textrm{for a.e.
}x\in(0,1)$ (1.3)

Denoting by $a_{k}(x)$ the Fourier coefficients in $y$ of $u(x,\cdot)$, we can
rewrite

$\mathcal{S}_{L}(u)=\int_{0}^{1}\sum_{k\in\frac{\pi\mathbb{Z}}{L}}(\dot{a}_{k}^{2}(x)+a_{k}^{2}(x)k^{4})\,{\rm
d}x,\quad\text{ and }\quad\fint_{-L}^{L}u_{,y}^{2}(x,y)\,{\rm
d}y=\sum_{k\in\frac{\pi\mathbb{Z}}{L}}a_{k}^{2}(x)k^{2}\,,$

where “dot” denotes the derivative. The main achievements of [9] was to show
that minima of $\mathcal{S}_{L}$ converge, and then to construct a recovery
sequence for the original energy $E_{h}$, including construction of the in-
plane displacement. Since the elastic energy $E_{h}$ includes all second
derivatives of $u$, and not only $u_{,yy}$ which appears in $\mathcal{S}_{L}$,
regularity statement for the minimizers of $\mathcal{S}_{L}$’s played a
crucial role for the construction of the recovery sequence.

The analysis of minima of $\mathcal{S}_{L}$ from [9] completely avoided the
notion of convergence of minimizers, which needs to be an integral part of a
$\Gamma$-convergence which we study here. To avoid the issue of nonlinear
constraint we use quadratic variables (i.e. monitoring $b_{k}:=a_{k}^{2}$
instead of $a_{k}$), which turns the constraint into a linear one. The second
term in the energy $\mathcal{S}_{L}$ becomes also linear, while the first term
can be rewritten as $\frac{(\dot{b}_{k})^{2}}{4b_{k}}$. One disadvantage of
this approach is the $L^{1}$-framework, which naturally leads the limit
functional to be defined on the space of measures. However, the constraint
provides a good pointwise control in $x$, in particular the limiting measure
can be written as a product of $\,{\rm d}x$ and $x$-dependent measures in $k$.
The lower bound argument (Proposition 4.2) is obtain using Reshetnyak Theorem.

The upper bound (construction of a recovery sequence) is much more tricky
since it needs to be done for any “limiting” measure with finite energy, in
contrast with [9], where it was done just for one (more regular) minimizer.
The proof of the upper bound (Proposition 5.1) consists of several steps:

  1. 1.

Given a limiting measure, to obtain $a_{k}$’s we will “discretize” the measure
in the $k$-variable (Lemma 5.3). Moreover, using smoothing of $a_{k}$’s (more
precisely of $a_{k}^{2}$), for which we need to extend the coefficients
$a_{k}$ from $[0,1]$ via dilation into larger interval $[0,1+]$, we get a good
starting point for the construction.

  2. 2.

Careful choice of the smoothing scale $\varepsilon(L)$ together with few other
parameters allow for definition of the out-of-plane displacement (see Lemma
5.4), which is then basis for the construction of the in-plane displacement as
well as estimates on the excess energy (Proposition 5.1).

The paper is organized as follows: in the next Chapter we provide a derivation
of the energy, including the functional-analytical framework in form of
measures with well-behaved distributional derivatives in $x$, as well as
rewriting the energy to a form compatible also with this framework. Finally,
we state the main $\Gamma$-convergence result of this paper Theorem 2.5. In
Chapter 3 we show how to disintegrate the limiting measures, while in the
subsequent Chapter we show the compactness for a sequence with excess energy
(Proposition 4.1). Subsequently we also show the lower bound (Proposition
4.2). The upper bound construction is content of Chapter 5. Eventually, in the
last Chapter we state and prove existence of minimizer (as a measure) for the
limiting energy as well as pointwise (in $k$) equipartion of the energy for
this minimizer (see Theorem 6.2). A finer analysis of this minimizer will be
pursued in a future work of the authors.

##  2 Setting of the problem and main results

We start by collecting some notation we will use throughout the paper.

Notation.

  1. $(a)$

$\chi_{A}$ denotes the characteristic function of the set $A$;

  2. $(b)$

$\mathcal{L}^{1}$ denotes the 1-dimensional Lebesgue measure;

  3. $(c)$

$\delta_{k}$ denotes the Dirac measure on $k\in\mathbb{R}$;

  4. $(d)$

$\mathcal{M}_{b}(A)$ denotes the space of bounded Radon measures on $A$ with
$A\subset\mathbb{R}^{2}$ Borel measurable;

  5. $(e)$

$\mathcal{M}^{+}_{b}(A)$ denotes the subspace of $\mathcal{M}_{b}(A)$ of
positive bounded Radon measures;

  6. $(f)$

For a function $f\colon A\subset\mathbb{R}\to\mathbb{R}$ we denote by
$\dot{f}(x)$ and $\ddot{f}(x)$ the first and the second derivative,
respectively;

  7. $(g)$

For a function $u\colon A\subset\mathbb{R}^{2}\to\mathbb{R}$ we denote by
$u_{,\small\underbrace{x\dots x}_{i\text{ times}}\small\underbrace{y\dots
y}_{j\text{ times}}}$ its partial derivative

$D^{i+j}u(x,y)=\frac{\,{\rm d}^{j}}{\,{\rm d}y^{j}}\frac{\,{\rm d}^{i}}{\,{\rm
d}x^{i}}u(x,y)\,,\quad i,j\in\mathbb{N},\,1\leq i+j\leq 3\,;$

  8. $(h)$

For a measure $\mu\in\mathcal{M}_{b}(A)$ we denote by $\mu_{,x}$ its
distributional derivative with respect to the first variable;

  9. $(i)$

For a measure $\mu\in\mathcal{M}_{b}(A)$ we denote by
$|\mu|\in\mathcal{M}^{+}_{b}(A)$ its total variation;

  10. $(j)$

For $\tilde{\mu}=(\mu_{1},\mu_{2})\in(\mathcal{M}_{b}(A))^{2}$ we analogously
denote by $|\tilde{\mu}|\in\mathcal{M}^{+}_{b}(A)$ its total variation;

  11. $(k)$

For $\mu_{1}\in\mathcal{M}_{b}(A)$, $\mu_{2}\in\mathcal{M}^{+}_{b}(A)$ we
write $\mu_{1}\ll\mu_{2}$ if $\mu_{1}$ is absolute continuous with respect to
$\mu_{2}$ and we indicate by $\frac{\,{\rm d}\mu_{1}}{\,{\rm d}\mu_{2}}\in
L^{1}(A,\mu_{2})$ the associated density (Radon-Nikodým derivative).

The Model. Let us now describe the model (energy) for the transition between
the flat and wrinkled region, which the first author already analyzed in [9].
Instead of considering the annular elastic sheet as in [5], we consider only a
rectangular piece (cut off from the sheet) near the transition region, in
particular simplifying the problem by avoiding the need to work in the radial
geometry. The annular sheet in [5] is stretched in the radial direction and
the concentric circles close to the transition region are stretched/compressed
proportional to the distance from the free boundary. We will model the radial
stretching by dead tension loads in the horizontal direction, while the
stretching/compression in the vertical direction will be modeled by
prescribing a non-euclidean metric of the form $\,{\rm d}x+(1+x)\,{\rm d}y$.
Moreover, the rectangle being part of the annulus, we prescribe periodic
boundary conditions in the vertical direction.

It is physically natural [22] and mathematically convenient to use “small-
slope” geometrically linear Föppl-von Kármán theory. In the membrane part of
the energy the in-plane displacement is represented via the _linear_ strain
while the out-of-plane displacement is kept non-linear (quadratic). The
bending part is modeled by simply $L^{2}$ norm of the Hessian of the out-of-
plane displacement instead of $L^{2}$ norm of the second fundamental form.
Denoting by $w=(w_{1},w_{2})$ and $u$ the in-plane and out-of-plane
displacement respectively, the elastic energy $E_{h}$ (normalized per unit
thickness) has the form

$\begin{split}E_{h}(w,u):=&\frac{1}{2}\int_{-1}^{1}\int_{-1}^{1}|e(w)+\frac{1}{2}\nabla
u\otimes\nabla u-xe_{2}\otimes e_{2}|^{2}\,{\rm d}x\,{\rm d}y\\\
&+\frac{1}{2}\int_{-1}^{1}\int_{-1}^{1}h^{2}|\nabla^{2}u|^{2}\,{\rm d}x\,{\rm
d}y-\int_{-1}^{1}(w_{1}(1,y)-w_{1}(-1,y))\,{\rm d}y.\end{split}$ (2.1)

Here $e(w):=(\nabla w+\nabla^{T}w)/2$ denotes the symmetric gradient of $w$
and $xe_{2}\otimes e_{2}$ is the deviation of the prescribed metric from the
euclidean one. The third integral models the applied tensile dead loads in the
horizontal direction. The factor $1/2$ in front of the elastic energy is
chosen for convenience, and can be changed to any factor using simple
rescaling of $w$ and $u$. Finally, we assume the displacement $(w,u)$ is
$2$-periodic in the second variable.

The behavior of $E_{h}$ as $h\to 0$ at the leading order is well understood
using relaxation techniques [40] (also called tension-field theory in the
mechanics community). Applied to $E_{h}$ from (2.1), in the limit $h\to 0$ the
bending term simply disappears, and the integrand in the membrane term changes
to $\min_{A\geq 0}|e(w)-xe_{2}\otimes e_{2}+A|^{2}$. Hence, one can explicitly
compute the (unique) minimizer of the relaxed energy ($w_{2}=0$ and $w_{1}=x$)
and its minimum $-2+\frac{1}{3}=-\frac{5}{3}=:\mathcal{E}_{0}$.

From [5] we know that the next term in the energy $E_{h}$ scales linearly in
$h$, hence the right quantity to look at is the rescaled _excess energy_
$\frac{E_{h}-\mathcal{E}_{0}}{h}$. For $x>0$ one expects that the sheet
wrinkles out-of-plane in the $y$-direction, in order to offset $-xe_{2}\otimes
e_{2}$ with $u_{,y}^{2}$. The linear scaling in $h$ predicts
$h^{2}|\nabla^{2}u|^{2}\sim h$, in particular $u_{,yy}$ (its largest
component) to be of order $h^{-1/2}$. In particular, the scale of wrinkles in
the bulk should be reciprocal of this value, i.e., $h^{1/2}$. Not
surprisingly, this is also the scale used in the upper bound construction in
[5].

In order to analyze limiting form of the wrinkles as $h\to 0$ we rescale the
$y$-variable by a factor $L:=h^{-1/2}$, so that the characteristic lengthscale
of wrinkles becomes $1$. Precisely, after performing the change of variables

$\hat{w}_{1}(x,y):=w_{1}(x,L^{-1}y),\quad\hat{w}_{2}(x,y):=Lw_{2}(x,L^{-1}y),\quad\hat{u}(x,y):=Lu(x,L^{-1}y),$

the energy $E_{h}$ becomes (see [9, page 630] for a straightforward algebraic
manipulation)

$\displaystyle\mathcal{E}_{L}(w,u):=$
$\displaystyle\fint_{-L}^{L}\int_{-1}^{1}\biggl{(}\biggl{(}w_{1,x}+\frac{u_{,x}^{2}}{2L^{2}}-1\biggr{)}^{2}-1\biggr{)}\,{\rm
d}x\,{\rm
d}y+\fint_{-L}^{L}\int_{-1}^{1}\biggl{(}w_{2,y}+\frac{u_{,y}^{2}}{2}-x\biggr{)}^{2}\,{\rm
d}x\,{\rm d}y$
$\displaystyle+\frac{1}{L^{2}}\fint_{-L}^{L}\int_{-1}^{1}{\Big{(}L^{2}w_{1,y}+w_{2,x}+u_{,x}u_{,y}\Big{)}^{2}}\,{\rm
d}x\,{\rm
d}y+\frac{1}{L^{2}}\fint_{-L}^{L}\int_{-1}^{1}\big{(}u^{2}_{,x}+u^{2}_{,yy}\big{)}\,{\rm
d}x\,{\rm d}y$ (2.2)
$\displaystyle+\frac{1}{L^{4}}\fint_{-L}^{L}\int_{-1}^{1}\Big{(}2u^{2}_{,xy}+\frac{1}{L^{2}}u^{2}_{,xx}\Big{)}\,{\rm
d}x\,{\rm d}y\,.$ (2.3)

Thus, the functional is defined as $\mathcal{E}_{L}\colon\mathcal{A}_{L}^{\rm
in}\times\mathcal{A}_{L}^{\rm out}\to[0,+\infty]$, where the function spaces
describing admissible deformations have the form

$\mathcal{A}_{L}^{\rm in}:=\Big{\\{}w=(w_{1},w_{2})\in
W^{1,2}((-1,1)\times\mathbb{R};\mathbb{R}^{2})\colon w(x,\cdot)\text{ is
$2L$-periodic $\forall x\in(-1,1)$}\Big{\\}},$ (2.4) $\mathcal{A}_{L}^{\rm
out}:=\Big{\\{}u\in W^{2,2}((-1,1)\times\mathbb{R})\colon u(x,\cdot)\text{ is
$2L$-periodic $\forall x\in(-1,1)$}\Big{\\}}.$ (2.5)

Furthermore, the $\frac{E_{h}-\mathcal{E}_{0}}{h}$ turns into
$\mathcal{F}_{L}\colon\mathcal{A}_{L}^{\rm in}\times\mathcal{A}_{L}^{\rm
out}\to\mathbb{R}$ defined as

$\mathcal{F}_{L}(w,u):=L^{2}(\mathcal{E}_{L}(w,u)-\mathcal{E}_{0})\,,$ (2.6)

where $\mathcal{E}_{0}=-\frac{5}{3}$ is as above the minimum of the relaxed
energy, so that

$\begin{split}\mathcal{F}_{L}(w,u)=L^{2}&\fint_{-L}^{L}\int_{-1}^{1}\biggl{(}w_{1,x}+\frac{u_{,x}^{2}}{2L^{2}}-1\biggr{)}^{2}\,{\rm
d}x\,{\rm
d}y-\frac{L^{2}}{3}+L^{2}\fint_{-L}^{L}\int_{-1}^{1}\biggl{(}w_{2,y}+\frac{u_{,y}^{2}}{2}-x\biggr{)}^{2}\,{\rm
d}x\,{\rm d}y\\\
&+\fint_{-L}^{L}\int_{-1}^{1}\Big{(}L^{2}w_{1,y}+w_{2,x}+u_{,x}u_{,y}\Big{)}^{2}\,{\rm
d}x\,{\rm
d}y+\fint_{-L}^{L}\int_{-1}^{1}\big{(}u^{2}_{,x}+u^{2}_{,yy}\big{)}\,{\rm
d}x\,{\rm d}y\\\
&+\frac{1}{L^{2}}\fint_{-L}^{L}\int_{-1}^{1}\Big{(}2u^{2}_{,xy}+\frac{1}{L^{2}}u^{2}_{,xx}\Big{)}\,{\rm
d}x\,{\rm d}y\,.\end{split}$

Before we rigorously proceed further, let us discuss heuristically the form of
functional $\mathcal{F}_{L}$ and its implications. Most of the terms in the
energy are of quadratic nature, and since in addition we are dealing with
oscillatory objects defined on longer and longer intervals, it is natural to
look at the problem in the Fourier space.

Expecting the limit of $\mathcal{F}_{L}$ to exists (in particular having the
minimizing sequence bounded as $L\to\infty$), both integrals on the first line
needs to (quickly) converge to $0$. The first integral can easily achieve that
by simply choosing $w_{1}\sim x+o(L^{-1})$ and $u_{,x}$ not too big, the
smallness of the second integral (after integration in $y$ and using
periodicity of $w$) implies the constrain $\fint_{-L}^{L}u_{,y}^{2}\,{\rm
d}y=2x+o(L^{-1})$.

In order to have continuity of the constraint in the limit $L\to\infty$, and
also for other reasons which will be apparent later, we will work with squares
of the Fourier coefficients and suitably defined measures as primary objects
of studies. In the following we denote by $k\in\mathbb{R}$ the variable
corresponding to the Fourier transform in the $y$-variable. Moreover, we use
the same notation to denote the second variable when working with measures.

######  Definition 2.1 (Measures $\mu^{L}$ and $\mu^{L}_{,x}$).

Let $u\in\mathcal{A}_{L}^{\rm out}$. We denote by
$\mu^{L}(u)\in\mathcal{M}_{b}^{+}((-1,1)\times\mathbb{R})$ the measure given
by

$\mu^{L}(u):=\sum_{k\in\frac{\pi\mathbb{Z}}{L}}a(x,k)\mathcal{L}^{1}\mathop{\hbox{\vrule
height=7.0pt,width=0.5pt,depth=0.0pt\vrule
height=0.5pt,width=6.0pt,depth=0.0pt}}\nolimits{(-1,1)}\times\delta_{k}\,,$

with

${a}(x,k):=k^{2}a_{k}^{2}(x)\,,$

and $a_{k}\in W^{2,2}(-1,1)$ being the $k$-th Fourier coefficient of
$u(x,\cdot)$ for all $x\in(-1,1)$, that is

$a_{k}(x):=\begin{cases}\displaystyle\sqrt{2}\fint_{-L}^{L}u(x,y)\sin(ky)\,{\rm
d}y&k\in\dfrac{\pi\mathbb{Z}}{L},k>0\,,\\\\[10.00002pt]
\displaystyle\sqrt{2}\fint_{-L}^{L}u(x,y)\cos(ky)\,{\rm
d}y&k\in\dfrac{\pi\mathbb{Z}}{L},k<0\,,\\\\[10.00002pt]
\displaystyle\fint_{-L}^{L}u(x,y)\,{\rm d}y&k=0\,.\end{cases}$ (2.7)

Moreover we denote by $\mu^{L}_{,x}(u)$ the distributional $x$-derivative of
$\mu^{L}(u)$.

######  Remark 2.2.

  1. $(i)$

The distributional $x$-derivative of a measure
$\mu\in\mathcal{M}_{b}^{+}((-1,1)\times\mathbb{R})$ is defined as follows: for
all $\varphi\in C^{\infty}_{c}((-1,1)\times\mathbb{R})$ we have

$\langle\mu_{,x},\varphi\rangle:=-\int_{(-1,1)\times\mathbb{R}}\varphi_{,x}\,{\rm
d}\mu\,.$

Moreover by a density argument $\mu_{,x}$ can be extended to functions
$\varphi(x,k)=\phi(x)\mathbbm{1}_{A}(k)$ with $\phi\in C^{\infty}_{c}(-1,1)$
and $A\subset\mathbb{R}$ bounded and measurable as

$\langle\mu_{,x},\phi(x)\mathbbm{1}_{A}(k)\rangle:=-\int_{(-1,1)\times
A}\dot{\phi}(x)\,{\rm d}\mu\,;$

  2. $(ii)$

Let $\mu\in\mathcal{M}_{b}^{+}((-1,1)\times\mathbb{R})$ be of the form

$\mu=\sum_{k\in K}a(x,k)\mathcal{L}^{1}\mathop{\hbox{\vrule
height=7.0pt,width=0.5pt,depth=0.0pt\vrule
height=0.5pt,width=6.0pt,depth=0.0pt}}\nolimits(-1,1)\times\delta_{k}\,,$

with $K\subset\mathbb{R}$ countable and $a(\cdot,k)\in W^{1,1}(-1,1)$ for all
$k\in K$. Then

$\mu_{,x}=\sum_{k\in K}a_{,x}(x,k)\mathcal{L}^{1}\mathop{\hbox{\vrule
height=7.0pt,width=0.5pt,depth=0.0pt\vrule
height=0.5pt,width=6.0pt,depth=0.0pt}}\nolimits(-1,1)\times\delta_{k}\,.$

Moreover as $a_{,x}(\cdot,k)=0$ a.e. in $\\{x\in(-1,1)\colon{a}(x,k)=0\\}$ it
follows $\mu_{,x}\in\mathcal{M}((-1,1)\times\mathbb{R})$ and $\mu_{,x}\ll\mu$.

######  Definition 2.3 (Convergence).

For $L>0$ let $(w^{L},u^{L})\in\mathcal{A}_{L}^{\rm
in}\times\mathcal{A}_{L}^{\rm out}$. We say a sequence $(w^{L},u^{L})$
converges as $L\to\infty$ to
$\mu\in\mathcal{M}_{b}^{+}((-1,1)\times\mathbb{R})$, if
$(\mu^{L}(u^{L}),\mu^{L}_{,x}(u^{L}))$ weakly-* converge to $(\mu,\mu_{,x})$.

We introduce the class of measures

$\begin{split}\mathcal{M}_{\infty}:=\bigg{\\{}\mu\in&\mathcal{M}_{b}^{+}((-1,1)\times\mathbb{R})\colon\mu((-1,0]\times\mathbb{R})=0,\
\mu_{,x}\in\mathcal{M}_{b}((-1,1)\times\mathbb{R})\,,\\\
&\mu_{,x}\ll\mu\,,\int_{(0,1)\times\mathbb{R}}\phi(x)\,{\rm
d}\mu(x,k)=\int_{0}^{1}2x\phi(x)\,{\rm d}x\quad\forall\phi\in
C_{c}^{\infty}(0,1)\bigg{\\}}\,,\end{split}$ (2.8)

and the functional
$\mathcal{F}_{\infty}\colon\mathcal{M}_{\infty}\to[0,+\infty]$

$\mathcal{F}_{\infty}(\mu)=\int_{(0,1)\times\mathbb{R}}\biggl{[}k^{2}+\frac{1}{4k^{2}}\Bigl{(}\frac{\,{\rm
d}\mu_{,x}}{\,{\rm d}\mu}\Bigr{)}^{2}\biggr{]}\,{\rm d}\mu\,.$ (2.9)

Here $\frac{\,{\rm d}\mu_{,x}}{\,{\rm d}\mu}$ denotes the Radon-Nikodym
derivative, existence of which follows from absolute continuity of $\mu_{,x}$
w.r.t. $\mu$. Moreover, if $\mu_{L}$ from definition 2.1 would be supported in
$(0,1]\times\mathbb{R}$, then $\mathcal{F}_{\infty}(\mu_{L})$ is simply equal
to
$\sum_{k\in\frac{\pi\mathbb{Z}}{L}}\int_{0}^{1}\dot{a}_{k}^{2}(x)+a_{k}^{2}(x)k^{4}\,{\rm
d}x$, i.e. via Plancherel equality (see equation (3.3)) it equals
$\mathcal{S}_{L}$ from (1.3).

