# A Minimization Problem with Free Boundary for $p$-Laplacian weakly coupled
System

Morteza Fotouhi and Henrik Shahgholian  Department of Mathematical Sciences,
Sharif University of Technology, Tehran, Iran<EMAIL_ADDRESS>Department
of Mathematics, Royal Institute of Technology, 100 44 Stockholm, Sweden
<EMAIL_ADDRESS>
###### Abstract.

In this paper we consider a weakly coupled $p$-Laplacian system of a Bernoulli
type free boundary problem, through minimization of a corresponding
functional. We prove various properties of any local minimizer and the
corresponding free boundary.

###### Key words and phrases:

p-Laplacian, minimizers, free boundary regularity, system

###### 1991 Mathematics Subject Classification:

35R35

This project was carried out during the program Geometric aspects of nonlinear
PDE at Institute Mittag Leffler, Stockholm, Sweden. H. Shahgholian was
supported by Swedish Research Council.

##  1\. Introduction

###  1.1. Problem setting

For $\Omega\subset{\mathbb{R}}^{n}$ ($n\geq 2$), we consider the problem of
minimizing the functional

(1) $J(\mathbf{u})=\int_{\Omega}\sum_{i=1}^{m}|\nabla
u^{i}|^{p}+Q^{p}\chi_{\\{|\mathbf{u}|>0\\}}dx,\qquad 1<p<\infty,$

in the class of vectorial functions

${\mathcal{K}}:=\\{\mathbf{u}=(u^{1},\dots,u^{m})\in
W^{1,p}(\Omega;{\mathbb{R}}^{m}):\mathbf{u}=\mathbf{g}\text{ on
}\partial\Omega\text{ and }u^{i}\geq 0\text{ for }i=1,\cdots,m\\},$

with a given boundary data $\mathbf{g}\in W^{1,p}(\Omega;{\mathbb{R}}^{m})$.
Here, $Q$ is a Hölder function satisfying

$0<Q_{\min}\leq Q\leq Q_{\max}<\infty,$

for some constants $Q_{\min}$ and $Q_{\max}$.

We are interested in regularity properties of minimizers $\mathbf{u}$, as well
as the free boundary $\Gamma=\partial\\{|\mathbf{u}|>0\\}\cap\Omega$. Any
local minimizer satisfies the $p$-Laplace equation

${\rm div\,}(|\nabla u^{i}|^{p-2}\nabla u^{i})=0,\qquad\hbox{ in }\
\\{u^{i}>0\\},$

also denoted by $\Delta_{p}u^{i}=0$. In fact, $\Delta_{p}u^{i}$ is a
nonnegative Radon measure with support on free boundary, $\Gamma$. The problem
is to find a reasonable representation of this measure and put it into some
pde-context for further analysis.

This problem is referred to as Bernoulli-type free boundary problem, and is
well studied in the literature, for the scalar case and for $p=2$, starting
with seminal work of H.W. Alt and L.A. Caffarelli [2], and also for any
$1<p<\infty$ in [8]. There are very few results for Bernoulli-type problems
that involve systems (see [7, 11, 20]). In [7], the authors study the minimum
problem (1) for $p=2$ and show the smoothness of the regular part of free
boundary as well as some partial result for Hausdorff dimension of singular
part. Indeed, they apply a reduction method to reduce the problem to its
scalar counterpart and the same result for the scalar case can be extended to
the vectorial problem. Also, a vectorial Bernoulli problem with no sign
assumption on the components is studied in [20]. In [11], the same result has
been obtained by the viscosity approach and improvement of flatness.

In this paper, we deal with a (weakly coupled) cooperative system for
$p$-laplacian version of Bernoulli-type problem, following similar procedure
as that in [7].

######  Remark 1.1.

It should be remarked that our approaches in this paper, with some extra
efforts, can be adapted to variable exponent case, as well as variable
coefficient one. Similar types of results are then expected.

###  1.2. Notation

For clarity of exposition we shall introduce some notations and definitions,
which are used frequently in this text.

Throughout this paper, ${\mathbb{R}}^{n}$ will be equipped with the Euclidean
inner product $x\cdot y$ and the induced norm $|x|$, $B_{r}(x_{0})$ will
denote the open $n$-dimensional ball with center $x_{0}$, radius $r$ and its
boundary with $\partial B_{r}(x_{0})$. In addition, $B_{r}=B_{r}(0)$ and
$\partial B_{r}=\partial B_{r}(0)$. For the target space, ${\mathbb{R}}^{m}$,
we use several norms

$\begin{split}&|\mathbf{u}|_{p}=((u^{1})^{p}+\dots+(u^{m})^{p})^{1/p},\\\
&|\mathbf{u}(x)|_{\infty}=\max_{1\leq j\leq m}|u^{j}(x)|.\end{split}$

For convenience, we denote the Euclidean norm without the index,
$|\mathbf{u}|=|\mathbf{u}|_{2}$. We also use the Euclidean norm in the
definition of $L^{\infty}$-norm, that is

$\Arrowvert{\mathbf{u}}\Arrowvert_{L^{\infty}}:=\sup_{x}|\mathbf{u}(x)|.$

###  1.3. Plan of the paper

The paper is organized as follows: In Section 2, we study the existence of
minimizer (Theorem 2.1) and show that minimizers are $p$-subharmonic (Lemma
2.3). Section 3 is devoted to the regularity property of solutions, including
the Hölder regularity (Lemma 3.2) and the Lipschitz regularity (Theorem 3.5).
Section 4 consists of the proof of nondegeneracy property (Lemma 4.1). Also,
in Theorem 4.2 an estimate for the density of the free boundary is obtained
which is enough to prove that the free boundary has zero Lebesgue measure. The
vector-valued measure $\Delta_{p}\mathbf{u}$ (Theorem 5.1) and
$(n-1)$-Hausdorff dimension of free boundary (Theorem 5.2) are discussed in
Section 5. The main result in Section 6 is the flatness of regular part of the
free boundary (Theorem 6.5). We prove a partial result for the regularity of
free boundary in Section 7 (Theorem 7.5), along with $C^{1,\alpha}$-regularity
of the free boundary when $p$ is sufficiently close to 2 (Theorem 7.6). In
Appendix A we deal with NTA properties of the free boundary. Also, in Appendix
B we present an auxiliary lemma to study the asymptotic behaviour of
$p$-harmonic functions.

##  2\. Existence of a minimizer

######  Theorem 2.1.

If $J(\mathbf{g})<\infty$, then there exists an absolute minimizer of $J$ over
class ${\mathcal{K}}$.

###### Proof.

Obviously the functional is nonnegative, and hence it takes an infimum value.
Let $\mathbf{u}_{k}$ be a minimizing sequence

$\inf_{\mathbf{v}\in{\mathcal{K}}}J(\mathbf{v})=\lim_{k\rightarrow\infty}J(\mathbf{u}_{k}).$

Then $\mathbf{u}_{k}-\mathbf{g}$ is bounded in
$W^{1,p}_{0}(\Omega;{\mathbb{R}}^{m})$ and up to a subsequence we can assume
that

$\begin{split}\mathbf{u}_{k}\rightharpoonup\mathbf{u},&\quad\text{ weakly in
}W^{1,p}(\Omega;{\mathbb{R}}^{m}),\\\
\mathbf{u}_{k}\rightarrow\mathbf{u},&\quad\text{ a.e. in }\Omega,\end{split}$

for some $\mathbf{u}\in{\mathcal{K}}$. The latter convergence implies

$\int_{\Omega}\chi_{\\{|\mathbf{u}_{k}|>0\\}}dx\rightarrow\int_{\Omega}\chi_{\\{|\mathbf{u}|>0\\}}dx,$

and the weakly lower semicontinuity of the norm implies that

$\int_{\Omega}\sum_{i=1}^{m}|\nabla
u^{i}|^{p}dx\leq\liminf_{k\rightarrow\infty}\int_{\Omega}\sum_{i=1}^{m}|\nabla
u_{k}^{i}|^{p}dx.$

These together show that
$J(\mathbf{u})\leq\lim_{k\rightarrow\infty}J(\mathbf{u}_{k})$ and hence
$\mathbf{u}\in{\mathcal{K}}$ is an absolute minimizer. ∎

######  Remark 2.2.

We say $\mathbf{u}\in{\mathcal{K}}$ is a local minimizer of $J$, if
$J(\mathbf{u})\leq J(\mathbf{v})$ for any $\mathbf{v}\in{\mathcal{K}}$ with

$\Arrowvert{\nabla\mathbf{u}-\nabla\mathbf{v}}\Arrowvert_{L^{p}(\Omega)}+\Arrowvert{\chi_{\\{|\mathbf{u}|>0\\}}-\chi_{\\{|\mathbf{v}|>0\\}}}\Arrowvert_{L^{1}(\Omega)}\leq\varepsilon,$

for some $\varepsilon>0$. Although, all results in this paper are proved for
local minimizers, for the sake of convenience we argue with absolute
minimizers.

######  Lemma 2.3.

If $\mathbf{u}$ is a (local) minimizer, then $u^{i}$ is p-subharmonic for all
$i=1,\dots,m$, i.e.,

$\int_{\Omega}|\nabla u^{i}|^{p-2}\nabla u^{i}\cdot\nabla\varphi\,dx\leq
0,\quad\text{ for all }\varphi\in C_{0}^{\infty}(\Omega),\,\varphi\geq 0.$

Moreover, in each component of $\\{|\mathbf{u}|>0\\}$ for $i=1,\dots,m$,
either $u^{i}$ is identically vanishing or it is positive. Hence,

(2) $\Delta_{p}u^{i}=0,\quad\text{ in }\\{|\mathbf{u}|>0\\},\,\text{ for all
}i=1,\dots,m,$

and consequently (by the maximum principle)

$\Arrowvert{\mathbf{u}}\Arrowvert_{L^{\infty}(\Omega)}\leq\Arrowvert{\mathbf{g}}\Arrowvert_{L^{\infty}(\Omega)}.$

###### Proof.

Let

$\varphi\in C_{0}^{\infty}(\Omega),\,\varphi\geq 0,$

$t>0$, and define $v^{j}=u^{j}$ for $j\neq i$ and
$v^{i}=(u^{i}-t\varphi)_{+}$. Then $\mathbf{v}\in{\mathcal{K}}$ and we can
choose $\mathbf{v}$ as a competitor, so

$\begin{split}0\leq\frac{1}{t}\left(J(\mathbf{v})-J(\mathbf{u})\right)=&\frac{1}{t}\int_{\Omega}|\nabla
v^{i}|^{p}-|\nabla
u^{i}|^{p}+Q^{p}\left(\chi_{\\{|\mathbf{v}|>0\\}}-\chi_{\\{|\mathbf{u}|>0\\}}\right)dx\\\
\leq&\frac{1}{t}\int_{\Omega}|\nabla(u^{i}-t\varphi)|^{p}-|\nabla
u^{i}|^{p}dx\\\ \rightarrow&\int_{\Omega}-p|\nabla u^{i}|^{p-2}\nabla
u^{i}\cdot\nabla\varphi\,dx.\end{split}$

The second statement in the lemma relies on the strong minimum principle for
$p$-harmonic functions. In fact, let $x\in\\{|\mathbf{u}|>0\\}$ but
$u^{i}(x)=0$ for some $i$. Choose an index $j$ such that $u^{j}>0$ inside
$B_{r}(x)$ where $r$ is small enough. Now denote by $\tilde{u}^{i}$ the
$p$-harmonic extension of $u^{i}$ inside $B_{r}(x)$ and consider the
competitor $\tilde{\mathbf{u}}$ by defining $\tilde{u}^{k}=u^{k}$ for $k\neq
i$. Since $\\{|\mathbf{u}|>0\\}=\\{|\tilde{\mathbf{u}}|>0\\}$, we get
$J(\tilde{\mathbf{u}})<J(\mathbf{u})$ unless $\tilde{u}^{i}=u^{i}$ and $u^{i}$
is $p$-harmonic in $B_{r}(x)$. But this violates the strong minimum principle
if $u^{i}(x)=0$. ∎

##  3\. Regularity of local minimizers

######  Lemma 3.1.

Let $\mathbf{u}$ be a (local) minimizer of $J$, and $v^{i}$ the harmonic
replacement (majorant) for $u^{i}$ in $B\subset\Omega$ (for $B$ a small ball).
Then there is a universal constant $C=C(n,p)$ such that

$\displaystyle\int_{B}|\nabla(u^{i}-v^{i})|^{p}\,dx\leq
CQ_{\max}^{p}|\\{|\mathbf{u}|=0\\}\cap B|,\qquad\text{ when }2\leq p,$
$\displaystyle\int_{B}|\nabla(u^{i}-v^{i})|^{p}\,dx\leq
C(Q_{\max})^{p^{2}/2}|\\{|\mathbf{u}|=0\\}\cap B|^{p/2}\left(\int_{B}|\nabla
u^{i}|^{p}\,dx\right)^{1-p/2},\text{ when }1<p\leq 2.$

###### Proof.

Let $v^{j}=u^{j}$ for all $j\neq i$, and extend $v^{i}$ by $u^{i}$ in
$\Omega\setminus B$. If $B$ is small enough (when $\mathbf{u}$ is absolute
minimizer, we do not need this assumption), then we have $J(\mathbf{u})\leq
J(\mathbf{v})$ and consequently

$\int_{B}|\nabla u^{i}|^{p}-|\nabla v^{i}|^{p}\,dx\leq
Q_{\max}^{p}\left|\\{|\mathbf{u}|=0\\}\cap B\right|.$

Set now $w_{s}(x)=su^{i}(x)+(1-s)v^{i}(x)$ for $0\leq s\leq 1$. Then

$\begin{split}\int_{B}|\nabla u^{i}|^{p}-|\nabla
v^{i}|^{p}\,dx=&\int_{B}|\nabla w_{1}|^{p}-|\nabla w_{0}|^{p}\,dx\\\
=&p\int_{0}^{1}ds\int_{B}|\nabla w_{s}|^{p-2}\nabla
w_{s}\cdot\nabla(u^{i}-v^{i})\,dx\\\ =&p\int_{0}^{1}ds\int_{B}\left(|\nabla
w_{s}|^{p-2}\nabla w_{s}-|\nabla v^{i}|^{p-2}\nabla
v^{i}\right)\cdot\nabla(u^{i}-v^{i})\,dx\\\
=&p\int_{0}^{1}\frac{ds}{s}\int_{B}\left(|\nabla w_{s}|^{p-2}\nabla
w_{s}-|\nabla v^{i}|^{p-2}\nabla
v^{i}\right)\cdot\nabla(w_{s}-v^{i})\,dx,\end{split}$

where for the third equality we have used $\Delta_{p}v^{i}=0$. Next using

$\left(|b|^{p-2}b-|a|^{p-2}a\right)\cdot(b-a)\geq\gamma\left\\{\begin{array}[]{ll}|b-a|^{2}(|b|+|a|)^{p-2},&1<p\leq
2,\\\\[8.0pt] |b-a|^{p},&2\leq p,\end{array}\right.$

we obtain for $p\geq 2$

$\int_{B}|\nabla u^{i}|^{p}-|\nabla v^{i}|^{p}\,dx\geq\gamma
p\int_{0}^{1}\frac{ds}{s}\int_{B}|\nabla(w_{s}-v^{i})|^{p}\,dx=\gamma
p\int_{0}^{1}s^{p-1}ds\int_{B}|\nabla(u^{i}-v^{i})|^{p}\,dx,$

which shows the desired estimate for $p\geq 2$.

