# Branch points for (almost-)Minimizers of Two-Phase Free Boundary Problems

Guy David, Max Engelstein, Mariana Smit Vega Garcia & Tatiana Toro
Laboratoire de Mathématiques d’Orsay, Univ. Paris-Sud, CNRS, Université Paris-
Saclay, 91405 Orsay, France<EMAIL_ADDRESS>School of Mathematics,
University of Minnesota,
Minneapolis, MN, 55455, USA<EMAIL_ADDRESS>Western Washington University
Department of Mathematics
BH 230
Bellingham, WA 98225, USA<EMAIL_ADDRESS>Department of
Mathematics
University of Washington
Box 354350
Seattle, WA 98195-4350<EMAIL_ADDRESS>
###### Abstract.

We study the existence and structure of branch points in two-phase free
boundary problems. More precisely, we construct a family of minimizers to an
Alt-Caffarelli-Friedman type functional whose free boundaries contain branch
points in the strict interior of the domain. We also give an example showing
that branch points in the free boundary of almost-minimizers of the same
functional can have very little structure. This last example stands in
contrast with recent results of De Philippis-Spolaor-Velichkov on the
structure of branch points in the free boundary of stationary solutions.

###### Key words and phrases:

branch points, almost-minimizers, two-phase free boundary problems Alt-
Caffarelli functional, Alt-Caffarelli-Friedman

###### 2020 Mathematics Subject Classification:

35R35,

G.D. was partially supported by the ANR, programme blanc GEOMETRYA,
ANR-12-BS01-0014, the European H2020 Grant GHAIA 777822, and the Simons
Collaborations in MPS Grant 601941, GD. M.E. has been partially supported by
the NSF grant DMS 2000288. M.S.V.G. has been partially supported by the NSF
grant DMS 2054282. T.T. was partially supported by the Craig McKibben & Sarah
Merner Professor in Mathematics, and by NSF grant DMS-1954545

##  1\. Introduction

In this paper we study the structure of “branch points” in the free boundary
of minimizers of Alt-Caffarelli-Friedman type functionals (see (1.1) below).
In particular, we show the existence of minimizers to the two-phase functional
whose zero set contains an open subset (of positive measure) which stays far
away from the fixed boundary of the domain. Relatedly, the free boundary of
this minimizer also contains branch or cusp points (c.f. (1.2)). We also show,
in contrast with recent results for critical points to (1.1) in [DPSV21a],
that the set of branch points in the free boundary of almost-minimizers to
(1.1) can have fractal like structure.

Alt, Caffarelli and Friedman, in [ACF84], gave the first rigorous mathematical
treatment of the two-phase energy

(1.1) $J_{\Omega}(u)=\int_{\Omega}|\nabla
u|^{2}dx+\lambda_{+}^{2}|\\{u>0\\}\cap\Omega|+\lambda_{-}^{2}|\\{u<0\\}\cap\Omega|$

where $\Omega\subset\mathbb{R}^{n}$ is a domain with locally Lipschitz
boundary, and $\lambda_{\pm}>0$111[ACF84] actually considered a related
functional which also weighs the zero set, but (1.1) captures the essence of
the more general functional, and for almost minimizers the general case can be
transformed to (1.1) through a standard change of notation. .

This is a two-phase analogue of the one-phase free boundary problem (also
called the Bernoulli problem) studied in [AC81], first introduced to model the
flow of two liquids in jets and cavities but later found to have applications
to a variety of problems including eigenvalue optimization, c.f. [DPSV21b,
Corollary 1.3].

We say that $u$ is a local minimizer of $J$ in $\Omega$ if $J_{D}(u)\leq
J_{D}(v)$ for all open $D$ with $\overline{D}\subset\Omega$ and all $v\in
W^{1,2}(\Omega)$ with $u=v$ in $\Omega\backslash\overline{D}$. Alternatively,
given some subset $S\subset\partial\Omega$ and continuous data $\varphi\in
C(S)$ we say that $u$ minimizes $J_{\Omega}$ for the data $\varphi$ if
$u=\varphi$ on $S$ and for any $v\in W^{1,2}(\Omega)$ with $v=\varphi$ on $S$
we have $J_{\Omega}(u)\leq J_{\Omega}(v)$. If the data $\varphi$ is not
important, we simply say that $u$ is a minimizer of $J_{\Omega}$. We note that
if $u$ minimizes $J_{\Omega}$ for some data $\varphi$, then $u$ is a local
minimizer of $J$ in $\Omega$.

Given a (local) minimizer, $u$, of particular interest are the free
boundaries, $\Gamma^{\pm}(u)=\partial\\{\pm u>0\\}$. When
$\Gamma^{+}\cap\Gamma^{-}=\emptyset$, each of $\Gamma^{\pm}$ is the free
boundary of a minimizer to an associated one-phase problem and thus have well
understood regularity (c.f. [AC81]). On the other hand, when
$\Gamma^{+}=\Gamma^{-}$, the free boundary regularity is also well understood,
first when $n=2$ in [ACF84], and later by Caffarelli ([Caf87, Caf89, Caf88];
see also the book [CS05]) and De Silva-Ferrari-Salsa (see e.g. [DSFS14a,
DSFS14b] and the recent survey article [DSFS19]). Until recently, the only
missing piece of the picture was the behavior of $\Gamma^{\pm}$ in
neighborhoods where the two sets are not disjoint but also not identical. To
be more precise, define the points in the intersection of $\Gamma^{\pm}$ as
two-phase points; $\Gamma_{\mathrm{TP}}(u):=\Gamma^{+}\cap\Gamma^{-}$. Points
which are in one of $\Gamma^{\pm}$ but not both are one-phase points;
$\Gamma_{\mathrm{OP}}(u):=\Gamma^{+}\cup\Gamma^{-}\setminus(\Gamma^{+}\cap\Gamma^{-})$.
It was a long open question how the free boundary behaved around branch
points, that is, points around which the free boundary contains both one-phase
points and two-phase points at every scale;

(1.2)
$\Gamma_{\mathrm{BP}}(u):=\Gamma_{\mathrm{TP}}(u)\cap\overline{\Gamma_{\mathrm{OP}}(u)}.$

This open question was finally resolved in the recent work of De Philippis,
Spolaor and Velichkov [DPSV21b] (see also [SV19] when $n=2$):

######  Theorem 1.1.

(Main Theorem in [DPSV21b]) Let $u$ be a (local) minimizer to the energy in
(1.1) in $\Omega$ with $\lambda_{\pm}>0$. Then for every
$x_{0}\in\Gamma^{+}\cap\Gamma^{-}\cap\Omega$ there exists an $r_{0}>0$ such
that both $\Gamma^{+}\cap B(x_{0},r_{0})$ and $\Gamma^{-}\cap B(x_{0},r_{0})$
are $C^{1,1/2}$-graphs.

We note that Theorem 1.1 is most interesting around branch points, i.e.
$x_{0}\in\Gamma_{\mathrm{BP}(u)}\cap\Omega$. However, left open in [DPSV21b]
is whether branch points actually exist in the strict interior of a domain, or
more precisely, does there exist a minimizer $u$ in $\Omega$ such that
$\Gamma_{\mathrm{BP}}(u)\cap\Omega\neq\emptyset$. Here we resolve that open
question when $\lambda_{+}=\lambda_{-}=1$:

######  Theorem 1.2.

(Main Theorem) There exists a domain $\Omega\subset\mathbb{R}^{2}$ and a
minimizer $u$ to $J_{\Omega}$ with $\lambda_{+}=\lambda_{-}=1$ such that
$\Gamma_{\mathrm{BP}}(u)\cap\Omega\neq\emptyset$. Even stronger, there exists
a “pool” of zeroes: a (nonempty) connected component $\mathcal{O}$ of
$\\{u=0\\}$ such that $\overline{\mathcal{O}}\subset\subset\Omega$ and
$\partial\mathcal{O}\cap\Gamma^{\pm}\neq\emptyset$.

Our tools are reminiscent of our previous studies on almost-minimizers with
free boundary, c.f. [DT15, DET19, DESVGT21]. In particular, we carefully
choose competitor functions and use ideas from harmonic analysis and geometric
measure theory. We further remark that our construction can be extended to
produce examples in dimensions $n\geq 3$, see Remark 4.5.

###  1.1. Comparison with other work on Branch Points

While we believe the question of whether branch points (or pools) in the free
boundary of minimizers of (1.1) exist has been open until now, there has been
substantial work on branch points for other related functionals and for
“critical points” of the functional (1.1).

In particular, branch points in the free boundaries of minimizers to a related
vectorial problem were constructed by Spolaor and Velichkov in [SV19].
Additionally, a related phenomena, when the free boundary of minimizers to a
one-phase version of (1.1) comes into contact with the fixed boundary (i.e.
$\partial\Omega$) resulting in branching like behavior, is well studied (e.g.
[CLS19, DPSV21a]).

The only other work we are aware of regarding branch points in the free
boundary of functions associated to the energy (1.1), is the recent preprint
[DPSV21a]. In this very nice work, the authors (amongst other things)
construct an infinite family of critical points to the functional (1.1) when
$n=2$ using (quasi-)conformal mappings (see [DPSV21a, Theorem 1.8]). Without
being precise, we recall that critical points to (1.1) satisfy the associated
Euler-Lagrange equations but do not necessarily (locally) minimize the
functional in any domain (e.g. $u(x)=|x|$ is a critical point of $J_{\Omega}$
with $\lambda_{+}=1$ but not a (local) minimizer).

######  Remark 1.3.

To be explicit, we note that (none of) the results of [DPSV21a] either imply
or are implied by our results here. In particular, our main theorem does not
analyze the rate at which $\Gamma^{+}$ and $\Gamma^{-}$ come together at the
cusp points, and thus does not produce examples with different rates. On the
other hand, it is not clear whether the examples produced in [DPSV21a] are
minimizers.

Furthermore, the methods of proof are very different, in so far as [DPSV21a]
draws an interesting connection with minimizers of a non-linear obstacle type
problem and uses (quasi-)conformal maps in their construction. We construct
the relevant boundary values and domains explicitly but do not have a closed
formula for our minimizer. Rather, we use tools from harmonic analysis and
geometric measure theory to constrain the behavior of the minimizer. In
particular, our methods extend to producing examples in dimension $n>2$; see
Remark 4.5, which presumably are out of reach of (quasi-)conformal methods.

We are not aware of any prior work on “pools” in the zero set of minimizers to
(1.1). We will limit ourselves to pointing out that it is easy to construct
examples of minimizers whose zero set has non-empty interior but that some
care is required to constrain this open component to the strict interior of
the domain.

###  1.2. Accumulation of Branch Points

Also of interest in [DPSV21a] is the fact that for certain, symmetric (in a
precise sense), critical points of (1.1) in two dimensions, the branch points
in the free boundary are locally isolated in (c.f. [DPSV21a, Theorem 1.6(a)]).
In fact, in analogy with area-minimizing surfaces (see, e.g. [Cha88, DLSS17,
DLSS18, DLSS20, DLMSV18]) one might conjecture the following:

######  Conjecture 1.4.

Let $u$ be a minimizer to (1.1) in some $\Omega\subset\mathbb{R}^{n}$. Then,
for any $D\subset\subset\Omega$ the set $D\cap\Gamma_{\mathrm{BP}}(u)$ is
locally contained in finitely many Lipschitz $(n-2)$-dimensional submanifolds.

In the second part of this paper we show that such a theorem fails for almost-
minimizers of (1.1). Recall that almost-minimizers minimize the energy (1.1)
up to some noise:

######  Definition 1.5.

We say that $u$ is an almost-minimizer to (1.1) in
$\Omega\subset\mathbb{R}^{n}$ if there is a $C>0$ and an $\alpha\in(0,1]$ such
that for every ball $B$ with $\overline{B}\subset\Omega$ and for every $v\in
W^{1,2}(\Omega)$ with $v=u$ in $\Omega\setminus\overline{B}$ we have

$J_{B}(u)\leq J_{B}(v)+Cr(B)^{n+\alpha}.$

Almost-minimizers arise naturally in constrained optimization (and thus in
eigenvalue-optimization problems see, e.g. [MTV17]). Almost-minimizers may not
satisfy an Euler-Lagrange equation, but the work of the authors in [DT15,
DET19, DESVGT21] show that the “first order” regularity of almost-minimizers
mimics that of minimizers; in particular, almost-minimizers are regular,
Lipschitz continuous, non-degenerate, $C^{1,\beta}$ up to their free boundary,
and, in the one-phase case, have free boundaries which are smooth up to a set
of $\mathcal{H}^{n-1}$-measure zero.

However in contrast to [DPSV21a, Theorem 1.6(a)] and Conjecture 1.4, in this
paper we prove that the set of branch points for almost-minimizers can be
essentially arbitrary:

######  Theorem 1.6.

(Corollary to Theorem 5.1) Let $E\subset\mathbb{R}^{n-1}$ be a compact set
with no interior point. Embed $E$ into $\mathbb{R}^{n}$ so that
$E\subset\\{(x^{\prime},0)\mid x^{\prime}\in\mathbb{R}^{n-1}\\}$ and let $R>0$
be so large that $E\subset B(0,R/10)$. Then there exists an almost-minimizer,
$u$, to $J_{B(0,R)}$ with $\lambda_{+}=\lambda_{-}=1$ such that
$\Gamma_{\mathrm{BP}}(u)=E$. Furthermore, we can take $u$ to be such that
$\Gamma^{+}$ is the reflection of $\Gamma^{-}$ across the hyper-plane
$\\{x_{n}=0\\}$ (so that this solution is “symmetric” in the sense of
[DPSV21a]).

For those familiar with almost-area minimizers, this theorem may seem trivial;
indeed any graph of a $C^{1,\alpha}$ function is almost-area minimizing.
However, it is not the case that if $\partial\\{u\neq 0\\}$ is locally given
by a (union of) smooth graphs and $u$ is smooth, except for jumps along
$\partial\\{u\neq 0\\}$, then $u$ is an almost-minimizer to the energy in
(1.1). Indeed, almost-minimizers must satisfy additional non-degeneracy and
boundedness conditions in addition to a condition on their normal derivative
at $\partial\\{u\neq 0\\}$ (c.f. Lemma 5.3 below).

The essence of Theorem 1.6 is that we are able to construct almost-minimizers
to (1.1) whose free boundaries are given by the graphs of any two smooth
functions $f^{-}\leq f^{+}$ over $\mathbb{R}^{n-1}\subset\mathbb{R}^{n}$.
These almost-minimizers are given by a regularized distance (see (5.9) below)
developed by the first author, with Feneuil and Mayboroda, to characterize the
geometry of sets of high co-dimension using degenerate PDE (see, e.g.
[DFM19]). This connection is surprising, but essentially is due to the fact
that regularized distances satisfy the same growth conditions as almost-
minimizers and, due to the results of [DEM21], one can prescribe their normal
derivatives along the set on which they vanish.

##  2\. Slice Minimizers and Uniform Lipschitz Continuity

Let $I_{N}:=[-3N,3N]$, with $N$ very large, and consider
$\overline{R}_{N}:=I_{N}\times[-1,1]\subset\mathbb{R}^{2}$. To define our
boundary conditions on $I_{N}\times\\{1\\}$ and $I_{N}\times\\{-1\\}$, we fix
some $\alpha\in(0,1)$ small but universal ($\alpha=1/10$ will do) and define

$\displaystyle f_{N}(x)=\begin{cases}1-\alpha,&\text{ if }|x|\leq N\\\
\frac{|x|(1+\alpha)}{N}-2\alpha&\text{ if }N\leq|x|\leq 2N\\\ 2&\text{ if
}2N\leq|x|\leq 3N\end{cases}$

(see Figure 1).