######  Remark 2.4.

  1. $(i)$

When convenient we will identify the class $\mathcal{M}_{\infty}$ with the
class of measures

$\begin{split}\bigg{\\{}\mu\in\mathcal{M}_{b}^{+}((0,1)\times\mathbb{R})&\colon\mu_{,x}\in\mathcal{M}_{b}((0,1)\times\mathbb{R})\,,\
\mu_{,x}\ll\mu\,,\\\ &\int_{(0,1)\times\mathbb{R}}\phi(x)\,{\rm
d}\mu(x,k)=\int_{0}^{1}2x\phi(x)\,{\rm d}x\quad\forall\phi\in
C_{c}^{\infty}(0,1)\bigg{\\}}\,;\end{split}$ (2.10)

  2. $(ii)$

For later convenience we observe that $\mathcal{F}_{\infty}$ can be rewritten
as follows

$\mathcal{F}_{\infty}(\mu)=\int_{(0,1)\times\mathbb{R}}k^{2}\,{\rm
d}\mu+\int_{(0,1)\times\mathbb{R}}\frac{1}{4k^{2}}\Bigl{(}\frac{\,{\rm
d}\mu}{\,{\rm d}|\tilde{\mu}|}\Bigr{)}^{-1}\Bigl{(}\frac{\,{\rm
d}\mu_{,x}}{\,{\rm d}|\tilde{\mu}|}\Bigr{)}^{2}\,{\rm d}|\tilde{\mu}|\,,$
(2.11)

where $\tilde{\mu}=(\mu,\mu_{,x})$ and $|\tilde{\mu}|$ denote its total
variation. Indeed, since $\mu_{,x}\ll\mu\ll|\tilde{\mu}|$, we have

$\frac{\,{\rm d}\mu_{,x}}{\,{\rm d}|\tilde{\mu}|}=\frac{\,{\rm
d}\mu_{,x}}{\,{\rm d}\mu}\frac{\,{\rm d}\mu}{\,{\rm d}|\tilde{\mu}|}\,,$

from which we deduce

$\begin{split}\int_{(0,1)\times\mathbb{R}}\frac{1}{4k^{2}}\Bigl{(}\frac{\,{\rm
d}\mu_{,x}}{\,{\rm d}\mu}\Bigr{)}^{2}\,{\rm
d}\mu&=\int_{(0,1)\times\mathbb{R}}\frac{1}{4k^{2}}\Bigl{(}\frac{\,{\rm
d}\mu_{,x}}{\,{\rm d}\mu}\Bigr{)}^{2}\frac{\,{\rm d}\mu}{\,{\rm
d}|\tilde{\mu}|}\,{\rm d}|\tilde{\mu}|\\\
&=\int_{(0,1)\times\mathbb{R}}\frac{1}{4k^{2}}\Bigl{(}\frac{\,{\rm
d}\mu}{\,{\rm d}|\tilde{\mu}|}\Bigr{)}^{-2}\Bigl{(}\frac{\,{\rm
d}\mu_{,x}}{\,{\rm d}|\tilde{\mu}|}\Bigr{)}^{2}\frac{\,{\rm d}\mu}{\,{\rm
d}|\tilde{\mu}|}\,{\rm d}|\tilde{\mu}|\\\
&=\int_{(0,1)\times\mathbb{R}}\frac{1}{4k^{2}}\Bigl{(}\frac{\,{\rm
d}\mu}{\,{\rm d}|\tilde{\mu}|}\Bigr{)}^{-1}\Bigl{(}\frac{\,{\rm
d}\mu_{,x}}{\,{\rm d}|\tilde{\mu}|}\Bigr{)}^{2}\,{\rm
d}|\tilde{\mu}|\,.\end{split}$

We are now ready to state our main result.

######  Theorem 2.5 ($\Gamma$-convergence).

Let $\mathcal{F}_{L}$ and $\mathcal{F}_{\infty}$ be as in (2.6) and (2.9)
respectively. Then the following holds:

  * $a)$

$($Compactness$)$. For $L>0$ let $(w^{L},u^{L})\in\mathcal{A}_{L}^{\rm
in}\times\mathcal{A}_{L}^{\rm out}$ be such that
$\sup_{L}\mathcal{F}_{L}(w^{L},u^{L})<+\infty$. Then there exists a
subsequence (not relabeled) and $\mu\in\mathcal{M}_{\infty}$ such that
$(w^{L},u^{L})$ converges as $L\to+\infty$ in the sense of Definition 2.3 to
$\mu$;

  * $b)$

$($$\Gamma$-convergence$)$. As $L\to+\infty$ the functionals $\mathcal{F}_{L}$
$\Gamma$-converge, with respect to the convergence in Definition 2.3, to the
functional $\mathcal{F}_{\infty}$.

##  3 Preliminaries

Let $u\in\mathcal{A}_{L}^{\rm out}$ and let $a_{k}(x)$ be defined as in (2.7).
Then we have

$\displaystyle u(x,y)$
$\displaystyle=a_{0}(x)+\sum_{k\in\frac{\pi\mathbb{Z}}{L},k>0}a_{k}(x)\sqrt{2}\sin(ky)+\sum_{k\in\frac{\pi\mathbb{Z}}{L},k<0}a_{k}(x)\sqrt{2}\cos(ky)$
(3.1) $\displaystyle=a_{0}(x)+\sum_{k\in\frac{\pi\mathbb{Z}}{L},k>0}{\rm
sign}(a_{k}(x))\frac{\sqrt{a(x,k)}}{k}\sqrt{2}\sin(ky)+\sum_{k\in\frac{\pi\mathbb{Z}}{L},k<0}{\rm
sign}(a_{k}(x))\frac{\sqrt{a(x,k)}}{-k}\sqrt{2}\cos(ky)\,.$

Then Plancherel equality yields

$\fint_{-L}^{L}u^{2}\,{\rm
d}y=a^{2}_{0}(x)+\sum_{k\in\frac{\pi\mathbb{Z}}{L},k\neq
0}a^{2}_{k}(x)=a^{2}_{0}(x)+\sum_{k\in\frac{\pi\mathbb{Z}}{L},k\neq
0}\frac{{a}(x,k)}{k^{2}}\,.$ (3.2)

The same holds for partial derivatives of $u$, that is

$\begin{split}\fint_{-L}^{L}(D^{\alpha}u)^{2}\,{\rm
d}y&=(D^{\alpha}a_{0}(x))^{2}+\sum_{k\in\frac{\pi\mathbb{Z}}{L},k\neq
0}\Big{(}\frac{\,{\rm d}^{\alpha_{1}}}{\,{\rm
d}x^{\alpha_{1}}}a_{k}(x)k^{\alpha_{2}}\Big{)}^{2}\\\
&=(D^{\alpha}a_{0}(x))^{2}+\sum_{k\in\frac{\pi\mathbb{Z}}{L},k\neq
0}\Big{(}\frac{\partial^{\alpha_{1}}}{\partial
x^{\alpha_{1}}}\big{(}\sqrt{a(x,k)}\big{)}k^{\alpha_{2}-1}\Big{)}^{2}\,,\end{split}$
(3.3)

with $\alpha=(\alpha_{1},\alpha_{2})$ multi-index with $|\alpha|\leq 2$. In
case $u$ has higher regularity, i.e., $u\in W^{k,2}((-1,1)\times\mathbb{R})$
with $k>2$, then the same applies for the higher derivatives, i.e., for
$|\alpha|\leq k$. For later convenience we also note that

$\frac{\partial}{\partial
x}\big{(}\sqrt{a(x,k)}\big{)}=\frac{a_{,x}(x,k)}{2\sqrt{a(x,k)}}\,,\quad\frac{\partial^{2}}{\partial
x^{2}}\big{(}\sqrt{a(x,k)}\big{)}=\frac{a_{,xx}(x,k)}{2\sqrt{a(x,k)}}-\frac{(a_{,x}(x,k))^{2}}{4\sqrt{a^{3}(x,k)}}\,.$
(3.4)

We now recall the definition of disintegration of measures only in a specific
case that we will be used throughout the paper, and we refer to [1] for a
complete treatment of the subject.

######  Definition 3.1 (Disintegration of measures in the $x$-variable).

Let $I\subset\mathbb{R}$ be an interval and let
$\mu\in\mathcal{M}_{b}(I\times\mathbb{R})$. We say that the family

$(\nu_{x},g(x))_{x\in I}\subset\mathcal{M}_{b}(\mathbb{R})\times\mathbb{R}$

is a disintegration of $\mu$ $($in the $x$-variable$)$ if $x\mapsto\nu_{x}$ is
Lebesgue measurable, $|\nu_{x}|(\mathbb{R})=1$ for every $x\in I$, $g\in
L^{1}(I)$, and

$\int_{I\times\mathbb{R}}f(x,k)\,{\rm
d}\mu=\int_{I}\int_{\mathbb{R}}f(x,k)\,{\rm d}\nu_{x}(k)g(x)\,{\rm d}x\,,$
(3.5)

for every $f\in L^{1}(I\times\mathbb{R};|\mu|)$.

Formally it simply means $\,{\rm d}\mu(x,k)=\,{\rm d}\nu_{x}(k)g(x)\,{\rm
d}x$.

######  Lemma 3.2.

Let $I\subset\mathbb{R}$ be an interval and let
$\mu\in\mathcal{M}^{+}_{b}(I\times\mathbb{R})$. Then

$\int_{I\times\mathbb{R}}\phi(x)\,{\rm d}\mu=\int_{I}g(x)\phi(x)\,{\rm
d}x\quad\forall\phi\in C_{c}^{\infty}(I)\,,$ (3.6)

for some non-negative $g\in L^{1}(I)$, if and only if there exists
$x\mapsto\nu_{x}\in\mathcal{M}^{+}_{b}(\mathbb{R})$ Lebesgue measurable such
that $(\nu_{x},g(x))_{x\in I}$ is a disintegration of $\mu$.

###### Proof.

Let $(\nu_{x},g(x))_{x\in
I}\subset\mathcal{M}^{+}_{b}(\mathbb{R})\times\mathbb{R}^{+}$ be a
disintegration of $\mu$. Then (3.5) holds with $f(x,k)=\phi(x)\in
C_{c}^{\infty}(I)$ and since $|\nu_{x}|(\mathbb{R})=\nu_{x}(\mathbb{R})=1$ we
readily deduce (3.6).
Assume instead that (3.6) holds true. Let $\pi_{1}\colon I\times\mathbb{R}\to
I$ be the canonical projection and let
$(\pi_{1})_{\sharp}\mu\in\mathcal{M}^{+}_{b}(I)$ be the push-forward of $\mu$
with respect to $\pi_{1}$. By the Disintegration Theorem (cf. [1, Theorem
2.28]) there exists $x\mapsto\nu_{x}\in\mathcal{M}^{+}_{b}(\mathbb{R})$
measurable with $\nu_{x}(\mathbb{R})=1$ such that

$\int_{I\times\mathbb{R}}f(x,k)\,{\rm
d}\mu(x,k)=\int_{I}\int_{\mathbb{R}}f(x,k)\,{\rm d}\nu_{x}(k)\,{\rm
d}(\pi_{1})_{\sharp}\mu(x)\,,$

for all $f\in L^{1}(I\times\mathbb{R};\mu)$. On the other hand (3.6) implies
that $(\pi_{1})_{\sharp}\mu(x)=g(x)\mathcal{L}^{1}\mathop{\hbox{\vrule
height=7.0pt,width=0.5pt,depth=0.0pt\vrule
height=0.5pt,width=6.0pt,depth=0.0pt}}\nolimits I$ and therefore
$(\nu_{x},g(x))_{x\in I}$ is a disintegration of $\mu$.
∎

######  Corollary 3.3 (Disintegration of $\mu\in\mathcal{M}_{\infty}$ in the
$x$-variable).

Let $\mu\in\mathcal{M}_{\infty}$. Then there exists
$x\mapsto\nu_{x}\in\mathcal{M}^{+}_{b}(\mathbb{R})$ measurable such that
$(\nu_{x},2x)_{x\in(0,1)}$ is a disintegration of $\mu$.

###### Proof.

The proof follows by Lemma 3.2 and from the fact that

$\int_{(0,1)\times\mathbb{R}}\phi(x)\,{\rm d}\mu=\int_{0}^{1}2x\phi(x)\,{\rm
d}x\quad\forall\phi\in C_{c}^{\infty}(0,1)\,.$ (3.7)

∎

##  4 Compactness and lower bound

In this section we prove compactness and the $\Gamma-\liminf$ inequality.

######  Proposition 4.1 (Compactness).

Let for $L>0$ be $(w^{L},u^{L})\in\mathcal{A}_{L}^{\rm
in}\times\mathcal{A}_{L}^{\rm out}$ such that
$\sup_{L}\mathcal{F}_{L}(w^{L},u^{L})<+\infty$. Then there exist a, not
relabeled, subsequence and $\mu\in\mathcal{M}_{\infty}$ such that
$(w^{L},u^{L})$ converges to $\mu$, as $L\to+\infty$, in the sense of
Definition 2.3.

###### Proof.

Let $(w^{L},u^{L})$ be as in the statement. Let $\mu^{L}:=\mu^{L}(u^{L})$ and
$\mu^{L}_{,x}:=\mu^{L}_{,x}(u^{L})$ be defined accordingly to Definition 2.1,
i.e., there exist ${a}^{L}(x,k)$ such that $a(\cdot,k)\in W^{1,1}(-1,1)$ and

$\mu^{L}=\sum_{k\in\frac{\pi\mathbb{Z}}{L}}a^{L}(x,k)\mathcal{L}^{1}\mathop{\hbox{\vrule
height=7.0pt,width=0.5pt,depth=0.0pt\vrule
height=0.5pt,width=6.0pt,depth=0.0pt}}\nolimits{(-1,1)}\times\delta_{k}\,,\quad\mu^{L}_{,x}=\sum_{k\in\frac{\pi\mathbb{Z}}{L}}a_{,x}^{L}(x,k)\mathcal{L}^{1}\mathop{\hbox{\vrule
height=7.0pt,width=0.5pt,depth=0.0pt\vrule
height=0.5pt,width=6.0pt,depth=0.0pt}}\nolimits{(-1,1)}\times\delta_{k}\,.$

Step 1: we show that there exists
$\mu\in\mathcal{M}_{b}^{+}((-1,1)\times\mathbb{R})$ with
$\mu_{,x}\in\mathcal{M}_{b}((-1,1)\times\mathbb{R})$ and such that
$(\mu^{L},\mu^{L}_{,x})\stackrel{{\scriptstyle*}}{{\rightharpoonup}}(\mu,\mu_{,x})$.
To this aim we observe that by taking
$0<C_{0}:=\sup_{L}\mathcal{F}_{L}(w^{L},u^{L})<+\infty$ we have

$\mathcal{F}_{L}(w^{L},u^{L})\leq C_{0}\,,$

so that in particular

$\begin{split}C_{0}\geq\mathcal{F}_{L}(w^{L},u^{L})\geq{L^{2}}\fint_{-L}^{L}\int_{-1}^{1}&\bigg{(}w^{L}_{2,y}+\frac{(u_{,y}^{L})^{2}}{2}-x\bigg{)}^{2}\,{\rm
d}x\,{\rm d}y-\frac{L^{2}}{3}\\\
&+\fint_{-L}^{L}\int_{-1}^{1}(u^{L}_{,x})^{2}+(u^{L}_{,yy})^{2}\,{\rm
d}x\,{\rm d}y\,.\end{split}$ (4.1)

By Fubini’s theorem, Jensen’s inequality and the fact that $w^{L}(x,\cdot)$ is
$2L$-periodic we get

$\begin{split}{L^{2}}\fint_{-L}^{L}\int_{-1}^{1}&\bigg{(}w^{L}_{2,y}+\frac{(u_{,y}^{L})^{2}}{2}-x\bigg{)}^{2}\,{\rm
d}x\,{\rm d}y\geq
L^{2}\int_{-1}^{1}\bigg{(}\fint_{-L}^{L}\Big{(}w^{L}_{2,y}+\frac{(u_{,y}^{L})^{2}}{2}-x\Big{)}\,{\rm
d}y\bigg{)}^{2}\,{\rm d}x\\\
&=L^{2}\int_{0}^{1}\bigg{(}\fint_{-L}^{L}\frac{(u_{,y}^{L})^{2}}{2}\,{\rm
d}y-x\bigg{)}^{2}\,{\rm
d}x+L^{2}\int_{-1}^{0}\bigg{(}\fint_{-L}^{L}\frac{(u_{,y}^{L})^{2}}{2}\,{\rm
d}y-x\bigg{)}^{2}\,{\rm d}x\\\ &\geq
L^{2}\int_{0}^{1}\bigg{(}\fint_{-L}^{L}\frac{(u_{,y}^{L})^{2}}{2}\,{\rm
d}y-x\bigg{)}^{2}\,{\rm
d}x+L^{2}\int_{-1}^{0}\bigg{(}\fint_{-L}^{L}\frac{(u_{,y}^{L})^{2}}{2}\,{\rm
d}y\bigg{)}^{2}\,{\rm d}x+L^{2}\int_{-1}^{0}x^{2}\,{\rm d}x\,,\end{split}$
(4.2)

where the last inequality follows by using that $(a+b)^{2}\geq a^{2}+b^{2}$
provided that $ab>0$ with $a=\frac{1}{2}\fint_{-L}^{L}(u_{,x}^{2})\,{\rm d}y$
and $b=-x$ for $x\in(-1,0)$. Combining (4.1) with (4.2) and using that
$\int_{-1}^{0}x^{2}\,{\rm d}x=\frac{1}{3}$ we find

$\begin{split}C_{0}\geq\mathcal{F}_{L}(w^{L},u^{L})\geq&L^{2}\int_{0}^{1}\bigg{(}\fint_{-L}^{L}\frac{(u_{,y}^{L})^{2}}{2}\,{\rm
d}y-x\bigg{)}^{2}\,{\rm d}x\\\
&+L^{2}\int_{-1}^{0}\bigg{(}\fint_{-L}^{L}\frac{(u_{,y}^{L})^{2}}{2}\,{\rm
d}y\bigg{)}^{2}\,{\rm
d}x+\fint_{-L}^{L}\int_{-1}^{1}(u^{L}_{,x})^{2}+(u^{L}_{,yy})^{2}\,{\rm
d}x\,{\rm d}y\,.\end{split}$ (4.3)

Thus from (3.3) it follows

$\begin{split}\frac{C}{L^{2}}\geq\int_{0}^{1}\bigg{(}\fint_{-L}^{L}\frac{(u^{L}_{,y})^{2}}{2}\,{\rm
d}y-x\bigg{)}^{2}\,{\rm
d}x&=\int_{0}^{1}\bigg{(}\frac{1}{2}\sum_{k\in\frac{\pi\mathbb{Z}}{L}}a^{L}(x,k)-x\bigg{)}^{2}\,{\rm
d}x\\\ &\geq
C\Bigg{(}\int_{0}^{1}\bigg{(}\frac{1}{2}\sum_{k\in\frac{\pi\mathbb{Z}}{L}}a^{L}(x,k)\bigg{)}^{2}\,{\rm
d}x-\frac{1}{3}\Bigg{)}\,,\end{split}$ (4.4)

and

$\frac{C}{L^{2}}\geq\int_{-1}^{0}\bigg{(}\fint_{-L}^{L}\frac{(u^{L}_{,y})^{2}}{2}\,{\rm
d}y\bigg{)}^{2}\,{\rm
d}x=\int_{-1}^{0}\bigg{(}\frac{1}{2}\sum_{k\in\frac{\pi\mathbb{Z}}{L}}a^{L}(x,k)\bigg{)}^{2}\,{\rm
d}x\,.$ (4.5)

Hence we obtain

$|\mu^{L}|((-1,1)\times\mathbb{R})=\mu^{L}((-1,1)\times\mathbb{R})=\int_{-1}^{1}\sum_{k\in\frac{\pi\mathbb{Z}}{L}}a^{L}(k,x)\,{\rm
d}x\leq C\,,$

from which we deduce the existence of a (not relabeled) subsequence and
$\mu\in\mathcal{M}^{+}_{b}((-1,1)\times\mathbb{R})$ such that
$\mu^{L}\stackrel{{\scriptstyle*}}{{\rightharpoonup}}\mu$. In addition (4.3)
together with (3.3) and (3.4) yield

$\begin{split}C&\geq\int_{-1}^{1}\fint_{-L}^{L}(u^{L}_{,x})^{2}+(u^{L}_{,yy})^{2}\,{\rm
d}y\,{\rm d}x\\\
&\geq\int_{-1}^{1}\Bigg{(}\sum_{k\in\frac{\pi\mathbb{Z}}{L}}a^{L}(x,k)k^{2}+\sum_{k\in\frac{\pi\mathbb{Z}}{L},k\neq
0}\frac{1}{4k^{2}}\frac{({a}^{L}_{,x}(x,k))^{2}}{a^{L}(x,k)}\Bigg{)}\,{\rm
d}x\\\
&\geq\int_{-1}^{1}\sum_{k\in\frac{\pi\mathbb{Z}}{L}}|{a}^{L}_{,x}(x,k)|\,{\rm
d}x=|\mu^{L}_{,x}|((-1,1)\times\mathbb{R})\,,\end{split}$ (4.6)

where the last inequality follows by Young’s inequality. Hence, up to
subsequence, we may deduce that there exists
$\tilde{\mu}\in\mathcal{M}_{b}(\times(-1,1)\times\mathbb{R})$ such that
$\mu^{L}_{,x}\stackrel{{\scriptstyle*}}{{\rightharpoonup}}\tilde{\mu}$.
Moreover given any $\varphi\in C^{\infty}_{c}((-1,1)\times\mathbb{R})$, it
holds

$\int_{(-1,1)\times\mathbb{R}}\varphi\,{\rm
d}\tilde{\mu}=\lim_{L\to+\infty}\int_{(-1,1)\times\mathbb{R}}\varphi\,{\rm
d}\mu^{L}_{,x}=-\lim_{L\to+\infty}\int_{(-1,1)\times\mathbb{R}}\varphi_{,x}\,{\rm
d}\mu^{L}=-\int_{(-1,1)\times\mathbb{R}}\varphi_{,x}\,{\rm d}\mu\,,$

which in turn implies $\tilde{\mu}=\mu_{,x}$.

Step 2: we show that $\mu_{,x}\ll\mu$. By Remark 2.2 $(ii)$ we have that
$\mu^{L}_{,x}\ll\mu^{L}$. Now let $N\in\mathbb{N}$ be fixed and let
$\mu^{L}_{N}:=\mu^{L}\mathop{\hbox{\vrule
height=7.0pt,width=0.5pt,depth=0.0pt\vrule
height=0.5pt,width=6.0pt,depth=0.0pt}}\nolimits(-1,1)\times(-N,N)$ and
$\mu_{N}:=\mu\mathop{\hbox{\vrule height=7.0pt,width=0.5pt,depth=0.0pt\vrule
height=0.5pt,width=6.0pt,depth=0.0pt}}\nolimits(-1,1)\times(-N,N)$. Then the
following properties hold:

$\mu^{L}_{N,x}:=\mu_{,x}^{L}\mathop{\hbox{\vrule
height=7.0pt,width=0.5pt,depth=0.0pt\vrule
height=0.5pt,width=6.0pt,depth=0.0pt}}\nolimits(-1,1)\times(-N,N)\,,\quad\mu_{N,x}:=\mu_{,x}\mathop{\hbox{\vrule
height=7.0pt,width=0.5pt,depth=0.0pt\vrule
height=0.5pt,width=6.0pt,depth=0.0pt}}\nolimits(-1,1)\times(-N,N)\,,$
$\mu^{L}_{N,x}\ll\mu^{L}_{N}\,,\quad(\mu^{L}_{N},\mu^{L}_{N,x})\stackrel{{\scriptstyle*}}{{\rightharpoonup}}(\mu_{N},\mu_{N,x})\,,$
(4.7)

and

$\frac{\,{\rm d}\mu^{L}_{N,x}}{\,{\rm d}\mu^{L}_{N}}(x,k)=\frac{\,{\rm
d}\mu^{L}_{,x}}{\,{\rm d}\mu^{L}}(x,k)\mathop{\hbox{\vrule
height=7.0pt,width=0.5pt,depth=0.0pt\vrule
height=0.5pt,width=6.0pt,depth=0.0pt}}\nolimits(-1,1)\times(-N,N)\,.$

Moreover recalling the definition of $\mu^{L}$ and (4.6) we have

$\begin{split}\int_{(-1,1)\times(-N,N)}\frac{1}{4N^{2}}\Big{(}\frac{\,{\rm
d}\mu^{L}_{N,x}}{\,{\rm d}\mu^{L}_{N}}(x,k)\Big{)}^{2}\,{\rm
d}\mu^{L}_{N}&\leq\int_{(-1,1)\times(-N,N)}\frac{1}{4k^{2}}\Big{(}\frac{\,{\rm
d}\mu^{L}_{,x}}{\,{\rm d}\mu^{L}}(x,k)\Big{)}^{2}\,{\rm d}\mu^{L}\\\
&\leq\int_{-1}^{1}\sum_{k\in\frac{\pi\mathbb{Z}}{L},k\neq
0}\frac{1}{4k^{2}}\frac{({a}^{L}_{,x}(x,k))^{2}}{a^{L}(x,k)}\,{\rm d}x\leq
C\,.\end{split}$ (4.8)

From (4.7), (4.8) and [1, Example 2.36 pg. 67, and discussion at pg. 66] we
deduce that $\mu_{N,x}\ll\mu_{N}$ for every $N\in\mathbb{N}$ and hence
$\mu_{,x}\ll\mu$.