In case $1<p\leq 2$, we have

$\begin{split}\int_{B}|\nabla u^{i}|^{p}-|\nabla v^{i}|^{p}\,dx&\geq\gamma
p\int_{0}^{1}\frac{ds}{s}\int_{B}|\nabla(w_{s}-v^{i})|^{2}\left(|\nabla
w_{s}|+|\nabla v^{i}|\right)^{p-2}\,dx\\\ &\geq\gamma
p\int_{0}^{1}sds\int_{B}|\nabla(u^{i}-v^{i})|^{2}\left(s|\nabla
u^{i}|+(2-s)|\nabla v^{i}|\right)^{p-2}\,dx,\\\ &\geq
C\int_{B}|\nabla(u^{i}-v^{i})|^{2}\left(|\nabla u^{i}|+|\nabla
v^{i}|\right)^{p-2}\,dx.\end{split}$

On the other hand, using the Hölder inequality, we have

$\begin{split}\int_{B}|\nabla(u^{i}-v^{i})|^{p}\,dx\leq\left(\int_{B}|\nabla(u^{i}-v^{i})|^{2}\left(|\nabla
u^{i}|+|\nabla v^{i}|\right)^{p-2}\,dx\right)^{p/2}\left(\int_{B}\left(|\nabla
u^{i}|+|\nabla v^{i}|\right)^{p}\right)^{1-p/2}.\end{split}$

We conclude the proof by applying $\int_{B}|\nabla
v^{i}|^{p}\leq\int_{B}|\nabla u^{i}|^{p}$ (since $v^{i}$ is $p$-harmonic). ∎

######  Lemma 3.2.

(Hölder regularity) Let $\mathbf{u}$ be a (local) minimizer of $J$ in $B_{1}$.
Then for some $\alpha=\alpha(n,p)$

$\Arrowvert{\mathbf{u}}\Arrowvert_{C^{\alpha}(B_{3/4})}\leq
C(n,p,\Arrowvert{\mathbf{u}}\Arrowvert_{L^{\infty}(B_{1})}).$

###### Proof.

Let $M=\Arrowvert{\mathbf{u}}\Arrowvert_{L^{\infty}(B_{1})}$ and
$B_{r}=B_{r}(y)$ for $y\in B_{3/4}$ and $r<1/8$. Since $u^{i}$ is a
$p$-subsolution, a Caccioppoli type inequality (see [16], Lemma 3.27) implies
that

$\int_{B_{r}}|\nabla
u^{i}|^{p}\,dx\leq\frac{C}{r^{p}}\int_{B_{2r}}(u^{i})^{p}\,dx\leq
CM^{p}r^{n-p}.$

On the other hand, if $v^{i}$ is the $p$-harmonic replacement of $u^{i}$
inside $B_{r}$, we have the gradient estimate (see [17])

$\sup_{B_{r/2}}|\nabla v^{i}|\leq\left(\frac{C}{r^{n}}\int_{B_{r}}|\nabla
v^{i}|^{p}\,dx\right)^{1/p}\leq\frac{CM}{r}.$

Now, let us take some $\rho<r/2$ which will be specified below and apply Lemma
3.1 in $B_{r}(y)$

$\displaystyle\Arrowvert{\nabla u^{i}}\Arrowvert_{L^{p}(B_{\rho})}\leq$
$\displaystyle\Arrowvert{\nabla(u^{i}-v^{i})}\Arrowvert_{L^{p}(B_{\rho})}+\Arrowvert{\nabla
v^{i}}\Arrowvert_{L^{p}(B_{\rho})}$ $\displaystyle\leq$
$\displaystyle\Arrowvert{\nabla(u^{i}-v^{i})}\Arrowvert_{L^{p}(B_{r})}+C\rho^{n/p}\Arrowvert{\nabla
v^{i}}\Arrowvert_{L^{\infty}(B_{r/2})}$ $\displaystyle\leq$ $\displaystyle
C\left\\{\begin{array}[]{ll}r^{n/p}+M\rho^{n/p}r^{-1}&\text{ for }2\leq
p,\\\\[7.0pt] M^{1-p/2}r^{n/p-1+p/2}+M\rho^{n/p}r^{-1}&\text{ for }1<p\leq
2.\end{array}\right.$

Thus for $r=\rho^{1-\alpha}$, if we take $\alpha=\alpha(n,p)$ sufficiently
small, we obtain

$\Arrowvert{\nabla u^{i}}\Arrowvert_{L^{p}(B_{\rho})}\leq
C(M,n,p,Q_{\max})\rho^{n/p-1+\alpha}.$

By virtue of Morrey’s theorem (see [19]) we conclude the proof of the lemma. ∎

The next lemma is essential to prove the Lipschitz regularity of the
minimizers.

######  Lemma 3.3.

Let $\mathbf{u}=(u^{1},\dots,u^{m})$ be a bounded minimizer in $B_{1}$ and
$u^{i}(0)=0$ for some $1\leq i\leq m$. Then there exists a constant
$C=C(n,p,Q_{\max})>0$ such that

$\Arrowvert{u^{i}}\Arrowvert_{L^{\infty}(B_{1/4})}\leq C.$

We need to remark that the constant $C$ is independent of the boundary values
of $\mathbf{u}$ on $\partial\Omega$. In other words when going away from a
free boundary, but staying uniformly inside the domain $\Omega$, the minimizer
cannot grow too large, regardless of the boundary values. In other words, for
large enough boundary values, the origin cannot be a free boundary point.

###### Proof.

For the sake of convenience consider $i=1$. Towards a contradiction, assume
that there is a sequence of bounded solutions $\mathbf{u}_{k}$ in $B_{1}$ such
that

$\Arrowvert{u^{1}_{k}}\Arrowvert_{L^{\infty}(B_{1/4})}>k.$

Set

$d_{k}(x):={\rm dist}(x,\\{u^{1}_{k}=0\\})\quad\text{ in }B_{1},$

and define

${\mathcal{O}}_{k}:=\left\\{x\in B_{1}:d_{k}(x)\leq(1-|x|)/3\right\\}.$

Obviously, $B_{1/4}\subset{\mathcal{O}}_{k}$. We have also

$m_{k}:=\sup_{{\mathcal{O}}_{k}}(1-|x|)u^{1}_{k}(x)\geq\frac{3}{4}\max_{B_{1/4}}u^{1}_{k}>\frac{3}{4}k,$

Since $u^{1}_{k}$ is bounded (for fixed $k$), we get
$(1-|x|)u^{1}_{k}(x)\rightarrow 0$ as $|x|\rightarrow 1$, and therefore
$m_{k}$ is attained at some point $x_{k}\in{\mathcal{O}}_{k}$. So,

$u^{1}_{k}(x_{k})=\frac{m_{k}}{1-|x_{k}|}\geq m_{k}>\frac{3}{4}k.$

Now let $y_{k}\in\partial\\{u^{1}_{k}>0\\}\cap B_{1}$ be such that
$|y_{k}-x_{k}|=d_{k}(x_{k})=:\delta_{k}$, which satisfies
$\delta_{k}\leq(1-|x_{k}|)/3$ due to $x_{k}\in{\mathcal{O}}_{k}$. This implies
that

$B_{2\delta_{k}}(y_{k})\subset B_{1}\qquad\text{ and }\qquad
B_{\delta_{k}/2}(y_{k})\subset{\mathcal{O}}_{k}.$

Indeed, if $z\in B_{2\delta_{k}}(y_{k})$,

$|z|\leq|z-y_{k}|+|y_{k}-x_{k}|+|x_{k}|\leq 2\delta_{k}+\delta_{k}+|x_{k}|\leq
1,$

and if $z\in B_{\delta_{k}/2}(y_{k})$,

$1-|z|\geq 1-|x_{k}|-|x_{k}-y_{k}|-|y_{k}-z|\geq
1-|x_{k}|-\delta_{k}-\delta_{k}/2\geq 3\delta_{k}/2\geq 3|z-y_{k}|\geq
3d_{k}(z).$

Also, we have $1-|z|\geq(1-|x_{k}|)/2$ for any $z\in B_{\delta_{k}/2}(y_{k})$.
Then

$\frac{1-|x_{k}|}{2}\max_{B_{\delta_{k}/2}(y_{k})}u^{1}_{k}\leq\max_{z\in
B_{\delta_{k}/2}(y_{k})}(1-|z|)u^{1}_{k}(z)\leq\max_{z\in{\mathcal{O}}_{k}}(1-|z|)u^{1}_{k}(z)=(1-|x_{k}|)u^{1}_{k}(x_{k})$

or

$\max_{B_{\delta_{k}/2}(y_{k})}u^{1}_{k}\leq 2u^{1}_{k}(x_{k}).$

Since $B_{\delta_{k}}(x_{k})\subset\\{u^{1}_{k}>0\\}$, then $u^{1}_{k}$ is
$p$-harmonic inside $B_{\delta_{k}}(x_{k})$, i.e. $\Delta_{p}u^{1}_{k}=0$. By
the Harnack inequality for $p$-harmonic functions, there is a constant
$c=c(n,p)$ such that

$\min_{B_{4\delta_{k}/5}(x_{k})}u^{1}_{k}\geq cu^{1}_{k}(x_{k}).$

In particular,

$\max_{B_{\delta_{k}/4}(y_{k})}u^{1}_{k}\geq cu^{1}_{k}(x_{k}).$

We define the sequence

$\mathbf{w}_{k}(x):=\frac{\mathbf{u}_{k}(y_{k}+(\delta_{k}/2)x)}{u^{1}_{k}(x_{k})},$

whose first component satisfies

(3) $\max_{B_{1}}w^{1}_{k}\leq 2,\qquad\max_{B_{1/2}}w^{1}_{k}\geq c>0,\qquad
w^{1}_{k}(0)=0.$

Moreover, $\mathbf{w}_{k}$ is a minimizer of

$J_{k}(\mathbf{w})=\int_{B_{1}}\sum_{i=1}^{m}|\nabla
w^{i}|^{p}+Q_{k}^{p}\chi_{\\{|\mathbf{w}|>0\\}}dx,$

where
$Q_{k}(x)=\frac{\delta_{k}Q(y_{k}+(\delta_{k}/2)x)}{2u_{k}^{1}(x_{k})}\rightarrow
0$. Now consider $v_{k}^{1}$ to be $p$-harmonic replacement of $u_{k}^{1}$ in
$B_{3/4}$ and apply Lemma 3.1

(4) $\int_{B_{3/4}}|\nabla(w_{k}^{1}-v_{k}^{1})|^{p}\,dx\leq C(\max
Q_{k})^{p}\rightarrow 0,$

when $2\leq p$. Similar statement holds for $1<p\leq 2$, we just need to note
that $\Arrowvert{\nabla w_{k}^{1}}\Arrowvert_{L^{p}}$ is uniformly bounded
($w_{k}^{1}$ is $p$-subsolution and uniformly bounded in $B_{1}$).
Furthermore, $w_{k}^{1}$ and $v_{k}^{1}$ are uniformly $C^{\alpha}$ in
$B_{5/8}$ and we can extract a subsequence (still denoted by $w_{k}^{1}$ and
$v_{k}^{1}$) such that $w_{k}^{1}\rightarrow w_{0}$ and $v_{k}^{1}\rightarrow
v_{0}$ uniformly in $B_{5/8}$. Observe that $\Delta_{p}v_{0}=0$ in $B_{5/8}$
and (4) implies that $w_{0}=v_{0}+c$. Hence, $w_{0}$ is also $p$-harmonic and
by the strong maximum principle, $w_{0}\equiv 0$ in $B_{5/8}$, since
$w_{0}\geq 0$ and $w_{0}(0)=0$. On the other hand, (3) necessitates

$\max_{B_{1/2}}w_{0}\geq c>0,$

which is a contradiction. ∎

A direct consequence of the above lemma is the following estimate.

######  Lemma 3.4.

Let $\mathbf{u}$ be a (local) minimizer in $\Omega$. If ${\rm
dist}(x_{0},\\{u^{i}=0\\})<\frac{1}{5}{\rm dist}(x_{0},\partial\Omega)$ then

$u^{i}(x_{0})\leq 4C{\rm dist}(x_{0},\\{u^{i}=0\\}),$

where $C$ is the constant defined in Lemma 3.3.

###### Proof.

Choose $y_{0}\in\\{u^{i}=0\\}$ such that ${\rm
dist}(x_{0},\\{u^{i}=0\\})=|x_{0}-y_{0}|=d_{0}$. Now apply Lemma 3.3 to

$\mathbf{v}(x):=\frac{\mathbf{u}(y_{0}+4d_{0}x)}{4d_{0}}$

to get

$u^{i}(x_{0})\leq 4Cd_{0}.$

∎

With the above two results we will obtain uniform Lipschitz regularity for
minimizers.