We then let

(2.1) $u_{N}=\mathrm{argmin}\\{J(u,\overline{R}_{N})\mid u_{N}(x,\pm 1)=\pm
f_{N}(x),x\in I_{N}\\},$

that is, a minimizer to the Alt-Caffarelli-Friedman function with boundary
values $\pm f_{N}(x)$. We should note that we are abusing the argmin notation
slightly, as this minimizer is not necessarily unique, but we can pick any
minimizer for the analysis below. We should also note that we are not
prescribing Dirichlet data on the “vertical” parts of the boundary. However,
an existence and regularity theory for minimizers given “partial Dirichlet
data” exists (see e.g. [ACF84]) or we could prescribe data on $\\{\pm
3N\\}\times[-1,1]$ that simply linearly interpolates between $f_{N}(\pm 3N)$
and $-f_{N}(\pm 3N)$.

When the precise value of $N$ is not important we will suppress it from the
notation.

$N$$\frac{(1+2\alpha)N}{1+\alpha}$$\frac{-(1+2\alpha)N}{1+\alpha}$$-N$$2N$$-2N$$3N$$-3N$$x$$f_{N}(x)$$1-\alpha$$1$$2$
Figure 1. Graph of $f_{N}$

In order to study $u_{N}$ we will introduce the “slice minimizer”, $v_{N}$,
defined as follows: for each $x\in I$, $v_{N}(x,-)$ is the unique222Our
computations in the Appendix show that this minimizer is unique minimizer of
the one-dimensional functional

(2.2) $H_{x}(w):=\int_{-1}^{1}(w^{\prime}(y))^{2}dy+|\\{w\neq 0\\}\cap[-1,1]|$

under the constraint $v_{N}(x,\pm 1)=\pm f_{N}(x)$.

Notice first that for $x\in I$, the set $\\{v_{N}(x,-)=0\\}$ is an interval.
Indeed, if $y_{1}$ and $y_{2}$ are the first/last points where $v_{N}$
vanishes, replacing $v$ by $0$ on the interval $[y_{1},y_{2}]$ yields an
admissible candidate $w$ with
$\int_{-1}^{1}(w^{\prime}(y))^{2}dy\leq\int_{-1}^{1}(v^{\prime})^{2}dy$ and
$|\\{w\neq 0\\}|\leq|\\{v\neq 0\\}|$, with strict inequality, unless
$v_{N}\equiv 0$ on $[y_{1},y_{2}]$. Moreover, $v_{N}$ is harmonic in the open
set $\\{v_{N}\neq 0\\}\cap(-1,1)$, that is, $v_{N}$ is locally affine. A
straightforward calculation, which we defer to the Appendix (c.f. Sections
6.1.1 and 6.1.2), allows us to explicitly calculate the slice minimizer and
its energy for each $x\in[-3N,3N]$.

######  Lemma 2.1.

Let $v(y)=v_{N}(x,y)$ be the minimizer of $H_{x}$

with $v(x,\pm 1)=\pm f_{N}(x)$.

  1. Case 1:

When $f_{N}(x)\geq 1$, $v(y)=yf_{N}(x)$, for $y\in[-1,1].$

Notice that in this case, $H_{x}(v)=2f_{N}(x)^{2}+2.$

  2. Case 2:

When $f_{N}(x)<1$, $v(y)=\text{sgn}(y)(|y|-1+f_{N}(x))_{+}.$

In this case, $H_{x}(v)=4f_{N}(x).$

$-1$$1$$1$$-1$$f_{N}(x)$$-f_{N}(x)$$y$$v_{N}(y)$$-1$$1$$1$$-1$$f_{N}(x)$$-f_{N}(x)$$y$$v_{N}(y)$

A crucial observation is that even though we built $v_{N}$ “slice by slice”,
its $x$-derivative still has small $L^{2}$-norm.

######  Lemma 2.2.

We have

(2.3) $\iint_{\overline{R}_{N}}\left|\frac{\partial v_{N}}{\partial
x}\right|^{2}\leq\frac{16}{N}.$

###### Proof.

Notice that $v_{N}(x,y)$ is not harmonic in $\\{v\neq 0\\}$, hence $v_{N}$ is
not a minimizer of $J(-,R_{N})$. However, $\frac{\partial v_{N}}{\partial x}$
exists a.e. in $(-3N,3N)\times(-1,1)$ and, where it exists,
$\left|\frac{\partial v_{N}}{\partial
x}\right|\leq|f_{N}^{\prime}(x)|\leq\frac{1+\alpha}{N}\leq\frac{2}{N}.$

Since $f(x)\equiv 1-\alpha$ for $|x|<N$ and $f(x)\equiv 2$ for $2N<|x|<3N$,
$\left|\frac{\partial v_{N}}{\partial x}\right|=0$ on $(-N,N)\times(-1,1)$, on
$(2N,3N)\times(-1,1)$ and on $(-3N,-2N)\times(-1,1)$. Consequently,

$\iint_{\overline{R}_{N}}\left|\frac{\partial v_{N}}{\partial
x}\right|^{2}=\int_{-2N}^{-N}\int_{-1}^{1}\left|\frac{\partial v_{N}}{\partial
x}\right|^{2}+\int_{N}^{2N}\int_{-1}^{1}\left|\frac{\partial v_{N}}{\partial
x}\right|^{2}\leq\left(\frac{2}{N}\right)^{2}(2N)\leq\frac{16}{N}.$

∎

Using the fact that $v_{N}$ has smaller energy than $u_{N}$ “slice by slice”,
but larger energy overall, we can transfer (2.3) to $u_{N}$:

######  Lemma 2.3.

Any $u_{N}$ which minimizes $J$ in $\overline{R}_{N}$ with $u_{N}(x,\pm 1)=\pm
f_{N}(x)$, satisfies

(2.4) $\iint_{\overline{R}_{N}}\left|\frac{\partial u_{N}}{\partial
x}\right|^{2}\leq\frac{16}{N}.$

###### Proof.

Since $u_{N}$ is a minimizer of $J$,

(2.5) $\iint_{\overline{R}_{N}}|\nabla u_{N}|^{2}+|\\{u_{N}\neq
0\\}|\leq\iint_{\overline{R}_{N}}|\nabla v_{N}|^{2}+|\\{v_{N}\neq 0\\}|.$

Moreover, for every $x\in[-3N,3N]$ fixed, $v_{N}(x,\cdot)$ is a minimizer of
$H_{x}$, hence

$\int_{-1}^{1}\left|\frac{\partial v_{N}}{\partial
y}\right|^{2}dy+|\\{v_{N}(x,\cdot)\neq
0\\}|\leq\int_{-1}^{1}\left|\frac{\partial u_{N}}{\partial
y}\right|^{2}dy+|\\{u_{N}(x,\cdot)\neq 0\\}|.$

Integrating the last inequality on $[-3N,3N]$ leads to

(2.6) $\iint_{\overline{R}_{N}}\left|\frac{\partial v_{N}}{\partial
y}\right|^{2}+|\\{v_{N}\neq
0\\}|\leq\iint_{\overline{R}_{N}}\left|\frac{\partial u_{N}}{\partial
y}\right|^{2}+|\\{u_{N}\neq 0\\}|.$

Combining (2.5) and (2.6) we conclude

$\iint_{\overline{R}_{N}}\left|\frac{\partial u_{N}}{\partial
x}\right|^{2}\leq\iint_{R}\left|\frac{\partial v_{N}}{\partial x}\right|^{2}.$

∎

Our next goal is to prove a uniform Lipschitz bound on $u_{N}$. To do so, we
will compare it with $v_{N}$. Since the latter minimizes the energy on each
slice, it will be convenient to integrate this “slice-by-slice” energy across
all values of $x$:

######  Definition 2.4.

Define the “total sliced energies” of a function $w$ by

(2.7) $S(w)=\int_{-3N}^{3N}\int_{-1}^{1}\left|\frac{\partial w}{\partial
y}\right|^{2}dydx+|\\{w\neq 0\\}|=\int_{-3N}^{3N}H_{x}(w(x,\cdot))dx,$

and with $Q=[a,b]\times[-1,1]\subset\overline{R}_{N}$,

(2.8) $S_{Q}(w)=\iint_{Q}\left|\frac{\partial w}{\partial
y}\right|^{2}dydx+|\\{w\neq 0\\}\cap Q|=\int_{a}^{b}H_{x}(w(x,\cdot))dx.$

The following lemma encapsulates the fact that $u_{N}$ is a minimizer and
$v_{N}$ is a slice minimizer, written in the language of total slice energy:

######  Lemma 2.5.

We have

(2.9) $S(v_{N})\leq S(u_{N})\leq J(u_{N})\leq S(v_{N})+\frac{16}{N}.$

###### Proof.

Let $w\in W^{1,2}(\overline{R}_{N})$. Notice that

(2.10) $S(w)=\iint_{\overline{R}_{N}}\left|\frac{\partial w}{\partial
y}\right|^{2}+|\\{w\neq 0\\}|\leq J(w),\quad
S(w)+\iint_{\overline{R}_{N}}\left|\frac{\partial w}{\partial
x}\right|^{2}=J(w).$

Using (2.10) and the fact that $H_{x}(v_{N}(x,\cdot))\leq
H_{x}(u_{N}(x,\cdot))$ a.e., we obtain

$S(v_{N})\leq S(u_{N})\leq J(u_{N}).$

Combining this with the equality in (2.10) and with (2.3), we conclude that if
$u_{N}$ is a minimizer of $J$, then

$S(v_{N})\leq S(u_{N})\leq J(u_{N})\leq
J(v_{N})=S(v_{N})+\iint_{\overline{R}_{N}}\left|\frac{\partial v_{N}}{\partial
x}\right|^{2}\leq S(v_{N})+\frac{16}{N}.$

∎

We can also localize these estimates to $Q=[a,b]\times[-1,1]$:

######  Lemma 2.6.

We also have

(2.11) $S_{Q}(u_{N})\leq S_{Q}(v_{N})+\frac{16}{N}.$

###### Proof.

Given a subset $X\subset[-3N,3N]\times[-1,1]$ and defining

$S_{X}(u_{N})=\iint_{X}\left|\frac{\partial u_{N}}{\partial
y}\right|^{2}dxdy+|\\{u_{N}\neq 0\\}\cap X|,$

we obtain $S(u_{N})=S_{Q}(u_{N})+S_{{\overline{R}_{N}}\setminus Q}(u_{N})$. If
we had $S_{Q}(u_{N})>S_{Q}(v_{N})+\frac{16}{N}$, then

(2.12) $S(u_{N})>S_{Q}(v_{N})+\frac{16}{N}+S_{\overline{R}_{N}\setminus
Q}(u_{N}).$

Since for a.e. $x\in[-3N,3N]$ we have $H_{x}(v_{N}(x,\cdot))\leq
H_{x}(u_{N}(x,\cdot))$, integrating this inequality we obtain
$S_{{\overline{R}_{N}}\setminus Q}(v_{N})\leq S_{{\overline{R}_{N}}\setminus
Q}(u_{N})$. Together with (2.12), this gives

$S(u_{N})>S_{Q}(v_{N})+\frac{16}{N}+S_{{\overline{R}_{N}}\setminus
Q}(v_{N})=S(v_{N})+\frac{16}{N},$

contradicting (2.9). ∎

With these estimates we are almost ready to prove the (uniform) Lipschitz
continuity of the $u_{N}$ on compact subsets of ${\overline{R}_{N}}$. We
introduce the following notation, by analogy with Definition 2.4, for
$Q=[a,b]\times[-1,1]$

(2.13) $J_{Q}(w):=\iint_{Q}|\nabla w|^{2}\,dA+|Q\cap\\{w\neq 0\\}|.$

Our first result is an immediate corollary of Lemma 2.6 and (2.4):

######  Corollary 2.7.

There exists a constant $C_{0}>0$ such that for any $N>0$ and any
$Q=[a,b]\times[-1,1]\subset\overline{R}_{N}$ we have

$J_{Q}(u_{N})\leq C_{0}|b-a|+\frac{32}{N}.$

###### Proof.

From Lemma 2.6 we have that $S_{Q}(u_{N})\leq S_{Q}(v_{N})+\frac{16}{N}$ and
from (2.4) we have $\iint_{Q}|\partial_{x}u_{N}|^{2}\,dA\leq\frac{16}{N}$.
Putting this together we get that

$J_{Q}(u_{N})\leq S_{Q}(v_{N})+\frac{32}{N}.$

Thus it suffices to show that $S_{Q}(v_{N})$ grows proportionally to $|b-a|$
with a constant of proportionality independent of $a,b,N$. Indeed, by Lemma
2.1 $H_{x}(v_{N})\leq\max\\{2f_{N}(x)^{2}+2,4f_{N}(x)\\}\leq 10$. Integrating
that across $[a,b]$ gives the desired result. ∎

From here we can conclude the main result of this section, the uniform
Lipschitz continuity:

######  Theorem 2.8.

For $0<\delta<1$, we can find constants $L\geq 1$ and $N_{0}>0$ such that

$\|u_{N}\|_{\mathrm{Lip}(\Omega_{N})}\leq L$

for $N\geq N_{0}$, where
$\Omega_{N}=(-3N+\delta,3N-\delta)\times(-1+\delta,1-\delta)$.

###### Proof.

As $u_{N}$ is a minimizer, we can apply [DT15, Theorem 8.1] (c.f. the
discussion at the bottom of page 504 in [DT15]) which gives a Lipschitz bounds
on almost-minimizers depending only on the distance from the boundary and the
$L^{2}$-norm of the gradient (see also [DET19, Remark 2.2]). Corollary 2.7
gives uniform bounds on the $L^{2}$-norm of the gradient of $u_{N}$ inside of
any rectangle and a covering argument finishes the proof. ∎

##  3\. Existence of a Zero set

The goals of this section are two fold. First to prove that
$|\\{u_{N}=0\\}|>0$ (which will follow from Lemma 3.5) and second to show that
the set $\\{u_{N}=0\\}$ does not get too close to the boundary of $R_{N}$
(Lemma 3.4 and Corollary 3.7).

Let us first describe the zero set of $v_{N}$, the “slice minimizer”:

######  Lemma 3.1.

The following holds regarding the set $\\{v_{N}(x,y)=0\\}$:

  * •

When $|x|\geq\frac{(1+2\alpha)N}{1+\alpha}$, $v_{N}(x,y)=0$ only when $y=0$.

  * •

When $|x|\leq N$, $v_{N}(x,y)=0$ for $|y|\leq\alpha$.

  * •

When $N<|x|<\frac{(1+2\alpha)N}{1+\alpha}$, $v_{N}(x,y)=0$ when $|y|\leq
2\alpha+1-\frac{|x|(1+\alpha)}{N}$.

###### Proof.

The result follows from the following simple observations:

  * •

When $|x|\geq\frac{(1+2\alpha)N}{1+\alpha}$, $f_{N}(x)\geq 1$. In this case
$v_{N}(x,y)=yf_{N}(x)$.

  * •

When $|x|\leq N$, $f_{N}(x)\equiv 1-\alpha$, and
$v_{N}(x,y)=\text{sgn}(y)(|y|-\alpha)_{+}$.

