# Effect of Boundary Conditions on Second-Order Singularly-Perturbed Phase
Transition Models on $\mathbb{R}$

Thomas Lam

###### Abstract.

The second-order singularly-perturbed problem concerns the integral functional
$\int_{\Omega}\varepsilon_{n}^{-1}W(u)+\varepsilon_{n}^{3}\|\nabla^{2}u\|^{2}\,dx$
for a bounded open set $\Omega\subseteq\mathbb{R}^{N}$, a sequence
$\varepsilon_{n}\to 0^{+}$ of positive reals, and a function
$W:\mathbb{R}\to[0,\infty)$ with exactly two distinct zeroes. This functional
is of interest since it models the behavior of phase transitions, and its
Gamma limit as $n\to\infty$ was studied by Fonseca and Mantegazza. In this
paper, we study an instance of the problem for $N=1$. We find a different form
for the Gamma limit, and study the Gamma limit under the addition of boundary
data.

##  1\. Introduction

For a bounded open set $\Omega\subseteq\mathbb{R}^{N}$, we may imagine
$\Omega$ as a container for a liquid whose density is given by
$u:\Omega\to\mathbb{R}$. The potential energy of the liquid can be measured by
the integral functional $u\mapsto\int_{\Omega}W(u)\,dx$ where
$W:\mathbb{R}\to[0,\infty)$ is the energy per unit volume.

Suppose that $W$ is a two-welled potential, so that $W$ has exactly two
distinct zeroes $z_{1},z_{2}$ with $z_{1}<z_{2}$, which are the phases of the
liquid. Then the liquid will tend to take on the two densities $z_{1}$ and
$z_{2}$, and in particular its density will simply take the form $u=z_{1}\cdot
1_{E}+z_{2}\cdot 1_{\Omega\setminus E}$ in order to minimize the potential
$\int_{\Omega}W(u)\,dx$, which is unexciting.

However, in practice, such rapid changes in density induces high interfacial
energy between phases. To account for this, the Van der Waals-Cahn-Hilliard
theory of phase transitions [7] models the potential energy via an integral
functional of the form

$J_{\varepsilon}(u):=\int_{\Omega}W(u)+\varepsilon^{2}\|\nabla
u\|^{2}\,dx,\qquad u\in W^{1,2}(\Omega).$ (1)

The problem of interest is to minimize $J_{\varepsilon}(u)$ subject to a mass
constraint $\int_{\Omega}u\,dx=m$. A minimizing function $u_{\varepsilon}$
gives a stable density distribution that the liquid would likely conform to.

What sort of stable density distribution is approached as $\varepsilon\to
0^{+}$? Specifically, suppose that we have a sequence $\varepsilon_{n}\to
0^{+}$ such that the sequence of minimizers $u_{\varepsilon_{n}}$ converges to
a function $u$ in some reasonable sense. What properties must be satisfied by
$u$?

Using Gamma convergence (described in Section 2.4), Modica [11] and Sternberg
[13] independently proved that such a $u$ satisfies $u\in\\{z_{1},z_{2}\\}$
almost everywhere, and minimizes the perimeter of $u^{-1}(z_{1})$. In
particular, they prove that for each sequence of positive reals
$\varepsilon_{n}$ with $\varepsilon_{n}\to 0^{+}$, we have that

$\Gamma\text{-}\lim_{n\to\infty}(J_{\varepsilon_{n}}/\varepsilon_{n})(u)=\begin{cases}2\displaystyle\int_{z_{1}}^{z_{2}}\sqrt{W(x)}\,dx\cdot\operatorname{Per}_{\Omega}(u^{-1}(z_{1})),&u\in
BV(\Omega;\\{z_{1},z_{2}\\})\\\ +\infty,&\text{otherwise}\end{cases}$

under $L^{1}(\Omega)$ convergence, where $\operatorname{Per}_{\Omega}(E)$
denotes the perimeter of a set $E\subseteq\Omega$, defined generally as

$\operatorname{Per}_{\Omega}(E):=\sup\left\\{\int_{E}\operatorname{div}\varphi:\varphi\in
C_{0}^{\infty}(\mathbb{R}^{N}),|\varphi|\leq 1\right\\}.$

Owen, Rubinstein and Sternberg [12] studied the family of functionals
$J_{\varepsilon}$ under boundary conditions instead of a mass constraint. They
prove that if $\Omega$ has $C^{2}$ boundary, $h_{\varepsilon}\in
L^{p}(\partial\Omega)\cap L^{\infty}(\partial\Omega)$ is the trace of a
function in $W^{1,2}(\Omega)$ for each $\varepsilon>0$, $h_{\varepsilon}\to
h\in L^{1}(\partial\Omega)\cap L^{\infty}(\partial\Omega)$ in
$L^{1}(\partial\Omega)$ as $\varepsilon\to 0$,
$\int_{\partial\Omega}\left|\frac{\partial
h_{\varepsilon}}{\partial\sigma}\right|$ is bounded in $\varepsilon$ and
$\left\|\frac{\partial
h_{\varepsilon}}{\partial\sigma}\right\|_{L^{\infty}(\partial\Omega)}\leq
C\varepsilon^{-1/4}$ for a constant $C>0$ where $\sigma$ is a surface
parameter on $\partial\Omega$, and
$K_{\varepsilon}:L^{1}(\Omega)\to\overline{\mathbb{R}}$ is defined as

$K_{\varepsilon}:=\begin{cases}\int_{\Omega}\varepsilon^{-1}W(u)+\varepsilon\|\nabla
u\|^{2}\,dx,&u\in W^{1,2}(\Omega)\text{ and
}\operatorname{Tr}u=h_{\varepsilon}\\\ +\infty,&\text{otherwise}\end{cases},$
(2)

where $\overline{\mathbb{R}}=[-\infty,\infty]$, then

$\displaystyle\Gamma\text{-}\lim_{n\to\infty}$ $\displaystyle
K_{\varepsilon_{n}}(u)$
$\displaystyle=\begin{cases}\int_{\Omega}|\nabla\chi(u)|+\int_{\partial\Omega}|\chi(h(x))-\chi(\operatorname{Tr}(u)(x))|\,d\mathcal{H}^{N-1}(x),&u\in
BV(\Omega;\\{z_{1},z_{2}\\})\\\ +\infty,&\text{otherwise}\end{cases},$

where

$\chi(t):=2\int_{z_{1}}^{t}\sqrt{W(z)}\,dz$

and $\varepsilon_{n}\to 0^{+}$.

Fonseca and Mantegazza [6] consider a second-order derivative. To be precise,
they define

$H_{\varepsilon}(u):=\int_{\Omega}\varepsilon^{-1}W(u)+\varepsilon^{3}\|\nabla^{2}u\|^{2}\,dx$
(3)

and proved that for every sequence of positive reals $\varepsilon_{n}\to
0^{+}$, we have

$\Gamma\text{-}\lim_{n\to\infty}H_{\varepsilon_{n}}(u)=\begin{cases}c\cdot\operatorname*{Per}_{\Omega}(u^{-1}(z_{1})),&u\in
BV(\Omega,\\{z_{1},z_{2}\\})\\\ +\infty,\text{otherwise}\end{cases}$

under $L^{1}(\Omega)$ convergence, where

$c:=\min\left\\{\int_{\mathbb{R}}W(u)+|u^{\prime\prime}|^{2}\,dt:u\in
W^{2,2}_{\text{loc}}(\mathbb{R}),\lim_{t\to-\infty}u(t)=z_{1},\lim_{t\to\infty}u(t)=z_{2}\right\\}.$
(4)

For a treatment of more general functionals, see [2], [3], and [4].

Fonseca and Mantegazza’s proofs appeal to rather sophisticated constructs such
as the Young measure. In this paper, we consider a 1-dimensional instance of
the problem solved by Fonseca and Mantegazza, which will allow for alternative
and more elementary methodologies.

Our goal will ultimately be to combine the efforts of Owen, Rubinstein, and
Sternberg with those of Fonseca and Mantegazza by considering the addition of
boundary conditions as in 2 to the second-order problem as in 3, which will be
possible due to our alternative methodologies.

To wit, define

$\Phi(u):=\left(\int_{0}^{1}(u(y)^{2}-1)^{2}\,dy\right)^{3/4}\left(\int_{0}^{1}|u^{\prime\prime}(y)|^{2}\,dy\right)^{1/4}$
(5)

for all $u\in W^{2,p}(0,1)$. Define the families

$\mathscr{J}:=\\{u\in
W^{2,p}(0,1):u(0_{+})=-1,u(1_{-})=1,u^{\prime}(0_{+})=u^{\prime}(1_{-})=0\\}$
(6)

and

$\mathscr{J}^{\prime}(t):=\\{u\in
W^{2,p}(0,1):u(0_{+})=-1,u(1_{-})=t,u^{\prime}(0_{+})=0\\}$ (7)

for each $t\in\mathbb{R}$. Let

$\alpha:=\frac{2}{3^{3/4}}\inf_{u\in\mathscr{J}}\Phi(u)$ (8)

and

$\beta(t):=\frac{4}{3^{3/4}}\inf_{u\in\mathscr{J}^{\prime}(t)}\Phi(u)$ (9)

for all $t\in\mathbb{R}$. Then our main result is the following theorem.

######  Theorem 1.1.

Let $\Omega=(a,b)$ and let $1\leq p\leq 4$. Let $a_{0},b_{0}\in\mathbb{R}$,
and for each $\varepsilon>0$, let
$a_{\varepsilon},b_{\varepsilon}\in\mathbb{R}$ be such that
$a_{\varepsilon}\to a_{0}$ and $b_{\varepsilon}\to b_{0}$ as $\varepsilon\to
0^{+}$. For each $\varepsilon>0$ define a functional
$G_{\varepsilon}(u):L^{p}(\Omega)\to\overline{\mathbb{R}}$ via

$G_{\varepsilon}(u):=\begin{cases}\int_{\Omega}\varepsilon^{-1}(u^{2}-1)^{2}+\varepsilon^{3}|u^{\prime\prime}|^{2}\,dx,&u\in
W^{2,p}(\Omega),u(a_{+})=a_{\varepsilon},u(b_{-})=b_{\varepsilon}\\\
+\infty,&\text{otherwise}\end{cases}.$

Let $\varepsilon_{n}\to 0^{+}$ be a sequence of positive reals. Then, under
strong $L^{p}(\Omega)$ convergence, we have that

$\displaystyle\Gamma\text{-}\lim_{n\to\infty}$ $\displaystyle
G_{\varepsilon_{n}}(u)$
$\displaystyle=\begin{cases}\alpha\operatorname*{essVar}_{\Omega}u+\beta(-a_{0}\operatorname{sgn}u(a_{+}))+\beta(-b_{0}\operatorname{sgn}u(b_{-})),&u\in\text{BPV}(\Omega;\\{-1,1\\})\\\
+\infty,&\text{otherwise}\end{cases}.$

Here, $\operatorname*{essVar}$ denotes essential variation, which is described
in Section 2.2. The addition of boundary conditions is of particular interest
here because they can ensure at least one phase transition by preventing the
existence of trivial minimizers.

The utility of this Gamma convergence result is justified by the following
compactness result.

######  Theorem 1.2 (Compactness for Second Order Problem).

Let $\Omega=(a,b)$ and let $1\leq p\leq 4$. For each $\varepsilon>0$, define
the functional $F_{\varepsilon}:L^{p}(\Omega)\to\overline{\mathbb{R}}$ via

$F_{\varepsilon}(u):=\begin{cases}\int_{\Omega}\varepsilon^{-1}(u^{2}-1)^{2}+\varepsilon^{3}|u^{\prime\prime}|^{2}\,dx,&u\in
W^{2,p}(\Omega)\\\ +\infty,&\text{otherwise}\end{cases}.$ (10)

Let $\varepsilon_{n}$ be a sequence of positive reals with $\varepsilon_{n}\to
0^{+}$. If we have a sequence $u_{n}\in L^{p}(\Omega)$ with
$\sup_{n\in\mathbb{N}}|F_{\varepsilon_{n}}(u_{n})|<\infty$, then there exists
a subsequence $u_{n_{k}}$ for which $u_{n_{k}}\to u$ in $L^{p}(\Omega)$ for
some $u\in BPV(\Omega;\\{-1,1\\})$.

Proofs of this result are given in [6], and [3]. We will provide yet another
proof.

The structure of this paper is as follows. In Section 2, we review pointwise
variation and results in Sobolev spaces. We then define Gamma convergence and
motivate its study.

In Section 3, we prove Theorem 1.2, and then give a more elementary proof for
Fonseca and Mantegazza’s results in the one-dimensional case. Specifically,
let $\Omega=(a,b)$ and define the integral functional
$F_{\varepsilon}:L^{p}(\Omega)\to\overline{\mathbb{R}}$ as in (10). Let
$\varepsilon_{n}$ be a sequence of positive reals with $\varepsilon_{n}\to
0^{+}$, and let $1\leq p\leq 4$. We first prove that if $u_{n}\in
L^{2}(\Omega)$ is a sequence for which
$\sup_{n\in\mathbb{N}}F_{\varepsilon_{n}}(u_{n})<\infty$, then we can find a
subsequence $u_{n_{k}}$ such that $u_{n_{k}}\to u$ in $L^{p}(\Omega)$ for some
$u\in BPV(\Omega;\pm 1)$. Then, we prove that for $1\leq p<\infty$, we have

$\Gamma\text{-}\lim_{n\to+\infty}F_{\varepsilon_{n}}(u)=\begin{cases}\alpha\operatorname*{essVar}_{\Omega}u,&u\in
BPV(\Omega;\pm 1)\\\ +\infty,&\text{otherwise}\end{cases}$

under $L^{p}(\Omega)$ convergence, where $\alpha$ is defined as in
$\eqref{eq:alpha}$.

Lastly, in Section 4, we build off of the work done in Section 3 to prove our
main result.

The current work, and a slicing methodology, will be used to extend to the
$N$-dimensional case. [8]

##  2\. Preliminaries

###  2.1. Pointwise Variation and Essential Variation

For a function $u:I\to\mathbb{R}$, where $I$ is an interval, we may define the
pointwise variation of $u$ as

$\operatorname*{Var}_{I}u:=\sup\left\\{\sum_{i=1}^{n}|u(x_{i})-u(x_{i-1})|,x_{i}\in
I,x_{0}<x_{1}<\ldots<x_{n}\right\\}.$

If $\operatorname*{Var}_{I}u<\infty$ then we write $u\in BPV(I)$. A useful
property is that if $u$ is absolutely continuous, then
$\operatorname*{Var}_{I}u=\int_{I}|u^{\prime}|\,dx$

A family of functions having uniformly bounded pointwise variation is a
powerful property.

######  Theorem 2.1 (Helly’s Selection Theorem).

Let $I$ be an interval and $\mathcal{F}\subseteq BPV(I)$ be an infinite family
of functions $u:I\to\mathbb{R}$ such that
$\sup_{u\in\mathcal{F}}\operatorname*{Var}_{I}u\leq C$ for a constant $C>0$.
Assume moreover that there exists $x_{0}\in I$ such that the set
$\\{u(x_{0}):u\in\mathcal{F}\\}$ is bounded. Then there exists a sequence
$\\{u_{n}\\}_{n}\in\mathcal{F}$ that converges pointwise to some $u\in
BPV(I)$.

This is given as Theorem 2.44 in [10], where a proof can be found.

In the case that such a pointwise convergence is obtained, we can moreover
obtain a bound on $\operatorname*{Var}_{I}u$.

######  Theorem 2.2.

If $u_{n},u\in BPV(I)$ and $u_{n}\to u$ pointwise, then

$\liminf_{n\to\infty}\operatorname*{Var}_{I}u_{n}\geq\operatorname*{Var}_{I}u.$

See Proposition 2.47 in [10] for a proof.

Now suppose, say, $u\in L^{p}(I)$. Then there is no sense in speaking of
$\operatorname*{Var}_{I}u$ because the values of $u$ are not well-defined
pointwise. The workaround is to define the essential pointwise variation of
$u$.

######  Definition 2.3 (Essential Pointwise Variation).

Let $I$ be an interval and $u\in L^{1}_{\text{loc}}(I)$. Then the essential
pointwise variation of $u$ over $I$ is given by

$\operatorname*{essVar}_{I}u:=\inf\left\\{\operatorname*{Var}_{I}\tilde{u}:\tilde{u}\text{
is a representative of }u\right\\}.$

Particularly, if $u\in W^{1,1}(I)$ then $u$ has an absolutely continuous
representative $\tilde{u}$ and we can show that
$\operatorname*{essVar}_{I}u=\int_{I}|\tilde{u}^{\prime}|\,dx$. A nice
property is that the infimum in the definition for
$\operatorname*{essVar}_{I}$ is obtained, i.e. there is always a
representative $\tilde{u}$ for $u$ such that
$\operatorname*{essVar}_{I}u=\operatorname*{Var}_{I}\tilde{u}$. This can be
proven using Helly’s Selection Theorem.

For $u\in L^{p}(I)$, we write $u\in BPV(I)$ if $u$ has a representative
$\tilde{u}$ with $\tilde{u}\in BPV(I)$. By the property from above, we see
that $u\in BPV(I)$ if and only if $\operatorname*{essVar}_{I}u<\infty$.

###  2.2. Sobolev Spaces

We begin by defining weak differentiation.

######  Definition 2.4 (Weak Derivative).

Let $\Omega\subseteq\mathbb{R}$ be an open set and $1\leq p\leq\infty$. For
$u\in L^{p}(\Omega)$, we say that $u$ admits a weak derivative of order
$k\in\mathbb{N}$ if there exists $v\in L^{p}(\Omega)$ satisfying

$\int_{\Omega}u\varphi^{(k)}\,dx=(-1)^{k}\int_{\Omega}v\varphi\,dx$

for all $\varphi\in C^{\infty}_{c}(\Omega)$.

It is not too difficult to verify that the weak derivative is unique up to
almost-everywhere equivalence. If $u$ has a differentiable representative, we
call its derivative (in the traditional sense) the strong derivative, so as to
distinguish the two notions of derivative.

The weak derivative is also notated in the same way as the strong derivative.
For example, if $\Omega=(-1,1)$ and we take $u(x)=|x|$, then $u$ admits a weak
derivative given by $u^{\prime}(x)=\operatorname{sgn}(x)$.

Sobolev spaces consist of $L^{p}$ functions that admit weak derivatives.

######  Definition 2.5 (Sobolev Space).

Let $\Omega\subseteq\mathbb{R}$ be open, $1\leq p\leq\infty$, and
$k\in\mathbb{N}$. Then the Sobolev space $W^{k,p}(\Omega)$ is the normed space
of all $u\in L^{p}(\Omega)$ that admit weak derivatives up to order $k$, such
that $u^{(l)}\in L^{p}(\Omega)$ for all $1\leq l\leq k$. We may endow
$W^{k,p}(\Omega)$ with the following norm:

$\|u\|_{W^{k,p}(\Omega)}:=\|u\|_{L^{p}(\Omega)}+\sum_{l=1}^{k}\|u^{(l)}\|_{L^{p}(\Omega)}.$

This definition may be unwieldy for showing that a function belongs to a
Sobolev space, so we often work with the following equivalent condition.