######  Theorem 3.5.

Let $\mathbf{u}$ be a (local) minimizer in $\Omega$, then $\mathbf{u}$ is
Lipschitz. Moreover, for every $K\Subset\Omega$ such that
$K\cap\partial\\{u^{i}>0\\}\neq\emptyset$ for some $1\leq i\leq m$, there is a
constant $C=C(n,p,Q_{\max},{\rm dist}(K,\partial\Omega),\Omega)>0$ such that

$\Arrowvert{\nabla u^{i}}\Arrowvert_{L^{\infty}(K)}\leq C.$

Once again we remark that the constant $C$ does not depend on the boundary
values of the minimizer, as long as we stay uniformly inside the domain.

###### Proof.

Step 1: We show that $u^{i}$ is bounded in $K$ with a universal constant $C$
depending on the following ingredients $n,p,Q_{\max},{\rm
dist}(K,\partial\Omega),\Omega$. Let $r_{0}=\frac{1}{5}{\rm
dist}(K,\partial\Omega)$ and for any arbitrary point $x\in K$ there is a
sequence of points $x=x_{0},\dots,x_{k}\in K$ with (we can assume $K$ is
connected, otherwise replace it with a bigger one which is connected)

$x_{j}\in B_{r_{0}/2}(x_{j-1}),\quad\text{ for }j=1,\dots,k,$

$B_{r_{0}}(x_{j})\subset\\{u^{i}>0\\}$ for $j=0,\dots,k-1$ and
$B_{r_{0}}(x_{k})\cap\\{u^{i}=0\\}\neq\emptyset$. Note that $k$, the number of
points, only depends on $\Omega$ and ${\rm dist}(K,\partial\Omega)$. From
Lemma 3.4, we get

$u^{i}(x_{k})\leq 4Cr_{0}.$

Since $u^{i}$ is $p$-harmonic in $B_{r_{0}}(x_{j})$, $j=0,\dots,k-1$, by
virtue of Harnack’s inequality, there is a constant $c$ such that

$u^{i}(x_{j+1})\geq cu^{i}(x_{j}).$

Thus

$u^{i}(x)\leq 4c^{-k}Cr_{0}.$

Step 2: Here we find a control on $\nabla u^{i}$ at points close to
$\\{u^{i}=0\\}$. If $d={\rm dist}(y,\\{u^{i}=0\\})<\frac{1}{11}{\rm
dist}(y,\partial\Omega)$, every points $x_{0}\in B_{d}(y)$ satisfy condition
Lemma 3.4. Then

$u^{i}(x_{0})\leq 4C{\rm dist}(x_{0},\\{u^{i}=0\\})\leq 8Cd.$

Let us define

$v(x):=\frac{u^{i}(y+dx)}{d}$

which is a $p$-harmonic in $B_{1}$ and
$\Arrowvert{v}\Arrowvert_{L^{\infty}(B_{1})}\leq 8C$. By $p$-Laplacian
estimate for gradient, we obtain

$|\nabla v(0)|\leq\tilde{C}(n,p,Q_{\max}),$

that is $|\nabla u^{i}(y)|\leq C(n,p,Q_{\max})$.

Step 3: Let $r_{1}=\frac{1}{11}{\rm dist}(K,\partial\Omega)$. If ${\rm
dist}(x,\\{u^{i}=0\\})\leq r_{1}$, by the result of Step 2 we have already
$|\nabla u^{i}(x)|\leq C$. If ${\rm dist}(x,\\{u^{i}=0\\})>r_{1}$, then
$u^{i}$ is $p$-harmonic inside $B_{r_{1}}(x)$ and
$\Arrowvert{u^{i}}\Arrowvert_{L^{\infty}(B_{r_{0}})}$ is universally bounded
by the result of Step 1. Thus $|\nabla u^{i}(x)|$ will be universally bounded.
∎

A straightforward corollary to this theorem, that can be useful later, is the
following

######  Corollary 3.6.

Let $\mathbf{u}$ be a (local) minimizer for our functional. For every
$K\Subset\Omega$ there exists constant $C=C(n,p,Q_{\max},{\rm
dist}(K,\partial\Omega),\Omega)$ such that

$\frac{1}{r}\mathchoice{{\vbox{\hbox{$\textstyle-$
}}\kern-7.83337pt}}{{\vbox{\hbox{$\scriptstyle-$
}}\kern-5.90005pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.75003pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.25003pt}}\\!\int_{\partial B_{r}}u^{i}\,dx>C\ \ \text{ implies }\ \
u^{i}>0\text{ in }B_{r}.$

###### Proof.

If $B_{r}\subset K$ contains a free boundary point, then by Theorem 3.5,
$u^{i}\leq Cr$ on $\partial B_{r}$. ∎

##  4\. Nondegeneracy

######  Lemma 4.1.

For any $0<\kappa<1$ there exists a constant $c=c(\kappa,n,m,p,Q_{\min})>0$
such that for every minimizer $\mathbf{u}$ and for any (small) ball
$B_{r}\subset\Omega$

$\Arrowvert{\mathbf{u}}\Arrowvert_{L^{\infty}(B_{r})}<cr\ \ \text{ implies }\
\ \mathbf{u}={\bf 0}\text{ in }B_{\kappa r}.$

###### Proof.

Without loss of generality, we may assume $r=1$. Let

$M=\Arrowvert{\mathbf{u}}\Arrowvert_{L^{\infty}(B_{\sqrt{\kappa}})}.$

Let $\phi(x)=\phi_{\kappa}(|x|)$ be the solution of

$\Delta_{p}\phi=0,\ \text{ in }B_{\sqrt{\kappa}}\setminus
B_{\kappa},\qquad\phi=0\ \text{ on }\partial B_{\kappa},\qquad\phi=1\ \text{
on }\partial B_{\sqrt{\kappa}}$

and extend $\phi=0$ in $B_{\kappa}$. Set $v=M\sqrt{\kappa}\phi$ and
$w^{i}=\min(u^{i},v)$ for all $i=1,\dots,m$. Since $v\geq u^{i}$ on $\partial
B_{\sqrt{\kappa}}$, so $w^{i}=u^{i}$ on $\partial B_{\sqrt{\kappa}}$.
Therefore $J(\mathbf{u})\leq J(\mathbf{w})$, or equivalently

$\int_{B_{\sqrt{\kappa}}}\sum_{i=1}^{m}|\nabla
u^{i}|^{p}+Q^{p}\chi_{\\{|\mathbf{u}|>0\\}}\,dx\leq\int_{B_{\sqrt{\kappa}}\setminus
B_{\kappa}}\sum_{i=1}^{m}|\nabla
w^{i}|^{p}+Q^{p}\chi_{\\{|\mathbf{w}|>0\\}}\,dx.$

Since $\\{|\mathbf{w}|>0\\}\subset\\{|\mathbf{u}|>0\\}$, we get

$\begin{split}\int_{B_{\kappa}}\sum_{i=1}^{m}|\nabla
u^{i}|^{p}+Q^{p}\chi_{\\{|\mathbf{u}|>0\\}}\,dx&\leq\int_{B_{\sqrt{\kappa}}\setminus
B_{\kappa}}\sum_{i=1}^{m}\left(|\nabla w^{i}|^{p}-|\nabla
u^{i}|^{p}\right)\,dx\\\ &\leq p\int_{B_{\sqrt{\kappa}}\setminus
B_{\kappa}}\sum_{i=1}^{m}|\nabla w^{i}|^{p-2}\nabla
w^{i}\cdot\nabla(w^{i}-u^{i})\,dx\\\ &=-p\int_{\partial
B_{\kappa}}\sum_{i=1}^{m}|\nabla w^{i}|^{p-2}(w^{i}-u^{i})(\nabla
w^{i}\cdot\nu)\,d{\mathcal{H}}^{n-1}\\\ &=p\int_{\partial
B_{\kappa}}\sum_{i=1}^{m}|\nabla v|^{p-2}u^{i}(\nabla
v\cdot\nu)\,d{\mathcal{H}}^{n-1}.\end{split}$

Since $|\nabla v|\leq C(p,\kappa,n)M$ on $\partial B_{\kappa}$, we find out
that

(5) $\int_{B_{\kappa}}\sum_{i=1}^{m}|\nabla
u^{i}|^{p}+Q^{p}\chi_{\\{|\mathbf{u}|>0\\}}\,dx\leq
CM^{p-1}\sum_{i=1}^{m}\int_{\partial B_{\kappa}}u^{i}\,d{\mathcal{H}}^{n-1}.$

On the other hand,

$\begin{split}\int_{\partial B_{\kappa}}u^{i}\,d{\mathcal{H}}^{n-1}&\leq
C(n,\kappa)\int_{B_{\kappa}}u^{i}+|\nabla u^{i}|\,dx\\\
&=C(n,\kappa)\int_{B_{\kappa}}\left(u^{i}+|\nabla
u^{i}|\right)\chi_{\\{u^{i}>0\\}}\,dx\\\ &\leq
C(n,\kappa,p,Q_{\min})\int_{B_{\kappa}}MQ^{p}\chi_{\\{u^{i}>0\\}}+|\nabla
u^{i}|^{p}+Q^{p}\chi_{\\{u^{i}>0\\}}\,dx\\\ &\leq
C(n,\kappa,p,Q_{\min})(1+M)\int_{B_{\kappa}}|\nabla
u^{i}|^{p}+Q^{p}\chi_{\\{u^{i}>0\\}}\,dx.\end{split}$

Comparing with (5), we we arrive at

$\int_{B_{\kappa}}\sum_{i=1}^{m}|\nabla
u^{i}|^{p}+Q^{p}\chi_{\\{|\mathbf{u}|>0\\}}\,dx\leq
CM^{p-1}(1+M)\int_{B_{\kappa}}\sum_{i=1}^{m}|\nabla
u^{i}|^{p}+Q^{p}\chi_{\\{|\mathbf{u}|>0\\}}\,dx.$

Therefore, if $M$ is small enough, we obtain that $\mathbf{u}={\bf 0}$ in
$B_{\kappa}$. ∎

An immediate consequence of the above lemma is the following. For any
$K\Subset\Omega$ there are positive constants $c_{0},C_{0}$ such that if
$B_{r}(x)\subset K\cap\\{|\mathbf{u}|>0\\}$ touches
$\partial\\{|\mathbf{u}|>0\\}$ then

(6) $c_{0}r\leq|\mathbf{u}(x)|\leq C_{0}r.$

######  Theorem 4.2.

For $K\Subset\Omega$ there exists constant $0<c=c(n,m,p,K,\Omega)<1$ such that
for any (local) minimizer $\mathbf{u}$ and for any (small) ball
$B_{r}(x)\subset K$ with $x\in\partial\\{|\mathbf{u}|>0\\}$,

(7)
$c<\frac{{\mathcal{L}}^{n}(B_{r}(x)\cap\\{|\mathbf{u}|>0\\})}{{\mathcal{L}}^{n}(B_{r}(x))}<1-c.$

###### Proof.

By Lemma 4.1, there exists $y\in B_{r/2}$ such that $|\mathbf{u}(y)|\geq
cr>0$. Using Lipschitz continuity we get

$\mathchoice{{\vbox{\hbox{$\textstyle-$
}}\kern-7.83337pt}}{{\vbox{\hbox{$\scriptstyle-$
}}\kern-5.90005pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.75003pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.25003pt}}\\!\int_{\partial B_{\kappa
r}(y)}|\mathbf{u}|\geq\frac{cr}{2},$

provided $\kappa$ is small enough. Hence

$\frac{1}{\kappa r}\mathchoice{{\vbox{\hbox{$\textstyle-$
}}\kern-7.83337pt}}{{\vbox{\hbox{$\scriptstyle-$
}}\kern-5.90005pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.75003pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.25003pt}}\\!\int_{\partial B_{\kappa
r}(y)}|\mathbf{u}|\geq\frac{c}{2\kappa},$

and also for at least one component $u^{i}$

$\frac{1}{\kappa r}\mathchoice{{\vbox{\hbox{$\textstyle-$
}}\kern-7.83337pt}}{{\vbox{\hbox{$\scriptstyle-$
}}\kern-5.90005pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.75003pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.25003pt}}\\!\int_{\partial B_{\kappa r}(y)}u^{i}\geq\frac{c}{2\kappa
m},$

which by Corollary 3.6 implies $|\mathbf{u}|>0$ in $B_{\kappa r}(y)$. This
gives the lower estimate in (7).

To prove the estimate from above we assume, for simplicity, $r=1$ and suppose
(towards a contradiction) that there is a sequence of minimizers
$\mathbf{u}_{k}$ in $B_{1}(0)$ such that
$0\in\partial\\{|\mathbf{u}_{k}|>0\\}$ and

${\mathcal{L}}^{n}(\\{|\mathbf{u}_{k}|=0\\})=:\varepsilon_{k}\rightarrow 0.$

Let $v_{k}^{i}$ be a $p$-harmonic function in $B_{1/2}$ with boundary data
$v_{k}^{i}=u_{k}^{i}$ on $\partial B_{1/2}$. From Lemma 3.1, we obtain that

(8) $\int_{B_{1/2}}|\nabla(v_{k}^{i}-u_{k}^{i})|^{p}\,dx\leq
C(\varepsilon_{k})\rightarrow 0.$

Since $u_{k}^{i}$ and $v_{k}^{i}$ are both uniformly Lipschitz in $B_{1/4}$,
we may assume that $u_{k}^{i}\rightarrow u_{0}^{i}$ and $v_{k}^{i}\rightarrow
v_{0}^{i}$ uniformly in $B_{1/4}$. Observe that $\Delta_{p}v_{0}^{i}=0$ and
(8) implies that $u_{0}^{i}=v_{0}^{i}+c$. Thus $\Delta_{p}u_{0}^{i}=0$ in
$B_{1/4}$ and from the strong minimum principle (since $u_{0}^{i}(0)=0$) it
follows $u_{0}^{i}\equiv 0$ in $B_{1/4}$, since $u_{0}^{i}\geq 0$ and
$u_{0}^{i}(0)=0$. On the other hand form nondegeneracy property, Lemma 4.1, we
know

$\Arrowvert{\mathbf{u}_{k}}\Arrowvert_{L^{\infty}(B_{1/2})}\geq c>0,$

which implies a similar inequality for $\mathbf{u}_{0}$, and hence a
contradiction. ∎

######  Remark 4.3.

Theorem 4.2, along with the Lebesgue density theorem implies that the free
boundary has zero Lebesgue measure

${\mathcal{L}}^{n}(\partial\\{|\mathbf{u}|>0\\})=0.$

##  5\. The vector-valued measure $\Delta_{p}\mathbf{u}$

Let $0\leq\zeta\in C^{\infty}_{0}(\Omega)$ be a test function, and define the
measure $\lambda^{i}$ by

$\int\zeta\,d\lambda^{i}=-\int|\nabla u^{i}|^{p-2}\nabla
u^{i}\cdot\nabla\zeta\,dx,$

which in virtue of Lemma 2.3 is a bounded non-negative measure, i.e. a Radon
measure. Obviously $\lambda^{i}$ is the formal way of expressing
$\Delta_{p}u^{i}$ in $\Omega$.

