text:
\section{Introduction}
We consider the nonlinear Neumann boundary value problem
\beq\label{N-prob}\left\{\bald
&F(D^2u,Du,u,x)=0 \ \ \IN \gO,
\\&H_1(Du)=0 \ \ \ON \pl_1\gO,\quad H_2(Du)=0 \ \ \ON \pl_2\gO,
\eald \right.
\eeq
where $\gO=\{x=(x_1,x_2)\in\R^2\mid x_1>0,\, x_2>0 \}$ (called the first quadrant),
$\pl_1\gO=\{x=(x_1,x_2)\in\R^2\mid x_1=0,\,x_2\geq 0\}$, and
$\pl_2\gO=\{x=(x_1,x_2)\in\R^2\mid x_1\geq 0,\,x_2=0\}$. Furthermore, $F$ is a fully
nonlinear, degenerate elliptic operator for functions $u$ on $\ol\gO$, and the Neumann type operators $H_i$ are given as follows.
\beq\tag{B1}\label{cond-H_i} \ \left\{\ \bald
& \text{For } i=1,2, \ H_i(p)=-p_i+h_i(p_j), \ \ \text{ where } j=2 \IF i=1 \AND j=1 \IF i=2,
\\& h_i \ \text{ convex},
\\ & h_i'(t)\geq 0\ \ \AE t\in\R, \quad \elim_{t\to \infty}h_i'(t)=\ga_i\in[0,\,1)
.
\eald \right.
\eeq

We remark that, since $h_i$ above are convex on $\R$, $h_i$ are differentiable
almost everywhere and $h_i'$ are ``nondecreasing" on $\R$, and
we have \ $\elim_{t\to \infty}h'_i(t)=\esup_{t\in \R}h_i'(t)$. Similarly, we have \
$\elim_{t\to 0+} h_i'(t)=\einf_{t>0}h_i'(t)$ and
$\elim_{t\to 0-} h_i'(t)=\esup_{t<0}h_i'(t)$, for instance.

The condition that
$h_i'(t)\geq 0$ a.e. on $\R$ can be stated that $h_i$ is nondecreasing on
$\R$.

In \cite{BISW2017}, the authors have studied the nonlinear elliptic partial
differential equation
\beq\label{n-dPDE}
F(D^2u,Du,u,x)=0 \ \ \IN  \gO:=(0,\infty)^n,
\eeq
with the nonlinear Neumann type
boundary condition, similar to the above, stated as
\beq\label{n-dBC}
H_i(Du)=0 \ \ \ON \pl_i \gO:=\{x=(x_1,\ldots,x_n)\in [0,\infty)^n\mid x_i=0\},
\eeq
where \ $H_i(p):=-p_i+h_i(p).$

Henceforth, let $\{e_1,\ldots,e_n\}$ denote the standard basis of $\R^n$
and the symbol $\,\cdot\,$ indicate the Euclidean inner product in $\R^n$.
For each $i\in\mathbb{I}:=\{1,\ldots,n\}$, the function $h_i$, defining \erf{n-dBC},
has the form
\beq\label{n-dh}
h_i(p):=\max_{R\in \mathbf{R}_i} (p\cdot Re_i-g_i(R))=\max_{R\in \mathbf{R}_i}
\left(\sum_{j\in \mathbb{I}\stm\{i\}}p_jR_{ji}-g_i(R)\right),
\eeq
where $\mathbf{R}_i\subset \mathbf{R}$, with $\mathbf{R}$ denoting
the set of $n\tim n$ real matrices $R=(R_{jk})\in \R^{n\tim n}$ such that  $R_{jj}=0$ and $R_{jk}\geq 0$ for all $j,k\in\mathbb{I}$,
\[
\mathbf{R}_i:=\{R\in\mathbf{R}\mid \sum_{j\in \mathbb{I}}R_{ji}\leq \gb_i\},
\]
for a given $\gb_i\in [0,1)$, and $g_i\in C(\mathbf{R}_i)$ is a given function.
Observe that the functions $h_i$ are convex functions on $\R^n$ and that for any $i\in\mathbb{I}$,
\[
p\cdot Re_i=\sum_{j\in \mathbb{I}\stm\{i\}}p_jR_{ji}
\]
and hence, $h_i(p)$ is independent of $p_i$. For any $p,q\in\R^n$, we have
\begin{align*}
&|p\cdot Re_i-q\cdot Re_i|\leq
|p-q||Re_i|,
\\&|Re_i|^2 =\sum_{j\in\mathbb{I}}R_{ji}^2\leq \gb_i\sum_{j\in\mathbb{I}}R_{ji}\leq \gb_i^2, \\
\intertext{and hence, }
&
|p\cdot Re_i-q\cdot Re_i|\leq \gb_i |p-q|,
\end{align*}
which implies that $h_i$ is Lipschitz continuous on $\R^n$ with $\gb_i$ as a Lipschitz
bound.
When $n=2$, with a slight abuse of notation, we may write $h_1(p)=h_1(p_2)$, $h_2(p)=h_2(p_1)$. The functions $h_i$ are obviously nondecreasing on $\R$ by \erf{n-dh}, and satisfy $\esup_{\R} h_i'\in [0,\gb_i]$. Thus, condition \erf{cond-H_i}
is satisfied in this case.