######  Theorem 2.6 (ACL Condition for $k=1$ on $\mathbb{R}$).

Let $\Omega\subseteq\mathbb{R}$ be an open set and $1\leq p<\infty$. Then
$u\in W^{1,p}(\Omega)$ if and only if $u$ is absolutely continuous and
$u,u^{\prime}\in L^{p}(\Omega)$. Moreover, if $u\in W^{1,p}(\Omega)$ then the
strong and weak derivatives of $u$ agree.

Absolute continuity of $L^{p}$ functions is discussed in the sense that there
exists an absolutely continuous representative, and similarly for
differentiability. A proof of this condition may be found in [10].

An analogue of this condition exists for $k=2$.

######  Theorem 2.7 (ACL Condition for $k=2$ on $\mathbb{R}$).

Let $\Omega\subseteq\mathbb{R}$ be an open set and $1\leq p<\infty$. Then
$u\in W^{2,p}(\Omega)$ if and only if $u\in C^{1}(\Omega)$, $u^{\prime}$ is
absolutely continuous, and $u,u^{\prime},u^{\prime\prime}\in L^{p}(\Omega)$.
Moreover, if $u\in W^{2,p}(\Omega)$ then the first and second-order strong
derivatives of $u$ agree with their weak analogues.

A proof of this may be found in [9].

Various algebraic properties satisfied by strong derivatives have analogues
for weak derivatives. For instance, we have the following chain rule.

######  Theorem 2.8 (Chain Rule).

Let $\Omega\subseteq\mathbb{R}$ be an open, bounded interval. Suppose
$f:\mathbb{R}\to\mathbb{R}$ is Lipschitz and $u\in W^{1,p}(\Omega)$. Then
$f(u)\in W^{1,p}(\Omega)$ and
$\frac{d}{dx}f(u)=f^{\prime}(\tilde{u})u^{\prime}$, where $\tilde{u}$ is the
absolutely continuous representative of $u$, and we take
$f^{\prime}(\tilde{u}(x))u^{\prime}(x)$ to be 0 whenever $u^{\prime}(x)=0$.

This is a consequence of Exercise 11.51 in [10].

It is useful to obtain a bound for $\int_{a}^{b}|u^{\prime}|^{2}\,dx$ in terms
of $\int_{a}^{b}|u|^{2}\,dx$ and $\int_{a}^{b}|u^{\prime\prime}|^{2}\,dx$ for
$u\in W^{2,p}(a,b)$. The following two results are special cases of Lemma 7.38
and Theorem 7.37 in [10].

######  Lemma 2.9.

Let $I=(a,b)$. Suppose $u\in W^{2,1}(I)$ such that $u^{\prime}$ has at least
one zero in $[a,b]$. Then there exists a universal constant $c>0$ such that

$\int_{a}^{b}|u^{\prime}|^{2}\,dx\leq
c\left(\int_{a}^{b}u^{2}\,dx\right)^{1/2}\left(\int_{a}^{b}|u^{\prime\prime}|^{2}\,dx\right)^{1/2}.$

######  Theorem 2.10.

For an open interval $I$ and $u\in W^{2,1}(I)$, there exists a universal
constant $c>0$ such that

$\left(\int_{I}|u^{\prime}|^{2}\,dx\right)^{1/2}\leq
cl^{-1}\left(\int_{I}u^{2}\,dx\right)^{1/2}+cl\left(\int_{I}|u^{\prime\prime}|^{2}\,dx\right)^{1/2}$

for every $l$ with $0<l<\mathcal{L}^{1}(I)$.

We note that by applying the QM-AM inequality to the above theorem, we may
obtain the inequality

$\int_{I}|u^{\prime}|^{2}\,dx\leq
c^{\prime}l^{-2}\int_{I}u^{2}\,dx+c^{\prime}l^{2}\int_{I}|u^{\prime\prime}|^{2}\,dx$

under the same conditions but for a different universal constant
$c^{\prime}>0$.

###  2.3. Weak Convergence

######  Definition 2.11 (Weak Convergence in $L^{p}$).

Let $1\leq p<\infty$. Let $E\subseteq\mathbb{R}$ be measurable, and let
$u_{n}\in L^{p}(E)$ for all $n\in\mathbb{N}$. For $u\in L^{p}(E)$, we say that
$u_{n}$ converges weakly to $u$ in $L^{p}(E)$ if

$\lim_{n\to\infty}\int_{E}u_{n}v\,dx=\int_{E}uv\,dx$

for all $v\in L^{p^{\prime}}(E)$, where $p^{\prime}\in[1,\infty]$ is such that
$\frac{1}{p}+\frac{1}{p^{\prime}}=1$, and we write $u_{n}\rightharpoonup u$ in
$L^{p}(E)$.

######  Definition 2.12 (Weak Convergence in $W^{k,p}$).

For $\Omega\subseteq\mathbb{R}$ open, $k\in\mathbb{N}$, $1\leq p<\infty$, and
$u_{n},u\in W^{k,p}(\Omega)$, we say that $u_{n}$ converges weakly in
$W^{k,p}(\Omega)$ if $u_{n}^{(l)}\rightharpoonup u^{(l)}$ in $L^{p}(\Omega)$
for every $0\leq l\leq k$, and we write $u_{n}\rightharpoonup u$ in
$W^{k,p}(\Omega)$.

It can be shown that weak limits are unique up to almost-everywhere
equivalence. Moreover, Hölder’s inequality implies that strong convergence in
$L^{p}(E)$ implies weak convergence in $L^{p}(E)$, and consequently strong
convergence in $W^{k,p}(\Omega)$ implies weak convergence in
$W^{k,p}(\Omega)$.

Proofs of the theorems that follow may be found in [5]. An important property
is that we may strengthen some convergences in the definition of weak
convergence in $W^{k,p}(\Omega)$.

######  Theorem 2.13.

Let $(a,b)$ be an interval, $k\in\mathbb{N}$, $1\leq p<\infty$, and suppose
$u_{n},u\in W^{k,p}(a,b)$ are such that $u_{n}\rightharpoonup u$ in
$W^{k,p}(a,b)$. Then $u_{n}^{(l)}\to u^{(l)}$ strongly in $L^{p}(a,b)$ for all
$0\leq l\leq k-1$.

Weak convergence enables us to consider a notion of weak compactness.

######  Theorem 2.14 (Weak Compactness in $W^{k,p}$).

Let $\Omega\subseteq\mathbb{R}$ be open, $k\in\mathbb{N}$, $1\leq p<\infty$,
and suppose that $u_{n}\in W^{k,p}(\Omega)$ is such that $\\{u_{n}\\}_{n}$ is
uniformly bounded on $W^{k,p}(\Omega)$. That is,
$\sup_{n\in\mathbb{N}}\|u_{n}\|_{W^{k,p}(\Omega)}<\infty$. Then there exist a
subsequence $\\{u_{n_{l}l}\\}_{l}$ and $u\in W^{k,p}(\Omega)$ such that
$u_{n_{l}}\rightharpoonup u$ in $W^{k,p}(\Omega)$.

Lastly, weak convergence is useful for obtaining a weak form of semi-
continuity for certain functions. A particular result we shall use is the
following.

######  Theorem 2.15.

Let $1\leq p<\infty$ and let $\Omega\subseteq\mathbb{R}^{N}$ be open and
bounded. Suppose that $f:\mathbb{R}\to\mathbb{R}$ is convex. Then the integral
functional $F:L^{p}(\Omega)\to\overline{\mathbb{R}}$ defined by

$F(u):=\int_{\Omega}f(u)\,dx$

is sequentially lower semi-continuous with respect to weak
$L^{p}_{\text{loc}}(\Omega)$ convergence. That is, we have

$F(u)\leq\liminf_{n\to\infty}F(u_{n})$

whenever $u_{n},u\in L^{p}(\Omega)$ such that $u_{n}\rightharpoonup u$ in
$L^{p}_{\text{loc}}(\Omega)$.

For a proof, see Theorem 5.14 in [5].

###  2.4. Gamma Limits

The Gamma limit is a notion of function convergence that is of interest
because it preserves minima.

######  Definition 2.16.

For a metric space $(X,d)$ and a sequence $f_{n}:X\to\overline{\mathbb{R}}$,
we say that $f_{n}$ Gamma converges to a function
$f:X\to\overline{\mathbb{R}}$ if for all $x_{0}\in X$:

  1. (1)

$\liminf_{n\to\infty}f_{n}(x_{n})\geq f(x_{0})$ for all sequences $x_{n}$ with
$x_{n}\stackrel{{\scriptstyle d}}{{\to}}x_{0}$, and

  2. (2)

$\limsup_{n\to\infty}f_{n}(x_{n})\leq f(x_{0})$ for some sequence $x_{n}$ with
$x_{n}\stackrel{{\scriptstyle d}}{{\to}}x_{0}$,

and we write $f=\Gamma\text{-}\lim_{n\to\infty}f_{n}$.

The two inequality conditions are referred to as the liminf inequality and
limsup inequality respectively.

The following properties are proven in Chapter 1 of [1].

  * •

The Gamma limit, if it exists, is unique pointwise. Moreover, it is given
precisely by

$\left(\Gamma\text{-}\lim_{n\to\infty}f_{n}\right)(x)=\inf\left\\{\liminf_{n\to\infty}f_{n}(x_{n}):x_{n}\stackrel{{\scriptstyle
d}}{{\to}}x\right\\}.$

  * •

It is not necessarily true that the Gamma limit coincides with the pointwise
limit, if both exist.

  * •

If $f=\Gamma\text{-}\lim f_{n}$ exists and $x_{n}$ is a minimizer for $f_{n}$,
then every accumulation point of the sequence $\\{x_{n}\\}_{n}$ is a minimizer
for $f$.

As an example, let us define $\Omega:=(0,1)$ and consider the functional
$F_{\varepsilon}:W^{1,2}(\Omega)\to\mathbb{R}$ defined by

$F_{\varepsilon}(u):=\begin{cases}\int_{0}^{1}(u^{2}-1)^{2}+\varepsilon^{2}|u^{\prime}|^{2}\,dx,&\int_{\Omega}u\,dx=0\\\
+\infty,&\text{otherwise}\end{cases}$

for $\varepsilon>0$. Although it is quite non-trivial, it can be shown that
$F_{\varepsilon}$ has a minimizer.

The $\varepsilon^{2}|u^{\prime}|^{2}$ term is a complication. Can it be
removed without much change to the minimizers? As before, let
$\varepsilon_{n}\to 0^{+}$ be a sequence. We can show that

$\Gamma\text{-}\lim_{n\to\infty}F_{\varepsilon_{n}}(u)=\int_{0}^{1}(u^{2}-1)^{2}\,dx$

under $W^{1,2}(\Omega)$ convergence. Now let us attempt to repeat the previous
logic. If we take $u_{n}$ to be a minimizer of $F_{\varepsilon_{n}}$, then any
subsequence of $u_{n}$ converging in $W^{1,2}(\Omega)$ must converge to a
minimizer of $\int_{0}^{1}(u^{2}-1)^{2}\,dx$. However, we can show that

$\inf\left\\{\int_{0}^{1}(u^{2}-1)^{2}\,dx:\int_{\Omega}u\,dx=0\text{ and
}u\in W^{2,1}(\Omega)\right\\}=0,$

hence a minimizer $u_{\text{min}}$ would have to take the form
$u_{\text{min}}=1\cdot 1_{E}+(-1)\cdot 1_{\Omega\setminus E}$ for some $E$
with $\mathcal{L}^{1}(E)=\frac{1}{2}$, which cannot be consistent with the
requirement that $u\in W^{2,1}(\Omega)$. Hence, there is no such minimizer!

Our failure to deduce anything meaningful from this Gamma convergence result
is a consequence of missing a compactness result that guarantees the existence
of converging subsequences of $u_{n}$. Fortunately, without computing the
exact minimizers, we can still recover a reasonable Gamma limit result by
weakening the convergence. Specifically, we may sacrifice the strong
$W^{1,2}(\Omega)$ convergence for a weaker convergence called weak
$L^{4}(\Omega)$ convergence (and written as “$u_{n}\rightharpoonup u$ in
$L^{4}(\Omega)$"). The advantage of this weakening is that if we take $u_{n}$
to be a minimizer of $F_{\varepsilon_{n}}$, then we can use weak compactness
to prove that there is a subsequence $u_{n_{k}}$ and some $u\in L^{4}(\Omega)$
for which $u_{n_{k}}\rightharpoonup u$ in $L^{4}(\Omega)$, which is precisely
the sort of compactness result we seek. Under weak $L^{4}(\Omega)$
convergence, the Gamma limit changes to

$\Gamma\text{-}\lim_{n\to\infty}F_{\varepsilon_{n}}(u)=\int_{0}^{1}f^{**}(u)\,dx,$

where $f^{**}(z):=\begin{cases}(z^{2}-1)^{2},&|z|>1\\\ 0,&|z|\leq
1\end{cases}$ denotes the convex envelope of $f(z):=(z^{2}-1)^{2}$. Due to the
existence of a compactness result, we are guaranteed the existence of a
subsequence $u_{n_{k}}$ converging weakly in $L^{2}(\Omega)$ to a minimizer of
$\int_{\Omega}f^{**}(u)\,dx$, and indeed there are many minimizers of this
functional.

This demonstrates the importance of choosing the correct metric of convergence
for the Gamma limit. If the convergence is too strong, there may be no
compactness result. On the other hand, the weaker the convergence, the weaker
the result we end up proving.

##  3\. The Second-Order Singularly-Perturbed Problem

Our first goal is to prove Theorem 1.2. As a steppingstone, we first prove a
similar compactness result in the context of a first order singularly-
perturbed problem, which is of interest in itself.

######  Theorem 3.1 (Compactness for First Order Problem).

Let $\Omega=(a,b)$ be a non-empty open interval, and let $1\leq p<\infty$. Let
$\varepsilon_{n}$ be a sequence of positive reals with $\varepsilon_{n}\to
0^{+}$.

  1. (i)

If we have a sequence $u_{n}\in W^{1,p}(\Omega)$ with

$C:=\sup_{n\in\mathbb{N}}\int_{\Omega}\varepsilon_{n}^{-1}(u_{n}^{2}-1)^{2}+\varepsilon_{n}|u_{n}^{\prime}|^{2}\,dx<\infty,$

then there exists a subsequence $u_{n_{k}}$ for which $u_{n_{k}}\to u$ in
$L^{p}(\Omega)$ for some $u\in BPV(\Omega;\\{-1,1\\})$.

  2. (ii)

Moreover, such a $u$ must satisfy
$\operatorname*{essVar}_{\Omega}u\leq\frac{3}{4}C.$

###### Proof.

Since $\int_{\Omega}(u_{n}^{2}-1)^{2}\,dx\leq C\varepsilon_{n}\to 0$, we see
that $u_{n}^{2}-1\to 0$ in $L^{2}(\Omega)$. So, by extraction of a
subsequence, we may assume that $u_{n}^{2}-1\to 0$ almost everywhere. In
particular, we get that $|u_{n}|\to 1$ almost everywhere.

Applying the AM-GM inequality inside the integral gives

$C\geq\int_{\Omega}\varepsilon_{n}^{-1}(u_{n}^{2}-1)^{2}+\varepsilon_{n}|u_{n}^{\prime}|^{2}\,dx\geq\int_{\Omega}2|u_{n}^{2}-1|\cdot|u_{n}^{\prime}|\,dx=\int_{\Omega}|U_{n}^{\prime}|\,dx$

for all $n$, where $U_{n}:=2u_{n}-\frac{2}{3}u_{n}^{3}$. Taking
$\tilde{U_{n}}$ to be the absolutely continuous representative of $U_{n}$, we
deduce that $\operatorname*{Var}_{\Omega}\tilde{U}_{n}\leq C$ for all $n$.
Moreover, since $|u_{n}|\to 1$ almost everywhere, we have that $|U_{n}|\to
2-\frac{2}{3}=\frac{4}{3}$ almost everywhere by continuity of $z\mapsto
2z-\frac{2}{3}z^{3}$, and so in particular there is some $x\in\Omega$ for
which $\tilde{U}_{n}(x)$ converges. From this, we may now apply Helly’s
Selection Theorem to find a subsequence $\tilde{U}_{n_{k}}$ that converges
pointwise to some $\tilde{U}\in BPV(\Omega)$, and
$C\geq\liminf_{n\to\infty}\operatorname*{Var}_{\Omega}\tilde{U}_{n}\geq\operatorname*{Var}\tilde{U}$
by Theorem 2.2.

It follows that $U_{n_{k}}$ converges to some $U$ almost everywhere that has
representative $\tilde{U}$, with $U=\pm\frac{4}{3}$ almost everywhere, and
moreover $\operatorname*{essVar}_{\Omega}U\leq\operatorname*{Var}\tilde{U}\leq
C<\infty$. Hence, by taking $u=\frac{3}{4}U$, we have that

$\operatorname*{essVar}_{\Omega}u\leq\frac{3}{4}C<\infty,$ (11)

and particularly $u$ has a representative $\tilde{u}\in BPV(\Omega;\pm 1)$.

To see that $u_{n_{k}}\to u$ almost everywhere, recall that $|u_{n_{k}}|\to 1$
almost everywhere. Taking representatives $\tilde{u}_{n_{k}}$, we have that
almost every $x_{0}\in\Omega$ satisfies $|\tilde{u}_{n_{k}}(x_{0})|\to 1$,
$2\tilde{u}_{n_{k}}-\frac{2}{3}\tilde{u}_{n_{k}}^{3}=\tilde{U}_{n_{k}}$ for
all $k$, $\tilde{U}_{n_{k}}(x_{0})\to\tilde{U}(x_{0})$, and
$\tilde{U}(x_{0})=\frac{4}{3}\tilde{u}(x_{0})$.

For every such $x_{0}$ we have that
$2\tilde{u}_{n_{k}}(x_{0})-\frac{2}{3}u_{n_{k}}(x_{0})^{3}\to\frac{4}{3}\tilde{u}(x_{0})$.
If $\tilde{u}(x_{0})=1$, then
$2\tilde{u}_{n_{k}}(x_{0})-\frac{2}{3}\tilde{u}_{n_{k}}(x_{0})^{3}-\frac{4}{3}\to
0$, and by factoring we obtain
$(\tilde{u}_{n_{k}}(x_{0})-1)^{2}(\tilde{u}_{n_{k}}(x_{0})+2)\to 0$. But
$|\tilde{u}_{n_{k}}(x_{0})|\to 1$, so
$\liminf_{k\to\infty}|\tilde{u}_{n_{k}}(x_{0})+2|\geq 1$, and hence it must
follow that $(\tilde{u}_{n_{k}}(x_{0})-1)^{2}\to 0$, i.e.
$\tilde{u}_{n_{k}}(x_{0})\to 1=\tilde{u}(x_{0})$. By a symmetrical argument we
see that if $\tilde{u}(x_{0})=-1$ then
$\tilde{u}_{n_{k}}(x_{0})\to-1=\tilde{u}(x_{0})$. This shows that
$u_{n_{k}}\to u$ almost everywhere.