Since each $u^{i}$ is $p$-subharmonic in $\Omega$ and $u^{i}\geq 0$ we have
that $\lambda^{i}$ is a positive Radon measure. Because $u^{i}$ is also
$p$-harmonic in $\\{u^{i}>0\\}$ we have that the support of $\lambda^{i}$ is
in
$\Omega\cap\partial\\{u^{i}>0\\}\subseteq\Omega\cap\partial\\{|\mathbf{u}|>0\\}$.111Observe
that $u^{i}$ may be zero in some component of $\\{|\mathbf{u}|>0\\}$. Let us
define

$\Lambda=\Lambda^{\mathbf{u}}:=\sum_{i=1}^{m}\lambda^{i}.$

######  Theorem 5.1.

For any $K\Subset\Omega$ there exist constants $c,C>0$ such that for any
(local) minimizer $\mathbf{u}$

$cr^{n-1}\leq\int_{B_{r}}d\Lambda\leq Cr^{n-1}$

for any ball $B_{r}\subset K$ with $x\in\partial\\{|\mathbf{u}|>0\\}$.

###### Proof.

Let $0\leq\zeta_{\epsilon}\in C^{\infty}_{0}(B_{r+\epsilon})$ be a suitable
test function, such that $\zeta_{\epsilon}=1$ on $B_{r}$ and
$|\nabla\zeta_{\epsilon}|\leq 2/\epsilon$. Then

$\int\zeta_{\epsilon}\,d\lambda^{i}=-\int|\nabla u^{i}|^{p-2}\nabla
u^{i}\cdot\nabla\zeta_{\epsilon}\,dx=-\int_{B_{r+\epsilon}\setminus
B_{r}}|\nabla u^{i}|^{p-2}\nabla u^{i}\cdot\nabla\zeta_{\epsilon}\,dx\leq
Cr^{n-1},$

where in the last inequality we have used that $u$ is Lipschitz. Letting
$\epsilon$ tend to zero, we arrive at

$\int_{B_{r}}\,d\lambda^{i}\leq Cr^{n-1}.$

To prove the estimate from below, we argue indirectly. It also suffices to
consider the case $r=1$. Assume there is a sequence of minimizers
$\mathbf{u}_{k}$ in the unit ball $B_{1}(0)$, such that
$0\in\partial\\{|\mathbf{u}_{k}|>0\\}$ and for the measures
$\Lambda_{k}:=\Lambda^{\mathbf{u}_{k}}$ we have

$\varepsilon_{k}:=\Lambda_{k}(B_{1})\rightarrow 0.$

Since the functions $\mathbf{u}_{k}$ are uniformly Lipschitz continuous, we
may assume that $\mathbf{u}_{k}\rightarrow\mathbf{u}_{0}$ in $B_{1/2}$, where
$\mathbf{u}_{0}$ is Lipschitz continuous as well. We may also extract a
subsequence (still denote by $\mathbf{u}_{k}$) such that $g_{k}^{i}:=|\nabla
u^{i}_{k}|^{p}\nabla u^{i}_{k}\rightarrow g_{0}^{i}$ weakly-$*$ in
$L^{\infty}(B_{1/2})$ for all $i=1,\dots,m$. We claim that

(9) $g_{0}^{i}=|\nabla u^{i}_{0}|^{p}\nabla u^{i}_{0},\quad\text{ in
}B_{1/2}.$

Suppose this is true, then for every positive test function $\zeta\in
C_{0}^{\infty}(B_{1/2})$ one has

$\begin{split}-\int_{B_{1/2}}|\nabla u^{i}_{0}|^{p-2}\nabla
u^{i}_{0}\cdot\nabla\zeta\,dx=&-\lim_{k\rightarrow\infty}\int_{B_{1/2}}|\nabla
u^{i}_{k}|^{p-2}\nabla u^{i}_{k}\cdot\nabla\zeta\,dx\\\
=&\lim_{k\rightarrow\infty}\int_{B_{1/2}}\zeta\,d\lambda^{i}_{k}\leq\Arrowvert{\zeta}\Arrowvert_{L^{\infty}(B_{1/2})}\lim_{k\rightarrow\infty}\varepsilon_{k}=0.\end{split}$

Thus $\lambda_{0}^{i}=0$ and $u^{i}_{0}$ is $p$-harmonic for all $i=1,\dots,m$
(note that $u^{i}_{0}$ is the limit of a sequence of $p$-subharmonic functions
and we already know that it is $p$-subharmonic). Since $u_{0}^{i}\geq 0$ and
$u_{0}^{i}(0)=0$, by the minimum principle, we have $u_{0}^{i}\equiv 0$ in
$B_{1/2}$.

On the other hand, by nondegeneracy property (Lemma 4.1) and that
$0\in\partial\\{|\mathbf{u}_{k}|>0\\}$ we have

$\Arrowvert{\mathbf{u}_{k}}\Arrowvert_{L^{\infty}(B_{1/4})}\geq c>0.$

Therefore, a similar inequality holds for $\mathbf{u}_{0}$ and we arrive at a
contradiction.

To close the argument we need to prove (9). In fact, if
$\overline{B_{\rho}}=\overline{B_{\rho}(y)}\subset\\{|\mathbf{u}_{0}|>0\\}$
then $B_{\rho}\subset\\{|\mathbf{u}_{k}|>0\\}$ for sufficiently large $k$ and
$u_{k}^{i}$ are $p$-harmonic in $B_{\rho}$ for all $i=1,\dots,m$, (see Lemma
2.3). Therefore, one can extract a subsequence of $\mathbf{u}_{k}$ locally
converging to $\mathbf{u}_{0}$ in $C^{1,\alpha}(B_{\rho})$. Hence,
$g_{0}^{i}=|\nabla u^{i}_{0}|^{p}\nabla u^{i}_{0}$ in $B_{\rho}$ for all
$i=1,\dots,m$. Next, if $B_{\rho}\subset\\{|\mathbf{u}_{0}|=0\\}$ then for any
$\kappa<1$ the nondegeneracy property entails that
$B_{\kappa\rho}\subset\\{|\mathbf{u}_{k}|=0\\}$ for sufficiently large
$k=k(\kappa)$. Thus $g_{0}^{i}=0=|\nabla u^{i}_{0}|^{p}\nabla u^{i}_{0}$. We
just need to show ${\mathcal{L}}^{n}(\partial\\{|\mathbf{u}_{0}|>0\\}\cap
B_{1/2})=0$. If $x_{0}\in\partial\\{|\mathbf{u}_{0}|>0\\}\cap B_{1/2}$, then
$\mathbf{u}_{0}(x_{0})=0$. Choose
$x_{k}\in\partial\\{|\mathbf{u}_{k}|>0\\}\cap B_{1}$ such that
$|x_{k}-x_{0}|={\rm dist}(x_{0},\partial\\{|\mathbf{u}_{k}|>0\\})$, then
relation (6) yields that $|x_{k}-x_{0}|\rightarrow 0$. Apply Lemma 4.1 to
obtain

$\Arrowvert{\mathbf{u}_{k}}\Arrowvert_{L^{\infty}(B_{2r}(x_{0}))}\geq\Arrowvert{\mathbf{u}_{k}}\Arrowvert_{L^{\infty}(B_{r}(x_{k}))}\geq
cr,$

for any ball $B_{r}(x_{0})\subset B_{1/2}$ and sufficiently large $k$. Passing
to the limit we get the same inequality for $\mathbf{u}_{0}$,

$\Arrowvert{\mathbf{u}_{0}}\Arrowvert_{L^{\infty}(B_{2r}(x_{0}))}\geq cr.$

This along with the Lipschitz continuity of $\mathbf{u}_{0}$ is enough to
prove that ${\mathcal{L}}^{n}(B_{r}(x_{0})\cap\\{|\mathbf{u}_{0}|>0\\})\geq
c{\mathcal{L}}^{n}(B_{r})$ for some $c>0$. (see the first part of the proof of
Theorem 4.2). This implies that
${\mathcal{L}}^{n}(\partial\\{|\mathbf{u}_{0}|>0\\}\cap B_{1/2})=0$ (see
Remark 4.3). ∎

The next theorem follows easily from Theorem 5.1. The proof is the same as the
proof of Theorem 4.5 in [2].

######  Theorem 5.2.

Let $\mathbf{u}$ be a (local) minimizer in $\Omega$. Then

  1. (i)

For every $K\Subset\Omega$ we have
${\mathcal{H}}^{n-1}(K\cap\partial\\{|\mathbf{u}|>0\\})<\infty$.

  2. (ii)

There exist nonnegative Borel functions $q^{i}$ such that

$\Delta_{p}u^{i}=q^{i}{\mathcal{H}}^{n-1}\lfloor\,\partial\\{|\mathbf{u}|>0\\},$

that is for every $\zeta\in C_{0}^{\infty}(\Omega)$

$-\int_{\Omega}|\nabla u^{i}|^{p-2}\nabla
u^{i}\cdot\nabla\zeta\,dx=\int_{\Omega\cap\partial\\{|\mathbf{u}|>0\\}}\zeta
q^{i}d{\mathcal{H}}^{n-1}.$

  3. (iii)

For any $K\Subset\Omega$ there exist constants $c,C>0$ such that

$c\leq\sum_{i=1}^{m}q^{i}\leq C,$

and for $B_{r}(x)\subset K$ with $x\in\partial\\{|\mathbf{u}|>0\\}$ we have

$cr^{n-1}\leq{\mathcal{H}}^{n-1}(B_{r}(x)\cap\partial\\{|\mathbf{u}|>0\\})\leq
Cr^{n-1}.$

######  Remark 5.3.

From $(i)$ in Theorem 5.2 it follows that, locally, the set
$A=\Omega\cap\\{|\mathbf{u}|>0\\}$ has finite perimeter in $\Omega$ in sense
of that $\mu_{\mathbf{u}}=-\nabla\chi_{A}$ is a Borel measure and the total
variation $|\mu_{\mathbf{u}}|$ is a Radon measure. We define the reduced
boundary of $A$ by

$\partial_{\rm{red}}A=\\{x\in\Omega:|\nu_{\mathbf{u}}|=1\\},$

where $\nu_{\mathbf{u}}(x)$ is the unique unit vector with

$\int_{B_{r}(x)}\left|\chi_{A}-\chi_{\\{y:(y-x)\cdot\nu_{\mathbf{u}}(x)<0\\}}\right|=o(r^{n}),\quad\text{
as }r\rightarrow 0,$

if such a vector exists, and $\nu_{\mathbf{u}}(x)=0$ otherwise. See [14],
Chapter 4, for more details.

##  6\. Local Analysis

To proceed, we will need some properties of the so-called blow-up limits.

######  Lemma 6.1.

Let $\mathbf{u}$ be a (local) minimizer in $\Omega$, $K\Subset\Omega$ and
$B_{r_{k}}(x_{k})\subset K$ be a sequence of balls with $r_{k}\rightarrow 0$,
$x_{k}\rightarrow x_{0}\in\Omega$, and $\mathbf{u}(x_{k})=0$. Consider the
blow-up sequence

(10) $\mathbf{u}_{k}(x)=\frac{1}{r_{k}}\mathbf{u}(x_{k}+r_{k}x).$

For a subsequence, there is a limit $\mathbf{u}_{0}$ such that

$\displaystyle\mathbf{u}_{k}\rightarrow\mathbf{u}_{0}$ $\displaystyle\ \text{
in }C^{0,\alpha}_{\rm{loc}}({\mathbb{R}}^{n};{\mathbb{R}}^{m})\text{ for every
}0<\alpha<1,$ (11)
$\displaystyle\nabla\mathbf{u}_{k}\rightarrow\nabla\mathbf{u}_{0}$
$\displaystyle\ \text{ a.e. in }{\mathbb{R}}^{n},$ (12)
$\displaystyle\partial\\{|\mathbf{u}_{k}|>0\\}\rightarrow\partial\\{|\mathbf{u}_{0}|>0\\}$
locally in the Hausdorff distance, (13)
$\displaystyle\chi_{\\{|\mathbf{u}_{k}|>0\\}}\rightarrow\chi_{\\{|\mathbf{u}_{0}|>0\\}}$
$\displaystyle\ \text{ in
}L^{1}_{\rm{loc}}({\mathbb{R}}^{n};{\mathbb{R}}^{m}),$ $\displaystyle\text{ if
}x_{k}\in\partial\\{|\mathbf{u}_{k}|>0\\}$ $\displaystyle\ \text{ then
}0\in\partial\\{|\mathbf{u}_{0}|>0\\}.$

###### Proof.

For the proof we refer to [2] and [3]. ∎

The following lemma shows that the blow-up limit is a minimizer in any ball.