Step 3: we show that $\mu\in\mathcal{M}^{+}_{b}((0,1)\times\mathbb{R})$, that
is, $\mu((-1,0]\times\mathbb{R})=0$, and that

$\int_{(0,1)\times\mathbb{R}}\phi(x)\,{\rm d}\mu=\int_{0}^{1}2x\phi(x)\,{\rm
d}x,$ (4.9)

for all $\phi\in C_{c}^{\infty}((0,1))$. To this purpose for fixed
$\delta\in(0,1)$ by (4.1) we have

$\begin{split}\mu^{L}((-1,\delta)\times\mathbb{R})&=\int_{-1}^{\delta}\sum_{k\in\frac{\pi\mathbb{Z}}{L}}a^{L}(x,k)\,{\rm
d}x\\\ &\leq
C\int_{-1}^{0}\bigg{(}\sum_{k\in\frac{\pi\mathbb{Z}}{L}}a^{L}(x,k)\bigg{)}^{2}\,{\rm
d}x+C\int_{0}^{\delta}\bigg{(}\sum_{k\in\frac{\pi\mathbb{Z}}{L}}a^{L}(x,k)-x\bigg{)}^{2}\,{\rm
d}x+C\int_{0}^{\delta}x^{2}\,{\rm d}x\\\
&\leq\frac{C}{L^{2}}+C\delta^{3}.\end{split}$

This together with the lower semicontinuity with respect to the weak*
convergence give

$\mu((-1,0]\times\mathbb{R})\leq\mu((-1,\delta)\times\mathbb{R})\leq\liminf_{L\to\infty}\mu^{L}((-1,\delta)\times\mathbb{R})\leq
C\delta^{3}.$

By sending $\delta\to 0$ we deduce $\mu((-1,0]\times\mathbb{R})=0.$ It remains
to show (4.9). Given $\phi\in C_{c}^{\infty}(0,1)$ it holds

$\int_{(0,1)\times\mathbb{R}}\phi(x)\,{\rm
d}\mu^{L}=\int_{0}^{1}\phi(x)\Big{(}\sum_{k\in\frac{\pi\mathbb{Z}}{L}}a^{L}(x,k)-2x\Big{)}\,{\rm
d}x+\int_{0}^{1}2x\phi(x)\,{\rm d}x.$

From (4.4) it follows that

$\int_{0}^{1}\Big{|}\phi(x)\sum_{k\in\frac{\pi\mathbb{Z}}{L}}a^{L}(x,k)-2x\Big{|}\,{\rm
d}x\leq
C\|\phi\|_{\infty}\bigg{(}\int_{0}^{1}\Big{(}\frac{1}{2}\sum_{k\in\frac{\pi\mathbb{Z}}{L}}a^{L}(x,k)-x\Big{)}^{2}\,{\rm
d}x\bigg{)}^{1/2}\leq\frac{C}{L}\to 0,$

as $L\to+\infty$, so that

$\lim_{L\to+\infty}\int_{(0,1)\times\mathbb{R}}\phi(x)\,{\rm
d}\mu^{L}=\int_{0}^{1}2x\phi(x)\,{\rm d}x.$ (4.10)

Next we fix $R\geq 1$ and take $\psi_{R}\in C^{\infty}_{c}(\mathbb{R})$ such
that $0\leq\psi_{R}\leq 1$, $\psi_{R}(k)\equiv 1$ if $|k|<R$ and
$\psi_{R}(k)\equiv 0$ if $|k|>R+1$. We have

$\int_{(0,1)\times\mathbb{R}}\phi(x)\,{\rm
d}\mu^{L}=\int_{(0,1)\times\mathbb{R}}\phi(x)\psi_{R}(k)\,{\rm
d}\mu^{L}+\int_{(0,1)\times\mathbb{R}}\phi(x)(1-\psi_{R}(k))\,{\rm
d}\mu^{L}\,.$ (4.11)

The weak* convergence yields

$\lim_{L\to+\infty}\int_{(0,1)\times\mathbb{R}}\phi(x)\psi_{R}(k)\,{\rm
d}\mu^{L}=\int_{(0,1)\times\mathbb{R}}\phi(x)\psi_{R}(k)\,{\rm d}\mu\,,$

whereas for the second term on the right hand-side of (4.11) we get

$\begin{split}\int_{(0,1)\times\mathbb{R}}|\phi(x)(1-\psi_{R}(k))|\,{\rm
d}\mu^{L}&\leq\int_{0}^{1}|\phi(x)|\Big{(}\sum_{k\in\frac{\pi\mathbb{Z}}{L},|k|>R}a^{L}(x,k)\Big{)}\,{\rm
d}x\\\
&\leq\frac{\|\phi\|_{\infty}}{R^{2}}\int_{0}^{1}\Big{(}\sum_{k\in\frac{\pi\mathbb{Z}}{L},|k|>R}a^{L}(x,k)k^{2}\Big{)}\,{\rm
d}x\\\
&\leq\frac{\|\phi\|_{\infty}}{R^{2}}\int_{0}^{1}\fint_{-L}^{L}(u_{,yy})^{2}\,{\rm
d}y\,{\rm d}x\leq\frac{C}{R^{2}}\,,\end{split}$

where the last to inequalities follow from (3.3) and (4.1). Thus passing to
the limit as $L\to+\infty$ in (4.11) we obtain

$\int_{(0,1)\times\mathbb{R}}\phi(x)\psi_{R}(k)\,{\rm
d}\mu-\frac{C}{R^{2}}\leq\lim_{L\to+\infty}\int_{(0,1)\times\mathbb{R}}\phi(x)\,{\rm
d}\mu^{L}\leq\int_{(0,1)\times\mathbb{R}}\phi(x)\psi_{R}(k)\,{\rm
d}\mu+\frac{C}{R^{2}}\,.$

Eventually by letting $R\to+\infty$ we deduce

$\lim_{L\to+\infty}\int_{(0,1)\times\mathbb{R}}\phi(x)\,{\rm
d}\mu^{L}=\int_{(0,1)\times\mathbb{R}}\phi(x)\,{\rm d}\mu\,,$

which together with (4.10) yield (4.9). ∎

######  Proposition 4.2 (Lower bound).

Let $\mathcal{F}_{L}$ and $\mathcal{F}_{\infty}$ be as in (2.6) and (2.9)
respectively. Let for $L>0$ be $(w^{L},u^{L})\subset\mathcal{A}_{L}^{\rm
in}\times\mathcal{A}_{L}^{\rm out}$ a sequence converging to
$\mu\in\mathcal{M}_{\infty}$ in the sense of Definition 2.3. Then there holds

$\liminf_{L\to\infty}\mathcal{F}_{L}(w^{L},u^{L})\geq\mathcal{F}_{\infty}(\mu).$
(4.12)

###### Proof.

Let $(w^{L},u^{L})$ be as in the statement and let $\mu^{L}:=\mu^{L}(u^{L})$
and $\mu^{L}_{,x}:=\mu^{L}_{,x}(u^{L})$ be defined accordingly to Definition
2.1, that is,

$\mu^{L}=\sum_{k\in\frac{\pi\mathbb{Z}}{L}}a^{L}(x,k)\mathcal{L}^{1}\mathop{\hbox{\vrule
height=7.0pt,width=0.5pt,depth=0.0pt\vrule
height=0.5pt,width=6.0pt,depth=0.0pt}}\nolimits{(-1,1)}\times\delta_{k}\,,\quad\mu^{L}_{,x}=\sum_{k\in\frac{\pi\mathbb{Z}}{L}}a_{,x}^{L}(x,k)\mathcal{L}^{1}\mathop{\hbox{\vrule
height=7.0pt,width=0.5pt,depth=0.0pt\vrule
height=0.5pt,width=6.0pt,depth=0.0pt}}\nolimits{(-1,1)}\times\delta_{k}\,.$

Recalling (4.3), (3.3) and (3.4) we have that

$\begin{split}\mathcal{F}_{L}(w^{L},u^{L})&\geq\int_{-1}^{1}\fint_{-L}^{L}(u^{L}_{,x})^{2}+(u^{L}_{,yy})^{2}\,{\rm
d}y\,{\rm d}x\\\
&\geq\int_{0}^{1}\Big{(}\sum_{k\in\frac{\pi\mathbb{Z}}{L},k\neq
0}\frac{1}{4k^{2}}\frac{({a}^{L}_{,x}(x,k))^{2}}{a^{L}(x,k)}+\sum_{k\in\frac{\pi\mathbb{Z}}{L}}a^{L}(x,k)k^{2}\Big{)}\,{\rm
d}x\\\
&=\int_{(0,1)\times\mathbb{R}}\bigg{(}k^{2}+\frac{1}{4k^{2}}\Big{(}\frac{\,{\rm
d}\mu_{,x}^{L}}{\,{\rm d}\mu^{L}}(x,k)\Big{)}^{2}\bigg{)}\,{\rm d}\mu^{L}\\\
&=\int_{(0,1)\times\mathbb{R}}k^{2}\,{\rm
d}\mu^{L}+\int_{(0,1)\times\mathbb{R}}\frac{1}{4k^{2}}\Bigl{(}\frac{\,{\rm
d}\mu^{L}}{\,{\rm d}|\tilde{\mu}^{L}|}(x,k)\Bigr{)}^{-1}\Bigl{(}\frac{\,{\rm
d}\mu^{L}_{,x}}{\,{\rm d}|\tilde{\mu}^{L}|}(x,k)\Bigr{)}^{2}\,{\rm
d}|\tilde{\mu}^{L}|\,,\end{split}$ (4.13)

where $\tilde{\mu}^{L}:=(\mu^{L},\mu^{L}_{,x})$, and the last equality follows
from Remark 2.4 $(ii)$. By Reshetnyak Theorem (cf. [1, Theorem 2.38]) there
hold

$\liminf_{L\to+\infty}\int_{(0,1)\times\mathbb{R}}k^{2}\,{\rm
d}\mu^{L}\geq\int_{(0,1)\times\mathbb{R}}k^{2}\,{\rm d}\mu\,,$ (4.14)

and

$\begin{split}\liminf_{L\to+\infty}&\int_{(0,1)\times\mathbb{R}}\frac{1}{4k^{2}}\Bigl{(}\frac{\,{\rm
d}\mu^{L}}{\,{\rm d}|\tilde{\mu}^{L}|}(x,k)\Bigr{)}^{-1}\Bigl{(}\frac{\,{\rm
d}\mu^{L}_{,x}}{\,{\rm d}|\tilde{\mu}^{L}|}(x,k)\Bigr{)}^{2}\,{\rm
d}|\tilde{\mu}^{L}|\\\
&\geq\int_{(0,1)\times\mathbb{R}}\frac{1}{4k^{2}}\Bigl{(}\frac{\,{\rm
d}\mu}{\,{\rm d}|\tilde{\mu}|}(x,k)\Bigr{)}^{-1}\Bigl{(}\frac{\,{\rm
d}\mu_{,x}}{\,{\rm d}|\tilde{\mu}|}(x,k)\Bigr{)}^{2}\,{\rm
d}|\tilde{\mu}|\,,\end{split}$ (4.15)

with $\tilde{\mu}:=(\mu,\mu_{,x})$. Gathering together (4.13), (4.14) and
(4.15) we find

$\begin{split}\liminf_{L\to\infty}\mathcal{F}_{L}(w^{L},u^{L})\geq&\int_{(0,1)\times\mathbb{R}}k^{2}\,{\rm
d}\mu+\int_{(0,1)\times\mathbb{R}}\frac{1}{4k^{2}}\Bigl{(}\frac{\,{\rm
d}\mu}{\,{\rm d}|\tilde{\mu}|}(x,k)\Bigr{)}^{-1}\Bigl{(}\frac{\,{\rm
d}\mu_{,x}}{\,{\rm d}|\tilde{\mu}|}(x,k)\Bigr{)}^{2}\,{\rm
d}|\tilde{\mu}|=\mathcal{F}_{\infty}(\mu)\,.\end{split}$

∎

##  5 Upper bound

In this section we prove the $\Gamma-\limsup$ inequality.

######  Proposition 5.1 (Upper bound).

Let $\mu\in\mathcal{M}_{\infty}$. Then for $L>0$ there exists a sequence
$(w^{L},u^{L})\in\mathcal{A}_{L}^{\rm in}\times\mathcal{A}_{L}^{\rm out}$ that
converges to $\mu\in\mathcal{M}_{\infty}$ in the sense of Definition 2.3 and
such that

$\limsup_{L\to\infty}\mathcal{F}_{L}(w^{L},u^{L})\leq\mathcal{F}_{\infty}(\mu)\,,$

with $\mathcal{F}_{L}$ and $\mathcal{F}_{\infty}$ defined as in (2.6) and
(2.9) respectively.

We divide the proof of Proposition 5.1 into a number of steps. For any
$\lambda\geq 1$ we define the following class of measures

$\begin{split}\mathcal{M}_{\infty}^{\lambda}:=\bigg{\\{}\mu\in&\mathcal{M}_{b}^{+}((0,\lambda)\times\mathbb{R})\colon\mu_{,x}\in\mathcal{M}_{b}((0,\lambda)\times\mathbb{R})\,,\quad\mu_{,x}\ll\mu\,,\\\
&\int_{0}^{\lambda}2x\phi(x)dx=\int_{(0,\lambda)\times\mathbb{R}}\phi(x)\,{\rm
d}\mu(k,x)\quad\forall\phi\in
C_{c}^{\infty}(0,\lambda)\bigg{\\}}\,,\end{split}$ (5.1)

and the functional
$\mathcal{F}_{\infty}^{\lambda}\colon\mathcal{M}_{\infty}^{\lambda}\to[0,+\infty]$

$\mathcal{F}_{\infty}^{\lambda}(\mu)=\int_{(0,\lambda)\times\mathbb{R}}\biggl{[}k^{2}+\frac{1}{4k^{2}}\Bigl{(}\frac{\,{\rm
d}\mu_{,x}}{\,{\rm d}\mu}\Bigr{)}^{2}\biggr{]}\,{\rm d}\mu\,.$ (5.2)

Then we have $\mathcal{M}_{\infty}=\mathcal{M}_{\infty}^{1}$ and
$\mathcal{F}_{\infty}=\mathcal{F}_{\infty}^{1}$.

######  Lemma 5.2 (Dilation of $\mu$).

Let $\mu\in\mathcal{M}_{\infty}$. Then for each $\lambda\in(1,2)$ there exists
$\mu_{\lambda}\in\mathcal{M}_{\infty}^{\lambda}$ such that

$(\mu_{\lambda},\mu_{\lambda,x})\mathop{\hbox{\vrule
height=7.0pt,width=0.5pt,depth=0.0pt\vrule
height=0.5pt,width=6.0pt,depth=0.0pt}}\nolimits((0,1)\times\mathbb{R})\stackrel{{\scriptstyle*}}{{\rightharpoonup}}(\mu,\mu_{,x})\quad\text{as
}\lambda\searrow 1\,,$ (5.3)

and

$\int_{(0,\lambda)\times\mathbb{R}}k^{2}\,{\rm
d}\mu_{\lambda}=\lambda^{2}\int_{(0,1)\times\mathbb{R}}k^{2}\,{\rm
d}\mu\,,\quad\int_{(0,\lambda)\times\mathbb{R}}\frac{1}{4k^{2}}\Bigl{(}\frac{\,{\rm
d}\mu_{\lambda,x}}{\,{\rm d}\mu_{\lambda}}\Bigr{)}^{2}\,{\rm
d}\mu_{\lambda}=\int_{(0,1)\times\mathbb{R}}\Bigl{(}\frac{\,{\rm
d}\mu_{,x}}{\,{\rm d}\mu}\Bigr{)}^{2}\,{\rm d}\mu$ (5.4)

so that, in particular

$\lim_{\lambda\searrow
1}\mathcal{F}_{\infty}^{\lambda}\big{(}\mu_{\lambda}\big{)}=\mathcal{F}_{\infty}(\mu)\,.$
(5.5)

Moreover $(\nu_{\frac{x}{\lambda}},2x)_{x\in(0,\lambda)}$ is a disintegration
of $\mu_{\lambda}$ where $\nu_{x}\in\mathcal{M}_{b}^{+}(\mathbb{R})$ for
$x\in(0,1)$ is the measure given by Corollary 3.3.

###### Proof.

Let $\mu_{\lambda}\in\mathcal{M}_{b}^{+}((0,\lambda)\times\mathbb{R})$ be
defined via duality as follows

$\int_{(0,\lambda)\times\mathbb{R}}\psi(x,k)\,{\rm
d}\mu_{\lambda}=\lambda^{2}\int_{(0,1)\times\mathbb{R}}\psi(\lambda x,k)\,{\rm
d}\mu\,,$ (5.6)

for every $\psi\in C((0,\lambda)\times\mathbb{R})$. Notice that
$\mu_{\lambda,x}\in\mathcal{M}_{b}((0,\lambda)\times\mathbb{R})$ and is given
by

$\int_{(0,\lambda)\times\mathbb{R}}\psi(x,k)\,{\rm
d}\mu_{\lambda,x}=\lambda\int_{(0,1)\times\mathbb{R}}\psi(\lambda x,k)\,{\rm
d}\mu_{,x}\,.$ (5.7)

Hence $\mu_{\lambda,x}\ll\mu_{\lambda}$ and

$\frac{\,{\rm d}\mu_{\lambda,x}}{\,{\rm
d}\mu_{\lambda}}(x,k)=\frac{1}{\lambda}\frac{\,{\rm d}\mu_{,x}}{\,{\rm
d}\mu}\Big{(}\frac{x}{\lambda},k\Big{)}\,.$

Moreover (5.6) together with the change of variable $x=\lambda\hat{x}$ imply

$\int_{0}^{\lambda}2x\phi(x)\,{\rm
d}x=\int_{(0,\lambda)\times\mathbb{R}}\phi(x)\,{\rm
d}\mu_{\lambda}\quad\forall\phi\in C_{c}^{\infty}(0,\lambda)\,.$

It follows that $\mu_{\lambda}\in\mathcal{M}_{\infty}^{\lambda}$. Moreover
from (5.6) and (5.7) we deduce that

$(\mu_{\lambda},\mu_{\lambda,x})\mathop{\hbox{\vrule
height=7.0pt,width=0.5pt,depth=0.0pt\vrule
height=0.5pt,width=6.0pt,depth=0.0pt}}\nolimits(\mathbb{R}\times(0,1))\stackrel{{\scriptstyle*}}{{\rightharpoonup}}(\mu,\mu_{,x})\quad\text{as
}\lambda\to 1\,,$

and (5.4) which implies (5.5). The fact that
$(\nu_{\frac{x}{\lambda}},2x)_{x\in(0,\lambda)}$ is a disintegration of
$\mu_{\lambda}$ follows again by a change of variable. ∎

######  Lemma 5.3 (Discretisation of $\mu$).

Let $\mu\in\mathcal{M}_{\infty}$ with $\mathcal{F}_{\infty}(\mu)<+\infty$ and
let $\lambda=\lambda(L)\searrow 1$ as $L\to\infty$. Then there exists
$(\mu^{L})\subset\mathcal{M}_{\infty}^{\lambda}$ with the following
properties:

  1. $(i)$

$\mu^{L}=\sum_{k\in\frac{\pi\mathbb{Z}}{L}}\overline{b}^{L}(x,k)\mathcal{L}^{1}\mathop{\hbox{\vrule
height=7.0pt,width=0.5pt,depth=0.0pt\vrule
height=0.5pt,width=6.0pt,depth=0.0pt}}\nolimits{(0,\lambda)}\times\delta_{k}$
with

$\overline{b}^{L}(\cdot,k)\in
W^{1,1}(0,\lambda)\quad\text{and}\quad\sum_{k\in\frac{\pi\mathbb{Z}}{L}}\overline{b}^{L}(x,k)=2x\,,\quad\forall
x\in(0,\lambda)\,,$ (5.8)
$\mathcal{F}_{\infty}^{\lambda}(\mu^{L})=\int_{0}^{\lambda}\sum_{k\in\frac{\pi\mathbb{Z}}{L}}k^{2}\overline{b}^{L}(x,k)\,{\rm
d}x+\int_{0}^{\lambda}\sum_{k\in\frac{\pi\mathbb{Z}}{L},k\neq
0}\frac{1}{4k^{2}}\frac{(\overline{b}^{L}_{,x}(x,k))^{2}}{\overline{b}^{L}(x,k)}\,{\rm
d}x\,;$ (5.9)

  2. $(ii)$

$(\mu^{L},\mu^{L}_{,x})\mathop{\hbox{\vrule
height=7.0pt,width=0.5pt,depth=0.0pt\vrule
height=0.5pt,width=6.0pt,depth=0.0pt}}\nolimits((0,1)\times\mathbb{R})\stackrel{{\scriptstyle*}}{{\rightharpoonup}}(\mu,\mu_{,x})$;

  3. $(iii)$

$\limsup_{L\to\infty}\mathcal{F}_{\infty}^{\lambda}(\mu^{L})\leq\mathcal{F}_{\infty}\big{(}\mu\big{)}$.