  * •

In the remaining interval, $f_{N}(x)=\frac{|x|(1+\alpha)}{N}-2\alpha$ and
$v_{N}(x,y)=\text{sgn}(y)(|y|-1+\frac{|x|(1+\alpha)}{N}-2\alpha)_{+}$.

∎

$N$$\frac{(1+2\alpha)N}{1+\alpha}$$\frac{-(1+2\alpha)N}{1+\alpha}$$-N$$2N$$-2N$$3N$$-3N$$x$$y$$1$$-1$$\alpha$$-\alpha$$v=0$

We expect a minimizer $u_{N}$ of $J$, taken among all functions $w\in
W^{1,2}({\overline{R}_{N}})$ with $w(x,\pm 1)=\pm f_{N}(x)$,333Recall we do
not need to specify the data on the “vertical” components of the boundary to
look similar to $v_{N}$. In particular, we want to extract information about
its zero set, and prove that $\\{u_{N}=0\\}$ has a “pool” close to $0$.

Before we can prove this closeness, we need to observe that our minimizer is
nice on “most” of the vertical slices.

######  Definition 3.2.

Let $X_{0}\subset I$ be the smallest set such that $x\notin X_{0}$ implies
that $u_{N}(x,\cdot)\in W^{1,2}([-1,1])$ and $\lim\limits_{y\rightarrow\pm
1}u_{N}(x,y)=\pm f_{N}(x)$.

Since $u_{N}\in W^{1,2}(R_{N})$ we note that $X_{0}$ has measure zero.

We now show that the zero set of $u_{N}$ does not get too close to the “top”
or “bottom” of the rectangle. We start by showing that if $u_{N}$ is small
close to the top or bottom of the rectangle, then that slice has large energy:

######  Lemma 3.3.

Let $\varepsilon\in(0,1),\delta\in(0,1/2)$ and assume that
$|u_{N}(x,y)|<\delta$ for some $x\in[-3N,3N]\backslash X_{0}$ and some $y$
with $1-|y|<\varepsilon$. Then
$H_{x}(u_{N}(x,-))\geq\frac{(f_{N}(x)-\delta)^{2}}{\varepsilon}$.

In particular, if $x\in I\setminus X_{0}$ and there exists $y$ such that
$1-|y|<\frac{1}{44}$ and $|u_{N}(x,y)|<\frac{1}{4}$, then
$H_{x}(u_{N}(x,-))\geq H_{x}(v_{N}(x,-))+1$.

###### Proof.

Without loss of generality, assume $u_{N}(x,-)$ is both equal to $f_{N}(x)$
and $\delta$ on an interval of length $\varepsilon$. The lowest energy way to
do this is assuming these values are achieved at the endpoints of the interval
and that $u_{N}$ interpolates between them linearly. Thus
$H_{x}(u_{N}(x,-))\geq\frac{(f_{N}(x)-\delta)^{2}}{\varepsilon}$. The second
result follows from the first once we remember that $f_{N}(x)\geq\frac{3}{4}$
for all $x$, and $H_{x}(v_{N}(x,-))\leq 10$ for all $x$. ∎

We are now ready to show the existence of a strip near the top and bottom of
$R$, on which $u$ cannot vanish. We actually show something stronger which is
that $u$ is quantitatively large in this strip:

######  Lemma 3.4.

Let $\delta>0$ and set

(3.1) $R_{\pm}=\left\\{(x,r)\in R_{N}\,;\,0\leq|x|\leq 3N-\delta\text{ and
}\pm y\in\left[1-\frac{1}{44},1-\frac{1}{88}\right]\right\\}$

(the two blue zones in the picture). Then there exists an
$N_{0}=N_{0}(\delta)>1$ such that if $N>N_{0}$ then

(3.2) $u_{N}(x,y)\geq 1/8\ \text{ on $R_{+}$ and $u_{N}(x,y)\leq-1/8$ on
}R_{-}.$

$N$$-N$$2N$$-2N$$3N$$-3N$$x$$y$$1$$-1$

###### Proof.

Fix $\delta>0$. Recall from Theorem 2.8 that there exists $L>0$ (independent
of $N$ but dependent on $\delta>0$) such that if

$\Omega=\left[-3N+\delta,3N-\delta\right]\times\left[-1+\frac{1}{100},1-\frac{1}{100}\right],$

then $\|u_{N}\|_{\mathrm{Lip}(\Omega)}\leq L$.

We first check that $|u_{N}(x_{0},y_{0})|<\frac{1}{8}$ on $R_{\pm}$. Note that
if $|u_{N}(x_{0},y_{0})|<\frac{1}{8}$ for some
$(x_{0},y_{0})\in[-3N+\delta,3N-\delta]\times([-1+\frac{1}{88},-1+\frac{1}{44}]\cup[1-\frac{1}{44},1-\frac{1}{88}])$,
then by Lipschitz continuity there exists an interval
$I\subset[-3N+\delta,3N-\delta]$ of length $\frac{1}{8L}$ such that
$|u(x,y_{0})|<\frac{1}{4}$ for all $x\in I$.

We apply Lemma 3.3 to conclude that for almost every $x\in I$ we have
$H_{x}(u_{N}(x,-))\geq H_{x}(v_{N}(x,-))+1$. If $Q=I\times[-1,1]$, then this
implies that $S_{Q}(u_{N})\geq S_{Q}(v_{N})+|I|=S_{Q}(v_{N})+\frac{1}{8L}.$ Of
course, this contradicts (2.11) as long as
$\frac{16}{N}<\frac{1}{8L}\Leftrightarrow 128L<N$.

Now we check that $u$ has the right sign on $R_{\pm}$. Suppose for instance
that $u(x,y)\leq-1/8$ somewhere on $R_{+}$. Since $u$ is continuous,
$u(x,y)\leq-1/8$ everywhere on $R_{+}$. Then for all
$x\in\left[-3N+\delta,3N-\delta\right]\setminus X_{0}$, $u(x,-)$ is a Sobolev
function that goes from $-1/8$ to at least $1/4$ in an interval of length at
most $1/88$, a direct computation shows that
$H_{x}(u_{N}(x,-))\geq\frac{3}{8}\times 88\geq 33$, and we reach a
contradiction as above. ∎

Now we show that $\\{u_{N}=0\\}$ must be contained in a strip around
$\\{y=0\\}$ when $|x|<N$. Actually we prove something more precise:

######  Lemma 3.5.

There exists an $N>1$ large enough such if $|x|<N-1$, and $u_{N}(x,y)>0$, then
$y>\alpha/8$. Similarly, if $u_{N}(x,y)<0$ then $y<-\alpha/8$.

###### Proof.

Assuming by contradiction this were not the case, without loss of generality
there would exist a point in $(x_{0},y_{0})$ with
$|x_{0}|<N-1,y_{0}\leq\alpha/8$ and $u_{N}(x_{0},y_{0})>0$.

By continuity of $u_{N}$, the connected component of $\\{u_{N}>0\\}$
containing $(x_{0},y_{0})$ must be seperated from $\\{y=-1\\}$ by
$\\{u_{N}=0\\}$. This implies that there is a connected subset of
$\\{(x,y)\mid|x|\leq N,u_{N}(x,y)=0\\}$ which touches the sets $\\{x=N\\}$ and
$\\{x=-N\\}$ and which separates $(x_{0},y_{0})$ from $\\{y=-1\\}$. By Lemma
3.4 this connected component cannot intersect the set
$\\{(x,y)\mid|x|<N,-1+1/44<y<-1+1/88\\}$. If $y_{0}<-1+1/44$ then this
connected component lies below $y=-1+1/88$ and the length of its projection
onto the $x$-axis is at least $2N$. Such a configuration has too much energy
by Lemma 3.3 and thus we can assume $y_{0}>-1+1/44$.

We can also assume that $x_{0}\notin X_{0}$ (since $\\{u_{N}>0\\}$ is open),
so by continuity of $u_{N}$ on the slice $\\{x=x_{0}\\}$ and Lemma 3.4 there
exists a point $(x_{0},\tilde{y})$ with $-1+1/44<\tilde{y}<y_{0}$ and
$u_{N}(x_{0},\tilde{y})=0$ and $(x_{0},\tilde{y})\in\partial\\{u_{N}>0\\}$.

We note that $\\{u_{N}>0\\}\cap([-N,N]\times[-1+1/88,1-1/88])$ is a locally
NTA domain, uniformly in $N$ (i.e. for any
$K\subset\subset([-N,N]\times[-1+1/88,1-1/88])$, $\\{u_{N}>0\\}$ satisfies the
corkscrew conditions at $Q\in\partial\\{u_{N}>0\\}\cap K$ with constants and
at scales that depend only on $K$ not $N$ c.f. [DET19, Theorem 2.3])). In
particular, there exists a point $(x_{1},y_{1})\in\\{u_{N}>0\\}$ such that
$\|(x_{1},y_{1})-(x_{0},\tilde{y})\|\leq r_{0}=r_{0}(K)\leq\alpha/8$ and
$\mathrm{dist}((x_{1},y_{1}),\\{u_{N}\leq 0\\})\geq r_{0}/M$ for some $M>1$
(where both $r_{0},M$ are independent of $N$ large).

By the non-degeneracy of $u_{N}$ (c.f. [AC81, Lemma 3.4]) and the Lipschitz
continuity of $u_{N}$, Theorem 2.8, there exists a constant $C>1$ (again
uniform for large $N$) such that $u_{N}\geq r_{0}/C$ in the ball
$B((x_{1},y_{1}),r_{0}/(3CL))$ (where $L$ is the Lipschitz constant).

Therefore, there exists an interval $I\subset[-N,N]$ of length $2r_{0}/(3CL)$
such that for each $x\in I$ there exists a $y<\alpha/2$ with
$u_{N}(x,y)>r_{0}/C$. Invoking the computations of Section 6.2 (c.f. Claim
6.1), we get that $H_{x}(u_{N}(x,-))\geq H_{x}(v_{N}(x,-))+\eta$ for every
$x\in I\backslash X_{0}$, where $\eta=\eta(\alpha,C,r_{0})>0$ is independent
of $N$.

Let $Q=I\times[-1,1]$; we get (from (2.11))

$S_{Q}(v_{N})+\frac{16}{N}\geq S_{Q}(u_{N})\geq
S_{Q}(v_{N})+\mathcal{H}^{1}(I)\eta.$

This gives a contradiction if $N>0$ is large enough (since
$|I|=2r_{0}/(3CL),\eta>0$ are independent of $N$). ∎

So we have a good control on where $\\{u_{N}=0\\}$ is in the central region.
Before we end this section, it behooves us to refine the result of Lemma 3.3,
with the goal of showing that when $f_{N}(x)\geq 1$, we can actually confine
$\\{u_{N}=0\\}$ to an arbitrarily thin strip around the line $\\{y=0\\}$. This
will be used in the next section to show that there are no one-phase points on
the sides of $R_{N}$. We first estimate the difference between $v_{N}(x,-)$
and near minimizers for $H_{x}$.

######  Lemma 3.6.

Let $x\in(-3N,3N)$ be such that $f_{N}(x)\geq 1$ and let
$\varepsilon\in(0,1)$. Let $w\in W^{1,2}([-1,1])$, with $w(\pm 1)=\pm
f_{N}(x)$ and assume $H_{x}(w)\leq H_{x}(v_{N}(x,-))+\varepsilon$. Then, there
exists a $C>0$ (uniform over the choice of $x,\varepsilon$ above), such that

$||w-v_{N}||_{L^{\infty}([-1,1])}\leq C\varepsilon^{1/2}.$

###### Proof.

Recall that for $x$ as in the statement of Lemma 3.6, we have
$v_{N}(x,y)=yf_{N}(x)$. Our plan is to first modify $w$, reducing energy, and
show the desired inequality for $w$, then estimate the $L^{\infty}$ distance
between the original $w$ and our modified functional. We know that $w\in
W^{1,2}([-1,1])$ so by Sobolev embedding we have that $w$ is Hölder continuous
and thus

$-1<\alpha:=\inf\\{t\mid w(t)=0\\}\leq\sup\\{t\mid w(t)=0\\}=:\beta<1.$

We construct

$\hat{w}(t)=\begin{cases}0&t\in[a,b]\\\
\frac{f_{N}(x)(t-a)}{1+a}&t\in[-1,a]\\\
\frac{f_{N}(x)(t-b)}{1-b}&t\in[b,1].\end{cases}$

We observe that $H_{x}(w)\geq H_{x}(\hat{w})$ (as we have enlarged the zero
set and minimized Dirichlet energy where $w$ is positive).

We can compute that

$H_{x}(\hat{w})=f_{N}^{2}(x)\left(\frac{1}{1-b}+\frac{1}{a+1}\right)+2-(b-a).$

Recall that $H_{x}(v_{N}(x,-))=2f_{N}^{2}(x)+2$ and rewrite

$\displaystyle H_{x}(\hat{w})-H_{x}(v_{N}(x,-))=$ $\displaystyle
f_{N}^{2}(x)\left(\frac{1}{1-b}+\frac{1}{a+1}-2\right)-(b-a)$
$\displaystyle\geq$
$\displaystyle\left(\frac{1}{1-b}+\frac{1}{a+1}-2\right)-(b-a)=:F(a,b),$

where the last inequality follows because $f_{N}(x)\geq 1$ in the salient
range and $\frac{1}{1-b}+\frac{1}{a+1}-2\geq 0$ as long as $1>b\geq a>-1$.

We compute that $F(0,0)=0,\nabla F(0,0)=(0,0)$ and that

$\nabla^{2}F(a_{0},b_{0})=\begin{pmatrix}\frac{2}{(1+a_{0})^{3}}&0\\\
0&\frac{2}{(1-b)^{3}}\end{pmatrix},$

is a diagonal matrix with entries between $10^{7}$ and $\frac{1}{10^{7}}$ as
long as $[a_{0},b_{0}]\in\left[-\frac{99}{100},\frac{99}{100}\right]$. If
either of $a_{0},b_{0}$ is outside that range, then Lemma 3.3 gives a
contradiction to the assumption on energy. Thus, by the Taylor remainder
theorem (and the fact that $F$ is $C^{2}$ as long as $a$ stays away $-1$ and
$b$ stays away from $1$) we have that

$|F(a,b)-F(0,0)|=(a,b)^{T}\nabla^{2}F(a_{0},b_{0})(a,b)\leq C(a^{2}+b^{2})$

for some $(a_{0},b_{0})$ on the segment connecting $(0,0)$ and $(a,b)$.

