# Regularity of Lipschitz free boundaries for a $p(x)$-Laplacian problem with
right hand side

Fausto Ferrari  Dipartimento di Matematica dell’Università di Bologna, Piazza
di Porta S. Donato, 5, 40126 Bologna, Italy<EMAIL_ADDRESS>and
Claudia Lederman  IMAS - CONICET and Departamento de Matemática, Facultad de
Ciencias Exactas y Naturales, Universidad de Buenos Aires, (1428) Buenos
Aires, Argentina<EMAIL_ADDRESS>
###### Abstract.

We continue our study in [FL] on viscosity solutions to a one-phase free
boundary problem for the $p(x)$-Laplacian with non-zero right hand side. We
first prove that viscosity solutions are locally Lipschitz continuous, which
is the optimal regularity for the problem. Then we prove that Lipschitz free
boundaries of viscosity solutions are $C^{1,\alpha}$. We also present some
applications of our results.

Moreover, we obtain new results for the operator under consideration that are
of independent interest, such as a Harnack inequality.

###### Key words and phrases:

free boundary problem, singular/degenerate operator, variable exponent spaces,
regularity of the free boundary, non-zero right hand side, viscosity
solutions, Harnack inequality, optimal regularity.
2020 Mathematics Subject Classification. 35R35, 35B65, 35J60, 35J70

F. F. was partially supported by INDAM-GNAMPA 2019 project: Proprietà di
regolarità delle soluzioni viscose con applicazioni a problemi di frontiera
libera and INDAM-GNAMPA 2022 project: Regolarità locale e globale per problemi
completamente non lineari.

C. L. was partially supported by the project GHAIA Horizon 2020 MCSA RISE 2017
programme grant 777822 and by the grants CONICET PIP 11220150100032CO
2016-2019, UBACYT 20020150100154BA and ANPCyT PICT 2019-00985. C. L. wishes to
thank the Department of Mathematics of the University of Bologna, Italy, for
the kind hospitality.

##  1\. Introduction and main results

In this paper we continue our study in [FL] on a one-phase free boundary
problem governed by the $p(x)$-Laplacian with non-zero right hand side. More
precisely, we denote by

$\Delta_{p(x)}u:=\mbox{div}(|\nabla u|^{p(x)-2}\nabla u),$

where $p$ is a function such that $1<p(x)<+\infty$. Then our problem is the
following:

(1.1) $\left\\{\begin{array}[]{ll}\Delta_{p(x)}u=f,&\hbox{in
$\Omega^{+}(u):=\\{x\in\Omega:u(x)>0\\}$},\\\ &\\\ |\nabla u|=g,&\hbox{on
$F(u):=\partial\Omega^{+}(u)\cap\Omega.$}\end{array}\right.$

Here $\Omega\subset\mathbb{R}^{n}$ is a bounded domain, $p\in C^{1}(\Omega)$
is a Lipschitz continuous function, $f\in C(\Omega)\cap L^{\infty}(\Omega)$
and $g\in C^{0,\beta}(\Omega)\cap L^{\infty}(\Omega),$ $g>0.$

This problem comes out naturally from limits of a singular perturbation
problem with forcing term as in [LW1], where solutions to (1.1), arising in
the study of flame propagation with nonlocal and electromagnetic effects, are
analyzed. On the other hand, (1.1) appears by minimizing the following
functional

(1.2) $J(v)=\int_{\Omega}\left(\frac{|\nabla
v|^{p(x)}}{p(x)}+\lambda(x)\chi_{\\{v>0\\}}+f(x)v\right)dx$

studied in [LW3], as well as in the seminal paper by Alt and Caffarelli [AC]
in the case $p(x)\equiv 2$ and $f\equiv 0.$ We refer also to [LW4], where
(1.1) appears in the study of an optimal design problem.

We are interested in the regularity of both the solutions and the free
boundaries of viscosity solutions of (1.1). This problem has been already
faced in [LW2] for weak solutions of (1.1), with the aid of the techniques
developed in [AC].

In the present work we are following the strategy introduced in the important
paper by De Silva [D], that was inspired by [S], for one-phase problems and
linear non-divergence operators. [D] was further extended to two-phase
problems in different settings see [DFS1, DFS2, DFS3]. The same technique was
applied to the $p$-Laplace operator ($p(x)\equiv p$ in (1.1)) for the one
phase case, with $p\geq 2$, in [LR].

In the linear homogeneous case, $f\equiv 0,$ (1.1) was studied for viscosity
solutions in the pioneer works by Caffarelli [C1, C2]. The results in [C1, C2]
have been widely generalized to different classes of homogeneous elliptic
problems. See for example [CFS, FS1, FS2] for linear operators, [AF, F1, F2,
Fe1, W1, W2] for fully nonlinear operators and [LN1, LN2] for the
$p$-Laplacian.

We recall that problem (1.1) was originally studied in the linear homogeneous
case in [AC], associated to (1.2). These techniques were generalized to the
linear case with $f\not\equiv 0$ in [GS, Le]. In the homogeneous case, to a
quasilinear uniformly elliptic situation [ACF], to the $p$-Laplacian [DP], to
an Orlicz setting [MW] and to the $p(x)$-Laplacian with $p(x)\geq 2$ [FMW].
Finally, (1.1) with $1<p(x)<\infty$ and $f\not\equiv 0$ was dealt with in
[LW2].

In [FL] we proved that flat free boundaries of viscosity solutions to (1.1)
are $C^{1,\alpha}.$ Here we first prove that viscosity solutions are locally
Lipschitz continuous, which is the optimal regularity for the problem. Then we
prove that Lipschitz free boundaries of viscosity solutions are
$C^{1,\alpha}.$

We devote this sequel to the study of these issues, which brought challenging
difficulties due to the nonlinear behavior of the $p(x)$-Laplacian and, as a
consequence, we present several novelties that are described in detail below.

Our main results are the following (for notation and the precise definition of
viscosity solution to (1.1) we refer to Section 2)

######  Theorem 1.1 (Optimal regularity).

Let $u$ be a viscosity solution to (1.1) in $B_{1}$. There exists a constant
$C>0$ such that

$\|\nabla u\|_{L^{\infty}(B_{1/2})}\leq C.$

######  Theorem 1.2 (Lipschitz implies $C^{1,\alpha}$).

Let $u$ be a viscosity solution to (1.1) in $B_{1}$, with $0\in F(u)$. If
$F(u)$ is a Lipschitz graph in a neighborhood of $0$, then $F(u)$ is
$C^{1,\alpha}$ in a (smaller) neighborhood of $0$.

In addition to the assumptions already stated above, we suppose that there
exist positive numbers $p_{\min},p_{\max}$ and $\gamma_{0}$ such that
$1<p_{\min}\leq p(x)\leq p_{\max}<\infty$ and $g(x)\geq\gamma_{0}>0.$

In Theorem 1.1 the constant $C$ depends only on $p_{\min}$, $p_{\max}$,
$\|\nabla p\|_{L^{\infty}(B_{3/4})}$, $\|f\|_{L^{\infty}(B_{3/4})}$,
$\|g\|_{C^{0,\beta}(\overline{B_{3/4}})}$, $\beta$,
$\|u\|_{L^{\infty}(B_{3/4})}$ and $n$ (the dimension of the space).

In Theorem 1.2 the constant $\alpha$ depends only on $p_{\min}$, $p_{\max}$,
$\beta$, $\|g\|_{L^{\infty}({B_{\rho}})}$, $\gamma_{0}$ and $n$, where $\rho$
is the radius of the ball $B_{\rho}$ where $F(u)$ is Lipschitz. Moreover, the
size of the neighborhood where $F(u)$ is $C^{1,\alpha}$ depends only on
$\rho$, $p_{\min}$, $p_{\max},$ $\|\nabla p\|_{L^{\infty}(B_{\rho})}$,
$\|f\|_{L^{\infty}(B_{\rho})}$, $\|g\|_{C^{0,\beta}(\overline{B_{\rho}})}$,
$\gamma_{0}$, $\beta$, $\|u\|_{L^{\infty}(B_{3\rho/4})}$, $n$ and the
Lipschitz constant of $F(u)$.

After we develop the necessary tools, Theorem 1.2 follows from Theorem 1.1 in
[FL] —where we proved that flat free boundaries are $C^{1,\alpha}$— and from
the main result in [LN1], via a blow-up argument.

As already mentioned, problem (1.1) was faced in [LW2] for weak solutions with
the techniques developed in [AC]. We want to emphasize at this point that the
approach in [AC] for weak solutions gives that flat free boundaries are
$C^{1,\alpha}$. Alt - Caffarelli’s approach does not include the result
Lipschitz free boundaries are $C^{1,\alpha}$. One of the consequences of our
Theorem 1.2 is an analogous result for weak solutions of (1.1) (see Corollary
7.3).

Among the novelties that our work presents, we also refer to Section 5 where
we prove some auxiliary results that are crucial in the proof of our main
theorem. In that section we revisit some lemmas that are well known in the
linear setting (see [CS] and the Appendix in [C2]), for the case of
$p_{0}$-harmonic functions (i.e., $\Delta_{p_{0}}u=0$, $p_{0}\in(1,\infty)$).
Our results concern the existence of first order expansions at one side
regular boundary points of positive Lipschitz $p_{0}$-harmonic functions,
vanishing at the boundary of a domain. This part required great effort and
passed through the equivalence of the notions of weak and viscosity solution
in the case of the $p_{0}$-Laplace operator. Moreover, our proof can be
applied to a general class of fully nonlinear degenerate elliptic operators
(see Remark 5.2). We strongly believe that these results are of independent
interest.

We remark that, as was the case in [FL], carrying out, for the inhomogeneous
$p(x)$-Laplace operator, the strategy devised in [D] required that we develop
new tools. In fact, the $p(x)$-Laplacian is a nonlinear operator that appears
naturally in divergence form from minimization problems, i.e., in the form
${\rm div}A(x,\nabla u)=f(x)$, with

$\lambda|\eta|^{p(x)-2}|\xi|^{2}\leq\sum_{i,j=1}^{n}\frac{\partial
A_{i}}{\partial\eta_{j}}(x,\eta)\xi_{i}\xi_{j}\leq\Lambda|\eta|^{p(x)-2}|\xi|^{2},\quad\xi\in\mathbb{R}^{n},$

where $0<\lambda\leq\Lambda$. This operator is singular in the regions where
$1<p(x)<2$ and degenerate in the ones where $p(x)>2$.

Let us stress that the main arguments in the approach introduced in [D] are
based on Harnack inequality. However, Harnack inequality for the
$p(x)$-Laplacian has a different form from the standard one —still valid for
the $p_{0}$-Laplace operator— even in the homogeneous case. Namely, Harnack
inequality for the inhomogeneous equation $\Delta_{p(x)}u=f$, with $f$
bounded, states that, for any nonnegative weak solution $u$ in $B_{4R},$ there
exist constants $C>0$ and $\mu\geq 0$ such that

(1.3) $\sup_{B_{R}}u\leq C(\inf_{B_{R}}{u}+R+\mu R),$

where $C$ depends on $u$, and $\mu$ depends on $||f||_{L^{\infty}(B_{4R})}$
(among other dependencies). We refer to [Wo] for the the proof and further
details on Harnack inequality for the inhomogeneous $p(x)$-Laplacian.

The presence of the extra term appearing in the right hand side of (1.3) —even
present when $f\equiv 0$, in which case $\mu=0$— brought a major difficulty in
the application of the strategy of [D] for problem (1.1), under a small
perturbation assumption. In order to successfully apply that strategy, we
proved a new Harnack inequality for the inhomogeneous $p(x)$-Laplacian
(Theorem 3.2) that is appropriate for small perturbation settings. Our result
roughly says that if $||f||_{L^{\infty}}$ is small and $p(x)$ is close to a
constant $p_{0}$, then the constant terms appearing in the right hand side of
(1.3) can be taken small.

This constitutes a key result in our proof of the nondegeneracy of viscosity
solutions of (1.1) with Lipschitz free boundaries, and it eventually leads to
our main Theorem 1.2. Let us emphasize that, in light of the discussion above
on inequality (1.3), our Harnack inequality for small perturbation settings is
indeed of independent interest.

Another important matter, not present in other free boundary problems treated
with the present approach, is the a priori control on the dependence on $u$ in
the constant $C$ appearing in Harnack inequality (1.3). This control is
required in order to perform iteration and blow-up arguments. This same fact
made the proof of Theorem 1.1 much more delicate.

As already mentioned, in Theorem 1.2 we make use of the main result in [LN1].
It is worth noticing that the application of this result to our problem
required nontrivial arguments due to the different notion of solution employed
in [LN1] (see Secton 6, Theorem 1.2 and Propositions 6.4 and 6.5).

Let us remark that, as a by-product of our theorems on the regularity of
$F(u)$, we get in Corollary 6.6 further regularity results for $F(u)$, under
additional regularity assumptions on the data $p,f$ and $g$.

We also discuss some applications of Theorem 1.1 in [FL] and Theorem 1.2 in
the present paper (see Section 8 and, in particular, Remark 8.4).

Let us mention as well that our results in Sections 7 and 8 are new even for
$p(x)\equiv p_{0}$, with $p_{0}$ a constant.

We finally point out that the $p(x)$-Laplacian is a particular case of
operator with nonstandard growth. Partial differential equations with
nonstandard growth have been receiving a lot of attention due to their wide
range of applications. Among them we mention the modeling of non-Newtonian
fluids, for instance, electrorheological [R] or thermorheological fluids [AR].
Other applications include non-linear elasticity [Z1], image reconstruction
[AMS, CLR] and the modeling of electric conductors [Z2], to cite a few.

Our work is organized as follows. In Section 2 we provide notation and basic
definitions. We also recall the relationship between the different notions of
solutions to $\Delta_{p(x)}u=f$ we are using. In Section 3 we obtain a Harnack
inequality for the inhomogeneous equation $\Delta_{p(x)}u=f$ (Theorem 3.2)
that is appropriate for small perturbation settings. Next, in Section 4 we
prove the local Lipschitz continuity of viscosity solutions of (1.1), Theorem
1.1. We then show the nondegeneracy of these solutions under the additional
assumption that $F(u)$ is a Lipschitz graph. In Section 5 we obtain a result
on asymptotic developments of positive Lipschitz $p_{0}$-harmonic functions at
one side regular boundary points that we use in Theorem 1.2 and in Section 7.
Then, in Section 6 we prove our main result, Theorem 1.2. In Section 7 we
discuss some consequences, and finally, in Section 8, we present some
applications of our results. For the sake of completeness, in Appendix A we
introduce the Sobolev spaces with variable exponent, which are the appropriate
spaces to work with weak solutions of the $p(x)$-Laplacian. We conclude the
paper with Appendix B, where we include a Liouville type result that we use in
our main theorem.

###  1.1. Assumptions

Throughout the paper we let $\Omega\subset\mathbb{R}^{n}$ be a bounded domain.

Assumptions on $p(x)$. We assume that the function $p(x)$ verifies

$p\in C^{1}(\Omega),\qquad 1<p_{\min}\leq p(x)\leq
p_{\max}<\infty,\qquad\nabla p\in L^{\infty}(\Omega),$

for some positive constants $p_{\min}$ and $p_{\max}$.

Assumptions on $f$. We assume that function $f$ verifies

$f\in C(\Omega)\cap L^{\infty}(\Omega).$

Assumptions on $g$. We assume that the function $g$ verifies

$g\in C^{0,\beta}(\Omega)\cap L^{\infty}(\Omega),\qquad g(x)\geq\gamma_{0}>0,$

for some positive constants $0<\beta<1$ and $\gamma_{0}$.

##  2\. Basic definitions, notation and preliminaries

In this section, we provide notation, basic definitions and some preliminaries
that will be relevant for our work.

Notation. For any continuous function
$u:\Omega\subset\mathbb{R}^{n}\to\mathbb{R}$ we denote

(2.1) $\Omega^{+}(u):=\\{x\in\Omega:u(x)>0\\},\qquad
F(u):=\partial\Omega^{+}(u)\cap\Omega.$

We refer to the set $F(u)$ as the free boundary of $u$, while $\Omega^{+}(u)$
is its positive phase (or side).

Throughout the paper, when we say that $F(u)$ is Lipschitz we are assuming
that

$\Omega^{+}(u)=\\{x=(x^{\prime},x_{n})\in\Omega:x_{n}>\psi(x^{\prime})\\},$

in an appropriate coordinate system, with $\psi$ Lipschitz on
$\mathbb{R}^{n-1}$.

We begin with some remarks on the $p(x)$-Laplacian. In particular, we recall
the relationship between the different notions of solutions to
$\Delta_{p(x)}u=f$ we are using, namely, weak and viscosity solutions. Then we
give the definition of viscosity solution to problem (1.1) and we deduce some
consequences. We here refer to the usual $C$-viscosity definition of
sub/supersolution and solution of an elliptic PDE, see e.g., [CIL].

We start by observing that direct calculations show that, for $C^{2}$
functions $u$ such that $\nabla u(x)\not=0$,

$\begin{split}&\Delta_{p(x)}u=\mbox{div}(|\nabla u|^{p(x)-2}\nabla u)\\\
&=|\nabla u(x)|^{p(x)-2}\left(\Delta
u+(p(x)-2)\Delta_{\infty}^{N}u+\langle\nabla p(x),\nabla
u(x)\rangle\log|\nabla u(x)|\right),\end{split}$

where

$\Delta_{\infty}^{N}u:=\Big{\langle}D^{2}u(x)\frac{\nabla u(x)}{|\nabla
u(x)|}\,,\,\frac{\nabla u(x)}{|\nabla u(x)|}\Big{\rangle}$

denotes the normalized $\infty$-Laplace operator.

We also deduce that

(2.2) $\begin{split}&|\nabla
u(x)|^{p(x)-2}\left(\mathcal{M}_{\lambda_{0},\Lambda_{0}}^{-}(D^{2}u(x))+\langle\nabla
p(x),\nabla u(x)\rangle\log|\nabla u(x)|\right)\\\
&\leq\Delta_{p(x)}u\leq|\nabla
u(x)|^{p(x)-2}\left(\mathcal{M}_{\lambda_{0},\Lambda_{0}}^{+}(D^{2}u(x))+\langle\nabla
p(x),\nabla u(x)\rangle\log|\nabla u(x)|\right),\end{split}$

where, $\lambda_{0}:=\min\\{1,p_{\min}-1\\}$ and
$\Lambda_{0}:=\max\\{1,p_{\max}-1\\}.$ As usual, if $0<\lambda\leq\Lambda$ are
numbers, and $e_{i}$ is the $i-$th eigenvalue of the $n\times n$ symmetric
matrix $M,$ then $\mathcal{M}_{\lambda,\Lambda}^{+}$ and
$\mathcal{M}_{\lambda,\Lambda}^{-}$ denote the extremal Pucci operators and
are defined (see [CC]) as

(2.3)
$\begin{split}\mathcal{M}_{\lambda,\Lambda}^{+}(M)&=\lambda\sum_{e_{i}<0}e_{i}+\Lambda\sum_{e_{i}>0}e_{i},\\\
\mathcal{M}_{\lambda,\Lambda}^{-}(M)&=\Lambda\sum_{e_{i}<0}e_{i}+\lambda\sum_{e_{i}>0}e_{i}.\end{split}$

First we need (see Appendix A for the definition of Sobolev spaces with
variable exponent)

######  Definition 2.1.

Assume that $1<p_{\min}\leq p(x)\leq p_{\max}<\infty$ with $p(x)$ Lipschitz
continuous in $\Omega$ and $f\in L^{\infty}(\Omega)$.

We say that $u$ is a weak solution to $\Delta_{p(x)}u=f$ in $\Omega$ if $u\in
W^{1,p(\cdot)}(\Omega)$ and, for every $\varphi\in C_{0}^{\infty}(\Omega)$,
there holds that

$-\int_{\Omega}|\nabla u(x)|^{p(x)-2}\nabla
u\cdot\nabla\varphi\,dx=\int_{\Omega}\varphi\,f(x)\,dx.$

We recall the following result we proved in [FL] (see [FL], Theorem 3.2)

######  Theorem 2.2.

Let $p$ and $f$ be as in Definition 2.1. Assume moreover that $f\in C(\Omega)$
and $p\in C^{1}(\Omega)$.

Let $u\in W^{1,p(\cdot)}(\Omega)\cap C(\Omega)$ be a weak solution to
$\Delta_{p(x)}u=f$ in $\Omega$. Then $u$ is a viscosity solution to
$\Delta_{p(x)}u=f$ in $\Omega$.

######  Remark 2.3.

We point out that the equivalence between weak and viscosity solutions to the
$p(x)$-Laplacian with right hand side $f\equiv 0$ was proved in [JLP]. On the
other hand, this equivalence, in case $p(x)\equiv p$ and $f\not\equiv 0$ was
dealt with in [JJ] and [MO]. See also [JLM] for the case $p(x)\equiv p$ and
$f\equiv 0$.

We need the following standard notion.

######  Definition 2.4.

Given $u,\varphi\in C(\Omega)$, we say that $\varphi$ touches $u$ from below
(resp. above) at $x_{0}\in\Omega$ if $u(x_{0})=\varphi(x_{0}),$ and

$u(x)\geq\varphi(x)\quad(\text{resp. $u(x)\leq\varphi(x)$})\quad\text{in a
neighborhood $O$ of $x_{0}$.}$

If this inequality is strict in $O\setminus\\{x_{0}\\}$, we say that $\varphi$
touches $u$ strictly from below (resp. above).

######  Definition 2.5.

Let $u$ be a continuous nonnegative function in $\Omega$. We say that $u$ is a
viscosity solution to (1.1) in $\Omega$, if the following conditions are
satisfied:

  1. (i)

$\Delta_{p(x)}u=f$ in $\Omega^{+}(u)$ in the weak sense of Definition 2.1.

  2. (ii)

For every $\varphi\in C(\Omega)$, $\varphi\in
C^{2}(\overline{\Omega^{+}(\varphi)})$. If $\varphi^{+}$ touches $u$ from
below (resp. above) at $x_{0}\in F(u)$ and $\nabla\varphi(x_{0})\not=0$, then

$|\nabla\varphi(x_{0})|\leq g(x_{0})\quad(\text{resp. $\geq g(x_{0})$)}.$

Next theorem follows as a consequence of our Theorem 2.2.

######  Theorem 2.6.

Let $u$ be a viscosity solution to (1.1) in $\Omega.$ Then the following
conditions are satisfied:

  1. (i)

$\Delta_{p(x)}u=f$ in $\Omega^{+}(u)$ in the viscosity sense, that is:

     * (ia)

for every $\varphi\in C^{2}(\Omega^{+}(u))$ and for every
$x_{0}\in\Omega^{+}(u),$ if $\varphi$ touches $u$ from above at $x_{0}$ and
$\nabla\varphi(x_{0})\not=0,$ then $\Delta_{p(x_{0})}\varphi(x_{0})\geq
f(x_{0}),$ that is, $u$ is a viscosity subsolution;

     * (ib)

for every $\varphi\in C^{2}(\Omega^{+}(u))$ and for every
$x_{0}\in\Omega^{+}(u),$ if $\varphi$ touches $u$ from below at $x_{0}$ and
$\nabla\varphi(x_{0})\not=0,$ then $\Delta_{p(x_{0})}\varphi(x_{0})\leq
f(x_{0}),$ that is, $u$ is a viscosity supersolution.

  2. (ii)

For every $\varphi\in C(\Omega)$, $\varphi\in
C^{2}(\overline{\Omega^{+}(\varphi)})$. If $\varphi^{+}$ touches $u$ from
below (resp. above) at $x_{0}\in F(u)$ and $\nabla\varphi(x_{0})\not=0$, then

$|\nabla\varphi(x_{0})|\leq g(x_{0})\quad(\text{resp. $\geq g(x_{0})$)}.$

######  Remark 2.7.

If $p(x)\equiv p$ or $f\equiv 0$, then any function satisfying the conditions
of Theorem 2.6 is a solution to (1.1) in the sense of Definition 2.5 (see
Remark 2.3).

