\documentclass[reqno]{amsart}
\usepackage{hyperref}
\usepackage{esint}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 123, pp. 1--12.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2016 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2016/123\hfil 
 Perron's method for $p$-harmonious functions]
{Perron's method for $p$-harmonious functions}

\author[D. Hartenstine, M. Rudd \hfil EJDE-2016/123\hfilneg]
{David Hartenstine, Matthew Rudd}

\address{David Hartenstine \newline
Department of Mathematics \\
Western Washington University \\
Bellingham, WA  98225, USA}
\email{david.hartenstine@wwu.edu}

\address{Matthew Rudd \newline
Department of Mathematics \\
The University of the South \\
Sewanee, TN 37383, USA}
\email{mbrudd@sewanee.edu}


\thanks{Submitted March 2, 2016. Published May 16, 2016.}
\subjclass[2010]{35J92, 39B22, 35A35, 35B05, 35D40}
\keywords{Mean-value property; median; $p$-harmonic functions; 
 $p$-Laplacian; \hfill\break\indent 
$p$-harmonious functions; Perron method}

\begin{abstract}
 We show that Perron's method produces continuous $p$-harmonious functions
 for $1<p<2$.  Such functions approximate $p$-harmonic functions and
 satisfy a functional equation involving a convex combination of the
 mean and median, generalizing the classical mean-value property
 of harmonic functions.  Simple sufficient conditions for the existence of
 barriers are given.  The $p=1$ situation, in which solutions to the
 Dirichlet problem may not be unique, is also considered.
 Finally, the relationship between $1$-harmonious functions and functions
 satisfying a local median value property is discussed.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{definition}[theorem]{Definition}
\allowdisplaybreaks

\section{Introduction}\label{intro}

Let $\Omega \subset \mathbb{R}^{N}$ be a bounded domain, and let $h > 0$,
$q \in (0,1)$, and $g \in C(\partial \Omega)$
be given.  Consider the Dirichlet problem
\begin{equation} \label{sfe}
\begin{gathered}
u  =  M^{h}_{q} u \quad \text{in }  \Omega \,, \\
u  =  g \quad \text{on }  \partial \Omega \,,
\end{gathered}
\end{equation}
where the statistical operator $M^{h}_{q} \colon C(\Omega) \to C(\Omega)$
is defined by
\begin{equation} \label{medianM}
\left( M^{h}_{q} \varphi \right)(x)
:=  (1-q)  \operatorname{median}_{\partial B^{h}_{x}}{ \{\varphi\} } +
q  \fint_{\partial B^{h}_{x}}{ \varphi(y) \, dy } \,, \quad \text{for }
 x \in \Omega ,
\end{equation}
and the open balls $B^{h}_{x} \subset \Omega$ are defined by
\begin{equation} \label{bh}
B^{h}_{x}  :=  B(x, r^{h}(x)), \quad \text{with }
r^{h}(x)  :=   \begin{cases}
\sqrt{2h} & \text{if }  \operatorname{dist}(x, \partial \Omega) \geq \sqrt{2h} \,, \\
\operatorname{dist}(x, \partial \Omega) & \text{otherwise} \,.
\end{cases}
\end{equation}
We seek a solution $u \in C(\overline{\Omega})$.
Because of their connection to $p$-harmonic functions, functions
satisfying $u=M_{q}^{h}$ are called $p$-harmonious.
In \cite{hartenstine:sfe13} we showed that $C(\overline{\Omega})$
solutions to \eqref{sfe} are unique, and that a solution exists when
there exist both a subsolution $v \in C(\overline{\Omega})$ with $v=g$ on
$\partial \Omega$ and a supersolution $w \in C(\overline{\Omega})$
with $w=g$ on $\partial \Omega$.  When $\Omega$ is strictly convex, such
a sub/supersolution pair can be found for any $g \in C(\partial \Omega)$
by solving Dirichlet problems for a Monge-Amp\`{e}re equation.
The main difficulty in applying this result when $\Omega$ is not strictly
convex is finding a subsolution and supersolution.  In this article,
we show that a $C(\overline{\Omega})$ solution can be produced using Perron's
 method under less restrictive hypotheses.

Note that if $q=1$ in \eqref{sfe}, a harmonic function $u \in C(\overline{\Omega})$
 with boundary values $g$ is a solution.  Thus, the equation $u = M^h_q u$ is
 a generalization of the well-known mean-value property of harmonic functions.
However, note that if $u = M^h_q u$, $u(x)$ need only equal its nonlinear
average over a single sphere centered at $x$, rather than all such
spheres contained in the domain.

We call solutions of \eqref{sfe} \textit{$p$-harmonious} because, as
$h \downarrow 0$, the solutions $u_{h}$ converge locally uniformly to the
unique $p$-harmonic function $u$ in $\Omega$ such that $u = g$ on
$\partial \Omega$,
with $p \in (1,2)$ determined by $q$ and $N$; the convergence proofs
in \cite{hartenstine:sfe13} and \cite{rudd:sef15} rely on averaging over
spheres with the particular function $r^{h}(\cdot)$ defined above.
The term \textit{$p$-harmonious} first appeared in \cite{manfredi:dpp12},
where the authors employed a convex combination of the mean and the midrange
of $\varphi$ over a ball $B$ (defined by $(1/2) (\sup_B \varphi + \inf_B \varphi)$))
instead of our convex combination of the median and mean as in \eqref{medianM},
thereby resulting in a different statistical functional equation.
Solutions of these equations approximate $p$-harmonic functions when $p>2$
(see also \cite{hartenstine:sfe13}).  These equations are closely related
to tug-of-war games, and have also been studied in \cite{luiro:oeu14}.