On the other hand, since $v_{N}(x,-)$ is linear in $y$ and $\hat{w}$ is
piecewise linear in $y$ we can see that

$\|v_{N}-\hat{w}\|_{L^{\infty}}=\max_{t=a,b}|v_{N}(x,t)-\hat{w}(t)|=f_{N}(x)\max_{t=a,b}|t|.$

Chaining everything together we get that

$\|v_{N}-\hat{w}\|_{L^{\infty}}\leq C\sqrt{H_{x}(\hat{w})-H_{x}(v_{N}(x,-))}.$

We now estimate $|\hat{w}-w|(t_{0})$ for $t_{0}\in[-1,1]$. We have two cases;
in the first, assume that $t_{0}\in[a,b]$. Then
$2|\hat{w}(t_{0})-w(t_{0})|=2|w(t_{0})|\leq\int_{a}^{t_{0}}|w_{y}|+\int_{t_{0}}^{b}|w_{y}|$.
Using Jensen’s inequality we get that

$|w(t_{0})|\leq\left(\frac{b-a}{4}\int_{a}^{b}|w_{y}|^{2}\,dy\right)^{1/2}\leq
C\sqrt{H_{x}(w)-H_{x}(\hat{w})}.$

Putting this together we have that

$|v_{N}(x,t_{0})-w(t_{0})|\leq
C\sqrt{H_{x}(w)-H_{x}(\hat{w})}+C\sqrt{H_{x}(\hat{w})-H_{x}(v_{N}(x,-))}\leq
C\varepsilon^{1/2}.$

In the second case we assume that $t_{0}\in[-1,a]$ (the case that
$t_{0}\in[b,1]$ works the same way). Since $w(-1)=\hat{w}(-1)$ and
$w(a)=\hat{w}(a)$ we get that
$2|w(t_{0})-\hat{w}(t_{0})|\leq\int_{-1}^{a}|\partial_{y}(w-\hat{w})|\,dy$.
Again applying Jensen’s inequality we get that

$|w(t_{0})-\hat{w}(t_{0})|\leq
C(a+1)\left(\int_{-1}^{a}\left|\partial_{y}w-\frac{f_{N}}{1+a}\right|^{2}\,dy\right)^{1/2}.$

Expanding out the integrand, and using the fact thatat $w(a)-w(-1)=f_{N}(x)$
we get

$|w(t_{0})-\hat{w}(t_{0})|\leq
C\left(\int_{-1}^{a}|\partial_{y}w|^{2}\,dy-2\frac{f_{N}(x)^{2}}{1+a}+\frac{f_{N}(x)^{2}}{1+a}\right)^{1/2}\leq
C\left(H_{x}(w)-H_{x}(\hat{w})\right)^{1/2}.$

We can chain the inequalities together as above to get the desired result. ∎

From here, we have an easy corollary: outside of the central box, the zero set
of $u_{N}$ is contained in a very thin strip around $\\{y=0\\}$.

######  Corollary 3.7.

Let $\delta,\theta>0$. There exists an $N_{0}=N_{0}(\delta,\theta)>0$ such
that for $N>N_{0}$ and every pair $(x,y)\in R_{N}$ such that $|x|<3N-\delta$,
$|y|\leq 1-\frac{1}{88}$, $f_{N}(x)\geq 1$, and $u_{N}(x,y)=0$, we have
$|y|<\theta$.

###### Proof.

Fix $\delta>0$ and let $N>0$ be big enough so that, invoking Theorem 2.8, we
can say $u_{N}$ is $L$-Lipschitz in
$\left[-3N+\delta,3N-\delta\right]\times\left[-1+\frac{1}{44},1-\frac{1}{44}\right]$.

Assume there exists a point $(x_{0},y_{0})$ such that $f_{N}(x_{0})\geq 1$,
$\theta\leq|y_{0}|\leq 1-\frac{1}{88}$, and $u_{N}(x_{0},y_{0})=0$. We know
that $|y_{0}|<1-\frac{1}{44}$, by Lemma 3.4. Then there exists an interval $I$
of length at least $\frac{\theta}{2L}$ such that
$|u_{N}(x,y_{0})-y_{0}f_{N}(x)|\geq\frac{\theta}{2}$ and $f_{N}(x)\geq 1$ on
all $x\in I$.

By Lemma 3.6, this implies that $H_{x}(u_{N}(x,-))\geq
H_{x}(v_{N}(x,-))+\frac{\theta^{4}}{4C}$ for almost every $x\in I$.
Integrating and letting $Q=I\times[-1,1]$ we get that $S_{Q}(u_{N})\geq
S_{Q}(v_{N})+C^{-1}\theta^{5}$ where $C=C(\delta)>0$ is independent of
$N,\theta$. This contradicts (2.11) as long as $N$ is large enough (depending
on $C,\theta$ and thus on $\delta,\theta$). ∎

##  4\. The Proof of Theorem 1.2: Ruling out One-Phase Points

The main goal of this section is to finish up the proof of our main Theorem
1.2, that $\Gamma_{\mathrm{BP}}(u_{N})\neq\emptyset$. In fact, we have the
following more precise description.

######  Theorem 4.1.

Let $u_{N},R_{N},f_{N}$ be as above. Then, for $N$ large enough, there exists
a “pool of zeroes”, i.e., a connected open set
$\mathcal{O}\subset\\{u_{N}=0\\}\cap\\{|x|<2N+1\\}\cap\\{|y|\leq
1-\frac{1}{44}\\}$ such that $|\mathcal{O}|>0$, $\partial{\mathcal{O}}$ is
contained in the free boundary $\Gamma^{+}\cup\Gamma^{-}$, and
$\partial{\mathcal{O}}$ meets $\Gamma^{+}$, $\Gamma^{-}$, and the set,
$\Gamma_{\mathrm{TP}}(u)$, of branch points.

The idea is that Lemma 3.5 guarantees the existence of a “pool of zeroes”
separating the positive and negative phases in the central part of $R_{N}$. We
then want to show that this pool does not “leak” to the sides of $R_{N}$. For
this we need the following lemma, whose proof will be the main goal of this
section.

######  Lemma 4.2.

Let

$R_{ext}=\big{\\{}(x,y)\in R_{N}\,\mid\,2N+1<|x|<3N-1\text{ and }|y|\leq
1-\frac{1}{88}\big{\\}}.$

There exists a $N_{0}>1$ such that if $N\geq N_{0}$, then every free boundary
point for $u_{N}$ in $R_{ext}$ is a two-phase point. Or, put another way:

$R_{ext}\cap(\Gamma^{+}(u_{N})\cup\Gamma^{-}(u_{N}))\subset\Gamma_{\mathrm{TP}}(u).$

Before we prove the lemma, let us see how its proof implies the theorem.

###### Proof of Theorem 4.1 assuming Lemma 4.2.

By Lemma 3.5, $u_{N}$ vanishes in the region where $|x|\leq N-1$ and $|y|\leq
1/8$. Denote by ${\mathcal{O}}_{0}$ the interior of $\\{u_{N}=0\\}$, and let
$\mathcal{O}$ the connected component of $B(0,1/8)$ in ${\mathcal{O}}_{0}$.

Let us first check that $\partial\mathcal{O}$ contains one-phase points of
both types. Consider the line segment $\ell$ from the origin to
$z_{+}=(0,1/44)$; we know that $u(z_{+})\geq 1/8$ (by (3.2)), so $\ell$ meets
$\partial\mathcal{O}$. At the first point of intersection $z=(0,y)$ (going up
from $0$), Lemma 3.5 says that $y>\alpha/8$, and then Lemma 3.5 says that
$u(w)\geq 0$ for $w$ near $z$; hence $z$ is a (positive) one-phase point.
Similarly, the first point of $\partial\mathcal{O}$ on the interval from $0$
to $z_{-}=(0,-1/44)$ is a negative one-phase point.

Next we want to show that ${\mathcal{O}}\cap R_{ext}=\emptyset$. Suppose not,
and let $\gamma$ be a path in ${\mathcal{O}}$ that goes from $0$ to some point
of $\mathcal{O}_{0}\cap R_{ext}$. Certainly $\gamma$ does not get close to the
top and bottom boundaries, i.e. where $|y|=1-\frac{1}{88}$, by (3.2). So there
is a point $(x_{0},y_{0})\in\gamma$ such that $|x_{0}|>2N+1$ and $|y_{0}|\leq
1-1/44$. Then, as above with the origin, we can find a point
$P=(x_{0},y_{0}^{\prime})$ above $(x_{0},y_{0})$, which lies in
$\partial\mathcal{O}$. This vertical segment lies inside of $\mathcal{O}$ (and
is non-empty by the openness of $\mathcal{O}_{0}$). By Lemma 4.2, this point
is a two-phase point. But in fact the proof of Lemma 4.2 will say more: near
$P$, the free boundary is a Lipschitz graph, with a small constant, and then
the non-degeneracy of $u$ shows that on the vertical line that goes through
$P$, $u$ is (strictly) positive on one side of $P$, and negative on the other
side; this contradicts the fact that the open segment between $(x_{0},y_{0})$
and $P$ lies in $\mathcal{O}$. Hence ${\mathcal{O}}\cap R_{ext}=\emptyset$.
Note this, with Lemma 3.4, implies that

$\mathcal{O}\subset R_{in}:=\\{(x,y)\in R_{N}\mid|x|\leq 2N+1,|y|\leq
1-\frac{1}{44}\\}.$

We still need to show that ${\mathcal{O}}$ contains a branch point. Suppose
not, and let $z\in\partial\mathcal{O}$ be given. Obviously $u$ takes nonzero
values near $z$, so $z$ lies in the free boundary, and by assumption $z$ is a
one-phase point (since only one phase points or branch points can be on the
boundary of an open subset of $\\{u=0\\}$). Suppose $z\in\Gamma^{+}(u)$. In
the present situation (and even in ambient dimension $3$), the free boundary
in a neighborhood of $z$ is a smooth hypersurface $\Gamma$, with $u>0$ on one
side of $\Gamma$, and $u=0$ on the other side. Thus
$B_{r}(z)\cap\partial\mathcal{O}\subset\Gamma^{+}(u)$ for some $r>0$ small
enough (depending on $z$).

More globally, the curve $\Gamma\subset\partial\mathcal{O}$ that contains $z$
is a Jordan curve (it is disjoint from $R_{ext}$, does not touch the boundary
of $R$, and is locally smooth), so $\mathcal{O}$, which is connected, is
contained in one of the two components of $U$ =
$\mathbb{R}^{2}\setminus\Gamma$. If $\mathcal{O}$ is contained in the
unbounded component of $U$, then we can replace $u$ with $0$ on the bounded
component of $U$ and keep $u$ a valid competitor (because
$\Gamma\subset\partial\mathcal{O}\subset R_{in}\subset\subset R_{N}$).
However, this will strictly decrease energy which is a contradiction. Thus
$\mathcal{O}$ is contained in the bounded component of $U$. Arguing as before,
if $\mathcal{O}$ is not the entirety of this bounded component, then we could
replace $u$ by $0$ on the rest of this bounded component and decrease energy.
So it must be that $\mathcal{O}$ is one of the connected components of
$\mathbb{R}^{2}\setminus\Gamma$.

This contradicts the fact that $\partial{\mathcal{O}}$ meets $\Gamma^{-}$ too.
So $\partial{\mathcal{O}}$ contains a branch point, and the theorem follows
from the proof of the lemma.

In ambient dimension $3$, the same proof would work, using the fact that a
connected smooth orientable hypersurface in $\mathbb{R}^{n}$ always separates
$\mathbb{R}^{n}$ in exactly two connected component, as in the Jordan curve
theorem. ∎

The rest of this section will be devoted to the proof of Lemma 4.2. We begin
by observing that when $|x|>2N$, $\partial_{y}u_{N}(x,-)$ must be close to
$\partial_{y}v_{N}(x,-)=2$ at most points:

######  Lemma 4.3.

For every $\varepsilon>0$ there exists an $N_{0}=N_{0}(\varepsilon)>0$ such
that if $N\geq N_{0}$ then

$\int_{\\{(x,y)\ |\ 2N<|x|<3N-\frac{1}{4},\ |y|\leq 1\\}}\left|\frac{\partial
u_{N}}{\partial y}-2\right|^{2}\,dxdy\leq\varepsilon.$

###### Proof.

Recall that $f_{N}(x)=2$ for $2N\leq|x|\leq 3N$, so $v_{N}(x,y)=2y$ and
$\partial_{y}v_{N}(x,-)=2$. Also, $y\mapsto v_{N}(x,y)$ is harmonic on
$[-1,1]$ and (when $x\notin X_{0}$) $y\mapsto u_{N}(x,y)$ is a $W^{1,2}$
function with the same trace as $v_{N}$. Fix such an $x$, and set
$I(w)=\int_{-1}^{1}|w_{y}|^{2}\,dy$ for $w\in W^{1,2}([-1,1])$. Amongst $w\in
W^{1,2}([-1,1])$ with the same boundary values $\pm 2$ as $v_{N}(x,-)$,
$v_{N}(x,-)$ minimizes $I$. Thus the Euler-Lagrange equation shows that
$\int_{-1}^{1}\partial_{y}v_{N}(\partial_{y}u_{N}-\partial_{y}v_{N})dy=0$.
Hence by Pythagorus

$\int_{-1}^{1}|\partial_{y}u_{N}|^{2}=\int_{-1}^{1}|\partial_{y}v_{N}|^{2}dy+\int_{-1}^{1}|\partial_{y}u_{N}-\partial_{y}v_{N}|^{2}dy.$

Now we also care about the functional $H_{x}$, so we need to add $|\\{y\mid
w(y)\neq 0\\}|$ to $I(w)$. For $v_{N}(x,-)$, we get $2$, since $v_{N}(x,y)=2y$
only vanishes at $0$. For $u_{N}(x,-)$, Corollary 3.7 says that for any given
$\theta>0$, and if $N$ is large enough $u_{N}(-)$ can only vanish for
$|y|<\theta$. Then $2-|\\{y\mid u_{N}(x,y)\neq 0\\}|\leq 2\theta$ and

$\displaystyle\int_{-1}^{1}|\partial_{y}u_{N}-2|^{2}dy$
$\displaystyle=\int_{-1}^{1}|\partial_{y}u_{N}-\partial_{y}v_{N}|^{2}dy=\int_{-1}^{1}|\partial_{y}u_{N}|^{2}\,dy-\int_{-1}^{1}|\partial_{y}v_{N}|^{2}\,dy$
$\displaystyle=H_{x}(u_{N}(x,-))-H_{x}(v_{N}(x,-))+2-|\\{y\mid u_{N}(x,y)\neq
0\\}|$ $\displaystyle\leq H_{x}(u_{N}(x,-))-H_{x}(v_{N}(x,-))+2\theta$

We now integrate over $x$ such that $2N<|x|<3N-\frac{1}{4}$, use (2.9), and
get the desired result for $N\geq N_{0}(\varepsilon)$. ∎

We now need to invoke the results of [DPSV21b] to show that the free boundary
is smooth and that the positive and negative parts of $u_{N}$ extend smoothly
to the free boundary. Recall that $u_{N}^{+}=\max\\{u_{N},0\\}$ and
$u_{N}^{-}=\max\\{-u_{N},0\\}$.

######  Lemma 4.4.

Let $\delta>0$. There exists $N_{0}=N_{0}(\delta)>1$ such that if $N\geq
N_{0}$, then each
$\partial\\{u^{\pm}_{N}>0\\}\cap(\\{2N<|x|<3N-1\\}\times[-1,1])$ is a
$C^{1,1/4}$-graph over the set $\\{y=0\\}$ with norm $\leq\delta$.
Furthermore,

$|\nabla u_{N}^{\pm}|\in
C^{0,1/4}\left(\overline{\\{u_{N}^{\pm}>0\\}}\cap(\\{2N\leq|x|\leq
3N-1-\delta\\}\times[-1,1])\right),$

with a $C^{0,1/4}$ seminorm less than $1$.

###### Proof.

Pick some $r_{0}>0$ small, but independent of $N,\delta$. For points $x_{0}$
such that $B(x_{0},r_{0})\cap\Gamma_{\mathrm{TP}}(u_{N})=\emptyset$, the
result follows from uniform Reifenberg flatness (Corollary 3.7) and standard
“flat-implies-smooth” regularity for the one-phase problem (c.f. [AC81]).