######  Lemma 6.2.

If $\mathbf{u}(x_{k})=0$ and $x_{k}\rightarrow x_{0}$, then any blow-up limit
$\mathbf{u}_{0}=\lim_{k}\mathbf{u}_{k}$ (see (10)) with respect to
$B_{r_{k}}(x_{k})$ is an absolute minimizer of $J_{0}$ in any ball
$B_{R}=B_{R}(0)$, where

$J_{0}(\mathbf{v}):=\int_{B_{R}}\sum_{i=1}^{m}|\nabla
v^{i}|^{p}+Q(x_{0})^{p}\chi_{\\{|\mathbf{v}|>0\\}}\,dx.$

###### Proof.

Let $\mathbf{w}\in W^{1,p}(B_{R};{\mathbb{R}}^{m})$ be such that $w^{i}\geq 0$
for $i=1,\dots,m$ and $\mathbf{w}=\mathbf{u}_{0}$ on $\partial B_{R}$. To show
$J_{0}(\mathbf{u}_{0})\leq J_{0}(\mathbf{w})$, we choose a cut of function
$\eta\in C_{0}^{\infty}(B_{R})$ with $0\leq\eta\leq 1$ and $\eta=1$ in $B_{r}$
for some $0<r<R$, and define

$\mathbf{w}_{k}=\left(\mathbf{w}+(1-\eta)(\mathbf{u}_{k}-\mathbf{u}_{0})\right)_{+},$

where the positive part is taken separately for each component. We also have
$\mathbf{w}_{k}=\mathbf{u}_{k}$ on $\partial B_{R}$. Since $\mathbf{u}$ is
(local) minimizer, for sufficiently large $k$ such that
$B_{Rr_{k}}(x_{k})\Subset\Omega$, we have

$\int_{B_{R}}\sum_{i=1}^{m}|\nabla
u^{i}_{k}|^{p}+Q_{k}^{p}\chi_{\\{|\mathbf{u}_{k}|>0\\}}\,dx\leq\int_{B_{R}}\sum_{i=1}^{m}|\nabla
w^{i}_{k}|^{p}+Q_{k}^{p}\chi_{\\{|\mathbf{w}_{k}|>0\\}}\,dx,$

where $Q_{k}(x):=Q(x_{k}+r_{k}x)$. Since $|\nabla\mathbf{u}_{k}|\leq C$ (due
to Lipschitz continuity of $\mathbf{u}$) and convergences (11) and (13), the
limit of the left hand side will be $J_{0}(\mathbf{u}_{0})$. Hence

$\displaystyle J_{0}(\mathbf{u}_{0})\leq$
$\displaystyle\liminf_{k\rightarrow\infty}\int_{B_{R}}\sum_{i=1}^{m}|\nabla
w_{k}^{i}|^{p}+Q_{k}^{p}\chi_{\\{|\mathbf{w}_{k}|>0\\}}\,dx$
$\displaystyle\leq$ $\displaystyle\int_{B_{R}}\sum_{i=1}^{m}|\nabla
w^{i}|^{p}\,dx+\int_{B_{r}}Q(x_{0})^{p}\chi_{\\{|\mathbf{w}|>0\\}}\,dx+\liminf_{k\rightarrow\infty}\int_{B_{R}\setminus
B_{r}}Q_{k}^{p}\chi_{\\{|\mathbf{w}_{k}|>0\\}}\,dx$ $\displaystyle\leq$
$\displaystyle\int_{B_{R}}\sum_{i=1}^{m}|\nabla
w^{i}|^{p}\,dx+\int_{B_{r}}Q(x_{0})^{p}\chi_{\\{|\mathbf{w}|>0\\}}\,dx+Q(x_{0})^{p}|B_{R}\setminus
B_{r}|.$

Now let $r\rightarrow R$, we get $J_{0}(\mathbf{u}_{0})\leq
J_{0}(\mathbf{w})$. ∎

######  Lemma 6.3.

Suppose $\nabla(Q^{p})\in L^{1}(\Omega)$ and $\mathbf{u}$ is an absolute
minimizer. Then

$\int_{|\mathbf{u}|>0}{\rm div\,}\left[\sum_{i=1}^{m}\left(|\nabla
u^{i}|^{p}\Psi-p|\nabla u^{i}|^{p-2}(\nabla u^{i}\cdot\Psi)\nabla
u^{i}\right)+Q^{p}\Psi\right]dx=0,$

for every $\Psi\in C_{c}^{\infty}(\Omega;{\mathbb{R}}^{n})$.

###### Proof.

Let us define

$\Phi_{t}(x)=x+t\Psi(x)\qquad\text{ and
}\qquad\mathbf{u}_{t}(x)=\mathbf{u}(\Phi_{t}(x)).$

One can show that for sufficiently small $|t|$,
$\Phi_{t}:\Omega\rightarrow\Omega$ is a diffeomorphism. We have
$D\Phi_{t}=I+tD\Psi$ and for $i=1,\dots,m$

$\nabla u^{i}_{t}=D\Phi_{t}(x)\nabla u^{i}(\Phi_{t}(x)).$

It follows that

$|\nabla u_{t}^{i}(x)|^{2}=(\nabla u^{i}(\Phi_{t}(x)))^{T}A_{t}(x)\nabla
u^{i}(\Phi_{t}(x)),$

where

$A_{t}=(D\Phi_{t})^{T}D\Phi_{t}=I+t((D\Psi)^{T}+D\Psi)+t^{2}(D\Psi)^{T}D\Psi.$

By a change of variables, we have

$\begin{split}J(\mathbf{u}_{t})=&\int_{\Omega}\sum_{i=1}^{m}|\nabla
u^{i}_{t}(x)|^{p}+Q^{p}\chi_{\\{|\mathbf{u}_{t}|>0\\}}\,dx\\\
=&\int_{\Omega}\sum_{i=1}^{m}\left((\nabla
u^{i}(\Phi_{t}(x)))^{T}A_{t}(x)\nabla
u^{i}(\Phi_{t}(x))\right)^{p/2}+Q^{p}(x)\chi_{\\{|\mathbf{u}|>0\\}}(\Phi_{t}(x))\,dx\\\
=&\int_{\Omega}\sum_{i=1}^{m}\left[\left((\nabla
u^{i}(y))^{T}A_{t}(\Phi_{t}^{-1}(y))\nabla
u^{i}(y)\right)^{p/2}+Q^{p}(\Phi_{t}^{-1}(y))\chi_{\\{|\mathbf{u}|>0\\}}(y)\right]\left|\det
D_{y}\Phi_{t}^{-1}(y)\right|\,dy\\\
=&\int_{\\{|\mathbf{u}|>0\\}}\sum_{i=1}^{m}\left[\left((\nabla
u^{i}(y))^{T}A_{t}(\Phi_{t}^{-1}(y))\nabla
u^{i}(y)\right)^{p/2}+Q^{p}(\Phi_{t}^{-1}(y))\right]\left|\det
D_{y}\Phi_{t}^{-1}(y)\right|\,dy.\end{split}$

We also have

$\frac{d}{dt}A_{t}(\Phi_{t}^{-1}(y))\Big{|}_{t=0}=D\Psi(y)+D\Psi(y)^{T}$

and

$\frac{d}{dt}\left|\det D_{y}\Phi_{t}^{-1}(y)\right|\Big{|}_{t=0}=-{\rm
div\,}\Psi(y).$

Now differentiate $J(\mathbf{u}_{t})$ with respect to $t$ and note that its
minimum is attained at $t=0$, then

$\begin{split}0=\frac{d}{dt}J(\mathbf{u}_{t})\Big{|}_{t=0}=&\int_{\\{|\mathbf{u}|>0\\}}\sum_{i=1}^{m}p|\nabla
u^{i}|^{p-2}(\nabla u^{i})^{T}D\Psi\nabla u^{i}-\Psi\cdot\nabla(Q^{p})dx\\\
&-\int_{\\{|\mathbf{u}|>0\\}}\left[\sum_{i=1}^{m}|\nabla
u^{i}|^{p}+Q^{p}\right]{\rm div\,}\Psi\,dx\\\
=&-\int_{\\{|\mathbf{u}|>0\\}}{\rm div\,}\left[\sum_{i=1}^{m}\left(|\nabla
u^{i}|^{p}\Psi-p|\nabla u^{i}|^{p-2}(\nabla u^{i}\cdot\Psi)\nabla
u^{i}\right)+Q^{p}\Psi\right]\\\ &\qquad\qquad+\sum_{i=1}^{m}p(\nabla
u^{i}\cdot\Psi)\Delta_{p}u^{i}\,dx.\end{split}$

Since each $u^{i}$ is $p$-harmonic in $\\{u^{i}>0\\}$, see (2), we arrive at
the desired claim, in the lemma. ∎

######  Definition 6.4.

The upper ${\mathcal{H}}^{n-1}$-density at any point
$x_{0}\in\partial\\{|\mathbf{u}|>0\\}$ is defined as

$\Theta^{*n-1}\left({\mathcal{H}}^{n-1}\big{\lfloor}\partial\\{|\mathbf{u}|>0\\},x_{0}\right):=\limsup_{r\rightarrow
0}\frac{{\mathcal{H}}^{n-1}(B_{r}(x_{0})\cap\partial\\{|\mathbf{u}|>0\\})}{\omega_{n-1}r^{n-1}},$

where $\omega_{n-1}$ denotes the volume of the unit sphere in
${\mathbb{R}}^{n-1}$. We already know (see for example Theorem 2.7 in [13])
that for ${\mathcal{H}}^{n-1}$-a.e. point
$x_{0}\in\partial\\{|\mathbf{u}|>0\\}$, their upper
${\mathcal{H}}^{n-1}$-density satisfy

$\Theta^{*n-1}\left({\mathcal{H}}^{n-1}\big{\lfloor}\partial\\{|\mathbf{u}|>0\\},x_{0}\right)\leq
1.$

######  Theorem 6.5.

Let $x_{0}\in\partial_{\rm red}\\{|\mathbf{u}|>0\\}$ and suppose that

$\Theta^{*n-1}\left({\mathcal{H}}^{n-1}\big{\lfloor}\partial\\{|\mathbf{u}|>0\\},x_{0}\right)\leq
1.$

Then ${\rm
Tan}(\partial\\{|\mathbf{u}|>0\\},x_{0})=\\{x:x\cdot\nu(x_{0})=0\\}$. If, in
addition, $x_{0}$ is a Lebesgue point for Radon measure
$q^{i}{\mathcal{H}}^{n-1}\big{\lfloor}\partial\\{|\mathbf{u}|>0\\}$, that is

(14)
$\int_{B_{r}(x_{0})\cap\partial\\{|\mathbf{u}|>0\\}}|q^{i}-q^{i}(x_{0})|\,d{\mathcal{H}}^{n-1}=o(r^{n-1}),\quad\text{
as }r\rightarrow 0,$

then $q^{i}(x_{0})=Q(x_{0})$ and

$\mathbf{u}(x_{0}+x)=\left(-x\cdot\nu(x_{0})\right)_{+}\mathbf{a}_{x_{0}}+o(|x|),\quad\text{
as }x\rightarrow 0,$

for some vector $\mathbf{a}_{x_{0}}=(\alpha^{1},\dots,\alpha^{m})$ that

(15)
$|\mathbf{a}_{x_{0}}|_{p}^{p}=(\alpha^{1})^{p}+\dots+(\alpha^{m})^{p}=\frac{1}{p-1}Q(x_{0})^{p}.$

###### Proof.

Without loss of generality assume that $\nu(x_{0})=\mathbf{e}^{n}$. Let
$\mathbf{u}_{k}$ be a blow-up sequence with respect to balls
$B_{r_{k}}(x_{0})$, with blow-up limit $\mathbf{u}_{0}$. Since $\nu(x_{0})$ is
the normal vector to $\partial\\{|\mathbf{u}|>0\\}$ at $x_{0}$,

$\int_{B_{r}(x_{0})}\left|\chi_{\\{|\mathbf{u}|>0|\\}}-\chi_{\\{x:(x-x_{0})\cdot\nu(x_{0})<0\\}}\right|\,dx=o(r^{n}),\quad\text{
as }r\rightarrow 0.$

This along with (13) implies
$\chi_{\\{|\mathbf{u}_{0}|>0\\}}=\chi_{\\{x_{n}<0\\}}$ almost every where in
${\mathbb{R}}^{n}$. By Lemma 6.2 we know that $\mathbf{u}_{0}$ is an absolute
minimizer of $J_{0}$ and so continuous. Then
$\\{|\mathbf{u}_{0}|>0\\}=\\{x_{n}<0\\}$. This proves that $\\{x_{n}=0\\}$ is
the topological tangent plane to $\partial\\{|\mathbf{u}|>0\\}$ at $x_{0}$.
Now let

$\phi(x)=\min\left(1,\max(0,2-|x_{n}|)\right)\eta(x^{\prime}),$

where $0\leq\eta\in C_{0}^{\infty}(B^{\prime}_{R})$ and $B^{\prime}_{R}$ is
$(n-1)$-dimensional ball with radius $R$ ($R$ is arbitrary and fixed). Denote
$\phi_{k}(x):=r_{k}\phi(\frac{x-x_{0}}{r_{k}})$ and write

$\displaystyle-\int_{{\mathbb{R}}^{n}}|\nabla u_{0}^{i}|^{p-2}\nabla
u_{0}^{i}\cdot\nabla\phi\,dx=-\lim_{k\rightarrow\infty}\int_{{\mathbb{R}}^{n}}|\nabla
u_{k}^{i}|^{p-2}\nabla u_{k}^{i}\cdot\nabla\phi\,dx$
$\displaystyle=-\lim_{k\rightarrow\infty}r_{k}^{-n}\int_{{\mathbb{R}}^{n}}|\nabla
u^{i}|^{p-2}\nabla
u^{i}\cdot\nabla\phi_{k}\,dx=\lim_{k\rightarrow\infty}r_{k}^{-n}\int_{{\mathbb{R}}^{n}}\Delta_{p}u^{i}\phi_{k}\,dx$
$\displaystyle=\lim_{k\rightarrow\infty}r_{k}^{-n}\int_{{\mathbb{R}}^{n}}q^{i}\phi_{k}\,d{\mathcal{H}}^{n-1}\lfloor\,\partial\\{|\mathbf{u}|>0\\}=\lim_{k\rightarrow\infty}\int_{{\mathbb{R}}^{n}}q^{i}(x_{0}+r_{k}x)\phi(x)\chi_{\partial\\{|\mathbf{u}_{k}|>0\\}}\,d{\mathcal{H}}^{n-1}$
$\displaystyle=\lim_{k\rightarrow\infty}\int_{{\mathbb{R}}^{n}}q^{i}(x_{0})\phi(x)\chi_{\partial\\{|\mathbf{u}_{k}|>0\\}}\,d{\mathcal{H}}^{n-1}=q^{i}(x_{0})\int_{\\{x_{n}=0\\}}\eta(x^{\prime})\,dx^{\prime},$

where we have used assumption (14) and property (12). Therefore, for any test
function $\zeta\in C_{0}^{\infty}(B_{R})$ we have