We introduce also the notion of comparison sub/supersolution.

######  Definition 2.8.

We say that $v\in C(\Omega)$ is a strict (comparison) subsolution (resp.
supersolution) to (1.1) in $\Omega$ if $v\in C^{2}(\overline{\Omega^{+}(v)})$,
$\nabla v\not=0$ in $\overline{\Omega^{+}(v)}$ and the following conditions
are satisfied:

  1. (i)

$\Delta_{p(x)}v>f$ (resp. $<f$) in $\Omega^{+}(v)$;

  2. (ii)

If $x_{0}\in F(v)$, then

$|\nabla v(x_{0})|>g(x_{0})\quad(\text{resp. $|\nabla v(x_{0})|<g(x_{0})$}).$

Notice that by the implicit function theorem, according to our definition, the
free boundary of a comparison sub/supersolution is $C^{2}$.

As a consequence of the previous discussion we have

######  Lemma 2.9.

Let $u$ be a viscosity solution to (1.1) in $\Omega$. If $v$ is a strict
(comparison) subsolution to (1.1) in $\Omega$ and $u\geq v^{+}$ in $\Omega$
then $u>v$ in $\Omega^{+}(v)\cup F(v)$. Analogously, if $v$ is a strict
(comparison) supersolution to (1.1) in $\Omega$ and $v\geq u$ in $\Omega$ then
$v>u$ in $\Omega^{+}(u)\cup F(u)$.

Notation. From now on $B_{\rho}(x_{0})\subset{\mathbb{R}}^{n}$ will denote the
open ball of radius $\rho$ centered at $x_{0}$, and $B_{\rho}=B_{\rho}(0)$. A
positive constant depending only on the dimension $n$, $p_{\min}$, $p_{\max}$
will be called a universal constant. We will use $c$, $c_{i}$ to denote small
universal constants and $C$, $C_{i}$ to denote large universal constants.

##  3\. A Harnack inequality for $\Delta_{p(x)}u=f$

In this section we prove a Harnack inequality for $\Delta_{p(x)}u=f$, under a
small perturbation assumption (Theorem 3.2).

We first prove

######  Lemma 3.1.

Assume that $1<p_{\min}\leq p(x)\leq p_{\max}<\infty$ with $p(x)$ Lipschitz
continuous in $B_{1}$ and $\|\nabla p\|_{L^{\infty}}\leq L$, for some $L>0$.
Let $p_{0}$ be such that $p_{\min}\leq p_{0}\leq p_{\max}$ and $f\in
L^{\infty}(B_{1})$.

Let $u\in W^{1,p(\cdot)}(B_{1})\cap L^{\infty}(B_{1})$ be a nonnegative
solution to

$\Delta_{p(x)}u=f\quad\mbox{in }B_{1},$

with $||u||_{L^{\infty}(B_{1})}\leq M$, for some $M>0$.

Given $\eta>0$, there exists
$\varepsilon_{0}=\varepsilon_{0}(\eta,n,p_{\min},p_{\max},M,L)>0$ such that if

$||f||_{L^{\infty}(B_{1})}\leq\varepsilon,\qquad||p-p_{0}||_{L^{\infty}(B_{1})}\leq\varepsilon,$

with $\varepsilon\leq\varepsilon_{0}$, then

(3.1) $||u-u_{0}||_{L^{\infty}(B_{3/4})}\leq\eta,$

for a suitable $u_{0}\in W^{1,\infty}(B_{3/4})$ nonnegative solution to

(3.2) $\Delta_{p_{0}}u_{0}=0\quad\mbox{in }B_{3/4}.$

###### Proof.

Let us suppose by contradiction that there exist $\eta_{0}>0$ and a sequence
of nonnegative functions $u_{k}\in W^{1,p_{k}(\cdot)}(B_{1})\cap
L^{\infty}(B_{1})$ with $p_{\min}\leq p_{k}(x)\leq p_{\max}$, $\|\nabla
p_{k}\|_{L^{\infty}}\leq L$, $||u_{k}||_{L^{\infty}(B_{1})}\leq M$, such that

$||f_{k}||_{L^{\infty}(B_{1})}\leq\frac{1}{k},\qquad||p_{k}-p_{0}||_{L^{\infty}(B_{1})}\leq\frac{1}{k},$
$\Delta_{p_{k}(x)}u_{k}=f_{k}\quad\mbox{in }B_{1},$

and such that

$||u_{k}-v||_{L^{\infty}(B_{3/4})}\geq\eta_{0},$

for every $v\in W^{1,\infty}(B_{3/4})$ nonnegative solution to
$\Delta_{p_{0}}v=0$ in $B_{3/4}$.

Then, by Theorem 1.1 in [Fa] we obtain that

$||u_{k}||_{C^{1,\alpha}(\overline{B_{3/4}})}\leq C\quad\mbox{ with }\quad
0<\alpha<1,$

where $C$ and $\alpha$ depend only on $n$, $p_{\min}$, $p_{\max}$, $L$ and
$M$. Therefore, there is a function $u_{0}\in
C^{1,\alpha}(\overline{B_{3/4}})$ such that, for a subsequence,

$u_{k}\rightarrow u_{0}\quad\mbox{and}\quad\nabla u_{k}\rightarrow\nabla
u_{0}\quad\mbox{uniformly in }\overline{B_{3/4}}.$

Since

$f_{k}\rightarrow 0\quad\mbox{and}\quad p_{k}\rightarrow
p_{0}\quad\mbox{uniformly in }{B_{1}},$

it follows that $u_{0}\in W^{1,\infty}(B_{3/4})$ is a nonnegative solution to

$\Delta_{p_{0}}u_{0}=0\quad\mbox{in }B_{3/4}$

and thus,

$0<\eta_{0}\leq||u_{k}-u_{0}||_{L^{\infty}(B_{3/4})}\to 0,$

which gives a contradiction and concludes the proof. ∎

As a consequence we get

######  Theorem 3.2.

Assume that $1<p_{\min}\leq p(x)\leq p_{\max}<\infty$ with $p(x)$ Lipschitz
continuous in $\Omega$ and $\|\nabla p\|_{L^{\infty}}\leq L$, for some $L>0$.
Let $p_{0}$ be such that $p_{\min}\leq p_{0}\leq p_{\max}$ and $f\in
L^{\infty}(\Omega)$. Let $x_{0}\in\Omega$ and $0<R_{1}\leq R\leq R_{2}$ such
that $B_{R}(x_{0})\subset\Omega$.

Let $u\in W^{1,p(\cdot)}(B_{R}(x_{0}))\cap L^{\infty}(B_{R}(x_{0}))$ be a
nonnegative solution to

$\Delta_{p(x)}u=f\quad\mbox{in }B_{R}(x_{0}),$

with $||u||_{L^{\infty}(B_{R}(x_{0}))}\leq M$, for some $M>0$.

Given $\sigma>0$, there exist positive constants
$\varepsilon_{1}=\varepsilon_{1}(\sigma,n,p_{\min},p_{\max},M,L,R_{1},R_{2})$
and $C=C(n,p_{\min},p_{\max})$ such that if

(3.3)
$||f||_{L^{\infty}(B_{R}(x_{0}))}\leq\varepsilon,\qquad||p-p_{0}||_{L^{\infty}(B_{R}(x_{0}))}\leq\varepsilon,$

with $\varepsilon\leq\varepsilon_{1}$, then

(3.4) $\sup_{{B_{R/2}(x_{0})}}u\leq C\inf_{B_{R/2}(x_{0})}u+\sigma.$

###### Proof.

We assume without loss of generality that $x_{0}=0$.

Case I. Suppose first that $R=1$.

We let $\eta>0$ to be precised later. We now take
$\varepsilon_{0}=\varepsilon_{0}(\eta,n,p_{\min},p_{\max},M,L)$ given by Lemma
3.1. Then, if (3.3) is satisfied with $\varepsilon\leq\varepsilon_{0}$, there
holds (3.1), for a suitable $u_{0}\in W^{1,\infty}(B_{3/4})$ nonnegative
solution to (3.2).

By Harnack’s inequality (Theorem 1.1 in [T]), there exists a positive constant
$C=C(n,p_{\min},p_{\max})$ such that

$\sup_{{B_{1/2}}}u_{0}\leq C\inf_{B_{1/2}}u_{0}.$

Since $||u-u_{0}||_{L^{\infty}(B_{3/4})}\leq\eta$, we obtain

$\displaystyle\sup_{{B_{1/2}}}u$
$\displaystyle\leq\sup_{{B_{1/2}}}u_{0}+\eta\leq C\inf_{B_{1/2}}u_{0}+\eta$
$\displaystyle\leq C\inf_{B_{1/2}}u+(C+1)\eta\leq C\inf_{B_{1/2}}u+\sigma,$

if we choose $\eta$ such that $(C+1)\eta<\sigma$. So (3.4) follows if (3.3) is
satisfied with $\varepsilon\leq\tilde{\varepsilon}_{0}$, where
$\tilde{\varepsilon}_{0}=\tilde{\varepsilon}_{0}(\sigma,n,p_{\min},p_{\max},M,L)$.

Case II. We now assume that $0<R_{1}\leq R\leq 1$ and consider
$\bar{u}(x)=\frac{u(Rx)}{R}$. Then, $\bar{u}\in
W^{1,\bar{p}(\cdot)}(B_{1})\cap L^{\infty}(B_{1})$ is a nonnegative solution
to

$\Delta_{\bar{p}(x)}\bar{u}=\bar{f}\quad\mbox{in }B_{1},$

with $||\bar{u}||_{L^{\infty}(B_{1})}\leq\overline{M}$, where
$\bar{p}(x)=p(Rx)$, $\|\nabla\bar{p}\|_{L^{\infty}}\leq L$,
$\bar{f}(x)=Rf(Rx)$ and $\overline{M}=\frac{M}{R_{1}}$.

Then, if

$||\bar{f}||_{L^{\infty}(B_{1})}\leq||{f}||_{L^{\infty}(B_{R})}\leq\varepsilon,\qquad||\bar{p}-p_{0}||_{L^{\infty}(B_{1})}=||{p}-p_{0}||_{L^{\infty}(B_{R})}\leq\varepsilon,$

for $\varepsilon\leq\tilde{\varepsilon}_{0}$, with
$\tilde{\varepsilon}_{0}=\tilde{\varepsilon}_{0}(\sigma,n,p_{\min},p_{\max},\overline{M},L)$
chosen as in Case I, we get

$\sup_{{B_{1/2}}}\bar{u}\leq C\inf_{B_{1/2}}\bar{u}+\sigma.$

That is,

$\sup_{{B_{R/2}}}{u}\leq C\inf_{B_{R/2}}{u}+R\sigma\leq
C\inf_{B_{R/2}}{u}+\sigma.$

Hence we get (3.4), if (3.3) is satisfied with
$\varepsilon\leq\varepsilon_{1}(\sigma,n,p_{\min},p_{\max},M,L,R_{1})$.

Case III. Finally, if we assume that $0<R_{1}\leq R\leq R_{2}$, we proceed as
in Case II and we obtain the desired result with
$\varepsilon_{1}=\varepsilon_{1}(\sigma,n,p_{\min},p_{\max},M,L,R_{1},R_{2})$.
∎

##  4\. Lipschitz continuity and nondegeneracy

In this section we prove Theorem 1.1, which gives the optimal regularity for
viscosity solutions to (1.1), i.e., the local Lipschitz continuity. We also
prove that if $F(u)$ is a Lipschitz graph, then viscosity solutions to (1.1)
are nondegenerate.

We recall the following result we proved in [FL]

######  Lemma 4.1.

Let $x_{0}\in B_{1}$ and $0<\bar{r}_{1}<\bar{r}_{2}\leq 1$. Assume that
$1<p_{\min}\leq p(x)\leq p_{\max}<\infty$ and $\|\nabla
p\|_{L^{\infty}}\leq\varepsilon^{1+\theta}$, for some $0<\theta\leq 1$. Let
$c_{1}$ and $c_{2}$ be positive constants.

There exist positive constants $\gamma\geq 1$, $\bar{c}$ and $\varepsilon_{0}$
such that the function

$w(x)=c_{1}|x-x_{0}|^{-\gamma}-c_{2},$

satisfies, for $\bar{r}_{1}\leq|x-x_{0}|\leq\bar{r}_{2}$,

$\Delta_{p(x)}w\geq\bar{c},\quad\text{for
}\,0<\varepsilon\leq\varepsilon_{0}.$

Here $\gamma=\gamma(n,p_{\min},p_{\max})$,
$\bar{c}=\bar{c}(p_{\min},p_{\max},c_{1})$ and
$\varepsilon_{0}=\varepsilon_{0}(n,p_{\min},p_{\max},\bar{r}_{1},c_{1})$.

###### Proof.

See Lemma 4.2 in [FL]. ∎

We will now prove two key estimates for viscosity solutions to (1.1). Estimate
(4.2) will imply that viscosity solutions are locally Lipschitz continuous
(see Theorem 1.1). If $F(u)$ is a Lipschitz continuous graph, we also obtain
estimate (4.3), which gives the nondegeneracy of $u$ close to $F(u)$.

We will use the notation $p_{+}^{r}=\sup_{B_{r}}p$ and
$p_{-}^{r}=\inf_{B_{r}}p$, for $r>0$ (see [Wo]).

######  Proposition 4.2.

Let $p_{\min}\leq p_{0}\leq p_{\max}$ and $0<\gamma_{0}\leq
g_{0}\leq\|g\|_{L^{\infty}(B_{2})}$. Let $u$ be a viscosity solution to (1.1)
in $B_{2}$ such that $0\in F(u)$. There exists a constant
$0<\tilde{\varepsilon}<1$ such that if

(4.1)
$\|f\|_{L^{\infty}(B_{2})}\leq\tilde{\varepsilon},\quad||g-g_{0}||_{L^{\infty}(B_{2})}\leq\tilde{\varepsilon},\quad||\nabla
p||_{L^{\infty}(B_{2})}\leq\tilde{\varepsilon},\quad||p-p_{0}||_{L^{\infty}(B_{2})}\leq\tilde{\varepsilon},$

then

(4.2) $u(x)\leq C_{0}\mbox{dist}(x,F(u)),\quad x\in B^{+}_{1/2}(u).$

Assume moreover that $F(u)$ is a Lipschitz continuous graph in $B_{2}$. Then

(4.3) $c_{0}\mbox{dist}(x,F(u))\leq u(x),\quad x\in B^{+}_{\rho_{0}}(u).$

The constants $\tilde{\varepsilon}$, $c_{0}$ and $C_{0}$ depend only on $n$,
$p_{\min}$, $p_{\max}$, $\|g\|_{L^{\infty}(B_{2})}$ and
$||u||_{L^{\infty}(B_{3/2})}^{p_{+}^{3/2}-p_{-}^{3/2}}$, where
$p_{+}^{3/2}=\sup_{B_{3/2}}p$ and $p_{-}^{3/2}=\inf_{B_{3/2}}p$. The constants
$\tilde{\varepsilon}$ and $c_{0}$ depend also on the Lipschitz constant of
$F(u)$ and on $\gamma_{0}$, and the constant $\rho_{0}$ depends only on the
Lipschitz constant of $F(u)$.

###### Proof.

Without loss of generality we will assume that $g_{0}=1$. We let $x_{0}\in
B_{1/2}^{+}(u)$ and we denote $d=\mbox{dist}(x_{0},F(u)).$ We consider the
rescaled function

(4.4) $\tilde{u}(x)=\frac{u(x_{0}+dx)}{d}.$

Then $\tilde{u}$ is a viscosity solution to (1.1) with right hand side
$\tilde{f}(x)=df(x_{0}+dx)$, exponent $\tilde{p}(x)=p(x_{0}+dx)$ and free
boundary condition $\tilde{g}(x)=g(x_{0}+dx).$ Since $d\leq 1$, the
assumptions (4.1) hold for the rescaled functions in $B_{3/2}$.

In particular, $\tilde{u}$ is well defined in the ball $\overline{B_{1}}$,
with $\tilde{u}>0$ in $B_{1}$, and it satisfies the equation

(4.5) $\Delta_{\tilde{p}(x)}\tilde{u}=\tilde{f}\quad\text{ in }B_{1}.$

We will show that

(4.6) $c_{0}\leq\tilde{u}(0)\leq C_{0},$

for suitable universal constants $C_{0},c_{0}>0$.

Step I: Upper bound. Let us prove the upper bound in (4.6). We will argue by
contradiction, assuming that $\tilde{u}(0)>C_{0}$, with $C_{0}\geq 1$ to be
precised later.

We will use a barrier like the one considered in Lemma 4.1, in the annulus
$B_{1}\setminus\overline{B}_{r}$, with $r$ suitably chosen.

We are going to fix $0<r<1$ in a universal way, keeping in mind the particular
form of Harnack’s inequality for the $p(x)$-Laplacian (see Theorem 2.1 [Wo]).
In fact, since there holds (4.5), it follows from [Wo] that there exists a
positive constant $C_{H}$ such that

(4.7) $\sup_{B_{r}}\tilde{u}\leq
C_{H}(\inf_{B_{r}}\tilde{u}+r({{||\tilde{f}||}_{L^{\infty}}}^{\frac{1}{p_{\max}-1}}+1)),$

if $r<\frac{1}{4}$. Using that ${||\tilde{f}||}_{L^{\infty}}\leq 1$ and
${||\nabla\tilde{p}||}_{L^{\infty}}\leq 1$, we obtain that the constant
$C_{H}$ depends only on $n$, $p_{\min}$, $p_{\max}$ and
$||\tilde{u}||_{L^{\infty}(B_{4r})}^{\tilde{p}_{+}^{4r}-\tilde{p}_{-}^{4r}}$,
where $\tilde{p}_{+}^{4r}=\sup_{B_{4r}}\tilde{p}$ and
$\tilde{p}_{-}^{4r}=\inf_{B_{4r}}\tilde{p}$.

We now notice that

(4.8)
${||\tilde{u}||}_{L^{\infty}(B_{4r})}^{\tilde{p}_{+}^{4r}-\tilde{p}_{-}^{4r}}\leq||u||_{L^{\infty}(B_{d}(x_{0}))}^{\tilde{p}_{+}^{4r}-\tilde{p}_{-}^{4r}}\Big{(}\frac{1}{d}\Big{)}^{\tilde{p}_{+}^{4r}-\tilde{p}_{-}^{4r}},$
(4.9)
${\tilde{p}_{+}^{4r}-\tilde{p}_{-}^{4r}}\leq{||\nabla\tilde{p}||}_{L^{\infty}(B_{4r})}8r\leq
d{||\nabla{p}||}_{L^{\infty}(B_{d}(x_{0}))}2\leq 2d,$

and also

(4.10) ${\tilde{p}_{+}^{4r}-\tilde{p}_{-}^{4r}}=\sup_{x\in
B_{4r}}{p(x_{0}+dx)}-\inf_{x\in
B_{4r}}{p(x_{0}+dx)}\leq\sup_{B_{d}(x_{0})}{p}-\inf_{B_{d}(x_{0})}{p}.$

Then, from (4.8), (4.9) and (4.10) and using that $B_{d}(x_{0})\subset
B_{3/2}$, we conclude that

${||\tilde{u}||}_{L^{\infty}(B_{4r})}^{\tilde{p}_{+}^{4r}-\tilde{p}_{-}^{4r}}\leq
c\max\Big{\\{}1,\
||u||_{L^{\infty}(B_{3/2})}^{{p}_{+}^{3/2}-{p}_{-}^{3/2}}\Big{\\}},$

where $c=\sup_{x\in(0,1)}\big{(}\frac{1}{x}\big{)}^{2x}$,
$p_{+}^{3/2}=\sup_{B_{3/2}}p$ and $p_{-}^{3/2}=\inf_{B_{3/2}}p$.

Hence, from (4.7) and the fact that ${||\tilde{f}||}_{L^{\infty}}\leq 1$, we
deduce that for every $x\in B_{r},$

(4.11)
$\frac{\tilde{u}(0)}{C_{H}}-2r\leq\inf_{B_{r}}\tilde{u}\leq\tilde{u}(x).$

We now fix $r=\min\\{\frac{1}{8},\frac{1}{4C_{H}}\\}$, and using that
$\tilde{u}(0)>C_{0}\geq 1,$ we get from (4.11)

$\tilde{u}(x)\geq\frac{\tilde{u}(0)}{2C_{H}},\quad x\in\overline{B}_{r}.$

Next let

$w(x)=|x|^{-\gamma}-1,$

where we fix $\gamma=\gamma(n,p_{\min},p_{\max})\geq 1$ given in Lemma 4.1.

We denote

(4.12) $G(x)=\bar{C}w(x)=\bar{C}\big{(}|x|^{-\gamma}-1\big{)}$

in $B_{1}\setminus\overline{B}_{r},$ where we fix
$\bar{C}=\bar{C}(r,\gamma)>0$ in such a way that $G=1$ on $\partial B_{r}.$

Let

$\bar{G}(x)=kG(x)=k\bar{C}\big{(}|x|^{-\gamma}-1\big{)},\quad\text{where}\
k=\frac{C_{0}}{2C_{H}}.$

Recalling that
$\tilde{u}(x)\geq\frac{\tilde{u}(0)}{2C_{H}}>\frac{C_{0}}{2C_{H}}$ in
$\bar{B}_{r}$, we get

(4.13) $\displaystyle\tilde{u}\geq 0=\bar{G},\quad\mbox{ on }\partial B_{1},$
$\displaystyle\tilde{u}\geq k=\bar{G},\quad\mbox{ on }\partial B_{r}.$

We claim that

(4.14) $\Delta_{\tilde{p}(x)}\bar{G}\geq\tilde{f}\quad\text{in
}B_{1}\setminus\bar{B}_{r},$

if $\tilde{\varepsilon}$ is suitably chosen.

In fact, by Lemma 4.1, we know that

$\Delta_{\tilde{p}(x)}\bar{G}\geq\bar{c}\quad\text{in
}B_{1}\setminus\bar{B}_{r},$

with $\bar{c}=\bar{c}(p_{\min},p_{\max},k,\bar{C})$, if
$\tilde{\varepsilon}\leq\bar{\varepsilon}_{0}(n,p_{\min},p_{\max},{r},k,\bar{C})$,
since $||\nabla\tilde{p}||_{L^{\infty}}\leq\tilde{\varepsilon}$. So, if we let
$\tilde{\varepsilon}\leq\bar{c}$, then
$||\tilde{f}||_{L^{\infty}}\leq\bar{c}$. That is, (4.14) holds.

Then, from (4.5), (4.14) and (4.13), we conclude that $\tilde{u}\geq\bar{G}$
in $\overline{B_{1}\setminus B_{r}}$, with $\bar{G}\in C^{2}$ and
$\nabla\bar{G}\neq 0$ in that set, and $\bar{G}$ touches $\tilde{u}$ from
below at some $z\in\partial B_{1}\cap F(\tilde{u}).$ Then

$2>1+\tilde{\varepsilon}\geq\tilde{g}(z)\geq|\nabla\bar{G}(z)|=\frac{C_{0}}{2C_{H}}|\nabla
G(z)|=\frac{C_{0}\gamma\bar{C}}{2C_{H}},$

so we obtain a contradiction if we choose
$C_{0}=\max\big{\\{}1,\frac{8C_{H}}{\gamma\bar{C}}\big{\\}}$. Hence (4.2)
follows.