###### Proof.

For each $L>1$ let $\mu_{\lambda}\in\mathcal{M}_{\infty}^{\lambda}$ be the
measure given by Lemma 5.2 and let
$(\nu_{\frac{x}{\lambda}},2x)_{x\in(0,\lambda)}$ be the corresponding
disintegration. We then define
$\mu^{L}\in\mathcal{M}_{b}^{+}((0,\lambda)\times\mathbb{R})$ as

$\mu^{L}:=\sum_{k\in\frac{\pi\mathbb{Z}}{L}}\overline{b}^{L}(x,k)\mathcal{L}^{1}\mathop{\hbox{\vrule
height=7.0pt,width=0.5pt,depth=0.0pt\vrule
height=0.5pt,width=6.0pt,depth=0.0pt}}\nolimits{(0,\lambda)}\times\delta_{k}\,,$
(5.10)

where for $(x,k)\in(0,\lambda)\times\frac{\pi\mathbb{Z}}{L}$ we set

$\overline{b}^{L}(x,k):=\begin{cases}0&\text{if }k=0\,,\\\\[10.00002pt]
2x\nu_{\frac{x}{\lambda}}(I_{k}^{L})&\text{if }k\neq
0\,,\end{cases}\quad\text{and}\quad
I^{L}_{k}:=\begin{cases}(k-\frac{\pi}{L},k]&\text{if }k>0\,,\\\\[10.00002pt]
[k,k+\frac{\pi}{L})&\text{if }k<0\,.\end{cases}$ (5.11)

Now, for each $k\in\frac{\pi\mathbb{Z}}{L}$, $\overline{b}^{L}(\cdot,k)\in
W^{1,1}(0,\lambda)$ with

$\overline{b}_{,x}^{L}(x,k)=\begin{cases}0&\text{if }k=0\,,\\\\[10.00002pt]
\displaystyle 2x\int_{I_{k}^{L}}\frac{\,{\rm d}\mu_{\lambda,x}}{\,{\rm
d}\mu_{\lambda}}(x,\hat{k})\,{\rm d}\nu_{\frac{x}{\lambda}}(\hat{k})&\text{if
}k\neq 0\,.\end{cases}$

Indeed if $k=0$ there is nothing to prove. If instead $k\neq 0$, for $\phi\in
C^{\infty}_{c}(0,\lambda)$ from the definition of $\nu_{\frac{x}{\lambda}}$
and recalling Remark 2.2 we get

$\begin{split}\int_{0}^{\lambda}&\overline{b}^{L}(x,k)\dot{\phi}(x)\,{\rm
d}x=\int_{0}^{\lambda}\int_{\mathbb{R}}\mathbbm{1}_{I_{k}^{L}}(\hat{k})\dot{\phi}(x)\,{\rm
d}\nu_{\frac{x}{\lambda}}(\hat{k})\,2x\,{\rm d}x\\\
&=\int_{(0,\lambda)\times\mathbb{R}}\mathbbm{1}_{I_{k}^{L}}(\hat{k})\dot{\phi}(x)\,{\rm
d}\mu_{\lambda}=-\int_{(0,\lambda)\times\mathbb{R}}\mathbbm{1}_{I_{k}^{L}}(\hat{k})\phi(x)\,{\rm
d}\mu_{\lambda,x}\\\
&=-\int_{(0,\lambda)\times\mathbb{R}}\mathbbm{1}_{I_{k}^{L}}(\hat{k})\phi(x)\frac{\,{\rm
d}\mu_{\lambda,x}}{\,{\rm d}\mu_{\lambda}}(x,\hat{k})\,{\rm
d}\mu_{\lambda}=-\int_{0}^{\lambda}\int_{I_{k}^{L}}\frac{\,{\rm
d}\mu_{\lambda,x}}{\,{\rm d}\mu_{\lambda}}(x,\hat{k})\,{\rm
d}\nu_{\frac{x}{\tilde{\lambda}}}(\hat{k})\,2x\phi(x)\,{\rm d}x\,;\end{split}$

furthermore by Young’s inequality and (5.5)

$\begin{split}\int_{0}^{\lambda}|\overline{b}_{,x}^{L}(x,k)|\,{\rm
d}x\leq\int_{I_{k}^{L}\times(0,\lambda)}\left|\frac{\,{\rm
d}\mu_{\lambda,x}}{\,{\rm d}\mu_{\lambda}}\right|\,{\rm
d}\mu_{\lambda}\leq\frac{1}{2}\int_{I_{k}^{L}\times(0,\lambda)}\left(k^{2}+\frac{1}{k^{2}}\left(\frac{\,{\rm
d}\mu_{\lambda,x}}{\,{\rm d}\mu_{\lambda}}\right)^{2}\right)\,{\rm
d}\mu_{\lambda}\leq C\,.\end{split}$

Thus in particular

$\int_{0}^{\lambda}\sum_{k\in\frac{\pi\mathcal{Z}}{L}}|\overline{b}_{,x}^{L}(x,k)|\,{\rm
d}x\leq\mathcal{F}_{\infty}^{\lambda}(\mu_{\lambda})\leq C\,.$

As a consequence we have that
$\mu_{,x}^{L}\in\mathcal{M}_{b}((0,\lambda)\times\mathbb{R})$, and

$\mu_{,x}^{L}=\sum_{k\in\frac{\pi\mathbb{Z}}{L}}\overline{b}_{,x}^{L}(x,k)\mathcal{L}^{1}\mathop{\hbox{\vrule
height=7.0pt,width=0.5pt,depth=0.0pt\vrule
height=0.5pt,width=6.0pt,depth=0.0pt}}\nolimits{(0,\lambda)}\times\delta_{k}\,,$

and by Remark 2.2 $(ii)$ $\tilde{\mu}_{,x}^{L}\ll\tilde{\mu}^{L}$. Moreover,
as $\nu_{\frac{x}{\lambda}}$ is a probability measure, there holds

$\sum_{k\in\frac{\pi\mathbb{Z}}{L}}\overline{b}^{L}(x,k)=2x\bigg{(}\sum_{k\in\frac{\pi\mathbb{Z}}{L},k\neq
0}\nu_{\frac{x}{\lambda}}(I_{k}^{L})\bigg{)}=2x\,,\quad\forall
x\in(0,\lambda)\,.$

Note in particular that $\mu^{L}\in\mathcal{M}_{\infty}^{\lambda}$. Moreover
(5.9) readily follows and $(i)$ is proved. We next show $(ii)$. Take
$\varphi\in C^{\infty}_{c}((0,1)\times\mathbb{R})$, so that from (5.11) we
obtain

$\begin{split}\int_{(0,1)\times\mathbb{R}}\varphi\,{\rm
d}\mu^{L}&=\int_{0}^{1}\sum_{k\in\frac{\pi\mathbb{Z}}{L},k\neq
0}\varphi(x,k)\nu_{\frac{x}{\lambda}}(I_{k}^{L})\,2x\,{\rm d}x\\\
&=\int_{0}^{1}\sum_{k\in\frac{\pi\mathbb{Z}}{L},k\neq
0}\int_{I_{k}^{L}}\big{(}\varphi(x,k)-\varphi(x,\hat{k})\big{)}\,{\rm
d}\nu_{\frac{x}{\lambda}}(\hat{k})\,2x\,{\rm
d}x+\int_{(0,1)\times\mathbb{R}}\varphi\,{\rm d}\mu_{\lambda}\,.\end{split}$
(5.12)

Since $\varphi$ is uniformly continuous for every $\varepsilon>0$ there is
$L_{0}>1$ such that for all $L\geq L_{0}$

$|\varphi(x,k)-\varphi(x,\hat{k})|<\varepsilon\quad\forall x\in(0,\lambda)\,,\
\forall k\in\frac{\pi\mathbb{Z}}{L},\ \forall\hat{k}\in I_{k}^{L}\,,$

from which we readily deduce that

$\int_{0}^{1}\sum_{k\in\frac{\pi\mathbb{Z}}{L},k\neq
0}\int_{I_{k}^{L}}\big{|}\varphi(x,k)-\varphi(x,\hat{k})\big{|}\,{\rm
d}\nu_{\frac{x}{\lambda}}(\hat{k})\,2x\,{\rm
d}x\leq\mu_{\lambda}((0,\lambda)\times\mathbb{R})\varepsilon=\lambda^{2}\mu((0,1)\times\mathbb{R})\varepsilon\,.$
(5.13)

From (5.12), (5.13),(5.3) and the arbitrariness of $\varepsilon$ we infer
$\mu^{L}\mathop{\hbox{\vrule height=7.0pt,width=0.5pt,depth=0.0pt\vrule
height=0.5pt,width=6.0pt,depth=0.0pt}}\nolimits((0,1)\times\mathbb{R})\stackrel{{\scriptstyle*}}{{\rightharpoonup}}\mu$
as $L\to+\infty$. By analogous arguments we get
$\mu_{,x}^{L}\mathop{\hbox{\vrule height=7.0pt,width=0.5pt,depth=0.0pt\vrule
height=0.5pt,width=6.0pt,depth=0.0pt}}\nolimits((0,1)\times\mathbb{R})\stackrel{{\scriptstyle*}}{{\rightharpoonup}}\mu_{,x}$
as $L\to+\infty$. It remains to prove $(iii)$. We start by observing that for
all $\delta>0$ and all $\hat{k}\in I_{k}^{L}$ we have

$k^{2}\leq\left(\hat{k}+\frac{\pi}{L}\right)^{2}\leq(1+\delta)\hat{k}^{2}+(1+\delta^{-1})\frac{\pi^{2}}{L^{2}}\,,$

so that

$\begin{split}\int_{0}^{\lambda}\sum_{k\in\frac{\pi\mathbb{Z}}{L}}k^{2}\overline{b}^{L}(x,k)\,{\rm
d}x&=\int_{0}^{\lambda}\sum_{k\in\frac{\pi\mathbb{Z}}{L},k\neq
0}\int_{I_{k}^{L}}k^{2}\,{\rm d}\nu_{\frac{x}{\lambda}}(\hat{k})\,2x\,{\rm
d}x\\\ &=\sum_{k\in\frac{\pi\mathbb{Z}}{L},k\neq 0}\int_{(0,\lambda)\times
I_{k}^{L}}k^{2}\,{\rm d}\mu_{\lambda}(x,\hat{k})\\\
&\leq\sum_{k\in\frac{\pi\mathbb{Z}}{L},k\neq 0}\int_{(0,\lambda)\times
I_{k}^{L}}\bigg{(}(1+\delta)\hat{k}^{2}+(1+\delta^{-1})\frac{\pi^{2}}{L^{2}}\bigg{)}\,{\rm
d}\mu_{\lambda}(x,\hat{k})\\\
&=(1+\delta)\int_{(0,\lambda)\times\mathbb{R}}\hat{k}^{2}\,{\rm
d}\mu_{\lambda}+\mu_{\lambda}((0,\lambda)\times\mathbb{R})(1+\delta^{-1})\frac{\pi^{2}}{L^{2}}\,.\end{split}$
(5.14)

Moreover there holds

$\begin{split}\int_{0}^{\lambda}\sum_{k\in\frac{\pi\mathbb{Z}}{L},k\neq
0}\frac{1}{4k^{2}}\frac{(\overline{b}^{L}_{,x}(x,k))^{2}}{\overline{b}^{L}(x,k)}\,{\rm
d}x=\int_{0}^{\lambda}\sum_{k\in\frac{\pi\mathbb{Z}}{L},k\neq
0}\frac{1}{4k^{2}}\bigg{(}\frac{\overline{b}_{,x}^{L}(x,k)}{\overline{b}^{L}(x,k)}\bigg{)}^{2}\overline{b}^{L}(x,k)\,{\rm
d}x\,.\end{split}$ (5.15)

Since $1/|k|\leq 1/|\hat{k}|$ for $\hat{k}\in I_{k}^{L}$ the following
inequality follows

$\frac{1}{2|k|}\frac{\overline{b}_{,x}^{L}(x,k)}{\overline{b}^{L}(x,k)}=\frac{1}{2|k|}\fint_{I_{k}^{L}}\frac{\,{\rm
d}\mu_{\lambda,x}}{\,{\rm d}\mu_{\lambda}}(x,\hat{k})\,{\rm
d}\nu_{\frac{x}{\lambda}}(\hat{k})\leq\fint_{I_{k}^{L}}\frac{1}{2|\hat{k}|}\frac{\,{\rm
d}\mu_{\lambda,x}}{\,{\rm d}\mu_{\lambda}}(x,\hat{k})\,{\rm
d}\nu_{\frac{x}{\lambda}}(\hat{k})\,.$ (5.16)

Now combining (5.15) with (5.16) we obtain

$\begin{split}\int_{0}^{\lambda}\sum_{k\in\frac{\pi\mathbb{Z}}{L},k\neq
0}\frac{1}{4k^{2}}\frac{(\overline{b}^{L}_{,x}(x,k))^{2}}{\overline{b}^{L}(x,k)}\,{\rm
d}x&=\int_{0}^{\lambda}\sum_{k\in\frac{\pi\mathbb{Z}}{L},k\neq
0}\bigg{(}\fint_{I_{k}^{L}}\frac{1}{2|\hat{k}|}\frac{\,{\rm
d}\mu_{\lambda,x}}{\,{\rm d}\mu_{\lambda}}(x,\hat{k})\,{\rm
d}\nu_{\frac{x}{\lambda}}(\hat{k})\bigg{)}^{2}\nu_{\frac{x}{\lambda}}(I_{k}^{L})\,2x\,{\rm
d}x\\\ &\leq\int_{0}^{\lambda}\sum_{k\in\frac{\pi\mathbb{Z}}{L},k\neq
0}\int_{I_{k}^{L}}\frac{1}{4\hat{k}^{2}}\bigg{(}\frac{\,{\rm
d}\mu_{\lambda,x}}{\,{\rm d}\mu_{\lambda}}(x,\hat{k})\bigg{)}^{2}\,{\rm
d}\nu_{\frac{x}{\lambda}}(\hat{k})\,2x\,{\rm d}x\\\
&=\int_{(0,\lambda)\times\mathbb{R}}\frac{1}{4\hat{k}^{2}}\bigg{(}\frac{\,{\rm
d}\mu_{\lambda,x}}{\,{\rm d}\mu_{\lambda}}(x,\hat{k})\bigg{)}^{2}\,{\rm
d}\mu_{\lambda}\\\
&=\int_{(0,1)\times\mathbb{R}}\frac{1}{4\hat{k}^{2}}\bigg{(}\frac{\,{\rm
d}\mu_{,x}}{\,{\rm d}\mu}(x,\hat{k})\bigg{)}^{2}\,{\rm d}\mu\,,\end{split}$
(5.17)

where the inequality follows by Jensen’s inequality. Finally gathering
together (5.14) and (5.17) and recalling (5.5) we infer

$\limsup_{L\to\infty}\mathcal{F}_{\infty}^{\lambda}(\mu^{L})\leq(1+\delta)\limsup_{L\to\infty}\mathcal{F}_{\infty}^{\lambda}\big{(}\mu_{\lambda}\big{)}\leq(1+\delta)\mathcal{F}_{\infty}(\mu)\,.$

By arbitrariness of $\delta$, $(iii)$ follows by letting $\delta\to 0$.
∎

######  Lemma 5.4 (Construction of $u$).

Let $\mu\in\mathcal{M}_{\infty}$ be such that
$\mathcal{F}_{\infty}(\mu)<+\infty$. Let $\varepsilon=\varepsilon(L)>0$ and
$n=n(L)\in\mathbb{N}$ be such that

$\lim_{L\to+\infty}\varepsilon(L)=0\,,\quad\lim_{L\to+\infty}n(L)=\lim_{L\to+\infty}\frac{L}{n(L)}=+\infty\,.$

Then there exists $\hat{u}^{L}\in\mathcal{A}_{L}^{\rm
out}\cap\mathcal{A}_{L_{0}}^{\rm out}$ with $L_{0}:=L/n(L)$ that satisfies the
following properties: let

$A^{L}(x):=\frac{1}{2}\fint_{-L}^{L}(\hat{u}^{L}_{,y}(x,\cdot))^{2}\,{\rm
d}y\quad\text{ and }\quad f^{L}(x):=\sqrt{\frac{x}{A^{L}(x)}}\,.$

Then for all $x\in(0,1)$ and $N\in\mathbb{N}$ there holds

$\max\\{x,\varepsilon\\}\lesssim A^{L}(x)\lesssim\max\\{x,\varepsilon\\}\,;$
(5.18)
$(f^{L}(x))^{2}\lesssim\frac{x}{\max\\{x,\varepsilon\\}}\,,\quad(f^{L}(x))^{2}\leq
1+o_{L}(1)\,;$ (5.19) $(f^{L}(x))^{2}\geq 1+o_{N}(1)\ \text{ if
}x\in(N\varepsilon,1)\,;$ (5.20)
$(\dot{f}^{L}(x))^{2}\lesssim\frac{\max\\{e^{-\frac{x}{\varepsilon}},e^{-\frac{1}{\sqrt{\varepsilon}}}\\}}{x\varepsilon}\,,\quad(\ddot{f}^{L}(x))^{2}\lesssim\frac{1}{x^{3}\varepsilon}\,;$
(5.21)

there exists a continuous increasing function
$\omega\colon[0,+\infty)\to[0,+\infty)$ with $\omega(0)=0$ such that

$\fint_{-L}^{L}(\hat{u}^{L}(x,\cdot))^{2}\,{\rm
d}y\lesssim{\begin{cases}\max\\{x,\varepsilon\\}(\omega(2N\varepsilon)+Ne^{-N})&\text{
if }x\in(0,N\varepsilon)\\\ x&\text{ if
}x\in(N\varepsilon,1)\end{cases}\quad}\,;$ (5.22)
$\fint_{-L}^{L}(\hat{u}^{L}(x,\cdot))^{4}\,{\rm d}y\lesssim
L_{0}^{2}(\max\\{\varepsilon,x\\})^{2}\,;$ (5.23)
$\fint_{-L}^{L}\left((\hat{u}_{,yy}^{L}(x,\cdot))^{2}+(\hat{u}_{,x}^{L}(x,\cdot))^{2}\right)\,{\rm
d}y\lesssim\frac{1}{\varepsilon}\,.$ (5.24)

Moreover

$(\mu^{L}(\hat{u}^{L}),\mu^{L}_{,x}(\hat{u}^{L}))\stackrel{{\scriptstyle*}}{{\rightharpoonup}}(\mu,\mu_{,x})\quad\text{in
}\mathcal{M}_{b}((-1,1)\times\mathbb{R})^{2}\,;$ (5.25)
$\limsup_{L\to\infty}\fint_{-L}^{L}\int_{-1}^{1}\big{(}(\hat{u}^{L}_{,x})^{2}+(\hat{u}^{L}_{,yy})^{2}\big{)}\,{\rm
d}x\,{\rm d}y\leq\mathcal{F}_{\infty}(\mu)\,;$ (5.26)
$\fint_{-L}^{L}\int_{-1}^{1}\left((\hat{u}^{L}_{,xy})^{2}+(\hat{u}^{L}_{,xx})^{2}+(\hat{u}_{,yyx}^{L})^{2}\right)\,{\rm
d}x\,{\rm d}y\lesssim\frac{1}{\varepsilon^{2}}\,;$ (5.27)
$\fint_{-L}^{L}\int_{-1}^{1}(\hat{u}^{L}_{,x})^{4}\,{\rm d}x\,{\rm
d}y\lesssim\frac{L_{0}^{2}}{\varepsilon^{2}}\,.$ (5.28)

Finally since $\hat{u}^{L}$ and all its derivatives are $2L_{0}$-periodic in
the $y$-variable the above estimates still hold true if we replace the average
integral on $[-L,L]$ with the average integral on $[-L_{0},L_{0}]$.

###### Proof.

Let $\mu\in\mathcal{M}_{\infty}$, $\varepsilon=\varepsilon(L)$, $n=n(L)$ and
$L_{0}$ be as in the statement. We set

$\lambda=\lambda(L):=(1+\sqrt{\varepsilon})\searrow 1\quad\text{as
}L\to+\infty\,,$

hence in particular $\lambda\searrow 1$ as $L_{0}\to+\infty$. We construct
$\hat{u}^{L}\in\mathcal{A}_{L_{0}}^{\rm out}$ and then we extend it
periodically in $[-1,1]\times[-L,L]$, without relabelling it. In this way,
since $L=n(L)L_{0}$ with $n(L)\in\mathbb{N}$, we have
$\hat{u}^{L}\in\mathcal{A}_{L}^{\rm out}$. The main idea is that of
discretizing the measure $\mu$ in the variable $k$ to get a measure
concentrated on lines $\mathbb{R}\times\\{k\\}$ with
$k\in\frac{\pi\mathbb{Z}}{L_{0}}$ where the weight on each line is a
coefficient $b^{L_{0}}(x,k)$. Afterwards we define $\hat{u}^{L}$ as in such a
way that its Fourier coefficients $a^{L}_{k}(x)$ are as close as possible to
${\sqrt{b^{L_{0}}(x,k)}}/k$ but at the same time have better regularity to
ensure $\hat{u}^{L}\in\mathcal{A}_{L}^{\rm out}$. In order to do that we first
dilate $\mu$ with a factor $\lambda$ in the $x$-variable as in Lemma 5.2, then
we discretize using Lemma 5.3, and finally we mollify $b^{L_{0}}(\cdot,k)$ at
scale $\varepsilon$ after a suitable extension in $\mathbb{R}$.

To this purpose we let $(\mu^{L_{0}})\subset\mathcal{M}_{\infty}^{\lambda}$ be
the sequence of Lemma 5.3 for the parameter $L_{0}$, which is of the form

$\mu^{L_{0}}=\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}}}\overline{b}^{L_{0}}(x,k)\mathcal{L}^{1}\mathop{\hbox{\vrule
height=7.0pt,width=0.5pt,depth=0.0pt\vrule
height=0.5pt,width=6.0pt,depth=0.0pt}}\nolimits{(0,\lambda)}\times\delta_{k}\,.$

By the mean value theorem, for each $\lambda$, we can find
$\bar{\lambda}\in(\frac{\lambda+1}{2},\lambda)$ such that

$\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}}}\overline{b}^{L_{0}}(\bar{\lambda},k)k^{2}\leq\fint_{\frac{\lambda+1}{2}}^{\lambda}\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}}}\overline{b}^{L_{0}}(x,k)k^{2}\,{\rm
d}x\,.$ (5.29)

By truncating $\bar{b}^{L_{0}}$ at $x=\bar{\lambda}$ we can define
$b^{L_{0}}\colon\mathbb{R}\times\frac{\pi\mathbb{Z}}{L_{0}}\to\mathbb{R}$ as

$b^{L_{0}}(k,x):=\begin{cases}0&\text{if }x\leq 0,\\\
\bar{b}^{L_{0}}(x,k)&\text{if }0<x<\bar{\lambda},\\\
\bar{b}^{L_{0}}(\bar{\lambda},k)&\text{if }x\geq\bar{\lambda}\,.\end{cases}$
(5.30)

Let
$\rho_{\varepsilon}(x):=\frac{1}{2\varepsilon}e^{\frac{-|x|}{\varepsilon}}$
and note that in particular

$|\dot{\rho}_{\varepsilon}(x)|=\frac{1}{\varepsilon}\rho_{\varepsilon}(x)\,.$
(5.31)

Finally we let
$a^{L}\colon\mathbb{R}\times\frac{\pi\mathbb{Z}}{L_{0}}\to\mathbb{R}$ be
defined as

$a^{L}(x,k):=b^{L_{0}}(\cdot,k)*\rho_{\varepsilon}(x)\,,$

and $\hat{u}^{L}\in\mathcal{A}_{L_{0}}^{\rm out}$ be the function

$\hat{u}^{L}(x,y):=\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}},k>0}\frac{\sqrt{a^{L}(x,k)}}{k}\sqrt{2}\sin(ky)+\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}},k<0}\frac{\sqrt{a^{L}(x,k)}}{k}\sqrt{2}\cos(ky)\,.$

Eventually we extend $\hat{u}^{L}$, without relabelling it, periodically in
$[-1,1]\times[-L,L]$.