$-\int_{B_{R}\cap\\{x_{n}<0\\}}|\nabla u_{0}^{i}|^{p-2}\nabla
u_{0}^{i}\cdot\nabla\zeta\,dx=q^{i}(x_{0})\int_{B_{R}^{\prime}}\zeta(x^{\prime},0)\,dx^{\prime}.$

Since $\Delta_{p}u^{i}_{0}=0$ in $\\{x_{n}<0\\}$, from boundary regularity it
follows that

$|\nabla u_{0}^{i}|^{p-2}\partial_{n}u_{0}^{i}=-q^{i}(x_{0})\quad\text{ on
}\\{x_{n}=0\\},$

in the classical sense. We need to show that

(16) $u_{0}^{i}(x)=\alpha^{i}(-x_{n})_{+},\quad\text{ where
}\alpha^{i}:=\left(q^{i}(x_{0})\right)^{1/(p-1)}.$

To see this, define $w_{0}$ by

$w_{0}(x):=\left\\{\begin{array}[]{ll}u_{0}^{i}(x),&\,\text{ in }x_{n}\leq
0,\\\\[8.0pt] -u_{0}^{i}(x^{\prime},-x_{n}),&\,\text{ in
}x_{n}>0.\end{array}\right.$

It is obvious that $w_{0}$ is $p$-harmonic in whole ${\mathbb{R}}^{n}$ as well
as

$\Arrowvert{\nabla
w_{0}}\Arrowvert_{L^{\infty}({\mathbb{R}}^{n})}=\Arrowvert{\nabla
u_{0}^{i}}\Arrowvert_{L^{\infty}({\mathbb{R}}^{n})}\leq\Arrowvert{\nabla
u^{i}}\Arrowvert_{L^{\infty}(B_{r}(x_{0}))},\quad\text{ for any }r>0.$

By Liouville’s theorem we conclude that $w_{0}$ is a linear function. The
boundary value on $x_{n}=0$, ($u_{0}^{i}=0$ and
$\partial_{n}u_{0}^{i}=-\alpha^{i}$) shows that $w_{0}(x)=-\alpha^{i}x_{n}$.
This proves (16) and shows that

$u^{i}(x_{0}+x)=\alpha^{i}(-x_{n})_{+}+o(|x|),\quad\text{ as }x\rightarrow 0.$

We just have to show (15). To do this, note that $\mathbf{u}_{0}$ is an
absolute minimizer of $J_{0}$. Apply Lemma 6.3 for $\mathbf{u}_{0}$ and some
$\Psi\in C_{c}^{\infty}({\mathbb{R}}^{n})$

$\begin{split}0=&\int_{\\{x_{n}<0\\}}{\rm
div\,}\left[\sum_{i=1}^{m}\left(|\nabla u_{0}^{i}|^{p}\Psi-p|\nabla
u_{0}^{i}|^{p-2}(\nabla u_{0}^{i}\cdot\Psi)\nabla
u_{0}^{i}\right)+Q(x_{0})^{p}\Psi\right]dx\\\
=&\int_{\\{x_{n}=0\\}}\sum_{i=1}^{m}\left(|\nabla
u_{0}^{i}|^{p}(\Psi\cdot\mathbf{e}^{n})-p|\nabla u_{0}^{i}|^{p-2}(\nabla
u_{0}^{i}\cdot\Psi)\partial_{n}u_{0}^{i}\right)+Q(x_{0})^{p}(\Psi\cdot\mathbf{e}^{n})\,d{\mathcal{H}}^{n-1}\\\
=&\int_{\\{x_{n}=0\\}}\left(\sum_{i=1}^{m}(1-p)(\alpha^{i})^{p}+Q(x_{0})^{p}\right)(\Psi\cdot\mathbf{e}^{n})\,d{\mathcal{H}}^{n-1}.\end{split}$

Thus (15) will be obtained. ∎

##  7\. Regularity of free boundary

######  Definition 7.1.

Let $\mathbf{u}\in C(\Omega,{\mathbb{R}}^{m})$. We say that the boundary
condition $\nabla|\mathbf{u}|_{p}=g$ on $\partial\\{|\mathbf{u}|>0\\}$ holds
in viscosity sense, if

  1. $\circ$

For every differentiable function
$\phi:{\mathbb{R}}^{n}\rightarrow{\mathbb{R}}$ that touches $|\mathbf{u}|_{p}$
from below in some $x_{0}\in\partial\\{|\mathbf{u}|>0\\}$, that is

$|\mathbf{u}(x_{0})|_{p}=\phi(x_{0}),\quad\text{ and
}\quad|\mathbf{u}|_{p}\geq\phi\ \text{ in }\\{|\mathbf{u}|>0\\}\cap
B_{r}(x_{0})$

for some $r>0$, we have $|\nabla\phi(x_{0})|\leq g(x_{0})$.

  2. $\circ$

For every differentiable function
$\phi:{\mathbb{R}}^{n}\rightarrow{\mathbb{R}}$ that touches $|\mathbf{u}|_{p}$
from above in some $x_{0}\in\partial\\{|\mathbf{u}|>0\\}$, that is

$|\mathbf{u}(x_{0})|_{p}=\phi(x_{0}),\quad\text{ and
}\quad|\mathbf{u}|_{p}\leq\phi\ \text{ in }\\{|\mathbf{u}|>0\\}\cap
B_{r}(x_{0})$

for some $r>0$, we have $|\nabla\phi(x_{0})|\geq g(x_{0})$.

######  Lemma 7.2.

Let $\mathbf{u}$ be a (local) minimizer, then the boundary condition

$\nabla|\mathbf{u}|_{p}=\frac{1}{(p-1)^{1/p}}Q,\quad\text{ on
}\partial\\{|\mathbf{u}|>0\\}.$

holds in the viscosity sense.

###### Proof.

We show that the boundary condition holds on every point
$x_{0}\in\partial\\{|\mathbf{u}|>0\\}$. Suppose $\phi$ touches
$|\mathbf{u}|_{p}$ from below at $x_{0}$. Consider the blow-up sequences

$\mathbf{u}_{k}(x)=\frac{\mathbf{u}(x_{0}+r_{k}x)}{r_{k}}\quad\text{ and
}\quad\phi_{k}(x)=\frac{\phi(x_{0}+r_{k}x)}{r_{k}},$

where $r_{k}\searrow 0$. Observe that

$\phi_{k}(0)=|\mathbf{u}_{k}(0)|_{p}=0,\quad\phi_{k}(x)\leq|\mathbf{u}_{k}(x)|_{p},$

and up to a subsequence we have

(17) $\nabla\phi(x_{0})\cdot
x=\lim_{k\rightarrow\infty}\phi_{k}(x)\leq\lim_{k\rightarrow\infty}|\mathbf{u}_{k}(x)|_{p}=|\mathbf{u}_{0}(x)|_{p}\qquad\text{
in }{\mathbb{R}}^{n}.$

If $\nabla\phi(x_{0})=0$, the viscosity condition holds trivially. Otherwise,
the non-coincidence set $\\{|\mathbf{u}_{0}|>0\\}$ contains half-space
$\\{x:\nabla\phi(x_{0})\cdot x>0\\}$. On the other hand, $\mathbf{u}_{0}$ is
minimizer of $J_{0}$ (Lemma 6.2) and by Lemma 2.3 every nontrivial component
of $\mathbf{u}_{0}$, say $u^{i}_{0}$, is positive in
$\\{x:\nabla\phi(x_{0})\cdot x>0\\}$. According to Lemma B.1,
$u^{i}_{0}(x)=\alpha^{i}(\nabla\phi(x_{0})\cdot x)+o(|x|)$ for some
$\alpha^{i}$. Thus any blowup of $\mathbf{u}_{0}$ at $x=0$ must be of the form
$\mathbf{u}_{00}(x)=\mathbf{a}_{0}(\nabla\phi(x_{0})\cdot x)$ where
$\mathbf{a}_{0}=(\alpha^{1},\cdots,\alpha^{m})$. Again apply Lemma 6.2 along
with Lemma 6.3, we get

$|\mathbf{a}_{0}|_{p}|\nabla\phi(x_{0})|=\frac{1}{(p-1)^{1/p}}Q(x_{0}).$

Thus (17) yields that

$\displaystyle\nabla\phi(x_{0})\cdot
x\leq|\mathbf{u}_{0}(x)|_{p}=|\mathbf{a}_{0}|_{p}|\nabla\phi(x_{0})\cdot
x|+o(|x|),$

and so

$|\nabla\phi(x_{0})|\leq\frac{1}{(p-1)^{1/p}}Q(x_{0}).$

The same argument holds when $\phi$ touches $|\mathbf{u}_{0}|_{p}$ from above.
∎

######  Definition 7.3.

A domain $\Omega\subset{\mathbb{R}}^{n}$ is called non-tangentially accessible
domain (NTA) with parameters $M\geq 1$ and $R_{0}>0$ if

  1. (i)

$\Omega$ satisfies the corkscrew condition, that is, for any
$x\in\partial\Omega$ and $r\in(0,R_{0})$ there exists $a_{r}(x)\in\Omega\cap
B_{r}(x)$ such that $M^{-1}r<{\rm dist}(a_{r}(x),\partial\Omega)$.

  2. (ii)

${\mathbb{R}}^{n}\setminus\Omega$ satisfies the corkscrew condition.

  3. (iii)

If $x\in\partial\Omega$ and $x_{1},x_{2}\in B_{r}(x)\cap\Omega$ for
$0<r<R_{0}$, then there exists a rectifiable curve
$\gamma:[0,1]\rightarrow\Omega$ with $\gamma(0)=x_{1}$ and $\gamma(1)=x_{2}$
such that ${\mathcal{H}}^{1}(\gamma)\leq M|x_{1}-x_{2}|$ and

$\min\left\\{{\mathcal{H}}^{1}(\gamma([0,t])),{\mathcal{H}}^{1}(\gamma([t,1]))\right\\}\leq
M{\rm dist}(\gamma(t),\partial\Omega),\quad\text{ for every }t\in[0,1],$

where ${\mathcal{H}}^{1}$ denotes length or the one-dimensional Hausdorff
measure.

######  Definition 7.4.

We say that $x_{0}\in\partial\\{|\mathbf{u}|>0\\}$ is a regular point of the
free boundary if for every $\epsilon>0$ there are $r<1$ and a vector
$\mathbf{a}\in{\mathbb{R}}^{m}$ and unit vector $\nu\in{\mathbb{R}}^{n}$ such
that

$\Arrowvert{\mathbf{u}_{r,x_{0}}-(x\cdot\nu)_{+}\mathbf{a}}\Arrowvert_{L^{\infty}(B_{1})}\leq\epsilon.$

We denote the set of all regular points by ${\mathcal{R}}_{\mathbf{u}}$.
Theorem 6.5 proves that
${\mathcal{H}}^{n-1}(\partial\\{|\mathbf{u}|>0\\}\setminus{\mathcal{R}}_{\mathbf{u}})=0$.

######  Theorem 7.5.

Let $\mathbf{u}=(u^{1},\dots,u^{m})$ be a (local) minimizer and
$x_{0}\in{\mathcal{R}}_{\mathbf{u}}$. Furthermore, assume that
$B_{r_{0}}(x_{0})\cap\\{|\mathbf{u}|>0\\}$ is NTA domain. Then
${\mathcal{R}}_{\mathbf{u}}\cap B_{r}(x_{0})$, for some $0<r\leq r_{0}$, is
$C^{1,\alpha}$ for a universal exponent $0<\alpha<1$.

###### Proof.

We may assume that $u^{1}>0$ in $B_{r_{0}}(x_{0})\cap\\{|\mathbf{u}|>0\\}$.
First we show that there is a Hölder function
$g:B_{r}(x_{0})\cap\partial\\{|\mathbf{u}|>0\\}\rightarrow[c,1]$ for some
$0<r\leq r_{0}$ and $0<c\leq 1$ such that $u^{1}$ is a viscosity solution to
the problem

(18) $\begin{split}\Delta_{p}u^{1}=0,&\quad\text{ in }\\{u^{1}>0\\}\cap
B_{r},\\\ |\nabla u^{1}|=\frac{gQ}{(p-1)^{1/p}},&\quad\text{ on
}\partial\\{u^{1}>0\\}\cap B_{r}.\end{split}$

Since $B_{r_{0}}(x_{0})\cap\\{|\mathbf{u}|>0\\}$ is NTA domain, the boundary
Harnack inequality (see [18]) implies that $g_{i}:=u^{i}/u^{1}$ is Hölder
continuous in $\overline{\\{|\mathbf{u}|>0\\}}\cap B_{r}$ for some $r\leq
r_{0}$. Define

$g:=(1+g_{2}^{p}+\dots+g_{m}^{p})^{-1/p},$

and observe that $u^{1}=g|\mathbf{u}|_{p}$. Suppose now the test function
$\phi$ is touching $u^{1}$ from below in a point
$y\in\partial\\{|\mathbf{u}|>0\\}$. For $\rho$ small enough, choose a constant
$C>0$ such that

$\frac{1}{g(x)}\geq\frac{1}{g(y)}-C|x-y|^{\mu}\geq 0,\ \ \text{ for every
}x\in\overline{\\{|\mathbf{u}|>0\\}}\cap B_{\rho},$

where $\mu$ is Hölder exponent of $g$. Set
$\psi(x)=\phi(x)(1/g(y)-C|x-y|^{\mu})$, we get $\psi(y)=|\mathbf{u}(y)|_{p}=0$
(note that since $\phi(y)=0$, so $\psi$ is differentiable at $y$) and

$\psi(x)\leq
u^{1}(x)(\frac{1}{g(y)}-C|x-y|^{\mu})=g(x)(\frac{1}{g(y)}-C|x-y|^{\mu})|\mathbf{u}(x)|_{p}\leq|\mathbf{u}(x)|_{p}.$

Therefore, $\psi$ touches $|\mathbf{u}|_{p}$ at $y$ from below. By Lemma 7.2

$|\nabla\psi(y)|\leq\frac{Q(y)}{(p-1)^{1/p}}.$

Note that $\nabla\psi(y)=\frac{1}{g(y)}\nabla\phi(y)$ to see the boundary
condition (18) in viscosity sense.