Step II: Lipschitz estimate. From (4.2) we deduce that $u$ is Lipschitz
continuous in $B_{1/4}$, with a Lipschitz constant depending only on $n$,
$p_{\min}$, $p_{\max}$ and $C_{0}$. In fact, this can be seen with similar
arguments as those in Theorem 1.1, Step III. When estimating the Lipschitz
constant, we use that, in the present case,
$\|f\|_{L^{\infty}(B_{2})}\leq\tilde{\varepsilon}<1$ and $||\nabla
p||_{L^{\infty}(B_{2})}\leq\tilde{\varepsilon}<1$.

Step III: Lower bound. Now we assume that $F(u)$ is a Lipschitz continuous
graph in $B_{2}$. Without loss of generality we assume that $F({u})$ is a
Lipschitz graph in the direction $e_{n}$ with Lipschitz constant $1$. We want
to prove that $\tilde{u}$ given by (4.4) satisfies the lower bound in (4.6).

We assume moreover that our point $x_{0}\in B_{1/2}^{+}(u)$ belongs to
$B_{\rho_{0}}$, with ${\rho_{0}}<1/5$. Then,
$d=\mbox{dist}(x_{0},F(u))<\rho_{0}<1/5$ so $\tilde{u}$ is well defined in the
ball $\overline{B_{5}}$.

Taking additionally $\rho_{0}<1/24$, we also obtain from the previous step
that $\tilde{u}$ is Lipschitz in ${B_{5}}$, with Lipschitz constant depending
only on $n$, $p_{\min}$, $p_{\max}$ and $C_{0}$. Moreover, since there exists
$\bar{x}\in\partial B_{1}\cap F(\tilde{u})$,
$||\tilde{u}||_{L^{\infty}(B_{5})}$ depends only on the Lipschitz constant of
$\tilde{u}$ in $B_{5}$.

Let us point out that also in this part of the proof we need to use more
delicate arguments than those in [D]. Thus, we first remark what does not
change. Since $F(\tilde{u})$ is a Lipschitz continuous graph, then
$\\{\tilde{u}>0\\}$ is a NTA domain, see [JK]. This fact implies that for
every couple of points $\delta$-away from $F(\tilde{u})$ in
$\\{\tilde{u}>0\\}$ such that they are contained in a ball of size
$\bar{M}\delta,$ there exists a Harnack chain of balls, whose length is of
order $\bar{M}$, contained in the domain, connecting the two points. In other
words, there exist $k$ balls in $\\{\tilde{u}>0\\}$ of radius comparable to
$\delta$ ($k$ depending only on $\bar{M}$), such that consecutive balls
intersect, connecting the two points.

As a consequence we will show that, in the present case, we can apply a
suitable Harnack inequality (Theorem 3.2) at each ball, and this will allow us
to estimate the value of $\tilde{u}$ at the first point with the value of
$\tilde{u}$ at the last one, times a universal constant, provided (4.1) holds,
for appropriate $\tilde{\varepsilon}$.

We start by considering, for $\eta>0$,

$\widetilde{G}(x)=\eta(1-{G}(x)),\quad\text{ in }B_{1}\setminus\bar{B_{r}}$

where ${G}$, as well as the constants $r$, $\gamma$ and $\bar{C}$, are defined
as in (4.12).

We observe that, $\nabla\widetilde{G}\neq 0$ in
$\overline{B_{1}\setminus{B_{r}}}$ and, on $\partial B_{r}$,

$|\nabla\widetilde{G}|=\eta\bar{C}|\nabla w|=\eta\bar{C}\gamma r^{-1-\gamma},$

then we can choose $\eta=\eta(r,\gamma),$ so that

$|\nabla\widetilde{G}|<\frac{1}{2}<1-\tilde{\varepsilon}\quad\text{ on
}\partial B_{r},$

if $\tilde{\varepsilon}<\frac{1}{2}$. Now, since

(4.15)
$\Delta_{\tilde{p}(x)}\widetilde{G}=-\Delta_{\tilde{p}(x)}\big{(}{\eta{G}}\big{)},\qquad\eta{G(x)}=\eta\bar{C}\big{(}|x|^{-\gamma}-1\big{)},$

we can apply Lemma 4.1 once more and deduce that

(4.16) $\Delta_{\tilde{p}(x)}\big{(}{\eta{G}}\big{)}\geq\hat{c}\quad\text{in
}B_{1}\setminus\bar{B_{r}},$

with $\hat{c}=\hat{c}(p_{\min},p_{\max},\eta,\bar{C})$, if
$\tilde{\varepsilon}\leq\hat{\varepsilon}_{0}(n,p_{\min},p_{\max},{r},\eta,\bar{C})$,
since $||\nabla\tilde{p}||_{L^{\infty}}\leq\tilde{\varepsilon}$. So, if we let
$\tilde{\varepsilon}<\hat{c}$, then
$||\tilde{f}||_{L^{\infty}(B_{5})}<\hat{c}$ and therefore, from (4.15) and
(4.16), we get

$\Delta_{\tilde{p}(x)}\widetilde{G}<-||\tilde{f}||_{L^{\infty}(B_{5})}\quad\text{in
}B_{1}\setminus\bar{B}_{r}.$

That is, $\widetilde{G}$ is a strict supersolution to the rescaled free
boundary problem in $B_{1}\setminus\bar{B}_{r}$.

Next, observe that from the assumptions we made, $F(\tilde{u})$ is a Lipschitz
graph in the direction $e_{n}$ with Lipschitz constant $1$ and consider the
function

$\widetilde{G}(x+4e_{n})$

in $B_{1}(-4e_{n})\setminus\overline{B_{r}}(-4e_{n})$, which is a strict
supersolution of our rescaled free boundary problem. There holds that
$\widetilde{G}(x+4e_{n})\geq 0$ as well as
$\widetilde{G}(x+4e_{n})\geq\tilde{u}(x)$ in
$B(-4e_{n})\setminus\overline{B_{r}}(-4e_{n})$, since $\tilde{u}\equiv 0$ in
$B_{1}(-4e_{n})$.

Now we move back the graph, by a translation depending on $t>0$, until the
graph of the function

$\widetilde{G}(x+(4-t)e_{n}):-(4-t)e_{n}+{(B_{1}\setminus\overline{B_{r}})}\to\mathbb{R}$

touches the graph of $\tilde{u}.$ Let say that the contact happens when
$t=t^{*}$ at a point $\tilde{z}$ such that
$\tilde{u}(\tilde{z})=\widetilde{G}(\tilde{z}+(4-t^{*})e_{n}).$

Since $\widetilde{G}(x+(4-t^{*})e_{n}))$ is a strict supersolution to the
rescaled free boundary problem, recalling the comparison result (see Lemma
2.9), we conclude that $\widetilde{G}(x+(4-t^{*})e_{n})$ cannot touch
$\tilde{u}$ from above at the common free boundary sets, neither at interior
of the annulus.

Then the contact point $\tilde{z}$ belongs to $-(4-t^{*})e_{n}+\partial
B_{1}.$ As a consequence $\eta=\tilde{u}(\tilde{z})$ and
$\tilde{d}=\mbox{dist}(\tilde{z},F(\tilde{u}))\leq 1.$ Since $\tilde{u}$ is
Lipschitz continuous with universal constant, then
$\eta=\tilde{u}(\tilde{z})\leq C\tilde{d}$ so that

(4.17) $C^{-1}\eta\leq\tilde{d}\leq 1.$

Hence, from (4.17) and by applying the cited result on NTA domains, we know
that we can construct a Harnack chain connecting $0$ and $\tilde{z}$, and the
length of this chain, let us say $m$, is bounded by a universal constant.

That is, we have balls $B_{r_{i}}(x_{i})$ with radius $r_{i}$ comparable to
$1$, $B_{2r_{i}}(x_{i})\subset\\{\tilde{u}>0\\}$, $0\leq i\leq m$, $x_{0}=0$,
$x_{m}=\tilde{z}$ and $y_{i}\in B_{r_{i-1}}(x_{i-1})\cap B_{r_{i}}(x_{i})$,
for $1\leq i\leq m$.

We can now apply Theorem 3.2 to $\tilde{u}$ at every ball $B_{2r_{i}}(x_{i})$.
That is, given $\sigma>0$ there exist
$\varepsilon_{1}=\varepsilon_{1}(\sigma)$ and $C^{*}$ universal such that if
$||\tilde{p}-p_{0}||_{L^{\infty}}\leq\varepsilon$ and
$||\tilde{f}||_{L^{\infty}}\leq\varepsilon,$ with
$\varepsilon\leq\varepsilon_{1}$, then

(4.18) $\sup_{B_{r_{i}}(x_{i})}\tilde{u}\leq
C^{*}\inf_{B_{r_{i}}(x_{i})}\tilde{u}+\sigma.$

For the application of Theorem 3.2 we need to recall that
${||\tilde{u}||}_{L^{\infty}(B_{5})}\leq M$, with $M$ universal.

Now, (4.18) implies that for any $x,y\in B_{r_{i}}(x_{i})$,

$c\tilde{u}(x)-c\sigma\leq\tilde{u}(y),$

where we have denoted $c=\frac{1}{C^{*}}$. Then, we obtain

$c\tilde{u}(y_{1})-c\sigma\leq\tilde{u}(x_{0}),$
$c\tilde{u}(y_{i+1})-c\sigma\leq\tilde{u}(y_{i}),\qquad 1\leq i\leq m-1,$

and

$c\tilde{u}(x_{m})-c\sigma\leq\tilde{u}(y_{m}).$

Then, iterating we deduce

$c^{m+1}\tilde{u}(x_{m})-\sigma\sum_{j=1}^{m+1}c^{j}\leq\tilde{u}(x_{0}).$

Thus, since $x_{0}=0$ and $x_{m}=\tilde{z},$ we have

$c^{m+1}\tilde{u}(\tilde{z})-\sigma
c\,\frac{1-c^{m+1}}{1-c}=c^{m+1}\tilde{u}(\tilde{z})-\sigma\sum_{j=1}^{m+1}c^{j}\leq\tilde{u}(0).$

Hence denoting $c_{1}=c^{m+1}$ and $c_{2}=c\,\frac{1-c^{m+1}}{1-c},$ we obtain

$c_{1}\eta-\sigma c_{2}=c_{1}\tilde{u}(\tilde{z})-\sigma
c_{2}\leq\tilde{u}(0),$

where $c_{1}$ and $c_{2}$ are universal constants. Now we fix $\sigma$
universal,

$\sigma=\frac{c_{1}\eta}{2c_{2}}.$

In this way we conclude that

$\tilde{u}(0)\geq c_{1}\eta-\frac{c_{1}\eta}{2}=\frac{c_{1}\eta}{2},$

if
$||\tilde{p}-p_{0}||_{L^{\infty}}\leq||p-p_{0}||_{L^{\infty}(B_{2})}\leq\tilde{\varepsilon}$
and
$||\tilde{f}||_{L^{\infty}}\leq||f||_{L^{\infty}(B_{2})}\leq\tilde{\varepsilon},$
with $\tilde{\varepsilon}\leq\varepsilon_{1}(\sigma)$. Since $\eta$ is
universal as well, we have finished the proof.∎

From Proposition 4.2, we can now obtain the proof of Theorem 1.1.

We recall again the notation we use: $p_{+}^{r}=\sup_{B_{r}}p$ and
$p_{-}^{r}=\inf_{B_{r}}p$, for $r>0$.

###### Proof of Theorem 1.1.

Let $u$ be a viscosity solution to (1.1) in $B_{1}$. We will divide the proof
into several steps.

Step I. Let us fix $z_{0}\in B_{5/8}\cap F(u)$. For $0<\rho\leq\frac{1}{16}$,
we consider the function

$\bar{u}(x)=\frac{1}{\rho}u(z_{0}+\rho x),\quad x\in B_{2}.$

Then $\bar{u}$ is a viscosity solution to (1.1) in $B_{2}$, with right hand
side $\bar{f}(x)=\rho f(z_{0}+\rho x)$, exponent $\bar{p}(x)=p(z_{0}+\rho x)$
and free boundary condition $\bar{g}(x)=g(z_{0}+\rho x)$. Moreover, $0\in
F(\bar{u})$.

Let us see that we can apply the first part of Proposition 4.2 to ${\bar{u}}$,
if $\rho$ is suitably chosen.

For that purpose, let us first show that the constants appearing in that
proposition can be taken independent of $\rho$. More precisely, we want to
find a bound independent of $\rho$ for

$||\bar{u}||_{L^{\infty}(B_{3/2})}^{\bar{p}_{+}^{3/2}-\bar{p}_{-}^{3/2}},\qquad\text{where}\quad\bar{p}_{+}^{3/2}=\sup_{B_{3/2}}\bar{p},\
\ \bar{p}_{-}^{3/2}=\inf_{B_{3/2}}\bar{p}.$

In fact, we have

(4.19)
${||\bar{u}||}_{L^{\infty}(B_{3/2})}^{\bar{p}_{+}^{3/2}-\bar{p}_{-}^{3/2}}\leq||u||_{L^{\infty}(B_{1/8}(z_{0}))}^{\bar{p}_{+}^{3/2}-\bar{p}_{-}^{3/2}}\Big{(}\frac{1}{\rho}\Big{)}^{\bar{p}_{+}^{3/2}-\bar{p}_{-}^{3/2}},$

and

(4.20) ${\bar{p}_{+}^{3/2}-\bar{p}_{-}^{3/2}}\leq
3{||\nabla\bar{p}||}_{L^{\infty}(B_{3/2})}\leq
3\rho{||\nabla{p}||}_{L^{\infty}(B_{1/8}(z_{0}))}.$

Then, from (4.19) and (4.20), we conclude that

$||\bar{u}||_{L^{\infty}(B_{3/2})}^{\bar{p}_{+}^{3/2}-\bar{p}_{-}^{3/2}}\leq
C=C\big{(}||u||_{L^{\infty}(B_{1/8}(z_{0}))},{||\nabla{p}||}_{L^{\infty}(B_{1/8}(z_{0}))}\big{)}.$

It follows that in order to apply the first part of Proposition 4.2 to
$\bar{u}$ we can take the constants $\tilde{\varepsilon}$ and $C_{0}$ in that
proposition depending only on $n$, $p_{\min}$, $p_{\max}$,
$||u||_{L^{\infty}(B_{1/8}(z_{0}))}$,
${||\nabla{p}||}_{L^{\infty}(B_{1/8}(z_{0}))}$ and
$||g||_{L^{\infty}(B_{1/8}(z_{0}))}$.

Then, if $\rho$ is small enough, there holds in $B_{2}$

$\displaystyle|\bar{f}(x)|$
$\displaystyle\leq||f||_{L^{\infty}(B_{1/8}(z_{0}))}\,\rho\leq\tilde{\varepsilon},$
$\displaystyle|\bar{g}(x)-g(z_{0})|$ $\displaystyle=|g(z_{0}+\rho
x)-g(z_{0})|\leq
2[g]_{C^{0,\beta}(B_{1/8}(z_{0}))}\,{\rho}^{\beta}\leq\tilde{\varepsilon},$
$\displaystyle|\nabla\bar{p}(x)|$ $\displaystyle\leq||\nabla
p||_{L^{\infty}(B_{1/8}(z_{0}))}\,\rho\leq\tilde{\varepsilon},$
$\displaystyle|\bar{p}(x)-p(z_{0})|$ $\displaystyle=|p(z_{0}+\rho
x)-p(z_{0})|\leq 2||\nabla
p||_{L^{\infty}(B_{1/8}(z_{0}))}\,\rho\leq\tilde{\varepsilon}.$

Hence, if $\rho\leq\rho_{0}$, $\rho_{0}$ depending only on
$\tilde{\varepsilon}$, $||f||_{L^{\infty}(B_{1/8}(z_{0}))}$,
$[g]_{C^{0,\beta}(B_{1/8}(z_{0}))}$, $\beta$ and $||\nabla
p||_{L^{\infty}(B_{1/8}(z_{0}))}$, then $\bar{u}$ satisfies

$\bar{u}(x)\leq C_{0}\mbox{dist}(x,F(\bar{u})),\quad x\in
B^{+}_{1/2}(\bar{u}).$

Step II. We deduce from the previous step that for every $z_{0}\in B_{5/8}\cap
F(u)$ there holds

(4.21) ${u}(x)\leq C_{0}\mbox{dist}(x,F({u})),\quad x\in
B_{\rho_{1}}(z_{0})\cap\\{u>0\\},$

for $C_{0}>0$ and $0<\rho_{1}<\frac{1}{32}$ constants depending only on $n$,
$p_{\min}$, $p_{\max}$, $||u||_{L^{\infty}(B_{3/4})}$,
${||\nabla{p}||}_{L^{\infty}(B_{3/4})}$, $||f||_{L^{\infty}(B_{3/4})}$,
$||g||_{C^{0,\beta}(\overline{B_{3/4}})}$ and $\beta$ (here we have used that
$B_{1/8}(z_{0})\subset B_{3/4}$ for every $z_{0}\in B_{5/8}\cap F(u)$).

Step III. Let $x_{0}\in B^{+}_{1/2}({u})$ such that
$\mbox{dist}(x_{0},F({u}))\leq{\rho_{1}/2}.$ We will show that

(4.22) $|\nabla u(x_{0})|\leq C_{1},$

for $C_{1}>0$ universal.

In fact, we denote $d_{0}=\mbox{dist}(x_{0},F({u}))$ and we define
$\tilde{u}(x)=\frac{1}{d_{0}}{u(x_{0}+d_{0}x)}$. Then, since
$B_{d_{0}}(x_{0})\subset\\{u>0\\}$,

$\Delta_{\tilde{p}(x)}\tilde{u}=\tilde{f}\text{ in }B_{1},$

with $\tilde{f}(x)=d_{0}f(x_{0}+d_{0}x)$ and $\tilde{p}(x)=p(x_{0}+d_{0}x)$
and therefore,

(4.23)
${||\nabla{\tilde{p}}||}_{L^{\infty}(B_{1})}\leq{||\nabla{p}||}_{L^{\infty}(B_{d_{0}}(x_{0}))},\qquad{||{\tilde{f}}||}_{L^{\infty}(B_{1})}\leq{||{f}||}_{L^{\infty}(B_{d_{0}}(x_{0}))}.$

Since $d_{0}=\mbox{dist}(x_{0},F({u}))$, there exists $z_{0}\in F(u)$ such
that $|x_{0}-z_{0}|=d_{0}$ and recalling that $d_{0}<1/8$ we see that
$z_{0}\in B_{5/8}\cap F(u)$.

Also $B_{d_{0}}(x_{0})\subset B_{2d_{0}}(z_{0})\subset B_{\rho_{1}}(z_{0})$.
Then (4.21) yields

${u}(x)\leq C_{0}\mbox{dist}(x,F({u}))\quad\text{ in }B_{d_{0}}(x_{0}).$

Moreover, if $x\in B_{d_{0}}(x_{0})$,

$\mbox{dist}(x,F({u}))\leq|x-z_{0}|<2d_{0}$

and then,

${u}(x)\leq C_{0}\mbox{dist}(x,F({u}))\leq C_{0}2d_{0}\quad\text{ in
}B_{d_{0}}(x_{0})$

which implies

(4.24)
${||{\tilde{u}}||}_{L^{\infty}(B_{1})}=\frac{1}{d_{0}}{||{u}||}_{L^{\infty}(B_{d_{0}}(x_{0}))}\leq
2C_{0}.$

Hence, from Theorem 1.1 in [Fa] we deduce that $\tilde{u}\in
C^{1,\alpha}(\overline{B_{1/2}})$ and
$||\nabla\tilde{u}||_{L^{\infty}(B_{1/2})}\leq C_{1}$. Taking into account
(4.23) and (4.24), we obtain that the constant $C_{1}>0$ can be taken
depending only on $n$, $p_{\min}$, $p_{\max}$,
${||\nabla{p}||}_{L^{\infty}(B_{3/4})}$, $||f||_{L^{\infty}(B_{3/4})}$ and
$C_{0}$. It follows that

$|\nabla u(x_{0})|=|\nabla\tilde{u}(0)|\leq C_{1},$

which proves (4.22).

Step IV. Let $x_{0}\in B^{+}_{1/2}({u})$ such that
$\mbox{dist}(x_{0},F({u}))>{\rho_{1}/2}.$ We will show that

(4.25) $|\nabla u(x_{0})|\leq C_{2},$

for $C_{2}>0$ universal.

In fact, there holds that $B_{\rho_{1}/2}(x_{0})\subset\\{u>0\\}$ and then,

$\Delta_{p(x)}u=f\text{ in }B_{\rho_{1}/2}(x_{0}).$

Now Theorem 1.1 in [Fa] implies that $u\in
C^{1,\alpha}(\overline{B_{\rho_{1}/4}(x_{0})})$ and $||\nabla
u||_{L^{\infty}(B_{\rho_{1}/4}(x_{0}))}\leq C_{2},$ where $C_{2}>0$ is a
constant that can be taken depending only on $n$, $p_{\min}$, $p_{\max}$,
$||u||_{L^{\infty}(B_{3/4})}$, ${||\nabla{p}||}_{L^{\infty}(B_{3/4})}$,
$||f||_{L^{\infty}(B_{3/4})}$ and $\rho_{1}$. This proves (4.25) and completes
the proof. ∎

##  5\. Asymptotic expansions

In this section we revisit some lemmas that are well known in the linear
setting (see [CS] and the Appendix in [C2]), for the case of $p_{0}$-harmonic
functions (i.e., $\Delta_{p_{0}}u=0$, $p_{0}\in(1,\infty)$). Our results —that
are used in Theorem 1.2 and Section 7— concern the existence of first order
expansions at one side regular boundary points of positive Lipschitz
$p_{0}$-harmonic functions, vanishing at the boundary of a domain. The proof
can be applied to a general class of fully nonlinear degenerate elliptic
operators (see Remark 5.2).

For the notion of solution we refer to Definition 2.1 and Remark 2.3. Our
result is the following

######  Lemma 5.1.

Let $1<p_{0}<\infty$ and let $u$ be a positive Lipschitz $p_{0}$-harmonic
function in a domain $\Omega\subset\mathbb{R}^{n}.$ Let
$x_{0}\in\partial\Omega$ and assume that $u$ vanishes continuously on
$\partial\Omega\cap B_{\rho}(x_{0}),$ for some $\rho>0.$

  * (a)

If there exists $B_{r}(y)\subset\Omega$ such that $x_{0}\in\partial B_{r}(y)$,
then

$u(x)=\alpha\langle x-x_{0},\nu\rangle^{+}+o(|x-x_{0}|),$

in the ball $B_{r}(y),$ with $\alpha>0$ and $\nu=\frac{y-x_{0}}{|y-x_{0}|}.$

  * (b)

If there exists a ball $B_{r}(y)\subset\Omega^{c}$ such that $x_{0}\in\partial
B_{r}(y)$, then

$u(x)=\beta\langle x-x_{0},\nu\rangle^{+}+o(|x-x_{0}|),$

with $\beta\geq 0$ and $\nu=\frac{x_{0}-y}{|x_{0}-y|}$. In addition, if
$\beta>0,$ then $B_{r}(y)$ is tangent to $\partial\Omega$ at $x_{0}.$

###### Proof.

We will assume, without loss of generality, that $x_{0}=0$, $\nu=e_{n}$ and
$\rho>1$. We will let $\lambda_{0}:=\min\\{1,p_{0}-1\\}$ and
$\Lambda_{0}:=\max\\{1,p_{0}-1\\}$.