Step 1: in this step we show (5.18)–(5.21). By (3.3) and (5.8) we have that

$\begin{split}2A^{L}(x)&=\fint_{-L}^{L}(\hat{u}_{,y})^{2}\,{\rm
d}y=\fint_{-L_{0}}^{L_{0}}(\hat{u}_{,y})^{2}\,{\rm d}y\\\
&=\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}}}{a}^{L}(x,k)=\Big{(}\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}}}{b}^{L_{0}}(\cdot,k)\Big{)}*\rho_{\varepsilon}(x)\\\
&=2\big{(}x\chi_{(0,\bar{\lambda})}+\bar{\lambda}\chi_{(\bar{\lambda},+\infty)}\big{)}*\rho_{\varepsilon}(x)=2x\chi_{\\{x\geq
0\\}}+\varepsilon\Big{(}e^{\frac{-|x|}{\varepsilon}}-e^{\frac{x-\bar{\lambda}}{\varepsilon}}\Big{)}\,,\end{split}$
(5.32)

for all $x\in[-1,1]$. Then for $\varepsilon$ small enough we have

$\frac{\varepsilon}{2e}\leq 2A^{L}(x)\leq{3\varepsilon}\quad\text{ if
}x\in(0,\varepsilon)\,,$ $x\leq
2x+\varepsilon\big{(}e^{\frac{-1}{\varepsilon}}-e^{\frac{1-\bar{\lambda}}{\varepsilon}}\big{)}\leq
2A^{L}(x)\leq{3x}\quad\text{ if }x\in(\varepsilon,1)\,,$

so that

$\frac{1}{2e}\max\\{x,\varepsilon\\}\leq
2A^{L}(x)=\fint_{-L}^{L}(\hat{u}_{,y})^{2}\,{\rm d}y\leq
3\max\\{x,\varepsilon\\}\,.$

We have that

$(f^{L}(x))^{2}=\frac{x}{A^{L}(x)}\,,$

and by estimates above

$(f^{L}(x))^{2}\lesssim\frac{x}{\max\\{x,\varepsilon\\}}\lesssim\frac{x}{\varepsilon}\quad\text{
if }x\in(0,\varepsilon)\,.$

Observing that
$(e^{\frac{-x}{\varepsilon}}-e^{\frac{x-\bar{\lambda}}{\varepsilon}})\geq 0$
if and only if $x\in(0,\bar{\lambda}/2)$ we have

$(f^{L}(x))^{2}=\frac{x}{x+\frac{\varepsilon}{2}(e^{\frac{-x}{\varepsilon}}-e^{\frac{x-\bar{\lambda}}{\varepsilon}})}\leq
1\quad\text{ in }(0,\bar{\lambda}/2)\,,$

and

$(f^{L}(x))^{2}\leq\frac{x}{x+\frac{\varepsilon}{2}(e^{\frac{-1}{\varepsilon}}-e^{\frac{1-\bar{\lambda}}{\varepsilon}})}\leq
1+\frac{\frac{\varepsilon}{2}|e^{\frac{-1}{\varepsilon}}-e^{\frac{1-\bar{\lambda}}{\varepsilon}}|}{\frac{1}{2}+\frac{\varepsilon}{2}(e^{\frac{-1}{\varepsilon}}-e^{\frac{1-\bar{\lambda}}{\varepsilon}})}=1+\frac{\varepsilon}{8}\quad\text{
for }x\in(\bar{\lambda}/2,1)\,.$

Moreover we have that

$(f^{L}(x))^{2}\geq\frac{x}{x+\frac{\varepsilon}{2}e^{{-N}}}=1-\frac{\frac{\varepsilon}{2}e^{{-N}}}{x+\frac{\varepsilon}{2}e^{{-N}}}\geq
1-\frac{\frac{\varepsilon}{2}e^{{-N}}}{N\varepsilon}=1+o_{N}(1)\quad\text{ for
}x\in[N\varepsilon,1)\,.$

A direct computation shows that

$(\dot{f}^{L}(x))^{2}=\frac{\left(A^{L}(x)-x\dot{A}^{L}(x)\right)^{2}}{4x(A^{L}(x))^{3}}\lesssim\frac{(\max\\{x,\varepsilon\\})^{2}\max\\{e^{-\frac{x}{\varepsilon}},e^{\frac{1-\bar{\lambda}}{\varepsilon}}\\}}{x(\max\\{x,\varepsilon\\})^{3}}\lesssim\frac{\max\\{e^{-\frac{x}{\varepsilon}},e^{\frac{1}{\sqrt{\varepsilon}}}\\}}{x\max\\{x,\varepsilon\\}}\lesssim\frac{\max\\{e^{-\frac{x}{\varepsilon}},e^{\frac{1}{\sqrt{\varepsilon}}}\\}}{x\varepsilon}\,,$

where the second inequality follows from
$\frac{1-\bar{\lambda}}{\varepsilon}\leq\frac{1-\frac{\lambda+1}{2}}{\varepsilon}=\frac{1}{2\sqrt{\varepsilon}}$.
Furthermore we have

$\begin{split}(\ddot{f}^{L}(x))^{2}&\lesssim\frac{x(\ddot{A}^{L}(x))^{2}}{(A^{L}(x))^{3}}+\frac{1}{x^{3}A^{L}(x)}+\frac{(\dot{A}^{L}(x))^{2}}{x(A^{L}(x))^{3}}+\frac{x(\dot{A}^{L}(x))^{4}}{(A^{L}(x))^{5}}\\\
&\lesssim\frac{x\varepsilon^{-2}e^{-\frac{x}{\varepsilon}}}{(\max\\{x,\varepsilon\\})^{3}}+\frac{1}{x^{3}\max\\{x,\varepsilon\\}}+\frac{1}{x(\max\\{x,\varepsilon\\})^{3}}+\frac{x}{(\max\\{x,\varepsilon\\})^{5}}\,.\end{split}$
(5.33)

Hence

$(\ddot{f}^{L}(x))^{2}\lesssim\frac{1}{\varepsilon^{4}}+\frac{1}{x^{3}\varepsilon}+\frac{1}{x\varepsilon^{3}}\lesssim\frac{1}{x^{3}\varepsilon}\quad\text{
if }x\in(0,\varepsilon)\,,$

and

$(\ddot{f}^{L}(x))^{2}\lesssim\frac{e^{-\frac{x}{\varepsilon}}}{x^{2}\varepsilon^{2}}+\frac{1}{x^{4}}\lesssim\frac{1}{x^{3}\varepsilon}\quad\text{
if }x\in(\varepsilon,1)\,.$

Step 2: in this step we show (5.22). By (3.2) it holds

$\begin{split}\fint_{-L}^{L}(\hat{u}^{L}(x,\cdot))^{2}\,{\rm
d}y=\fint_{-L_{0}}^{L_{0}}(\hat{u}^{L}(x,\cdot))^{2}\,{\rm
d}y&=\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}},k\neq
0}\frac{{a^{L}(x,k)}}{k^{2}}\\\
&=\bigg{(}\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}},k\neq
0}\frac{{b^{L_{0}}(\cdot,k)}}{k^{2}}\bigg{)}*\rho_{\varepsilon}(x)\\\
&=\int_{0}^{+\infty}\bigg{(}\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}},k\neq
0}\frac{{b^{L_{0}}(z,k)}}{k^{2}}\bigg{)}\rho_{\varepsilon}(x-z)\,{\rm
d}z\,.\end{split}$ (5.34)

Moreover by the fundamental theorem of calculus and Hölder’s inequality we
have

$b^{L_{0}}(z,k)=\left(\sqrt{b^{L_{0}}(z,k)}\right)^{2}=\left(\int_{0}^{z}\frac{b^{L_{0}}_{,x}(\hat{z},k)}{2\sqrt{b^{L_{0}}(\hat{z},k)}}\,{\rm
d}\hat{z}\right)^{2}\leq
z\int_{0}^{z}\frac{(b^{L_{0}}_{,x}(\hat{z},k))^{2}}{4{b^{L_{0}}(\hat{z},k)}}\,{\rm
d}\hat{z}\,.$ (5.35)

Combining (5.34) with (5.35) we find

$\begin{split}\fint_{-L}^{L}(\hat{u}^{L}(x,\cdot))^{2}\,{\rm
d}y&\leq\int_{0}^{+\infty}\bigg{(}\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}},k\neq
0}\int_{0}^{z}\frac{(b^{L_{0}}_{,x}(\hat{z},k))^{2}}{4k^{2}{b^{L_{0}}(\hat{z},k)}}\,{\rm
d}\hat{z}\bigg{)}z\rho_{\varepsilon}(x-z)\,{\rm d}z\end{split}$

By definition of $b^{L_{0}}$ it follows that

$\begin{split}\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}},k\neq
0}\int_{0}^{z}\frac{(b^{L_{0}}_{,x}(\hat{z},k))^{2}}{4k^{2}{b^{L_{0}}(\hat{z},k)}}\,{\rm
d}\hat{z}&=\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}},k\neq
0}\int_{0}^{z\wedge\bar{\lambda}}\frac{(\bar{b}^{L_{0}}_{,x}(\hat{z},k))^{2}}{4k^{2}{\bar{b}^{L_{0}}(\hat{z},k)}}\,{\rm
d}\hat{z}\\\
&\leq\int_{(0,z\wedge\bar{\lambda})\times\mathbb{R}}\frac{1}{4k^{2}}\bigg{(}\frac{\,{\rm
d}\mu^{L_{0}}_{,x}}{\,{\rm d}\mu^{L_{0}}}\bigg{)}^{2}\,{\rm
d}\mu^{L_{0}}=:\omega(z)\,,\end{split}$ (5.36)

where the last inequality can be obtained by arguing exactly as in (5.17).
Therefore we deduce that

$\fint_{-L_{0}}^{L_{0}}(\hat{u}^{L}(x,\cdot))^{2}\,{\rm
d}y\leq\int_{0}^{+\infty}\omega(z)z\rho_{\varepsilon}(x-z)\,{\rm d}z\,.$

Note that $\omega(z)\to 0$ as $z\to 0$ and
$\omega(z)\leq\omega(\bar{\lambda})\leq(\lambda^{2})\mathcal{F}_{\infty}(\mu)\leq
C$. Let $N\geq 2$ be a natural number. Assume $x\in(0,N\varepsilon]$. Since
$\omega$ is increasing we have

$\begin{split}\int_{0}^{+\infty}\omega(z)z\rho_{\varepsilon}(x-z)\,{\rm
d}z&\leq\omega(2N\varepsilon)\int_{0}^{2N\varepsilon}z\rho_{\varepsilon}(x-z)\,{\rm
d}z+\omega(\bar{\lambda})\int_{2N\varepsilon}^{+\infty}z\rho_{\varepsilon}(x-z)\,{\rm
d}z\\\
&\leq\omega(2N\varepsilon)\max\\{x,\varepsilon\\}+CNe^{-N}\varepsilon\\\
&\lesssim\max\\{x,\varepsilon\\}(\omega(2N\varepsilon)+Ne^{-N})\,.\end{split}$
(5.37)

If instead $x\in(N\varepsilon,1)$, we get

$\begin{split}\int_{0}^{+\infty}\omega(z)z\rho_{\varepsilon}(x-z)\,{\rm
d}z&\leq\omega(\bar{\lambda})\int_{0}^{+\infty}z\rho_{\varepsilon}(x-z)\,{\rm
d}z\lesssim x.\end{split}$ (5.38)

Step 3: in this step we show (5.23). By the mean value theorem and the fact
that $u^{L}(x,\cdot)$ is $2L_{0}$-periodic, for fixed $x$, we can find
$y_{0}=y_{0}(x)\in[-L_{0},L_{0}]$ such that

$\hat{u}^{L}(x,y_{0})=\fint_{-L_{0}}^{L_{0}}\hat{u}^{L}(x,\hat{y})\,{\rm
d}\hat{y}=0\,,$

where the second equality follows by the definition of $\hat{u}^{L}$ and the
fact that

$\begin{split}&\fint_{-L_{0}}^{L_{0}}\sin(k\hat{y})\,{\rm
d}\hat{y}=\frac{k}{2L_{0}}(-\cos(kL_{0})+\cos(kL_{0}))=0\,,\\\
&\fint_{-L_{0}}^{L_{0}}\cos(k\hat{y})\,{\rm
d}\hat{y}=\frac{k}{2L_{0}}(\sin(kL_{0})-\sin(kL_{0}))=0\,,\end{split}$

for all $k\in\frac{\pi\mathcal{Z}}{L_{0}}$. Thus by the fundamental theorem of
calculus, Hölder’s inequality and Plancherel it holds

$\begin{split}|\hat{u}^{L}(x,y)|=\left|\int_{y_{0}}^{y}\hat{u}_{,y}(x,y^{\prime})\,{\rm
d}y^{\prime}\right|&\leq\sqrt{L_{0}}\left(\int_{-L_{0}}^{L_{0}}(u^{L}_{,y})^{2}\,{\rm
d}y^{\prime}\right)^{\frac{1}{2}}\\\ &=L_{0}\sqrt{2A^{L}(x)}\,.\end{split}$

This together with steps 1 and 2 yield

$\fint_{-L}^{L}\int_{-1}^{1}{(\hat{u}^{L})^{4}}\,{\rm d}x\,{\rm
d}y=\fint_{-L_{0}}^{L_{0}}\int_{-1}^{1}{(\hat{u}^{L})^{4}}\,{\rm d}x\,{\rm
d}y\lesssim
L_{0}^{2}A^{L}(x)\fint_{-L_{0}}^{L_{0}}\int_{-1}^{1}{(\hat{u}^{L})^{2}}\,{\rm
d}x\,{\rm d}y\lesssim L_{0}^{2}(\max\\{x,\varepsilon\\})^{2}\,.$

Step 4: in this step we show (5.24). By (3.3) the definition of $a^{L}$ and
(5.29)

$\begin{split}\fint_{-L}^{L}(u_{,yy}^{L})^{2}\,{\rm
d}y&=\fint_{-L_{0}}^{L_{0}}(u_{,yy}^{L})^{2}\,{\rm
d}y=\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}}}k^{2}a^{L}(x,k)=\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}}}k^{2}b^{L_{0}}(x,\cdot)*\rho_{\varepsilon}(x)\\\
&=\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}}}\biggl{(}k^{2}\int_{0}^{\bar{\lambda}}\bar{b}^{L_{0}}(z,k)\rho_{\varepsilon}(x-z)\,{\rm
d}z\biggr{)}+\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}}}\bar{b}^{L_{0}}(\bar{\lambda},k)k^{2}\int_{\bar{\lambda}}^{+\infty}\rho_{\varepsilon}(x-z)\,{\rm
d}z\\\
&\leq\|\rho_{\varepsilon}\|_{\infty}\int_{0}^{\bar{\lambda}}\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}}}\overline{b}^{L_{0}}(x,k)k^{2}\,{\rm
d}x+e^{\frac{x-\bar{\lambda}}{\varepsilon}}\fint_{\frac{\lambda+1}{2}}^{\lambda}\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}}}\overline{b}^{L_{0}}(x,k)k^{2}\,{\rm
d}x\\\
&\lesssim\left(\frac{1}{\varepsilon}+\frac{1}{\lambda-1}e^{\frac{x-\bar{\lambda}}{\varepsilon}}\right)\int_{0}^{{\lambda}}\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}}}\overline{b}^{L_{0}}(x,k)k^{2}\,{\rm
d}x\\\
&\lesssim\left(\frac{1}{\varepsilon}+\frac{1}{\lambda-1}e^{\frac{x-\bar{\lambda}}{\varepsilon}}\right)\mathcal{F}_{\infty}^{\lambda}(\mu^{L_{0}})\lesssim\frac{1}{\varepsilon}\,.\end{split}$
(5.39)

Since the function $(z_{1},z_{2})\mapsto{z_{1}^{2}}/{z_{2}}$ is convex by
Jensen’s inequality we have

$\frac{(a^{L}_{,x}(x,k))^{2}}{a^{L}(x,k)}=\frac{\big{(}b^{L_{0}}_{,x}(\cdot,k)*\rho_{\varepsilon}(x)\big{)}^{2}}{b^{L_{0}}(\cdot,k)*\rho_{\varepsilon}(x)}\leq\frac{(b^{L_{0}}_{,x}(\cdot,k))^{2}}{b^{L_{0}}(\cdot,k)}*\rho_{\varepsilon}(x)\,.$

This together with (3.3) and (3.4) imply

$\begin{split}\fint_{-L}^{L}(u_{,x}^{L})^{2}\,{\rm
d}y=\fint_{-L_{0}}^{L_{0}}(u_{,x}^{L})^{2}\,{\rm
d}y&=\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}},k\neq
0}\frac{1}{4k^{2}}\frac{(a^{L}_{,x}(x,k))^{2}}{a^{L}(x,k)}\leq\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}},k\neq
0}\frac{1}{4k^{2}}\frac{(b^{L_{0}}_{,x}(\cdot,k))^{2}}{b^{L_{0}}(\cdot,k)}*\rho_{\varepsilon}(x)\\\
&=\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}},k\neq
0}\bigg{(}\frac{1}{4k^{2}}\int_{0}^{\bar{\lambda}}\frac{(b^{L_{0}}_{,x}(z,k))^{2}}{b^{L_{0}}(z,k)}\rho_{\varepsilon}(x-z)\,{\rm
d}z\bigg{)}\\\
&\leq\|\rho_{\varepsilon}\|_{\infty}\int_{0}^{\bar{\lambda}}\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}},k\neq
0}\frac{1}{4k^{2}}\frac{(b^{L_{0}}_{,x}(x,k))^{2}}{b^{L_{0}}(x,k)}\,{\rm
d}x\leq\frac{1}{\varepsilon}\mathcal{F}_{\infty}^{\lambda}(\mu^{L_{0}})\lesssim\frac{1}{\varepsilon}\,.\end{split}$
(5.40)

Thus combining (5.39) with (5.40) we infer (5.24).

Step 5: in this step we show (5.25). By recalling Definition 2.1 we have

$\hat{\mu}^{L}:=\mu^{L}(\hat{u}^{L})=\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}}}a^{L}(x,k)\mathcal{L}^{1}\mathop{\hbox{\vrule
height=7.0pt,width=0.5pt,depth=0.0pt\vrule
height=0.5pt,width=6.0pt,depth=0.0pt}}\nolimits(-1,1)\times\delta_{k}\,,$

and

$\hat{\mu}^{L}_{,x}=\mu^{L}_{,x}(\hat{u}^{L})=\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}}}a^{L}_{,x}(x,k)\mathcal{L}^{1}\mathop{\hbox{\vrule
height=7.0pt,width=0.5pt,depth=0.0pt\vrule
height=0.5pt,width=6.0pt,depth=0.0pt}}\nolimits(-1,1)\times\delta_{k}\,,$

Let $\varphi\in C^{\infty}_{c}((-1,1)\times\mathbb{R})$. Then

$\begin{split}\int_{(-1,1)\times\mathbb{R}}\varphi\,{\rm
d}\hat{\mu}^{L}&=\int_{(-1,1)\times\mathbb{R}}\varphi\,{\rm
d}\hat{\mu}^{L}-\int_{(0,1)\times\mathbb{R}}\varphi\,{\rm
d}\mu^{L_{0}}+\int_{(0,1)\times\mathbb{R}}\varphi\,{\rm
d}\mu^{L_{0}}\,.\end{split}$

By Lemma 5.3 we have that

$\lim_{L\to\infty}\int_{(0,1)\times\mathbb{R}}\varphi\,{\rm
d}\mu^{L_{0}}=\int_{(0,1)\times\mathbb{R}}\varphi\,{\rm d}\mu\,,$

hence it suffices to show that

$\lim_{L\to+\infty}\biggl{(}\int_{(-1,1)\times\mathbb{R}}\varphi\,{\rm
d}\hat{\mu}^{L}-\int_{(0,1)\times\mathbb{R}}\varphi\,{\rm
d}\mu^{L_{0}}\biggr{)}=0\,.$

Indeed by (5.32) and (5.8) we have

$\begin{split}\biggl{|}\int_{(-1,1)\times\mathbb{R}}\varphi\,{\rm
d}\hat{\mu}^{L}-&\int_{(0,1)\times\mathbb{R}}\varphi\,{\rm
d}\mu^{L_{0}}\biggr{|}\leq\|\varphi\|_{\infty}\left|\int_{(-1,1)\times\mathbb{R}}\,{\rm
d}\hat{\mu}^{L}-\int_{(0,1)\times\mathbb{R}}\,{\rm d}\mu^{L_{0}}\right|\\\
&=\|\varphi\|_{\infty}\Bigg{|}\int_{-1}^{1}\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}}}a^{L}(x,k)\,{\rm
d}x-\int_{0}^{1}\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}}}\bar{b}^{L}(x,k)\,{\rm
d}x\Bigg{|}\\\
&=\|\varphi\|_{\infty}\left|\int_{-1}^{1}\varepsilon\Big{(}e^{-\frac{|x|}{\varepsilon}}-e^{\frac{x-\bar{\lambda}}{\varepsilon}}\Big{)}\,{\rm
d}x\right|\leq C\varepsilon^{2}\to 0\quad\text{as }L\to+\infty\,.\end{split}$

Moreover we have

$\int_{(-1,1)\times\mathbb{R}}\varphi\,{\rm
d}\hat{\mu}^{L}_{,x}=-\int_{(-1,1)\times\mathbb{R}}\varphi_{,x}\,{\rm
d}\hat{\mu}^{L}\to-\int_{(-1,1)\times\mathbb{R}}\varphi_{,x}\,{\rm
d}\mu=\int_{(-1,1)\times\mathbb{R}}\varphi\,{\rm d}\mu_{,x}\,.$

Step 6: in this step we show (5.26). By (3.3), (5.30) we have

$\begin{split}&\fint_{-L_{0}}^{L_{0}}\int_{-1}^{1}(\hat{u}^{L}_{,yy})^{2}\,{\rm
d}x\,{\rm
d}y=\int_{-1}^{1}\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}}}a^{L}(x,k)k^{2}\,{\rm
d}x=\int_{-1}^{1}\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}}}b^{L_{0}}(\cdot,k)*\rho_{\varepsilon}(x)k^{2}\,{\rm
d}x\\\
&=\int_{-1}^{1}\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}}}\biggl{(}k^{2}\int_{0}^{\bar{\lambda}}\bar{b}^{L_{0}}(z,k)\rho_{\varepsilon}(x-z)\,{\rm
d}z\biggr{)}\,{\rm
d}x+\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}}}\bar{b}^{L_{0}}(\bar{\lambda},k)k^{2}\int_{-1}^{1}\int_{\bar{\lambda}}^{+\infty}\rho_{\varepsilon}(x-z)\,{\rm
d}z\,.\end{split}$ (5.41)

Fubini’s theorem yields

$\int_{-1}^{1}\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}}}\biggl{(}k^{2}\int_{0}^{\bar{\lambda}}\bar{b}^{L_{0}}(z,k)\rho_{\varepsilon}(x-z)\,{\rm
d}z\biggr{)}\,{\rm
d}x\leq\int_{0}^{\bar{\lambda}}\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}}}\bar{b}^{L_{0}}(x,k)k^{2}\,{\rm
d}x\,.$ (5.42)

while from (5.29) we deduce

$\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}}}\bar{b}^{L_{0}}(\bar{\lambda},k)k^{2}\int_{-1}^{1}\int_{\bar{\lambda}}^{+\infty}\rho_{\varepsilon}(x-z)\,{\rm
d}z\,{\rm
d}x\leq\frac{\varepsilon}{2}\left(e^{\frac{1-\bar{\lambda}}{\varepsilon}}-e^{\frac{-1-\bar{\lambda}}{\varepsilon}}\right)\fint_{\frac{\lambda+1}{2}}^{\lambda}\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}}}\overline{b}^{L_{0}}(x,k)k^{2}\,{\rm
d}x\,.$ (5.43)

Analogously from (3.3), (3.4) and Jensen’s inequality it holds

$\begin{split}\fint_{-L_{0}}^{L_{0}}\int_{-1}^{1}(\hat{u}^{L}_{,x})^{2}\,{\rm
d}x\,{\rm d}y&=\int_{-1}^{1}\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}},k\neq
0}\frac{1}{4k^{2}}\frac{(a^{L}_{,x}(x,k))^{2}}{a^{L}(x,k)}\,{\rm d}x\\\
&\leq\int_{-1}^{1}\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}},k\neq
0}\frac{1}{4k^{2}}\frac{(b^{L_{0}}_{,x}(\cdot,k))^{2}}{b^{L_{0}}(\cdot,k)}*\rho_{\varepsilon}(x)\,{\rm
d}x\\\
&\leq\int_{0}^{\bar{\lambda}}\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}},k\neq
0}\frac{1}{4k^{2}}\frac{(\bar{b}^{L_{0}}_{,x}(x,k))^{2}}{\bar{b}^{L_{0}}(x,k)}\,{\rm
d}x\,.\end{split}$ (5.44)

Gathering together (5.41)–(5.44) we obtain

$\fint_{-L}^{L}\int_{-1}^{1}\big{(}(\hat{u}^{L}_{,x})^{2}+(\hat{u}^{L}_{,yy})^{2}\big{)}\,{\rm
d}x\,{\rm
d}y\leq\biggl{(}1+\frac{C\varepsilon(e^{\frac{1-\bar{\lambda}}{\varepsilon}-e^{\frac{-1-\bar{\lambda}}{\varepsilon}}})}{1-\lambda}\biggr{)}\mathcal{F}^{\lambda}_{\infty}(\mu^{L_{0}})\,,$
(5.45)

and hence by letting $L\to+\infty$ and recalling Lemma 5.3 $(iii)$ we deduce
(5.26).