Lastly, to obtain $L^{p}(\Omega)$ convergence, we may use the fact that
$\operatorname*{essVar}_{\Omega}U_{n_{k}}\leq C$ and that $U_{n_{k}}$
converges almost everywhere to deduce that $U_{n_{k}}$ is uniformly bounded by
some constant $C^{\prime}$. Then the inverse image of $[-C,C]$ under $z\mapsto
2z-\frac{2}{3}z^{3}$ is compact by continuity, thus
$|U_{n_{k}}|=|2u_{n_{k}}-\frac{2}{3}u_{n_{k}}^{3}|\leq C^{\prime}$ implies
that $|u_{n_{k}}|\leq C^{\prime\prime}$ for some constant $C^{\prime\prime}$.
In particular, we see that $u_{n_{k}}$ is uniformly bounded, so we may apply
dominated convergence to obtain the desired $L^{p}(\Omega)$ convergence. This
proves item (i), and the bound claimed in item (ii) was acquired in (11). ∎

Next, we will prove an essential interpolation inequality.

######  Theorem 3.2.

Let $\Omega=(a,b)$ be a non-empty open interval, let
$\Omega_{0}=(a_{0},b_{0})\Subset(a,b)$, and let $1\leq p\leq 4$. Let
$\varepsilon_{n}$ be a sequence of positive reals with $\varepsilon_{n}\to
0^{+}$. If we have a sequence $u_{n}\in W^{2,p}(\Omega)$ with

$\sup_{n\in\mathbb{N}}\int_{\Omega}\varepsilon_{n}^{-1}(u_{n}^{2}-1)^{2}+\varepsilon_{n}^{3}|u_{n}^{\prime\prime}|^{2}\,dx<\infty,$

then for a universal constant $C^{\prime}>0$ we have that

$\int_{\Omega_{0}}\varepsilon_{n}|u_{n}^{\prime}|^{2}\,dx\leq
C^{\prime}\int_{\Omega}\varepsilon_{n}^{-1}(u_{n}^{2}-1)^{2}+\varepsilon_{n}^{3}|u_{n}^{\prime\prime}|^{2}\,dx$

for all large enough $n$.

We begin by proving two lemmas.

######  Lemma 3.3.

Let $\Omega$ be an open interval, $1\leq p\leq 4$, and $u_{n}\in
L^{p}(\Omega)$ such that
$\lim_{n\to\infty}\int_{a}^{b}(u_{n}^{2}-1)^{2}\,dx=0$. Then
$\\{|u_{n}|^{p}\\}_{n}$ is uniformly integrable.

###### Proof.

First, find some $c>0$ so small that $cz^{p}\leq(z^{2}-1)^{2}$ for all $z$
large enough; say, $z\geq M_{0}$. This is possible by virtue of order of
growth, since $1\leq p\leq 4$. Now

$\int_{\Omega}|u_{n}|^{p}\,dx=\int_{\\{|u_{n}|<M_{0}\\}}|u_{n}|^{p}\,dx+\int_{\\{|u_{n}|\geq
M_{0}\\}}|u_{n}|^{p}\,dx\leq M_{0}^{p}+\int_{\\{|u_{n}|\geq
M_{0}\\}}\frac{1}{c}(u_{n}^{2}-1)^{2}\,dx<\infty,$

so $|u_{n}|^{p}$ is integrable for all $n$.

Fix $\eta>0$. Find $N_{\eta}$ such that
$\int_{\Omega}(u_{n}^{2}-1)^{2}\,dx\leq c\eta$ for all $n>N_{\eta}$. Lastly,
by integrability of $|u_{n}|^{p}$ for $n=1,2,\cdots,N_{\eta}$, we choose
$M_{\eta}\geq M_{0}$ so large that $\int_{|u_{n}|\geq
M_{\eta}}|u_{n}|^{p}\,dx<\eta$ for all such $n$. This $M_{\eta}$ witnesses the
uniform integrability since for all $n\geq N_{\eta}$ we have

$\int_{|u_{n}|\geq M_{\eta}}|u_{n}|^{p}\,dx\leq\frac{1}{c}\int_{|u_{n}|\geq
M_{\eta}}(u_{n}^{2}-1)^{2}\,dx\leq\eta.$

∎

######  Lemma 3.4.

Let $\Omega$ be an open interval. Then for every $u\in W^{2,p}(\Omega)$, the
set $Z=\\{x\in\Omega:u(x)=0,u^{\prime}(x)\neq 0\\}$ is countable.

###### Proof.

Let $x_{0}\in Z$. It suffices to show that $x_{0}$ is an isolated point of
$Z$. We may assume without loss of generality that $u^{\prime}(x_{0})>0$.

Since $u\in W^{2,p}(\Omega)$ we know that $u^{\prime}$ is continuous (See
Theorem 2.7). Thus, there exists $\delta>0$ such that $u^{\prime}(x)>0$ for
all $x\in(x_{0}-\delta,x_{0}+\delta)$. We conclude that there is no
$x\in(x_{0}-\delta,x_{0}+\delta)$ for which $u(x)=0$ (so that in particular,
$x\not\in Z$), otherwise we obtain a contradiction from Rolle’s Theorem. ∎

Although it is not worth stating as a lemma, we will be using the inequality
$(|z|-1)^{2}\leq(z^{2}-1)^{2}$ quite liberally, which follows from factoring
$(z^{2}-1)^{2}$ and taking cases on the sign of $z$.

We may now prove Theorem 1.2.

###### Proof.

Let

$C=\sup_{n\in\mathbb{N}}\int_{a}^{b}\varepsilon_{n}^{-1}(u_{n}^{2}-1)^{2}+\varepsilon_{n}^{3}|u_{n}^{\prime\prime}|^{2}\,dx.$
(12)

Since $(a_{0},b_{0})\Subset(a,b)$ and $\varepsilon_{n}\to 0$, it is sufficient
to obtain an interpolation inequality of the form

$\int_{\Omega_{\varepsilon_{n}}}\varepsilon_{n}|u_{n}^{\prime}|^{2}\,dx\leq
C^{\prime}\int_{\Omega}\varepsilon_{n}^{-1}(u_{n}^{2}-1)^{2}+\varepsilon_{n}^{3}|u_{n}^{\prime\prime}|^{2}\,dx,$
(13)

where $C^{\prime}$ is a universal constant and
$\Omega_{\varepsilon}:=(a+\varepsilon,b-\varepsilon)$ for all $\varepsilon>0$.

Note that if $\int_{\Omega}\varepsilon_{n}^{-1}(u_{n}^{2}-1)^{2}\,dx=0$, then
we must have either $u_{n}\equiv 1$ or $u_{n}\equiv-1$. Hence
$\int_{\Omega_{\varepsilon_{n}}}\varepsilon_{n}|u_{n}^{\prime}|^{2}\,dx=0$, so
the inequality (13) holds and there is nothing more to show. Thus, we may
assume the contrary, so that

$\int_{\Omega}\varepsilon_{n}^{-1}(u_{n}^{2}-1)^{2}+\varepsilon_{n}^{3}|u_{n}^{\prime\prime}|^{2}\,dx>0.$
(14)

It is clear from (12) that

$\int_{\Omega}|u_{n}^{\prime\prime}|^{2}\,dx\leq\frac{C}{\varepsilon_{n}^{3}}$
(15)

and

$\int_{\Omega}(u_{n}^{2}-1)^{2}\,dx\leq\varepsilon_{n}C$

for all $n$.

We divide our analysis of $u_{n}$ into three sets: Where $u_{n}=0$, where
$u_{n}$ is near 0, and where $u_{n}$ is far from 0.

Firstly, the argument in Lemma 3.3 implies that $u_{n}\in L^{2}(\Omega)$, and
(15) implies that $u_{n}^{\prime\prime}\in L^{2}(\Omega)$. It follows by
Theorem 2.10 that $u_{n}^{\prime}\in L^{2}(\Omega)$. Hence, by this fact and
the assumption (14), there must exist $\alpha_{n}\in(0,\tfrac{1}{2})$ so small
that

$\int_{0<|u_{n}|\leq\alpha_{n}}|u_{n}^{\prime}|^{2}\,dx\leq\int_{\Omega}\varepsilon_{n}^{-1}(u_{n}^{2}-1)^{2}+\varepsilon_{n}^{3}|u_{n}^{\prime\prime}|^{2}\,dx.$
(16)

Our three sets in question are then $u_{n}^{-1}(\\{0\\})$,
$u_{n}^{-1}\big{(}[-\alpha_{n},0)\cup(0,\alpha_{n}]\big{)}$, and
$u_{n}^{-1}\big{(}(-\infty,-\alpha_{n})\cup(\alpha_{n},+\infty)\big{)}$.

For the set $u_{n}^{-1}(\\{0\\})$, we claim that

$\int_{u_{n}=0}|u_{n}^{\prime}|\,dx=0.$ (17)

To prove this, it suffices to show that

$\int_{\\{y:u_{n}(y)=0,u_{n}^{\prime}(y)\neq 0\\}}|u_{n}^{\prime}|\,dx=0$

since the integral over the set where $u_{n}^{\prime}=0$ is clearly 0. But by
Lemma 3.4, the domain over which we integrate here is countable, so the
integral is indeed 0.

The set $u_{n}^{-1}\big{(}[-\alpha_{n},0)\cup(0,\alpha_{n}]\big{)}$ is already
handled by (16).

It remains to handle
$u_{n}^{-1}((-\infty,-\alpha_{n})\cup(\alpha_{n},+\infty))$. This set is open
by continuity, so it is the at-most countable union of disjoint open
intervals. Let those intervals that do not have either $a$ or $b$ as an
endpoint be enumerated as $\\{(a_{i},b_{i})\\}_{i\in I}$ for a countable,
possibly empty index set $I$. Let the union of these intervals be $E_{n}$, so
$E_{n}$ is where $|u_{n}|>\alpha_{n}$ except possibly near the endpoints of
$(0,1).$

Then for all $i\in I$ we have that $u_{n}(a_{i})=u_{n}(b_{i})$ by a continuity
argument. Thus by Rolle’s Theorem there is $c_{i}\in(a_{i},b_{i})$ for which
$u_{n}^{\prime}(c_{i})=0$. Moreover, $u_{n}\neq 0$ over $(a_{i},b_{i})$, thus
$v_{n}:=|u_{n}|-1$ is differentiable over $(a_{i},b_{i})$. If we assume
without loss of generality that $u_{n}>0$ over $(a_{i},b_{i})$, then in fact
$v_{n}=u_{n}-1$, so it is clear that $v_{n}\in W^{2,p}(a_{i},b_{i})$, with
$v_{n}^{\prime}=u_{n}^{\prime}$ and
$v_{n}^{\prime\prime}=u_{n}^{\prime\prime}$. By Lemma 2.9, it follows that

$\displaystyle\int_{a_{i}}^{b_{i}}|u_{n}^{\prime}|^{2}\,dx$ $\displaystyle\leq
c_{1}\left(\int_{a_{i}}^{b_{i}}(|u_{n}|-1)^{2}\,dx\right)^{1/2}\left(\int_{a_{i}}^{b_{i}}|u_{n}^{\prime\prime}|^{2}\,dx\right)^{1/2}$
$\displaystyle\leq
c_{1}\left(\int_{a_{i}}^{b_{i}}(u_{n}^{2}-1)^{2}\,dx\right)^{1/2}\left(\int_{a_{i}}^{b_{i}}|u_{n}^{\prime\prime}|^{2}\,dx\right)^{1/2}$

for a constant $c_{1}>0$. Multiplying by $\varepsilon_{n}$ gives

$\int_{a_{i}}^{b_{i}}\varepsilon_{n}|u_{n}^{\prime}|^{2}\,dx\leq
c_{1}\left(\int_{a_{i}}^{b_{i}}\varepsilon_{n}^{-1}(u_{n}^{2}-1)^{2}\,dx\right)^{1/2}\left(\int_{a_{i}}^{b_{i}}\varepsilon_{n}^{3}|u_{n}^{\prime\prime}|^{2}\,dx\right)^{1/2},$

then applying the AM-GM inequality gives

$\int_{a_{i}}^{b_{i}}\varepsilon_{n}|u_{n}^{\prime}|^{2}\,dx\leq\frac{c_{1}}{2}\int_{a_{i}}^{b_{i}}\varepsilon_{n}^{-1}(u_{n}^{2}-1)^{2}+\varepsilon_{n}^{3}|u_{n}^{\prime\prime}|^{2}\,dx.$

Summing this inequality over all $i\in I$, we conclude that

$\int_{E_{n}}\varepsilon_{n}|u_{n}^{\prime}|^{2}\,dx\leq\frac{c_{1}}{2}\int_{E_{n}}\varepsilon_{n}^{-1}(u_{n}^{2}-1)^{2}+\varepsilon_{n}^{3}|u_{n}^{\prime\prime}|^{2}\,dx.$
(18)

We are almost done. Let
$S_{n}=\sup\\{x\in\Omega:|u_{n}(y)|\geq\alpha_{n}\,\forall a<y\leq x\\}$ and
$T_{n}=\inf\\{x\in\Omega:|u_{n}(y)|\geq\alpha_{n}\,\forall x\leq y<b\\}$. If
$\\{x\in\Omega:|u_{n}(y)|\geq\alpha_{n}\,\forall a<y\leq x\\}$ is empty then
we take $S_{n}=a$, and similarly for $T_{n}$. The intervals $(a,S_{n})$ and
$(T_{n},b)$, if distinct and non-empty, are the two intervals that we omitted
from $\\{x\in\Omega:u_{n}(x)>\alpha_{n}\\}$ to obtain $E_{n}$, so that
$E_{n}\cup(a,S_{n})\cup(T_{n},b)=\\{x\in\Omega:u_{n}(x)>\alpha_{n}\\}$. It
remains to handle these two intervals.

If $S_{n}-a\leq\varepsilon_{n}$, then clearly
$\int_{(a,S_{n})\cap\Omega_{\varepsilon_{n}}}\varepsilon_{n}|u_{n}^{\prime}|^{2}\,dx=0$.
If otherwise $S_{n}-a>\varepsilon_{n}$ then we may apply Theorem 2.10 to
$|u_{n}|-1\in W^{2,p}(a,S_{n})$ to get

$\displaystyle\int_{a}^{S_{n}}|u_{n}^{\prime}|^{2}\,dx$ $\displaystyle\leq
c_{2}\varepsilon_{n}^{-2}\int_{a}^{S_{n}}(|u_{n}|-1)^{2}\,dx+c_{1}\varepsilon_{n}^{2}\int_{a}^{S_{n}}|u_{n}^{\prime\prime}|^{2}\,dx$
$\displaystyle\leq
c_{2}\varepsilon_{n}^{-2}\int_{a}^{S_{n}}(u_{n}^{2}-1)^{2}\,dx+c_{1}\varepsilon_{n}^{2}\int_{a}^{S_{n}}|u_{n}^{\prime\prime}|^{2}\,dx,$

for a constant $c_{2}>0$. Multiplying by $\varepsilon_{n}$ gives

$\int_{(a,S_{n})\cap\Omega_{\varepsilon_{n}}}\varepsilon_{n}|u_{n}^{\prime}|^{2}\,dx\leq\int_{a}^{S_{n}}\varepsilon_{n}|u_{n}^{\prime}|^{2}\,dx\leq
c_{2}\int_{a}^{S_{n}}\varepsilon_{n}^{-1}(u_{n}^{2}-1)^{2}+\varepsilon_{n}^{3}|u_{n}^{\prime\prime}|^{2}\,dx.$
(19)

Thus the inequality (19) holds in either the cases
$S_{n}-a\leq\varepsilon_{n}$ and $S_{n}-a>\varepsilon_{n}$. Applying the same
arguments above to the interval $(T_{n},b)$, we see that in either of the
cases $b-T_{n}\leq\varepsilon_{n}$ and $b-T_{n}>\varepsilon_{n}$ we may obtain
the inequality

$\int_{(T_{n},b)\cap\Omega_{\varepsilon_{n}}}\varepsilon_{n}|u_{n}^{\prime}|^{2}\,dx\leq
c_{2}\int_{T_{n}}^{b}\varepsilon_{n}^{-1}(u_{n}^{2}-1)^{2}+\varepsilon_{n}^{3}|u_{n}^{\prime\prime}|^{2}\,dx.$
(20)

By summing (18), (19), and (20), we get the bound

$\displaystyle\int_{\\{|u_{n}|>\alpha_{n}\\}\cap\Omega_{\varepsilon_{n}}}\varepsilon_{n}|u_{n}^{\prime}|^{2}\,dx$
$\displaystyle\leq\int_{E_{n}\cup((a,S_{n})\cap\Omega_{\varepsilon_{n}})\cup((T_{n},b)\cap\Omega_{\varepsilon_{n}})}\varepsilon_{n}|u_{n}^{\prime}|^{2}\,dx$
$\displaystyle\leq\max(c_{1}/2,c_{2})\int_{E_{n}\cup(a,S_{n})\cup(T_{n},b)}\varepsilon_{n}^{-1}(u_{n}^{2}-1)^{2}+\varepsilon_{n}^{3}|u_{n}^{\prime\prime}|^{2}\,dx$
$\displaystyle=\max(c_{1}/2,c_{2})\int_{|u_{n}|>\alpha_{n}}\varepsilon_{n}^{-1}(u_{n}^{2}-1)^{2}+\varepsilon_{n}^{3}|u_{n}^{\prime\prime}|^{2}\,dx.$
(21)

At last, once we add up (16), (17), and (21), we can arrive at the inequality

$\int_{\Omega_{\varepsilon_{n}}}\varepsilon_{n}|u_{n}^{\prime}|^{2}\,dx\leq
c_{3}\int_{\Omega}\varepsilon_{n}^{-1}(u_{n}^{2}-1)^{2}+\varepsilon_{n}^{3}|u_{n}^{\prime\prime}|^{2}\,dx,$

where $c_{3}=1+\max(c_{1}/2,c_{2})$. This is of the form (13), so we are done.
∎

We now turn to the proof of Theorem 1.2.

###### Proof.

Let $C=\sup_{n\in\mathbb{N}}F_{\varepsilon_{n}}(u_{n})$. We first claim that
for each $\delta>0$ there exists a subsequence $u_{n_{k}}$ of $u_{n}$ such
that $u_{n_{k}}\to v$ in $L^{p}(a+\delta,b-\delta)$ for some function $v\in
BPV((a+\delta,b-\delta);\pm 1)$, with
$\operatorname*{essVar}_{(a+\delta,b-\delta)}v\leq C_{3}$ for a constant
$C_{3}$ not depending on $\delta$.

To see this, we apply Theorem 3.2 to the interval $(a+\delta,b-\delta)$ to
obtain the bound

$\int_{a+\delta}^{b-\delta}\varepsilon_{n}|u_{n}^{\prime}|^{2}\,dx\leq\int_{\Omega}\varepsilon_{n}^{-1}(u_{n}^{2}-1)^{2}+\varepsilon_{n}^{3}|u_{n}^{\prime\prime}|^{2}\,dx\leq
C$

for all large enough $n$. It follows that

$\int_{a+\delta}^{b-\delta}\varepsilon_{n}^{-1}(u_{n}^{2}-1)^{2}+\varepsilon_{n}|u_{n}^{\prime}|^{2}\,dx\leq
2C$

for all such $n$, and so by item (i) of Theorem 3.1 applied to the interval
$(a+\delta,b-\delta)$, we may find a subsequence $u_{n_{k}}$ with
$u_{n_{k}}\to v$ in $L^{p}(a+\delta,b-\delta)$ for some $v\in
BPV((a+\delta,b-\delta);\pm 1)$, and moreover by item (ii) of Theorem 3.1 we
must have that
$\operatorname*{essVar}_{(a+\delta,b-\delta)}v\leq\frac{3}{2}C$. Hence the
claim is proven with $C_{3}=\frac{3}{2}C$.