If $B(x_{0},r_{0})\cap\Gamma_{\mathrm{TP}}(u_{N})\neq\emptyset$, then we can
consider $y\in B(x_{0},r_{0})\cap\Gamma_{\mathrm{TP}}(u_{N})$ and look at
$B(y,2r_{0})$. Then the regularity follows from the $\theta$-Reifenberg
flatness at scale 1 (Corollary 3.7), the uniform Lipschitz continuity (Theorem
2.8) and [DPSV21b, Theorem 3.1].

In both instances, the dependence on $r_{0},\theta$ is such that the
$C^{1,1/4}$-norm of the graph(s) goes to zero as $r_{0}>0$ stays constant but
the Reifenberg flatness parameter, $\theta$, goes to zero. Corollary 3.7 says
we can take $\theta$ arbitrarily small, at the expends of making $N$ larger.
The regularity of the gradient is a consequence of standard elliptic
regularity once we know the regularity of the free boundary. ∎

We are finally ready to prove Lemma 4.2:

###### Proof.

Let $\varepsilon>0$ be small, to be chosen later and assume by contradition
that there exists a one phase point
$(x_{0},y_{0})\in\partial\\{u_{N}>0\\}\cap\\{2N\leq|x|\leq 3N-1\\}$. By Lemma
4.3 we may choose $N$ large enough (depending only on $\varepsilon$) such that
for any square, $Q$, of side length $\ell(Q)\leq 3/4$ centered at the point
$(x_{0},y_{0})$ we have

(4.1) $\int_{Q}\left|\frac{\partial u_{N}}{\partial
y}-2\right|^{2}\,dxdy\leq\varepsilon.$

Using Fubini we also get that

$\int_{0}^{3/4}\int_{\partial Q}\left|\frac{\partial u_{N}}{\partial
y}-2\right|^{2}\,d\mathcal{H}^{1}d\ell(Q)\leq\varepsilon,$

where the integration is occurring over squares all centered at
$(x_{0},y_{0})$ with increasing side lengths. Thus we can pick a cube $Q_{0}$
(which will be fixed going forward) with $1/2\leq\ell(Q_{0})\leq 3/4$ such
that

(4.2) $\int_{Q_{0}}\left|\frac{\partial u_{N}}{\partial
y}-2\right|^{2}\,dxdy,\int_{\partial Q_{0}}\left|\frac{\partial
u_{N}}{\partial y}-2\right|^{2}\,d\mathcal{H}^{1}\leq 4\varepsilon.$

Note that $u_{N}|_{Q_{0}}$ is $L$-Lipschitz (by Theorem 2.8), where $L$ is
independent of $N,Q_{0}$.

Let $\tilde{Q}=Q_{0}\cap\\{u_{N}>0\\}$. By Lemma 4.4, the domain $\tilde{Q}$
is piece-wise $C^{1,1/4}$ and is an NTA domain, with constants uniform in $N$
(c.f. [JK82] for definitions and details). Let us say a bit more about this;
the NTA constants depend on how the vertical edges of $Q$ touch the smooth
graph $\partial\\{u_{N}>0\\}$. However, this graph over $\\{y=0\\}$ has very
small norm (uniform in $N$) so this intersection happens (quantitatively)
transversely and thus the NTA constants are also uniform in $N$. These bounds
on norm of the graph which gives $\partial\\{u_{N}>0\\}$ also imply that
$4\geq\mathcal{H}^{1}(\tilde{\partial Q})\geq 1$ and
$|\tilde{Q}|\geq\frac{1}{8}$. From this information, using (4.2) and
Chebyshev, there exists a $A\in\tilde{Q}$ with
$\mathrm{dist}(A,\partial\tilde{Q})>\frac{1}{100}$ and
$|\partial_{y}u(A)-2|^{2}<8\varepsilon$ (this will work as long as
$\varepsilon>0$ is small enough).

Let $\omega_{N}^{A}$ be the harmonic measure of $\tilde{Q}$ with a pole at $A$
(where the notation emphasizes the dependence on $N$). Since $\partial_{y}u$
is a harmonic function in $\tilde{Q}$ we have the following integral
representation:

$\partial_{y}u_{N}(A)=\int_{\partial\tilde{Q}}\partial_{y}u_{N}(P)d\omega_{N}^{A}(P).$

Finally, it will be convenient to let
$\partial\tilde{Q}=\Gamma_{1}\cup\Gamma_{2}$ where
$\Gamma_{1}=\partial\\{u_{N}>0\\}\cap\overline{Q}_{0}$ and
$\Gamma_{2}=\partial Q_{0}\cap\overline{\\{u_{N}>0\\}}$, see the picture
below.

$\Gamma_{2}$$\Gamma_{1}$$\partial\\{u_{N}>0\\}$$\tilde{Q}$

Recall that $(x_{0},y_{0})$ is a one-phase point for $u_{N}$. Thus we can
compute
$|\partial_{y}u_{N}(x_{0},y_{0})|\leq|\partial_{\nu}u_{N}(x_{0},y_{0})|=1$,
where the last equality holds due to the free boundary condition at (regular)
one-phase points. Because the derivative of $u_{N}$ restricted to
$\partial\\{u_{N}>0\\}$ has $C^{0,1/4}$-semi-norm less than 1, we also have
that

$\displaystyle|\partial_{y}u_{N}(x_{0},y_{0})-\partial_{y}u_{N}(P)|<$
$\displaystyle\left(\frac{3}{4}\right)^{1/4}<\frac{94}{100}\ \text{ for all
}P\in\Gamma_{1}$ $\displaystyle\Rightarrow\partial_{y}u_{N}(P)\leq$
$\displaystyle 2-\frac{1}{20}\ \ \text{ for all }P\in\Gamma_{1}.$

Putting things together, we have that

(4.3) $\displaystyle 2-3\varepsilon^{1/2}\leq\partial_{y}u_{N}(A)\leq$
$\displaystyle\left(2-\frac{1}{20}\right)\omega_{N}^{A}(\Gamma_{1})+\int_{\Gamma_{2}}\partial_{y}u_{N}(P)d\omega_{N}^{A}(P)$
$\displaystyle\Rightarrow
2(1-\omega_{N}^{A}(\Gamma_{1}))+\frac{\omega_{N}^{A}(\Gamma_{1})}{20}\leq$
$\displaystyle\int_{\Gamma_{2}}\partial_{y}u_{N}(P)d\omega_{N}^{A}(P)+3\varepsilon^{1/2}$
$\displaystyle\Rightarrow 2+c\leq$
$\displaystyle\fint_{\Gamma_{2}}\partial_{y}u_{N}(P)d\omega_{N}^{A}(P),$

where in the last line we used that
$\omega_{N}^{A}(\Gamma_{2})=1-\omega_{N}^{A}(\Gamma_{1})$ and also Bourgain’s
Lemma (c.f. [JK82, Lemma 4.2]), which implies that there is a constant
$\tilde{c}>0$ (independent of $\varepsilon,N$) such that
$\frac{\omega_{N}^{A}(\Gamma_{1})}{1-\omega_{N}^{A}(\Gamma_{1})}\geq\tilde{c}$
and $\tilde{c}^{-1}\geq\frac{1}{1-\omega_{N}^{A}(\Gamma_{1})}$.

Recall that $u$ is $L-$Lipschitz and write
$\Gamma_{2}=\Gamma_{2,+}\cup\Gamma_{2,-}$, with

$\Gamma_{2,+}=\\{P\in\Gamma_{2}\mid L\geq|\partial_{y}u_{N}(P)|>2+c/2\\}\
\text{ and }\Gamma_{2,-}=\\{P\in\Gamma_{2}\mid|\partial_{y}u_{N}(P)|\leq
2+c/2\\}$

After overestimating $\fint_{\Gamma_{2}}\partial_{y}u_{N}d\omega_{N}^{A}$ over
each set, (4.3), gives us

(4.4) $\frac{c}{2}\leq
L\,\frac{\omega_{N}^{A}(\Gamma_{2,+})}{\omega_{N}^{A}(\Gamma_{2})}.$

Once more invoking Bourgain’s theorem we have that

$\theta:=\frac{c\omega_{N}^{A}(\Gamma_{2})}{2L}$

is bounded strictly away from zero, independently of $N$ (large enough) or
$\varepsilon>0$.

Using the condition on the integral of $|\partial_{y}u_{N}-2|$ on $\partial
Q_{0}$ (i.e. (4.2)) we see that

$\mathcal{H}^{1}(\Gamma_{2,+})<\frac{16\varepsilon}{c^{2}}.$

Recall that $\mathcal{H}^{1}(\partial\tilde{Q})\geq 1$ and we get that

(4.5)
$\frac{\mathcal{H}^{1}(\Gamma_{2,+})}{\mathcal{H}^{1}(\partial\tilde{Q})}<\frac{16\varepsilon}{c^{2}}\ll\theta\equiv\frac{\omega_{N}^{A}(\Gamma_{2,+})}{\omega_{N}^{A}(\partial\tilde{Q})}.$

We now recall that in $\tilde{Q}$ we have that the harmonic measure
$\omega_{N}^{A}\in A_{\infty}(\mathcal{H}^{1})$ (see e.g. [DJ90, Theorem 2]).
In fact, the $A_{\infty}$-constants depend on the NTA constants of
$\tilde{Q}$, the $1$-Ahlfors regularity of $\partial\tilde{Q}$, the distance
from $A$ to $\partial\tilde{Q}$ and the diameter of $\tilde{Q}$ (for more
details and definitions of the relevant terms, see [DJ90]). As discussed
above, all of these quantities can be taken uniform for $N$ large enough. Thus
we can take $\varepsilon>0$ small until (4.5) contradicts $\omega_{N}^{A}\in
A_{\infty}(\mathcal{H}^{1})$ and we are done. ∎

######  Remark 4.5.

The arguments above can be adapted to produce cusp points in
$\mathbb{R}^{2+1}$, but not directly in ambient dimensions larger than $3$. In
the setting of $\mathbb{R}^{3}$, our domain is given by
$R_{N}:=B^{\prime}(0,3N)\times[-1,1]\subset\mathbb{R}^{2+1}$ where
$B^{\prime}$ is a ball in $\mathbb{R}^{2}$. Then
$f_{N}(r,\theta)=f_{N}(r):B^{\prime}(0,3N)\rightarrow\mathbb{R}$ depends only
on the radial variable. Now we define $f_{N}$ piecewise:

$\displaystyle f_{N}(r)=\begin{cases}1-\alpha,&\text{ if }r\leq 1\\\
\left(\frac{\log(r)}{\log(2N)}+1\right)(1+\alpha)-2\alpha&\text{ if
}1\leq|x|\leq 2N\\\ 2,&\text{ if }2N\leq r\leq 3N.\end{cases}$

We can then define the slice minimizer similarly, where $v_{N}(r,\theta,-)\in
W^{1,2}([-1,1])$ minimizes with the boundary values $v_{N}(r,\theta,\pm 1)=\pm
f_{N}(r)$. In particular when $f_{N}(r)\geq 1$ we have
$v_{N}(r,\theta,y)=yf_{N}(r)$ and when $f_{N}(r)\leq 1$ we have
$v_{N}(r,\theta,y)=\mathrm{sgn}(y)\left(|y|-1+f_{N}(r)\right)_{+}$. In either
setting we have $|\partial_{r}v_{N}|\leq|\partial_{r}f_{N}|$.

Computing just like in Lemma 2.2, we get that

(4.6) $\iint_{\overline{R}_{N}}\left|\frac{\partial v_{N}}{\partial
r}\right|^{2}\simeq\int_{1}^{2N}\frac{1}{r^{2}\log(2N)^{2}}r\,dr=\frac{1}{\log(2N)}\stackrel{{\scriptstyle
N\rightarrow\infty}}{{\rightarrow}}0.$

From here we argue identically as above, noting that we never use the precise
bound on $\iint_{\overline{R}_{N}}\left|\frac{\partial v_{N}}{\partial
r}\right|^{2}$, just that it goes to zero as $N\rightarrow\infty$, and every
other quantity stays bounded.

For example, the contradiction in the proof of Lemma 3.5 now comes when
$\eta\leq\frac{C}{\eta\log(2N)}$, which is not true for $N>1$ large enough.

##  5\. Accumulating Cusps for Almost Minimizers

In this section we prove that the cusp set for almost-minimizers to (1.1) can
be essentially arbitrary. To state our results in maximum generality we
introduce the variable coefficient version of (1.1):

(5.1) $J_{\Omega}(u):=\int_{\Omega}|\nabla
u|^{2}+q_{+}^{2}(x)\chi_{\\{u>0\\}}(x)+q_{-}^{2}(x)\chi_{\\{u<0\\}}\,dx.$

Throughout this section we will assume that $q_{\pm}\in
C^{0,\alpha}(\overline{\Omega})$ for some $\alpha\in(0,1)$ and that the
weights satisfy the non-degeneracy condition; $q_{\pm}\geq c_{0}>0$ in all of
the domain. Clearly we recover the original functional (1.1) by letting
$q_{\pm}\equiv\lambda_{\pm}$ in (5.1).

We now state our main result:

######  Theorem 5.1.

Let $f^{-}\leq f^{+}\in C^{2}(\mathbb{R}^{n})$ such that $f^{+}=f^{-}=0$
outside of $B(0,R/10)$ for some large $R>0$. Let $\Gamma^{\pm}$ be the graphs
of $f^{\pm}$. For any $q_{\pm}\in C^{0,\alpha}$, with $c_{0}^{-1}\geq
q_{\pm}\geq c_{0}>0$, there exists an almost-minimizer $u$ to the energy
$J_{B(0,R)}$ such that $\Gamma^{\pm}=\partial\\{\pm u>0\\}$.

Notice that for $u$ as in Theorem 5.1 we have
$\Gamma_{\mathrm{BP}}(u)=\partial_{\mathbb{R}^{n-1}}\\{f^{-}(x)=f^{+}(x)\\}$.
Recall that any closed set can be the zero set of a $C^{2}$ function (take a
smoothing of the distance function to the given set, c.f. [Ste70, VI, Theorem
2]). As such, we have the following corollary (compare to Theorem 1.6 from the
introduction):

######  Corollary 5.2.

Let $E\subset\mathbb{R}^{n-1}$ be any compact set with no interior point in
$R^{n-1}$ and let $q_{\pm}\in C^{0,\alpha}(\mathbb{R}^{n})$ be non-degenerate
and bounded. Then, for some $R>0$ large enough depending on $E$, there exists
an almost-minimizer $u$ to $J_{B(0,R)}$ with weights $q_{\pm}$, such that
$\Gamma_{\mathrm{BP}}(u)=E$.

Furthermore, we can take $\Gamma^{+}:=\partial\\{u>0\\}$ to be the reflection
of $\Gamma^{-}:=\partial\\{u<0\\}$ around
$\\{x_{n}=0\\}\subset\mathbb{R}^{n}$.

Theorem 5.1 will follow from two lemmas, the first a general result about what
functions are almost-minimizers to the two-phase functional. The second, a
construction of such functions.