We define

(5.1) $\tilde{u}=\left\\{\begin{array}[]{l}u\quad
x\in\bar{\Omega}\cap\overline{B_{1}},\\\ 0\quad
x\in\bar{\Omega}^{c}\cap\overline{B_{1}}.\\\ \end{array}\right.$

Hence $\tilde{u}$ is Lipschitz in $\overline{B_{1}}$. To simplify the notation
we will denote $\tilde{u}$ as $u.$

Case (a). Let $h$ be the solution of

$\left\\{\begin{array}[]{l}\mathcal{M}_{\lambda_{0},\Lambda_{0}}^{-}(D^{2}h)=0,\quad
B_{r}(y)\setminus\overline{B_{r/2}}(y)\\\ h=0,\quad\mbox{on}\>\>\partial
B_{r}(y)\\\ h=\min_{\overline{B_{r/2}}(y)}{u},\quad\mbox{on}\>\>\partial
B_{r/2}(y).\end{array}\right.$

Let $h\equiv\min_{\overline{B_{r/2}}(y)}u$ in $B_{r/2}(y)$ and $h\equiv 0$ in
$B^{c}_{r}(y)$. Then, $h\geq 0$, $h\in C^{2}(\overline{B_{r}(y)\setminus
B_{r/2}(y)})$, see [CC], and

(5.2) $h(x)=cx_{n}^{+}+o(|x|),\quad c>0.$

In addition, recalling (2.2), we have in $B_{r}(y)$, in the viscosity sense,

$\begin{split}0=\Delta_{p_{0}}{u}(x)&=|\nabla{u}(x)|^{p_{0}-2}\left(\Delta{u}+(p_{0}-2)\langle
D^{2}{u}(x)\frac{\nabla{u}(x)}{|\nabla{u}(x)|},\frac{\nabla{u}(x)}{|\nabla{u}(x)|}\rangle\right)\\\
&\geq|\nabla{u}(x)|^{p_{0}-2}\mathcal{M}^{-}_{\lambda_{0},\Lambda_{0}}(D^{2}u(x)).\end{split}$

Hence, applying Lemma 6 in [IS], we conclude that
$\mathcal{M}^{-}_{\lambda_{0},\Lambda_{0}}(D^{2}u(x))\leq 0$, in the viscosity
sense, in $B_{r}(y).$ Since $u\geq 0$ in $B_{r}(y)$, we deduce that $u\geq h$
in $B_{r}(y).$ We define now

$\alpha_{0}=\sup\\{m:\quad{u}(x)\geq mh(x),\quad B_{1}\cap B_{r}(y)\\}$

and for $k\in\mathbb{N}$

$\alpha_{k}=\sup\\{m:\quad{u}(x)\geq mh(x),\quad B_{2^{-k}}\cap B_{r}(y)\\}.$

In particular these sets are well defined and not empty, since $m=1$ belongs
to all of them. The sequence $\\{\alpha_{k}\\}_{k\in\mathbb{N}}$ is increasing
and bounded because $u$ is Lipschitz. Let

$\tilde{\alpha}=\lim_{k\to\infty}\alpha_{k}.$

From the definition of $\tilde{\alpha}$ there holds that $\tilde{\alpha}>0$
and

(5.3) $\liminf_{x\to 0,\>\>x\in
B_{r}(y)}\frac{u(x)-\tilde{\alpha}h(x)}{|x|}\geq 0.$

Let us show that

(5.4) $\limsup_{x\to 0,\>\>x\in
B_{r}(y)}\frac{u(x)-\tilde{\alpha}h(x)}{|x|}\leq 0.$

Then, (5.2), (5.3) and (5.4) will give the desired result.

We argue by contradiction assuming that

$\limsup_{x\to 0,\>\>x\in
B_{r}(y)}\frac{u(x)-\tilde{\alpha}h(x)}{|x|}=2\delta>0.$

Hence, there exists a sequence $x^{k}\in B_{r}(y),$ $|x^{k}|=r_{k}\to 0$ such
that for every $k$

$\frac{u(x^{k})-\tilde{\alpha}h(x^{k})}{|x^{k}|}\geq\delta.$

We define $r_{k}=|x^{k}|,$ $y^{k}=\frac{x^{k}}{r_{k}},$ so that $r_{k}\to 0$
and $|y^{k}|=1.$ Moreover, we denote

$u_{k}(x):=\frac{u(r_{k}x)}{r_{k}},\quad h_{k}(x):=\frac{h(r_{k}x)}{r_{k}}.$

Since $u$ and $h$ are Lipschitz in $\overline{B}_{1}$ and $u(0)=h(0)=0$ then,
there exists $v$ Lipschitz continuous in $\mathbb{R}^{n}$ such that, for a
subsequence,

${u}_{k}-\tilde{\alpha}h_{k}\to v$

uniformly on compact sets and such that $y^{k}\to y^{0},$ $|y^{0}|=1,$
$y^{0}_{n}\geq 0.$ Since

${u}_{k}(y^{k})-\tilde{\alpha}h_{k}(y^{k})\geq\delta,$

as a consequence $v(y^{0})\geq\delta.$ Then there exists $z^{0}$ with
$|z^{0}|=1$, $z^{0}_{n}>0$ and
$\overline{B_{\varepsilon}(z^{0})}\subset\\{x_{n}>0\\}$ such that

$v(x)\geq\frac{\delta}{2}\qquad\text{ in }B_{\varepsilon}(z^{0})$

and

(5.5)
$\begin{split}{u}_{k}(x)-\tilde{\alpha}h_{k}(x)\geq\frac{\delta}{2}\qquad\text{
in }B_{\varepsilon}(z^{0}).\end{split}$

We know that in $B_{r}(y)\cap B_{2^{-k}}$

$u(x)\geq\alpha_{k}h(x),$

and $r_{k}\to 0.$ We take a sequence $j_{k}\to+\infty$ such that
$r_{k}<2^{-j_{k}}$ and then

${u}(x)\geq\alpha_{j_{k}}h(x)\quad\text{ in }B_{r}(y)\cap B_{2^{-{j_{k}}}}.$

Hence

${u}(x)\geq\alpha_{j_{k}}h(x)\quad\text{ if }|x|<r_{k},\ x\in B_{r}(y).$

As a consequence,

${u}_{k}(x)\geq\alpha_{j_{k}}h_{k}(x)\quad\text{ if }|x|<1,\
|x-\frac{y}{r_{k}}|<\frac{r}{r_{k}},$

and, recalling (5.5), we have
${u}_{k}(x)-\alpha_{j_{k}}h_{k}(x)\geq\frac{\delta}{2}$ in
$B_{\varepsilon}(z^{0}).$ We also observe that

(5.6) $\displaystyle\mathcal{M}_{\lambda_{0},\Lambda_{0}}^{-}(D^{2}h_{k})$
$\displaystyle=0\qquad\mbox{ in
}B_{\frac{r}{r_{k}}}(\frac{y}{r_{k}})\cap\overline{B}_{\frac{r}{2r_{k}}}^{c}(\frac{y}{r_{k}}),$
$\displaystyle\mathcal{M}_{\lambda_{0},\Lambda_{0}}^{-}(D^{2}{u}_{k})$
$\displaystyle\leq 0\qquad\mbox{ in
}B_{1}\cap{B}_{\frac{r}{r_{k}}}(\frac{y}{r_{k}}).$

Hence, in the viscosity sense, if $k$ is large,

$\mathcal{M}_{\lambda_{0},\Lambda_{0}}^{-}(D^{2}({u}_{k}-\alpha_{j_{k}}h_{k}))\leq
0\quad\mbox{in}\>\>\>B_{1}\cap{B}_{\frac{r}{r_{k}}}(\frac{y}{r_{k}}).$

In fact, since $\alpha_{j_{k}}h_{k}\in C^{2},$ we get from (5.6), reasoning as
in Proposition 2.13 in [CC],

$\mathcal{M}_{\lambda_{0},\Lambda_{0}}^{-}(D^{2}({u}_{k}-\alpha_{j_{k}}h_{k}))\leq-\mathcal{M}_{\lambda_{0},\Lambda_{0}}^{-}(D^{2}(\alpha_{j_{k}}h_{k}))=0$

in $B_{1}\cap{B}_{\frac{r}{r_{k}}}(\frac{y}{r_{k}})$. We now consider, for
large $k$, $w_{k}$ satisfying

$\begin{split}\left\\{\begin{array}[]{l}\mathcal{M}_{\lambda_{0},\Lambda_{0}}^{-}(D^{2}w_{k})=0\quad\text{
in }D_{k}:=B_{1}\cap{B}_{\frac{r}{r_{k}}}(\frac{y}{r_{k}}),\\\
w_{k}=\frac{\delta}{4}\varphi\quad\text{ on
}B_{\varepsilon}(z^{0})\cap\partial D_{k},\\\ w_{k}=0\quad\text{ on }\partial
D_{k}\setminus B_{\varepsilon}(z^{0}),\end{array}\right.\end{split}$
$\varphi\in C_{0}^{\infty}(B_{\varepsilon}(z^{0})),\quad 0\leq\varphi\leq
1,\quad\varphi\equiv 1\text{ in }B_{\varepsilon/2}(z^{0}).$

Then $w_{k}\in C(\overline{D_{k}})\cap C^{2}(\overline{{D}_{k}\cap B_{1/2}}),$
$w_{k}\geq 0$ in $\overline{D_{k}}$ and

${u}_{k}-\alpha_{j_{k}}h_{k}\geq w_{k}\quad\mbox{ in }\overline{D_{k}}.$

We claim that there exist $\mu>0$ and $\tilde{\rho}_{0}>0$ such that, for
large $k$,

(5.7) $\frac{w_{k}(x)}{h_{k}(x)}\geq\mu\quad\text{ in
}B_{\tilde{\rho}_{0}}\cap{B}_{\frac{r}{r_{k}}}(\frac{y}{r_{k}}).$

In fact, we consider $\varphi_{k}$ a $C^{2}$ diffeomorphism which maps, for
$\rho_{0}$ small, $B_{\rho_{0}}\cap{B}_{\frac{r}{r_{k}}}(\frac{y}{r_{k}})$ in
$B_{1}^{+}:=B_{1}\cap\\{x_{n}>0\\}$, with
$\varphi_{k}(B_{\rho_{0}}\cap\partial{B}_{\frac{r}{r_{k}}}(\frac{y}{r_{k}}))=B_{1}\cap\\{x_{n}=0\\}$
and $\varphi_{k}(0)=0$. We choose $\varphi_{k}$ with uniformly bounded $C^{2}$
norms. Then, we define

$\tilde{w}_{k}(x)=w_{k}(\varphi_{k}^{-1}(x)),\quad\tilde{h}_{k}(x)=h_{k}(\varphi_{k}^{-1}(x))\quad\text{
for }x\in B_{1}^{+}.$

We first observe that, for every $M,N\in\mathcal{S}^{n\times n}$, the
following inequalities hold (see Lemma 2.10 in [CC])

$\mathcal{M}_{\lambda_{0},\Lambda_{0}}^{+}(M-N)\geq\mathcal{M}_{\lambda_{0},\Lambda_{0}}^{-}(M)-\mathcal{M}_{\lambda_{0},\Lambda_{0}}^{-}(N)\geq\mathcal{M}_{\lambda_{0},\Lambda_{0}}^{-}(M-N).$

Then, we can apply Proposition 2.1 in [SS] with
$F(M):=\mathcal{M}_{\lambda_{0},\Lambda_{0}}^{-}(M)$ and we obtain that

$\tilde{F}_{k}(D^{2}\tilde{w}_{k}(x),D\tilde{w}_{k}(x),x)=0\quad\mbox{in}\>\>\>B_{1}^{+},$

where, for $M\in\mathcal{S}^{n\times n}$, $q\in\mathbb{R}^{n}$ and $x\in
B_{1}^{+}$,

$\tilde{F}_{k}(M,q,x):=\mathcal{M}_{\lambda_{0},\Lambda_{0}}^{-}(D\varphi^{T}_{k}(\varphi^{-1}_{k}(x))MD\varphi_{k}(\varphi^{-1}_{k}(x))+qD^{2}\varphi_{k}(\varphi^{-1}_{k}(x))),$

with $\tilde{F}_{k}$ satisfying for every $M,N\in\mathcal{S}^{n\times n}$,
$p,q\in\mathbb{R}^{n}$ and $x\in B_{1}^{+}$,

$\begin{split}\mathcal{M}_{\lambda_{0},\Lambda_{0}}^{+}(M-N)+K|p-q|&\geq\tilde{F}_{k}(M,p,x)-\tilde{F}_{k}(N,q,x)\\\
&\geq\mathcal{M}_{\lambda_{0},\Lambda_{0}}^{-}(M-N)-K|p-q|.\end{split}$

Here $K$ is a fixed constant depending only on the uniform bound of the
$C^{2}$ norms of $\varphi_{k}.$

As a consequence, $\tilde{w}_{k}$ satisfy in the viscosity sense, the
following set of inequalities

(5.8)
$\begin{split}\left\\{\begin{array}[]{l}\mathcal{M}_{\lambda_{0},\Lambda_{0}}^{+}(D^{2}\tilde{w}_{k})+K|\nabla\tilde{w}_{k}|\geq
0\quad\text{ in }B_{1}^{+},\\\
\mathcal{M}_{\lambda_{0},\Lambda_{0}}^{-}(D^{2}\tilde{w}_{k})-K|\nabla\tilde{w}_{k}|\leq
0\quad\text{ in }B_{1}^{+}.\end{array}\right.\end{split}$

With similar arguments we obtain that $\tilde{h}_{k}$ satisfy in the viscosity
sense the inequalities in (5.8) in $B_{1}^{+}$, as well.

We also notice that, since $h_{k}(x)\to cx_{n}^{+}$ uniformly on compact sets
of $\mathbb{R}^{n},$ with $c>0,$ then,
$\tilde{h}_{k}(\frac{1}{2}e_{n})=h_{k}(\varphi_{k}^{-1}(\frac{1}{2}e_{n}))\to\tilde{c}>0$.

Hence, we can apply Proposition 2.4 of [SS] and we get

(5.9) $\tilde{h}_{k}(x)\leq C\tilde{h}_{k}(\frac{1}{2}e_{n})x_{n}\leq
C_{0}x_{n}\quad\text{ in }B_{1/2}^{+},$

for a positive constant $C_{0}$ and large $k.$

On the other hand, for $k_{1}$ large and fixed, there holds $w_{k}\geq
w_{k_{1}}$ in $D_{k_{1}}$, for $k\geq k_{1}$. We remark that $w_{k_{1}}>0$ in
$D_{k_{1}}.$ Thus, for any $0<r_{0}<1$,

(5.10)
$\tilde{w}_{k}(\frac{r_{0}}{2}e_{n})=w_{k}(\varphi^{-1}_{k}(\frac{r_{0}}{2}e_{n}))\geq
w_{k_{1}}(\varphi^{-1}_{k}(\frac{r_{0}}{2}e_{n}))\to\tilde{c}_{r_{0}}>0.$

Now the application of Proposition 2.5 in [SS] to
$\tilde{w}_{k}^{r_{0}}(x):=\tilde{w}_{k}(r_{0}x)$, for $r_{0}>0$ universal and
small, gives

$\tilde{w}_{k}(x)\geq c_{0}\tilde{w}_{k}(\frac{r_{0}}{2}e_{n})x_{n}\quad\mbox{
in }B^{+}_{\frac{r_{0}}{2}},$

for a positive constant $c_{0}$. Hence, using (5.10) with this choice of
$r_{0}$, we get

(5.11) $\tilde{w}_{k}(x)\geq c_{1}x_{n}\quad\text{ in
}B^{+}_{\frac{r_{0}}{2}},$

for $c_{1}$ a positive constant and large $k$. Thus, from (5.9) and (5.11), we
obtain

$\frac{\tilde{w}_{k}(x)}{\tilde{h}_{k}(x)}\geq\frac{c_{1}}{C_{0}}:=\mu\quad\text{
in }B_{\rho_{1}}^{+},$

for $\rho_{1}>0$ small and large $k.$ Now, going back to the original
variables, we conclude that

$\frac{w_{k}(x)}{h_{k}(x)}\geq\mu\quad\text{ in
}B_{\tilde{\rho}_{0}}\cap{B}_{\frac{r}{r_{k}}}(\frac{y}{r_{k}}),$

for some constants $\tilde{\rho}_{0}>0$ and $\mu>0,$ and large $k.$ That is,
(5.7) holds.

Finally, since

$u_{k}-\alpha_{j_{k}}h_{k}\geq w_{k}\quad\text{ in
}B_{1}\cap{B}_{\frac{r}{r_{k}}}(\frac{y}{r_{k}}),$

we get

$u_{k}-\alpha_{j_{k}}h_{k}\geq w_{k}=\frac{w_{k}}{h_{k}}h_{k}\geq\mu
h_{k}\quad\text{ in
}B_{\tilde{\rho}_{0}}\cap{B}_{\frac{r}{r_{k}}}(\frac{y}{r_{k}}).$

As a consequence,

$u_{k}-(\alpha_{j_{k}}+\mu)h_{k}\geq 0\quad\text{ in
}B_{\tilde{\rho}_{0}}\cap{B}_{\frac{r}{r_{k}}}(\frac{y}{r_{k}}).$

Then, in the original variables, we have

$u(r_{k}x)-(\alpha_{j_{k}}+\mu)h(r_{k}x)\geq 0\quad\text{ when
}|x|\leq\tilde{\rho}_{0},\ |x-\frac{y}{r_{k}}|<\frac{r}{r_{k}},$

or, equivalently, when $|r_{k}x|\leq r_{k}\tilde{\rho}_{0},$ $|r_{k}x-y|<r$.
Since $\alpha_{j_{k}}+\mu\to\tilde{\alpha}+\mu$, there holds that
$\alpha_{j_{k}}+\mu\geq\tilde{\alpha}+{\mu/2},$ if $k$ is large enough. Hence,

$u-(\tilde{\alpha}+\mu/2)h\geq 0\quad\text{ in
}B_{r_{k_{0}}\tilde{\rho}_{0}}\cap B_{r}(y),$

for some suitable $k_{0}$. As a consequence, if $2^{-k}\leq
r_{k_{0}}\tilde{\rho}_{0}$,

$u-(\alpha_{k}+\mu/2)h\geq u-(\tilde{\alpha}+\mu/2)h\geq 0\quad\text{ in
}B_{2^{-k}}\cap B_{r}(y),$

but this contradicts the definition of $\alpha_{k}$ and completes the proof.

Case (b). Recalling (5.1), we have that $\tilde{u}$ is Lipschitz in
$\overline{B_{1}}$, satisfies $\Delta_{p_{0}}\tilde{u}\geq 0$ in $B_{1}$ in
the sense of Definition 2.2 in [JLM] and, by Theorem 2.5 of that paper, in the
viscosity sense. We again denote $\tilde{u}$ as $u$. Without loss of
generality we may suppose that $B_{2r}(y)\subset B_{1}.$

Let $h$ be the solution of

$\left\\{\begin{array}[]{l}\mathcal{M}_{\lambda_{0},\Lambda_{0}}^{+}(D^{2}h)=0,\quad
B_{2r}(y)\setminus\overline{B_{r}}(y)\\\ h=0,\quad\mbox{on}\>\>\partial
B_{r}(y)\\\ h=\max_{\partial B_{2r}(y)}{u},\quad\mbox{on}\>\>\partial
B_{2r}(y),\end{array}\right.$

and define $h\equiv 0$ in $B_{r}(y)$. Then, $h\geq 0$, $h\in
C^{2}(\overline{B_{2r}(y)\setminus B_{r}(y)}),$ see [CC], and

(5.12) $h(x)=cx_{n}^{+}+o(|x|),\quad c>0.$

In addition, recalling (2.2), we have in $B_{1}$, in the viscosity sense,

$\begin{split}0\leq\Delta_{p_{0}}{u}(x)&=|\nabla{u}(x)|^{p_{0}-2}\left(\Delta{u}+(p_{0}-2)\langle
D^{2}{u}(x)\frac{\nabla{u}(x)}{|\nabla{u}(x)|},\frac{\nabla{u}(x)}{|\nabla{u}(x)|}\rangle\right)\\\
&\leq|\nabla{u}(x)|^{p_{0}-2}\mathcal{M}^{+}_{\lambda_{0},\Lambda_{0}}(D^{2}u(x)).\end{split}$

Hence, applying Lemma 6 in [IS], we conclude that
$\mathcal{M}^{+}_{\lambda_{0},\Lambda_{0}}(D^{2}u(x))\geq 0$, in the viscosity
sense, in $B_{1}.$ Since $u=0$ on $\partial B_{r}(y)$, then $u\leq h$ on
$\partial(B_{2r}(y)\setminus\overline{B}_{r}(y))$, thus we deduce that $u\leq
h$ in $B_{2r}(y)\setminus\overline{B}_{r}(y).$ We define now

$\beta_{0}=\inf\\{m:\quad mh(x)\geq{u}(x),\quad B_{1}\cap B_{r}^{c}(y)\\}$

and for $k\in\mathbb{N}$

$\beta_{k}=\inf\\{m:\quad mh(x)\geq{u}(x),\quad B_{2^{-k}}\cap
B_{r}^{c}(y)\\}.$

In particular these sets are well defined and not empty, for $k\geq k_{0}$,
since $m=1$ belongs to all of them. The sequence
$\\{\beta_{k}\\}_{k\in\mathbb{N}}$ is monotone decreasing, so that

$\tilde{\beta}:=\inf_{k\in\mathbb{N}}\beta_{k}\geq 0,$

because $\beta_{k}\geq 0$ for $k\in\mathbb{N}.$ There holds that

(5.13) $\limsup_{x\to 0,\>\>x\in
B_{r}^{c}(y)}\frac{{u}(x)-\tilde{\beta}h(x)}{|x|}\leq 0.$

We will show that

(5.14) $\liminf_{x\to 0,\>\>x\in
B_{r}^{c}(y)}\frac{{u}(x)-\tilde{\beta}h(x)}{|x|}\geq 0.$

Then, (5.12), (5.13) and (5.14) will give the desired result.