The game-theoretic approach to the $p$-Laplacian ($1 \leq p \leq \infty$),
pioneered by Kohn and Serfaty \cite{kohn:dam06}, Peres and
 Sheffield \cite{peres:twn08}, and Peres, Schramm, Sheffield and Wilson
\cite{peres:til09}, has led to much recent activity and new insights
about nonlinear elliptic and parabolic equations, with potentially
far-reaching consequences.  See, for example,
\cite{antunovic:til12,armstrong:fda12,charro:mpi09,luiro:hip13,manfredi:dpp12b,
gomez:tgi13,lopezsoriano:ilt13,llorente:nuc14,llorente:mvp15,ferrari:cpf14,
daniel:gin13,delpezzo:tgp14,rossi:tgp11}.


Recall that $u$ is $p$-harmonic in $\Omega$ if and only if it is a solution
of the $p$-Laplace equation \cite{heinonen:npt93,lindqvist:npl06},
\begin{equation} \label{p-lap-eqn}
-\Delta_{p} u = 0 \quad \text{in }  \Omega\,;
\end{equation}
for $p \in (1,\infty)$, the $p$-Laplacian $-\Delta_{p}$ is the operator
defined formally by
\begin{equation} \label{p-lap}
-\Delta_{p} \varphi  :=  -\text{div}\left( |D \varphi |^{p-2} D \varphi \right) \,.
\end{equation}
Since the $p$-Laplacian is singular when $p < 2$ and degenerate when $p > 2$,
solutions of \eqref{p-lap-eqn} must be defined
in a weak sense; one can choose either the viscosity or the variational definition,
as Juutinen et al \cite{julin:npe12,juutinen:evs01}
showed that these definitions are equivalent.

For a smooth function $u$ with nonvanishing gradient, we define the
 $1$-Laplacian $\Delta_1$ by
 $\Delta_1 u = |Du|\operatorname{div}{ \left( Du/|Du| \right) }$.
 Note that this definition of $\Delta_1 u$ differs by a factor of $|Du|$
from what is obtained in \eqref{p-lap} with $p=1$.
 With this definition, a calculation shows that
\begin{equation}\label{p-lap-decomp1}
\Delta_p u = |Du|^{p-2} \left( \, (p-1) \Delta u + (2-p) \Delta_{1} u \, \right)
\end{equation}
for any smooth $u$ with nonvanishing gradient.
A normalized 1-homogeneous $p$-Laplacian defined by
$\Delta^N_p u = |Du|^{2-p}\Delta_p u$ is thus a linear combination of
$\Delta u$ and $\Delta_1 u$.  This combination is convex when $1 \leq p \leq 2$.
 Note that if $Du$ is never zero, $\Delta_p u= 0$ if and only if $\Delta_p^N u = 0$.
 Furthermore, this equivalence holds in the weak sense as well as the viscosity
sense  by the results of \cite{juutinen:evs01}.  The connection between $p$-harmonic
functions and the statistical functional equations defining $p$-harmonious
functions arises from the relationship between $\Delta u(x)$ and
$\fint_{\partial B_h^x} u$ and that between $\Delta_1 u(x)$ and
$\operatorname{median}_{\partial B^{h}_{x}}\{ u \}$, both of which follow 
from Taylor's theorem.
 More precisely, as shown in \cite{hartenstine:asc11}, a function is
$p$-harmonic in two dimensions if and only if it satisfies an asymptotic
mean/median value property; combining the calculations in \cite{hartenstine:asc11}
 with those in \cite{rudd:sef15} shows that this result holds in all dimensions
$N \geq 2$.

This asymptotic description of $p$-harmonic functions was motivated by other
asymptotic statistical characterizations of $p$-harmonic functions
\cite{manfredi:amv10e}, which rely on a decomposition of the $p$-Laplacian
that is different than \eqref{p-lap-decomp1}.
For a smooth function with nonzero gradient, we define the
$\infty$-Laplacian $\Delta_{\infty}$ by
\[
\Delta_{\infty} u = |Du|^{-2} \Sigma_{i,j} u_i u_j u_{ij}.
\]
 Then the equation
\begin{equation}\label{p-lap-decomp2}
\Delta_p u = |Du|^{p-2} ( \Delta u + (p-2) \Delta_{\infty}u)
\end{equation}
holds for any smooth $u$ with nonvanishing gradient.
Again dividing by $|Du|^{p-2}$ results in a normalized $p$-Laplacian,
equal to the one given above.  Exploiting the connection between
$\Delta_{\infty} u (x)$ and the midrange of $u$ over a ball centered at
 $x$, Manfredi, Parviainen and Rossi obtained that $u$ is $p$-harmonic
if and only if it is equal, in an asymptotic sense, to a linear combination
of its mean and midrange.  This combination is convex when $p>2$.
Thus, the median and midrange play similar roles in the $1<p<2$ and $p>2$
realms respectively.  We remark that the normalized $p$-Laplacian can
also be decomposed as $\Delta_p^N u = \Delta_1 u + (p-1) \Delta_{\infty} u$.
This identity was used to obtain characterizations of $p$-harmonic functions
for $1 \leq p \leq \infty$ in terms of lower dimensional averages
in \cite{kawohl:snp12}.

If $q$ is taken to be zero in \eqref{sfe}, the situation is different.
 In \cite{rudd:mvh13}, it was shown that functions satisfying a local median
value property are $1$-harmonic in the viscosity sense
(this is using the definition of $1$-harmonic that has the gradient factor
in front: $\Delta_1 u = |Du| \mathrm{div}(Du/|Du|)$).
It was also demonstrated that solutions to the Dirichlet problem for the
local median value property may not exist, and when they do exist,
may fail to be unique. Elementary geometric reasoning shows that the
1-parameter family of functions $u_{\alpha}$ in \cite{rudd:mvh13} that
satisfy the local median value property are also $1$-harmonious,
at least for $h$ sufficiently small.  These functions all agree on
the boundary of their domain, so $1$-harmonious functions are not uniquely
determined by their boundary values. This implies that a comparison
principle for the classes $S_g$ and $S^g$ as used below in
Section 3 cannot hold when $q=0$ (since it would immediately imply uniqueness).
 However, in this case, the arguments used in Section~\ref{proof} do
lead to the existence of weak solutions that bound any $C(\overline{\Omega})$
solutions (Theorem~\ref{perronmed}).