######  Lemma 5.3.

Let, $n\geq 2,\alpha\in(0,1),f^{-}\leq f^{+}\in C^{2}(\mathbb{R}^{n})$ such
that $f^{-}=f^{+}=0$ outside of $B(0,R/10)$ for some $R>1$, and let
$\Gamma^{\pm}$ be the graph of $f^{\pm}$. Let $\Omega^{+}$ (resp.
$\Omega^{-}$) be the part of $B(0,4R)$ that lies above (resp. below) the graph
of $f^{+}$ (resp. $f^{-}$) for some $R>0$ large. Let $u^{\pm}\in
C^{1,\alpha}(\overline{\Omega}^{\pm})$ be such that there exists a constant
$C_{1}>0$ such that for $x\in\Omega^{\pm}\cap B(0,2R)$ we have that

(5.2) $\displaystyle C_{1}^{-1}\mathrm{dist}(x,\Gamma^{\pm})\leq u^{\pm}\leq$
$\displaystyle C_{1}\mathrm{dist}(x,\Gamma^{\pm})$ $\displaystyle|\nabla
u^{\pm}|\leq$ $\displaystyle C_{2}$ $\displaystyle\|D^{2}u^{\pm}\|\leq$
$\displaystyle C_{1}\mathrm{dist}(x,\Gamma^{\pm})^{-1}.$

Then $u=u^{+}-u^{-}$ is an almost-minimizer to (5.1) inside of $B(0,R)$ where
$q_{\pm}$ are the $C^{0,\alpha}$ functions which agree with $|\nabla u^{\pm}|$
on $\Gamma^{\pm}$.

More precisely, there exists a constant
$C=C(C_{1},\|f^{\pm}\|_{C^{2}},\|u^{\pm}\|_{C^{1,\alpha}(\overline{\Omega^{\pm}})})>0$
and
$1>r_{0}=r_{0}(C_{1},\|f^{\pm}\|_{C^{2}},\|u^{\pm}\|_{C^{1,\alpha}(\overline{\Omega^{\pm}})})>0$
such that for any ball $B$ satisfying $\overline{B}\subset B(0,R)$, and
$r(B)\leq r_{0}$ we have

(5.3) $J_{B}(u)\leq J_{B}(v)+Cr^{n+\frac{\alpha}{4n+2\alpha}}$

for any $v=u$ on $B(0,R)\backslash B$.

Key to the proof of Lemma 5.3 is the following result which is adapted from
[DSJ09]:

######  Lemma 5.4.

Let $v$ be a critical point to $J_{B}$ (associated to $q_{\pm}$). Assume there
exists, parameterized by $t\in[a,b]$, a family of
$\phi_{t}:\overline{B}\rightarrow\mathbb{R}$ (continuous in both variables)
that satisfy the following properties:

  1. (1)

$\Delta\phi_{t}=0$ in $\\{\phi_{t}\neq 0\\}\cap B$.

  2. (2)

$\\{\phi_{t}=0\\}=\partial\\{\phi_{t}>0\\}=\partial\\{\phi_{t}<0\\}$.
Furthermore $t\mapsto\\{\phi_{t}=0\\}$ is continuous in the Hausdorff distance
sense.

  3. (3)

At every point on $\partial\\{\pm\phi_{t}>0\\}$ there exists a ball inside
$\\{\pm\phi_{t}>0\\}$ which touches the free boundary at that point.

  4. (4)

At every $x_{1}\in\partial\\{\pm\phi_{t}>0\\}$ we satisfy

$(\partial_{\nu^{+}}\phi_{t})^{2}(x_{1})-(\partial_{\nu^{-}}\phi_{t})^{2}(x_{1})\geq
q^{2}_{+}(x_{1})-q^{2}_{-}(x_{1})$

(respectively $\leq q^{2}_{+}-q^{2}_{-}$ for supersolutions)

  5. (5)

$\phi_{b}\leq v$ in $\overline{B}$ (respectively $\phi_{b}\geq v$ for
supersolutions).

  6. (6)

For all $\rho\in[a,b]$, $\phi_{\rho}\leq v$ on $\partial B$ and
$\phi_{\rho}<v$ on $\partial B\cap\overline{\\{v\neq 0\\}}$ (with the
inequalities reversed for supersolutions)

then we have that $\phi_{a}\leq v$ in $\overline{B}$ (respectively
$\phi_{a}\geq v$ in $\overline{B}$ for super solutions).

###### Proof of Lemma 5.3, assuming Lemma 5.4.

We first check the almost-minimization condition (5.3) for
$x_{0}\in\Gamma^{+}\cap\Gamma^{-}$:

Case 1: Let $x_{0}\in\Gamma^{+}\cap\Gamma^{-}$ and $r>0$ small enough and let
$v=u$ on $B(0,R)\backslash B(x_{0},r)$.

Since $u^{\pm}\in C^{1,\alpha}(\overline{\Omega^{\pm}})$ and
$\Gamma^{+}\cap\Gamma^{-}$ is closed and smooth, there exists an $r_{0}$ such
that if $r<r_{0}$ then, for $x\in B(x_{0},r)$

(5.4) $\displaystyle q_{+}(x_{0})((x-x_{0})\cdot e)^{+}-$ $\displaystyle
q_{-}(x_{0})((x-x_{0})\cdot e)^{-}-r^{1+7\alpha/8}\leq u(x)$ $\displaystyle
u(x)\leq q_{+}(x_{0})((x-x_{0})\cdot e)^{+}-$ $\displaystyle
q_{-}(x_{0})((x-x_{0})\cdot e)^{-}+r^{1+7\alpha/8}.$

Recall that $q_{\pm}(x_{0})=|\nabla u^{\pm}|(x_{0})$ by definition.

Recall the functional proved to be monotone by Weiss [Wei99]:

$W(u,x_{0},r)\equiv\frac{1}{r^{n}}\int_{B(x_{0},r)}|\nabla
u|^{2}+q^{2}_{+}(x)\chi_{\\{u>0\\}}+q_{-}^{2}(x)\chi_{\\{u<0\\}}\,dx-\frac{1}{r^{n+1}}\int_{\partial
B(x_{0},r)}u^{2}\,d\sigma.$

It follows from (5.4) and the $C^{0,\alpha}$-character of $|\nabla u^{\pm}|$
that

(5.5) $W(u,x_{0},r)\leq
W(u,x_{0},0)+Cr^{7\alpha/8}=\frac{1}{2}\mathrm{Vol}(B(0,1))(q^{2}_{+}(x_{0})+q^{2}_{-}(x_{0}))+r^{7\alpha/8},$

where $C>0$ depends on the $C^{0,\alpha}$-norm of $|\nabla u^{\pm}|$
restricted to $\overline{\Omega}^{\pm}$.

We now want to show that for any minimizer $v$ to $J_{B(x_{0},r)}$ with $v=u$
in $\mathbb{R}^{n}\backslash B(x_{0},r)$ we have

(5.6) $(\partial\\{v>0\\}\cup\partial\\{v<0\\})\cap B(x_{0},r)\subset\\{x\in
B(x_{0},r)\mid|(x-x_{0})\cdot e|<r^{1+\alpha/2}\\}$

Assume that (5.6) holds. We would like to compare $W(v,x_{0},r)$ to
$W(u,x_{0},r)$ but naïvely underestimating $W(v,x_{0},r)$ by $W(v,x_{0},0)$ is
problematic as $x_{0}$ may not be in the free boundary of $v$. To combat this,
let $x^{+}$ be the closest point to $x_{0}$ in $\partial\\{v>0\\}$ and $x^{-}$
the closest point to $x_{0}$ in $\partial\\{v<0\\}$. We note that (5.6)
implies that $|x_{0}-x^{\pm}|<r^{1+\alpha/2}$. Let
$\rho=r/2-\max\\{|x_{+}-x_{0}|,|x_{-}-x_{0}|\\}$. Note that
$B(x^{\pm},2\rho)\subset B(x_{0},r)$ and $r>2\rho>r(1-r^{\alpha/2})$. Also,
$B(x_{0},r/2)\supset B(x^{\pm},\rho)$. Hence

(5.7) $W(v,x_{0},r/2)\geq
W(v^{+},x^{+},\rho)+W(v^{-},x^{-},\rho)-Cr^{\alpha/2},$

because $|2\rho/r-1|<r^{\alpha/2}$ and $v$ is Lipschitz in $B(x_{0},r/2)$ with
a constant controlled by $\frac{1}{r^{n}}\int_{B(x_{0},r)}|\nabla
v|^{2}\leq\frac{1}{r^{n}}J_{B(x_{0},r)}(u)$, the latter of which is bounded by
the Lipschitz norm of $u$ and the supremums of $q_{\pm}$.

To estimate each term in the summand on (5.7), we think of $v^{+},v^{-}$ is
being separate critical points to the _one-phase problems_ associated to
weights
$\tilde{q}_{\pm}|_{\partial\\{v^{\pm}>0\\}}:=\partial_{\nu^{\pm}}v^{\pm}|_{\partial\\{v^{\pm}>0\\}}$.
By the regularity theory of the two-phase problem in [DPSV21b] and (5.6) we
know that $\partial\\{\pm v>0\\}$ are $C^{1,\alpha}$ in $B(x_{0},r/2)$. Thus
$\tilde{q}_{\pm}\in C^{0,\alpha}(\partial\\{v^{\pm}>0\\})$ and can be extended
Hölder continuously to all of $B(x_{0},r/2)$ (with norm uniform in the
constants we care about).

The free boundary condition satisfied by $v$ being a minimizer to the two-
phase problem tells us that
$\tilde{q}_{\pm}:=\partial_{\nu^{\pm}}v^{\pm}=q_{\pm}$ at one-phase points but
at two-phase points the free boundary condition only implies that
$\tilde{q}_{\pm}:=\partial_{\nu^{\pm}}v^{\pm}\geq q_{\pm}$. These observations
tell us that $\|\tilde{q}_{\pm}-q_{\pm}\|_{L^{\infty}(B(x_{0},r/2))}\leq
Cr^{\alpha}$.

By monotonicity, we can underestimate each

$\displaystyle W(v^{\pm},x^{\pm},\rho)\geq$
$\displaystyle\frac{1}{\rho^{n}}\int_{B(x^{\pm},\rho)}|\nabla
v^{\pm}|^{2}+\tilde{q}^{2}_{\pm}(x)\chi_{\\{\pm
v>0\\}}\,dx-\frac{1}{\rho^{n+1}}\int_{\partial
B(x^{\pm},\rho)}(v^{\pm})^{2}\,d\sigma-Cr^{\alpha}$ $\displaystyle\geq$
$\displaystyle\frac{1}{2}\mathrm{Vol}(B(0,1))\tilde{q}^{2}_{\pm}(x^{\pm})-Cr^{\alpha}$
$\displaystyle\geq$
$\displaystyle\frac{1}{2}\mathrm{Vol}(B(0,1))q^{2}_{\pm}(x^{\pm})-Cr^{\alpha}$
$\displaystyle\geq$
$\displaystyle\frac{1}{2}\mathrm{Vol}(B(0,1))q_{\pm}^{2}(x_{0})-Cr^{\alpha}.$

Putting this back into (5.7), using monotonicity and (5.5) we get that

$\displaystyle W(v,x_{0},r)$ $\displaystyle\geq
W(v,x_{0},r/2)-Cr^{\alpha/2}\geq\frac{1}{2}\mathrm{Vol}(B(0,1))(q_{+}^{2}(x_{0})+q_{-}^{2}(x_{0}))-Cr^{\alpha/2}$
$\displaystyle\geq W(u,x_{0},r)-Cr^{\alpha/2}.$

Multiplying through by $r^{n}$ gives the almost-minimization inequality.

So to finish Case 1, it suffices to prove (5.6). Here is where we apply Lemma
5.4. We do this on the interval $[a,b]=[r^{1+\alpha/2},3r]$ (recall that we
can take $r$ small so that $3r<1$). To simplify notation, let us assume that
$x_{0}=0$ and $e=e_{n}$. We then create the family of sub/super-solutions to
the two phase problem in $B(x_{0},r)$ defined by

$\overline{w}_{t}=M_{+}(x_{n}-t)^{+}-m_{-}(x_{n}-t)^{-},\qquad\underline{w}_{t}=m_{+}(x_{n}+t)^{+}-M_{-}(x_{n}+t)^{-},$

where $m_{\pm}=\min_{B(x_{0},r)}q_{\pm}$ and
$M_{\pm}=\max_{B(x_{0},r)}q_{\pm}$. Let $v$ be a minimizer to the two phase
problem in $B(x_{0},r)$ with boundary values $u$.

We first verify condition (5) in Lemma 5.4. We note that for any $x\in
B(x_{0},r)$

$\overline{w}_{3r}(x)\leq-2m_{-}r\leq-2q_{-}(x_{0})r+Cr^{1+\alpha}\stackrel{{\scriptstyle\eqref{e:blowuptp}}}{{\leq}}\inf_{\partial
B(x_{0},r)}u=\inf_{\partial B(x_{0},r)}v\leq v(x)$

where the last inequality follows from the minimum principle applied to the
superharmonic function $-v^{-}$ and the middle inequalities follow from the
$C^{0,\alpha}$ continuity of $q_{-}$ and the fact that we can take
$r_{0}\ll\inf_{\overline{B(0,R)}}q_{-}$.

To check condition (6) in Lemma 5.4 we will show that
$\overline{w}_{t}|_{\partial B(x_{0},r)}<u|_{\partial B(x_{0},r)}$ for all
$r^{1+7\alpha/8}<r^{1+\alpha/2}\leq t\leq 3r$. When $x_{n}>t$ we have

$\displaystyle\overline{w}_{t}(x)=M_{+}(x_{n}-t)\leq$ $\displaystyle
q_{+}(x_{0})(x_{n}-t)+(\mathrm{osc}q_{+})(|x_{n}|+|t|)$ $\displaystyle\leq$
$\displaystyle
q_{+}(x_{0})x_{n}-q_{+}(x_{0})r^{1+\alpha/2}+Cr^{1+\alpha}<q_{+}(x_{0})x_{n}-r^{1+7\alpha/8}\stackrel{{\scriptstyle\eqref{e:blowuptp}}}{{\leq}}u(x),$

again assuming $r_{0}$ is small enough.

When $0\leq x_{n}<t$ we note that $\overline{w}_{t}(x)<0$. So there is only
something to prove if
$u(x)<0\stackrel{{\scriptstyle\eqref{e:blowuptp}}}{{\Rightarrow}}x_{n}<Cr^{1+7\alpha/8}$.
In this case $x_{n}<t/2$ (for $r$ small enough) and we have

$\overline{w}_{t}(x)\leq
m_{-}(x_{n}-t)\leq-\frac{m_{-}}{2}t\leq-\frac{m_{-}}{2}r^{1+\alpha/2}<-r^{1+7\alpha/8}\stackrel{{\scriptstyle\eqref{e:blowuptp}}}{{\leq}}u(x).$

When $x_{n}<0$ we have

$\overline{w}_{t}(x)=m_{-}(x_{n}-t)\leq
q_{-}(x_{0})(x_{n}-t)+Cr^{1+\alpha}\leq
q_{-}(x_{0})x_{n}-Cr^{1+\alpha/2}\stackrel{{\scriptstyle\eqref{e:blowuptp}}}{{<}}u(x).$

Condition (1) in Lemma 5.4 follows from the fact that $\overline{w}_{t}$ is
linear where it is not zero. Conditions (2) and (3) follow from the
observation that $\\{\overline{w}_{t}=0\\}=\\{x_{n}=t\\}$ which moves
continuously with $t$. Finally the free boundary condition, Condition (4),
follows from the definition of $M_{\pm},m_{\pm}$.