Step 7: in this step we show (5.27). By (3.3), (3.4)

$\fint_{-L}^{L}\int_{-1}^{1}(\hat{u}^{L}_{,xy})^{2}\,{\rm d}x\,{\rm
d}y=\fint_{-L_{0}}^{L_{0}}\int_{-1}^{1}(\hat{u}^{L}_{,xy})^{2}\,{\rm d}x\,{\rm
d}y=\int_{-1}^{1}\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}},k\neq
0}\frac{1}{4}\frac{(a^{L}_{,x}(x,k))^{2}}{a^{L}(x,k)}\,{\rm d}x\,.$ (5.46)

Next we observe that (5.31) yields

${(a^{L}_{,x}(x,k))^{2}}={(b^{L}(\cdot,k)*\dot{\rho}_{\varepsilon}(x))^{2}}\leq{(b^{L}(\cdot,k)*|\dot{\rho}_{\varepsilon}|(x))^{2}}=\frac{1}{\varepsilon^{2}}{(b^{L}(\cdot,k)*\rho_{\varepsilon}(x))^{2}}\leq\frac{1}{\varepsilon^{2}}(a^{L}(x,k))^{2}\,,$
(5.47)

so that combining (5.46) with (5.47) and recalling (5.32) we obtain

$\fint_{-L}^{L}\int_{-1}^{1}(\hat{u}^{L}_{,xy})^{2}\,{\rm d}x\,{\rm
d}y\leq\frac{1}{4\varepsilon^{2}}\int_{-1}^{1}\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}}}(a^{L}(x,k))^{2}\,{\rm
d}x\lesssim\frac{1}{\varepsilon^{2}}\,.$ (5.48)

In a similar way (3.3) and (3.4) give

$\begin{split}\fint_{-L}^{L}\int_{-1}^{1}(\hat{u}_{,xx}^{L})^{2}\,{\rm
d}x\,{\rm
d}y&=\fint_{-L_{0}}^{L_{0}}\int_{-1}^{1}(\hat{u}_{,xx}^{L})^{2}\,{\rm
d}x\,{\rm d}y=\int_{-1}^{1}\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}},k\neq
0}\frac{1}{k^{2}}\Big{[}\Big{(}\sqrt{a^{L}(x,k)}\Big{)}_{,xx}\Big{]}^{2}\,{\rm
d}x\\\ &\lesssim\int_{-1}^{1}\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}},k\neq
0}\frac{1}{k^{2}}\frac{(a^{L}_{,xx}(x,k))^{2}}{a^{L}(x,k)}\,{\rm
d}x+\int_{-1}^{1}\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}},k\neq
0}\frac{1}{k^{2}}\frac{(a^{L}_{,x}(x,k))^{4}}{(a^{L}(x,k))^{3}}\,{\rm d}x\\\
&\lesssim\frac{1}{\varepsilon^{2}}\int_{-1}^{1}\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}},k\neq
0}\frac{1}{4k^{2}}\frac{(a^{L}_{,x}(x,k))^{2}}{a^{L}(x,k)}\,{\rm d}x\\\
&\lesssim\frac{1}{\varepsilon^{2}}\int_{-1}^{1}\fint_{-L_{0}}^{L_{0}}(\hat{u}^{L}_{,x})^{2}\,{\rm
d}y\,{\rm
d}x\lesssim\frac{1}{\varepsilon^{2}}\mathcal{F}^{\lambda}_{\infty}(\mu^{L_{0}})\lesssim\frac{1}{\varepsilon^{2}}\,,\end{split}$
(5.49)

where the second inequality follows from

${(a^{L}_{,x}(x,k))^{4}}=(a^{L}_{,x}(x,k))^{2}{(b^{L}(\cdot,k)*\dot{\rho}_{\varepsilon}(x))^{2}}\leq\frac{1}{\varepsilon^{2}}(a^{L}_{,x}(x,k))^{2}(a^{L}(x,k))^{2}\,,$

and

$\begin{split}{(a^{L}_{,xx}(x,k))^{2}}={(b^{L}_{,x}(\cdot,k)*\dot{\rho}_{\varepsilon}(x))^{2}}\leq\frac{1}{\varepsilon^{2}}(a^{L}_{,x}(x,k))^{2}\,.\end{split}$

Moreover appealing again to (5.47) we find

$\begin{split}\fint_{-L}^{L}\int_{-1}^{1}(\hat{u}_{,yyx}^{L})^{2}\,{\rm
d}x\,{\rm
d}y&=\fint_{-L_{0}}^{L_{0}}\int_{-1}^{1}(\hat{u}_{,yyx}^{L})^{2}\,{\rm
d}x\,{\rm d}y=\int_{-1}^{1}\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}},k\neq
0}{k^{2}}\Big{[}\Big{(}\sqrt{a^{L}(x,k)}\Big{)}_{,x}\Big{]}^{2}\,{\rm d}x\\\
&=\int_{-1}^{1}\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}},k\neq
0}\frac{k^{2}}{4}\frac{(a^{L}_{,x}(x,k))^{2}}{a^{L}(x,k)}\,{\rm
d}x\lesssim\frac{1}{\varepsilon^{2}}\int_{-1}^{1}\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}},k\neq
0}{k^{2}}{a^{L}(x,k)}\,{\rm d}x\\\
&\lesssim\frac{1}{\varepsilon^{2}}\int_{-1}^{1}\fint_{-L_{0}}^{L_{0}}(\hat{u}^{L}_{,yy})^{2}\,{\rm
d}y\,{\rm
d}x\lesssim\frac{1}{\varepsilon^{2}}\mathcal{F}^{\lambda}_{\infty}(\mu^{L_{0}})\lesssim\frac{1}{\varepsilon^{2}}\,.\end{split}$
(5.50)

Eventually gathering together (5.48)–(5.50) we deduce (5.27).

Step 8: in this step we show (5.28). Analogously to step 3 we can find
$y_{0}\in[-L_{0},L_{0}]$ such that

$\hat{u}^{L}_{,x}(x,y_{0})=\fint_{-L_{0}}^{L_{0}}\hat{u}^{L}_{,x}(x,\hat{y})\,{\rm
d}\hat{y}=0\,,$

so that by the fundamental theorem of calculus, Hölder’s inequality it holds

$\begin{split}|\hat{u}_{,x}^{L}(x,y)|=\left|\int_{y_{0}}^{y}\hat{u}_{,xy}(x,y^{\prime})\,{\rm
d}y^{\prime}\right|&\leq\sqrt{L_{0}}\left(\int_{-L_{0}}^{L_{0}}(u^{L}_{,xy})^{2}\,{\rm
d}y^{\prime}\right)^{\frac{1}{2}}\\\
&=\sqrt{2}L_{0}\Biggl{(}\sum_{k\in\frac{\pi\mathbb{Z}}{L},k\neq
0}\frac{1}{4}\frac{(a^{L}_{,x}(x,k))^{2}}{a^{L}(x,k)}\Biggr{)}^{\frac{1}{2}}\\\
&\lesssim
L_{0}\Bigg{(}\frac{1}{4\varepsilon^{2}}\sum_{k\in\frac{\pi\mathbb{Z}}{L},k\neq
0}a^{L}(x,k)\Bigg{)}^{\frac{1}{2}}\lesssim\frac{L_{0}}{\varepsilon}\,,\end{split}$
(5.51)

where the last two inequalities follow from (5.47) and (5.32). Therefore
(5.51) gives

$\begin{split}\fint_{-L_{0}}^{L_{0}}\int_{-1}^{1}{(\hat{u}^{L}_{,x})^{4}}\,{\rm
d}x\,{\rm
d}y&\lesssim\frac{L_{0}^{2}}{\varepsilon^{2}}\fint_{-L_{0}}^{L_{0}}\int_{-1}^{1}{(\hat{u}^{L}_{,x})^{2}}\,{\rm
d}x\,{\rm d}y\lesssim\frac{L_{0}^{2}}{\varepsilon^{2}}\,,\end{split}$

and the proof is concluded.

∎

We are now in a position to prove Proposition 5.1.

###### Proof of Proposition 5.1.

Let $\mu\in\mathcal{M}_{\infty}$ be as in the statement. Let
$\varepsilon=\varepsilon(L)>0$ and $n=n(L)\in\mathbb{N}$ be such that

$\lim_{L\to+\infty}\varepsilon(L)=0\,,\quad\lim_{L\to+\infty}n(L)=\lim_{L\to+\infty}\frac{L}{n(L)}=+\infty\,,$
(5.52)

to be chosen later. Let $\hat{u}^{L}\in\mathcal{A}_{L}^{\rm
out}\cap\mathcal{A}_{L_{0}}^{\rm out}$ with $L_{0}:=L/n(L)$ be the function
given by Lemma 5.4. Recall that

$A^{L}(x)=\frac{1}{2}\fint_{-L}^{L}(\hat{u}^{L}_{,y})^{2}\,{\rm d}y\ \text{
and }\ f^{L}(x)=\sqrt{\frac{x}{A^{L}(x)}}\ \text{ for }\ x\geq 0\,.$

Furthermore we let $M=M(L)\in\mathbb{N}$, $M\geq 2$ to be chosen later such
that setting $\delta=\delta(L):=\frac{\varepsilon(L)}{M(L)}<\varepsilon(L)$ we
have

$\lim_{L\to+\infty}M(L)=+\infty\,,\quad\text{and}\quad\lim_{L\to+\infty}\delta(L)=\lim_{L\to+\infty}M^{2}(L)\delta(L)=0\,.$
(5.53)

We consider $\psi_{\delta}\in C^{\infty}(\mathbb{R})$ such that

$\psi_{\delta}\equiv 0\ \text{ in }(-\infty,\delta]\,,\quad\psi_{\delta}\equiv
1\ \text{ in }[2\delta,+\infty)\,,\quad|\dot{\psi}_{\delta}(x)|\leq
C\delta^{-1}\,,\quad|\ddot{\psi}_{\delta}(x)|\leq C\delta^{-2}\,.$

Note that $\dot{\psi}_{\delta}=\ddot{\psi}_{\delta}=0$ in
$(\delta,2\delta)^{c}$. We next define
$(w^{L},u^{L})=\big{(}(w^{L}_{1},w^{L}_{2}),u^{L}\big{)}$ as follows:

$\begin{split}&u^{L}(x,y):=\psi_{\delta}(x)f^{L}(x)\hat{u}^{L}(x,y)\,,\\\
&w_{2}^{L}(x,y):=\psi_{\delta}^{2}(x)xy+B^{L}(x)-\frac{1}{2}\int_{0}^{y}(u^{L}_{,y})^{2}\,{\rm
d}y^{\prime}\,,\\\
&w^{L}_{1}(x,y):=x-\frac{1}{L^{2}}\int_{0}^{y}(w^{L}_{2,x}+u^{L}_{,x}u^{L}_{,y})\,{\rm
d}y^{\prime}\,,\end{split}$

where

$B^{L}(x):=\frac{1}{2}\fint_{-L_{0}}^{L_{0}}\int_{0}^{y}(u^{L}_{,y})^{2}\,{\rm
d}y^{\prime}\,{\rm
d}y-\fint_{-L_{0}}^{L_{0}}\int_{0}^{x}u^{L}_{,x}u^{L}_{,y}\,{\rm
d}x^{\prime}\,{\rm d}y\,.$

Clearly $u^{L}\in\mathcal{A}_{L}^{\rm out}\cap\mathcal{A}_{L_{0}}^{\rm out}$.
We show that $w^{L}\in\mathcal{A}_{L}^{\rm in}\cap\mathcal{A}_{L_{0}}^{\rm
in}$. Precisely to see that $w^{L}(x,\cdot)$ is $2L_{0}$ periodic we use the
following fact:

A differentiable function $h$ is $T$-periodic if $h^{\prime}$ is $T$-periodic
and $h(t)=h(t+T)$ for some $t$.

The function $w^{L}_{2,y}(x,\cdot)$ is $2L_{0}$-periodic, since $u^{L}_{,y}$
is, and from (3.3) satisfies

$\begin{split}w_{2}^{L}(x,L_{0})-w_{2}^{L}(x,-L_{0})&=2L_{0}\psi_{\delta}^{2}(x)x-\int_{-L_{0}}^{L_{0}}(u^{L}_{,y})^{2}\,{\rm
d}y\\\ &=2L_{0}\psi_{\delta}^{2}(x)(x-(f^{L}(x))^{2}A^{L}(x))=0\,,\end{split}$

from which we deduce $w^{L}_{2}(x,\cdot)$ is $2L_{0}$-periodic. Using this
periodicity, and in particular also of $w^{L}_{2,x}$, we see that
$w^{L}_{1,y}(x,\cdot)$ is $2L_{0}$-periodic. Moreover we have

$\begin{split}w_{1}^{L}(x,L_{0})&-w_{1}^{L}(x,-L_{0})=-\frac{1}{L^{2}}\int_{-L_{0}}^{L_{0}}(w^{L}_{2,x}+u^{L}_{,x}u^{L}_{,y})\,{\rm
d}y\\\
&=-\frac{1}{L^{2}}\left(2L_{0}\dot{B}^{L}(x)-\int_{-L_{0}}^{L_{0}}\int_{0}^{y}u_{,y}^{L}u^{L}_{,xy}\,{\rm
d}y^{\prime}\,{\rm d}y+\int_{-L_{0}}^{L_{0}}u_{,x}^{L}u_{,y}^{L}\,{\rm
d}y\right)=0\,,\end{split}$

where the second and the third equalities follow from

$w^{L}_{2,x}=(\psi_{\delta}^{2}(x)x)^{\prime}y+\dot{B}^{L}(x)-\int_{0}^{y}u^{L}_{,y}u^{L}_{,xy}\,{\rm
d}y^{\prime}\,,$

and

$2L_{0}\dot{B}^{L}(x)=2L_{0}\left(\fint_{-L_{0}}^{L_{0}}u^{L}_{,y}u^{L}_{,xy}\,{\rm
d}y-\fint_{-L_{0}}^{L_{0}}u^{L}_{,x}u^{L}_{,y}\,{\rm d}y\right)\,.$

Thus we deduce that $w^{L}_{1,y}$ is $2L_{0}$-periodic. For the reader
convenience we divide the rest of the proof into several steps. We will
repeatedly use that the averaged integral over $(-L,L)$ of a $2L_{0}$-periodic
function is equal to the averaged integral over $(-L_{0},L_{0})$ of the same
function.

Step 1: we show that $(w^{L},u^{L})$ converges to $\mu$ in the sense of
Definition 2.3. We have that

$\hat{u}^{L}(x,y)=\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}},k>0}a^{L}_{k}(x)\sin(ky)+\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}},k<0}a^{L}_{k}(x)\cos(ky)\,,$

so that

$u^{L}(x,y)=\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}},k>0}\psi_{\delta}(x)f^{L}(x)a^{L}_{k}(x)\sin(ky)+\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}},k<0}\psi_{\delta}(x)f^{L}(x)a^{L}_{k}(x)\cos(ky)\,.$

Therefore we get

$\begin{split}\mu^{L}:=\mu^{L}(u^{L})&=\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}}}\psi^{2}_{\delta}(x)(f^{L}(x))^{2}(a^{L}_{k}(x))^{2}k^{2}\mathcal{L}^{1}\mathop{\hbox{\vrule
height=7.0pt,width=0.5pt,depth=0.0pt\vrule
height=0.5pt,width=6.0pt,depth=0.0pt}}\nolimits(-1,1)\times\delta_{k}\\\
&=\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}}}\psi_{\delta}^{2}(x)(f^{L}(x))^{2}(a^{L}_{k}(x))^{2}k^{2}\mathcal{L}^{1}\mathop{\hbox{\vrule
height=7.0pt,width=0.5pt,depth=0.0pt\vrule
height=0.5pt,width=6.0pt,depth=0.0pt}}\nolimits(\delta,1)\times\delta_{k}\,.\end{split}$

We show $\mu^{L}\stackrel{{\scriptstyle*}}{{\rightharpoonup}}\mu$. We fix
$\varphi\in C_{c}^{\infty}((-1,1)\times\mathbb{R})$ and we write

$\int_{(-1,1)\times\mathbb{R}}\varphi\,{\rm
d}\mu^{L}=\left(\int_{(-1,1)\times\mathbb{R}}\varphi\,{\rm
d}\mu^{L}-\int_{(-1,1)\times\mathbb{R}}\varphi\,{\rm
d}\mu^{L}(\hat{u}^{L})\right)+\int_{(-1,1)\times\mathbb{R}}\varphi\,{\rm
d}\mu^{L}(\hat{u}^{L})\,.$

By Lemma 5.4 it holds

$\lim_{L\to+\infty}\int_{(-1,1)\times\mathbb{R}}\varphi\,{\rm
d}\mu^{L}(\hat{u}^{L})=\int_{(-1,1)\times\mathbb{R}}\varphi\,{\rm d}\mu\,.$

Therefore it is sufficient to show that

$\lim_{L\to+\infty}\left(\int_{(-1,1)\times\mathbb{R}}\varphi\,{\rm
d}\mu^{L}-\int_{(-1,1)\times\mathbb{R}}\varphi\,{\rm
d}\mu^{L}(\hat{u}^{L})\right)=0\,.$

Recalling that $\psi_{\delta}=0$ in $(-1,\delta)$, $\psi_{\delta}=1$ in
$(2\delta,1)$ and $M\varepsilon=M\delta^{2}\geq 2\delta$, we have

$\begin{split}\bigg{|}\int_{(-1,1)\times\mathbb{R}}\varphi\,{\rm
d}\mu^{L}-\int_{(0,1)\times\mathbb{R}}\varphi\,{\rm
d}\mu^{L}(\hat{u}^{L})\bigg{|}&=\bigg{|}\int_{0}^{1}\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}}}\varphi(x,k)(a^{L}_{k}(x))^{2}k^{2}\Big{(}\psi^{2}_{\delta}(x)(f^{L}(x))^{2}-1\Big{)}\,{\rm
d}x\bigg{|}\\\
&\leq\|\varphi\|_{\infty}\bigg{|}\int_{0}^{M\varepsilon}\Big{(}\psi^{2}_{\delta}(x)(f^{L}(x))^{2}-1\Big{)}\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}}}(a^{L}_{k}(x))^{2}k^{2}\,{\rm
d}x\bigg{|}\\\
&+\|\varphi\|_{\infty}\bigg{|}\int_{M\varepsilon}^{1}\Big{(}(f^{L}(x))^{2}-1\Big{)}\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}}}(a^{L}_{k}(x))^{2}k^{2}\,{\rm
d}x\bigg{|}\,.\end{split}$

Since

$\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}}}(a^{L}_{k}(x))^{2}k^{2}=A^{L}(x)=\frac{1}{2}\fint_{-L}^{L}(\hat{u}_{,y}(x,\cdot))^{2}\,{\rm
d}y\,,$

by (5.18) and (5.19) we have

$\Big{(}\psi^{2}_{\delta}(x)(f^{L}(x))^{2}-1\Big{)}\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}}}(a^{L}_{k}(x))^{2}k^{2}\leq\Big{(}\psi^{2}_{\delta}(x)(1+o_{L}(1))-1\Big{)}\max\\{x,\varepsilon\\}$

which together with (5.53) imply

$\|\varphi\|_{\infty}\bigg{|}\int_{0}^{M\varepsilon}\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}}}(a^{L}_{k}(x))^{2}k^{2}(\psi^{2}_{\delta}(x)(f^{L}(x))^{2}-1)\,{\rm
d}x\bigg{|}\leq CM\varepsilon=CM^{2}\delta\to 0\quad\text{ as }L\to+\infty\,.$

Whereas (5.19), (5.20) with $N=M$ and the fact that $M=M(L)\to+\infty$ imply

$(f^{L}(x))^{2}=1+o_{L}(1)\quad\text{ in }\quad[M\varepsilon,1)\,,$

so that

$\|\varphi\|_{\infty}\bigg{|}\int_{M\varepsilon}^{1}\sum_{k\in\frac{\pi\mathbb{Z}}{L_{0}}}a^{L}_{k}(x)(f^{L}(x)-1)\,{\rm
d}x\bigg{|}\leq o_{L}(1)\to 0\quad\text{ as }L\to+\infty\,.$

Eventually by duality we have

$\int_{\mathbb{R}\times(-1,1)}\varphi\,{\rm
d}\mu^{L}_{,x}=-\int_{\mathbb{R}\times(-1,1)}\varphi_{,x}\,{\rm
d}\mu^{L}\to-\int_{\mathbb{R}\times(-1,1)}\varphi_{,x}\,{\rm
d}\mu=\int_{\mathbb{R}\times(-1,1)}\varphi\,{\rm d}\mu_{,x}\,,$

which in turn implies
$\mu^{L}_{,x}\stackrel{{\scriptstyle*}}{{\rightharpoonup}}\mu_{,x}$.