We finish with a diagonalization argument. Let
$\Omega_{m}=\left(a+\delta_{m},b-\delta_{m}\right)$ for a sequence
$\delta_{m}\to 0^{+}$. For each $m$ we define a subsequence $u_{m,n}$ of
$u_{n}$ recursively as follows:

  * •

$u_{1,n}=u_{n}$ for all $n$.

  * •

For $m\geq 2$, we take $\\{u_{m,n}\\}_{n}$ to be a subsequence of
$\\{u_{m-1,n}\\}_{n}$ that converges in $L^{p}(\Omega_{m})$ to some
$v_{m}:\Omega_{m}\to\\{-1,1\\}$ with
$\operatorname*{essVar}_{\Omega_{m}}v_{m}\leq C_{3}$, which exists by the
claim.

Let $w_{n}=u_{n,n}$. We claim that for some $u\in L^{p}(\Omega;\\{-1,1\\})$ we
have that $w_{n}\to u$ in $L^{p}(\Omega_{m})$ for every $m$, so that in
particular we have $w_{n}\to u$ in $L^{p}_{\text{loc}}(\Omega)$.

To see this, consider $m_{1},m_{2}$ with $m_{2}>m_{1}$, so that
$\\{u_{m_{2},n}\\}_{n}$ is a subsequence of $\\{u_{m_{1},n}\\}_{n}$. Then we
may write $u_{m_{2},k}=u_{m_{1},n_{k}}$ for some sequence $n_{k}$, and so

$\|v_{m_{2}}-v_{m_{1}}\|_{L^{p}(\Omega_{m_{1}})}\leq\|v_{m_{2}}-u_{m_{2},k}\|_{L^{p}(\Omega_{m_{1}})}+\|u_{m_{1},n_{k}}-v_{m_{1}}\|_{L^{p}(\Omega_{m_{1}})}.$

Sending $k\to+\infty$ we deduce that $v_{m_{2}}=v_{m_{1}}$ almost everywhere.
over $\Omega_{m_{1}}$, so in general $v_{m_{2}}$ is an extension of
$v_{m_{1}}$ for all $m_{2}$ and $m_{1}$ with $m_{2}>m_{1}$. Take $u$ to be the
maximal such extension, defined over all of $\Omega$. Then for any $m$, we
have that

$\|w_{n}-u\|_{L^{p}(\Omega_{m})}=\|u_{n,n}-v_{m}\|_{L^{p}(\Omega_{m})}\xrightarrow{n\to+\infty}0$

because $\\{u_{n,n}\\}_{n=m}^{\infty}$ is a subsequence of
$\\{u_{m,n}\\}_{n}$, which converges to $v_{m}$ in $L^{p}(\Omega_{m})$.

This proves the claim. In particular, we see that $w_{n}\to u$ in measure.
Moreover the family $\\{|w_{n}|^{p}\\}_{n}$ is uniformly integrable by Lemma
3.3. Thus by Vitali, we get that in fact, $w_{n}\to u$ in $L^{p}(\Omega)$.

Lastly, to see that $u\in BPV(\Omega)$, note that

$\operatorname*{essVar}_{\Omega}u=\lim_{m\to\infty}\operatorname*{essVar}_{\Omega_{m}}u=\lim_{m\to\infty}\operatorname*{essVar}_{\Omega_{m}}v_{m}\leq
C_{3},$

so we are done. ∎

We now find the Gamma-limit under $L^{p}(\Omega)$ convergence.

######  Theorem 3.5.

Let $\Omega=(a,b)$ be a non-empty open interval, and let $1\leq p<\infty$. For
each $\varepsilon>0$, define a functional
$F_{\varepsilon}:L^{p}(\Omega)\to\overline{\mathbb{R}}$ as in (10). Let
$\varepsilon_{n}$ be a sequence of positive reals with $\varepsilon_{n}\to
0^{+}$. For all $u\in L^{p}(\Omega)$, we have

$\Gamma\text{-}\lim_{n\to+\infty}F_{\varepsilon_{n}}(u)=\begin{cases}\alpha\operatorname*{essVar}_{\Omega}u,&u\in
BPV(\Omega;\pm 1)\\\ +\infty,&\text{otherwise}\end{cases}$

under $L^{p}(\Omega)$ convergence, where $\alpha$ is defined as in (8).

We begin with a useful interpolation result.

######  Lemma 3.6.

Let $T>0$ and $A,m\in\mathbb{R}$ with $m\neq 0$. Then there exists
$f:[0,T]\to\mathbb{R}$ satisfying the following properties:

  1. (i)

$f\in C^{\infty}([0,T])$,

  2. (ii)

$f(0)=A$, $f^{\prime}(0)=m$, $f(T)=0$, and $f^{\prime}(T)=0$,

  3. (iii)

$\int_{0}^{T}|f^{(k)}(x)|^{2}\,dx\leq C_{k}(A^{2}+m^{2}T^{2})T^{1-2k}$ for a
constant $C_{k}$ depending only on $k$,

  4. (iv)

$\sup_{[0,T]}|f|\leq|A|+\frac{1}{2}|m|\cdot|T|$.

###### Proof.

We take $g:\mathbb{R}\to\mathbb{R}$ with

$g(x)=\begin{cases}mx+A,&-T/2<x<T/2\\\ 0,&\text{otherwise}\end{cases}.$

Now let $\varphi\in C^{\infty}_{c}((-1,1))$ be a non-negative and symmetric
mollifier, so that $\varphi(x)=\varphi(-x)$ for all $x\in(-1,1)$. For each
$\varepsilon>0$ we define
$\varphi_{\varepsilon}(x):=\frac{1}{\varepsilon}\varphi\left(\frac{x}{\varepsilon}\right)$
and the mollification

$g_{\varepsilon}(x):=\int_{x-\varepsilon}^{x+\varepsilon}g(y)\varphi_{\varepsilon}(x-y)\,dy.$

For $\varepsilon<T/2$ we recover the properties $g_{\varepsilon}(0)=A$,
$g^{\prime}_{\varepsilon}(0)=m$, $g_{\varepsilon}(T)=0$, and
$g_{\varepsilon}^{\prime}(T)=0$, because by symmetry of
$\varphi_{\varepsilon}$ we have that $g_{\varepsilon}(x)=mx+A$ for $x$ near
$0$ and $g_{\varepsilon}(x)=0$ for $x$ near $T$. Note that for each $k$, we
have that
$|g(y)\varphi_{\varepsilon}^{(k)}(x-y)|\leq\varepsilon^{-k}\|\varphi^{(k)}\|_{\infty}|g(y)|$
for all $x,y\in\mathbb{R}$. As $g$ is integrable, we may use dominated
convergence to obtain

$\displaystyle g^{(k)}_{\varepsilon}(x)$
$\displaystyle=\frac{d^{k}}{{dx}^{k}}\int_{x-\varepsilon}^{x+\varepsilon}g(y)\varphi_{\varepsilon}(x-y)\,dy=\frac{d^{k}}{{dx}^{k}}\int_{\mathbb{R}}g(y)\varphi_{\varepsilon}(x-y)\,dy$
$\displaystyle=\int_{\mathbb{R}}g(y)\varphi_{\varepsilon}^{(k)}(x-y)\,dy=\int_{x-\varepsilon}^{x+\varepsilon}g(y)\varphi_{\varepsilon}^{(k)}(x-y)\,dy.$

Now we integrate in $x$ and apply Minkowski’s inequality for integrals to get

$\displaystyle\int_{0}^{T}g_{\varepsilon}^{(k)}(x)^{2}\,dx$
$\displaystyle=\int_{0}^{T}\left(\int_{x-\varepsilon}^{x+\varepsilon}g(y)\varphi_{\varepsilon}^{(k)}(x-y)\,dy\right)^{2}\,dx$
$\displaystyle\leq\left(\int_{-\varepsilon}^{T+\varepsilon}\left(\int_{(y-\varepsilon,y+\varepsilon)\cap[0,T]}|g(y)|^{2}|\varphi_{\varepsilon}^{(k)}(x-y)|^{2}\,dx\right)^{1/2}\,dy\right)^{2}$
$\displaystyle\leq\left(\int_{-\varepsilon}^{T+\varepsilon}|g(y)|\left(\int_{y-\varepsilon}^{y+\varepsilon}|\varphi_{\varepsilon}^{(k)}(x-y)|^{2}\,dx\right)^{1/2}\,dy\right)^{2}$
$\displaystyle=\left(\int_{-\varepsilon}^{T+\varepsilon}|g(y)|\,dy\right)^{2}\int_{-\varepsilon}^{\varepsilon}|\varphi_{\varepsilon}^{(k)}(z)|^{2}\,dz.$

Handling the first integral is a matter of computation, which gets us the
bound

$\int_{-\varepsilon}^{T+\varepsilon}|g(y)|\,dy\leq\int_{-\varepsilon}^{T+\varepsilon}|m|\cdot|y|+|A|\,dy=\frac{|m|}{2}\cdot\left((T+\varepsilon)^{2}+\varepsilon^{2}\right)+(T+2\varepsilon)|A|.$

As for the second integral, we may write
$\varphi_{\varepsilon}^{(k)}(x)=\frac{d^{k}}{{dx}^{k}}\frac{1}{\varepsilon}\varphi\left(\frac{x}{\varepsilon}\right)=\frac{1}{\varepsilon^{k+1}}\varphi^{(k)}\left(\frac{x}{\varepsilon}\right)$,
so that

$\int_{-\varepsilon}^{\varepsilon}|\varphi_{\varepsilon}^{(k)}(z)|^{2}\,dz=\int_{-\varepsilon}^{\varepsilon}\frac{1}{\varepsilon^{2k+2}}|\varphi^{(k)}(z/\varepsilon)|^{2}\,dz=\frac{1}{\varepsilon^{2k+1}}\int_{-1}^{1}|\varphi^{(k)}(z)|^{2}\,dz.$

Altogether, we get that

$\int_{0}^{T}g_{\varepsilon}^{(k)}(x)^{2}\,dx\leq\frac{\left(\frac{|m|}{2}\cdot\left((T+\varepsilon)^{2}+\varepsilon^{2}\right)+(T+2\varepsilon)|A|\right)^{2}}{\varepsilon^{2k+1}}\cdot\|\varphi^{(k)}\|_{L^{2}(\mathbb{R})}^{2}.$

Choosing $\varepsilon=T/4$, and applying the inequality $(a+b)^{2}\leq
2a^{2}+2b^{2}$, we see that

$\int_{0}^{T}g_{T/4}^{(k)}(x)^{2}\,dx\leq\frac{|m|^{2}T^{4}+|A|^{2}T^{2}}{T^{2k+1}}\cdot
C_{k}=C_{k}(|A|^{2}+|m|^{2}T^{2})T^{1-2k}$

for a constant $C_{k}$ depending only on $k$.

Lastly, since $|g(x)|\leq|m|\cdot|x|+|A|\leq|m|\cdot\frac{|T|}{2}+|A|$ for all
$x$, we must have

$|g_{T/4}(x)|\leq\int_{x-\varepsilon}^{x+\varepsilon}|g(y)|\varphi_{\varepsilon}(x-y)\,dy\leq\left(|m|\cdot\frac{|T|}{2}+|A|\right)\int_{x-\varepsilon}^{x+\varepsilon}\varphi_{\varepsilon}(x-y)\,dy=|m|\cdot\frac{|T|}{2}+|A|$

for all $x$, so that in particular
$\sup_{[0,T]}|g_{T/4}(x)|\leq|m|\cdot\frac{|T|}{2}+|A|$. We conclude that
taking $f\equiv g_{T/4}$ satisfies the required properties. ∎

Now we prove the theorem.

###### Proof.

Step 1

If $u\not\in L^{p}(\Omega;\pm 1)$ then for some $\varepsilon>0$, the set
$E=\\{x\in\Omega:||u(x)|-1|\geq\varepsilon\\}$ has positive measure. We now
claim that the inequality $(z^{2}-1)^{2}\geq\varepsilon^{2-p}||z|-1|^{p}$
holds for all $z$ with $||z|-1|\geq\varepsilon$. To see this, note that for
all such $z$ with $z\geq 0$ we have

$|z+1|^{2}\cdot|z-1|^{2-p}\geq\varepsilon^{2-p}$

which rearranges to the claimed inequality, and similarly for $z\leq 0$. Hence
we may use this and Minkowski’s inequality to write

$\displaystyle\int_{E}\varepsilon_{n}^{-1}(u_{n}^{2}-1)^{2}\,dx$
$\displaystyle\geq\int_{E}\varepsilon_{n}^{-1}\varepsilon^{2-p}||u_{n}|-1|^{p}\,dx$
$\displaystyle\geq\varepsilon_{n}^{-1}\varepsilon^{2-p}\left(\left(\int_{E}||u|-1|^{p}\,dx\right)^{1/p}-\left(\int_{E}\big{|}|u|-|u_{n}|\big{|}^{p}\,dx\right)^{1/p}\right)^{p}$
$\displaystyle\geq\varepsilon_{n}^{-1}\varepsilon^{2-p}\left(\varepsilon\mathcal{L}^{1}(E)^{1/p}-\left(\int_{E}|u-u_{n}|^{p}\,dx\right)^{1/p}\right)^{p}.$

Now it is clear that $F_{\varepsilon_{n}}(u_{n})\to+\infty$ because
$\|u-u_{n}\|_{L^{p}(E)}\to 0$ and $\varepsilon_{n}^{-1}\to+\infty$. Hence
$\Gamma\text{-}\lim_{n\to\infty}F_{\varepsilon_{n}}(u)=+\infty$.

Step 2

We now consider the case $u\in BPV(\Omega;\pm 1)$.

Suppose $u$ jumps between $-1$ and $1$ or vice versa at the $J$ points
$x_{1},x_{2},\cdots,x_{J}\in\Omega$. Find $\delta_{1},\cdots,\delta_{J}>0$ so
small that the intervals $(a_{i},b_{i})$ are disjoint, where
$(a_{i},b_{i}):=(x_{i}-\delta_{i},x_{i}+\delta_{i})$.

Let $u_{n}\to u$ in $L^{p}(\Omega)$. Suppose
$L=\displaystyle\liminf_{n\to\infty}F_{\varepsilon_{n}}(u_{n})$. For the
liminf inequality, we need to prove that $L\geq\alpha\operatorname*{essVar}u$.

By extraction of a subsequence, assume that actually
$L=\displaystyle\lim_{n\to\infty}F_{\varepsilon_{n}}(u_{n})$. Fix an arbitrary
$\eta\in(0,1)$. Then we may find $N_{\eta}$ so large that

$L+\eta\geq F_{\varepsilon_{n}}(u_{n})$

for all $n\geq N_{\eta}$.

Pick some $x_{i}$ and, for simplicity, assume without loss of generality that
$u$ jumps from $-1$ to $1$ at $x_{i}$.

The key claim that we shall prove in this step is that, for all $n$ large
enough, we may modify $u_{n}$ in $(a_{i},b_{i})$ by “anchoring" it to $-1$ at
$a_{i}$ and to $1$ at $b_{i}$ with a derivative of $0$ at both points, without
spending more than $C\eta$ potential energy for a constant $C$ not depending
on $\eta$. More concretely, there exists $v_{n,i}:[a_{i},b_{i}]\to\mathbb{R}$
such that:

  * •

$v_{n,i}\in W^{2,p}((a_{i},b_{i}))$,

  * •

$v_{n,i}(a_{i})=-1$, $v_{n,i}(b_{i})=1$,

  * •

$v_{n,i}^{\prime}(a_{i})=v_{n,i}^{\prime}(b_{i})=0$,

  * •

$\displaystyle\int_{a_{i}}^{b_{i}}\varepsilon_{n}^{-1}(v_{n,i}^{2}-1)^{2}+\varepsilon_{n}^{3}|v_{n,i}^{\prime\prime}|^{2}\,dx\leq\int_{a_{i}}^{b_{i}}\varepsilon_{n}^{-1}(u_{n}^{2}-1)^{2}+\varepsilon_{n}^{3}|u_{n}^{\prime\prime}|^{2}\,dx+C\eta$.

To do this, we consider only $n\geq N_{\eta}$, and it suffices to modify
$u_{n}$ in $(x_{i},b_{i})$ by affixing to $1$ in the manner described. First,
observe that since $u_{n}\to u$ in $L^{p}(\Omega)$, we have in particular that
$u_{n}\to 1$ in $L^{p}((x_{i},b_{i}))$, so
$\|u_{n}-1\|_{L^{p}((x_{i},b_{i}))}\to 0$ as $n\to+\infty$. Putting this
aside, note that we also have
$\int_{x_{i}}^{b_{i}}\frac{|u_{n}-1|^{p}}{\|u_{n}-1\|_{L^{p}((x_{i},b_{i}))}^{p}}\,dx=1$.
Combining this with $F_{\varepsilon_{n}}(u_{n})\leq L+\eta$, it follows that

$\int_{x_{i}}^{b_{i}}\varepsilon_{n}^{-1}(u_{n}^{2}-1)^{2}+\varepsilon_{n}^{3}|u_{n}^{\prime\prime}|^{2}+\frac{|u_{n}-1|^{p}}{\|u_{n}-1\|_{L^{p}((x_{i},b_{i}))}^{p}}\,dx\leq
L+\eta+1.$

Now let $K_{n}=\lceil\varepsilon_{n}^{-1}\rceil$ and subdivide the interval
$(x_{i},b_{i})$ into $K_{n}$ same-length intervals $(y_{j-1},y_{j})$, so that:

$x_{i}=y_{0}<y_{1}<\cdots<y_{K_{n}}=b_{i}$

By a “pigeonhole principle-like argument", there exists $1\leq j\leq K_{n}$
such that

$\int_{y_{j-1}}^{y_{j}}\varepsilon_{n}^{-1}(u_{n}^{2}-1)^{2}+\varepsilon_{n}^{3}|u_{n}^{\prime\prime}|^{2}+\frac{|u_{n}-1|^{p}}{\|u_{n}-1\|_{L^{p}((x_{i},b_{i}))}^{p}}\,dx\leq\frac{L+\eta+1}{K_{n}}\leq\varepsilon_{n}(L+\eta+1).$
(22)

A particular consequence is the bound
$\int_{y_{j-1}}^{y_{j}}\varepsilon_{n}^{-1}(u_{n}^{2}-1)^{2}+\varepsilon_{n}^{3}|u_{n}^{\prime\prime}|^{2}\,dx\leq\frac{L+\eta+1}{K_{n}}\leq\varepsilon_{n}(L+\eta+1)$,
which implies by Theorem 3.2 (applied with the intervals in the inclusion
$(\frac{2}{3}y_{j-1}+\frac{1}{3}y_{j},\frac{1}{3}y_{j-1}+\frac{2}{3}y_{j})\Subset(y_{j-1},y_{j})$)
that

$\int_{\frac{2}{3}y_{j-1}+\frac{1}{3}y_{j}}^{\frac{1}{3}y_{j-1}+\frac{2}{3}y_{j}}\varepsilon_{n}|u_{n}^{\prime}|^{2}\,dx\leq
c\int_{y_{j-1}}^{y_{j}}\varepsilon_{n}^{-1}(u_{n}^{2}-1)^{2}+\varepsilon_{n}^{3}|u_{n}^{\prime\prime}|^{2}\,dx\leq\varepsilon_{n}c(L+\eta+1)$

for all sufficiently large $n$, for a universal constant $c$. Combining this
with (22), we obtain