We will proceed by contradiction. In fact, assume that there exists $\delta>0$
such that

$\liminf_{x\to 0,\>\>x\in
B_{r}^{c}(y)}\frac{{u}(x)-\tilde{\beta}h(x)}{|x|}=-2\delta.$

Then, there exists a sequence $\\{x^{k}\\}_{k\in\mathbb{N}}\subset
B_{r}^{c}(y)$, $x^{k}\to 0$, such that

$\frac{{u}(x^{k})-\tilde{\beta}h(x^{k})}{|x^{k}|}\leq-\delta.$

We define $r_{k}=|x^{k}|,$ $y^{k}=\frac{x^{k}}{r_{k}},$ so that $r_{k}\to 0$
and $|y^{k}|=1.$ Moreover, we denote

$u_{k}(x):=\frac{u(r_{k}x)}{r_{k}},\quad h_{k}(x):=\frac{h(r_{k}x)}{r_{k}}.$

Since $u$ is Lipschitz in $B_{1}$, $h\in C^{2}(\overline{B_{2r}(y)\setminus
B_{r}(y)})$ and $u(0)=h(0)=0$ then, there exists $v$ Lipschitz continuous in
$\mathbb{R}^{n}$ such that, for a subsequence,

${u}_{k}-\tilde{\beta}h_{k}\to v$

uniformly on compact sets and such that $y^{k}\to y^{0},$ $|y^{0}|=1,$
$y^{0}_{n}\geq 0.$ Since

${u}_{k}(y^{k})-\tilde{\beta}h_{k}(y^{k})\leq-\delta,$

as a consequence $v(y^{0})\leq-\delta.$ Then there exists $z^{0}$ with
$|z^{0}|=1$, $z^{0}_{n}>0$ and
$\overline{B_{\varepsilon}(z^{0})}\subset\\{x_{n}>0\\}$ such that

$v(x)\leq-\frac{\delta}{2}\qquad\text{ in }B_{\varepsilon}(z^{0})$

and

(5.15)
$\begin{split}{u}_{k}(x)-\tilde{\beta}h_{k}(x)\leq-\frac{\delta}{2}\qquad\text{
in }B_{\varepsilon}(z^{0}).\end{split}$

We know that in $B_{r}^{c}(y)\cap B_{2^{-k}}$

$u(x)\leq\beta_{k}h(x),$

and $r_{k}\to 0.$ We take a sequence $j_{k}\to+\infty$ such that
$r_{k}<2^{-j_{k}}$ and then

${u}(x)\leq\beta_{j_{k}}h(x)\quad\text{ in }B_{r}^{c}(y)\cap
B_{2^{-{j_{k}}}}.$

Hence

${u}(x)\leq\beta_{j_{k}}h(x)\quad\text{ if }|x|<r_{k},\ x\in B_{r}^{c}(y).$

As a consequence,

${u}_{k}(x)\leq\beta_{j_{k}}h_{k}(x)\quad\text{ if }|x|<1,\
|x-\frac{y}{r_{k}}|>\frac{r}{r_{k}},$

and, recalling (5.15), we have
${u}_{k}(x)-\beta_{j_{k}}h_{k}(x)\leq-\frac{\delta}{2}$ in
$B_{\varepsilon}(z^{0}).$ We also observe that

(5.16) $\displaystyle\mathcal{M}_{\lambda_{0},\Lambda_{0}}^{+}(D^{2}h_{k})$
$\displaystyle=0\qquad\mbox{ in
}B_{\frac{2r}{r_{k}}}(\frac{y}{r_{k}})\cap\overline{B}_{\frac{r}{r_{k}}}^{c}(\frac{y}{r_{k}}),$
$\displaystyle\mathcal{M}_{\lambda_{0},\Lambda_{0}}^{+}(D^{2}{u}_{k})$
$\displaystyle\geq 0\qquad\mbox{ in
}B_{1}\cap\overline{B}_{\frac{r}{r_{k}}}^{c}(\frac{y}{r_{k}}).$

Hence, in the viscosity sense,

$\mathcal{M}_{\lambda_{0},\Lambda_{0}}^{+}(D^{2}({u}_{k}-\beta_{j_{k}}h_{k}))\geq
0\quad\mbox{in}\>\>\>B_{1}\cap\overline{B}_{\frac{r}{r_{k}}}^{c}(\frac{y}{r_{k}}).$

In fact, since $\beta_{j_{k}}h_{k}\in C^{2},$ we get from (5.16), reasoning as
in Proposition 2.13 in [CC],

$\mathcal{M}_{\lambda_{0},\Lambda_{0}}^{+}(D^{2}({u}_{k}-\beta_{j_{k}}h_{k}))\geq-\mathcal{M}_{\lambda_{0},\Lambda_{0}}^{+}(D^{2}(\beta_{j_{k}}h_{k}))=0$

in $B_{1}\cap\overline{B}_{\frac{r}{r_{k}}}^{c}(\frac{y}{r_{k}})$. Thus, we
deduce that

$\mathcal{M}_{\lambda_{0},\Lambda_{0}}^{-}(D^{2}(\beta_{j_{k}}h_{k}-{u}_{k}))\leq
0\quad\mbox{ in
}B_{1}\cap\overline{B}_{\frac{r}{r_{k}}}^{c}(\frac{y}{r_{k}}).$

We now consider $w_{k}$ satisfying

$\begin{split}\left\\{\begin{array}[]{l}\mathcal{M}_{\lambda_{0},\Lambda_{0}}^{-}(D^{2}w_{k})=0\quad\text{
in }D_{k}:=B_{1}\cap\overline{B}_{\frac{r}{r_{k}}}^{c}(\frac{y}{r_{k}}),\\\
w_{k}=\frac{\delta}{4}\varphi\quad\text{ on
}B_{\varepsilon}(z^{0})\cap\partial D_{k},\\\ w_{k}=0\quad\text{ on }\partial
D_{k}\setminus B_{\varepsilon}(z^{0}),\end{array}\right.\end{split}$ (5.17)
$\varphi\in C_{0}^{\infty}(B_{\varepsilon}(z^{0})),\quad 0\leq\varphi\leq
1,\quad\varphi\equiv 1\text{ in }B_{\varepsilon/2}(z^{0}).$

Then $w_{k}\in C(\overline{D_{k}})\cap C^{2}(\overline{{D}_{k}\cap B_{1/2}}),$
$w_{k}\geq 0$ in $\overline{D_{k}}$ and

$\beta_{j_{k}}h_{k}-{u}_{k}\geq w_{k}\quad\mbox{ in }\overline{D_{k}}.$

We claim that there exist $\mu>0$ and $\tilde{\rho}_{0}>0$ such that, for
large $k$,

(5.18) $\frac{w_{k}(x)}{h_{k}(x)}\geq\mu\quad\text{ in
}B_{\tilde{\rho}_{0}}\cap\overline{B}_{\frac{r}{r_{k}}}^{c}(\frac{y}{r_{k}}).$

In fact, we consider $\varphi_{k}$ a $C^{2}$ diffeomorphism which maps, for
$\rho_{0}$ small,
$B_{\rho_{0}}\cap\overline{B}_{\frac{r}{r_{k}}}^{c}(\frac{y}{r_{k}})$ in
$B_{1}^{+}:=B_{1}\cap\\{x_{n}>0\\}$, with
$\varphi_{k}(B_{\rho_{0}}\cap\partial{B}_{\frac{r}{r_{k}}}(\frac{y}{r_{k}}))=B_{1}\cap\\{x_{n}=0\\}$
and $\varphi_{k}(0)=0$. We choose $\varphi_{k}$ with uniformly bounded $C^{2}$
norms. Then, we define

$\tilde{w}_{k}(x)=w_{k}(\varphi_{k}^{-1}(x)),\quad\tilde{h}_{k}(x)=h_{k}(\varphi_{k}^{-1}(x))\quad\text{
for }x\in B_{1}^{+}.$

Reasoning as in Case a), we get that $\tilde{w}_{k}$ satisfy in the viscosity
sense, the following set of inequalities

(5.19)
$\begin{split}\left\\{\begin{array}[]{l}\mathcal{M}_{\lambda_{0},\Lambda_{0}}^{+}(D^{2}\tilde{w}_{k})+K|\nabla\tilde{w}_{k}|\geq
0\quad\text{ in }B_{1}^{+},\\\
\mathcal{M}_{\lambda_{0},\Lambda_{0}}^{-}(D^{2}\tilde{w}_{k})-K|\nabla\tilde{w}_{k}|\leq
0\quad\text{ in }B_{1}^{+},\end{array}\right.\end{split}$

where $K$ is a fixed constant depending only on the uniform bound of the
$C^{2}$ norms of $\varphi_{k}.$ With similar arguments we obtain that
$\tilde{h}_{k}$ satisfy in the viscosity sense the inequalities in (5.19) in
$B_{1}^{+}$, as well.

We also notice that, since $h_{k}(x)\to cx_{n}^{+}$ uniformly on compact sets
of $\mathbb{R}^{n},$ with $c>0,$ then,
$\tilde{h}_{k}(\frac{1}{2}e_{n})=h_{k}(\varphi_{k}^{-1}(\frac{1}{2}e_{n}))\to\tilde{c}>0$.

Hence, we can apply Proposition 2.4 of [SS] and we get

(5.20) $\tilde{h}_{k}(x)\leq C\tilde{h}_{k}(\frac{1}{2}e_{n})x_{n}\leq
C_{0}x_{n}\quad\text{ in }B_{1/2}^{+},$

for a positive constant $C_{0}$ and large $k.$

On the other hand, let $w_{0}$ satisfying

$\begin{split}\left\\{\begin{array}[]{l}\mathcal{M}_{\lambda_{0},\Lambda_{0}}^{-}(D^{2}w_{0})=0\quad\text{
in }B_{1}^{+},\\\ w_{0}=\frac{\delta}{4}\varphi\quad\text{ on
}B_{\varepsilon}(z^{0})\cap\partial B_{1}^{+},\\\ w_{0}=0\quad\text{ on
}\partial B_{1}^{+}\setminus
B_{\varepsilon}(z^{0}),\end{array}\right.\end{split}$

with $\varphi$ as in (5.17). Then $w_{k}\geq w_{0}$ in $B_{1}^{+}.$ We remark
that $w_{0}>0$ in $B_{1}^{+}.$ Thus, for any $0<r_{0}<1$,

(5.21)
$\tilde{w}_{k}(\frac{r_{0}}{2}e_{n})=w_{k}(\varphi^{-1}_{k}(\frac{r_{0}}{2}e_{n}))\geq
w_{0}(\varphi^{-1}_{k}(\frac{r_{0}}{2}e_{n}))\to\tilde{c}_{r_{0}}>0.$

Now the application of Proposition 2.5 in [SS] to
$\tilde{w}_{k}^{r_{0}}(x):=\tilde{w}_{k}(r_{0}x)$, for $r_{0}>0$ universal and
small, gives

$\tilde{w}_{k}(x)\geq c_{0}\tilde{w}_{k}(\frac{r_{0}}{2}e_{n})x_{n}\quad\mbox{
in }B^{+}_{\frac{r_{0}}{2}},$

for a positive constant $c_{0}$. Hence, using (5.21) with this choice of
$r_{0}$, we get

(5.22) $\tilde{w}_{k}(x)\geq c_{1}x_{n}\quad\text{ in
}B^{+}_{\frac{r_{0}}{2}},$

for $c_{1}$ a positive constant and large $k$. Thus, from (5.20) and (5.22),
we obtain

$\frac{\tilde{w}_{k}(x)}{\tilde{h}_{k}(x)}\geq\frac{c_{1}}{C_{0}}:=\mu\quad\text{
in }B_{\rho_{1}}^{+},$

for $\rho_{1}>0$ small and large $k.$ Now, going back to the original
variables, we conclude that

$\frac{w_{k}(x)}{h_{k}(x)}\geq\mu\quad\text{ in
}B_{\tilde{\rho}_{0}}\cap\overline{B}_{\frac{r}{r_{k}}}^{c}(\frac{y}{r_{k}}),$

for some constants $\tilde{\rho}_{0}>0$ and $\mu>0,$ and large $k.$ That is,
(5.18) holds.

Finally, since

$\beta_{j_{k}}h_{k}-u_{k}\geq w_{k}\quad\text{ in
}B_{1}\cap\overline{B}_{\frac{r}{r_{k}}}^{c}(\frac{y}{r_{k}}),$

we get

$\beta_{j_{k}}h_{k}-u_{k}\geq w_{k}=\frac{w_{k}}{h_{k}}h_{k}\geq\mu
h_{k}\quad\text{ in
}B_{\tilde{\rho}_{0}}\cap\overline{B}_{\frac{r}{r_{k}}}^{c}(\frac{y}{r_{k}}).$

As a consequence,

$(\beta_{j_{k}}-\mu)h_{k}-u_{k}\geq 0\quad\text{ in
}B_{\tilde{\rho}_{0}}\cap\overline{B}_{\frac{r}{r_{k}}}^{c}(\frac{y}{r_{k}}).$

Then, in the original variables, we have

$(\beta_{j_{k}}-\mu)h(r_{k}x)-u(r_{k}x)\geq 0\quad\text{ when
}|x|\leq\tilde{\rho}_{0},\ |x-\frac{y}{r_{k}}|>\frac{r}{r_{k}},$

or, equivalently, when $|r_{k}x|\leq r_{k}\tilde{\rho}_{0},$ $|r_{k}x-y|>r$.
Since $\beta_{j_{k}}-\mu\to\tilde{\beta}-\mu$, there holds that
$\beta_{j_{k}}-\mu\leq\tilde{\beta}-\frac{\mu}{2},$ if $k$ is large enough.
Hence,

$(\tilde{\beta}-\mu/2)h-u\geq 0\quad\text{ in
}B_{r_{k_{0}}\tilde{\rho}_{0}}\cap B_{r}^{c}(y),$

for some suitable $k_{0}$. As a consequence, if $2^{-k}\leq
r_{k_{0}}\tilde{\rho}_{0}$,

$(\beta_{k}-\mu/2)h-u\geq(\tilde{\beta}-\mu/2)h-u\geq 0\quad\text{ in
}B_{2^{-k}}\cap B_{r}^{c}(y),$

but this contradicts the definition of $\beta_{k}$ and completes the proof. ∎

######  Remark 5.2.

Lemma 5.1 also holds if we replace in the statement the $p_{0}$-Laplace
operator by a general class of fully nonlinear degenerate elliptic operators.
More precisely, we can consider $u$ a Lipschitz viscosity solution of an
equation of the form

$F(D^{2}u(x),Du(x),x)=0\quad\text{ in }\Omega,$

with ${F}$ satisfying for every $M\in\mathcal{S}^{n\times n}$,
$q\in\mathbb{R}^{n}$ and $x\in\Omega$,

$|q|^{\sigma}\mathcal{M}_{\lambda,\Lambda}^{-}(M)\leq{F}(M,q,x)\leq|q|^{\sigma}\mathcal{M}_{\lambda,\Lambda}^{+}(M),$

for some $0<\lambda\leq\Lambda$ and $\sigma\in\mathbb{R}$, and the same proof
applies.

##  6\. Regularity of the free boundary

In this section we prove our main result, namely, Theorem 1.2.

Since we will apply a result of [LN1], we include first the definition of
viscosity solution employed in that paper in case of nonnegative solutions.
These are solutions of problem (1.1) with $p(x)\equiv p_{0}$, $f\equiv 0$ and
$g\equiv 1$.

######  Definition 6.1 (Definition 1.4 in [LN1]).

Let $D\subset\mathbb{R}^{n}$ be a domain, $u\in C(D)$ be nonnegative and
$1<p_{0}<\infty$. $u$ is a viscosity (or weak) solution of

(6.1) $\left\\{\begin{array}[]{ll}\Delta_{p_{0}}u=0&\hbox{in
$D^{+}(u):=\\{x\in D:u(x)>0\\}$},\\\ &\\\ |\nabla u|=1&\hbox{on
$F(u):=\partial D^{+}(u)\cap D,$}\end{array}\right.$

if there holds that $u$ is $p_{0}$-harmonic in $D^{+}(u)$, in the sense that
$u\in W^{1,p_{0}}(D^{+}(u))$ and

$\int_{D^{+}(u)}|\nabla u|^{p_{0}-2}\nabla
u\cdot\nabla\varphi\,dx=0\quad\text{for every }\varphi\in
W_{0}^{1,p_{0}}(D^{+}(u)),$

and the free boundary condition in (6.1) is satisfied in the following sense.
Assume that $x_{0}\in F(u)$ and there exists a ball $B_{r}(y)\subset D$, with
$x_{0}\in\partial B_{r}(y)$. If $\nu=\frac{y-x_{0}}{|y-x_{0}|}$, then the
following holds, as $x\to x_{0}$ non-tangentially, for $\alpha=1$,

  * (i)

if $B_{r}(y)\subset D^{+}(u)$, then $u(x)=\alpha\langle
x-x_{0},\nu\rangle^{+}+o(|x-x_{0}|)$,

  * (ii)

if $B_{r}(y)\subset D^{+}(u)^{c}$, then $u(x)=\alpha\langle
x_{0}-x,\nu\rangle^{+}+o(|x-x_{0}|)$.

We next extend the result of Lemma 6.2 in [DFS1] to the global homogenous
$p_{0}$-Laplacian free boundary problem (i.e., to problem (1.1) in
$\Omega=\mathbb{R}^{n}$ with $p(x)\equiv p_{0}$, $f\equiv 0$ and $g\equiv 1$).
This result is valid for globally Lipschitz continuous functions. The notion
of viscosity solution we employ in Lemma 6.2 is the one in [LN1] (see
Definition 6.1 above).

######  Lemma 6.2.

Let $1<p_{0}<\infty$. Let $v:\mathbb{R}^{n}\to\mathbb{R}$ be a nonnegative
Lipschitz viscosity solution (in the sense of Definition 6.1) to

(6.2) $\left\\{\begin{array}[]{ll}\Delta_{p_{0}}v=0,&\hbox{in $\\{v>0\\}$},\\\
&\\\ |\nabla v|=1,&\mbox{on}\quad F(v):=\partial\\{v>0\\}.\end{array}\right.$

Assume that

$\\{v>0\\}=\\{(x^{\prime},x_{n})\in\mathbb{R}^{n}:\quad
x^{\prime}\in\mathbb{R}^{n-1},\quad x_{n}>h(x^{\prime})\\},$

with $h$ a Lipschitz continuous function, $h(0)=0$ and $\mbox{Lip}(h)\leq M.$
Then $h$ is linear and, after a rotation,

$v(x)=x_{n}^{+}.$

###### Proof.

We will denote $B^{\prime}_{r}$ the ball of radius $r$ centered at $0$ in
$\mathbb{R}^{n-1}$.

We follow the idea of the proof in Lemma 6.2 in [DFS1], coupled with results
about the regularity of the free boundary in the homogeneous two phase problem
associated with the $p_{0}$-Laplace operator. In fact, if $v$ is a viscosity
solution of (6.2) and its free boundary is a Lipschitz graph, from the
regularity results in [LN1], we know that the free boundary $F(v)$ is
$C^{1,\alpha}$ in $B_{1},$ with a bound $C$ depending only on $n,p_{0}$ and on
the Lipschitz constant $M$ of $h.$ Then,

(6.3) $|h(x^{\prime})-h(0)-\langle\nabla h(0),x^{\prime}\rangle|\leq
C|x^{\prime}|^{1+\alpha}$

in $B^{\prime}_{1},$ where $C=C(n,p_{0},M).$ Moreover, since $v$ is a global
solution to (6.2), considering the rescaled function
$v_{R}(x)=\frac{v(Rx)}{R},$ we still obtain a solution to problem (6.2) whose
free boundary is the graph of the function
$h_{R}(x^{\prime})=\frac{h(Rx^{\prime})}{R}$ for
$x^{\prime}\in\mathbb{R}^{n-1}.$ This function preserves the same Lipschitz
constant and then satisfies the inequality (6.3). That is,

$|h_{R}(x^{\prime})-h_{R}(0)-\langle\nabla h_{R}(0),x^{\prime}\rangle|\leq
C|x^{\prime}|^{1+\alpha}$

for $x^{\prime}\in B^{\prime}_{1}.$ This fact can be read as

$|h(Rx^{\prime})-h(0)-\langle\nabla h(0),Rx^{\prime}\rangle|\leq
CR|x^{\prime}|^{1+\alpha}$

for $x^{\prime}\in B^{\prime}_{1}.$ Then,

$|h(y^{\prime})-h(0)-\langle\nabla h(0),y^{\prime}\rangle|\leq
C\frac{|y^{\prime}|^{1+\alpha}}{R^{\alpha}}$

for $y^{\prime}\in B^{\prime}_{R}.$ Hence, passing to the limit $R\to\infty,$
we conclude that $h$ is linear in $\mathbb{R}^{n-1}$. Since $v$ is Lipschitz,
then Lemma B.1 in Appendix B applies and, up to a proper rotation,
$v(x)=x_{n}^{+}.$ ∎

For the sake of completeness we recall the following theorem we proved in [FL]

######  Theorem 6.3 (Theorem 1.1 in [FL]).

Let $u$ be a viscosity solution to (1.1) in $B_{1}$. Assume that $0\in F(u),$
$g(0)=1$ and $p(0)=p_{0}.$ There exists a universal constant
$\bar{\varepsilon}>0$ such that, if the graph of $u$ is
$\bar{\varepsilon}-$flat in $B_{1},$ in the direction $e_{n},$ that is

$(x_{n}-\bar{\varepsilon})^{+}\leq u(x)\leq(x_{n}+\bar{\varepsilon})^{+},\quad
x\in B_{1},$

and

(6.4) $\|\nabla
p\|_{L^{\infty}(B_{1})}\leq\bar{\varepsilon},\quad\|f\|_{L^{\infty}(B_{1})}\leq\bar{\varepsilon},\quad[g]_{C^{0,\beta}(B_{1})}\leq\bar{\varepsilon},$

then $F(u)$ is $C^{1,\alpha}$ in $B_{1/2}$.

The constants $\bar{\varepsilon}$ and $\alpha$ depend only on $p_{\min}$,
$p_{\max}$, $\beta$ and $n$.

In the proof of Theorem 1.2 we will also use

######  Proposition 6.4.

Let $u_{k}$ be a sequence of viscosity solutions to (1.1) in $B_{2}$, with
right hand side $f_{k}$, exponent $p_{k}$ and free boundary condition $g_{k}$,
where $f_{k}$, $p_{k}$ and $g_{k}$ are as in Subsection 1.1. Assume that
$u_{k}$ are uniformly Lipschitz and that, for some $\alpha>0$ and
$\nu\in\mathbb{R}^{n}$ with $|\nu|=1$, $u_{k}\to u_{0}(x)=\alpha\langle
x,\nu\rangle^{+}$, $f_{k}\to 0$, $p_{k}\to p_{0}$, $\nabla p_{k}\to 0$ and
$g_{k}\to 1$ uniformly in $B_{2}$. Assume moreover that $F(u_{k})$ are uniform
Lipschitz graphs and $F(u_{k})\to F(u_{0})$ in Hausdorff distance in $B_{2}$.
Then $\alpha\geq 1$.

###### Proof.

Without loss of generality we assume that $\nu=e_{n}$. Suppose by
contradiction that $0<\alpha<1$. We take $\varphi\in
C^{\infty}(\mathbb{R}^{n}),$ with $0\leq\varphi\leq 1,$ $\varphi\equiv 0$ in
${B^{c}_{1/2}}$ and $\varphi\equiv 1$ in $B_{1/4}.$ For $0<\xi<1/4$ depending
on $\alpha$, to be fixed later, and $0<\varepsilon<1,$ we define

$D_{\varepsilon}=D_{\varepsilon}^{\xi}=B_{1}\cap\\{x_{n}>-\xi+\varepsilon\varphi(x)\\}.$

Let $\lambda_{0},\Lambda_{0}$ as in (2.2). For $\rho>0$ fixed and depending on
$\alpha,$ to be precised later, we consider $v_{\varepsilon}$ such that

$\begin{split}\left\\{\begin{array}[]{l}\mathcal{M}^{+}_{\lambda_{0},\Lambda_{0}}(D^{2}v_{\varepsilon})=-\rho,\quad\mbox{in}\>\>D_{\varepsilon}\\\
v_{\varepsilon}=\alpha(x_{n}+\xi),\quad\mbox{on}\>\>\partial
B_{1}\cap\\{x_{n}\geq-\xi\\},\\\
v_{\varepsilon}=0,\quad\mbox{on}\>\>B_{1}\cap\\{x_{n}=-\xi+\varepsilon\varphi(x)\\},\end{array}\right.\end{split}$

$v_{\varepsilon}\equiv 0$ on
$\overline{B_{1}}\setminus\overline{D_{\varepsilon}}$.

Step I.  We will show that, in $B_{3/4}$, $v_{\varepsilon}$ is a strict
supersolution to problem (1.1) with right hand side $f_{k}$, exponent $p_{k}$
and free boundary condition $g_{k}$, for $\rho$, $\varepsilon$ and $\xi$
suitably chosen and large $k$.