Step 2: we show that

$\limsup_{L\to+\infty}\fint_{-L}^{L}\int_{-1}^{1}((u^{L}_{,x})^{2}+(u^{L}_{,yy})^{2})\,{\rm
d}x\,{\rm
d}y\leq\mathcal{F}_{\infty}(\mu)+C\lim_{L\to+\infty}\omega(2M^{2}\delta)\log
M\,.$ (5.54)

To this purpose we note that

$u^{L}_{,x}(x,y)=\psi_{\delta}(x)f^{L}(x)\hat{u}^{L}_{,x}(x,y)+\dot{\psi}_{\delta}(x)f^{L}(x)\hat{u}^{L}(x,y)+\psi_{\delta}(x)\dot{f}^{L}(x)\hat{u}^{L}(x,y)\,.$
(5.55)

Therefore by Young’s inequality

$\begin{split}\fint_{-L}^{L}\int_{-1}^{1}(u^{L}_{,x})^{2}\,{\rm d}x\,{\rm
d}y&\leq(1+\alpha)\fint_{-L}^{L}\int_{\delta}^{1}(f^{L}(x))^{2}(\hat{u}^{L}_{,x})^{2}\,{\rm
d}x\,{\rm d}y\\\
&+2(1+\alpha^{-1})\frac{1}{\delta^{2}}\fint_{-L}^{L}\int_{\delta}^{2\delta}(f^{L}(x))^{2}(\hat{u}^{L})^{2}\,{\rm
d}x\,{\rm d}y\\\
&+2(1+\alpha^{-1})\fint_{-L}^{L}\int_{\delta}^{1}(\dot{f}^{L}(x))^{2}(\hat{u}^{L})^{2}\,{\rm
d}x\,{\rm d}y\,,\end{split}$ (5.56)

for any $\alpha>0$. In this way by recalling (5.19) (5.22) and (5.21) we have

$\fint_{-L}^{L}\int_{-1}^{1}(f^{L}(x))^{2}(\hat{u}^{L}_{,x})^{2}\,{\rm
d}x\,{\rm
d}y\leq(1+o_{L}(1))\fint_{-L}^{L}\int_{-1}^{1}(\hat{u}^{L}_{,x})^{2}\,{\rm
d}x\,{\rm d}y\,,$ (5.57)
$\frac{1}{\delta^{2}}\fint_{-L}^{L}\int_{\delta}^{2\delta}(f^{L}(x))^{2}(\hat{u}^{L})^{2}\,{\rm
d}x\,{\rm
d}y\lesssim\frac{1}{\delta^{2}}{\delta}(\omega(2M\varepsilon)+Me^{-M})\int_{\delta}^{2\delta}\,{\rm
d}x\lesssim(\omega(2M\varepsilon)+Me^{-M})\,,$ (5.58)

and

$\begin{split}\fint_{-L}^{L}\int_{\delta}^{1}(\dot{f}^{L}(x))^{2}(\hat{u}^{L})^{2}\,{\rm
d}x\,{\rm
d}y&\lesssim\int_{\delta}^{\varepsilon}\frac{1}{x\varepsilon}\varepsilon(\omega(2M\varepsilon)+Me^{-M})\,{\rm
d}x\\\
&+\int_{\varepsilon}^{\sqrt{\varepsilon}}\frac{e^{-\frac{x}{\varepsilon}}}{x\varepsilon}x(\omega(2M\varepsilon)+Me^{-M})\,{\rm
d}x+\int_{\sqrt{\varepsilon}}^{1}\frac{e^{-\frac{1}{\sqrt{\varepsilon}}}}{x\varepsilon}x\,{\rm
d}x\\\
&\lesssim(\log\frac{\varepsilon}{\delta}+1)(\omega(2M\varepsilon)+Me^{-M})+\frac{e^{-\frac{1}{\sqrt{\varepsilon}}}}{\varepsilon}\,.\end{split}$
(5.59)

Gathering together (5.56)–(5.59) and recalling that $\delta=\varepsilon/M$ we
deduce

$\fint_{-L}^{L}\int_{-1}^{1}(u^{L}_{,x})^{2}\,{\rm d}x\,{\rm
d}y\leq(1+\bar{\alpha})\fint_{-L}^{L}\int_{-1}^{1}(\hat{u}^{L}_{,x})^{2}\,{\rm
d}x\,{\rm d}y+C(\log M+1)(\omega(2M^{2}\delta)+Me^{-M})+o_{L}(1)\,,$ (5.60)

with $\bar{\alpha}:=\alpha+\alpha o_{L}(1)+o_{L}(1)$. Moreover as
$u^{L}_{,yy}=\psi_{\delta}(x)f^{L}(x)\hat{u}^{L}_{,yy}$ we get

$\fint_{-L}^{L}\int_{-1}^{1}(u^{L}_{,yy})^{2}\,{\rm d}x\,{\rm
d}y\leq\fint_{-L}^{L}\int_{-1}^{1}(f^{L}(x))^{2}(\hat{u}^{L}_{,x})^{2}\,{\rm
d}x\,{\rm d}y\leq\fint_{-L}^{L}\int_{-1}^{1}(\hat{u}^{L}_{,x})^{2}\,{\rm
d}x\,{\rm d}y\,.$ (5.61)

By (5.60), (5.61), (5.26) and the fact that $M^{2}\delta\to 0$ we finally
deduce

$\limsup_{L\to+\infty}\fint_{-L}^{L}\int_{-1}^{1}((u^{L}_{,x})^{2}+(u^{L}_{,yy})^{2})\,{\rm
d}x\,{\rm
d}y\leq(1+\alpha)\mathcal{F}_{\infty}(\mu)+C\lim_{L\to+\infty}\omega(2M^{2}\delta)\log
M\,.$

Eventually by the arbitrariness of $\alpha$ we infer the desired estimate.

Step 3: we show that

$L^{2}\fint_{-L}^{L}\int_{-1}^{1}\biggl{(}w^{L}_{1,x}+\frac{(u^{L}_{,x})^{2}}{2L^{2}}-1\biggr{)}^{2}\,{\rm
d}x\,{\rm
d}y\lesssim\frac{L_{0}^{4}}{L^{2}}\frac{1}{\delta^{2}\varepsilon}\lesssim\frac{L_{0}^{4}}{L^{2}}\frac{1}{\delta^{3}M}\,.$
(5.62)

By Young’s inequality we have

$\begin{split}L^{2}\fint_{-L}^{L}\int_{-1}^{1}\biggl{(}w^{L}_{1,x}+\frac{(u^{L}_{,x})^{2}}{2L^{2}}-1\biggr{)}^{2}\,{\rm
d}x\,{\rm
d}y\lesssim\fint_{-L}^{L}\int_{-1}^{1}\frac{(u^{L}_{,x})^{4}}{L^{2}}\,{\rm
d}x\,{\rm
d}y+L^{2}\fint_{-L}^{L}\int_{-1}^{1}\bigl{(}w^{L}_{1,x}-1\bigr{)}^{2}\,{\rm
d}x\,{\rm d}y\,.\end{split}$ (5.63)

We estimate the first term on the right hand-side of (5.63). By (5.55) it
follows

$\begin{split}\fint_{-L}^{L}\int_{-1}^{1}&{(u^{L}_{,x})^{4}}\,{\rm d}x\,{\rm
d}y\lesssim\fint_{-L}^{L}\int_{-1}^{1}(f^{L}(x))^{4}{(\hat{u}^{L}_{,x})^{4}}\,{\rm
d}x\,{\rm d}y\\\
&+\frac{1}{\delta^{4}}\fint_{-L}^{L}\int_{\delta}^{2\delta}(f^{L}(x))^{4}{(\hat{u}^{L})^{4}}\,{\rm
d}x\,{\rm
d}y+\fint_{-L}^{L}\int_{\delta}^{1}(\dot{f}^{L}(x))^{4}{(\hat{u}^{L})^{4}}\,{\rm
d}x\,{\rm d}y\,.\end{split}$ (5.64)

By (5.19) and (5.28) we have

$\fint_{-L}^{L}\int_{-1}^{1}(f^{L}(x))^{4}{(\hat{u}^{L}_{,x})^{4}}\,{\rm
d}x\,{\rm d}y\lesssim\frac{L_{0}^{2}}{\varepsilon^{2}}\,,$ (5.65)

whereas from (5.19), (5.23), and the fact that $x\in(\delta,2\delta)$ we get

$\begin{split}\frac{1}{\delta^{4}}\fint_{-L}^{L}\int_{\delta}^{2\delta}(f^{L}(x))^{4}{(\hat{u}^{L})^{4}}\,{\rm
d}x\,{\rm
d}y&\lesssim\frac{1}{\delta^{4}}\frac{\delta^{2}}{\varepsilon^{2}}L_{0}^{2}\int_{\delta}^{2\delta}(\max\\{x,\varepsilon\\})^{2}\,{\rm
d}x\lesssim\frac{L_{0}^{2}}{\delta}\,.\end{split}$ (5.66)

Finally by (5.21) and (5.23)

$\begin{split}\fint_{-L}^{L}\int_{\delta}^{1}(\dot{f}^{L}(x))^{4}{(\hat{u}^{L})^{4}}&\,{\rm
d}x\,{\rm d}y\lesssim
L_{0}^{2}\int_{\delta}^{1}\frac{1}{x^{2}\varepsilon^{2}}(\max\\{x,\varepsilon\\})^{2}\,{\rm
d}x\lesssim
L_{0}^{2}\left(\frac{1}{\delta}+\frac{1}{\varepsilon^{2}}\right)\,.\end{split}$
(5.67)

Thus gathering together (5.64)–(5.67) we infer

$\fint_{-L}^{L}\int_{-1}^{1}\frac{(u^{L}_{,x})^{4}}{L^{2}}\,{\rm d}x\,{\rm
d}y\lesssim\frac{L_{0}^{2}}{L^{2}}\left(\frac{1}{\delta}+\frac{1}{\varepsilon^{2}}\right)\,.$
(5.68)

We now pass to estimate the second term on the right hand side of (5.63). To
this aim we observe that integrating by parts it holds

$\begin{split}w_{2,x}^{L}+u^{L}_{,x}u^{L}_{,y}&=(x\psi_{\delta}^{2}(x))^{\prime}y+\dot{B}^{L}(x)-\int_{0}^{y}u^{L}_{,y}u^{L}_{,xy}\,{\rm
d}y^{\prime}+u^{L}_{,x}u^{L}_{,y}(x,y)\\\
&=(x\psi_{\delta}^{2}(x))^{\prime}y+\dot{B}^{L}(x)+\int_{0}^{y}u_{,yy}u_{,x}\,{\rm
d}y^{\prime}+u^{L}_{,x}u^{L}_{,y}|_{y=0}\\\
&=(x\psi_{\delta}^{2}(x))^{\prime}y+\int_{0}^{y}u_{,yy}u_{,x}\,{\rm
d}y^{\prime}+C^{L}(x)\,,\end{split}$ (5.69)

where

$C^{L}(x):=-\fint_{-L}^{L}\int_{0}^{y}u_{,yy}u_{,x}\,{\rm d}y^{\prime}\,{\rm
d}y=-\fint_{-L_{0}}^{L_{0}}\int_{0}^{y}u_{,yy}u_{,x}\,{\rm d}y^{\prime}\,{\rm
d}y\,,$

and the last equality is a consequence of the following identity

$\dot{B}^{L}(x)+u^{L}_{,x}u^{L}_{,y}|_{y=0}=\fint_{-L}^{L}\biggl{(}\int_{0}^{y}u^{L}_{,y}u^{L}_{,xy}\,{\rm
d}y^{\prime}-u^{L}_{,x}u^{L}_{,y}+u^{L}_{,x}u^{L}_{,y}|_{y=0}\biggr{)}\,{\rm
d}y=C^{L}(x)\,.$

Using (5.69) and the definition of $\omega$ we get

$\begin{split}w^{L}_{1,x}-1&=-\frac{1}{L^{2}}\int_{0}^{y}\left((x\psi_{\delta}^{2}(x))^{\prime\prime}y^{\prime}+\int_{0}^{y^{\prime}}(u^{L}_{,yy}u^{L}_{,xx}+u^{L}_{,yyx}u^{L}_{,x})\,{\rm
d}y^{\prime\prime}+\dot{C}^{L}(x)\right)\,{\rm d}y^{\prime}\\\
&=-\frac{1}{L^{2}}\left((x\psi_{\delta}^{2}(x))^{\prime\prime}\frac{y^{2}}{2}+\int_{0}^{y}\int_{0}^{y^{\prime}}(u^{L}_{,yy}u^{L}_{,xx}+u^{L}_{,yyx}u^{L}_{,x})\,{\rm
d}y^{\prime\prime}\,{\rm d}y^{\prime}+\dot{C}^{L}(x)y\right)\,.\end{split}$

This together with Young’s inequality give

$\begin{split}L^{2}\fint_{-L}^{L}\int_{-1}^{1}\bigl{(}w^{L}_{1,x}-1\bigr{)}^{2}\,{\rm
d}x\,{\rm
d}y&=L^{2}\fint_{-L_{0}}^{L_{0}}\int_{-1}^{1}\bigl{(}w^{L}_{1,x}-1\bigr{)}^{2}\,{\rm
d}x\,{\rm d}y\\\
&\lesssim\frac{1}{L^{2}}\fint_{-L_{0}}^{L_{0}}\int_{-1}^{1}((x\psi_{\delta}^{2}(x))^{\prime\prime})^{2}y^{4}\,{\rm
d}x\,{\rm d}y\\\
&+\frac{1}{L^{2}}\fint_{-L_{0}}^{L_{0}}\int_{-1}^{1}\biggl{(}\int_{0}^{y}\int_{0}^{y^{\prime}}(u^{L}_{,yy}u^{L}_{,xx}+u^{L}_{,yyx}u^{L}_{,x})\,{\rm
d}y^{\prime\prime}\,{\rm d}y^{\prime}\biggr{)}^{2}\,{\rm d}x\,{\rm d}y\\\
&+\frac{1}{L^{2}}\fint_{-L_{0}}^{L_{0}}\int_{-1}^{1}(\dot{C}^{L}(x))^{2}y^{2}\,{\rm
d}x\,{\rm d}y\,.\end{split}$ (5.70)

As
$(x\psi_{\delta}^{2}(x))^{\prime\prime}=4\psi_{\delta}(x)\dot{\psi}_{\delta}(x)+2x\psi_{\delta}(x)\ddot{\psi}_{\delta}(x)+2x(\dot{\psi}_{\delta}(x))^{2}$
in $(\delta,2\delta)$ and $(x\psi_{\delta}^{2}(x))^{\prime\prime}=0$
otherwise, we have

$\begin{split}\frac{1}{L^{2}}\fint_{-L_{0}}^{L_{0}}\int_{-1}^{1}((x\psi_{\delta}^{2}(x))^{\prime\prime})^{2}y^{4}\,{\rm
d}x\,{\rm
d}y\lesssim\frac{L_{0}^{4}}{L^{2}}\int_{\delta}^{2\delta}((x\psi_{\delta}^{2}(x))^{\prime\prime})^{2}\,{\rm
d}x\lesssim\frac{L_{0}^{4}}{L^{2}}\frac{1}{\delta}\,.\end{split}$ (5.71)

We now estimate the second term on the right hand-side of (5.70). We first
observe that if $a,b,c,d$ are $2L_{0}$-periodic then by applying in order
Hölder, Young and Jensen inequalities we have

$\begin{split}\bigg{[}\int_{0}^{y}\int_{0}^{y^{\prime}}&(ab+cd)\,{\rm
d}y^{\prime\prime}\,{\rm
d}y^{\prime}\bigg{]}^{2}\leq\left[\int_{0}^{y}\|a\|_{L^{2}(0,y^{\prime})}\|b\|_{L^{2}(0,y^{\prime})}+\|c\|_{L^{2}(0,y^{\prime})}\|d\|_{L^{2}(0,y^{\prime})}\,{\rm
d}y^{\prime}\right]^{2}\\\
&\lesssim\left[\int_{0}^{y}\|a\|_{L^{2}(0,y^{\prime})}\|b\|_{L^{2}(0,y^{\prime})}\,{\rm
d}y^{\prime}\right]^{2}+\left[\int_{0}^{y}\|c\|_{L^{2}(0,y^{\prime})}\|d\|_{L^{2}(0,y^{\prime})}\,{\rm
d}y^{\prime}\right]^{2}\\\ &\lesssim
L_{0}^{2}\fint_{-L_{0}}^{L_{0}}\|a\|^{2}_{L^{2}(0,y^{\prime})}\|b\|^{2}_{L^{2}(0,y^{\prime})}\,{\rm
d}y^{\prime}+L_{0}^{2}\fint_{-L_{0}}^{L_{0}}\|c\|^{2}_{L^{2}(0,y^{\prime})}\|d\|^{2}_{L^{2}(0,y^{\prime})}\,{\rm
d}y^{\prime}\\\ &\lesssim
L_{0}^{4}\left[\bigg{(}\fint_{-L_{0}}^{L_{0}}a^{2}\,{\rm
d}y^{\prime\prime}\bigg{)}\bigg{(}\fint_{-L_{0}}^{L_{0}}b^{2}\,{\rm
d}y^{\prime\prime}\bigg{)}+\bigg{(}\fint_{-L_{0}}^{L_{0}}c^{2}\,{\rm
d}y^{\prime\prime}\bigg{)}\bigg{(}\fint_{-L_{0}}^{L_{0}}d^{2}\,{\rm
d}y^{\prime\prime}\bigg{)}\right]\,.\end{split}$

Therefore it follows that

$\begin{split}\frac{1}{L^{2}}\fint_{-L_{0}}^{L_{0}}\int_{-1}^{1}\biggl{(}\int_{0}^{y}&\int_{0}^{y^{\prime}}(u^{L}_{,yy}u^{L}_{,xx}+u^{L}_{,yyx}u^{L}_{,x})\,{\rm
d}y^{\prime\prime}\,{\rm d}y^{\prime}\biggr{)}^{2}\,{\rm d}x\,{\rm d}y\\\
&=\frac{1}{L^{2}}\fint_{-L_{0}}^{L_{0}}\int_{\delta}^{1}\biggl{(}\int_{0}^{y}\int_{0}^{y^{\prime}}(u^{L}_{,yy}u^{L}_{,xx}+u^{L}_{,yyx}u^{L}_{,x})\,{\rm
d}y^{\prime\prime}\,{\rm d}y^{\prime}\biggr{)}^{2}\,{\rm d}x\,{\rm d}y\\\
&\lesssim\frac{L_{0}^{4}}{L^{2}}\int_{\delta}^{1}\bigg{(}\fint_{-L_{0}}^{L_{0}}(u^{L}_{,yy})^{2}\,{\rm
d}y^{\prime\prime}\bigg{)}\bigg{(}\fint_{-L_{0}}^{L_{0}}(u^{L}_{,xx})^{2}\,{\rm
d}y^{\prime\prime}\bigg{)}\,{\rm d}x\\\
&+\frac{L_{0}^{4}}{L^{2}}\int_{\delta}^{1}\bigg{(}\fint_{-L_{0}}^{L_{0}}(u^{L}_{,yyx})^{2}\,{\rm
d}y^{\prime\prime}\bigg{)}\bigg{(}\fint_{-L_{0}}^{L_{0}}(u^{L}_{,x})^{2}\,{\rm
d}y^{\prime\prime}\bigg{)}\,{\rm d}x\,.\end{split}$ (5.72)

Next we show separately that:

$\fint_{-L_{0}}^{L_{0}}(u^{L}_{,yy})^{2}\,{\rm
d}y^{\prime\prime}\lesssim\frac{1}{\varepsilon}\,,\qquad\qquad\fint_{-L_{0}}^{L_{0}}(u^{L}_{,x})^{2}\,{\rm
d}y^{\prime\prime}\lesssim\frac{1}{x\varepsilon}\max\\{x,\varepsilon\\}\,,$
(5.73) $\int_{\delta}^{1}\fint_{-L_{0}}^{L_{0}}(u^{L}_{,xx})^{2}\,{\rm
d}y^{\prime\prime}\,{\rm
d}x\lesssim\frac{1}{\delta^{2}}\,,\qquad\qquad\int_{\delta}^{1}\fint_{-L_{0}}^{L_{0}}(u^{L}_{,yyx})^{2}\,{\rm
d}y^{\prime\prime}\,{\rm d}x\lesssim\frac{1}{\delta\varepsilon}\,.$ (5.74)

Since $u^{L}_{,yy}=\psi_{\delta}(x)f^{L}(x)\hat{u}^{L}_{,yy}$ by (5.19) and
(5.24) we have

$\fint_{-L_{0}}^{L_{0}}(\hat{u}_{,yy}^{L})^{2}\,{\rm
d}y\lesssim\frac{1}{\varepsilon}\,.$

By (5.55), (5.19), (5.24), (5.22) and (5.21) we have

$\begin{split}\fint_{-L_{0}}^{L_{0}}(u^{L}_{,x})^{2}\,{\rm
d}y\lesssim\fint_{-L_{0}}^{L_{0}}(f^{L}(x))^{2}(\hat{u}^{L}_{,x})^{2}\,{\rm
d}y&+\frac{1}{\delta^{2}}\fint_{-L_{0}}^{L_{0}}\chi_{(\delta,2\delta)}(f^{L}(x))^{2}(\hat{u}^{L})^{2}\,{\rm
d}y+\fint_{-L_{0}}^{L_{0}}(\dot{f}^{L}(x))^{2}(\hat{u}^{L})^{2}\,{\rm d}y\\\
&\lesssim\frac{1}{\varepsilon}+\frac{1}{\delta^{2}}\frac{\delta}{\varepsilon}\max\\{x,\varepsilon\\}\chi_{(\delta,2\delta)}+\frac{1}{x\varepsilon}\max\\{x,\varepsilon\\}\\\
&\lesssim\frac{1}{\varepsilon}+\frac{1}{\delta}\chi_{(\delta,2\delta)}+\frac{1}{x\varepsilon}\max\\{x,\varepsilon\\}\lesssim\frac{1}{x\varepsilon}\max\\{x,\varepsilon\\}\,.\end{split}$

From (5.55) it follows

$\begin{split}u^{L}_{,xx}(x,y)&=\psi_{\delta}(x)f^{L}(x)\hat{u}^{L}_{,xx}(x,y)+\psi_{\delta}(x)\ddot{f}^{L}(x)\hat{u}^{L}(x,y)+\ddot{\psi}_{\delta}(x)f^{L}(x)\hat{u}^{L}(x,y)\\\
&+2\dot{\psi}_{\delta}(x)f^{L}(x)\hat{u}^{L}_{,x}(x,y)+2\dot{\psi}_{\delta}(x)\dot{f}^{L}(x)\hat{u}^{L}(x,y)+2\psi_{\delta}(x)\dot{f}^{L}(x)\hat{u}^{L}_{,x}(x,y)\,.\end{split}$

Hence by (5.19), (5.27), (5.21), (5.22) and (5.26) we have

$\begin{split}\int_{\delta}^{1}\fint_{-L_{0}}^{L_{0}}(u_{,xx}^{L})^{2}\,{\rm
d}y\,{\rm
d}x&\lesssim\int_{\delta}^{1}\fint_{-L_{0}}^{L_{0}}(f^{L}(x))^{2}(\hat{u}_{,xx}^{L})^{2}\,{\rm
d}y\,{\rm
d}x+\int_{\delta}^{1}\fint_{-L_{0}}^{L_{0}}(\ddot{f}^{L}(x))^{2}(\hat{u}^{L})^{2}\,{\rm
d}y\,{\rm d}x\\\
&+\int_{\delta}^{1}\fint_{-L_{0}}^{L_{0}}(\dot{f}^{L}(x))^{2}(\hat{u}_{,x}^{L})^{2}\,{\rm
d}y\,{\rm
d}x+\frac{1}{\delta^{4}}\int_{\delta}^{2\delta}\fint_{-L_{0}}^{L_{0}}(f^{L}(x))^{2}(\hat{u}^{L})^{2}\,{\rm
d}y\,{\rm d}x\\\
&+\frac{1}{\delta^{2}}\int_{\delta}^{2\delta}\fint_{-L_{0}}^{L_{0}}(\dot{f}^{L}(x))^{2}(\hat{u}^{L})^{2}\,{\rm
d}y\,{\rm
d}x+\frac{1}{\delta^{2}}\int_{\delta}^{2\delta}\fint_{-L_{0}}^{L_{0}}(f^{L}(x))^{2}(\hat{u}_{,x}^{L})^{2}\,{\rm
d}y\,{\rm d}x\\\
&\lesssim\frac{1}{\varepsilon^{2}}+\int_{\delta}^{1}\frac{1}{x^{3}\varepsilon}\max\\{x,\varepsilon\\}\,{\rm
d}x+\frac{1}{\delta\varepsilon}+\frac{1}{\delta^{4}}\int_{\delta}^{2\delta}\frac{x}{\varepsilon}\max\\{x,\varepsilon\\}\,{\rm
d}x\\\
&+\frac{1}{\delta^{2}}\int_{\delta}^{2\delta}\frac{1}{x\varepsilon}\max\\{x,\varepsilon\\}\,{\rm
d}x+\frac{1}{\delta^{2}}\lesssim\frac{1}{\varepsilon^{2}}+\frac{1}{\delta^{2}}+\frac{1}{\delta\varepsilon}\lesssim\frac{1}{\delta^{2}}\,.\end{split}$
(5.75)