The regularity of free boundary follows by the known results on the regularity
of the one-phase scalar problem (18); see [15]. ∎

Our next result improves the theorem above, in the sense that we can remove
the NTA conditions, for $p$ close to 2. The main reason for not being able to
handle the NTA, is the lack of ACF-monotonicity formula, use in classical
paper; see proof of Proposition A.3, to prove the connectivity argument for
the NTA.

######  Theorem 7.6.

Let $\mathbf{u}=(u^{1},\dots,u^{m})$ be a (local) minimizer. Then there is
$\epsilon_{0}>0$, such that for any $p\in(2-\epsilon_{0},2+\epsilon_{0})$ we
have

  * i)

The regular set ${\mathcal{R}}_{\mathbf{u}}$, is locally $C^{1,\alpha}$.

  * ii)

In dimensions $2,3,4$ the free boundary is $C^{1,\alpha}$.

Here $0<\alpha<1$ is is a universal exponent.

###### Proof.

To prove (i) it suffices, in virtue of Theorem 7.5, to show that the free
boundary is NTA, when $\epsilon_{0}$ is small enough. This, however, is a
consequence of Proposition A.5 in Appendix A, by choosing $\epsilon_{0}$
accordingly.

Turning to sstatement (ii) we recall that in these dimensions, and for $p=2$,
free boundaries are locally $C^{1,\alpha}$; for $n=2$ this was shown in [2],
for $n=3$ see [6], for $n=4$ see [12]. Now for $p\approx 2$, the free boundary
has to be close to that of the case $p=2$, and hence flat. More specifically,
given a free boundary point $x_{0}$ for the $p$-problem, and $p$ close enough
to 2, we have that in $B_{r}(x_{0})$ the free boundaries of both $p$ and 2,
have Huasdorff distance $\delta\ll r$, and in particular the $p$-free boundary
is $(\delta/2)$-flat. ∎

##  Appendix A Non-tangentially accessible domain (NTA)

We note that in Definition 7.3, the condition $(iii)$ can be replaced by
Harnack chain condition (see [4]), i.e.,

  1. (iii)’

Given $\varepsilon>0$, $x_{1},x_{2}\in\Omega$ such that ${\rm
dist}(x_{i},\partial\Omega)\geq\varepsilon$, $i=1,2$ and
$|x_{1}-x_{2}|\geq\tilde{C}\varepsilon$, we can find points
$x_{1}=y_{1},y_{2},\cdots,y_{\ell}=x_{2}$ for which

    1. (a)

$B_{\varepsilon}(y_{i})\subset\Omega$ for $i=1,\cdots,\ell$.

    2. (b)

$B_{\varepsilon}(y_{i})\cap B_{\varepsilon}(y_{i+1})\neq\emptyset$ for
$i=1,\cdots,\ell-1$.

    3. (c)

The length of chain, $\ell$, depends on $\tilde{C}$ but not on $\varepsilon$.

We need an analogue of Theorem 4.1 in [1] to show that the non-coincidence set
is NTA.

######  Lemma A.1.

Let $\mathbf{u}$ be a minimizer and suppose $0<|\mathbf{u}(x_{0})|$. Define
$\delta={\rm dist}(x_{0},\Gamma)$, $\delta_{1}={\rm
dist}(x_{0},\\{|\mathbf{u}|\leq\frac{1}{2}|\mathbf{u}(x_{0})|_{\infty}\\})$,
and suppose $B(x_{0},\delta)\subset\Omega$. Then, there exist universal
constants $\lambda>1>\sigma$ such that

  1. (i)

$\sigma\delta\leq\delta_{1}\leq\delta$.

  2. (ii)

For some $y\in\partial B_{\delta_{1}}(x_{0})$,
$|\mathbf{u}(y)|_{\infty}\geq\lambda|\mathbf{u}(x_{0})|_{\infty}$.

###### Proof.

According to relation (6),

$c_{0}\delta\leq|\mathbf{u}(x_{0})|\leq C_{0}\delta,$

and if $u^{i}(x_{0})=\max_{1\leq j\leq
m}u^{j}(x_{0})=|\mathbf{u}(x_{0})|_{\infty}$,

$\frac{c_{0}\delta}{\sqrt{m}}\leq u^{i}(x_{0})\leq C_{0}\delta.$

If $z\in\partial\\{|\mathbf{u}|\leq\frac{1}{2}|\mathbf{u}(x_{0})|_{\infty}\\}$
such that $|z-x_{0}|=\delta_{1}$, then

$\frac{1}{2}u^{i}(x_{0})=\frac{1}{2}|\mathbf{u}(x_{0})|_{\infty}\leq|\mathbf{u}(x_{0})|-|\mathbf{u}(z)|\leq|\mathbf{u}(x_{0})-\mathbf{u}(z)|\leq
C_{0}\delta_{1},$

where we have used
$|\mathbf{u}(z)|=\frac{1}{2}|\mathbf{u}(x_{0})|_{\infty}\leq\frac{1}{2}|\mathbf{u}(x_{0})|$.
Thus $(i)$ holds for $\sigma=\frac{c_{0}}{2C_{0}\sqrt{m}}$.

To see $(ii)$, note that $v(x):=u^{i}(x_{0}+\delta_{1}x)/u^{i}(x_{0})$ is a
$p$-harmonic function in $B_{\delta/\delta_{1}}$ such that $v(0)=1$,
$v(\hat{z})\leq\frac{1}{2}$ for some $\hat{z}\in\partial B_{1}$. In addition,
the Lipschitz constant of $v$ is bounded by $C_{0}\delta_{1}/u^{i}(x_{0})\leq
C_{0}\delta/u^{i}(x_{0})\leq C_{0}\sqrt{m}/c_{0}=1/2\sigma$. We claim that
there is a universal constant $\lambda>1$ such that $v(\hat{y})\geq\lambda$
for some $\hat{y}\in\partial B_{1}$. Otherwise, we find a sequence of
$p$-harmonic functions $v_{k}$ with

$\Arrowvert{v_{k}}\Arrowvert_{L^{\infty}(B_{1})}\leq
1+\frac{1}{k},\quad\Arrowvert{\nabla
v_{k}}\Arrowvert_{L^{\infty}(B_{1})}\leq\frac{1}{2\sigma},\quad
v_{k}(0)=1,\quad v_{k}(z_{k})\leq\frac{1}{2},$

for some $|z_{k}|=1$. Then there is a subsequence converging in
$C^{1,\alpha}(\overline{B}_{1})$ to a $p$-harmonic $v_{0}$ where $v_{0}(0)=1$,
$v_{0}(z_{0})\leq 1/2$, $\Arrowvert{v_{0}}\Arrowvert_{L^{\infty}(B_{1})}=1$.
This is a contradiction with the maximum principle. Therefore, we have found
$y\in\partial B_{\delta_{1}}(x_{0})$ such that

$|\mathbf{u}(y)|_{\infty}\geq u^{i}(y)\geq\lambda
u^{i}(x_{0})=\lambda|\mathbf{u}(x_{0})|_{\infty}.\qed$

######  Lemma A.2.

There exists a universal constant $\tilde{c}_{0}$ such that if
$x_{0}\in\partial\\{|\mathbf{u}|>0\\}$, $x_{1}\in B_{r/2}(x_{0})$ and $A_{r}$
is the connected component of
$\\{|\mathbf{u}|>\frac{1}{2}|\mathbf{u}(x_{1})|_{\infty}\\}\cap B_{r}(x_{0})$
containing $x_{1}$, then

$\Arrowvert{\mathbf{u}}\Arrowvert_{L^{\infty}(A_{r})}\geq\tilde{c}_{0}r.$

###### Proof.

We use Lemma A.1 to inductively define a sequence of points
$x_{1},x_{2},\dots,x_{k},x_{k+1}$ so that for $j=1,\dots,k$,

  1. $(i)$

$|x_{j+1}-x_{j}|=\delta_{j}={\rm
dist}(x_{j},\\{|\mathbf{u}|\leq\frac{1}{2}|\mathbf{u}(x_{j})|_{\infty}\\}),$

  2. $(ii)$

$|\mathbf{u}(x_{j+1})|_{\infty}\geq\lambda|\mathbf{u}(x_{j})|_{\infty},$

  3. $(iii)$

$B_{\delta_{j}}(x_{j})\subset\\{|\mathbf{u}|>\frac{1}{2}|\mathbf{u}(x_{1})|_{\infty}\\}.$

By $(ii)$, we know that this process cannot continue indefinitely without
stepping out of $B_{r}(x_{0})$. So, we stop at the first $k$ for which
$B_{\delta_{k+1}}(x_{k+1})\not\subset B_{r}(x_{0})$. Also, by Lemma A.1, we
know that

$\delta_{j}\leq{\rm dist}(x_{j},\Gamma)\leq\frac{\delta_{j}}{\sigma},$

and therefore by (6),

$c_{0}\delta_{j}\leq|\mathbf{u}(x_{j})|\leq\frac{C_{0}\delta_{j}}{\sigma}.$

Now applying $(ii)$, we obtain (recall that
$\sigma=\frac{c_{0}}{2C_{0}\sqrt{m}}$)

$\delta_{j}\leq\frac{\sqrt{m}}{c_{0}}|\mathbf{u}(x_{j})|_{\infty}\leq\frac{\sqrt{m}\lambda^{-\ell}}{c_{0}}|\mathbf{u}(x_{j+\ell})|_{\infty}\leq\frac{\lambda^{-\ell}}{2\sigma^{2}}\delta_{j+\ell}.$

Therefore,

$|x_{k}-x_{0}|\leq|x_{1}-x_{0}|+\sum_{j=1}^{k-1}|x_{j+1}-x_{j}|\leq\frac{r}{2}+\sum_{j=1}^{k-1}\delta_{j}\leq\frac{r}{2}+\frac{\delta_{k}}{2\sigma^{2}}\sum_{j=1}^{k-1}\lambda^{j-k}\leq\frac{r}{2}+\frac{\delta_{k}}{2\sigma^{2}(\lambda-1)}.$

On the other hand,

$\begin{split}\delta_{k+1}=&{\rm
dist}(x_{k+1},\\{|\mathbf{u}|\leq\frac{1}{2}|\mathbf{u}(x_{k+1})|_{\infty}\\})\\\
\leq&\delta_{k}+{\rm
dist}(x_{k},\\{|\mathbf{u}|\leq\frac{1}{2}|\mathbf{u}(x_{k+1})|_{\infty}\\})\\\
\leq&\delta_{k}+{\rm
dist}(x_{k},\\{|\mathbf{u}|\leq\frac{1}{2}|\mathbf{u}(x_{k})|_{\infty}\\})=2\delta_{k}.\end{split}$

Since $B_{\delta_{k+1}}(x_{k+1})\not\subset B_{r}(x_{0})$, we get

$r\leq|x_{k+1}-x_{0}|+\delta_{k+1}\leq|x_{k}-x_{0}|+3\delta_{k}\leq\frac{r}{2}+c\delta_{k}.$

Thus $\gamma r\leq\delta_{k}$ for the universal constant $\gamma$. It
necessitates that $B_{\gamma r}(x_{k})\subset A_{r}$ and also

$|\mathbf{u}(x_{k})|\geq c_{0}{\rm dist}(x_{k},\Gamma)\geq c_{0}\gamma r.$

In particular,

$\max_{A_{r}}|\mathbf{u}|\geq c_{0}\gamma r.\qed$

######  Proposition A.3.

There is $\varepsilon_{0}>0$ such that for any
$p\in(2-\varepsilon_{0},2+\varepsilon_{0})$ there is no global minimizer of
$J_{0}$ (Lemma 6.2) that $\mathbf{u}(0)=0$ and $\\{|\mathbf{u}|>0\\}\cap
B_{R}$ is disconnected for every $R>0$.

###### Proof.

First, for $p=2$. We apply the monotonicity formula and Lemma 4.4 in [1]. Let
$A_{1}$ and $A_{2}$ two different connected components of
$\\{|\mathbf{u}_{\infty}|>0\\}$. According to Lemma 2.3, we may choose a
positive component of vector $\mathbf{u}_{\infty}$ for each set $A_{i}$,
$i=1,2$. Thus we find a function $v$ which is harmonic in $A_{1}\cup A_{2}$
and vanishes in $\\{|\mathbf{u}_{\infty}|=0\\}$. Also, since
$\mathbf{u}_{\infty}$ is a minimizer, we know that (Theorem 4.2)

$|B_{r}\setminus(A_{1}\cup A_{2})|\geq c|B_{r}|.$

From here we infer that $\Phi(r)/r^{\beta}$ is a non-decreasing function of
$r$ for some positive constant $\beta>0$ [1, Lemma 4.4], where

$\Phi(r)=\frac{1}{r^{4}}\left(\int_{B_{r}\cap A_{1}}|\nabla
v|^{2}|x|^{2-n}\,dx\right)\left(\int_{B_{r}\cap A_{2}}|\nabla
v|^{2}|x|^{2-n}\,dx\right).$

Since $v$ is Lipschitz, say with constant $C_{0}$, we have the bound

$\Phi(r)\leq C_{0}^{4},$

and thus

$\Phi(1)\leq r^{-\beta}\Phi(r)\leq C_{0}^{4}r^{-\beta},$

which can not be valid for sufficiently large value of $r$. The contradiction
proves the proposition when $p=2$.