We first observe that $v_{\varepsilon}>0$ in $D_{\varepsilon}.$ In addition,
$v_{\varepsilon}\in C^{2,{\tilde{\alpha}}}(D_{\varepsilon})$ and
$v_{\varepsilon}\in C^{1,{\tilde{\alpha}}}(\overline{D_{\varepsilon}\cap
B_{3/4}})$. We define

$w_{\varepsilon}=v_{\varepsilon}-\alpha(x_{n}+\xi),$

that satisfies

$\begin{split}\left\\{\begin{array}[]{l}\mathcal{M}^{+}_{\lambda_{0},\Lambda_{0}}(D^{2}w_{\varepsilon})=-\rho,\quad\mbox{in}\>\>D_{\varepsilon}\\\
w_{\varepsilon}=0,\quad\mbox{on}\>\>\partial B_{1}\cap\\{x_{n}\geq-\xi\\},\\\
w_{\varepsilon}=-\alpha(x_{n}+\xi)=-\alpha\varepsilon\varphi(x),\quad\mbox{on}\>\>B_{1}\cap\\{x_{n}=-\xi+\varepsilon\varphi(x)\\}.\\\
\end{array}\right.\end{split}$

Hence, using ABP estimate (Theorem 3.6 in [CC]), we obtain that
$||w_{\varepsilon}||_{L^{\infty}(D_{\varepsilon})}\leq
C_{0}(\rho+\varepsilon)$, with $C_{0}>0$ independent of $\varepsilon$ and
$\xi$ universal. Then from the inner estimates in Corollary 5.7, [CC] and the
boundary estimates in Theorem 1.4, [SS] we deduce that there exist positive
constants $C_{1}$ and ${C_{2}}$ such that

(6.5)
$\begin{split}||w_{\varepsilon}||_{C^{1,{\tilde{\alpha}}}(\overline{D_{\varepsilon}\cap
B_{3/4}})}&\leq
C_{1}\left(||w_{\varepsilon}||_{{L^{\infty}}(D_{\varepsilon})}+\rho+\alpha\varepsilon||\varphi||_{C^{1,{\tilde{\alpha}}}(B_{1})}\right)\leq{C_{2}}(\rho+\varepsilon),\end{split}$

where $C_{1}$ and ${C_{2}}$ depend on the maximal curvature of
$\\{x_{n}=-\xi+\varepsilon\varphi(x)\\},$ see [SS], and can be chosen
universal independent of $\varepsilon$ and $\xi$. Then

$||\nabla w_{\varepsilon}||_{L^{\infty}(\overline{D_{\varepsilon}\cap
B_{3/4}})}\leq{C_{2}}(\rho+\varepsilon)$

and

$||\nabla v_{\varepsilon}|-\alpha|\leq|\nabla v_{\varepsilon}-\alpha
e_{n}|\leq{C_{2}}(\rho+\varepsilon)\qquad\mbox{ in
}\overline{D_{\varepsilon}\cap B_{3/4}}.$

We now fix

(6.6) $\begin{array}[]{l}\rho=\frac{c(\alpha)}{{C_{2}}},\qquad
0<\varepsilon\leq\min\\{\frac{c(\alpha)}{{C_{2}}},\frac{1}{4}\\},\qquad
2\xi=\min\\{\frac{c(\alpha)}{{C_{2}}},\frac{1}{4}\\},\\\ \\\ \text{with }\quad
c(\alpha)=\frac{1}{2}\min\\{\frac{1-\alpha}{2},\frac{\alpha}{2}\\}\quad\text{and
}C_{2}\quad\text{as in}\quad\eqref{wep-c1alpha},\\\ \end{array}$

and then,

(6.7) $\frac{\alpha}{2}\leq|\nabla
v_{\varepsilon}|\leq\frac{1+\alpha}{2}\qquad\mbox{ in
}\overline{D_{\varepsilon}\cap B_{3/4}}.$

Recalling (2.2), we obtain, in $D_{\varepsilon}\cap B_{3/4},$

$\Delta_{p_{k}(x)}v_{\varepsilon}\leq I+II$

where

$\begin{split}I=|\nabla
v_{\varepsilon}|^{p_{k}(x)-2}\mathcal{M}^{+}_{\lambda_{0},\Lambda_{0}}(D^{2}v_{\varepsilon})\end{split}$

and

$\begin{split}II=|\nabla v_{\varepsilon}|^{p_{k}(x)-2}\langle\nabla
p_{k}(x),\nabla v_{\varepsilon}\rangle\log|\nabla
v_{\varepsilon}|.\end{split}$

Then

$\begin{split}I=|\nabla v_{\varepsilon}|^{p_{k}(x)-2}(-\rho)\leq-
c_{1}\rho,\end{split}$

since $|\nabla v_{\varepsilon}|^{p_{k}(x)-2}\geq c_{1}=c_{1}(\alpha,p_{\max})$
because of (6.7). Moreover,

$\begin{split}II\leq|\nabla v_{\varepsilon}|^{p_{k}(x)-1}|\nabla
p_{k}(x)|\left|\log|\nabla v_{\varepsilon}|\right|\leq|\nabla
p_{k}(x)|c_{2},\end{split}$

where we have used that $|t|^{p_{k}(x)-1}\left|\log|t|\right|\leq
c_{2}=c_{2}(p_{\min})$ for $|t|\leq 1$ and (6.7). Then,

(6.8) $\begin{split}\Delta_{p_{k}(x)}v_{\varepsilon}\leq-c_{1}\rho+{||\nabla
p_{k}||}_{L^{\infty}}c_{2}\leq-
c_{1}\rho+c_{2}\frac{c_{1}}{c_{2}}\frac{\rho}{2}=-\frac{c_{1}}{2}\rho\end{split}$

if $k$ is large so that ${||\nabla
p_{k}||}_{L^{\infty}}\leq\frac{c_{1}}{c_{2}}\frac{\rho}{2}.$

On the other hand, for $k$ large, we have
${||f_{k}||}_{L^{\infty}}<\frac{c_{1}}{2}\rho$ and
${g_{k}(x)}>\frac{1+\alpha}{2}$ in $B_{1}$ and then,

(6.9)
$\left\\{\begin{array}[]{l}\Delta_{p_{k}(x)}v_{\varepsilon}\leq-\frac{c_{1}}{2}\rho<-{||f_{k}||}_{L^{\infty}}\leq
f_{k},\quad\mbox{in}\>\>D_{\varepsilon}\cap B_{3/4}\\\ |\nabla
v_{\varepsilon}|\geq\frac{\alpha}{2},\quad\mbox{in}\>\>\overline{D_{\varepsilon}\cap
B_{3/4}}\quad(\mbox{this implies}\>\>\>\nabla v_{\varepsilon}\not=0)\\\
|\nabla
v_{\varepsilon}|\leq\frac{1+\alpha}{2}<g_{k},\quad\mbox{in}\>\>\overline{D_{\varepsilon}\cap
B_{3/4}}.\end{array}\right.$

Hence, for our choice of $\rho$, $\varepsilon$ and $\xi$ done in (6.6),
$v_{\varepsilon}$ is a strict supersolution to problem (1.1) in $B_{3/4}$ with
right hand side $f_{k}$, exponent $p_{k}$ and free boundary condition $g_{k}$,
for large $k$, as claimed.

Step II.  We will now get some uniform bounds for the functions $u_{k}$. In
fact, let $v$ be such that

$\begin{split}\left\\{\begin{array}[]{l}\mathcal{M}^{+}_{\lambda_{0},\Lambda_{0}}(D^{2}v)=-\rho,\quad\mbox{in}\>\>\\{x_{n}>-{\xi},\
5/8<|x|<1\\},\\\ v=\alpha(x_{n}+\xi),\quad\mbox{on}\>\>\partial
B_{1}\cap\\{x_{n}\geq-\xi\\},\\\ v=0,\quad\mbox{on}\>\>\\{x_{n}=-\xi,\
5/8<|x|<1\\},\\\ v=0,\quad\mbox{on}\>\>\partial
B_{5/8}\cap\\{x_{n}\geq-\xi\\}.\\\ \end{array}\right.\end{split}$

Then, $v_{\varepsilon}>v>0$ in $\\{x_{n}>-{\xi},\ 5/8<|x|<1\\}$. Moreover,
there exists $c_{3}>0$ universal such that

(6.10) $v>c_{3}\quad\text{ in }\\{x_{n}\geq-\frac{\xi}{2},\ 3/4\leq|x|\leq
1\\}.$

Let us now fix $\delta$ universal such that

(6.11) $0<\delta<\frac{\xi}{2}\quad\text{ and }\quad 2L\delta<c_{3},$

where $L$ is the uniform Lipschitz constant of the functions $u_{k}$.

We will first show that, if $k$ is large,

(6.12) $u_{k}=0\qquad\text{ in }\overline{B_{1}}\cap\\{x_{n}\leq-\delta\\},$
(6.13) $u_{k}\leq 2L\delta\qquad\text{ in
}\overline{B_{1}}\cap\\{|x_{n}|\leq\delta\\},$ (6.14)
$\frac{\alpha}{2}\leq|\nabla u_{k}|\leq L\qquad\text{ in
}\overline{B_{1}}\cap\\{x_{n}\geq\delta\\}.$

In fact, since $\partial\\{u_{k}>0\\}\to\partial\\{u_{0}>0\\}=\\{x_{n}=0\\}$
in the Hausdorff distance in $B_{2}$, then
$\partial\\{u_{k}>0\\}\subset\\{|x_{n}|<\delta\\}$ in $B_{2}$, if $k$ is
large. Hence (6.12) and (6.13) follow.

Since $u_{k}\to u_{0}=\alpha x_{n}^{+}$ uniformly in $B_{2}$, then for large
$k$,

(6.15) $\left\\{\begin{array}[]{l}u_{k}\geq\alpha\frac{\delta}{4}\quad\text{in
}B_{2}\cap\\{x_{n}>\frac{\delta}{2}\\},\\\
\Delta_{p_{k}(x)}u_{k}=f_{k}\quad\text{in
}B_{2}\cap\\{x_{n}>\frac{\delta}{2}\\},\\\ \end{array}\right.$

and then, by the $C^{1,\bar{\alpha}}$ estimates (Theorem 1.1 in [Fa]),

$\nabla u_{k}\to\alpha e_{n}\quad\text{ uniformly in
}\overline{B_{1}}\cap\\{x_{n}\geq{\delta}\\},$

which gives (6.14) for $k$ large.

We now observe that $v-u_{0}\geq\frac{\alpha}{2}\xi$ on $\partial
B_{1}\cap\\{x_{n}\geq-\frac{\xi}{2}\\}$ and

$v-u_{0}\geq\frac{\alpha\xi}{4}\quad\text{ in }\\{x_{n}\geq-\frac{\xi}{2},\
1-\sigma\leq|x|\leq 1\\},$

for some universal $0<\sigma<1/4$. Here we have used that $v\in
C^{\bar{\alpha}}(\\{x_{n}\geq-{\xi},\ 5/8\leq|x|\leq 1\\})$ (see, for
instance, Theorem 2 in [Si]). Then,

$v-u_{k}\geq\frac{\alpha\xi}{8}\quad\text{ in }\\{x_{n}\geq-\frac{\xi}{2},\
1-\sigma\leq|x|\leq 1\\},$

for large $k$. Recalling (6.10), (6.11) and (6.13), we obtain

(6.16) $v_{\varepsilon}>u_{k}\qquad\text{ in }\\{|x_{n}|\leq\delta,\
3/4\leq|x|\leq 1\\},$ (6.17)
$v_{\varepsilon}-u_{k}\geq\frac{\alpha\xi}{8}\quad\text{ in
}\\{x_{n}\geq-\frac{\xi}{2},\ 1-\sigma\leq|x|\leq 1\\}.$

Step III.  We will show that

(6.18) $v_{\varepsilon}\geq u_{k}\quad\text{in }\overline{B_{1}},\quad\text{
for every }0<\varepsilon<\frac{\xi}{2},$

if $k$ is large.

If the result is not true, then

$\max_{\overline{B_{1}}}(u_{k}-v_{\varepsilon})=(u_{k}-v_{\varepsilon})(\tilde{x}_{k})>0\quad\text{
for some }\tilde{x}_{k}\in\overline{B_{1}}.$

If $|\tilde{x}_{k}|\geq 3/4$, then (6.12), (6.16) and (6.17) imply that

$\tilde{x}_{k}\in\\{x_{n}>\delta,\ 3/4\leq|x|\leq 1-\sigma\\}.$

From (6.14) and (6.15) we get

$\frac{\alpha}{2}\leq|\nabla v_{\varepsilon}(\tilde{x}_{k})|\leq L$

and

(6.19) $\Delta_{p_{k}(\tilde{x}_{k})}v_{\varepsilon}(\tilde{x}_{k})\geq
f_{k}(\tilde{x}_{k}).$

Now, the uniform $C^{1,\bar{\alpha}}$ estimates for $v_{\varepsilon}$ in
$\overline{B_{1}}\cap\\{x_{n}\geq 0\\}$ give

$\frac{\alpha}{4}\leq|\nabla v_{\varepsilon}|\leq 2L\quad\text{ in
}B_{\mu}(\tilde{x}_{k}),$

for some $\mu>0$ universal. Then, proceeding as in the computations leading to
(6.8), we get

$\Delta_{p_{k}(x)}v_{\varepsilon}\leq-\bar{c}\rho\quad\text{ in
}B_{\mu}(\tilde{x}_{k}),$

with $\bar{c}>0$ universal, if $k$ is large. Therefore,

$\Delta_{p_{k}(x)}v_{\varepsilon}<f_{k}\quad\text{ in
}B_{\mu}(\tilde{x}_{k}),$

for large $k$, which contradicts (6.19). Then $\tilde{x}_{k}\in B_{3/4}$.

Since $\varepsilon<\frac{\xi}{2}$ and $\delta<\frac{\xi}{2}$, we have

$\partial\\{u_{k}>0\\}\subset\\{|x_{n}|<\delta\\}\subseteq\\{x_{n}>-\xi+\varepsilon\varphi(x)\\},$

and then $\tilde{x}_{k}\in\\{u_{k}>0\\}\cap B_{3/4}\subset D_{\varepsilon}\cap
B_{3/4}$.

So (6.9) implies that, for large $k$,

$\nabla v_{\varepsilon}(\tilde{x}_{k})\neq 0$

and

$f_{k}(\tilde{x}_{k})>\Delta_{p_{k}(\tilde{x}_{k})}v_{\varepsilon}(\tilde{x}_{k})\geq
f_{k}(\tilde{x}_{k}),$

a contradiction. This shows (6.18).

Step IV.  We will finally show that, for some $\varepsilon_{k}>0$, we have
$v_{\varepsilon_{k}}\geq u_{k}$ in $B_{3/4}$ and $F(u_{k})\cap
F(v_{\varepsilon_{k}})\cap B_{3/4}\neq\emptyset$, if $k$ is large enough. This
will contradict that $v_{\varepsilon}$ is a strict supersolution to problem
(1.1) in $B_{3/4}$ and concludes the proof.

In fact, from (6.18) we know that

$v_{\varepsilon}\geq
u_{k}\quad\mbox{in}\>\>\overline{\\{u_{k}>0\\}}\quad\text{ for
}0<\varepsilon<\frac{\xi}{2}.$

Let

$\varepsilon_{k}=\sup\left\\{\varepsilon>0:\>\>v_{\varepsilon}\geq
u_{k}\>\>\>\mbox{in}\>\>\overline{\\{u_{k}>0\\}}\right\\}.$

Since $\xi\leq 1/4$, if we consider $\varepsilon=2\xi$, then
$B_{\xi}\subset\\{x_{n}<-\xi+\varepsilon\varphi(x)\\}$ because, in $B_{1/4}$,
$x_{n}=-\xi+\varepsilon\varphi(x)=-\xi+2\xi=\xi.$

Moreover, $0\in\partial\\{u_{0}>0\\}$ and
$\partial\\{u_{k}>0\\}\to\partial\\{u_{0}>0\\}$ in the Hausdorff distance,
then for $k$ large, there exist $\hat{x}_{k}\in
B_{\xi}\cap\partial\\{u_{k}>0\\}$ and $\bar{x}_{k}\in B_{\xi}$ such that
$u_{k}(\bar{x}_{k})>0=v_{\varepsilon}(\bar{x}_{k})$, with
$\bar{x}_{k}\in\overline{\\{u_{k}>0\\}}$. Then, $0<\varepsilon_{k}<2\xi$.

Therefore, there holds $v_{\varepsilon_{k}}\geq u_{k}$ in
$\overline{\\{u_{k}>0\\}}$ and then,

$v_{\varepsilon_{k}}\geq u_{k}\quad\text{ in }\overline{B_{1}},$
$v_{\varepsilon_{k}}(x_{k})=u_{k}(x_{k})\quad\text{ for some
}x_{k}\in\overline{\\{u_{k}>0\\}}.$

Proceeding exactly as in Step III we obtain that

$x_{k}\in\overline{\\{u_{k}>0\\}}\cap B_{3/4}.$

If $x_{k}\in\\{u_{k}>0\\}\cap B_{3/4},$ then
$v_{\varepsilon_{k}}(x_{k})=u_{k}(x_{k})>0.$ Since $v_{\varepsilon_{k}}\geq
u_{k}>0$ in a neighborhood of $x_{k}$, this produces a contradiction because
$v_{\varepsilon_{k}}$ is a strict supersolution to problem (1.1) in $B_{3/4}$.

As a consequence $x_{k}\in\partial\\{u_{k}>0\\}\cap B_{3/4}$ and
$v_{\varepsilon_{k}}(x_{k})=u_{k}(x_{k})=0$ and there exist $x_{k_{j}}\to
x_{k}$ such that $u_{k}(x_{k_{j}})>0$. Then
$v_{\varepsilon_{k}}(x_{k_{j}})\geq u_{k}(x_{k_{j}})>0$ and therefore
$x_{k}\in\partial\\{v_{\varepsilon_{k}}>0\\}.$ Hence $x_{k}\in F(u_{k})\cap
F(v_{\varepsilon_{k}})\cap B_{3/4},$ which gives a contradiction again. This
shows that $\alpha\geq 1$ and completes the proof. ∎

We will also need

######  Proposition 6.5.

Let $u_{k}$ be a sequence of viscosity solutions to (1.1) in $B_{2}$, with
right hand side $f_{k}$, exponent $p_{k}$ and free boundary condition $g_{k}$,
where $f_{k}$, $p_{k}$ and $g_{k}$ are as in Subsection 1.1. Assume that, for
some $\alpha\geq 0$ and $\nu\in\mathbb{R}^{n}$ with $|\nu|=1$, $u_{k}\to
u_{0}(x)=\alpha\langle x,\nu\rangle^{+}$, $f_{k}\to 0$, $p_{k}\to p_{0}$,
$\nabla p_{k}\to 0$ and $g_{k}\to 1$ uniformly in $B_{2}$. Then $\alpha\leq
1$.

###### Proof.

Without loss of generality we assume that $\nu=e_{n}$. Suppose by
contradiction that $\alpha=1+\eta,$ with $\eta>0$, then

$u_{0}(x)=(1+\eta)x_{n}^{+}.$

For $\delta>0$ and $\varepsilon>0$ small, to be precised later, we define

${Q}(x):=(1+\frac{\eta}{2})x_{n}+\delta
x_{n}^{2}-\varepsilon|x^{\prime}|^{2},$

where we denote $x=(x^{\prime},x_{n})$, $x^{\prime}\in\mathbb{R}^{n}$.

Let us show that

(6.20) $\left\\{\begin{aligned} &u_{0}>Q\quad\text{in
}B_{\rho_{0}}\setminus\\{0\\},\\\ &u_{0}(0)=Q(0),\end{aligned}\right.$

for some $\rho_{0}=\rho_{0}(\delta,\eta)>0$. In fact, there holds that

$\displaystyle u_{0}(x)=(1+\eta)x_{n}^{+}>(1+\frac{\eta}{2})x_{n}+\delta
x_{n}^{2}\geq Q(x)\qquad\text{ for }\,0<|x_{n}|<\frac{\eta}{2\delta},$
$\displaystyle
u_{0}(x^{\prime},0)=0>-\varepsilon|x^{\prime}|^{2}=Q(x^{\prime},0)\qquad\text{
for }\,x^{\prime}\neq 0,$

so (6.20) follows for $\rho_{0}=\min\\{1,\frac{\eta}{2\delta}\\}$.

Claim. We claim that, in $B_{1}$, ${Q}$ is a strict subsolution to problem
(1.1) with right hand side $f_{k}$, exponent $p_{k}$ and free boundary
condition $g_{k}$, for large $k$.

Indeed, we have

$\nabla Q=(1+\frac{\eta}{2})e_{n}+2Mx,\qquad D^{2}Q=2M,$

where $M\in R^{n\times n}$ is given by

(6.21) $M_{ij}=0\,\text{ for }\,i\neq j\qquad M_{ii}=-\varepsilon\,\text{ for
}\,i\neq n,\qquad M_{nn}=\delta.$

Then,

(6.22) $1+\frac{\eta}{4}\leq|\nabla Q|\leq 1+{\eta}\quad\text{ in }B_{1},$

if $\delta\leq{\eta}/{8}$ and $\varepsilon\leq{\eta}/{8}$.

Moreover, applying the lower bound in (2.2), we obtain

(6.23) $\begin{split}&\Delta_{p_{k}(x)}{Q}(x)\geq|\nabla
Q|^{p_{k}(x)-2}\mathcal{M}_{\lambda_{0},\Lambda_{0}}^{-}(D^{2}Q)+|\nabla
Q|^{p_{k}(x)-2}\langle\nabla Q,\nabla p_{k}(x)\rangle\log|\nabla Q|\\\
&\geq|\nabla
Q|^{p_{k}(x)-2}2\mathcal{M}_{\lambda_{0},\Lambda_{0}}^{-}(M)-|\nabla
p_{k}(x)||\nabla Q|^{p_{k}(x)-1}\log|\nabla Q|.\end{split}$

We also observe that (6.22) implies

(6.24) $|\nabla Q|^{p_{k}(x)-2}\geq c_{1}\qquad|\nabla
Q|^{p_{k}(x)-1}\log|\nabla Q|\leq c_{2},$

in $B_{1}$, where $c_{1}=c_{1}(\eta,p_{\min})>0$ and
$c_{2}=c_{2}(\eta,p_{\max})>0$.

Now, from (6.21) and (2.3) it is not hard to see that

(6.25)
$\mathcal{M}_{\lambda_{0},\Lambda_{0}}^{-}(M)=-\Lambda_{0}\varepsilon(n-1)+\lambda_{0}\delta\geq\frac{\lambda_{0}\delta}{2},$

if $\varepsilon\leq\frac{\lambda_{0}\delta}{2\Lambda_{0}(n-1)}$. We next take
$k$ large enough so that

(6.26) $|\nabla p_{k}|\leq\frac{\lambda_{0}\delta
c_{1}}{2c_{2}},\qquad|f_{k}|\leq\frac{c_{1}\lambda_{0}\delta}{4}\quad\text{
for }x\in B_{1}.$

Putting together (6.23), (6.24), (6.25) and (6.26), we obtain in $B_{1}$

$\begin{split}&\Delta_{p_{k}(x)}{Q}(x)\geq
c_{1}2\mathcal{M}_{\lambda_{0},\Lambda_{0}}^{-}(M)-|\nabla p_{k}(x)|c_{2}\\\
&\geq c_{1}\lambda_{0}\delta-\frac{\lambda_{0}\delta
c_{1}}{2c_{2}}c_{2}>f_{k}.\end{split}$

If, additionally, $k$ is large so that

$g_{k}\leq 1+\frac{\eta}{8},\quad\text{ for }x\in B_{1},$

we obtain from (6.22) that

$|\nabla Q|>g_{k}\quad\text{ in }B_{1},$

thus proving our claim.

We finally deduce from (6.20) that there exist a sequence $\sigma_{k}\to 0$
and points $x_{k}\in B_{\rho_{0}}$ such that, denoting $Q_{k}=Q+\sigma_{k}$,
we get

$\left\\{\begin{aligned} &u_{k}\geq Q_{k}\quad\text{in }B_{\rho_{0}},\\\
&u_{k}(x_{k})=Q(x_{k}),\end{aligned}\right.$

if $k$ is large. We notice that if $u_{k}(x_{k})>0$, then $Q_{k}(x_{k})>0$.
Otherwise $u_{k}(x_{k})=0=Q_{k}(x_{k})$, and since $\nabla Q_{k}(x_{k})\neq
0$, then $x_{k}\in F(Q_{k})$.