Analogously by (5.55) it follows

$u^{L}_{,yyx}(x,y)=\psi_{\delta}(x)f^{L}(x)\hat{u}^{L}_{,yyx}(x,y)+\dot{\psi}_{\delta}(x)f^{L}(x)\hat{u}^{L}_{,yy}(x,y)+\psi_{\delta}(x)\dot{f}^{L}(x)\hat{u}^{L}_{,yy}(x,y)\,,$

from which together with (5.27), (5.19), (5.24) and (5.21)

$\begin{split}\int_{\delta}^{1}\fint_{-L_{0}}^{L_{0}}(u_{,yyx}^{L})^{2}\,{\rm
d}y\,{\rm
d}x&\lesssim\int_{\delta}^{1}\fint_{-L_{0}}^{L_{0}}(f^{L}(x))^{2}(\hat{u}_{,yyx}^{L})^{2}\,{\rm
d}y\,{\rm
d}x+\frac{1}{\delta^{2}}\int_{\delta}^{2\delta}\fint_{-L_{0}}^{L_{0}}(f^{L}(x))^{2}(\hat{u}_{,yy}^{L})^{2}\,{\rm
d}y\,{\rm d}x\\\
&+\int_{\delta}^{1}\fint_{-L_{0}}^{L_{0}}(\dot{f}^{L}(x))^{2}(\hat{u}_{,yy}^{L})^{2}\,{\rm
d}y\,{\rm
d}x\lesssim\frac{1}{\varepsilon^{2}}+\frac{1}{\delta^{2}}\frac{\delta}{\varepsilon}\frac{\delta}{\varepsilon}+\frac{1}{\delta\varepsilon}\lesssim\frac{1}{\delta\varepsilon}\,.\end{split}$

Now (5.73) and (5.74) yield

$\begin{split}\frac{L_{0}^{4}}{L^{2}}\int_{\delta}^{1}\bigg{(}\fint_{-L_{0}}^{L_{0}}(u^{L}_{,yy})^{2}\,{\rm
d}y^{\prime\prime}\bigg{)}\bigg{(}\fint_{-L_{0}}^{L_{0}}(u^{L}_{,xx})^{2}\,{\rm
d}y^{\prime\prime}\bigg{)}\,{\rm
d}x\lesssim\frac{L_{0}^{4}}{L^{2}}\frac{1}{\delta^{2}\varepsilon}\,,\end{split}$
(5.76)

and

$\begin{split}\frac{L_{0}^{4}}{L^{2}}\int_{\delta}^{1}\bigg{(}\fint_{-L_{0}}^{L_{0}}(u^{L}_{,yyx})^{2}\,{\rm
d}y^{\prime\prime}\bigg{)}\bigg{(}\fint_{-L_{0}}^{L_{0}}(u^{L}_{,x})^{2}\,{\rm
d}y^{\prime\prime}\bigg{)}\,{\rm
d}x\lesssim\frac{L_{0}^{4}}{L^{2}}\left(\frac{1}{\delta}+\frac{1}{\varepsilon}\right)\frac{1}{\delta\varepsilon}\lesssim\frac{L_{0}^{4}}{L^{2}}\frac{1}{\delta^{2}\varepsilon}\,.\end{split}$
(5.77)

Gathering together (5.72), (5.76) and (5.77) we find

$\frac{1}{L^{2}}\fint_{-L_{0}}^{L_{0}}\int_{-1}^{1}\biggl{(}\int_{0}^{y}\int_{0}^{y^{\prime}}(u^{L}_{,yy}u^{L}_{,xx}+u^{L}_{,yyx}u^{L}_{,x})\,{\rm
d}y^{\prime\prime}\,{\rm d}y^{\prime}\biggr{)}^{2}\,{\rm d}x\,{\rm
d}y\lesssim\frac{L_{0}^{4}}{L^{2}}\frac{1}{\delta^{2}\varepsilon}\,.$ (5.78)

It remains to estimate the third term on the right hand-side of (5.70). In a
similar way, see in particular (5.76) and (5.77), we have

$\begin{split}\frac{1}{L^{2}}\fint_{-L_{0}}^{L_{0}}\int_{-1}^{1}(\dot{C}^{L}(x))^{2}y^{2}\,{\rm
d}x\,{\rm
d}y&\lesssim\frac{L_{0}^{2}}{L^{2}}\int_{-1}^{1}\bigg{(}\fint_{-L_{0}}^{L_{0}}\int_{0}^{y}(u^{L}_{,yy}u^{L}_{,xx}+u^{L}_{,yyx}u^{L}_{,x})\,{\rm
d}y^{\prime}\,{\rm d}y\bigg{)}^{2}\,{\rm d}x\\\
&\lesssim\frac{L_{0}^{4}}{L^{2}}\int_{-1}^{1}\fint_{-L_{0}}^{L_{0}}\bigg{(}\fint_{-L_{0}}^{L_{0}}(u^{L}_{,yy})^{2}\,{\rm
d}y^{\prime\prime}\bigg{)}\bigg{(}\fint_{-L_{0}}^{L_{0}}(u^{L}_{,xx})^{2}\,{\rm
d}y^{\prime\prime}\bigg{)}\,{\rm d}y\,{\rm d}x\\\
&+\frac{L_{0}^{4}}{L^{2}}\int_{-1}^{1}\fint_{-L_{0}}^{L_{0}}\bigg{(}\fint_{-L_{0}}^{L_{0}}(u^{L}_{,yyx})^{2}\,{\rm
d}y^{\prime\prime}\bigg{)}\bigg{(}\fint_{-L_{0}}^{L_{0}}(u^{L}_{,x})^{2}\,{\rm
d}y^{\prime\prime}\bigg{)}\,{\rm d}y\,{\rm d}x\\\
&\lesssim\frac{L_{0}^{4}}{L^{2}}\frac{1}{\delta^{2}\varepsilon}\,.\end{split}$
(5.79)

Gathering together (5.70), (5.71), (5.78) and (5.79) we infer

$L^{2}\fint_{-L}^{L}\int_{-1}^{1}\bigl{(}w^{L}_{1,x}-1\bigr{)}^{2}\,{\rm
d}x\,{\rm
d}y\lesssim\frac{L_{0}^{4}}{L^{2}}\frac{1}{\delta^{2}\varepsilon}\,,$

which together with (5.68) and (5.63) implies

$\begin{split}L^{2}\fint_{-L}^{L}\int_{-1}^{1}\biggl{(}w^{L}_{1,x}+\frac{(u^{L}_{,x})^{2}}{2L^{2}}-1\biggr{)}^{2}\,{\rm
d}x\,{\rm
d}y\lesssim\frac{L_{0}^{2}}{L^{2}}\left(\frac{1}{\delta}+\frac{1}{\varepsilon^{2}}\right)+\frac{L_{0}^{4}}{L^{2}}\frac{1}{\delta^{2}\varepsilon}\lesssim\frac{L_{0}^{4}}{L^{2}}\frac{1}{\delta^{2}\varepsilon}\,.\end{split}$

Step 4: we show that

$L^{2}\fint_{-L}^{L}\int_{-1}^{1}\Big{(}w^{L}_{2,y}+\frac{(u_{,y}^{L})^{2}}{2}-x\Big{)}^{2}\,{\rm
d}x\,{\rm d}y\leq L^{2}\left(\frac{1}{3}+C\delta^{3}\right)\,.$ (5.80)

Recalling the definition of $w_{2}^{L}$ it holds
$w^{L}_{2,y}+\frac{(u_{,y}^{L})^{2}}{2}-x=\psi_{\delta}^{2}(x)x-x$, so that

$\begin{split}\fint_{-L}^{L}\int_{-1}^{1}\Big{(}w^{L}_{2,y}+\frac{(u_{,y}^{L})^{2}}{2}-x\Big{)}^{2}\,{\rm
d}x\,{\rm d}y=\int_{-1}^{1}(\psi_{\delta}^{2}(x)x-x)^{2}\,{\rm
d}x\leq\int_{-1}^{2\delta}x^{2}\,{\rm
d}x=\frac{8}{3}\delta^{3}+\frac{1}{3}\,.\end{split}$

Step 5: we show that

$\fint_{-L}^{L}\int_{-1}^{1}\Big{(}L^{2}w_{1,y}^{L}+w_{2,x}^{L}+u_{,x}^{L}u_{,y}^{L}\Big{)}^{2}\,{\rm
d}x\,{\rm d}y=0\,.$ (5.81)

This is a trivial consequence of the following identity

$w^{L}_{1,y}=-\frac{1}{L^{2}}(w_{2,x}^{L}+u_{,x}^{L}u_{,y}^{L})\,,$

which follows from the definitions of $w_{1}^{L}$ and $w_{2}^{L}$.

Step 6: we show that

$\frac{1}{L^{2}}\fint_{-L}^{L}\int_{-1}^{1}\bigg{(}2(u^{L}_{,xy})^{2}+\frac{1}{L^{2}}(u^{L}_{,xx})^{2}\bigg{)}\,{\rm
d}x\,{\rm
d}y\lesssim\left(\frac{1}{L^{2}}+\frac{1}{L^{4}}\right)\frac{1}{\delta^{2}}\,.$
(5.82)

We have

$u^{L}_{,xy}(x,y)=\psi_{\delta}(x)f^{L}(x)\hat{u}^{L}_{,xy}(x,y)+\dot{\psi}_{\delta}(x)f^{L}(x)\hat{u}^{L}_{,y}(x,y)+\psi_{\delta}(x)\dot{f}^{L}(x)\hat{u}^{L}_{,y}(x,y)\,.$

Thus by (5.18), (5.19), (5.27), (5.26) and (5.21)

$\begin{split}\fint_{-L}^{L}\int_{-1}^{1}(u^{L}_{,xy})^{2}\,{\rm d}x\,{\rm
d}y&\lesssim\fint_{-L}^{L}\int_{\delta}^{1}(f^{L}(x))^{2}(\hat{u}^{L}_{,xy})^{2}\,{\rm
d}x\,{\rm d}y\\\
&+\frac{1}{\delta^{2}}\fint_{-L}^{L}\int_{\delta}^{2\delta}(f^{L}(x))^{2}(\hat{u}^{L}_{,y})^{2}\,{\rm
d}x\,{\rm d}y\\\
&+\fint_{-L}^{L}\int_{\delta}^{1}(\dot{f}^{L}(x))^{2}(\hat{u}^{L}_{,y})^{2}\,{\rm
d}x\,{\rm
d}y\lesssim\frac{1}{\varepsilon^{2}}+1+\frac{1}{\varepsilon}+\frac{\varepsilon}{\delta}\lesssim\frac{1}{\delta\varepsilon}\,.\end{split}$

This together with (5.75) imply (5.82).

Conclusions. By Step 1 we have that $(\omega^{L},u^{L})$ converges to $\mu$ in
the sense of Definition 2.3. Moreover by collecting the estimates showed in
Steps 2–6, i.e., (5.54), (5.62), (5.80), (5.81) and (5.82) we find

$\begin{split}\limsup_{L\to+\infty}L^{2}(R_{L}(w^{L},u^{L})-\mathcal{E}_{0})&\leq\mathcal{F}_{\infty}(\mu)+C\lim_{L\to+\infty}\omega(2M^{2}\delta)\log
M\\\
&+C\lim_{L\to+\infty}\Big{(}\frac{L_{0}^{4}}{L^{2}}\frac{1}{\delta^{3}M}+L^{2}\delta^{3}+\frac{1}{L^{2}\delta^{2}}\Big{)}\,.\end{split}$
(5.83)

We now proceed with the choice of the parameters. We start by noticing that
for every $\overline{M}\in\mathbb{N}$ there exists
$L_{\overline{M}}\geq\overline{M}^{1/2}$ such that

$\omega(2\overline{M}^{2}L^{-2/3})\leq{\overline{M}}^{-1}\quad\forall L\geq
L_{\overline{M}}\,.$ (5.84)

Since $L_{\overline{M}+1}\geq L_{\overline{M}}$ we set

$M=M(L):=\overline{M}\quad\text{ if
}L\in[L_{\overline{M}},L_{\overline{M}+1})$

Next we define

$\delta:=\frac{\varepsilon}{M}=L^{-2/3}M^{-1/8}\,\iff\,\varepsilon=L^{-2/3}M^{7/8}\,,$

and we choose

$L_{0}:=\frac{L}{n}\in[M^{1/8},2M^{1/8})\,\iff\,n\in\Big{[}\frac{L}{2M^{1/8}},\frac{L}{M^{1/8}}\Big{)}\,.$

These choices ensures the validity of (5.52) and (5.53). Indeed we have

$\varepsilon=\frac{1}{L^{2/3}M^{-7/8}}=\frac{1}{L^{2/3}\overline{M}^{-7/8}}\leq\frac{1}{\overline{M}^{3}\overline{M}^{-7/8}}\quad\text{
if }L\in[L_{\overline{M}},L_{\overline{M}+1})\,,$

where the last inequality follows from the fact that $L\geq
L_{\overline{M}}\geq\overline{M}^{1/2}$. Hence $\varepsilon\to 0$ and
$\delta\to 0$ as $L\to+\infty$. In a similar way we have $M^{2}\delta\to 0$
and $L_{0},n\to+\infty$ as $L\to+\infty$.
Recalling that $\omega$ is monotone we find

$\omega(2M^{2}\delta)=\omega(2M^{2}L^{-2/3}M^{-1/8})\leq\omega(2M^{2}L^{-2/3})=\omega(2\overline{M}^{2}L^{-2/3})\quad\text{
if }L\in[L_{\overline{M}},L_{\overline{M}+1})\,,$

which together with (5.84) imply

$\log M\omega(2M^{2}\delta)\leq\log(\overline{M})\overline{M}^{-1}\quad\text{
if }L\in[L_{\overline{M}},L_{\overline{M}+1})\,.$ (5.85)

Moreover if $L\in[L_{\overline{M}},L_{\overline{M}+1})$ it holds

$L^{2}\delta^{3}=L^{2}L^{-2}M^{-3/8}=\overline{M}^{-3/8}\,,$ (5.86)
$\frac{1}{L^{2}\delta^{2}}=\frac{L^{4/3}M^{1/4}}{L^{2}}=\frac{M^{1/4}}{L^{2/3}}=\frac{\overline{M}^{1/4}}{L^{2/3}}\leq\frac{\overline{M}^{1/4}}{(\overline{M}^{1/2})^{2/3}}=\overline{M}^{-1/12}\,,$
(5.87)

and

$\frac{L_{0}^{4}}{L^{2}}\frac{1}{\delta^{3}M}\leq\frac{16M^{1/2}}{MM^{-3/8}}=16M^{-1/8}=16\overline{M}^{-1/8}\,,$
(5.88)

Eventually collecting (5.83)–(5.88) we infer

$\begin{split}\limsup_{L\to+\infty}L^{2}(R_{L}(w^{L},u^{L})-\mathcal{E}_{0})\leq\mathcal{F}_{\infty}(\mu)\,.\end{split}$

∎

##  6 Existence and regularity of minimizers of $\mathcal{F}_{\infty}$

In this section we address the existence of minimizers of the limiting
functional $\mathcal{F}_{\infty}$ and we discuss some properties such as
equipartition of the energy. In order to do that we need to introduce the
definition of disintegration of measures in the $k$-variable, which is
slightly different from the disintegration in the $x$-variable introduced in
Section 3.

In the following for a given interval $I\subset\mathbb{R}$ we denote by
$L^{0}(I)$ the space of functions $g\colon I\to\mathbb{R}$ that are Lebesgue
measurable. Moreover the map $\pi_{2}\colon I\times\mathbb{R}\to\mathbb{R}$
denotes the canonical projection, and for any
$\mu\in\mathcal{M}_{b}(I\times\mathbb{R})$ we indicate by
$(\pi_{2})_{\sharp}\mu\in\mathcal{M}_{b}^{+}(\mathbb{R})$ its push-forward
with respect to the map $\pi_{2}$.

######  Definition 6.1 (Disintegration of measures in the $k$-variable).

Let $I\subset\mathbb{R}$ be an interval and let
$\mu\in\mathcal{M}_{b}(I\times\mathbb{R})$. We say that the family

${\big{(}\lambda\,,\,(g_{k})_{k\in\mathbb{R}}\big{)}}\quad\text{ with
}\quad\lambda\in\mathcal{M}_{b}(\mathbb{R})\quad\text{ and }\quad g_{k}\in
L^{0}(I)\quad\forall k\in\mathbb{R}\,,$

is a disintegration of $\mu$ $($in the $k$-variable$)$ if $k\mapsto g_{k}$ is
$\lambda$-measurable, $\int_{0}^{1}g_{k}\,{\rm d}x=1$ for $\lambda$-a.e.
$k\in\mathbb{R}$ and

$\int_{I\times\mathbb{R}}f(x,k)\,{\rm
d}\mu=\int_{\mathbb{R}}\int_{I}f(x,k)g_{k}(x)\,{\rm d}x\,{\rm d}\lambda(k)\,,$
(6.1)

for every $f\in L^{1}(I\times\mathbb{R};|\mu|)$.

With this definition at hand we can state the main result of this section.

######  Theorem 6.2 (Minimizers of $\mathcal{F}_{\infty}$).

Let $\mathcal{M}_{\infty}$ and $\mathcal{F}_{\infty}$ be as in (2.8) and (2.9)
respectively. Then there exists $\hat{\mu}\in\mathcal{M}_{\infty}$ such that

$\mathcal{F}_{\infty}(\hat{\mu})=\inf_{\mu\in\mathcal{M}_{\infty}}\mathcal{F}_{\infty}(\mu)\,.$

Moreover, every minimizer $\hat{\mu}$ satisfies the following properties:
there exist a constant $C>0$ and a $(\pi_{2})_{\sharp}\hat{\mu}$-measurable
map $k\mapsto g_{k}$ with $g_{k}\in BV(0,1)$ for $(\pi_{2})_{\sharp}\hat{\mu}$
a.e. $k\in\mathbb{R}$, such that

${\big{(}(\pi_{2})_{\sharp}\hat{\mu}\,,\,(g_{k})_{k\in\mathbb{R}}\big{)}}\quad\text{
is a disintegration of }\ \hat{\mu}\,,$ $\int_{0}^{1}k^{2}g_{k}(x)\,{\rm
d}x=\int_{0}^{1}\frac{1}{4k^{2}}\left(\frac{\,{\rm d}\hat{\mu}_{,x}}{\,{\rm
d}\hat{\mu}}\right)^{2}g_{k}(x)\,{\rm d}x\quad\text{for }\
(\pi_{2})_{\sharp}\hat{\mu}\ \text{ a.e. $k\in\mathbb{R}$}\,,$ (6.2)

and

$(\pi_{2})_{\sharp}\hat{\mu}\Big{(}\big{\\{}|k|<C\big{\\}}\Big{)}=0\,.$ (6.3)

As a direct consequence minimizers of $\mathcal{F}_{\infty}$ satisfy
equipartition of the energy.

######  Corollary 6.3 (Equipartition of the energy).

Let $\hat{\mu}\in\mathcal{M}_{\infty}$ be a minimizers of
$\mathcal{F}_{\infty}$. Then it holds

$\int_{(0,1)\times\mathbb{R}}k^{2}\,{\rm
d}\hat{\mu}=\int_{(0,1)\times\mathbb{R}}\frac{1}{4k^{2}}\Big{(}\frac{\,{\rm
d}\hat{\mu}_{,x}}{\,{\rm d}\hat{\mu}}\Big{)}^{2}\,{\rm d}\hat{\mu}\,.$

###### Proof.

By Theorem 6.2 it holds

$\begin{split}\int_{(0,1)\times\mathbb{R}}k^{2}\,{\rm
d}\hat{\mu}&=\int_{\mathbb{R}}\int_{0}^{1}k^{2}g_{k}(x)\,{\rm d}x\,{\rm
d}(\pi_{2})_{\sharp}\hat{\mu}\\\
&=\int_{\mathbb{R}}\int_{0}^{1}\frac{1}{4k^{2}}\left(\frac{\,{\rm
d}\hat{\mu}_{,x}}{\,{\rm d}\hat{\mu}}\right)^{2}g_{k}(x)\,{\rm d}x\,{\rm
d}(\pi_{2})_{\sharp}\hat{\mu}=\int_{(0,1)\times\mathbb{R}}\frac{1}{4k^{2}}\Big{(}\frac{\,{\rm
d}\hat{\mu}_{,x}}{\,{\rm d}\hat{\mu}}\Big{)}^{2}\,{\rm
d}\hat{\mu}\,.\end{split}$

∎

We divide the proof of Theorem 6.2 into several steps. Precisely we need to
show that the functional $\mathcal{F}_{\infty}$ is convex and lower semi-
continuous and that the class of measures $\mathcal{M}_{\infty}$ admits a
disintegration in the $k$-variable of the form
${\big{(}(\pi_{2})_{\sharp}\hat{\mu}\,,\,(g_{k})_{k\in\mathbb{R}}\big{)}}$.
First of all we recall that by Remark 2.4 $(ii)$ we have

$\mathcal{F}_{\infty}(\mu)=\int_{(0,1)\times\mathbb{R}}k^{2}\,{\rm
d}\mu+\int_{(0,1)\times\mathbb{R}}\frac{1}{4k^{2}}\Bigl{(}\frac{\,{\rm
d}\mu}{\,{\rm d}|\tilde{\mu}|}\Bigr{)}^{-1}\Bigl{(}\frac{\,{\rm
d}\mu_{,x}}{\,{\rm d}|\tilde{\mu}|}\Bigr{)}^{2}\,{\rm d}|\tilde{\mu}|\,,$

with $\tilde{\mu}=(\mu,\mu_{,x})$ and $|\tilde{\mu}|$ its total variation.
This alternative formulation turns out to be more convenient, in particular
the term

$\int_{(0,1)\times\mathbb{R}}\frac{1}{4k^{2}}\Bigl{(}\frac{\,{\rm
d}\mu}{\,{\rm d}|\tilde{\mu}|}\Bigr{)}^{-1}\Bigl{(}\frac{\,{\rm
d}\mu_{,x}}{\,{\rm d}|\tilde{\mu}|}\Bigr{)}^{2}\,{\rm d}|\tilde{\mu}|$

is reminiscent of the Benamou-Brenier functional used in optimal transport
which enjoys nice properties such as lower semicontinuity and convexity. Here
we consider a specific case and we refer to [45, Section 5.3.1] for a general
treatment of this topic.

For any $\rho,E\in\mathcal{M}_{b}((0,1)\times\mathbb{R})$ the Benamou-Brenier
functional is defined as

$\mathscr{B}_{2}(\rho,E):=\sup\left\\{\int_{(0,1)\times\mathbb{R}}a(x,k)\,{\rm
d}\rho+\int_{(0,1)\times\mathbb{R}}b(x,k)\,{\rm d}E\colon(a,b)\in
C_{b}((0,1)\times\mathbb{R};K_{2})\right\\}\,,$ (6.4)

where

$K_{2}:=\left\\{(z_{1},z_{2})\in\mathbb{R}^{2}\colon
z_{1}+\frac{1}{2}z_{2}^{2}\leq 0\right\\}\,.$

We next recall some properties of $\mathscr{B}_{2}$ which follow from [45,
Proposition 5.18]. Then we state and prove two intermediate Lemmas (cf. Lemma
6.5 and Lemma 6.6) which will be used to show the validity of Theorem 6.2.

######  Proposition 6.4 (Properties of $\mathscr{B}_{2}$).

The functional $\mathscr{B}_{2}$ is convex and lower semi-continuous on the
space $(\mathcal{M}_{b}((0,1)\times\mathbb{R}))^{2}$. Moreover, the following
property hold: if both $\rho$ and $E$ are absolutely continuous w.r.t. a same