Thus by Lemma 5.4 above (which will be checked below) it must be the case that
$\overline{w}_{r^{1+\alpha/2}}\leq v$ in all of $B(x_{0},r)$. A similar
argument shows that $\underline{w}_{r^{1+\alpha/2}}\geq v$ in all of
$B(x_{0},r)$. Thus $\partial\\{v>0\\},\partial\\{v<0\\}\subset\\{x\in
B(x_{0},r)\mid|x_{n}|\leq r^{1+\alpha/2}\\}$.

Case 2: Now we assume that $x_{0}\in\Gamma^{+}\cup\Gamma^{-}$. Let $M>1$ to be
chosen later. If $R$ is such that
$B(x_{0},Mr)\cap(\Gamma^{+}\cap\Gamma^{-})\neq\emptyset$ then note that
$B(x_{0},r)\subset B(y,2Mr)$, for some $y\in\Gamma^{+}\cap\Gamma^{-}$. As
such, if $u=v$ in $B(0,R)\backslash B(x_{0},r)$ then $u=v$ in
$B(0,R)\backslash B(y,2Mr)$. So Case 1 tell us that $u$ satisfies the almost-
minimization criterion in these balls with some $\tilde{C}=C(2M)^{n+\alpha/2}$
(where $C>0$ comes from Case 1).

If $B(x_{0},Mr)\cap(\Gamma^{+}\cap\Gamma^{-})=\emptyset$ we assume, without
loss of generality, that $x_{0}\in\Gamma^{+}\backslash\Gamma^{-}$ and we
assume that $B(x_{0},r)\cap\Gamma^{-}\neq\emptyset$ (otherwise $u^{-}$ is
identically zero on $B(x_{0},r)$ and the analysis is even easier). Pick $y\in
B(x_{0},r)\cap\Gamma^{-}$. As long as $M>3$ then
$B(y,2r)\cap(\Gamma^{+}\cap\Gamma^{-})=\emptyset$ and $u^{+},u^{-}$ satisfy
one-phase versions of (5.4) in $B(x_{0},2r)$ and $B(y,2r)$ respectively. To
show that $u^{+}$ and $u^{-}$ each satisfy the correct energy inequalities for
the one-phase versions of $J$ inside of $B(x_{0},r)$ and $B(y,r)$ we proceed
as in Case 1 (but the argument is simpler because we need only care about one-
phase functionals). The almost-minimization property for $u$ in $B(x_{0},r)$
for the two-phase functional then follows since the two-phase energy is the
sum of the one-phase energies. We omit these details.

Case 3: Assume that $x_{0}\notin\Gamma^{+}\cup\Gamma^{-}$. Let $M>1$ be as
above (perhaps enlarge a bit, to be determined later) and let $\varepsilon\ll
1$ (to be determined later). If
$B(x_{0},Mr^{1-\varepsilon})\cap(\Gamma^{+}\cup\Gamma^{-})\neq\emptyset$ then
the almost minimization criterion in $B(y,2Mr^{1-\varepsilon})$444It may be
surprising that we break scaling by considering a radius of size
$r^{1-\varepsilon}$ but we are merely taking advantage of the lack of scale
invariance in the characterization of almost-minimizers for some
$y\in\Gamma^{+}\cup\Gamma^{-}$ shows that

$J_{B(x_{0},r)}(u)-J_{B(x_{0},r)}(v)\leq
C(2Mr^{1-\varepsilon})^{n+\alpha/2}=C_{M}r^{(1-\varepsilon)(n+\alpha/2)}\leq
C_{M}r^{n+\alpha/4},$

for any $u=v$ outside of $B(x_{0},r)$. Note this last inequality above holds
for $r<1$ as long as $(1-\varepsilon)(n+\alpha/2)\geq
n+\alpha/4\Leftrightarrow\frac{\alpha}{2(2n+\alpha)}\geq\varepsilon$.

Thus we may assume that
$B(x_{0},Mr^{1-\varepsilon})\cap(\Gamma^{+}\cup\Gamma^{-})=\emptyset$. If
$B(x_{0},r)\subset\\{u=0\\}$ the almost-minimization criterion follows from
the fact that $u\equiv 0$ in $B(x_{0},r)$. So we may assume that
$B(x_{0},Mr^{1-\varepsilon})\subset\\{u>0\\}$ (the negative phase follows
similarly).

Let $v$ minimize $J_{B(x_{0},r)}$ with boundary values $u|_{\partial
B(x_{0},r)}$. Let us first assume that $v$ is simply a harmonic function in
$B(x_{0},r)$ (this will be the case when $v>0$). We compute that

$\int_{B(x_{0},r)}|\nabla u|^{2}-|\nabla
v|^{2}=-\int_{B(x_{0},r)}|\nabla(u-v)|^{2}=\int_{B(x_{0},r)}\Delta u(u-v).$

We note that

$u-v<\max_{B(x_{0},r)}u-\min_{B(x_{0},r)}v=\max_{B(x_{0},r)}u-\min_{\partial
B(x_{0},r)}u<\mathrm{osc}_{B(x_{0},r)}u<Cr,$

where we used the maximum principle and the assumption on $|\nabla u|$.
Similarly with a lower bound so that $\sup_{B(x_{0},r)}|u-v|\leq Cr$. By our
assumptions

$|\Delta u(z)|\leq\|\nabla^{2}u(z)\|\leq C\mathrm{dist}(z,\\{u=0\\})^{-1}\leq
Cr^{\varepsilon-1},\qquad\forall z\in B(x_{0},r).$

Note the last inequality above is because
$B(x_{0},Mr^{1-\varepsilon})\subset\\{u>0\\}$. Putting all of this together we
get that

(5.8) $\int_{B(x_{0},r)}|\nabla u|^{2}-|\nabla v|^{2}\leq Cr^{n+\varepsilon}.$

If $v$ does vanish, the maximum principle still says that $v\geq 0$. Let $h$
be the harmonic extension of $u|_{B(x_{0},r)}$ into $B:=B(x_{0},r)$ and
compute, recalling that $h>0$,

$\displaystyle J_{B}(v)-J_{B}(h)$
$\displaystyle\geq-(\sup_{B}q^{2}_{+})|\\{v=0\\}\cap B|+\int_{B}|\nabla
v|^{2}-|\nabla h|^{2}$ $\displaystyle=-(\sup_{B}q^{2}_{+})|\\{v=0\\}\cap
B|+\int_{B}|\nabla(v-h)|^{2}$ $\displaystyle=$
$\displaystyle-(\sup_{B}q^{2}_{+})|\\{v=0\\}\cap
B|+\int_{\partial\\{v>0\\}\cap B}\partial_{\nu}vh\,d\mathcal{H}^{n-1}$
$\displaystyle\geq$ $\displaystyle-(\sup_{B}q^{2}_{+})|\\{v=0\\}\cap
B|+(\inf_{B}h)(\inf_{B}q_{+})\mathcal{H}^{n-1}(\partial\\{v=0\\}\cap B)$
$\displaystyle=$ $\displaystyle-(\mathrm{osc}_{B}q_{+}^{2})|\\{v=0\\}\cap B|$
$\displaystyle+$ $\displaystyle(\inf_{B}q_{+})\left((\inf_{\partial
B}u)\mathcal{H}^{n-1}(\partial\\{v=0\\}\cap B)-\inf_{B}q_{+}|\\{v=0\\}\cap
B|\right).$

We apply the isoperimetric inequality to $\\{v=0\\}\cap B$ (recall that $v>0$
on $\partial B$) and estimate $\inf_{\partial B}u\geq
C^{-1}(\mathrm{dist}(x_{0},\partial\\{u>0\\})-r)\geq\tilde{C}^{-1}r^{1-\varepsilon}$
to get

$J_{B}(v)-J_{B}(h)\geq-Cr^{\alpha+n}+C|\\{v=0\\}\cap
B|^{\frac{n-1}{n}}(r^{1-\varepsilon}-c|\\{v=0\\}\cap B|^{1-\frac{1}{n}}).$

As long as $\varepsilon<1/n$ and $r<r_{0}$ is small enough we have
$J_{B}(v)-J_{B}(h)\geq-Cr^{\alpha+n}$, which combined with (5.8) (which
compares $u$ to its harmonic extension) gives us the inequality we want.

∎

Although Lemma 5.4 is really a minor modification of [DSJ09, Lemma 2.3] (see
also [Caf87]), we give a proof for the convenience of the reader.

###### Proof of Lemma 5.4.

Let $E=\\{\rho\in[a,b]\mid\phi_{\rho}\leq v\\}$. Since
$\rho\mapsto\phi_{\rho}(x_{0})$ is continuous for every $x_{0}\in\overline{B}$
it must be the case that $E$ is a closed subset of $[a,b]$. Furthermore $b\in
E$ by assumption. Thus we will have shown that $E=[a,b]$ (and we will be done)
if we can show that $E$ is (relatively) open in $[a,b]$.

We first observe that for all $t\in E$ we have
$\overline{\\{\phi_{t}>0\\}}\cap\overline{B}\subset\overline{B}\cap\\{v>0\\}$.
Indeed, otherwise the zero sets of $\phi_{t}$ and $v$ would touch in the
interior of $B$. If $\phi_{t}(x_{1})=v(x_{1})=0$ then it must be the case that
$x_{1}\in\partial\\{v>0\\}$ (since $\phi_{t}\leq v$ and
$x_{1}\in\partial\\{\phi_{t}>0\\}$ by the condition (2) on the zero set of
$\phi_{t}$). If $x_{1}$ is a two-phase point for $v$ then Condition (4)
implies that $\partial_{\nu^{+}}\phi_{t}(x_{1})\geq\partial_{\nu^{+}}v(x_{1})$
or $\partial_{\nu^{-}}\phi_{t}(x_{1})\leq\partial_{\nu^{-}}v(x_{1})$. Either
way we get a contradiction to the Hopf maximum principle. If $x_{1}$ is a one-
phase point for $v$ then $\partial_{\nu^{+}}\phi_{t}(x_{1})\geq
q_{+}(x_{1})=\partial_{\nu^{+}}v(x_{1})$ and we again get a contradiction by
the Hopf maximum principle. Note in both cases we use the condition that
$\phi_{t}<v$ on the boundary whenever they are both non-zero to conclude that
$v-\phi_{t}\neq 0$ and thus the normal derivatives above must have a strict
sign.

Since for all $t_{0}\in E$ we have
$\overline{\\{\phi_{t_{0}}>0\\}}\cap\overline{B}\subset\overline{B}\cap\\{v>0\\}$
and since $t\mapsto\partial\\{\phi_{t}>0\\}$ is continuous there exists a
$\varepsilon>0$ such that if $|\tilde{t}-t_{0}|<\varepsilon$ we have
$\overline{\\{\phi_{\tilde{t}}>0\\}}\cap\overline{B}\subset\overline{B}\cap\\{v>0\\}$.

Thus in $\\{\phi_{\tilde{t}}>0\\}$ we have that both functions are harmonic
and $v\geq\phi_{\tilde{t}}$ on $\partial(\\{\phi_{\tilde{t}}>0\\}\cap
B)=(\partial
B\cap\\{\phi_{\tilde{t}}>0\\})\cup(\partial\\{\phi_{\tilde{t}}>0\\}\cap\overline{B})$.
Thus $v\geq\phi_{\tilde{t}}$ on this set. Furthermore $v>0=\phi_{\tilde{t}}$
on $\\{\phi_{\tilde{t}}=0\\}\cap B$ (by the assumption that the zero set is
the free boundary).

Finally, on the set $\\{v\leq 0\\}$ we have $\phi_{\tilde{t}}<0$ so the
difference $v-\phi_{\tilde{t}}$ is superharmonic on $\\{v\leq 0\\}$. On the
boundary of this set we have $v-\phi_{\tilde{t}}\geq 0$ so by the maximum
principle we have $v\geq\phi_{\tilde{t}}$ on this set and putting this
together with the previous parts on all $\overline{B}$. This shows $E$ is open
and we are done. ∎

We are ready to finish by constructing the functions satisfying the hypothesis
of Lemma 5.3. First we recall the definition of the regularized distance
function, introduced in [DFM19]:

(5.9)
$D_{\mu,\beta}(x):=\left(\int\frac{1}{|x-y|^{d+\beta}}d\mu(y)\right)^{-1/\beta}.$

We use the following facts about this regularized distance, see [DFM19,
DEM21]:

######  Lemma 5.5.

Let $\beta>0$ and let $\mu$ be a $d$-Ahlfors regular measure. Then

$D_{\mu,\beta}\in C^{\infty}(\mathbb{R}^{n}\backslash\mathrm{spt}\mu)\cap
C(\mathbb{R}^{n})$

satisfies the following estimates with $C_{1}>0$ depending on $n,\beta,d$ and
the Ahlfors regularity character of $\mu$:

(5.10) $\displaystyle C_{1}^{-1}\mathrm{dist}(x,\mathrm{spt}\mu)\leq
D_{\mu,\beta}(x)\leq$ $\displaystyle C_{1}\mathrm{dist}(x,\mathrm{spt}\mu)$
$\displaystyle\|\nabla D_{\mu,\beta}(x)\|\leq$ $\displaystyle C_{1}$
$\displaystyle\|D^{2}D_{\mu,\beta}(x)\|\leq$ $\displaystyle
C_{1}\mathrm{dist}(x,\mathrm{spt}\mu)^{-1}.$

Furthermore, there is some dimensional constant $c=c(n,d,\beta)>0$ such that
if $\mathrm{spt}\mu$ is given by a $C^{2}$-submanifold and
$\Theta^{d}(\mu,Q):=\mathrm{lim}_{r\downarrow 0}\frac{\mu(B(Q,r))}{r^{d}}$
satisfies $\Theta^{d}(\mu,Q)^{-1/\beta}\in C^{0,\alpha}(\mathrm{spt}(\mu))$
then $|\nabla D_{\mu,\beta}|\in C^{0,\alpha}(\mathbb{R}^{n})$ and on the
support of $\mu$ we have $|\nabla
D_{\mu,\beta}|=c\Theta^{d}(\mu,Q)^{-1/\beta}$.

###### Proof.

The estimates in (5.10) are straightfoward applications of the Ahlfors
regularity condition (c.f. [DFM19, DEM21]). If $\mathrm{spt}\mu$ has a tangent
at $x_{0}$ and if $\Theta^{d}(\mu,x_{0})$ exists, the fact that the non-
tangential limit of $|\nabla D_{\mu,\beta}|$ exists at $x_{0}$ and is equal to
$c\Theta^{d}(\mu,Q)^{-1/\beta}$ is contained in the proof of [DEM21, Theorem
5.3].

Thus to complete the proof of the lemma, it suffices to show that $|\nabla
D_{\mu,\beta}|$ extends to $\mathrm{spt}(\mu)$ in a Hölder continuous fashion.
This follows from the $C^{2}$-character of $\mathrm{spt}(\mu)$ and the
estimates in [DEM21, Section 2]; to be slightly more precise the difference
between $\nabla D_{\mu,\beta}$ at a point
$x\in\mathbb{R}^{n}\setminus\mathrm{spt}(\mu)$ and the closest point
$Q\in\mathrm{spt}(\mu)$ is controlled by the $\alpha$ numbers (in the sense of
Tolsa [Tol09]) of $\mu$ at the point $Q$ and the scale $|x-Q|$ (c.f. [DEM21,
equation (2.19)]). Since, $\Theta^{d}(\mu,Q)$ and $\mathrm{spt}(\mu)$ are
regular, these $\alpha$-numbers decay like $r^{\alpha}$, which gives the
desired result. ∎

We are now ready to compute the proof of Theorem 5.1:

###### Proof of Theorem 5.1.

Let $d\mu^{\pm}(Q)=q_{\pm}^{-1}(Q)d\mathcal{H}^{n-1}|_{\Gamma^{\pm}}(Q)$ and
let $u^{\pm}=D_{\mu^{\pm},1}$ in $\Omega^{\pm}$ and identically equal to zero
elsewhere. Note that $\mu^{\pm}$ are Ahlfors regular by the boundedness and
non-degeneracy of $q_{\pm}$ and the $C^{2}$-character of $f^{\pm}$ (we don’t
need to worry about issues at infinity because $f^{\pm}$ are constant outside
of a large ball).