That is, for large $k$, ${Q}_{k}$ is a strict subsolution in $B_{\rho_{0}}$ to
problem (1.1), with right hand side $f_{k}$, exponent $p_{k}$ and free
boundary condition $g_{k}$, touching $u_{k}$ from below at $x_{k}\in
B^{+}_{\rho_{0}}(Q_{k})\cup F(Q_{k})$, a contradiction. Then $\alpha\leq 1.$ ∎

We are now in a position to prove Theorem 1.2.

###### Proof of Theorem 1.2.

Let $u$ be a viscosity solution to (1.1) in $B_{1}$ such that $0\in F(u)$ and
such that $F(u)$ is a Lipschitz graph in $B_{r_{0}}$, for some $0<r_{0}\leq
1$. Without loss of generality we assume that $g(0)=1$ and we denote
$p(0)=p_{0}.$

We will divide the proof into several steps.

Step I. Lipschitz continuity and nondegeneracy. Let us first show that $u$ is
Lipschitz and nondegenerate in a neighborhood of $0$.

In fact, for $0<r\leq\frac{r_{0}}{2}\leq\frac{1}{2}$, we consider the function

$\bar{u}(x)=\frac{1}{r}u(rx),\quad x\in B_{2}.$

Then $\bar{u}$ is a viscosity solution to (1.1) in $B_{2}$, with right hand
side $\bar{f}(x)=rf(rx)$, exponent $\bar{p}(x)=p(rx)$ and free boundary
condition $\bar{g}(x)=g(rx)$. Moreover, $0\in F(\bar{u})$.

From Theorem 1.1 we know that $\bar{u}$ is Lipschitz continuous in $B_{1/2}$
with a Lipschitz constant depending only on $n$, $p_{\min}$, $p_{\max},$
$\|\nabla p\|_{L^{\infty}(B_{3r_{0}/8})}$, $\|f\|_{L^{\infty}(B_{3r_{0}/8})}$,
$\beta$, $\|g\|_{C^{0,\beta}(\overline{B_{3r_{0}/8}})}$ and
$\|u\|_{L^{\infty}(B_{3r_{0}/8})}$.

In order to prove the nondegeneracy, let us see that we can apply the second
part of Proposition 4.2 to ${\bar{u}}$, if $r$ is suitably chosen.

For that purpose, let us first show that the constants appearing in that
proposition can be taken independent of $r$. More precisely, we want to find a
bound independent of $r$ for

$||\bar{u}||_{L^{\infty}(B_{3/2})}^{\bar{p}_{+}^{3/2}-\bar{p}_{-}^{3/2}},\qquad\text{where}\quad\bar{p}_{+}^{3/2}=\sup_{B_{3/2}}\bar{p},\
\ \bar{p}_{-}^{3/2}=\inf_{B_{3/2}}\bar{p}.$

In fact, we have

(6.27)
${||\bar{u}||}_{L^{\infty}(B_{3/2})}^{\bar{p}_{+}^{3/2}-\bar{p}_{-}^{3/2}}\leq||u||_{L^{\infty}(B_{3r_{0}/4})}^{\bar{p}_{+}^{3/2}-\bar{p}_{-}^{3/2}}\Big{(}\frac{1}{r}\Big{)}^{\bar{p}_{+}^{3/2}-\bar{p}_{-}^{3/2}},$

and

(6.28) ${\bar{p}_{+}^{3/2}-\bar{p}_{-}^{3/2}}\leq
3{||\nabla\bar{p}||}_{L^{\infty}(B_{3/2})}\leq
3r{||\nabla{p}||}_{L^{\infty}(B_{3r_{0}/4})}.$

Then, from (6.27) and (6.28), we conclude that

$||\bar{u}||_{L^{\infty}(B_{3/2})}^{\bar{p}_{+}^{3/2}-\bar{p}_{-}^{3/2}}\leq
C=C\big{(}||u||_{L^{\infty}(B_{3r_{0}/4})},{||\nabla{p}||}_{L^{\infty}(B_{3r_{0}/4})}\big{)}.$

It follows that in order to apply the second part of Proposition 4.2 to
$\bar{u}$ we can take the constants $\tilde{\varepsilon}$ and $c_{0}$ in that
proposition depending only on $n$, $p_{\min}$, $p_{\max}$,
$||u||_{L^{\infty}(B_{3r_{0}/4})}$,
${||\nabla{p}||}_{L^{\infty}(B_{3r_{0}/4})}$, $||g||_{L^{\infty}(B_{r_{0}})}$,
$\gamma_{0}$ and on the Lipschitz constant of $F(u)$.

Then, if $r$ is small enough, there holds in $B_{2}$

$\displaystyle|\bar{f}(x)|$ $\displaystyle\leq
r||f||_{L^{\infty}(B_{r_{0}})}\leq\tilde{\varepsilon},$
$\displaystyle|\bar{g}(x)-1|$ $\displaystyle=|g(rx)-g(0)|\leq
2{r}^{\beta}[g]_{C^{0,\beta}(B_{r_{0}})}\leq\tilde{\varepsilon},$
$\displaystyle|\nabla\bar{p}(x)|$ $\displaystyle\leq r||\nabla
p||_{L^{\infty}(B_{r_{0}})}\leq\tilde{\varepsilon},$
$\displaystyle|\bar{p}(x)-p_{0}|$ $\displaystyle=|p(rx)-p(0)|\leq 2r||\nabla
p||_{L^{\infty}(B_{r_{0}})}\leq\tilde{\varepsilon}.$

Hence, for $r$ small enough, $\bar{u}$ is nondegenerate in $B_{\rho_{0}}$, for
$\rho_{0}>0$ depending only on the Lipschitz constant of $F(u)$.

That is, $u$ is Lipschitz continuous and nondegenerate in
$B_{\hat{\rho_{0}}}$, for a suitable universal $\hat{\rho_{0}}>0$, with a
universal Lipschitz constant $L_{0}$.

Step II. Blow up limit. We now consider the blow up sequence

(6.29)
$u_{k}(x)=u_{\delta_{k}}(x)=\frac{u(\delta_{k}x)}{\delta_{k}},\quad\text{
where }\delta_{k}\to 0,$

$\delta_{k}>0$. As before, each $u_{k}$ is a viscosity solution to (1.1) with
right hand side $f_{k}(x)=\delta_{k}f(\delta_{k}x)$, exponent
$p_{k}(x)=p(\delta_{k}x)$ and free boundary condition
$g_{k}(x)=g(\delta_{k}x)$.

Our goal is to apply Theorem 6.3 to $u_{k}$, for large $k$. We will first
observe that, taking $k$ sufficiently large, the assumption (6.4) in that
theorem is satisfied for the universal constant $\bar{\varepsilon}$. In fact,
in $B_{1}$,

(6.30)
$\begin{split}&|f_{k}(x)|=\delta_{k}|f(\delta_{k}x)|\leq\delta_{k}||f||_{L^{\infty}(B_{r_{0}})}\leq\bar{\varepsilon},\\\
&|\nabla p_{k}(x)|\leq\delta_{k}||\nabla
p||_{L^{\infty}(B_{r_{0}})}\leq\bar{\varepsilon},\\\
&|p_{{k}}(x)-p_{0}|=|p(\delta_{k}x)-p(0)|\leq\delta_{k}||\nabla
p||_{L^{\infty}(B_{r_{0}})}\leq\bar{\varepsilon},\\\
&[g_{k}]_{C^{0,\beta}(B_{1})}\leq\delta_{k}^{\beta}[g]_{C^{0,\beta}(B_{r_{0}})}\leq\bar{\varepsilon},\\\
&|g_{{k}}(x)-1|=|g(\delta_{k}x)-g(0)|\leq\delta_{k}^{\beta}[g]_{C^{0,\beta}(B_{r_{0}})}\leq\bar{\varepsilon}.\end{split}$

On the other hand, since $u$ is Lipschitz and nondegenerate in
$B_{\hat{\rho_{0}}}$, with Lipschitz constant $L_{0}$, then, for every $R>0$,
$u_{k}$ are Lipschitz and uniformly nondegenerate in $B_{R}$, with Lipschitz
constant $L_{0}$, if $k\geq k_{0}(R)$. Then, standard arguments (see for
instance, [AC], 4.7) imply that (up to a subsequence), there holds that

(6.31) $\displaystyle u_{k}\to u_{0}\ \text{in}\ C_{\rm
loc}^{0,\gamma}(\mathbb{R}^{n}),\ \text{for all}\ 0<\gamma<1,$
$\displaystyle\partial\\{u_{k}>0\\}\to\partial\\{u_{0}>0\\}\ \text{locally in
Hausdorff distance},$

to a function $u_{0}:\mathbb{R}^{n}\to\mathbb{R}$, which is globally Lipschitz
with constant $L_{0}$ and nondegenerate in $\mathbb{R}^{n}$. Moreover,
$F(u_{0})$ is a global Lipschitz graph.

We also observe that the estimates in (6.30) also imply that

$f_{k}\to 0,\quad\nabla p_{k}\to 0,\quad p_{k}\to p_{0},\quad g_{k}\to
1,\quad\text{ uniformly on compacts of }\ \mathbb{R}^{n}.$

Step III. Limit equation. Since $u$ satisfies in the viscosity sense
$\Delta_{p(x)}u=f$ in $\\{u>0\\}$, then every $u_{k}$ satisfies in the
viscosity sense $\Delta_{p_{k}(x)}u_{k}=f_{k}$ in $\\{u_{k}>0\\}$. We claim
that the blow up limit $u_{0}$ is a viscosity solution to
$\Delta_{p_{0}}u_{0}=0$ in $\\{u_{0}>0\\}$.

In fact, let us see that $u_{0}$ is a viscosity subsolution to
$\Delta_{p_{0}}u_{0}=0$ in $\\{u_{0}>0\\}$.

Let $x_{0}\in\\{u_{0}>0\\}$ and let $P$ be a quadratic polynomial such that
$P\geq u_{0}$ in $B_{\sigma}(x_{0}),$ $P(x_{0})=u(x_{0})$ and $\nabla
P(x_{0})\not=0$. We can assume that $|\nabla P|\geq c>0$ in
$B_{\sigma}(x_{0})$ and $B_{\sigma}(x_{0})\subset\\{u_{k}>0\\}$ for $k$ large,
so $\Delta_{p_{k}(x)}u_{k}(x)=f_{k}(x)$ in $B_{\sigma}(x_{0})$. We want to
prove that $\Delta_{p_{0}}P(x_{0})\geq 0$. We argue by contradiction assuming
that there exists $\rho>0$ such that $\Delta_{p_{0}}P(x_{0})<-\rho<0.$ For
$\varepsilon>0$, we define
$\tilde{P}(x)=\tilde{P}_{\varepsilon}(x)=P(x)+\varepsilon|x-x_{0}|^{2}.$ Hence
$\nabla\tilde{P}=\nabla P+2\varepsilon(x-x_{0})$ and

(6.32) $|\nabla\tilde{P}|\geq\frac{c}{2}\quad\text{ in }B_{{\sigma}}(x_{0}),$

if $\varepsilon$ is sufficiently small. Letting $\varepsilon\to 0$, we get

$\Delta_{p_{0}}\tilde{P}(x_{0})\to\Delta_{p_{0}}P(x_{0})<-\rho$

and then, if $\varepsilon$ is small enough, we obtain

(6.33) $\Delta_{p_{0}}\tilde{P}(x_{0})<-\frac{\rho}{2}.$

We now fix $\varepsilon>0$ small such that (6.32) and (6.33) hold. We have

(6.34) $\tilde{P}(x)>u_{0}(x)\ \text{ in
}\overline{B}_{{\sigma}}(x_{0})\setminus\\{x_{0}\\},\ \text{ and
}\tilde{P}(x_{0})=u_{0}(x_{0}).$

Moreover, since $u_{k}\to u_{0}$ uniformly in $B_{\sigma}(x_{0}),$ then
$|u_{k}-u_{0}|<\gamma_{k}$ in $B_{\sigma}(x_{0})$ with $\gamma_{k}\to 0.$
Hence, from

$\tilde{P}(x)\geq u_{0}(x)>u_{k}(x)-\gamma_{k}\quad\text{ in
}B_{{\sigma}}(x_{0}),$

it follows

$\tilde{P}(x)+\gamma_{k}>u_{k}(x)\quad\text{ in }B_{{\sigma}}(x_{0}).$

Let

$t_{k}=\sup\\{t\geq 0:\quad\tilde{P}(x)+\gamma_{k}\geq
u_{k}(x)+t\quad\mbox{in}\>\>B_{{\sigma}}(x_{0})\\}.$

Since $\gamma_{k}\to 0$ and $\tilde{P}(x)+\gamma_{k}$ is bounded in
$B_{{\sigma}}(x_{0}),$ then $t_{k}$ is finite so that

$\tilde{P}(x)+\gamma_{k}\geq u_{k}(x)+t_{k}\quad\text{ in
}B_{{\sigma}}(x_{0})$

and there exists $x_{k}\in\overline{B}_{{\sigma}}(x_{0})$ such that

$\tilde{P}(x_{k})+\gamma_{k}=u_{k}(x_{k})+t_{k}.$

Then

$\begin{split}u_{k}(x_{0})+\gamma_{k}+\gamma_{k}\geq
u_{0}(x_{0})+\gamma_{k}=\tilde{P}(x_{0})+\gamma_{k}\geq
u_{k}(x_{0})+t_{k}.\end{split}$

As a consequence

$t_{k}\leq 2\gamma_{k}\to 0$

and $t_{k}\to 0.$ Let $\tilde{P}_{k}(x)=\tilde{P}(x)+\gamma_{k}-t_{k}.$

Then

$\tilde{P}_{k}(x)\geq u_{k}(x)\quad\text{ in }B_{{\sigma}}(x_{0})$

and $\tilde{P}_{k}(x_{k})=u_{k}(x_{k})$ for
$x_{k}\in\overline{B}_{{\sigma}}(x_{0}).$

Since $\tilde{P}(x)>u_{0}(x)$ on $\partial B_{{\sigma}}(x_{0})$ then

$\tilde{P}(x)-u_{0}(x)\geq\bar{c}>0$

on $\partial B_{{\sigma}}(x_{0})$ and

$\begin{split}&\tilde{P}_{k}(x)-u_{k}(x)=\tilde{P}(x)+\gamma_{k}-t_{k}-u_{k}(x)\geq\tilde{P}(x)+\gamma_{k}-t_{k}-u_{0}(x)-\gamma_{k}\\\
&=\tilde{P}(x)-t_{k}-u_{0}(x)\geq\bar{c}-t_{k}\geq\frac{\bar{c}}{2}\end{split}$

on $\partial B_{{\sigma}}(x_{0})$ if $k\geq k_{0},$ since $t_{k}\to 0.$ We
recall here that $u_{k}\leq u_{0}+\gamma_{k}.$ Hence $x_{k}\not\in\partial
B_{{\sigma}}(x_{0})$ if $k\geq k_{0}.$ Then

$\tilde{P}_{k}(x)\geq u_{k}(x)\quad\text{ in }B_{{\sigma}}(x_{0}),$

$\tilde{P}_{k}(x_{k})=u_{k}(x_{k})$ for $x_{k}\in B_{{\sigma}}(x_{0})$, and
$\nabla\tilde{P}_{k}\not=0$ in $B_{{\sigma}}(x_{0})$ and thus,

(6.35)
$\Delta_{p_{k}(x_{k})}\tilde{P}(x_{k})=\Delta_{p_{k}(x_{k})}\tilde{P}_{k}(x_{k})\geq
f_{k}(x_{k}).$

Since $x_{k}\in B_{{\sigma}}(x_{0})$ then, for a subsequence,
$x_{k}\to\bar{x}\in\overline{B}_{{\sigma}}(x_{0})$. Hence, using that
$\gamma_{k}\to 0$, $t_{k}\to 0$ and

$\tilde{P}({x_{k}})+\gamma_{k}-t_{k}=\tilde{P}_{k}(x_{k})=u_{k}(x_{k}),$

we obtain that $\tilde{P}(\bar{x})=u_{0}(\bar{x}).$ Then $\bar{x}=x_{0},$
because (6.34) holds.

Now, letting $k\to\infty$ in (6.35), we get

$\Delta_{p_{0}}\tilde{P}(x_{0})\geq 0,$

which gives a contradiction to (6.33). Hence $\Delta_{p_{0}}P(x_{0})\geq 0.$

Arguing in a similar way, we deduce that $u_{0}$ is a viscosity supersolution
to $\Delta_{p_{0}}u_{0}=0$ in $\\{u_{0}>0\\}$ as well.

Step IV. Limit free boundary problem. We want to show that $u_{0}$ is a
viscosity solution (in the sense of Definition 6.1) to problem

(6.36) $\left\\{\begin{array}[]{ll}\Delta_{p_{0}}u_{0}=0,&\hbox{in
$\\{u_{0}>0\\}$},\\\ &\\\ |\nabla u_{0}|=1,&\hbox{on
$F(u_{0}).$}\end{array}\right.$

Hence we have to check that free boundary condition is satisfied in the sense
of (i) and (ii) of that definition. We divide our analysis into two cases.

Case (a). Let $x_{0}\in F(u_{0})$ such that there exists a ball
$B_{r}(y)\subset\\{u_{0}>0\\}$, with $x_{0}\in\partial B_{r}(y)$. We denote
$\nu=\frac{y-x_{0}}{|y-x_{0}|}$. Then, by Case (a) in Lemma 5.1,

(6.37) $u_{0}(x)=\alpha\langle x-x_{0},\nu\rangle^{+}+o(|x-x_{0}|)\quad\text{
in }B_{r}(y),$

with $\alpha>0.$

We now consider a sequence $\lambda_{j}\to 0,$ $\lambda_{j}>0$. Since $u_{0}$
is Lipschitz in $\mathbb{R}^{n}$ then there exists a function $u_{00}$ such
that, for a subsequence,

$\frac{u_{0}(x_{0}+\lambda_{j}x)}{\lambda_{j}}\to
u_{00}(x)\quad\text{uniformly on compact sets of }\mathbb{R}^{n}.$

From (6.37) we know that $u_{00}(x)=\alpha\langle x,\nu\rangle^{+}$ in
$\\{\langle x,\nu\rangle\geq 0\\}.$ Then $\\{\langle x,\nu\rangle=0\\}\subset
F(u_{00})$. Since $F(u_{0})$ is a Lipschitz graph, also $F(u_{00})$ is a
Lipschitz graph, so we have $\\{\langle x,\nu\rangle=0\\}=F(u_{00}).$ Hence,

$u_{00}(x)=\alpha\langle x,\nu\rangle^{+}\quad\text{ in }\mathbb{R}^{n}.$

This result holds for any sequence $\lambda_{j}\to 0,$ $\lambda_{j}>0,$
therefore

(6.38) $u_{0}(x)=\alpha\langle x-x_{0},\nu\rangle^{+}+o(|x-x_{0}|)\quad\text{
in }\mathbb{R}^{n},$

with $\alpha>0.$

We want to show that $\alpha=1$.

Since $x_{0}\in F(u_{0})$ and recalling (6.29) and (6.31), we know that there
exists, up to a subsequence,

$x_{k}\in F(u_{k}),\quad|x_{k}-x_{0}|<1/k.$

We fix $R>0$ such that $|x_{0}|<R$ and let $\mu_{j}=1/\sqrt{j}$.

For each $j$ there exists $k_{j}\geq j$ such that

$|u_{{k_{j}}}(x)-u_{0}(x)|\leq\frac{\mu_{j}}{j}\qquad\text{for }x\in B_{R+2}.$

We now define

$(u_{k_{j}})_{\mu_{j}}(x)=\frac{1}{\mu_{j}}{u_{k_{j}}(x_{k_{j}}+\mu_{j}x)},\quad(u_{0})_{\mu_{j}}(x)=\frac{1}{\mu_{j}}{u_{0}(x_{k_{j}}+\mu_{j}x)}.$

Then, if $j\geq j_{0}$,

$|(u_{k_{j}})_{\mu_{j}}(x)-(u_{0})_{\mu_{j}}(x)|=\frac{|u_{k_{j}}(x_{k_{j}}+\mu_{j}x)-u_{0}(x_{k_{j}}+\mu_{j}x)|}{\mu_{j}}\leq\frac{1}{j}\qquad\text{for
}x\in B_{2}.$

We now observe that

$\frac{|x_{k_{j}}-x_{0}|}{\mu_{j}}<\frac{1/k_{j}}{\mu_{j}}\leq\frac{1}{\sqrt{j}}\quad\to
0,$

and, recalling (6.38), we obtain

$(u_{0})_{\mu_{j}}(x)\to u_{00}(x)=\alpha\langle x,\nu\rangle^{+}\quad\text{
uniformly in }B_{2}.$

Then,

$\displaystyle|(u_{k_{j}})_{\mu_{j}}(x)-u_{00}(x)|\leq$
$\displaystyle|(u_{k_{j}})_{\mu_{j}}(x)-(u_{0})_{\mu_{j}}(x)|$
$\displaystyle+|(u_{0})_{\mu_{j}}(x)-u_{00}(x)|\to 0\quad\text{ uniformly in
}B_{2}.$

Denoting $\rho_{j}=\delta_{k_{j}}\mu_{j}$,
$\bar{x}_{j}=\delta_{k_{j}}x_{k_{j}}$ and
$u_{\rho_{j}}(x)=\frac{1}{\rho_{j}}u(\bar{x}_{j}+\rho_{j}x)=(u_{k_{j}})_{\mu_{j}}(x)$,
we get

$\displaystyle\rho_{j}\to 0,$ $\displaystyle\bar{x}_{j}\in
F(u),\quad\bar{x}_{j}\to 0,$ $\displaystyle
u_{\rho_{j}}(x)=\frac{1}{\rho_{j}}u(\bar{x}_{j}+\rho_{j}x)$
$\displaystyle\to\alpha\langle x,\nu\rangle^{+}\quad\text{ uniformly in
}B_{2}.$

Reasoning as in Step II, we see that each $u_{\rho_{j}}$ is a viscosity
solution to (1.1) in $B_{2}$ with right hand side
$\bar{f}_{j}(x)=\rho_{j}f(\bar{x}_{j}+\rho_{j}x)$, exponent
$\bar{p}_{j}(x)=p(\bar{x}_{j}+\rho_{j}x)$ and free boundary condition
$\bar{g}_{j}(x)=g(\bar{x}_{j}+\rho_{j}x)$,

$\bar{f}_{j}\to 0,\quad\nabla\bar{p}_{j}\to 0,\quad\bar{p}_{j}\to
p_{0},\quad\bar{g}_{j}\to 1,\quad\text{ uniformly in }B_{2}.$

Moreover, $u_{\rho_{j}}$ are uniformly Lipschitz and nondegenerate in $B_{2}$
for $j\geq j_{0}$, $\partial\\{u_{\rho_{j}}>0\\}$ are uniform Lipschitz graphs
and $\partial\\{u_{\rho_{j}}>0\\}\to\\{\langle x,\nu\rangle=0\\}$ in Hausdorff
distance in $B_{2}$.

Now, applying Propositions 6.4 and 6.5 to the sequence $u_{\rho_{j}}$, we
deduce that $\alpha=1$. Then, (i) in Definition 6.1 is satisfied in this case.