$\int_{\frac{2}{3}y_{j-1}+\frac{1}{3}y_{j}}^{\frac{1}{3}y_{j-1}+\frac{2}{3}y_{j}}\varepsilon_{n}|u_{n}^{\prime}|^{2}+\frac{|u_{n}-1|^{p}}{\|u_{n}-1\|_{L^{p}((x_{i},b_{i}))}^{p}}\,dx\leq\varepsilon_{n}(c+1)(L+\eta+1).$

Let the integrand
$\varepsilon_{n}|u_{n}^{\prime}|^{2}+\frac{|u_{n}-1|^{p}}{\|u_{n}-1\|_{L^{p}((x_{i},b_{i}))}^{p}}$
be $H$. If there does not exist
$x\in(\frac{2}{3}y_{j-1}+\frac{1}{3}y_{j},\frac{1}{3}y_{j-1}+\frac{2}{3}y_{j})$
for which $H(x)\leq\frac{6K_{n}\varepsilon_{n}(c+1)(L+\eta+1)}{b_{i}-x_{i}}$,
then

$\varepsilon_{n}(c+1)(L+\eta+1)\geq\int_{\frac{2}{3}y_{j-1}+\frac{1}{3}y_{j}}^{\frac{1}{3}y_{j-1}+\frac{2}{3}y_{j}}H\,dx\geq\frac{6K_{n}\varepsilon_{n}(c+1)(L+\eta+1)}{b_{i}-x_{i}}\cdot\frac{y_{j}-y_{j-1}}{3}$
$=\frac{6K_{n}\varepsilon_{n}(c+1)(L+\eta+1)}{b_{i}-x_{i}}\cdot\frac{b_{i}-x_{i}}{3K_{n}}=2\varepsilon_{n}(c+1)(L+\eta+1),$

which is a contradiction. We conclude that for all large enough $n$, there
exists
$x_{0}\in(\frac{2}{3}y_{j-1}+\frac{1}{3}y_{j},\frac{1}{3}y_{j-1}+\frac{2}{3}y_{j})$,
depending on $n$, such that
$H(x_{0})\leq(C_{1}/2)K_{n}\varepsilon_{n}\leq(C_{1}/2)\cdot\frac{\varepsilon_{n}+1}{\varepsilon_{n}}\cdot\varepsilon_{n}=(C_{1}/2)(\varepsilon_{n}+1)\leq
C_{1}$ for a constant $C_{1}$ that does not depend on $n$. In particular, we
now know that such an $x_{0}$ satisfies:

  1. (1)

$\varepsilon_{n}^{2}|u_{n}^{\prime}(x_{0})|^{2}\leq\varepsilon_{n}C_{1}$,

  2. (2)

$|u_{n}(x_{0})-1|\leq C_{1}^{1/p}\|u_{n}-1\|_{L^{p}((x_{i},b_{i}))}$.

These properties, combined with the fact that $u_{n}\to 1$ in
$L^{p}(x_{i},b_{i})$, imply that, for all large enough $n$, we may obtain
$\varepsilon_{n}^{2}|u_{n}^{\prime}(x_{0})|^{2}+|u_{n}(x_{0})-1|^{2}\leq\eta$.

Let $A=u_{n}(x_{0})-1$, $m=u_{n}^{\prime}(x_{0})$, and $T=y_{j}-x_{0}$. Using
these constants, we may find a smooth $f:[0,T]\to\mathbb{R}$ as described in
Lemma 3.6. We are now ready to define $v_{n,i}$ over $(x_{i},b_{i})$ as

$v_{n,i}(x):=\begin{cases}u_{n}(x),&x_{i}<x\leq x_{0}\\\
f(x-x_{0})+1,&x_{0}<x\leq y_{j}\\\ 1,&y_{j}<x<b_{i}\end{cases}.$

By virtue of $f$ being a smooth connector, we must have $v_{n,i}\in
W^{2,p}(x_{i},b_{i})$. Moreover, by property (iii) in Lemma 3.6, we have the
bounds

$\int_{x_{i}}^{b_{i}}(v_{n,i}-1)^{2}\,dx=\int_{x_{0}}^{y_{j}}|f(x-x_{0})|^{2}\,dx\leq
C_{0}(A^{2}+m^{2}T^{2})T$ (23)

and

$\int_{x_{i}}^{b_{i}}|v_{n,i}^{\prime\prime}|^{2}\,dx=\int_{x_{0}}^{y_{j}}|f^{\prime\prime}(x-x_{0})|^{2}\,dx\leq
C_{2}(A^{2}+m^{2}T^{2})T^{-3}$ (24)

for universal constants $C_{0},C_{2}>0$.

Since (23) is not quite a bound on the integral of $(v_{n,i}^{2}-1)^{2}$, we
will need to prove the inequality

$\int_{x_{i}}^{b_{i}}(v_{n,i}^{2}-1)^{2}\,dx\leq
C^{\prime}\int_{x_{i}}^{b_{i}}(v_{n,i}-1)^{2}\,dx$ (25)

for all $n$ large enough, for some constant $C^{\prime}>0$. Indeed, observe
that for all $x\in(x_{i},b_{i})$, we have by property (iv) of Lemma 3.6 that

$|v_{n,i}(x)+1|\leq 2+|v_{n,i}-1|\leq 2+\sup_{[0,T]}|f|\leq
2+|A|+\frac{1}{2}|m|\cdot|T|.$

Now, $T\leq\varepsilon_{n}$, and from
$\varepsilon_{n}^{2}|u_{n}^{\prime}(x_{0})|^{2}+|u_{n}(x_{0})-1|^{2}\leq\eta\leq
1$ we have that $|A|=|u_{n}(x_{0})-1|\leq 1$ and
$|m|\cdot|T|=|u_{n}^{\prime}(x_{0})|\cdot(y_{j}-x_{0})\leq\varepsilon_{n}|u_{n}^{\prime}(x_{0})|\leq
1$. We hence obtain $|v_{n,i}(x)+1|\leq 4$ for all $x\in(x_{i},b_{i})$, and so
taking $C^{\prime}=4^{2}$ we get

$\int_{x_{i}}^{b_{i}}(v_{n,i}^{2}-1)^{2}\,dx=\int_{x_{i}}^{b_{i}}(v_{n,i}-1)^{2}(v_{n,i}+1)^{2}\,dx\leq
C^{\prime}\int_{x_{i}}^{b_{i}}(v_{n,i}-1)^{2},$

as we wanted.

We may now add (23) to (24) and apply (25) to obtain

$\displaystyle\phantom{{}={}}\int_{x_{0}}^{b_{i}}\varepsilon_{n}^{-1}(v_{n,i}^{2}-1)^{2}+\varepsilon_{n}^{3}|v_{n,i}^{\prime\prime}|^{2}\,dx$
$\displaystyle\leq\max(C_{0},C_{2})\max(1,C^{\prime})\left[\frac{T}{\varepsilon_{n}}(A^{2}+m^{2}T^{2})+\frac{\varepsilon_{n}^{3}}{T^{3}}(A^{2}+m^{2}T^{2})\right]$
$\displaystyle=C_{3}(A^{2}+m^{2}T^{2})\left(\frac{T}{\varepsilon_{n}}+\frac{\varepsilon_{n}^{3}}{T^{3}}\right)$

for some constant $C_{3}>0$.

Since
$\frac{2}{3}y_{j-1}+\frac{1}{3}y_{j}<x_{0}<\frac{1}{3}y_{j-1}+\frac{2}{3}y_{j}$,
we have that
$(b_{i}-x_{i})\varepsilon_{n}\geq\frac{b_{i}-x_{i}}{K_{n}}=y_{j}-y_{j-1}>y_{j}-x_{0}=T>y_{j}-(\frac{1}{3}y_{j-1}+\frac{2}{3}y_{j})=\frac{1}{3}(y_{j}-y_{j-1})=\frac{b_{i}-x_{i}}{3K_{n}}\geq\frac{b_{i}-x_{i}}{3}\cdot\frac{\varepsilon_{n}}{1+\varepsilon_{n}}$.
So we may use this to write

$\displaystyle\phantom{{}={}}\int_{x_{0}}^{b_{i}}\varepsilon_{n}^{-1}(v_{n,i}^{2}-1)^{2}+\varepsilon_{n}^{3}|v_{n,i}^{\prime\prime}|^{2}\,dx$
$\displaystyle\leq
C_{3}((u_{n}(x_{0})-1)^{2}+|u_{n}^{\prime}(x_{0})|^{2}T^{2})\left(\frac{T}{\varepsilon_{n}}+\frac{\varepsilon_{n}^{3}}{T^{3}}\right)$
$\displaystyle\leq
C_{3}((u_{n}(x_{0})-1)^{2}+|u_{n}^{\prime}(x_{0})|^{2}(b_{i}-x_{i})^{2}\varepsilon_{n}^{2})\left(\frac{(b_{i}-x_{i})\varepsilon_{n}}{\varepsilon_{n}}+\varepsilon_{n}^{3}\cdot\frac{27(1+\varepsilon_{n})^{3}}{(b_{i}-x_{i})^{3}\varepsilon_{n}^{3}}\right)$
$\displaystyle\leq
C_{3}((u_{n}(x_{0})-1)^{2}+|u_{n}^{\prime}(x_{0})|^{2}(b_{i}-x_{i})^{2}\varepsilon_{n}^{2})\left(b_{i}-x_{i}+\frac{27\cdot
8}{(b_{i}-x_{i})^{3}}\right)$ $\displaystyle\leq
C_{4}((u_{n}(x_{0})-1)^{2}+|u_{n}^{\prime}(x_{0})|^{2}\varepsilon_{n}^{2})\leq
C_{4}\eta,$

Where $C_{4}>0$ is a constant with no dependence on $n$, and we have applied
our choice of $x_{0}$. At last, it follows that

$\displaystyle\phantom{{}={}}\int_{x_{i}}^{b_{i}}\varepsilon_{n}^{-1}(v_{n,i}^{2}-1)^{2}+\varepsilon_{n}^{3}|v_{n,i}^{\prime\prime}|^{2}\,dx$
$\displaystyle=\int_{x_{i}}^{x_{0}}\varepsilon_{n}^{-1}(v_{n,i}^{2}-1)^{2}+\varepsilon_{n}^{3}|v_{n,i}^{\prime\prime}|^{2}\,dx+\int_{x_{0}}^{b_{i}}\varepsilon_{n}^{-1}(v_{n,i}^{2}-1)^{2}+\varepsilon_{n}^{3}|v_{n,i}^{\prime\prime}|^{2}\,dx$
$\displaystyle=\int_{x_{i}}^{x_{0}}\varepsilon_{n}^{-1}(u_{n}^{2}-1)^{2}+\varepsilon_{n}^{3}|u_{n}^{\prime\prime}|^{2}\,dx+\int_{x_{0}}^{b_{i}}\varepsilon_{n}^{-1}(v_{n,i}^{2}-1)^{2}+\varepsilon_{n}^{3}|v_{n,i}^{\prime\prime}|^{2}\,dx$
$\displaystyle=\int_{x_{i}}^{x_{0}}\varepsilon_{n}^{-1}(u_{n}^{2}-1)^{2}+\varepsilon_{n}^{3}|u_{n}^{\prime\prime}|^{2}\,dx+C_{4}\eta$
$\displaystyle\leq\int_{x_{i}}^{b_{i}}\varepsilon_{n}^{-1}(u_{n}^{2}-1)^{2}+\varepsilon_{n}^{3}|u_{n}^{\prime\prime}|^{2}\,dx+C_{4}\eta.$

Doing the same thing for the interval $(a_{i},x_{i})$, we finally obtain

$\int_{a_{i}}^{b_{i}}\varepsilon_{n}^{-1}(v_{n,i}^{2}-1)^{2}+\varepsilon_{n}^{3}|v_{n,i}^{\prime\prime}|^{2}\,dx\leq\int_{a_{i}}^{b_{i}}\varepsilon_{n}^{-1}(u_{n}^{2}-1)^{2}+\varepsilon_{n}^{3}|u_{n}^{\prime\prime}|^{2}\,dx+2C_{4}\eta.$
(26)

Thus the key claim has been proven.

Step 3

We may now complete the liminf argument. We recall the definitions of the
family $\mathscr{J}$ and the constant $\alpha$ from $\eqref{eq:famJ}$ and
$\eqref{eq:alpha}$ respectively.

Let us first write

$L+\eta\geq
F_{\varepsilon_{n}}(u_{n})\geq\sum_{i=1}^{J}\int_{a_{i}}^{b_{i}}\varepsilon_{n}^{-1}(u_{n}^{2}-1)^{2}+\varepsilon_{n}^{3}|u_{n}^{\prime\prime}|^{2}\,dx$

for all large enough $n$. We now apply the key claim (26) to every interval
$(a_{i},b_{i})$ to get

$L+\eta\geq-2C_{4}J\eta+\sum_{i=1}^{J}\int_{a_{i}}^{b_{i}}\varepsilon_{n}^{-1}(v_{n,i}(x)^{2}-1)^{2}+\varepsilon_{n}^{3}|v_{n,i}^{\prime\prime}(x)|^{2}\,dx.$

We now apply the change of variables $y=\frac{x-a_{i}}{b_{i}-a_{i}}$. Defining
$w_{n,i}:[0,1]\to\mathbb{R}$ via
$w_{n,i}(y)=v_{n,i}\left((b_{i}-a_{i})y+a_{i}\right)$, we have that
$w_{n,i}^{\prime\prime}(y)=(b_{i}-a_{i})^{2}v_{n,i}^{\prime\prime}\left((b_{i}-a_{i})y+a_{i}\right)$,
so that we get

$\displaystyle L+\eta$
$\displaystyle\geq-2C_{4}J\eta+\sum_{i=1}^{J}(b_{i}-a_{i})\int_{0}^{1}\varepsilon_{n}^{-1}(w_{n,i}(y)^{2}-1)^{2}+\frac{\varepsilon_{n}^{3}}{(b_{i}-a_{i})^{4}}|w_{n,i}^{\prime\prime}(y)|^{2}\,dy$
$\displaystyle=-2C_{4}J\eta+\sum_{i=1}^{J}\frac{b_{i}-a_{i}}{\varepsilon_{n}}\int_{0}^{1}(w_{n,i}(y)^{2}-1)^{2}\,dy+\frac{\varepsilon_{n}^{3}}{(b_{i}-a_{i})^{3}}\int_{0}^{1}|w_{n,i}^{\prime\prime}(y)|^{2}\,dy,$

and by applying the weighted AM-GM inequality we may go down again to get

$\displaystyle L+\eta$
$\displaystyle\geq-2C_{4}J\eta+\sum_{i=1}^{J}\frac{4}{3^{3/4}}\left(\int_{0}^{1}(w_{n,i}(y)^{2}-1)^{2}\,dy\right)^{3/4}\left(\int_{0}^{1}|w_{n,i}^{\prime\prime}(y)|^{2}\,dy\right)^{1/4}$
$\displaystyle=-2C_{4}J\eta+\sum_{i=1}^{J}\frac{4}{3^{3/4}}\Phi(w_{n,i}).$

Lastly, as $w_{n,i}\in\mathscr{J}$ for all $i$, we have that
$\frac{4}{3^{3/4}}\Phi(w_{n,i})\geq\frac{4}{3^{3/4}}\inf_{w\in\mathscr{J}}\Phi(w)=2\alpha$,
thus

$\displaystyle L+\eta$ $\displaystyle\geq-2C_{4}J\eta+\sum_{i=1}^{J}2\alpha$
$\displaystyle=-2C_{4}J\eta+2J\alpha=-2C_{4}J\eta+\alpha\operatorname*{essVar}_{\Omega}u.$

Hence $L+\eta\geq-2C_{4}J\eta+\alpha\operatorname*{essVar}_{\Omega}u$. As
$\eta$ was arbitrary, we conclude that
$L\geq\alpha\operatorname*{essVar}_{\Omega}u$, as needed. This proves the
liminf inequality for $u\in BPV(\Omega;\pm 1)$.

Step 4

In this step, we briefly resolve the case $u\in L^{2}(\Omega;\pm 1)\setminus
BPV(\Omega;\pm 1)$.

Showing that $F_{\varepsilon_{n}}(u_{n})\to+\infty$ in this case (which is
sufficient to conclude the Gamma limit) requires modification of the argument
in Steps 2 and 3. We sketch the proof: Find $2N$ Lebesgue points
$a<x_{1}<y_{1}<\cdots<x_{N}<y_{N}<b$ for which $u(x_{i})=-1$ and $u(y_{i})=1$.
For each $i$ we may find an interval $(x_{i},x_{i}+\delta)$ for which $u=-1$
in most of $(x_{i},\delta)$ and an interval $(y_{i}-\delta,y_{i})$ for which
$u=1$ in most of $(y_{i}-\delta,y_{i})$. We can then apply the pigeonhole
argument in these intervals by using the fact that
$\|u_{n}+1\|_{L^{p}((x_{i},x_{i}+\delta)\cap\\{u=-1\\}}\to 0$ and
$\|u_{n}-1\|_{L^{p}((y_{i}-\delta,y_{i})\cap\\{u=1\\}}\to 0$. In the end, we
obtain a lower bound for $F_{\varepsilon_{n}}(u_{n})$ that tends to $+\infty$
when we send $N\to+\infty$.

Step 5

The remaining thing to prove is the limsup inequality for $u\in BPV(\Omega;\pm
1)$. This entails finding a sequence $u_{n}\in W^{2,p}(\Omega)$ for which
$u_{n}\to u$ in $L^{p}(\Omega)$ and
$\limsup_{n\to\infty}F_{\varepsilon_{n}}(u_{n})\leq\alpha\operatorname*{Var}_{\Omega}u$.

First, recall the definition of $\Phi$ as in (5) and the family $\mathscr{J}$
as in (6). Define the subfamily

$\mathscr{J}_{n}:=\left\\{h\in\mathscr{J}:\frac{\int_{0}^{1}|h^{\prime\prime}|^{2}\,dx}{\int_{0}^{1}(h^{2}-1)^{2}\,dx}\leq\frac{1}{\varepsilon_{n}}\right\\}.$

Since $\varepsilon_{n}\to 0^{+}$, it is clear that
$\bigcup_{n=1}^{\infty}\mathscr{J}_{n}=\mathscr{J}$. Thus if we define the
sets

$S_{n}:=\left\\{\Phi(h):h\in\mathscr{J}_{n}\right\\},$
$S:=\left\\{\Phi(h):h\in\mathscr{J}\right\\},$

then we have $\bigcup_{n=1}^{\infty}S_{n}=S$, and it is not hard to show that
$\lim_{n\to\infty}\inf S_{n}=\inf S$. We note also that $\alpha$, as defined
in (8), may be written as $\alpha=\frac{2}{3^{3/4}}\inf S$.