The rest of the paper is organized as follows.
In Section \ref{prelim}, we recall the definition and some fundamental properties
of the median of a function and establish properties of additional
 mean/median operators needed for the proofs of the main results,
Theorems~\ref{perronsol} and \ref{perronmed}. In the following section,
 the definitions of sub- and supersolutions are expanded from those
used in \cite{hartenstine:sfe13}, the upper and lower Perron solutions are defined,
and Theorem~\ref{perronsol},  stating that when $q>0$ and  the upper and
lower Perron solutions, $u^P$ and $u_P$ respectively, are equal on
$\partial \Omega$ they are equal throughout $\overline{\Omega}$ and
solve \eqref{sfe}, is proved.  As in the classical Perron method, the equality
of $u^P$ and $u_P$ follows from the existence of barriers.  However, it is not
obvious how to construct barriers for the problem considered here,
due to the nonlocal nature of the operator $M^h_q$.
This issue is also discussed in Section \ref{proof}. Section~\ref{q=0} concerns
the situation when $q=0$.  The relationship between $1$-harmonious functions
and those satisfying the local median value property is explored in Section 5,
 where an example of a function satisfying the seemingly stronger requirements
of the local median value property but is not $1$-harmonious for any $h>0$ is given.

\section{Preliminaries}\label{prelim}

Throughout this article $\Omega \subset \mathbb{R}^N$ is a bounded domain.
We begin by reviewing some properties of medians.  Recall the following
definition \cite{hartenstine:sfe13,ziemer:wdf89}.

\begin{definition} \label{median-defn} \rm
If $u$ is a real-valued integrable function on the measurable set
$E \subset \mathbb{R}^{k}$ and $0<|E| < \infty$,
then $m$ is a \emph{median} of $u$ over $E$
if and only if
\[
|  \{x\in E  :  u(x) < m \}  |  \leq  \frac{1}{2}  | E  |
\quad  \text{and} \quad
|  \{x \in E  :  u(x) > m \} |  \leq  \frac{1}{2}  |  E  | \,.
\]
\end{definition}

The set ${ \operatorname{median}_{E}{ \{ u \} } } $ of all medians of $u$ over $E$
is a non-empty compact interval,
\begin{equation} \label{homo-trans}
\operatorname{median}_{E}{ \{ \alpha u + \beta \} } = \alpha \operatorname{median}_{E}{ \{ u \} } + \beta \, ,
\quad \text{for }  \alpha, \beta \in \mathbb{R} \, ,
\end{equation}
and it is easy to construct examples of semicontinuous functions with multiple
medians over a given set.
This does not happen when working with continuous functions, since continuous
functions have unique medians on compact connected sets \cite{hartenstine:sfe13}.

To deal with non-continuous functions, which do not necessarily have unique medians,
we introduce two new mean/median operators: $\underline{M}_{q,h}$ and
$\overline{M}_{q,h}$.  For an integrable function
$u:\overline{\Omega} \to \mathbb{R}$, $q \in [0,1]$ and $h>0$, let
\begin{equation}
\begin{gathered}
\underline{M}_{q,h} u(x) = q \operatorname{mean}_{\partial B_x^h} u + (1-q)
\min \operatorname{median}_{\partial B_x^h} u ,\\
\overline{M}_{q,h} u(x) = q \operatorname{mean}_{\partial B_x^h} u + (1-q) \max
\operatorname{median}_{\partial B_x^h} u
\end{gathered}
\end{equation}
for $x \in \Omega$, and for $x \in \partial \Omega$,
let $\underline{M}_{q,h} u(x) = \overline{M}_{q,h} u(x) = u(x)$.
These operators satisfy a natural monotonicity property.

\begin{proposition} \label{mono}
Let $u$ and $v$ be real-valued integrable functions on $\Omega$ with $u \leq v$.
Then for any $q \in [0,1]$, $\underline{M}_{q,h} u(x) \leq \underline{M}_{q,h} v(x)$
and $\overline{M}_{q,h} u(x) \leq \overline{M}_{q,h} v(x)$ for all $x \in \Omega$.
\end{proposition}

\begin{proof}
We prove the claim for $\underline{M}_{q,h}$; the one for $\overline{M}_{q,h}$
is similar.  Since $u\leq v$,
 $\operatorname{mean}_{\partial B_x^h} u \leq \operatorname{mean}_{\partial B_x^h} v$ for any $x$ and $h$,
so the claim holds if $\min \operatorname{median}_{\partial B_x^h} u \leq \min \operatorname{median}_{\partial B_x^h} v$.
If this is not true, then there exist $x \in \Omega$ and a number $m$ that
is a median of $v$ that is too small to be a median of $u$ on $\partial B_x^h$.
 Because $m$ is a median of $v$,
\[
|\{y \in \partial B_x^h: v(y) \leq m\}| \geq (1/2)|\partial B_x^h|.
\]
Since $u \leq v$,
$\{y \in \partial B_x^h: v(y) \leq m\} \subset \{y \in \partial B_x^h: u(y) \leq m\}$,
 so
\[
|\{y \in \partial B_x^h: u(y) \leq m\}| \geq (1/2)|\partial B_x^h|.
\]
On the other hand, since $m$ is too small to be a median of $u$ in
$\partial B_x^h$, we must have that
\[
|\{y \in \partial B_x^h: u(y) \leq m\}| < (1/2)|\partial B_x^h|.
\]
This is a clear contradiction and the claim holds.
\end{proof}