Case (b). Let $x_{0}\in F(u_{0})$ such that there exists a ball
$B_{r}(y)\subset\\{u_{0}\equiv 0\\}$, with $x_{0}\in\partial B_{r}(y)$. We
denote $\nu=\frac{x_{0}-y}{|x_{0}-y|}$. Then, from the proof of Case (b) in
Lemma 5.1, we get

(6.39) $u_{0}(x)=\alpha\langle x-x_{0},\nu\rangle^{+}+o(|x-x_{0}|)\quad\text{
in }B^{c}_{r}(y),$

with $\alpha\geq 0.$

We now consider a sequence $\lambda_{j}\to 0,$ $\lambda_{j}>0$. Then, for a
subsequence and a function $u_{00}$,

$\frac{u_{0}(x_{0}+\lambda_{j}x)}{\lambda_{j}}\to
u_{00}(x)\quad\text{uniformly on compact sets of }\mathbb{R}^{n}.$

From (6.39) we know that $u_{00}(x)=\alpha\langle x,\nu\rangle^{+}$ in
$\\{\langle x,\nu\rangle\geq 0\\}.$ Since $B_{r}(y)\subset\\{u_{0}\equiv
0\\}$, we have $u_{00}(x)=0$ in $\\{\langle x,\nu\rangle\leq 0\\}.$ Hence,

$u_{00}(x)=\alpha\langle x,\nu\rangle^{+}\quad\text{ in }\mathbb{R}^{n}.$

Now, if $\alpha=0$, then $u_{00}\equiv 0$ in $\mathbb{R}^{n}$. This
contradicts that $F(u_{00})$ is a Lipschitz graph and shows that $\alpha>0$.

Since this result holds for any sequence $\lambda_{j}\to 0,$ $\lambda_{j}>0,$
we conclude that

$u_{0}(x)=\alpha\langle x-x_{0},\nu\rangle^{+}+o(|x-x_{0}|)\quad\text{ in
}\mathbb{R}^{n},$

with $\alpha>0.$ Now proceeding as in Case (a), we obtain that $\alpha=1$.
Then, (ii) in Definition 6.1 is satisfied in the present case.

This shows that $u_{0}$ is a viscosity solution to problem (6.36) in the sense
of Definition 6.1.

Step V. Conclusion. We have proved that $u_{0}$ is a viscosity solution (in
the sense of Definition 6.1) to (6.36) that is Lipschitz continuous and
$F(u_{0})$ is a Lipschitz graph.

Thus, from Lemma 6.2 it follows that, up to a rotation, $u_{0}(x)=x_{n}^{+}.$
Then for sufficiently large $k$ we have that, in $B_{1}$,

(6.40) $(x_{n}-\bar{\varepsilon})^{+}\leq
u_{k}(x)\leq(x_{n}+\bar{\varepsilon})^{+},$

for $\bar{\varepsilon}$ the universal constant in Theorem 6.3. Recalling
(6.30), we deduce that Theorem 6.3 applies and, as a consequence, we conclude
that the free boundaries of $u_{k}$ as well as that of $u$ are $C^{1,\alpha}$,
in a neighborhood of $0$. ∎

As a by-product of Theorems 6.3 and 1.2, we obtain further regularity results
for $F(u)$ under additional regularity assumptions on the data.

######  Corollary 6.6.

Let $u$ be as in Theorem 6.3 or as in Theorem 1.2. Assume moreover that $p\in
C^{2}(B_{1})$, $f\in C^{1}(B_{1})$ and $g\in C^{2}(B_{1})$, then there exists
$\delta>0$ such that $B_{\delta}\cap F(u)\in C^{2,\sigma}$ for every
$0<\sigma<1$. If $p\in C^{m+1,\sigma}(B_{1})$, $f\in C^{m,\sigma}(B_{1})$ and
$g\in C^{m+1,\sigma}(B_{1})$ for some $0<\sigma<1$ and $m\geq 1$, then
$B_{\delta}\cap F(u)\in C^{m+2,\sigma}$.

Finally, if $p$, $f$ and $g$ are analytic in $B_{1}$, then $B_{\delta}\cap
F(u)$ is analytic.

###### Proof.

The result follows from the application of Theorem 2 in [KN]. ∎

##  7\. Some consequences

In this section we discuss some consequences of our results.

As already mentioned, in [LW2] problem (1.1) was considered for weak
solutions, which is a different notion of solution from the one we are
considering here (see Definition 7.1 below). One of the consequences of our
Theorem 1.2 is an analogous result for weak solutions (Corollary 7.3).

The notation and the assumptions on $\Omega,p,f$ and $g$ will be the same as
in the rest of the paper (see Subsection 1.1 and Section 2). In particular we
will use the notation $\Omega^{+}(u)$ and $F(u)$ in (2.1).

We first have

######  Definition 7.1 (Definition 2.2 in [LW2]).

We call $u$ a weak solution of (1.1) in $\Omega$ if

  1. (i)

$u$ is continuous and nonnegative in $\Omega$, $u\in W^{1,p(\cdot)}(\Omega)$
and $\Delta_{p(x)}u=f$ in $\Omega^{+}(u)$ (in the sense of Definition 2.1).

  2. (ii)

For $D\subset\subset\Omega$ there are constants $c_{\min}=c_{\min}(D)$,
$C_{\max}=C_{\max}(D)$, $r_{0}=r_{0}(D)$, $0<c_{\min}\leq C_{\max}$,
$r_{0}>0$, such that for balls $B_{r}(x)\subset D$ with $x\in F(u)$ and
$0<r\leq r_{0}$

$c_{\min}\leq\frac{1}{r}\sup_{B_{r}(x)}u\leq C_{\max}.$

  3. (iii)

For $\mathcal{H}^{n-1}$ a.e. $x_{0}\in\partial_{\rm{red}}\\{u>0\\}$ (that is,
for ${\mathcal{H}}^{n-1}$-almost every point $x_{0}\in F(u)$ such that $F(u)$
has an exterior unit normal $\nu(x_{0})$ in the measure theoretic sense) $u$
has the asymptotic development

$u(x)=g(x_{0})\langle x-x_{0},\nu(x_{0})\rangle^{-}+o(|x-x_{0}|).$

  4. (iv)

For every $x_{0}\in F(u)$,

$\displaystyle\limsup_{\stackrel{{\scriptstyle x\to x_{0}}}{{u(x)>0}}}|\nabla
u(x)|\leq g(x_{0}).$

If there is a ball $B\subset\\{u=0\\}$ touching $F(u)$ at $x_{0}$ then,

$\limsup_{\stackrel{{\scriptstyle x\to
x_{0}}}{{u(x)>0}}}\frac{u(x)}{\mbox{dist}(x,B)}\geq g(x_{0}).$

Then we prove

######  Proposition 7.2.

Let $u$ be a weak solution to (1.1) in $\Omega$ in the sense of Definition
7.1. Then $u$ is a viscosity solution to (1.1) in $\Omega$ in the sense of
Definition 2.5.

###### Proof.

Let $u$ be as in the statement. Then $u$ is continuous and nonnegative in
$\Omega$ and satisfies condition (i) in Definition 2.5. In order to show that
it verifies condition (ii) in that definition, we divide the analysis into two
cases.

Case (a). Let $\varphi\in C(\Omega)$, $\varphi\in
C^{2}(\overline{\Omega^{+}(\varphi)})$ be such that $\varphi^{+}$ touches $u$
from below at $x_{0}\in F(u)$ and $\nabla\varphi(x_{0})\not=0$. We want to
show that

(7.1) $|\nabla\varphi(x_{0})|\leq g(x_{0}).$

We first observe that, under the present assumptions, Proposition 2.1 in [LW2]
applies, so $u$ is locally Lipschitz in $\Omega$.

Also there holds that $\varphi^{+}$ has a $C^{2}$ extension $\tilde{\varphi}$
in a neigborhood $\mathcal{O}$ of $x_{0}$ ($\tilde{\varphi}=\varphi^{+}$ in
$\overline{\Omega^{+}(\varphi)}\cap\mathcal{O}$, $\tilde{\varphi}<0$ otherwise
in $\mathcal{O}$), that to simplify the notation we still denote $\varphi$.

Moreover, $\varphi$ touches $u$ from below at $x_{0}\in F(u)$ as well.

By the implicit function theorem, $F(\varphi)$ is a $C^{2}$ hypersurface in a
neighborhood of $x_{0}$. Then, $F(\varphi)$ has a tangent ball $B$ at $x_{0}$,
with $B\subset\Omega^{+}(\varphi)$ and also with $B\subset\Omega^{+}(u)$ and
$x_{0}\in F(u)\cap\partial B$.

We now consider a sequence $\lambda_{j}\to 0,$ $\lambda_{j}>0$. Since $u$ and
$\varphi$ are Lipschitz in a neighborhood of $x_{0}$, then there exist
Lipschitz functions $u_{0}$ and $\varphi_{0}$ such that, for a subsequence,

$u_{\lambda_{j}}(x)=\frac{u(x_{0}+\lambda_{j}x)}{\lambda_{j}}\to
u_{0}(x),\qquad\frac{\varphi(x_{0}+\lambda_{j}x)}{\lambda_{j}}\to\varphi_{0}(x),$

uniformly on compact sets of $\mathbb{R}^{n}$. For simplicity we assume that
the interior normal to $\partial B$ at $x_{0}$ is $e_{n}$. Then

$u_{0}(x)\geq\varphi_{0}(x)=|\nabla\varphi(x_{0})|x_{n}^{+}\ \text{ in
}\\{x_{n}\geq 0\\},$ $\Delta_{p_{0}}u_{0}=0\ \text{ in
}\\{u_{0}>0\\}\supset\\{x_{n}>0\\},\ \text{ with }p_{0}=p(x_{0}).$

Then, the application of Lemma 5.1, Case (a), at the origin, gives

$u_{0}(x)=\gamma x_{n}^{+}+o(|x|)\ \text{ in }B_{1}(e_{n}),\text{ with
}\gamma>0.$

We now consider a sequence $\mu_{j}\to 0,$ $\mu_{j}>0$. Then, there exist
Lipschitz functions $u_{00}$ and $\varphi_{00}$ such that, for a subsequence,

$({u_{0}})_{\mu_{j}}(x)=\frac{u_{0}(\mu_{j}x)}{\mu_{j}}\to
u_{00}(x),\qquad\frac{\varphi_{0}(\mu_{j}x)}{\mu_{j}}\to\varphi_{00}(x),$

uniformly on compact sets of $\mathbb{R}^{n}$. There holds that

$u_{00}(x)=\gamma
x_{n}^{+}\geq\varphi_{00}(x)=|\nabla\varphi(x_{0})|x_{n}^{+}\ \text{ in
}\\{x_{n}\geq 0\\},$

and

(7.2) $|\nabla u_{00}(x)|=\gamma\geq|\nabla\varphi(x_{0})|\ \text{ in
}\\{x_{n}>0\\}.$

Now let

$\alpha:=\limsup_{\stackrel{{\scriptstyle x\to x_{0}}}{{u(x)>0}}}|\nabla
u(x)|.$

Then, by (iv) in Definition 7.1, we have

(7.3) $g(x_{0})\geq\alpha.$

Let us see that

(7.4) $|\nabla u_{00}|\leq\alpha\ \text{ in }\mathbb{R}^{n}.$

In fact, let $R>0$ and $\epsilon>0$. Then, there exists $\lambda_{0}>0$ such
that $|\nabla u(x)|\leq\alpha+\epsilon$ in $B_{\lambda_{0}R}(x_{0})$. We thus
have $|\nabla u_{\lambda_{j}}(x)|\leq\alpha+\epsilon$ in $B_{R}$ for $j$
large. Passing to the limit, we obtain $|\nabla u_{0}|\leq\alpha+\epsilon$ in
$B_{R}$ and then $|\nabla u_{0}|\leq\alpha$ in $\mathbb{R}^{n}$. Now also
$|\nabla({u_{0}})_{\mu_{j}}|\leq\alpha$ in $\mathbb{R}^{n}$. Passing to the
limit again, we obtain (7.4).

Then, (7.3), (7.4) and (7.2) give
$g(x_{0})\geq\alpha\geq\gamma\geq|\nabla\varphi(x_{0})|$. That is, (7.1)
holds.

Case (b). Now let $\varphi\in C(\Omega)$, $\varphi\in
C^{2}(\overline{\Omega^{+}(\varphi)})$ such that $\varphi^{+}$ touches $u$
from above at $x_{0}\in F(u)$ and $\nabla\varphi(x_{0})\not=0$. We want to
show that

(7.5) $|\nabla\varphi(x_{0})|\geq g(x_{0}).$

Also in this case there holds that $\varphi^{+}$ has a $C^{2}$ extension
$\tilde{\varphi}$ in a neighborhood of $x_{0}$, that to simplify the notation
we still denote $\varphi$.

By the implicit function theorem, $F(\varphi)$ is a $C^{2}$ hypersurface in a
neighborhood of $x_{0}$. Then, $F(\varphi)$ has a tangent ball $B$ at $x_{0}$,
with $B\subset\Omega\setminus\overline{\Omega^{+}(\varphi)}$ and also with
$B\subset\\{u=0\\}$ and $x_{0}\in F(u)\cap\partial B$.

Now let

$\alpha:=\limsup_{\stackrel{{\scriptstyle x\to
x_{0}}}{{u(x)>0}}}\frac{u(x)}{\mbox{dist}(x,B)}.$

Then, by (iv) in Definition 7.1, we have

(7.6) $g(x_{0})\leq\alpha.$

Let $x_{k}\to x_{0}$ with $u(x_{k})>0$ be such that

(7.7) $\frac{u(x_{k})}{\mbox{dist}(x_{k},B)}\to\alpha.$

Since $\varphi^{+}\geq u$ in a neighborhood of $x_{0}$, then
$\varphi(x_{k})>0$. Now let $y_{k}\in\partial B$ such that
$\mbox{dist}(x_{k},B)=|x_{k}-y_{k}|$. Then $\varphi(y_{k})\leq 0$ and

(7.8)
$\frac{\varphi(x_{k})-\varphi(y_{k})}{|x_{k}-y_{k}|}\geq\frac{\varphi(x_{k})}{\mbox{dist}(x_{k},B)}\geq\frac{u(x_{k})}{\mbox{dist}(x_{k},B)}.$

But, for a subsequence,

(7.9)
$\frac{\varphi(x_{k})-\varphi(y_{k})}{|x_{k}-y_{k}|}=\nabla\varphi(\xi_{k})\cdot\frac{(x_{k}-y_{k})}{|x_{k}-y_{k}|}\to\nabla\varphi(x_{0})\cdot\frac{\nabla\varphi(x_{0})}{|\nabla\varphi(x_{0})|},$

where for every $k$, $\xi_{k}$ is a point in the segment joining $x_{k}$ and
$y_{k}$. Putting (7.7), (7.8) and (7.9) together we get
$|\nabla\varphi(x_{0})|\geq\alpha$. Now recalling (7.6), we get (7.5) which
completes the proof. ∎

Then, we obtain

######  Corollary 7.3.

Let $u$ be a weak solution to (1.1) in $B_{1}$ in the sense of Definition 7.1,
with $0\in F(u)$. If $F(u)$ is a Lipschitz graph in a neighborhood of $0$,
then $F(u)$ is $C^{1,\alpha}$ in a (smaller) neighborhood of $0$.

###### Proof.

The result is an immediate application of Theorem 1.2 and Proposition 7.2. ∎

##  8\. Some applications

In this section we discuss some applications of both the results obtained in
the present paper and in [FL], and we draw some conclusions on them (see
Remark 8.4).

The applications of our results discussed here correspond to three different
minimization problems that were already studied in [LW1], [LW3] and [LW4]. Our
results below rely on the thorough understanding of the properties of
nonnegative local minimizers achieved in those papers. We also refer to them
for the motivation and related literature.

The notation and the assumptions on $\Omega,p$ and $f$ will be the same as in
the rest of the paper (see Subsection 1.1 and Section 2). In particular we
will use the notation $\Omega^{+}(u)$ and $F(u)$ in (2.1).

Our first application is

######  Proposition 8.1.

Let $\Omega$, $p$ and $f$ be as above. Let
$0<{\lambda_{\min}}\leq\lambda(x)\leq{\lambda_{\max}}<\infty$ with $\lambda\in
C^{0,\beta}(\Omega)$. Let $u\in W^{1,p(\cdot)}(\Omega)\cap L^{\infty}(\Omega)$
be a nonnegative local minimizer of the energy functional
$J(v)={\displaystyle\int_{\Omega}\Big{(}\frac{|\nabla
v|^{p(x)}}{p(x)}+\lambda(x)\chi_{\\{v>0\\}}+fv\Big{)}\,dx}$ in $\Omega$.

Then, $u$ is a viscosity solution to (1.1) in $\Omega$ with
$g(x)=(\frac{p(x)}{p(x)-1}\,\lambda(x))^{1/p(x)}$.

Let $x_{0}\in F(u)$ be such that $F(u)$ is a Lipschitz graph in a neighborhood
of $x_{0}$, then $F(u)$ is $C^{1,\alpha}$ in a (smaller) neighborhood of
$x_{0}$.

Let $x_{0}\in F(u)$ be such that $F(u)$ has a normal in the measure theoretic
sense, then $F(u)$ is $C^{1,\alpha}$ in a neighborhood of $x_{0}$.

Moreover, there is a subset $\mathcal{R}$ of $F(u)$ which is locally a
$C^{1,\alpha}$ surface. The set $\mathcal{R}$ is open and dense in $F(u)$ and
the remainder of the free boundary has $(n-1)-$dimensional Hausdorff measure
zero.

###### Proof.

By Theorem 5.1 in [LW3], $u$ is a weak solution to (1.1) in $\Omega$ with
$g(x)=(\frac{p(x)}{p(x)-1}\,\lambda(x))^{1/p(x)}$ in the sense of Definition
7.1. Then by Proposition 7.2, $u$ is a viscosity solution to (1.1) in $\Omega$
in the sense of Definition 2.5, with the same $g$.

Let $x_{0}\in F(u)$ be such that $F(u)$ is a Lipschitz graph in a neighborhood
of $x_{0}$. Then, from the application of Theorem 1.2, $F(u)$ is
$C^{1,\alpha}$ in a smaller neighborhood of $x_{0}$.

Let $x_{0}\in F(u)$ be such that $F(u)$ has a normal in the measure theoretic
sense. Without loss of generality we assume that $x_{0}=0$, $g(0)=1$ and that
the inward unit normal to $F(u)$ at $0$ in the measure theoretic sense is
$e_{n}$. Also we denote $p(0)=p_{0}$.

Then, by Theorem 3.9 in [LW3] there holds that

(8.1) $u(x)=x_{n}^{+}+o(|x|)\ \text{ in }\mathbb{R}^{n}.$

By Corolary 3.2 and Theorem 3.5 in [LW3] we know that $u$ is Lipschitz and
nondegenerate in some ball $B_{r_{0}}$, with $0<r_{0}<1$.

Then, as in Step II in the proof of Theorem 1.2, we take $\delta_{k}>0$,
$\delta_{k}\to 0$, and consider a blow up sequence $u_{k}$ as in (6.29). As in
that theorem, our goal is to apply Theorem 6.3 to $u_{k}$, for large $k$. We
first observe that, taking $k$ sufficiently large, the assumption (6.4) in
that theorem is satisfied for the universal constant $\bar{\varepsilon}$. In
fact, in $B_{1}$, (6.30) holds.

Arguing again as in Theorem 1.2, we see that (6.31) holds with
$u_{0}(x)=x_{n}^{+}\ \text{ in }\mathbb{R}^{n},$ because of (8.1). Then,
reasoning as in this same theorem and using Theorem 3.6 in [LW3], we obtain
for $k$ sufficiently large that (6.40) holds in $B_{1}$, for
$\bar{\varepsilon}$ the universal constant in Theorem 6.3. Therefore, Theorem
6.3 applies to $u_{k}$, and as a consequence, $F(u)$ is $C^{1,\alpha}$ in a
neighborhood of $0$.

Finally, denoting $\mathcal{R}$ the set of points in $F(u)$ such that $F(u)$
has a normal in the measure theoretic sense, we argue as in Theorem 5.2 in
[LW3] and obtain that $\mathcal{R}$ is dense in $F(u)$ and
${\mathcal{H}}^{n-1}(F(u)\setminus\mathcal{R})=0$. ∎

Our next application is

######  Proposition 8.2.

For $\varepsilon>0$, let
$B_{\varepsilon}(s)=\int_{0}^{s}\tilde{\beta}_{\varepsilon}(\tau)\,d\tau$
where
${\tilde{\beta}}_{\varepsilon}(s)={1\over\varepsilon}\tilde{\beta}({s\over\varepsilon})$,
with $\tilde{\beta}$ a Lipschitz function satisfying $\tilde{\beta}>0$ in
$(0,1)$, $\tilde{\beta}\equiv 0$ outside $(0,1)$. Let $\Omega$, $p$ and $f$ be
as above, $1<p_{\min}\leq p_{\varepsilon_{j}}(x)\leq p_{\max}<\infty$ and
$\|\nabla p_{\varepsilon_{j}}\|_{L^{\infty}}\leq L$. Let
$u^{\varepsilon_{j}}\in W^{1,p_{\varepsilon_{j}}(\cdot)}(\Omega)$ be a family
of nonnegative local minimizers of the energy functional
$J_{\varepsilon_{j}}(v)={\displaystyle\int_{\Omega}\Big{(}\frac{|\nabla
v|^{p_{\varepsilon_{j}}(x)}}{p_{\varepsilon_{j}}(x)}+B_{\varepsilon_{j}}(v)+{f^{{\varepsilon}_{j}}}v\Big{)}\,dx}$
in $\Omega$ such that $u^{\varepsilon_{j}}\rightarrow u$ uniformly on compact
subsets of $\Omega$, ${f^{{\varepsilon}_{j}}}\rightharpoonup f$ $*-$weakly in
$L^{\infty}(\Omega)$, $p_{\varepsilon_{j}}\to p$ uniformly on compact subsets
of $\Omega$ and ${\varepsilon_{j}}\to 0$.

Then, $u$ is a viscosity solution to (1.1) in $\Omega$ with
$g(x)=(\frac{p(x)}{p(x)-1}\,M)^{1/p(x)}$ and $M=\int\tilde{\beta}(s)\,ds$.

Let $x_{0}\in F(u)$ be such that $F(u)$ is a Lipschitz graph in a neighborhood
of $x_{0}$, then $F(u)$ is $C^{1,\alpha}$ in a (smaller) neighborhood of
$x_{0}$.

Let $x_{0}\in F(u)$ be such that $F(u)$ has a normal in the measure theoretic
sense, then $F(u)$ is $C^{1,\alpha}$ in a neighborhood of $x_{0}$.

Moreover, there is a subset $\mathcal{R}$ of $F(u)$ which is locally a
$C^{1,\alpha}$ surface. The set $\mathcal{R}$ is open and dense in $F(u)$ and
the remainder of the free boundary has $(n-1)-$dimensional Hausdorff measure
zero.

###### Proof.

We argue exactly as in the proof of Proposition 8.1. We apply again our
results in Proposition 7.2 and Theorems 1.2 and 6.3, and in this case we make
use of Theorems 5.3, 4.3, 4.4 and Remark 4.2 in [LW3], and Theorem 5.3 in
[LW1]. ∎

We also obtain

######  Remark 8.3.

In [LW4] an optimization problem with volume constraint for an energy
associated to the inhomogeneous $p(x)$-Laplacian was considered. By means of a
penalization technique, it was shown that nonnegative minimizers $u$ are weak
solutions to (1.1) in a bounded domain $\Omega$ in the sense of Definition 7.1
with $g(x)=(\frac{p(x)}{p(x)-1}\,\lambda_{u})^{1/p(x)}$, where $\lambda_{u}>0$
is a constant.

Under the assumptions we made on $p$ and $f$ at the beginning of present
section, by combining our results with those in [LW4], we can argue as in
Propositions 8.1 and 8.2 and obtain the same conclusions for $u$ and $F(u)$.

######  Remark 8.4.

In Propositions 8.1 and 8.2 and Remark 8.3, our $C^{1,\alpha}$ regularity
results on $F(u)$ under the Lipschitz assumption on $F(u)$ follow from the
application of Theorem 1.2 in the present paper and are new.

We want to point out that the rest our $C^{1,\alpha}$ regularity results on