To prove the proposition for $p$ close to 2 we argue by contradiction. Assume
that there is a sequence of global minimizers $\mathbf{u}_{i}$ for $p_{i}\to
2$ that $\mathbf{u}(0)=0$ and $\\{|\mathbf{u}_{i}|>0\\}\cap B_{R}$ is not
connected for any $R>0$.

We remark that all our results are stated in a slightly more general form,
with constants depending uniformly for all $p\in[1/2,3]$; see for the details
[9]. In fact, we will have a uniform Lipschitz constant for all $p$ in this
compact interval as well as the nondegeneracy constant. Also, the constant $c$
in Theorem 4.2 will be uniform. Therefore, we may choose a convergence
subsequence $\mathbf{u}_{i}\rightarrow\tilde{\mathbf{u}}_{0}$. By the same
reasoning as the proof of Lemma 6.2 we get that $\tilde{\mathbf{u}}_{0}$ is a
minimizer for $p=2$. On the other hand, $\\{|\mathbf{u}_{0}|>0\\}\cap B_{R}$
is not connected for every $R$ which contradicts the later part of the proof.
∎

######  Lemma A.4.

Let $\varepsilon_{0}>0$ be the constant defined in Proposition A.3. Then for
any $p\in(2-\varepsilon_{0},2+\varepsilon_{0})$ there are constants $M\geq 1$
and $R_{0}>0$ such that if $x_{0}\in\partial\\{|\mathbf{u}|>0\\}$, $y\in
B_{R_{0}}(x_{0})\cap\partial\\{|\mathbf{u}|>0\\}$ and $x_{1},x_{2}\in
B_{r}(y)$ for some $r<R_{0}$, then $\\{|\mathbf{u}|>d\\}\cap B_{Mr}(y)$ has a
connected component containing $x_{1}$ and $x_{2}$, where
$d=\frac{1}{2}\min(|\mathbf{u}(x_{1})|_{\infty},|\mathbf{u}(x_{2})|_{\infty})$.

###### Proof.

Fix constant $p$ and assume the contrary there are sequences
$\partial\\{|\mathbf{u}|>0\\}\ni y_{i}\to x_{0}$, $x_{1}^{i},x_{2}^{i}\in
B_{r_{i}}(y_{i})$, $r_{i}\to 0$ and $M_{i}\to\infty$ such that $x_{1}^{i}$ and
$x_{2}^{i}$ are not connected in $\\{|\mathbf{u}|>d_{i}\\}\cap
B_{M_{i}r_{i}}(y_{i})$ where
$d_{i}=\frac{1}{2}\min(|\mathbf{u}(x_{1}^{i})|_{\infty},|\mathbf{u}(x_{2}^{i})|_{\infty})$.
Consider the blowup sequence

$\mathbf{u}(y_{i}+r_{i}x)/r_{i}\to\mathbf{u}_{0}(x),\qquad d_{i}/r_{i}\to a,$

Lemma A.2 yields that $\\{|\mathbf{u}_{0}|>a\\}\cap B_{R}$ has at least two
connected components for any $R>0$. Note that $a<\infty$ due to the Lipschitz
regularity. Now consider a blowdown of $\mathbf{u}_{0}$

$\mathbf{u}_{0}(R_{i}x)/R_{i}\to\mathbf{u}_{\infty}(x),\qquad R_{i}\to\infty,$

then $\\{|\mathbf{u}_{\infty}|>0\\}\cap B_{R}$ must have at least two
connected components for any $R>0$. On the other hand, $\mathbf{u}_{\infty}$
is a minimizer of $J_{0}$; Lemma 6.2. This contradicts Proposition A.3. ∎

######  Proposition A.5.

Let $\mathbf{u}$ be a minimizer when
$p\in(2-\varepsilon_{0},2+\varepsilon_{0})$ and $\varepsilon_{0}$ is the
constant defined in Proposition A.3. Suppose
$x_{0}\in\partial\\{|\mathbf{u}|>0\\}$. Then $\\{|\mathbf{u}|>0\\}$ is NTA in
a neighborhood of $x_{0}$.

###### Proof.

Step 1: (Property $(i)$, the corkscrew condition for $\\{|\mathbf{u}|>0\\}$.)
Assume that $M>\Arrowvert{\nabla\mathbf{u}}\Arrowvert_{L^{\infty}(B_{1})}/c$
where $c$ is the nondegeneracy constant defined in Lemma 4.1. If the condition
fails at a point $x\in\partial\\{|\mathbf{u}|>0\\}$, then for any $y\in
B_{r}(x)\cap\\{|\mathbf{u}|>0\\}$ we must have ${\rm
dist}(y,\partial\\{|\mathbf{u}|>0\\})\leq M^{-1}r$. Thus

$|\mathbf{u}(y)|\leq
M^{-1}r\Arrowvert{\nabla\mathbf{u}}\Arrowvert_{L^{\infty}(B_{1})}\leq cr$

and by nondegeneracy, Lemma 4.1, $\mathbf{u}=0$ in $B_{\kappa r}(x)$. It
contradicts that $x\in\partial\\{|\mathbf{u}|>0\\}$.

Step 2: (Property $(ii)$, the corkscrew condition for $\\{|\mathbf{u}|=0\\}$.)
Assume the contrary, $\\{|\mathbf{u}|=0\\}$ does not satisfy the corkscrew
condition in any neighborhood of $x_{0}$ for any constant $M$. Thus there is a
sequence $x_{j}\rightarrow x_{0}$ and $r_{j}\rightarrow 0$ such that we can
not find point $a_{r_{j}}(x_{j})$ with the desired property. On the other
hand, Theorem 4.2 infer that the interior of $\\{|\mathbf{u}|=0\\}$ is
nonempty. Let $B_{\tau_{j}}(y_{j})$ be the biggest ball inside
$B_{r_{j}}(x_{j})\cap\\{|\mathbf{u}|=0\\}$, then we must have
$\tau_{j}/r_{j}\rightarrow 0$. Now consider the blowup
$\mathbf{u}_{j}(x):=\mathbf{u}(x_{j}+r_{j}x)/r_{j}\rightarrow\mathbf{u}_{0}(x)$,
it will be a minimizer whose coincidence set has no interior in $B_{1}$. This
contradicts Theorem 4.2.

Step 3: (Harnack chain condition.)
Suppose that $x_{1}$ and $x_{2}$ are such that for some $\tilde{C}>0$ and
$\varepsilon>0$ we have

$|x_{1}-x_{2}|<\tilde{C}\varepsilon,\qquad
B_{\varepsilon}(x_{i})\subset\\{|\mathbf{u}|>0\\},\ i=1,2.$

We may assume without loss of generality, ${\rm
dist}(x_{1},\partial\\{|\mathbf{u}|>0\\})\leq{\rm
dist}(x_{2},\partial\\{|\mathbf{u}|>0\\})=\delta_{0}$. If
$\delta_{0}\geq\tilde{C}\varepsilon$, then $x_{1}\in
B_{\tilde{C}\varepsilon}(x_{2})\subset\\{|\mathbf{u}|>0\\}$ and we can easily
find the Harnack chain. So, consider the case
$\delta_{0}<\tilde{C}\varepsilon$ and choose
$x\in\partial\\{|\mathbf{u}|>0\\}$ such that $|x-x_{2}|=\delta_{0}$. Then
$x_{1},x_{2}\in B_{r}(x)$ for $r=2\tilde{C}\varepsilon$.

By Lemma A.4, $\\{|\mathbf{u}|>d\\}\cap B_{Mr}(x)$ has a connected component
containing $x_{1}$ and $x_{2}$, where
$d=\frac{1}{2}\min(|\mathbf{u}(x_{1})|_{\infty},|\mathbf{u}(x_{2})|_{\infty})$.
Now we have a curve $\gamma:[0,1]\rightarrow\\{|\mathbf{u}|>d\\}\cap
B_{Mr}(x)$ having $x_{1}$ and $x_{2}$ as end point. For every $t\in[0,1]$ we
know that

$|\mathbf{u}(\gamma(t))|\geq d\geq\frac{c_{0}\varepsilon}{2\sqrt{m}},$

where $c_{0}$ comes from (6). Hence,

${\rm
dist}(\mathbf{u}(\gamma(t)),\partial\\{|\mathbf{u}|>0\\})\geq\sigma\varepsilon,$

where $\sigma=\frac{c_{0}}{2C_{0}\sqrt{m}}$. Now we can find a sequence
$y_{1},\cdots,y_{\ell}$ on the image of $\gamma$ such that

$\gamma[0,1]\subset\bigcup_{i=1}^{\ell}B_{\sigma\varepsilon}(y_{i})\subset\\{|\mathbf{u}|>0\\}\cap
B_{Mr+\sigma\varepsilon}(x).$

Since $Mr+\sigma\varepsilon=(2M\tilde{C}+\sigma)\varepsilon$, the number of
balls in covering, $\ell$ can be bounded by a constant depending only on the
dimension and $(2M\tilde{C}+\sigma)/\sigma$, but not on $x_{1},x_{2}$ or
$\varepsilon$. ∎

##  Appendix B  An approximation lemma

Here we prove a lemma, which we used in Lemma 7.2, and is generalization of
Lemma A.1 in [5] to any $1<p<\infty$ (see also Lemma A.1 in [10]).

######  Lemma B.1.

Let $u$ be a nonnegative Lipschitz function in $B_{1}^{+}$ and assume that it
is $p$-harmonic in $\\{u>0\\}$ and $u(0)=0$ ($1<p<\infty$). Then it has the
asymptotic development

$u(x)=\alpha x_{n}+o(|x|),\ \ \text{ as }x\rightarrow 0,$

for some $\alpha\geq 0$, if either

  1. (i)

$u$ vanishes on $\\{x_{n}=0\\}$, or

  2. (ii)

$\\{x_{n}>0\\}\subset\\{u>0\\}$.

###### Proof.

Part $(i)$ is Lemma A.1 in [10]. The proof of $(ii)$ is also similar by a
slight modification. Let $\ell_{k}:=\sup\\{l:lx_{n}\leq u(x)\ \text{ in }\
B_{2^{-k}}^{+}\\}$. Since $\ell_{k}$ is a nondecreasing sequence and bounded
by the Lipschitz constant of $u$. Suppose
$\alpha=\lim_{k\rightarrow\infty}\ell_{k}$, then

$u(x)\geq\alpha x_{n}+o(|x|).$

If the claim fails there exists a sequence $x^{k}\rightarrow 0$ such that

$u(x^{k})\geq\alpha x_{n}^{k}+\delta_{0}|x^{k}|,$

for some $\delta_{0}>0$. Define $u_{k}(x):=u(r_{k}x)/r_{k}$ where
$r_{k}=|x^{k}|\rightarrow 0$. We may also assume that $r_{k}\leq 2^{-k}$.
Since $u_{k}$ are uniformly Lipschitz, we may consider the blowup
$u_{0}=\lim_{k\rightarrow\infty}u_{k}$, as well as $x^{k}/r_{k}\rightarrow
x^{0}$, $|x^{0}|=1$. From the construction we will have $\alpha x_{n}\leq
u_{0}(x)$ in $B_{1}^{+}$, and

$\frac{\delta_{0}}{2}+\alpha x_{n}\leq u_{0}(x)\quad\text{ and
}\quad\frac{\delta_{0}}{2}+\ell_{k}x_{n}\leq u_{k}(x)\quad\text{ in
}B_{\varepsilon}(x^{0}),$

for a sufficiently small $\varepsilon>0$ and large $k$. Let now $w_{k}$ be a
$p$-harmonic function in $B_{1}^{+}$ with smooth boundary values

$\begin{array}[]{cl}w_{k}=\ell_{k}x_{n}&\ \text{ on }\partial
B_{1}^{+}\setminus B_{\varepsilon/2}(x^{0}),\\\\[8.0pt]
w_{k}=\ell_{k}x_{n}+\frac{\delta_{0}}{4}&\ \text{ on }\partial B_{1}^{+}\cap
B_{\varepsilon/4}(x^{0}),\\\\[8.0pt] \ell_{k}x_{n}\leq
w_{k}\leq\ell_{k}x_{n}+\frac{\delta_{0}}{4}&\ \text{ on }\partial
B_{1}^{+}\cap B_{\varepsilon/2}(x^{0}),\\\\[8.0pt] w_{k}=0&\ \text{ on
}\\{x_{n}=0\\}\cap B_{1}.\end{array}$

From the comparison principle we will have $w_{k}\leq u_{k}$ in $B_{1}^{+}$.
(Note that $u_{k}(x)\geq\ell_{k}x_{n}$, since $r_{k}\leq 2^{-k}$).
Furthermore, $w_{k}\rightarrow w_{0}$ in $C^{1,\sigma}(B_{1/2}^{+})$ where
$w_{0}$ is $p$-harmonic with boundary data $w_{0}=0$ on $\\{x_{n}=0\\}$ and
$w_{0}\geq\alpha x_{n}$ on $\partial B_{1}^{+}$. By Hopf boundary principle,

$w_{0}(x)\geq(\alpha+\mu)x_{n}\quad\text{ in }B_{\gamma}^{+},$

for some small $\mu$ and $\gamma$. Thus for $x\in B_{\gamma}^{+}$,

$\begin{split}u_{k}(x)\geq w_{k}(x)&\geq
w_{0}(x)-x_{n}\Arrowvert{\nabla(w_{k}-w_{0})}\Arrowvert_{L^{\infty}(B_{1/2}^{+})}\\\
&\geq\left(\alpha+\mu-\Arrowvert{\nabla(w_{k}-w_{0})}\Arrowvert_{L^{\infty}(B_{1/2}^{+})}\right)x_{n}\\\
&\geq\left(\alpha+\mu/2\right)x_{n}\end{split}$

Now returning to $u$ we get $u(x)\geq(\alpha+\mu/2)x_{n}$ in
$B_{\gamma/r_{k}}^{+}$. This is a contradiction with the definition of
$\ell_{k}$ when $k$ is sufficiently large. ∎

## Declarations

Data availability statement: All data needed are contained in the manuscript.

Funding and/or Conflicts of interests/Competing interests: The authors declare
that there are no financial, competing or conflict of interests.

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