For all $n$, find $h_{n}\in\mathscr{J}_{n}$ for which

$\inf S_{n}\leq\Phi(h_{n})\leq\frac{1}{n}+\inf S_{n}.$ (27)

As in Step 2, suppose $u$ “jumps" at the points
$x_{1},x_{2},\cdots,x_{J}\in\Omega$, and find pairwise disjoint intervals
$(a_{i},b_{i})\subseteq\Omega$ with $x_{i}\in(a_{i},b_{i})$. Consider some
$x_{i}$, and without loss of generality assume that $u$ jumps from $-1$ to $1$
at $x_{i}$. Let $\zeta_{n}\to 0$ be a sequence of positive reals that we shall
choose later. We now define $u_{n}$ over $(a_{i},b_{i})$ as

$u_{n}(x):=\begin{cases}-1,&a_{i}<x<x_{i}-\zeta_{n}/2\\\
h_{n}\left(\frac{x-x_{i}+\zeta_{n}/2}{\zeta_{n}}\right),&x_{i}-\zeta_{n}/2\leq
x\leq x_{i}+\zeta_{n}/2\\\ 1,&x_{i}+\zeta_{n}/2<x<b_{i}\end{cases}.$

We define $u_{n}$ similarly over all other intervals $(a_{j},b_{j})$, and in
$\Omega\setminus\bigcup_{j=1}^{J}(a_{j},b_{j})$ we let $u_{n}$ agree with $u$.
Since $h_{n}(0_{+})=-1$, $h_{n}(1_{-})=1$, and
$h_{n}^{\prime}(0_{+})=h_{n}^{\prime}(1_{-})=0$, we have that $u_{n}\in
W^{2,p}(\Omega)$. Moreover, $u_{n}\to u$ almost everywhere as a consequence of
$\zeta_{n}\to 0^{+}$, and $\\{|u_{n}|\\}_{n}$ is uniformly integrable, so by
Vitali we have $u_{n}\to u$ in $L^{p}(\Omega)$. It remains to verify the
limsup inequality. We may compute

$\displaystyle F_{\varepsilon_{n}}(u_{n})$
$\displaystyle=\sum_{i=1}^{J}\int_{a_{i}}^{b_{i}}\varepsilon_{n}^{-1}(u_{n}^{2}-1)^{2}+\varepsilon_{n}^{3}|u_{n}^{\prime\prime}|^{2}\,dx$
$\displaystyle=\sum_{i=1}^{J}\int_{a_{i}}^{b_{i}}\varepsilon_{n}^{-1}\left(h_{n}\left(\frac{x-x_{i}+\zeta_{n}/2}{\zeta_{n}}\right)^{2}-1\right)^{2}+\frac{\varepsilon_{n}^{3}}{\zeta_{n}^{4}}\left|h_{n}^{\prime\prime}\left(\frac{x-x_{i}+\zeta_{n}/2}{\zeta_{n}}\right)\right|^{2}\,dx,$

and after changing variables we get

$\displaystyle F_{\varepsilon_{n}}(u_{n})$
$\displaystyle=\sum_{i=1}^{J}\zeta_{n}\int_{0}^{1}\varepsilon_{n}^{-1}\left(h_{n}\left(y\right)^{2}-1\right)^{2}+\frac{\varepsilon_{n}^{3}}{\zeta_{n}^{4}}\left|h_{n}^{\prime\prime}\left(y\right)\right|^{2}\,dy$
$\displaystyle=\sum_{i=1}^{J}\frac{\zeta_{n}}{\varepsilon_{n}}\int_{0}^{1}\left(h_{n}\left(y\right)^{2}-1\right)^{2}\,dy+\frac{\varepsilon_{n}^{3}}{\zeta_{n}^{3}}\int_{0}^{1}\left|h_{n}^{\prime\prime}\left(y\right)\right|^{2}\,dy.$
(28)

We now choose $\zeta_{n}$ so that we obtain the equality case in the AM-GM
inequality. Specifically, we choose

$\zeta_{n}=\varepsilon_{n}\left(\frac{3\displaystyle\int_{0}^{1}\left|h_{n}^{\prime\prime}\left(y\right)\right|^{2}\,dy}{\displaystyle\int_{0}^{1}\left(h_{n}\left(y\right)^{2}-1\right)^{2}\,dy}\right)^{1/4}.$

Showing that this choice is valid for all sufficiently large $n$ amounts to
proving that $\zeta_{n}\to 0^{+}$. Fortunately, since
$h_{n}\in\mathscr{J}_{n}$, we have that

$\zeta_{n}\leq\varepsilon_{n}\left(\frac{3}{\varepsilon_{n}}\right)^{1/4}=3^{1/4}\varepsilon_{n}^{3/4}\to
0^{+}.$

With the choice of $\zeta_{n}$ justified, we may now continue the computation
in (28) by using the choice of $\zeta_{n}$ and (27) to write

$F_{\varepsilon_{n}}(u_{n})=\sum_{i=1}^{J}\frac{4}{3^{3/4}}\left(\int_{0}^{1}(h_{n}^{2}-1)^{2}\,dy\right)^{3/4}\left(\int_{0}^{1}|h_{n}^{\prime\prime}|^{2}\,dy\right)^{1/4}\leq\frac{4J}{3^{3/4}}\left(\frac{1}{n}+\inf
S_{n}\right).$

Taking the limsup, we obtain

$\limsup_{n\to\infty}F_{\varepsilon_{n}}(u_{n})\leq\limsup_{n\to\infty}\frac{4J}{3^{3/4}}\left(\frac{1}{n}+\inf
S_{n}\right)=\frac{2\operatorname*{Var}_{\Omega}u}{3^{3/4}}\left(\inf
S\right)=\alpha\operatorname*{Var}_{\Omega}u.$

This completes the proof. ∎

##  4\. Boundary Conditions

The Gamma limit will change upon restricting to the boundary conditions
$u(a_{+})=a_{\varepsilon}$ and $u(b_{-})=b_{\varepsilon}$ for
$a_{\varepsilon}\to-1$ and $b_{\varepsilon}\to 1$ as $\varepsilon\to 0^{+}$. A
portion of the work needed to account for the boundary conditions has already
been done in the proof of Theorem 3.5, but there is still much to be done.
Intuitively, every jump in the interior of $(a,b)$ induces a factor of
$\alpha$, whereas a jump at the boundary induces a factor of $\beta(t)$
depending on the height of the jump. Recall that $\beta(t)$ is defined as in
the statement of Theorem 1.1, and we will again reference the family
$\mathscr{J}^{\prime}(t)$ and the functional $\Phi$.

We first introduce a new family of functions $\mathscr{J}^{\prime}(t)$,
defined as

$\mathscr{J}^{\prime}_{\infty}(t):=\left\\{u\in
W^{2,\infty}(-\infty,0):u(0)=t\text{ and }u(x)=-1\ \forall x\leq-L\text{ for
some }L>0\right\\}$

for a parameter $t\in\mathbb{R}$. We associate with each
$u\in\mathscr{J}^{\prime}_{\infty}(t)$ the constant $L_{u}$, where $-L_{u}$ is
the first time that $u$ reaches $-1$ and remains at this value indefinitely.
That is,

$L_{u}:=\inf\left\\{L>0:u(x)=-1\ \forall x\leq-L\right\\}.$

We also define a new functional $\Psi:W^{2,\infty}(-\infty,0)\to\mathbb{R}$
via

$\Psi(u):=\int_{-\infty}^{0}(u^{2}-1)^{2}+|u^{\prime\prime}|^{2}\,dx.$

The relevance of these constructs is as follows.

######  Lemma 4.1.

$\beta(t)=\frac{4}{3^{3/4}}\inf_{u\in\mathscr{J}^{\prime}(t)}\Phi(u)=\inf_{v\in\mathscr{J}^{\prime}_{\infty}(t)}\Psi(v).$

###### Proof.

We first note that if $t=-1$ then clearly
$\inf_{u\in\mathscr{J}^{\prime}(t)}\Phi(u)=0$ and
$\inf_{u\in\mathscr{J}^{\prime}_{\infty}(t)}\Psi(u)=0$, so we may assume that
$t\neq-1$. Now consider $u\in\mathscr{J}^{\prime}(t)$. Then take $L>0$
depending only on $u$ that we shall choose later, and define
$v\in\mathscr{J}^{\prime}_{\infty}(t)$ via

$v(x)=\begin{cases}-1,&x\leq-L\\\ u(x/L+1),&-L<x<0\end{cases}.$

Then

$\displaystyle\Psi(v)$
$\displaystyle=\int_{-L}^{0}(v(x)^{2}-1)^{2}+|v^{\prime\prime}(x)|^{2}\,dx$
$\displaystyle=\int_{-L}^{0}(u(x/L+1)^{2}-1)^{2}+\frac{1}{L^{4}}|u^{\prime\prime}(x/L+1)|^{2}\,dx$
$\displaystyle=\int_{0}^{1}L(u(y)^{2}-1)^{2}+\frac{1}{L^{3}}|u^{\prime\prime}(y)|^{2}\,dy.$
(29)

By examination of the equality case of AM-GM, this is precisely
$\frac{4}{3^{3/4}}\Phi(u)$ for a proper choice of $L$. Specifically, one may
take
$L=\left(\frac{3\int_{0}^{1}|u^{\prime\prime}|^{2}\,dx}{\int_{0}^{1}(u^{2}-1)^{2}\,dx}\right)^{1/4}$,
which is well-defined from the assumption that $t\neq-1$. This gives
$\frac{4}{3^{3/4}}\Phi(u)=\Psi(v)\geq\inf_{v\in\mathscr{J}_{\infty}^{\prime}(t)}\Psi(v)$,
and taking the infimum gives
$\inf_{u\in\mathscr{J}^{\prime}(t)}\Phi(u)\geq\inf_{v\in\mathscr{J}_{\infty}^{\prime}(t)}\Psi(v)$.

On the other hand, if $v\in\mathscr{J}^{\prime}_{\infty}(t)$, then from taking
$u(x):=v\left(L_{v}(x-1)\right)$ we get from using the same computations done
in (29) and applying AM-GM that

$\displaystyle\Psi(v)$
$\displaystyle=\int_{0}^{1}\frac{L_{v}}{3}(u(y)^{2}-1)^{2}+\frac{1}{L_{v}^{3}}|u^{\prime\prime}(y)|^{2}\,dy$
$\displaystyle\geq\frac{4}{3^{3/4}}\left(\int_{0}^{1}(u(y)^{2}-1)^{2}\,dy\right)^{3/4}\left(\int_{0}^{1}|u^{\prime\prime}(y)|^{2}\,dy\right)^{1/4}$
$\displaystyle=\frac{4}{3^{3/4}}\Phi(u),$

so that
$\Psi(v)\geq\frac{4}{3^{3/4}}\inf_{u\in\mathscr{J}^{\prime}(t)}\Phi(u)$. Thus
$\inf_{v\in\mathscr{J}^{\prime}_{\infty}(t)}\Psi(v)\geq\frac{4}{3^{3/4}}\inf_{u\in\mathscr{J}^{\prime}(t)}\Phi(u)$,
finishing the proof. ∎

We will now prove several crucial results concerning the family
$\mathscr{J}^{\prime}_{\infty}(t)$ and the functional $\Psi$. The first is a
compactness result.

######  Lemma 4.2.

Let $\Omega\subseteq\mathbb{R}$ be a non-empty bounded open set and $u_{n}\in
w^{2,\infty}(\Omega)$ such that

$M:=\sup_{n\in\mathbb{N}}\int_{\Omega}(u_{n}^{2}-1)^{2}+|u_{n}^{\prime\prime}|^{2}\,dx<\infty.$

There exists a subsequence $u_{n_{k}}$ and $u\in W^{2,2}(\Omega)$ such that

  1. (1)

$u_{n_{k}}\rightharpoonup u$ in $W^{2,2}(\Omega)$,

  2. (2)

$u_{n_{k}}^{\prime}\to u^{\prime}$ almost everywhere and in $L^{2}(\Omega)$,
and

  3. (3)

$u_{n_{k}}\to u$ almost everywhere and in $L^{2}(\Omega)$

###### Proof.

Find $A>0$ large enough so that $z^{2}\leq(z^{2}-1)^{2}$ for all $|z|\geq A$.
Then

$\int_{\Omega}u_{n}^{2}\,dx\leq\int_{\\{|u_{n}|<A\\}}u_{n}^{2}\,dx+\int_{\\{|u_{n}|\geq
A\\}}u_{n}^{2}\,dx\leq
A^{2}\mathcal{L}^{1}(\Omega)+A\int_{\Omega}(u_{n}^{2}-1)^{2}\leq
A^{2}\mathcal{L}^{1}(\Omega)+AM,$

so $\\{u_{n}\\}_{n}$ is uniformly bounded in $L^{2}(\Omega)$. Since
$\\{u_{n}^{\prime\prime}\\}_{n}$ is uniformly bounded in $L^{2}(\Omega)$ as
well, we may apply Theorem 2.10 with some $l<\mathcal{L}^{1}(\Omega)$ to
deduce that $\\{u_{n}^{\prime}\\}_{n}$ is uniformly bounded in
$L^{2}(\Omega)$. Hence $\\{u_{n}\\}_{n}$ is uniformly bounded in
$W^{2,2}(\Omega)$.

By Theorem 2.14, there exists a subsequence $u_{n_{k}}$ and $u\in
W^{2,2}(0,T)$ such that $u_{n_{k}}\rightharpoonup u$ in $W^{2,2}(0,T)$.
Moreover, by 2.13, we have that $u_{n_{k}}\to u$ strongly in $L^{2}(0,T)$ and
$u_{n_{k}}^{\prime}\to u^{\prime}$ strongly in $L^{2}(0,T)$. By extracting a
subsequence, we may also get $u_{n_{k}}\to u$ and $u_{n_{k}}^{\prime}\to
u^{\prime}$ almost everywhere. ∎

Next, we show that we may “force" a bound on $L_{u}$ for functions
$u\in\mathscr{J}^{\prime}_{\infty}(t)$.

######  Lemma 4.3.

Let $M,\eta>0$. Then there exists a constant $L_{M,\eta}>0$ depending only on
$M$ and $\eta$ such that for every $t\in\mathbb{R}$ and
$u\in\mathscr{J}^{\prime}_{\infty}(t)$ with $\Psi(u)\leq M<\infty$, there
exists $v\in\mathscr{J}^{\prime}_{\infty}(t)$ such that
$\Psi(v)\leq\Psi(u)+O(\eta)$ and $v(x)=-1$ for all $x\leq-L_{M,\eta}$. That
is, $L_{v}\leq L_{M,\eta}$.

###### Proof.

We begin by constructing a function $v_{1}\in\mathscr{J}^{\prime}_{\infty}(t)$
such that $\Psi(v_{1})\leq\Psi(u)+O(\eta)$ and
$\mathcal{L}^{1}(v_{1}^{-1}((0,2)))\leq C$ for a constant $C$ depending only
on $M$ and $\eta$.

Let $K=\frac{M}{\eta}$. We will take $C=2K+2$. Let $E:=u^{-1}((0,2))$. If we
have that $\mathcal{L}^{1}(E)\leq 2K+2$, then we simply take $v_{1}=u$.
Otherwise, we may let $E_{1}:=(-\infty,y)\cap E$, $F:=[y,z]\cap E$, and
$E_{2}:=(z,0)\cap E$, where $y,z$ are chosen such that
$\mathcal{L}^{1}(E_{1})=\mathcal{L}^{1}(E_{2})=K$ and
$\mathcal{L}^{1}(F)=\mathcal{L}^{1}(E)-2K>2$.

Since $u>0$ over $E_{1}$, we may write

$M\geq\int_{E_{1}}(u^{2}-1)^{2}+|u^{\prime\prime}|^{2}\,dx\geq\int_{E_{1}}(u-1)^{2}+|u^{\prime\prime}|^{2}\,dx,$

so there exists $x_{1}\in E_{1}$ such that
$(u(x_{1})-1)^{2}+u^{\prime\prime}(x_{1})^{2}\leq\frac{M}{\mathcal{L}^{1}(E_{1})}=\frac{M}{K}=\eta$.
It follows by Lemma 3.6 that there exists $\tilde{u}_{1}\in
C^{\infty}([x_{1},x_{1}+1])$ such that $\tilde{u}_{1}(x_{1})=u(x_{1})$,
$\tilde{u}_{1}^{\prime}(x_{1})=u^{\prime}(x_{1})$, $\tilde{u}_{1}(x_{1}+1)=1$,
and $\tilde{u}_{1}^{\prime}(x_{1}+1)=0$, with

$\displaystyle\int_{x_{1}}^{x_{1}+1}(\tilde{u}_{1}^{2}-1)^{2}+|\tilde{u}^{\prime\prime}|^{2}\,dx$
$\displaystyle\leq\int_{x_{1}}^{x_{1}+1}9(\tilde{u}_{1}-1)^{2}\,dx+\int_{x_{1}}^{x_{1}+1}|\tilde{u}^{\prime\prime}|^{2}\,dx$
$\displaystyle\leq
9C^{\prime}(u(x_{1})-1)^{2}+C^{\prime}|u^{\prime}(x_{1})|^{2}$
$\displaystyle\leq 9C^{\prime}\eta=O(\eta)$

for some universal constant $C^{\prime}>0$. We construct $x_{2}\in E_{2}$ and
$\tilde{u}_{2}\in C^{\infty}([x_{2}-1,x_{2}])$ in a similar manner so that
$\int_{x_{2}-1}^{x_{2}}(\tilde{u}_{2}-1)^{2}+|\tilde{u}_{2}^{\prime\prime}|^{2}\,dx=O(\eta)$.

Now define $\tilde{u}$ as

$\tilde{u}(x):=\begin{cases}u(x),&-\infty<x<x_{1}\\\
\tilde{u}_{1}(x),&x_{1}\leq x\leq x_{1}+1\\\ 1,&x_{1}+1<x<x_{2}-1\\\
\tilde{u}_{2}(x),&x_{2}-1\leq x\leq x_{2}\\\ u(x),&x_{2}<x<0\end{cases}.$

Note that this is well-defined in the sense that $x_{1}+1<x_{2}-1$ because
$x_{2}-x_{1}>z-y\geq\mathcal{L}^{1}(F)\geq 2$. Moreover our choices for
$\tilde{u}_{1}$ and $\tilde{u}_{2}$ ensure that $u\in
W^{2,\infty}(-\infty,0)$, and

$\displaystyle\Psi(\tilde{u})$
$\displaystyle=\int_{(-\infty,x_{1})\cup(x_{2},0)}(\tilde{u}^{2}-1)^{2}+|\tilde{u}^{\prime\prime}|^{2}\,dx+\int_{[x_{1},x_{1}+1]\cup[x_{2}-1,x_{2}]}(\tilde{u}^{2}-1)^{2}+|\tilde{u}^{\prime\prime}|^{2}\,dx$
$\displaystyle\leq\int_{(-\infty,x_{1})\cup(x_{2},0)}(u^{2}-1)^{2}+|\tilde{u}^{\prime\prime}|^{2}\,dx+\int_{[x_{1},x_{1}+1]\cup[x_{2}-1,x_{2}]}(u^{2}-1)^{2}+|u^{\prime\prime}|^{2}\,dx+O(\eta)$
$\displaystyle\leq\Psi(u)+O(\eta).$

We may now take $v_{1}$ to be

$v_{1}(x):=\begin{cases}\tilde{u}(x-x_{2}+x_{1}+2),&-\infty<x<x_{2}-1\\\
\tilde{u}(x),&x_{2}-1<x<0\end{cases}.$

In essence, we have “deleted" an interval in which $\tilde{u}=1$. We still
have $v\in W^{2,\infty}(-\infty,0)$ and $v(0_{+})=u(0_{+})=t$, with
$\lim_{x\to-\infty}v(x)=-1$, so that $v\in\mathscr{J}^{\prime}_{\infty}(t)$.
Furthermore, $\Psi(v)=\Psi(\tilde{u})\leq\Psi(u)+O(\eta)$. Lastly, we see that

$\displaystyle\mathcal{L}^{1}(\\{x<0:0<v_{1}(x)<2\\})$
$\displaystyle\leq\mathcal{L}^{1}(\\{x\in(-\infty,x_{1}+1)\cup(x_{2}-1,0):0<\tilde{u}(x)<2\\})$
$\displaystyle\leq\mathcal{L}^{1}(\\{x\in E_{1}\cup E_{2}:0<u(x)<2\\})+2$
$\displaystyle=2K+2=C,$

which was our goal.