The next result shows that these operators preserve semicontinuity.

\begin{proposition}\label{semicont}
If $u \in LSC (\bar \Omega) \cap L^{\infty}(\Omega)$,  then
$\underline{M}_{q,h} u \in LSC (\bar \Omega)$, and if
$v \in USC (\bar \Omega) \cap L^{\infty} (\Omega)$, then
$\overline{M}_{q,h} v \in USC (\bar \Omega)$.
\end{proposition}

\begin{proof}
Let $x \in \Omega$, and suppose $\{x_n\} \subset \Omega$ is such that $x_n \to x$.
Define the map $\rho_n: \partial B_x^h \to \partial B_{x_n}^h$ by
 $\rho_n(y) = x_n + r^h (x_n)/r^h(x) (y -x)$.  Define $u_n (y) = u (\rho_n(y))$
for $y \in \partial B_x^h$.  Since $\rho_n(y) \to y$ as $n \to \infty$ and
$u$ is lower semicontinuous at $y$, we have that
$u(y) \leq \liminf_{n \to \infty} u_n(y)$ for all $y \in \partial B_x^h$.
Integrating this last inequality and using Fatou's Lemma (note that because $u$
is bounded, we may assume that $u$ (and therefore $u_n$) is non-negative), we obtain
\begin{equation}
\int_{\partial B_x^h} u(y) \, dy
\leq \int_{\partial B_x^h} \liminf_{n \to \infty} u_n (y) \, dy
\leq \liminf_{n \to \infty} \int_{\partial B_x^h} u_n (y) \, dy.
\end{equation}
This implies that the function $z \to \operatorname{mean}_{\partial B_z^h} u$ is lower
semicontinuous at $x$, and hence in $\Omega$.

We now turn to the minimal median.
 We begin by establishing that for any $\alpha \in \mathbb{R}$,
\begin{equation}\label{measest}
|\partial B_x^h \cap \{u > \alpha\}| \leq \liminf_{n \to \infty}
|\partial B_x^h \cap \{u_n > \alpha\}|.
\end{equation}
Let $\alpha \in \mathbb{R}$, and let $f_{\alpha} = \chi_{\partial B_x^h \cap \{u > \alpha\}}$
and $f_{\alpha}^n = \chi_{\partial B_x^h \cap \{u_n > \alpha\}}$.
We have that $f_{\alpha} (y) \leq \liminf_{n \to \infty} f_{\alpha}^n (y)$
for all $y \in \partial B_x^h$, so \eqref{measest} follows again by Fatou.

Now let $\alpha = \min \operatorname{median}_{\partial B_x^h} u$.
Then for any $\varepsilon > 0$, $\alpha - \varepsilon$ is too small to be a median of $u$
on $\partial B_x^h$, so
$|\partial B_x ^h \cap \{u > \alpha - \varepsilon\}| > (1/2) |\partial B_x^h|$
and by \eqref{measest},
\[
\liminf_{n \to \infty} |\partial B_x^h \cap \{u_n > \alpha - \varepsilon\}|
> (1/2)|\partial B_x^h|.
\]
So for all but finitely many $n$, $|\partial B_x^h \cap
\{u_n > \alpha - \varepsilon\}| > (1/2)|\partial B_x^h|$. For each such $n$,
 $\alpha - \varepsilon$ is too small to be a median, so for these $n$,
 $\min \operatorname{median}_{\partial B_x^h} u_n \geq \alpha - \varepsilon$. Since this holds for
all but finitely many $n$, we obtain
\[
\liminf_{n \to \infty} \min \operatorname{median}_{\partial B_x^h} u_n \geq \alpha - \varepsilon.
\]
Since $\varepsilon$ is arbitrary, we conclude
\[
\liminf_{n \to \infty} \min \operatorname{median}_{\partial B_x^h} u_n
= \liminf_{n \to \infty} \min \operatorname{median}_{\partial B_{x_n}^h} u \geq \alpha.
\]
Recalling the definition of $\alpha$, we see that the minimal median operator
is lower semicontinuous at $x$, and therefore in $\Omega$.

To show $\underline{M}_{q,h} u \in LSC(\overline{\Omega})$, we show that for
any $\alpha \in \mathbb{R}$,
$\{z \in \overline{\Omega}: \underline{M}_{q,h} u (z) > \alpha\}$
is open relative to $\overline{\Omega}$. We already have that
$\underline{M}_{q,h}u$ is LSC in $\Omega$, so if $x \in \Omega$ is such that
$\underline{M}_{q,h}u(x) > \alpha$, the same is true in a neighborhood of
 $x$ in $\Omega$. Now suppose $x \in \partial \Omega$ is such that
$u(x) = \underline{M}_{q,h}u (x) >\alpha$. Because $u \in LSC(\overline{\Omega})$,
$u > \alpha$ in some neighborhood $\mathcal{N}$ of $x$ in $\overline{\Omega}$.
Then for all $y \in \Omega$ sufficiently close to $x$,
$\partial B_y^h \subset \mathcal{N}$, so $\underline{M}_{q,h} u(y) >\alpha$.
If $y \in \mathcal{N} \cap \partial \Omega$,
$u(y) = \underline{M}_{q,h}u(y) > \alpha$, so we see that $x$ is an interior
point of $\{z \in \overline{\Omega}: \underline{M}_{q,h} u (z) > \alpha\}$,
and the proof is complete.
\end{proof}

We also need the following result concerning the continuity of the minimal
and maximal mean and median operators with respect to monotone sequences
of functions.