The result immediately follows from Lemmas 5.5 and 5.3. ∎

##  6\. Appendix

###  6.1. Proof of Lemma 2.1

Fix $-3N\leq x\leq 3N$. In this section we will study minimizers of (2.2) with
boundary data $v_{N}(\pm 1)=\pm f_{N}(x)$.

We know that the set $\\{v_{N}=0\\}$ is an interval, and that $v_{N}$ is
harmonic (i.e. affine) in $\\{v_{N}\neq 0\\}\cap(-1,1)$. Hence we can search
for $v_{N}$ among the function $v_{N}$ with boundary data $v_{N}(\pm 1)=\pm
f_{N}(x)$, with $v_{N}=0$ on $[a,b]$, and which are linear on $[-1,a]$ and
$[b,1]$:

$\displaystyle v_{N}(x)=\begin{cases}\frac{f_{N}(x)}{a+1}(y-a),&\text{ if
}-1\leq y\leq a\\\ 0,&\text{ if }a\leq y\leq b\\\
\frac{f_{N}(x)}{1-b}(y-b),&\text{ if }b\leq y\leq 1.\end{cases}$

Then

$H(v_{N})=\frac{(f_{N}(x))^{2}}{1-b}+\frac{(f_{N}(x))^{2}}{a+1}+2-(b-a).$

We will minimize
$G(a,b)=\frac{(f_{N}(x))^{2}}{1-b}+\frac{(f_{N}(x))^{2}}{a+1}+2-(b-a)$ given
the constraint $-1<a\leq b<1$.

####  6.1.1. Assume $f_{N}(x)\geq 1$

We claim that $G(0,0)\leq G(a,b)$ for any $-1<a\leq b<1$, hence the minimizer
is $v_{N}(x)=yf_{N}(x)$.

Notice that $G(0,0)=2(f_{N}(x))^{2}+2$, hence it suffices to show

$2(f_{N}(x))^{2}+2\leq(f_{N}(x))^{2}\left(\frac{1}{1-b}+\frac{1}{a+1}\right)+2-b+a.$

This is equivalent to

$b-a\leq(f_{N}(x))^{2}\left(\frac{1}{1-b}-1+\frac{1}{a+1}-1\right)=(f_{N}(x))^{2}\left(\frac{b}{1-b}-\frac{a}{a+1}\right)$

Since $f_{N}(x)\geq 1$, it suffices to show that

$b-a\leq\frac{b}{1-b}-\frac{a}{a+1}$

which holds since $-1<a$ and $b<1$.

####  6.1.2. Assume $0\leq f_{N}(x)<1$

Looking for critical points of $G$, one looks for solutions of

$\frac{\partial G}{\partial a}(a,b)=-\frac{(f_{N}(x))^{2}}{(1+a)^{2}}+1=0,\
\frac{\partial G}{\partial b}(a,b)=\frac{(f_{N}(x))^{2}}{(1-b)^{2}}-1=0,$

leading to $(1+a)^{2}=(1-b)^{2}=(f_{N}(x))^{2}$. Now $1+a=-f_{N}(x)$ and
$1-b=-f_{N}(x)$ are impossible (unless $f_{N}(x)=0$) because $-1<a\leq b<1$,
so $a=f_{N}(x)-1$ and $b=1-f_{N}(x)$. Notice that $a\leq b$ is only satisfied
when $f_{N}(x)\leq 1$.

We have $G(f_{N}(x)-1,1-f_{N}(x))=4f_{N}(x)$. Let us now show $4f_{N}(x)\leq
G(a,b)$ for any $-1<a\leq b<1$. It suffices to show

$2f_{N}(x)\leq\frac{(f_{N}(x))^{2}}{1-b}+1-b,\ \ \text{ and }\ \
2f_{N}(x)\leq\frac{(f_{N}(x))^{2}}{a+1}+1+a,$

which are both true, as the minimizers of the right-hand sides occur when
$b=1-f_{N}(x)$ and $a=f_{N}(x)-1$.

Hence the minimizer is $v_{N}(y)=\text{sgn}(y)(|y|-1+f_{N}(x))_{+}.$

###  6.2. Energy bounds when $x\in[-N,N]$

In this subsection we want to provide some crucial bounds on
$H_{x}(u_{N}(x,-))$ when $x\in[-N,N]$. Notice that if $x\in[-N,N]$,
$v_{N}(x,y)=\text{sgn}(y)(|y|-\alpha)_{+}$. Therefore
$H_{x}(v_{N}(x,\cdot))=4(1-\alpha)$. We want to show that if $u_{N}$ is large
and $v_{N}$ is zero, then the slice energy of $u_{N}$ is quantitatively bigger
than that of $v_{N}$.

More precisely we have the following claim:

######  Claim 6.1.

Let $x\in[-N,N]\backslash X_{0}$ and assume that $u_{N}(x,y)>\beta>0$ (or
$u_{N}(x,y)<-\beta<0$) for some $y<\alpha/2$. Then there exists
$\eta=\eta(\alpha,\beta)>0$ such that $H(u_{N}(x,\cdot))\geq
H(v_{N}(x,\cdot))+\eta$.

We first start with estimates on the energy depending on the size of the zero
set of $u_{N}$: call $A=\\{y\in[-1,1]\ :\ u_{N}(x,y)=0\\}$ and $\delta=|A|$,
so $0\leq\delta<2$. We have

$H(u_{N}(x,\cdot))=2-\delta+\int_{-1}^{1}|\partial_{y}u_{N}(x,\cdot)|^{2}dy$

Now, since $f_{N}(x)=1-\alpha$ for $|x|<N$, by comparing with a linear
function we have

$\displaystyle\int_{-1}^{1}|\partial_{y}u_{N}(x,\cdot)|^{2}dy$
$\displaystyle\geq\frac{(1-\alpha)^{2}}{\inf A+1}+\frac{(1-\alpha)^{2}}{1-\sup
A}=(1-\alpha)^{2}\left(\frac{1}{1-\sup A}+\frac{1}{\inf A+1}\right).$

Recall that for $x,y>0$, $\frac{1}{x}+\frac{1}{y}\geq\frac{4}{x+y}$, with
equality only if $x=y$. Applying this inequality with $x=1-\sup A$ and $y=\inf
A+1$ we conclude

$\displaystyle\int_{-1}^{1}|\partial_{y}u_{N}(x,\cdot)|^{2}dy$
$\displaystyle\geq(1-\alpha)^{2}\frac{4}{2-\sup A+\inf
A}=\frac{4(1-\alpha)^{2}}{2-(\sup A-\inf A)}\geq\frac{4(1-\alpha)^{2}}{2-|A|}$
$\displaystyle=\frac{4(1-\alpha)^{2}}{2-\delta}.$

Therefore

$H(u_{N}(x,\cdot))\geq
2-\delta+\frac{4(1-\alpha)^{2}}{2-\delta}=4(1-\alpha)+\frac{(2(1-\alpha)-(2-\delta))^{2}}{2-\delta}.$

Rearranging we get by Lemma 2.1:

(6.1) $H(u_{N}(x,\cdot))\geq
H(v_{N}(x,\cdot))+\frac{(\delta-2\alpha)^{2}}{2-\delta}.$

Notice that if $\delta\leq\alpha/2$, (6.1) gives $H(u_{N}(x,\cdot))\geq
H(v_{N}(x,\cdot))+\frac{\alpha^{2}}{2}$, hence we can assume
$\delta>\alpha/2$.

Now we assume there is a point $(x,y_{0})$ with $y_{0}<\alpha/2$ and
$u_{N}(x,y_{0})>\beta>0$ (the case where it is negative follows similarly). We
have two cases:

Case 1: Assume there exists $\tilde{y}>y_{0}$ with $u_{N}(x,\tilde{y})=0$. Let
$A_{+}=\sup\\{y\mid u_{N}(x,y)=0\\}$, be the largest such value of $y$, and
let $A_{-}=\inf\\{y\mid y>y_{0},u_{N}(x,y)=0\\}$ be the smallest such value of
$y$. Furthermore let $B_{+}=\sup\\{y<\tilde{y}\mid u_{N}(x,y)=0\\}$ and
$B_{-}=\inf\\{y\mid u_{N}(x,y)=0\\}$. We then have that

$H(u_{N}(x,y))\geq
2-(A_{+}-A_{-}+B_{+}-B_{-})+\frac{\beta^{2}}{A_{-}-\tilde{y}}+\frac{\beta^{2}}{\tilde{y}-B_{+}}+(1-\alpha)^{2}\left(\frac{1}{B_{-}+1}+\frac{1}{1-A_{+}}\right).$

Using again $1/x+1/y\geq\frac{4}{x+y}$ we have

$H(u_{N}(x,y))\geq
2-(A_{+}-A_{-}+B_{+}-B_{-})+\frac{4\beta^{2}}{A_{-}-B_{+}}+\frac{4(1-\alpha)^{2}}{2-(A_{+}-B_{-})}.$

Let $z=(A_{-}-B_{+})$ and $w=A_{+}-B_{-}$. We notice that
$w-z=A_{+}-A_{-}+B_{+}-B_{-}\geq\delta$. Let
$G(z,w):=2-(w-z)+\frac{4\beta^{2}}{z}+\frac{4(1-\alpha)^{2}}{2-w}$. We want to
find the minimum of $G(z,w)$ on the domain
$(z,w)\in[0,2]^{2}\cap\\{w-z\geq\delta\\}$ as this gives a lower bound for
$H(u_{N}(x,y))$. Straight forward calculus shows that the only critical point
of $G(z,w)$ is a local minimum at $(z,w)=(2\beta,2\alpha)$ which gives the
lower bound $H(u_{N}(x,-))\geq 4(1-\alpha)+4\beta=H(v_{N}(x,-))+4\beta$. On
the boundary of the domain, it is easy to see that $G(z,w)=+\infty$ when $z=0$
or $w=2$. When $w-z=\delta$ we rewrite the offset equation above,

$\displaystyle H(u_{N}(x,y))\geq$ $\displaystyle
2-\delta+\frac{4\beta^{2}}{z}+\frac{4(1-\alpha)^{2}}{2-(z+\delta)}$
$\displaystyle\geq$ $\displaystyle
2-\delta-z+z+\frac{4\beta^{2}}{z}+\frac{4(1-\alpha)^{2}}{2-(z+\delta)}$
$\displaystyle\geq$ $\displaystyle 4(1-\alpha)+z+\frac{4\beta^{2}}{z}$
$\displaystyle+$
$\displaystyle\frac{-4(1-\alpha)(2-\delta-z)+(2-\delta-z)^{2}+4(1-\alpha)^{2}}{2-\delta-z}$
$\displaystyle\geq$ $\displaystyle
H(v_{N}(x,-))+\frac{4\beta^{2}}{z}+\frac{(\delta+z-2\alpha)^{2}}{2-\delta-z}+z$
$\displaystyle\geq$ $\displaystyle H(v_{N}(x,-))+\frac{4\beta^{2}}{z}+z.$

Minimizing this in $z$ we get that $z=2\beta$, giving us the same lower bound
as above, $H(u_{N}(x,-))\geq H(v_{N}(x,-))+4\beta$.

Case 2: Here we have that the zero set of $u_{N}(x,-)$ lies totally below
$y_{0}$, i.e. if $A=\\{y\mid u_{N}(x,y)=0\\}$, then $\sup A<y_{0}$.
Calculating as above we have

$H(u_{N}(x,-))\geq 2-\delta+(1-\alpha)^{2}\left(\frac{1}{\inf
A+1}+\frac{1}{1-\sup A}\right).$

Note that $\inf A\leq\sup A\leq y_{0}\leq\alpha/2$ and $\sup A-\inf
A\geq|A|=\delta$.

Calling $x=\inf A$ and $y=\sup A$ we obtain the bounds $x+\delta\leq
y\leq\frac{\alpha}{2},$ $-1<x\leq\frac{\alpha}{2}-\delta$. We want to minimize
the function $G(a,b)=\frac{1}{a+1}+\frac{1}{1-b}$ in the domain given by those
bounds, recalling that $\frac{\alpha}{2}<\delta$. Notice that as $a$
approaches $-1$, $G(a,b)$ goes to infinity. When $b=\frac{\alpha}{2}$, we
minimize $G(a,b)$ when $a=\frac{\alpha}{2}-\delta$. Finally, when
$b=a+\delta$, we obtain the function
$g(a)=\frac{1}{a+1}+\frac{1}{1-a-\delta}$, with
$-1<a\leq\frac{\alpha}{2}-\delta$. This function has a minimum at
$a=-\frac{\delta}{2}$.

Case a: when $\frac{\alpha}{2}-\delta<-\frac{\delta}{2}$ (that is,
$\frac{\delta}{2}<\frac{\alpha}{2}$), one compares
$G(\alpha/2-\delta,\alpha/2)=\frac{1}{1+\frac{\alpha}{2}-\delta}+\frac{1}{1-\frac{\alpha}{2}}$
with $G(-\delta/2,\delta/2)=\frac{2}{1-\frac{\delta}{2}}$ to find

$G(a,b)\geq G(-\delta/2,\delta/2)=\frac{4}{2-\delta}.$

In this case we follow the computations of 6.1, recalling we are now under the
case $\delta<\alpha$, and notice that

$H(u_{N}(x,\cdot))\geq
H(v_{N}(x,\cdot))+\frac{(\delta-2\alpha)^{2}}{2-\delta}\geq
H(v_{N}(x,\cdot))+\frac{(\delta-2\alpha)^{2}}{2-\delta}\geq
H(v_{N}(x,\cdot))+\frac{\alpha^{2}}{2}.$

Case b: when $\frac{\alpha}{2}-\delta\leq-\frac{\delta}{2}$ (that is,
$\frac{\alpha}{2}\leq\frac{\delta}{2}$), we conclude the minimum of $G$ is
attained when $a=\inf A=\frac{\alpha}{2}-\delta$ and $a=\sup
A=\frac{\alpha}{2}$. Note that
$\frac{1}{w}+\frac{1}{z}=\frac{4}{w+z}+\frac{(w-z)^{2}}{wz(w+z)}$ and apply
this to $w=1-\alpha/2,z=1+\alpha/2-\delta$. Using the same arguments as above
we get that,

$H(u_{N}(x,-))\geq
H(v_{N}(x,-))+\frac{(\delta-2\alpha)^{2}}{2-\delta}+\frac{(\alpha-\delta)^{2}}{(1-\delta)(1-\alpha/2)(1+\alpha/2-\delta)}.$

A straightforward calculation shows that $H(u_{N}(x,-))\geq
H(v_{N}(x,-))+\frac{\alpha^{2}}{10^{4}}$.

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