Now we can construct $v$. Observe that

$M\geq\int_{\\{\min(|v-1|,|v+1|)>1\\}}(v_{1}^{2}-1)^{2}+|v_{1}^{\prime\prime}|^{2}\,dx\geq\mathcal{L}^{1}(\\{x<0:\min(|v(x)-1|,|v(x)+1|)>1\\}).$

Thus, if we take $L_{M,\eta}=C+M+K+1$, then the set

$G:=\\{-L_{M,\eta}+1<x<0:-2<v(x)<0\\}$

satisfies $\mathcal{L}^{1}(G)\geq K$. Since
$\int_{G}(v_{1}+1)^{2}+|v^{\prime\prime}|^{2}\,dx\leq\int_{G}(v_{1}^{2}-1)^{2}+|v^{\prime\prime}|^{2}\,dx\leq
M$, there exists $x_{3}\in G$ such that

$(v_{1}(x_{3})+1)^{2}+|v_{1}^{\prime\prime}(x_{3})|^{2}\leq\frac{M}{\mathcal{L}^{1}(G)}\leq
M/K=\eta.$

Hence, as we did before to $u$, we may use Lemma 3.6 to modify $v_{1}$ in the
interval $(x_{3}-1,x_{3})$ and hence obtain a function $v$ for which $v(x)=-1$
for all $x<x_{3}-1$, $v(x)=v_{1}(x)$ for $x_{3}<x<0$, and
$\int_{x_{3}-1}^{x_{3}}(v_{1}^{2}-1)^{2}+|v_{1}^{\prime\prime}|^{2}\,dx\leq
O(\eta)$.

This function $v$ satisfies
$\Psi(v)\leq\Psi(v_{1})+O(\eta)\leq\Psi(u)+O(\eta)$ and, since
$-L_{M,\eta}+1<x_{3}$, we have that $v(x)=-1$ for all $x<-L_{M,\eta}$, where
$L_{M,\eta}$ depends only on $M$ and $\eta$, as desired. ∎

Now we arrive at our first major result.

######  Lemma 4.4.

Let $t_{0}\in\mathbb{R}$. Then we have that

$\lim_{t\to t_{0}}\beta(t)=\beta(t_{0}),$

where $\beta(t)$ is defined as in (9). That is, $\beta$ is continuous.

###### Proof.

Step 1

We claim that $\beta(t)$ is decreasing over $t\leq-1$ and increasing over
$t\geq-1$.

First note that $\beta(t)\geq 0$ for all $t$ and $\beta(-1)=0$. Now suppose
$s,t\in\mathbb{R}$ satisfy either $-1<s\leq t$ or $t\leq s<-1$. We show that
$\beta(s)\leq\beta(t)$ by proving that for all $v\in\mathscr{J}^{\prime}(t)$
we may find $u\in\mathscr{J}^{\prime}(s)$ for which $\Phi(u)\leq\Phi(v)$.

Indeed, since $v(1_{-})=t$ and $v(0_{+})=-1$, there must exist $T\in(0,1)$ for
which $v(T)=s$. Now take $u(x):=v(Tx)$. Evidently,
$u\in\mathscr{J}^{\prime}(s)$. Moreover we have

$\int_{0}^{1}(u^{2}-1)^{2}\,dx=\int_{0}^{T}T^{-1}(v^{2}-1)^{2}\,dx\leq\int_{0}^{1}T^{-1}(v^{2}-1)^{2}\,dx$

and

$\int_{0}^{1}|u^{\prime\prime}|^{2}\,dx=\int_{0}^{T}T^{3}|v^{\prime\prime}|^{2}\,dx\leq\int_{0}^{1}T^{3}|v^{\prime\prime}|^{2}\,dx.$

Thus

$\displaystyle\Phi(u)$
$\displaystyle=\left(\int_{0}^{1}(u^{2}-1)^{2}\,dx\right)^{3/4}\left(\int_{0}^{1}|u^{\prime\prime}|^{2}\,dx\right)^{1/4}$
$\displaystyle\leq\left(\int_{0}^{1}T^{-1}(v^{2}-1)^{2}\,dx\right)^{3/4}\left(\int_{0}^{1}T^{3}|v^{\prime\prime}|^{2}\,dx\right)^{1/4}$
$\displaystyle=\left(\int_{0}^{1}(v^{2}-1)^{2}\,dx\right)^{3/4}\left(\int_{0}^{1}|v^{\prime\prime}|^{2}\,dx\right)^{1/4}$
$\displaystyle=\Phi(v),$

as needed.

Step 2

Take $t_{0}\in\mathbb{R}$. We show that $\limsup_{t\to
t_{0}}\beta(t)\leq\beta(t_{0})$. It is sufficient to prove that for any
$v\in\mathscr{J}^{\prime}(t_{0})$, we may pick
$u_{t}\in\mathscr{J}^{\prime}(t)$ for each $t\in\mathbb{R}$ such that
$\lim_{t\to t_{0}}\Phi(u_{t})=\Phi(v)$. This is because if we fix $\eta>0$,
then we may take $v\in\mathscr{J}^{\prime}(t_{0})$ such that
$\beta(t_{0})\leq\frac{4}{3^{3/4}}\Phi(v)\leq\beta(t_{0})+\eta$, take $u_{t}$
for each $t\in\mathbb{R}$ as above, and then choose $\delta>0$ so small that
$|\Phi(v)-\Phi(u_{t})|<\eta$ for all $t$ with $|t-t_{0}|<\delta$, so that

$\beta(t)\leq\frac{4}{3^{3/4}}\Phi(u_{t})\leq\frac{4}{3^{3/4}}(\Phi(v)+\eta)\leq\beta(t_{0})+\eta+\frac{4}{3^{3/4}}\eta,$

which is enough.

For $v\in\mathscr{J}^{\prime}(t_{0})$, we take
$u_{t}\in\mathscr{J}^{\prime}(t)$ to be $u_{t}:=\frac{1+t}{1+t_{0}}(v+1)-1$.
Then

$\int_{0}^{1}(u_{t}^{2}-1)^{2}\,dx=\left(\frac{1+t}{1+t_{0}}\right)^{4}\int_{0}^{1}(v+1)^{2}\left(v+1-2\cdot\frac{1+t_{0}}{1+t}\right)^{2}\,dx.$

Taking the limit, we obtain

$\lim_{t\to t_{0}}\int_{0}^{1}(u_{t}^{2}-1)^{2}\,dx=\lim_{t\to
t_{0}}\int_{0}^{1}(v+1)^{2}\left(v+1-2\cdot\frac{1+t_{0}}{1+t}\right)^{2}\,dx=\int_{0}^{1}(v^{2}-1)^{2}\,dx$

because $(v+1)^{2}\left(v+1-2\cdot\frac{1+t_{0}}{1+t}\right)^{2}$ converges to
$(v^{2}-1)^{2}$ pointwise as $t\to t_{0}$ and, since $v$ is bounded, we have
that $(v+1)^{2}\left(v+1-2\cdot\frac{1+t_{0}}{1+t}\right)^{2}$ is uniformly
bounded for all $t$ sufficiently near $t$, which is enough to apply dominated
convergence.

We also have

$\lim_{t\to t_{0}}\int_{0}^{1}|u_{t}^{\prime\prime}|^{2}\,dx=\lim_{t\to
t_{0}}\left(\frac{1+t}{1+t_{0}}\right)^{2}\int_{0}^{1}|v^{\prime\prime}|^{2}\,dx=\int_{0}^{1}|v^{\prime\prime}|^{2}\,dx.$

Altogether, we see that

$\displaystyle\lim_{t\to t_{0}}\Phi(u_{t})$ $\displaystyle=\left(\lim_{t\to
t_{0}}\int_{0}^{1}(u_{t}^{2}-1)^{2}\,dx\right)^{3/4}\left(\lim_{t\to
t_{0}}\int_{0}^{1}|u_{t}^{\prime\prime}|^{2}\,dx\right)^{1/4}$
$\displaystyle=\left(\int_{0}^{1}(v^{2}-1)^{2}\,dx\right)^{3/4}\left(\int_{0}^{1}|v^{\prime\prime}|^{2}\,dx\right)^{1/4}$
$\displaystyle=\Phi(v).$

This completes the proof that $\beta$ is upper semi-continuous.

Step 3

We now define yet another family $\overline{\mathscr{J}^{\prime}_{\infty}(t)}$
via

$\overline{\mathscr{J}^{\prime}_{\infty}(t)}:=\left\\{u\in
W^{2,\infty}(-\infty,0):u(0_{-})=t\text{ and
}\lim_{x\to-\infty}u(x)=-1\right\\}.$

We claim that

$\beta(t)=\inf_{u\in\mathscr{J}^{\prime}_{\infty}(t)}\Psi(u)=\inf_{u\in\overline{\mathscr{J}^{\prime}_{\infty}(t)}}\Psi(u).$

Since
$\mathscr{J}^{\prime}_{\infty}(t)\subseteq\overline{\mathscr{J}^{\prime}_{\infty}(t)}$,
it is clear that
$\inf_{u\in\mathscr{J}^{\prime}_{\infty}(t)}\Psi(u)\geq\inf_{u\in\overline{\mathscr{J}^{\prime}_{\infty}(t)}}\Psi(u)$.
To show that
$\inf_{u\in\mathscr{J}^{\prime}_{\infty}(t)}\Psi(u)\leq\inf_{u\in\overline{\mathscr{J}^{\prime}_{\infty}(t)}}\Psi(u)$,
we fix $\varepsilon>0$ and $u\in\overline{\mathscr{J}^{\prime}_{\infty}(t)}$.
It is then sufficient to find
$v\in\overline{\mathscr{J}^{\prime}_{\infty}(t)}$ for which
$\Psi(v)\leq\Psi(u)+O(\varepsilon)$.

Find $M$ so large that
$\int_{-\infty}^{-M}|u^{\prime\prime}|^{2}\,dx<\varepsilon^{2}$ and
$|u(x)+1|<\varepsilon$ for all $x\leq-M$. Then

$\int_{-M-1}^{-M}(u+1)^{2}\,dx\leq\varepsilon^{2},$

and so by Theorem 2.10 applied to $u+1$ with $l=1$, we see that

$\left(\int_{-M-1}^{-M}|u^{\prime}|^{2}\,dx\right)^{1/2}\leq
c\left(\int_{-M-1}^{-M}(u+1)^{2}\,dx\right)^{1/2}+c\left(\int_{-M-1}^{-M}|u^{\prime\prime}|^{2}\,dx\right)^{1/2}\leq
2c\varepsilon$

for a universal constant $c>0$, so that
$\int_{-M-1}^{-M}|u^{\prime}|^{2}\,dx\leq 4c^{2}\varepsilon^{2}$. It follows
that there exists $T\in(M,M+1)$ such that $|u^{\prime}(-T)|\leq
2c\varepsilon$. Since $|u(-T)+1|<\varepsilon$, we may apply Lemma 3.6 to find
$f\in C^{\infty}([-T-1,-T])$ such that

  1. (i)

$f(-T-1)=f^{\prime}(-T-1)=0$, $f(-T)=u(-T)$, $f^{\prime}(-T)=u^{\prime}(-T)$,

  2. (ii)

for a universal constant $C_{0}$ we have

$\int_{-T-1}^{-T}|f+1|^{2}\,dx\leq
C_{0}(|u(-T)+1|^{2}+|u^{\prime}(-T)|^{2})=C_{0}(\varepsilon^{2}+4c^{2}\varepsilon^{2}),$

  3. (iii)

for a universal constant $C_{2}$ we have

$\int_{-T-1}^{-T}|f^{\prime\prime}|^{2}\,dx\leq
C_{2}(|u(-T)+1|^{2}+|u^{\prime}(-T)|^{2})=C_{2}(\varepsilon^{2}+4c^{2}\varepsilon^{2}),$

and

  4. (iv)

$\sup_{[-T-1,-T]}|f|\leq|u(-T)+1|+\frac{1}{2}|u^{\prime}(-T)|=(1+2c)\varepsilon$.

Naturally we now take

$v(x):=\begin{cases}0,&x<-T-1\\\ f(x),&-T-1\leq x\leq-T\\\
u(x),&-T<x<0\end{cases}.$

By item (i), we see that $v\in W^{2,\infty}(-\infty,0)$, and so
$v\in\mathscr{J}^{\prime}_{\infty}(t)$.

For $\varepsilon$ sufficiently small, we have $(1+2c)\varepsilon<1$, which
ensures that $|f(x)-1|\leq 2$ for all $x\in[-T-1,-T]$ by item (iv). This bound
and item (ii) gives

$\displaystyle\int_{-\infty}^{0}(v^{2}-1)^{2}\,dx$
$\displaystyle\leq\int_{-\infty}^{0}(u^{2}-1)^{2}\,dx+\int_{-T-1}^{-T}(f^{2}-1)^{2}\,dx$
$\displaystyle\leq\int_{-\infty}^{0}(u^{2}-1)^{2}\,dx+4\int_{-T-1}^{-T}(f+1)^{2}\,dx$
$\displaystyle\leq\int_{-\infty}^{0}(u^{2}-1)^{2}\,dx+4C_{0}(\varepsilon^{2}+4c^{2}\varepsilon^{2})\,dx,$
(30)

and item (iii) gives

$\displaystyle\int_{-\infty}^{0}|v^{\prime\prime}|^{2}\,dx$
$\displaystyle\leq\int_{-\infty}^{0}|u^{\prime\prime}|^{2}\,dx+\int_{-T-1}^{-T}|f^{\prime\prime}|^{2}\,dx$
$\displaystyle\leq\int_{-\infty}^{0}|u^{\prime\prime}|^{2}\,dx+C_{2}(\varepsilon^{2}+4c^{2}\varepsilon^{2}).$
(31)

Adding (30) and (31) gives

$\Psi(v)\leq\Psi(u)+(4C_{0}+C_{2})(\varepsilon^{2}+4c^{2}\varepsilon^{2})=O(\varepsilon),$

which is enough. This proves the claim.

Step 4

In this step, we show that if $u\in W^{2,\infty}(-\infty,0)$ such that
$\Psi(u)<\infty$, then $\lim_{x\to-\infty}u(x)\in\\{-1,1\\}$.

It is easy to see that this is true provided that the limit
$\lim_{x\to-\infty}u(x)$ exists. To show that the limit exists, let
$l_{1}:=\liminf_{x\to-\infty}u(x)$ and $l_{2}:=\limsup_{x\to-\infty}u(x)$, and
suppose for contradiction that $l_{1}<l_{2}$. Then we may find $k_{1},k_{2}$
with $l_{1}<k_{1}<k_{2}<l_{2}$ such that $k_{1},k_{2}\not\in\\{-1,1\\}$. The
continuity of $u$ ensures that $u^{-1}(k_{1})$ and $u^{-1}(k_{2})$ are
unbounded.

Now consider the quantity

$\gamma:=\inf\left\\{\int_{0}^{T}(v^{2}-1)^{2}+|v^{\prime\prime}|^{2}\,dx:v\in
W^{2,\infty}(0,T),T>1,v(0_{+})=k_{1},v(T_{-})=k_{2}\right\\}.$

We claim that $\gamma=0$. Indeed, we may find sequences $x_{n},y_{n}$ such
that $u(-x_{n})=k_{1}$, $u(-y_{n})=k_{2}$, and $x_{n}+1<y_{n}<x_{n+1}$ for all
$n$. But now

$\infty>\Psi(u)\geq\sum_{n=1}^{\infty}\int_{-y_{n}}^{-x_{n}}(u^{2}-1)^{2}+|u^{\prime\prime}|^{2}\,dx\geq\sum_{n=1}^{\infty}\gamma,$

which can only be possible if $\gamma=0$.

Now find sequences $T_{n}>1$ and $v_{n}\in W^{2,\infty}(0,T_{n})$ for which
$v_{n}(0_{+})=k_{1}$, $v_{n}({T_{n}}_{-})=k_{2}$, and
$\int_{0}^{T_{n}}(v_{n}^{2}-1)^{2}+|v_{n}^{\prime\prime}|^{2}\,dx\to 0$.

Since $T_{n}>1$, we may obtain the uniform bound
$\int_{0}^{1}(v_{n}^{2}-1)^{2}+|v_{n}^{\prime\prime}|^{2}\,dx\leq M$ for some
$M>0$. Also,

$0=\lim_{n\to\infty}\int_{0}^{1}(v_{n}^{2}-1)^{2}+|v_{n}^{\prime\prime}|^{2}\,dx\geq\lim_{n\to\infty}\int_{0}^{1}|v_{n}^{\prime\prime}|^{2}\,dx,$

so we have that $v_{n}^{\prime\prime}\to 0$ in $L^{2}(0,1)$. Moreover, by
Lemma 4.2, there exists a subsequence $v_{n_{k}}$ and $v\in W^{2,2}(0,1)$ such
that $v_{n_{k}}^{\prime\prime}\rightharpoonup v^{\prime\prime}$ in
$L^{2}(0,1)$, $v_{n_{k}}^{\prime\prime}\to 0$ in $L^{2}(0,1)$, and
$v_{n_{k}}\to v$ pointwise almost everywhere. Now, notice that both
$v_{n_{k}}^{\prime\prime}\rightharpoonup 0$ and
$v_{n_{k}}^{\prime\prime}\rightharpoonup v^{\prime\prime}$ in $L^{2}(0,1)$, so
$v^{\prime\prime}=0$ almost everywhere. It follows that $v$ is affine, so that
$v(x)=mx+b$ for some $m,b\in\mathbb{R}$.

Now, Fatou’s Lemma gives that

$0=\lim_{n\to\infty}\int_{0}^{1}(v_{n}^{2}-1)^{2}+|v_{n}^{\prime\prime}|^{2}\,dx\geq\lim_{n\to\infty}\int_{0}^{1}(v_{n}^{2}-1)^{2}\,dx\geq\int_{0}^{1}(v^{2}-1)^{2}\,dx,$

which implies that $m=0$ and $b\in\\{-1,1\\}$.