\begin{proposition}\label{monoconv}
 Suppose that $\{v_n\} \subset LSC (\overline{\Omega})$ is a nondecreasing
sequence of bounded functions  converging pointwise to the bounded function
$v \in LSC(\overline{\Omega})$.  Then for all $x \in \Omega$,
$\underline{M}_q^h v_n (x) \to \underline{M}_q^h v(x)$.
Suppose that $\{w_n\} \subset USC (\overline{\Omega})$ is a nonincreasing
sequence of bounded functions  converging pointwise to the bounded
function $w \in USC(\overline{\Omega})$.  Then for all $x \in \Omega$,
 $\overline{M}_q^h w_n (x) \to \overline{M}_q^h w (x)$.
\end{proposition}

\begin{proof}
Again, we prove the claim for $\underline{M}_q^h$. By Proposition~\ref{mono},
for each $x \in \Omega$, the sequence $\{\underline{M}_q^h v_n (x)\}$
is nondecreasing and bounded above by $\underline{M}_q^h v(x)$.
Therefore $\lim_{n \to \infty} \underline{M}_q^h v_n (x)$ exists and is
less than or equal to $\underline{M}_q^h v(x)$.  Fix $x \in \Omega$ and
let $m = \min \operatorname{median}_{\partial B_x^h}v$.  The means of $v_n$ over
$\partial B_x^h$ converge to the mean of $v$ on $\partial B_x^h$
by the dominated convergence theorem
(see the proof of \cite[Theorem 3.2]{hartenstine:sfe13}),
so the claim holds if
 $\min \operatorname{median}_{\partial B_x^h} v_n \to \min \operatorname{median}_{\partial B_x^h} v$.
Suppose that this is not the case.  Then there exists $\varepsilon > 0$ such
that $m- \varepsilon > m_* := \lim_{n \to \infty} \min \operatorname{median}_{\partial B_x^h} v_n$.
 The number $m - \varepsilon$ is too small to be a median of $v$ on $\partial B_x^h$, so
\[
|\partial B_x^h \cap \{v \leq m- \varepsilon\}| < (1/2) |\partial B_x^h|.
\]
However, $m - \varepsilon$ is not too small to be a median of $v_n$ on
$\partial B_x^h$ so
\[
|\partial B_x^h \cap \{v_n \leq m- \varepsilon\}| \geq (1/2) |\partial B_x^h|.
\]
Note that this is true for all $n$, since
$\min \operatorname{median}_{\partial B_x^h} v_n \leq m_*$ for all $n$ by the monotonicity of
$\{v_n\}$.   Therefore,
\[
\lim_{n \to \infty} |\partial B_x^h \cap \{v_n \leq m- \varepsilon\}|
\geq (1/2) |\partial B_x^h|.
\]
But, again using the monotonicity of the sequence and continuity from above, we have
\begin{align*}
\lim_{n \to \infty} |\partial B_x^h \cap \{v_n \leq m- \varepsilon\}| 
&= |\cap_{n=1}^{\infty} (\partial B_x^h \cap \{v_n \leq m - \varepsilon\}) | \\
&= |\partial B_x^h \cap \{v \leq m-\varepsilon\}| < (1/2) |\partial B_x^h|,
\end{align*}
contradicting the previous line.
\end{proof}

\section{Perron method and barriers}\label{proof}

In this section we assume that $0<q \leq 1$.  
We begin by defining \emph{subsolutions}
 and \emph{supersolutions} for the problem \eqref{sfe} and the Perron solutions. 
 Let
\begin{gather*}
S_g = \{v \in C(\overline{\Omega}): v \leq M^{h}_{q} v  \text{ and } 
 v \leq g  \text{ on } \partial \Omega\} , \\ 
S^g = \{w \in C(\overline{\Omega}): w \geq M^{h}_{q} w  \text{ and } 
 w \geq g  \text{ on } \partial \Omega\}.
\end{gather*}
The \emph{lower Perron solution} is then
\[
u_P(x) = \sup \{v(x): v \in S_g\}
\]
\noindent
and the \emph{upper Perron solution} is
\[
u^P(x) = \inf \{w(x): w \in S^g\}.
\]

The Perron solutions are well-defined and finite at all $x \in \Omega$.  
First note that the constant functions $\min_{\partial \Omega} g$ and 
$\max_{\partial \Omega} g$ belong to $S_g$ and $S^g$ respectively so 
these sets of functions are nonempty.  By a simple modification of the 
proof of the comparison principle (Theorem~3.1) in \cite{hartenstine:sfe13}, 
we have that $v \leq w$ for all $v \in S_g$ and $w \in S^g$.  
It follows that $u_P \leq u^P$ in $\overline{\Omega}$ and that both $u_P$ and 
$u^P$ are bounded (by $\min_{\partial \Omega} g$ and $\max_{\partial \Omega} g$). 
 By construction we have that $u_P \in LSC (\overline{\Omega})$ and
 $u^P \in USC (\overline{\Omega})$.

Observe that if \eqref{sfe} has a solution $u \in C(\overline{\Omega})$, then 
$u \in S_g \cap S^g$, and that any function in $S_g \cap S^g$ is a 
$C(\overline{\Omega})$ solution.  Also note that if \eqref{sfe} has a solution, 
then that solution must coincide with $u_P$ and $u^P$.  
To see this, suppose $u \in C(\overline{\Omega})$ is a solution of \eqref{sfe}.  
Then $u \in S_g$, so that $u \leq u_P$.  Similarly, $u \in S^g$, so $u \geq u^P$, 
but $u_P \leq u^P$, so we see $u=u_P=u^P$.

The following theorem is the main result of this article.

\begin{theorem} \label{perronsol}
If $u_P = u^P$ $(=g)$ on $\partial \Omega$, then $u_P = u^P$ in $\overline{\Omega}$ 
and the Perron solutions solve \eqref{sfe}.
\end{theorem}

\begin{proof}
By assumption, $u_P =g$ on $\partial \Omega$.  By Proposition~\ref{mono} 
and the definition of $u_P$, 
$v \leq M_q^h v = \underline{M}_q^h v \leq \underline{M}_q^h u_P$ for any 
$v \in S_g$.  Taking the sup over all $v \in S_g$, we get 
$u_P \leq \underline{M}_q^h u_P$.  We now define a sequence of 
$LSC(\overline{\Omega})$ functions by taking
\[
u_1 = u_P \quad \text{and} \quad u_{j+1} = \underline{M}_q^h u_j, \quad j \geq 1.
\]
Each $u_j$ belongs to $LSC(\overline{\Omega})$ by Proposition~\ref{semicont},
 $u_j = g$ on $\partial \Omega$ by construction, and the sequence $\{u_j\}$ 
is nondecreasing by Proposition~\ref{mono}.  Furthermore, since
 $u_1 \leq \max_{\partial \Omega} g$, we get that 
$u_j \leq \max_{\partial \Omega} g$ for all $j$.  
Therefore $u_j$ converges pointwise to some bounded function 
$u \in LSC(\overline{\Omega})$ with $u=g$ on $\partial \Omega$.  
By Proposition~\ref{monoconv}, we have that 
$\underline{M}_q^h u_j \to \underline{M}_q^h u$ pointwise, so that
\[
u(x) = \lim_{j \to \infty} u_{j+1} (x) 
= \lim_{j \to \infty} \underline{M}_q^h u_j (x) = \underline{M}_q^h u(x)
\]
for all $x \in \Omega$.  Thus $u(x) = \underline{M}_q^h u (x)$ for all 
$x \in \Omega$.  Similarly, define a sequence $U_j \in USC(\overline{\Omega})$ by
\[
U_1 = u^P \quad \text{and} \quad U_{j+1} = \overline{M}_q^h U_j, \quad j \geq 1.
\]
Letting $U (x) = \lim_{j \to \infty} U_j(x)$, we get that 
$U \in USC(\overline{\Omega})$, $U = \overline{M}_q^h U$ in $\Omega$ and 
$U = g$ on $\partial \Omega$.  Note that by construction, $u_j \leq U_j$ 
for all $j$, so that $u \leq U$.  We are now in exactly the same situation 
as in the proof of \cite[Theorem 3.2]{hartenstine:sfe13}, 
following equation (3.18) in that paper. Running precisely the same argument
 as in that proof, we obtain $u = U$ in $\overline{\Omega}$, 
which implies that they belong to $C(\overline{\Omega})$,
 and therefore that $u = M_q^h u = M_q^h U = U$ in $\Omega$, so that 
$u$ solves \eqref{sfe}.  Finally, this implies that $u \in S_g$, 
and since $u_P \leq u$, we obtain that $u=u_P$ so that $u_P$ is in fact 
a solution of \eqref{sfe}.  We note also that $u_P = u^P$.
\end{proof}

A natural question then is: Under what conditions can it be guaranteed 
that $u_P = u^P$ on $\partial \Omega$? As in the classical Perron method, 
the existence of a solution in $C(\overline{\Omega})$ can be reduced to 
the existence of barriers at each point.  Let $x \in \partial \Omega$. 
 A \emph{lower barrier at $x$} is a function $v \in S_g$ with $v(x) = g(x)$, 
and an \emph{upper barrier at $x$} is a function $w \in S^g$ with $w(x) = g(x)$. 
 If at each point of $\partial \Omega$, there exist both an upper and a 
lower barrier, then Theorem~\ref{perronsol} can be applied to produce a 
$C(\overline{\Omega})$ solution to \eqref{sfe}.  
Due to the nonlocal nature of $M_q^h$, it is unclear how to produce barriers 
in the standard way (by solving PDE problems). Note that while solutions of 
the functional equation $u = M_q^h u$ are related to $p$-harmonic functions, 
$p$-harmonic functions in general do not solve this functional equation 
\cite{hartenstine:asc11}.  Furthermore, it is not obvious how to use the 
solvability of \eqref{sfe} in balls contained in $\Omega$, since the change 
in domain required to do so also changes the operators.

While necessary and sufficient conditions on $g$ and $\Omega$ for the 
solvability of \eqref{sfe} are not known at this time, it is possible 
to give some simple sufficient conditions.  Any linear function $L$ 
satisfies $L = M_q^h L$ for any $h>0$ and any $q \in [0,1]$, so linear 
functions are possible barriers. If $g$ is such that at every 
$x \in \partial \Omega$ we can find upper and lower linear barriers, 
then $u_P = u^P = g$ on $\partial \Omega$, and \eqref{sfe} has a solution. 
More generally, if the convex and concave envelopes of $g$ coincide on 
$\partial \Omega$, \eqref{sfe} has a solution.  The convex envelope of $g$, 
defined for $x \in \overline{\Omega}$, is
\[
g_* (x) = \sup \{v(x) \in C(\overline{\Omega}): v  \text{ is convex, } 
v \leq g  \text{ on } \partial \Omega\},
\]
and the concave envelope $g^*$ is defined similarly.  
If $\Omega$ is not convex, we consider $v \in C(\overline{\Omega})$ 
to be convex if $v$ is the restriction of a convex function defined on 
a convex domain to $\overline{\Omega}$.  It was shown in \cite{hartenstine:sfe13} 
that a convex (concave) function $u$ ($v$) satisfies $u \leq M_q^h u$ 
($v \geq M_q^h v$) for any $h>0$ and any $q \in [0,1]$.  
Thus $g_* \leq u_P$ and $g^* \geq u^P$, so that if $g_* = g^*$ on 
$\partial \Omega$ the hypotheses of Theorem~\ref{perronsol} are met.   
 We also see that regardless of whether $g_* = g^*$ on $\partial \Omega$, 
if $u \in C(\overline{\Omega})$ is a solution of \eqref{sfe}, 
$g_* \leq u \leq g^*$ in $\overline{\Omega}$.