However, we claim that $v$ must satisfy $v(0_{+})=k_{1}$, which would imply
that $k_{1}\in\\{-1,1\\}$, resulting in a contradiction. To see this, use the
fact that $v_{n}(0_{+})=k_{1}$ for all $n$, the Fundamental Theorem of
Calculus, and Hölder’s inequality to write

$|v_{n}(x)-k_{1}|\leq\int_{0}^{x}|v_{n}^{\prime}(t)|\,dt\leq\left(\int_{0}^{x}|v_{n}^{\prime}(t)|^{2}\,dt\right)^{1/2}\sqrt{x}\leq\left(\int_{0}^{1}|v_{n}^{\prime}(t)|^{2}\,dt\right)^{1/2}\sqrt{x}.$

Then since $v_{n}^{\prime}$ converges in $L^{p}(0,1)$ we must have that
$\int_{0}^{1}|v_{n}^{\prime}(t)|^{2}\,dt$ is bounded, so $|v_{n}(x)-k_{1}|\leq
M_{1}\sqrt{x}$ for a constant $M_{1}>0$. Sending $n\to+\infty$ we then have
$|v(x)-k_{1}|\leq M_{1}\sqrt{x}$ for almost every $x\in(0,1)$, so we may send
$x\to 0^{+}$ along an appropriate sequence to deduce that
$|v(0_{+})-k_{1}|=0$, as needed.

Step 5

We are now ready to show that $\liminf_{t\to t_{0}}\beta(t)\geq\beta(t_{0})$.

Letting $t_{n}\to t_{0}$ be arbitrary, we just need to show that
$\liminf_{n\to\infty}\beta(t_{n})\geq\beta(t_{0})$.

By the monotone properties that we have proven about $\beta$ in Step 1, we see
that $\\{\beta(t_{n}):n\in\mathbb{N}\\}$ is bounded by a constant $M$.
Specifically, we may take

$M=\max(\beta(\sup_{n\in\mathbb{N}}t_{n}),\beta(\inf_{n\in\mathbb{N}}t_{n})).$

Next, by extraction of a subsequence, let us assume that there exists the
limit $L:=\lim_{n\to\infty}\beta(t_{n})$.

Now fix $\eta>0$. Select
$\tilde{u}_{n}\in\mathscr{J}^{\prime}_{\infty}(t_{n})$ such that
$\beta(t_{n})\leq\Psi(\tilde{u}_{n})\leq\beta(t_{n})+\frac{1}{n}$. Since
$\Psi(\tilde{u}_{n})\leq M+1$ for all $n$, we may use Step 6 to find
$u_{n}\in\mathscr{J}^{\prime}_{\infty}(t_{n})$ such that
$\mathcal{L}^{1}(\\{x<0:0<u_{n}(x)<2\\})\leq C_{M,\eta}$ where $C_{M,\eta}$
depends only on $M$ and $\eta$, and
$\Psi(u_{n})\leq\Psi(\tilde{u}_{n})+O(\eta)$, so that

$\beta(t_{n})\leq\Psi(u_{n})\leq\beta(t_{n})+\frac{1}{n}+O(\eta).$ (32)

Note that $\\{u_{n}\\}$ is uniformly bounded in $\Psi$. In particular,
$\sup_{n\in\mathbb{N}}\int_{-m}^{0}|u_{n}^{\prime\prime}|^{2}\,dx<\infty$ for
each $m\in\mathbb{N}$.

Consider $m=2$. By Lemma 4.2 we may extract a subsequence $\\{u_{2,n}\\}_{n}$
of $\\{u_{n}\\}_{n}$ for which $u_{2,n}\rightharpoonup v_{2}$ in
$W^{2,2}(-2,0)$ for some $v_{2}\in W^{2,2}(-2,0)$. and $u_{2,n}\to v_{2}$
almost everywhere.

Inductively, for $m\geq 3$ we take $\\{u_{m,n}\\}_{n}$ to be a subsequence of
$\\{u_{m-1,n}\\}_{n}$ such that $u_{m,n}\rightharpoonup v_{m}$ in
$W^{2,2}(-m,0)$ for some $v_{m}\in W^{2,2}(-m,0)$ and such that $u_{m,n}\to
v_{m}$ almost everywhere.

We claim that $v_{m}$ extends $v_{m-1}$ for all $m\geq 3$. Indeed,
$u_{m-1,n}\to v_{m-1}$ almost everywhere, and since $\\{u_{m,n}\\}_{n}$ is a
subsequence of $\\{u_{m-1,n}\\}_{n}$, we have that $u_{m,n}\to v_{m-1}$ almost
everywhere over $(-m+1,0)$. Since $u_{m,n}\to v_{m}$ almost everywhere, it
follows that $v_{m-1}$ and $v_{m}$ agree over $(-m+1,0)$ as needed.

It follows that there exists a unique $u:(-\infty,0)\to\mathbb{R}$ extending
each $v_{m}$. Now consider $\\{u_{n,n}\\}$. For each $m$, $\\{u_{n,n}\\}_{n}$
may be viewed as a subsequence of $\\{u_{m,n}\\}$ for $n$ large enough, so
$u_{n,n}\to v_{n}=u$ almost everywhere in $(-m,0)$. Thus $u_{n,n}\to v$ over
$(-\infty,0)$ almost everywhere. Similarly, we see that
$u_{n,n}\rightharpoonup v$ in $W^{2,2}(-m,0)$ for all $m>0$. In particular,
$u_{n,n}\rightharpoonup v$ in $W^{2,2}_{\text{loc}}(-m,0)$.

Since $\\{u_{n,n}\\}_{n}$ is a subsequence of $\\{u_{n}\\}_{n}$, we let
$n_{k}$ be such that $\\{u_{n_{k}}\\}_{k}=\\{u_{n,n}\\}_{n}$.

To finish we use the fact that $u_{n_{k}}\to u$ almost everywhere in
$(-\infty,0)$ to obtain the inequality

$\int_{-\infty}^{0}(u^{2}-1)^{2}\,dx\leq\liminf_{k\to\infty}\int_{0}^{T}(u_{n_{k}}^{2}-1)^{2}\,dx$

by Fatou’s Lemma. Next, we use the property that
$u_{n_{k}}^{\prime\prime}\rightharpoonup u^{\prime\prime}$ in
$L^{2}_{\text{loc}}(0,T)$ and the fact that $z\mapsto z^{2}$ is convex,
together with Theorem 2.15, to conclude that

$\int_{-\infty}^{0}|u^{\prime\prime}|^{2}\,dx\leq\liminf_{k\to\infty}\int_{0}^{T}|u_{n_{k}}^{\prime\prime}|^{2}\,dx.$

Combining these two inequalities gives
$\Psi(u)\leq\liminf_{k\to\infty}\Psi(u_{n_{k}})$

Now, on one hand, we have from (32) that
$\Psi(u_{n_{k}})\leq\beta(t_{n_{k}})+\frac{1}{n_{k}}+O(\eta)$, and taking the
liminf we have

$L+O(\eta)=\liminf_{k\to\infty}\beta(t_{n_{k}})+O(\eta)\geq\liminf_{k\to\infty}\Psi(u_{n_{k}}).$

On the other hand, we claim that $\Psi(u)\geq\beta(t_{0})$. It suffices to
prove that $u\in\overline{\mathscr{J}^{\prime}_{\infty}(t_{0})}$.

Since $u_{n_{k}}\to u$ almost everywhere, and $u_{n_{k}}(0_{-})=t_{n_{k}}$
with $t_{n_{k}}\to t_{0}$, we must have $u(0_{-})=t_{0}$. Moreover, since
$\Psi(u)<\infty$, we have by Step 5 that the limit $\lim_{x\to-\infty}u(x)$
exists, and is either $1$ or $-1$. However,

$\mathcal{L}^{1}(\\{u>0\\})=\mathcal{L}^{1}\left(\liminf_{k\to\infty}\\{u_{n_{k}}>0\\}\right)\leq\liminf_{k\to\infty}\mathcal{L}^{1}(\\{u_{n_{k}}>0\\})\leq
C_{M,\eta}<\infty,$

which eliminates the possibility that the limit is 1. Thus
$\lim_{x\to-\infty}u(x)=-1$, so we may conclude that
$u\in\overline{\mathscr{J}^{\prime}_{\infty}(t_{0})}$.

With $\Psi(u)\geq\beta(t_{0})$, we have that

$L+O(\eta)\geq\liminf_{k\to\infty}\Psi(u_{n_{k}})\geq\Psi(u)\geq\beta(t_{0}).$

But $\eta>0$ was arbitrary, so $L\geq\beta(t_{0})$, which is what we wanted to
show. ∎

We now prove our main result, Theorem 1.1.

###### Proof.

The cases in which $u\not\in L^{p}(\Omega,\\{-1,1\\})$ or $u\in
L^{p}(\Omega,\\{-1,1\\})\setminus BPV(\Omega;\\{-1,1\\})$ are handled as in
Steps 1 and 4 of the proof of Theorem 3.5, so let us assume that $u\in
BPV(\Omega;\\{-1,1\\})$.

For the liminf inequality, let $u_{n}\to u$ in $L^{p}(\Omega)$. We may assume
that $u_{n}(a_{+})=a_{\varepsilon_{n}}$ and $u_{n}(b_{-})=b_{\varepsilon_{n}}$
for all $n$.

Fix $\eta>0$, and suppose that $u(b_{-})=-1$. Then we may find an interval
$(b-\delta,b)$ over which $u=-1$, and now for all large enough $n$ we follow
Step 2 of the proof of Theorem 3.5 in which we “modify" $u_{n}$ in
$(b,b-\delta)$ so that it becomes “affixed" to $-1$ at $b-\delta$. That is, we
find $v_{n}:(b-\delta,b)\to\mathbb{R}$ such that $v_{n}\in
W^{2,p}(b-\delta,b)$, $v_{n}(b-\delta_{-})=-1$,
$v_{n}^{\prime}(b-\delta_{-})=0$,
$v_{n}(b_{-})=u_{n}(b_{-})=b_{\varepsilon_{n}}$, and

$\int_{b-\delta}^{b}\varepsilon_{n}^{-1}(v_{n}^{2}-1)^{2}+\varepsilon_{n}^{3}|v_{n}^{\prime\prime}|^{2}\,dx\leq\int_{b-\delta}^{b}\varepsilon_{n}^{-1}(u_{n}^{2}-1)^{2}+\varepsilon_{n}^{3}|u_{n}^{\prime\prime}|^{2}\,dx+\eta.$

Now, as in Step 3 of the proof of Theorem 3.5, we may change variables and
apply the AM-GM inequality to eventually obtain the bound

$\int_{b-\delta}^{b}\varepsilon_{n}^{-1}(v_{n}^{2}-1)^{2}+\varepsilon_{n}^{3}|v_{n}^{\prime\prime}|^{2}\,dx\geq\frac{4}{3^{3/4}}\inf_{u\in\mathscr{J}^{\prime}(b_{\varepsilon_{n}})}\Phi(u)=2\beta(b_{\varepsilon_{n}}).$

If instead $u(b_{-})=1$, then by a symmetrical argument, we instead obtain the
term $2\beta(-b_{\varepsilon_{n}})$. Both of these terms may be written as
$2\beta(-\operatorname{sgn}(u(b_{-}))b_{\varepsilon_{n}})$.

Using the same argument, we may obtain the term
$2\beta(-\operatorname{sgn}(u(a_{+}))a_{\varepsilon_{n}})$, and we recover the
term $\alpha\operatorname*{essVar}_{\Omega}u$ as in Steps 2 and 3 of the proof
of Theorem 3.5.

Overall, we see that the inequality

$\liminf_{n\to\infty}G_{\varepsilon_{n}}(u_{n})\geq\alpha\operatorname*{essVar}_{\Omega}u+\liminf_{n\to\infty}2\beta(-\operatorname{sgn}(u(a_{+}))a_{\varepsilon_{n}})+2\beta(-\operatorname{sgn}(u(b_{-}))b_{\varepsilon_{n}})$

may be obtained, and so we get

$\liminf_{n\to\infty}G_{\varepsilon_{n}}(u_{n})\geq\alpha\operatorname*{essVar}_{\Omega}u+2\beta(-\operatorname{sgn}(u(a_{+}))a_{0})+2\beta(-\operatorname{sgn}(u(b_{-}))b_{0})$

by continuity of $\beta$.

For the limsup inequality, we use a construction similar to that done in Step
5 of the proof of Theorem 3.5. We begin by strengthening the continuity result
on $\beta(t)$. Define the constant

$\beta_{\varepsilon}(t):=\inf\left\\{\Psi(v):v\in\mathscr{J}^{\prime}_{\infty}(t),L_{v}\leq\frac{1}{\sqrt{\varepsilon}}\right\\}$

and let $t_{0}\in\mathbb{R}$ with $t_{n}\to t_{0}$. We claim that
$\lim_{n\to\infty}\beta_{\varepsilon_{n}}(t_{n})=\beta(t_{0})$. To see this,
fix $\eta>0$ and for each $n\in\mathbb{N}$ take
$v_{n}\in\mathscr{J}^{\prime}_{\infty}(t_{n})$ for which
$\beta(t_{n})\leq\Psi(v_{n})\leq\beta(t_{n})+\frac{1}{n}$. The continuity of
$\beta$ ensures that $\\{\Psi(v_{n})\\}_{n}$ is bounded by a constant $M>0$.
By Lemma 4.3 we may construct $\tilde{v}_{n}$ for which
$\Psi(\tilde{v}_{n})\leq\Psi(v_{n})+O(\eta)\leq\beta(t_{n})+\frac{1}{n}+O(\eta)$
and $L_{\tilde{v}_{n}}\leq L_{M,\eta}$ where $L_{M,\eta}$ depends only on $M$
and $\eta$. Particularly $L_{M,\eta}$ has no dependence on $n$, thus
$L_{\tilde{n}_{n}}\leq\frac{1}{\sqrt{\varepsilon_{n}}}$ for all $n$ large
enough. For all such $n$ we may write

$\beta_{\varepsilon_{n}}(t_{n})\leq\Psi(\tilde{v}_{n})\leq\beta(t_{n})+\frac{1}{n}+O(\eta),$

and so by taking the limsup we may obtain

$\limsup_{n\to\infty}\beta_{\varepsilon_{n}}(t_{n})\leq\beta(t_{0})+O(\eta)$

by continuity of $\beta$. As $\eta>0$ was arbitrary, we get
$\limsup_{n\to\infty}\beta_{\varepsilon_{n}}(t_{n})\leq\beta(t_{0})$. But we
clearly also have $\beta_{\varepsilon_{n}}(t_{n})\geq\beta(t_{n})$, and taking
the liminf finishes the proof of the claim.

We now turn to the proof of the limsup inequality. As we did for the liminf
inequality, assume the case $u(b_{-})=-1$ and find $\delta>0$ small enough so
that $u=-1$ over the interval $(b-\delta,b)$. We will define $u_{n}$ over
$(b-\delta,b)$. Take
$v_{n}\in\mathscr{J}^{\prime}_{\infty}(b_{\varepsilon_{n}})$ satisfying
$L_{v_{n}}\leq\frac{1}{\sqrt{\varepsilon_{n}}}$ such that

$\beta_{\varepsilon_{n}}(b_{\varepsilon_{n}})\leq\Psi(v_{n})\leq\beta_{\varepsilon_{n}}(b_{\varepsilon_{n}})+\frac{1}{n}.$
(33)

Since $\varepsilon_{n}L_{v_{n}}\leq\sqrt{\varepsilon_{n}}\to 0$, we have
$\varepsilon_{n}L_{v_{n}}<\delta$ for all $n$ large enough. For all such $n$,
we define

$u_{n}(x):=\begin{cases}-1,&b-\delta<x\leq b-\varepsilon_{n}L_{v_{n}}\\\
v_{n}\left(\frac{x-b}{\varepsilon_{n}}\right),&b-\varepsilon_{n}L_{v_{n}}<x<b\end{cases}.$

This satisfies $u_{n}(b-\delta)=-1$, $u_{n}^{\prime}(b-\delta)=0$, and the
boundary condition $u_{n}(b_{-})=b_{\varepsilon_{n}}$. Moreover

$\displaystyle\int_{b-\delta}^{b}\varepsilon_{n}^{-1}(u_{n}^{2}-1)^{2}$
$\displaystyle+\varepsilon_{n}^{3}|u_{n}^{\prime\prime}|^{2}\,dx$
$\displaystyle=\int_{b-\varepsilon_{n}L_{v_{n}}}^{b}\varepsilon_{n}^{-1}\left(v_{n}\left(\frac{x-b}{\varepsilon_{n}}\right)^{2}-1\right)^{2}+\frac{\varepsilon_{n}^{3}}{\varepsilon_{n}^{4}}v_{n}^{\prime\prime}\left(\frac{x-b}{\varepsilon_{n}}\right)^{2}\,dx$
$\displaystyle=\int_{-L_{v_{n}}}^{0}(v_{n}^{2}-1)^{2}+|v_{n}^{\prime\prime}|^{2}\,dx$
$\displaystyle=\Psi(v_{n})\leq\beta_{\varepsilon_{n}}(b_{\varepsilon_{n}})+\frac{1}{n},$

thus

$\limsup_{n\to\infty}\int_{b-\delta}^{b}\varepsilon_{n}^{-1}(u_{n}^{2}-1)^{2}+\varepsilon_{n}^{3}|u_{n}^{\prime\prime}|^{2}\,dx\leq\limsup_{n\to\infty}\beta_{\varepsilon_{n}}(b_{\varepsilon_{n}})+\frac{1}{n}=\beta(b_{0})$

by the claim. It remains to show that $u_{n}\to-1$ in $L^{p}(b-\delta,b)$.

Since $\varepsilon_{n}L_{v_{n}}\to 0$, we have that $u_{n}\to-1$ almost
everywhere in $(b-\delta,b)$. By Vitali, it suffices to prove that
$\\{|u_{n}|^{p}\cdot 1_{(b-\delta,b)}\\}_{n}$ is uniformly integrable. Indeed,
we have

$\int_{b-\delta}^{b}(u_{n}^{2}-1)^{2}\,dx=\varepsilon_{n}\int_{-L_{v_{n}}}^{0}(v_{n}^{2}-1)^{2}\,dx=\varepsilon_{n}\Psi(v_{n}),$

and since
$\limsup_{n\to\infty}\Psi(v_{n})\leq\lim_{n\to\infty}\beta_{\varepsilon_{n}}(b_{\varepsilon_{n}})+\frac{1}{n}<\infty$,
we have that $\lim_{n\to\infty}\int_{b-\delta}^{b}(u_{n}^{2}-1)^{2}\,dx=0$, so
uniform integrability follows from Lemma 3.3.

Our construction for $u_{n}$ over $(b-\delta,b)$ provides us with the term
$\beta(b_{0})$ for the Gamma limit in the case that $u(b_{-})=-1$. If instead
$u(b_{-})=1$, we may negate our construction to instead acquire the term
$\beta(-b_{0})$. The term in both cases is equal to
$\beta(b_{0}\operatorname{sgn}u(b_{-}))$. A symmetrical construction near the
endpoint $a$ yields the term $\beta(a_{0})\operatorname{sgn}u(a_{+}))$. Lastly
we may define $u_{n}$ away from the endpoints as in Step 5 of the proof of
Theorem 3.5 in order to recover the term
$\alpha\operatorname*{essVar}_{\Omega}u$, completing the proof. ∎

##  5\. Acknowledgments

This research was partially supported by the National Science Foundation under
grants No. DMS 2108784 and DMS 1714098. I am also indebted to my advisor
Giovanni Leoni, whose guidance and feedback made this paper possible.

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