\begin{remark}\label{weakPerron} \rm
We note that if $u_P \ne u^P$, the above method still produces weak solutions 
to \eqref{sfe}.  Namely by iterating $u_P$ as in the proof of
 Theorem~\ref{perronsol}, we produce $u \in LSC(\overline{\Omega})$ with
 $u = \underline{M}_q^h u$ in $\Omega$ and $u(x) = \sup_{ v \in S_g} v (x)$ 
for $x \in  \partial \Omega$.  Similarly, there exists 
$U \in USC(\overline{\Omega})$ with $U = \overline{M}_q^h U$ in $\Omega$ 
and $U(x) = \inf_{ w \in S^g} w (x)$ for $x \in  \partial \Omega$.
\end{remark}


\section{Perron method when $q=0$}\label{q=0}

While the results of Section~\ref{prelim} apply when $q=0$, some of the 
arguments in Section~\ref{proof} do not.  As indicated near the end of 
Section~\ref{intro}, $C(\overline{\Omega})$ solutions of \eqref{sfe} in 
the $q=0$ case, when they exist, need not be unique.  
This non-uniqueness demonstrates the impossibility of a comparison principle 
of the type used in Section~\ref{proof}.  This comparison principle was 
used to obtain the boundedness of the Perron solutions as well as the 
inequality $u_P \leq u^P$.  Thus, while the Perron solutions can be defined 
in the $q=0$ case in precisely the same way as in the previous section 
(note that $S_g$ and $S^g$ are nonempty since they contain the constant 
functions $\min_{\partial \Omega}g$ and $\max_{\partial \Omega} g$ respectively), 
we do not  know \emph{a priori} that they are finite.  
In this section, we show that the Perron solutions are bounded and that the
 method used in Section~\ref{proof} produces weak solutions to \eqref{sfe} 
that bound all possible $C(\overline{\Omega})$ solutions.  
These weak solutions need only be semicontinuous and satisfy functional 
equations involving the minimal or maximal median operator.

The following maximum and minimum principle immediately implies the finiteness 
of the Perron solutions.

\begin{proposition}\label{medianmaxmin}
If $v \in C(\overline{\Omega})$ satisfies $v \leq M_0^h v$ in $\Omega$, 
then $\max_{\overline{\Omega}} v = \max_{\partial \Omega} v$. 
If $w \in C(\overline{\Omega})$ satisfies $w \geq M_0^h w$ in $\Omega$, 
then $\min_{\overline{\Omega}} w = \min_{\partial \Omega} w$.
\end{proposition}

We remark that the corresponding result also holds when $q>0$ and can be 
proved more easily by a standard argument.

\begin{proof}
We prove the statement for a subsolution $v$ by contradiction.  
If the claim is not true, there exists $v \in C(\overline{\Omega})$ 
that  attains a maximum $T$, where $T > \max_{\partial \Omega} v$, 
and satisfies $v \leq M_0^h v$ in $\Omega$.  The set 
$\Omega_T = \{x \in  \overline{\Omega}: v(x) = T\}$ is nonempty, 
closed and disjoint from $\partial \Omega$.  Consider the convex hull of 
$\Omega_T$, denoted $\operatorname{conv}\Omega_T$.  
This set is compact, convex, and equal to the convex hull of its extreme points.  
Note that $\operatorname{conv}\Omega_T$ need not be a subset of
 $\overline{\Omega}$ if $\Omega$ is not convex.  
Clearly $\Omega_T \subset (\operatorname{conv}\Omega_T) \cap \Omega$.

Let $x$ be an extreme point of $\operatorname{conv}\Omega_T$.  
Then $x$ cannot be written as a nontrivial convex combination of points in
 $\operatorname{conv}\Omega_T$ (in other words, $x \ne ty + (1-t)z$ for any 
$0<t<1$ and $y$, $z \in \operatorname{conv} \Omega_T$).  
By the definition of convex hull, $x$ must be a convex combination of elements 
of $\Omega_T$. But, because this convex combination cannot be nontrivial 
(because $\Omega_T$ is contained in $\operatorname{conv}\Omega_T$), 
$x$ must belong to $\Omega_T$.  Because $x$ is an extreme point of 
$\operatorname{conv}\Omega_T$, there exists a supporting hyperplane 
$H$ to $\operatorname{conv}\Omega_T$ such that 
$H \cap \operatorname{conv}\Omega_T = \{x\}$. 
 This implies that the functional inequality $v \leq M_0^h v$ cannot hold at $x$. 
This is not hard to see: because $\operatorname{conv}\Omega_T \setminus \{x\}$ 
lies (strictly) on one side of $H$, the measure of 
$\partial B_x^h \cap \operatorname{conv}\Omega_T$ is strictly less than half 
the measure of $\partial B_x^h$. Recalling that 
$\Omega_T \subset\operatorname{conv}\Omega_T$, we see that $v< T$ on more 
than half of $\partial B_x^h$ so that $\operatorname{median}_{\partial B_x^h} v < T =v(x)$.
For the facts and definitions from convex geometry used above,  we refer to 
the books \cite{gardner:gt06} and \cite{schneider:cbb93}.
\end{proof}

Consider the problem \eqref{sfe} when $q=0$ and define the classes $S_g$ and $S^g$ 
as in the previous section.  As in the $q>0$ case, any function in $S_g \cap S^g$ 
is a $C(\overline{\Omega})$ solution to the Dirichlet problem \eqref{sfe},
 and any $C(\overline{\Omega})$ solution must belong to $S_g \cap S^g$.  
However, as mentioned in Section~\ref{intro}, we don't generally expect 
$S_g \cap S^g$ to consist of a single element when nonempty.  
The next result shows that when the upper and lower Perron solutions
 coincide on the boundary, they solve \eqref{sfe} in a weak sense and bound
 all possible $C(\overline{\Omega})$ solutions.

\begin{theorem}\label{perronmed}
 If $u_P = u^P$ $(=g)$ on $\partial \Omega$, then there exist 
$u \in LSC(\overline{\Omega})$ and $U \in USC(\overline{\Omega})$, both equal
 to $g$ on $\partial \Omega$,  such that $u = \underline{M}_0^h u$ in $\Omega$ 
and  $U = \overline{M}_0^h U$ in $\Omega$.  Furthermore, if $\overline{u}$ 
is a $C(\overline{\Omega})$ solution of \eqref{sfe} with $q=0$, then 
$U \leq \overline{u} \leq u$.
\end{theorem}

\begin{proof} 
The proof of Theorem~\ref{perronsol} produces the functions $u$ and $U$ 
with the claimed properties.  The part of the argument establishing that 
$u=U$ does not apply when $q=0$ however.  Suppose that $\overline{u}$ is a 
$C(\overline{\Omega})$ solution of \eqref{sfe}.  
Then $\overline{u} \in S_g \cap S^g$.  It follows that $\overline{u} \leq u_P$ 
by the definition of $u_P$ and since $u$ is produced by monotone iteration 
starting with $u_P$ (see the proof of Theorem~\ref{perronsol}), we have that 
$\overline{u} \leq u$.  Similarly, since $\overline{u} \in S^g$, we get that
$\overline{u} \geq U$, proving the last claim.
\end{proof}

If \eqref{sfe} has multiple $C(\overline{\Omega})$ solutions, the inequality 
in Theorem~\ref{perronmed} shows that $u$ and $U$ (and therefore $u_P$ and $u^P$) 
do not coincide.  An analogous statement to Remark~\ref{weakPerron} also 
holds when $q=0$.

It was also shown in \cite{rudd:mvh13} that the Dirichlet problem for the local 
median value property may not have a $C(\overline{\Omega})$ solution.  
$1$-harmonious functions need not satisfy the local median value property 
(since the defining property for $1$-harmonious functions requires that $u(x)$ 
equal its median over a single sphere centered at $x$ and not all such spheres
 with sufficiently small radius as is the case for the local median value property, 
see the next section).  However, this suggests that it may be possible for 
$S_g \cap S^g$ to be empty, so that there is no $C(\overline{\Omega})$ 
1-harmonious function in that domain equal to $g$ on the boundary.


\section{Local median value property and $1$-harmonious functions} \label{lmvp}

Functions satisfying a local median value property were studied in
\cite{rudd:mvh13}.  It was shown there that such functions are 
(normalized) $1$-harmonic in the viscosity sense.  A function 
$u \in C(\Omega)$ satisfies the local median value property if $u(x)$ 
is equal to its median on $\partial B_{\epsilon}(x)$ for all $\varepsilon < R(x)$ where 
$R: \Omega \to \mathbb{R}$ is positive, continuous and satisfies 
$R(x) \leq \mathrm{dist}(x, \partial \Omega)$. This property appears to be 
stronger than the condition defining $1$-harmonious functions since it requires 
that $u(x)$ equal its median over all sufficiently small spheres centered at $x$, 
whereas to satisfy $u =M_{0}^{h} u$, $u(x)$ need only equal its median on the 
single sphere $\partial B_x^h$.  However, for any given $h>0$, 
a function can satisfy the local median value property without satisfying 
$u =M_{0}^{h} u$, since $R(x)$ could be smaller than $r^h(x)$ defined in \eqref{bh}.  
Thus, while these conditions are clearly related, there is no obvious connection 
between them.  Furthermore, as the following example demonstrates, a function $u$ 
can possess the local median value property without satisfying $u=M_0^h u$ 
for \emph{any} $h>0$.

Let $\rho>0$ be small $(\rho<1$ is sufficient, but it is easier to see the claim
 when $\rho \approx 0$).  Let $\Omega = B_1((1,0)) \subset \mathbb{R}^2$. 
 Define $u \in C(\overline{\Omega})$ by
\[
u(x,y) =  \begin{cases} 
0 & \text{if } -\rho x \leq y \leq \rho x, \\
\varepsilon & \text{if }  y = \rho x + \varepsilon  \text{ for }  \varepsilon >0, \\
-\varepsilon & \text{if }  y = -\rho x + \varepsilon \text{ for } \varepsilon < 0.
\end{cases} 
\]
In other words, $u$ is zero in the part of $\overline{\Omega}$ between the 
lines $y = \rho x$ and $y = -\rho x$.  Outside of this sector, $u>0$ and 
is constant along lines parallel to the line $y = \rho x$ above $y=\rho x$ 
and along lines parallel to the line $y = -\rho x$ below $y= -\rho x$.  
It is not hard to see that $u$ satisfies the local median value property 
since its level sets are either line segments or the region bounded  
by the lines $y = \rho x$ and $y = -\rho x$.  Given any $h>0$, 
there exists $\varepsilon>0$ such that $B := B_{(\varepsilon,0)}^h$ is tangent to $\Omega$ 
at the origin.  Then $u(\varepsilon,0) = 0$, but because $\rho < 1$, 
$|\partial B \cap \{u>0\}| > (1/2)|\partial B|$, so that $\operatorname{median}_{\partial B} u > 0$,
and $u$ is not $1$-harmonious.


\subsection*{Acknowledgements}
Much of this work was done during the authors' Research in Pairs stay
at the Mathematisches Forschungsinstitut Oberwolfach in November 2013.
 We thank the MFO for their support and hospitality and for providing such
a stimulating environment.


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