As shown in \cite{BISW2017}, the boundary value problem \erf{n-dPDE}--\erf{n-dBC} arises in a scaling limit of stochastic control of queuing systems.
A main contribution in \cite{BISW2017} is that, under suitable hypotheses,
the value function of the stochastic control problem, obtained in the scaling limit,
is identified as the unique viscosity solution of \erf{n-dPDE} and \erf{n-dBC},
with an appropriate choice of $F$. This identification result heavily depends
on the uniqueness or comparison theorem for viscosity solutions,
subsolutions, and supersolutions of \erf{n-dPDE}--\erf{n-dBC}.
The comparison theorem in \cite{BISW2017} is basically valid only for the $H_i$ having
the positive homogeneity of degree one.

The aim of this paper is to attempt to remove the above requirement that the $H_i$ should be positively homogeneous of degree one. We are successful to remove the homogeneity requirement on the $H_i$, but only in dimension two. The comparison principle
for \erf{n-dPDE}--\erf{n-dBC}, with $h_i$ given by \erf{n-dh},
is still largely open in higher dimensions and even in dimension two (see \erf{cond-slope}).

We remark that there are a great amount of contributions  to the Neumann
boundary value problem on relatively regular domains even in the framework of
viscosity solutions. We refer for some general results to
\cites{CIL1992, I1991, B1999}, and for some results on domains with corners
to \cites{DI1991, BISW2017} and the references therein.

The crucial part of the proof of our comparison principle is to build an appropriate
convex test function $f$ on $\R^2$.  As described in Corollary \ref{exist-f} below,
the required properties on $f$ are that $f$ is a $C^{1,1}$ function, has a
quadratic growth, and  satisfies
\beq\label{cond-f}
H_1(Df(x))
\bcases
\geq 0\ \ &\IF\ x_1\leq 0, \\
\leq 0 &\IF\ x_1\geq 0,
\ecases
\quad\AND \quad
H_2(Df(x))
\bcases
\geq 0 \ \ &\IF\ x_2\leq 0, \\
\leq 0 &\IF\ x_2\geq 0.
\ecases
\eeq
Here and henceforth, we say that a function $g$ defined on an open subset $U$ of $\R^2$ is of class $C^{1,1}$ if the first derivatives of $g$ is bounded on $U$.

Our strategy of building such a test function $f$, a version of which
appears as $\phi_\ep$
in Section 5,  is first to define
a convex function $\psi$ (see \erf{def-psi} below) on $\R^2$ associated with the curves $H_i=0$, $i=1,2$,
having a linear growth, and then to take the convex conjugate of the function
$\psi^2$ plus a small quadratic function $q_\ep\mid x\mapsto \ep|x|^2$, with
$\ep>0$.  Roughly speaking, the addition of $q_\ep$ before taking the convex conjugate corresponds to the regularization by the inf-convolution
(called also the Yosida approximation) of the convex function $\psi^2$.
See \erf{inf-conv} below for this regularization.)

The paper is organized as follows. Let $f$, $\psi$, and $q_\ep$ be the functions
appearing in the explanation above. Throughout Sections 2--4, we
assume that $H_i$, $i=1,2$, vanish at the origin. This normalization condition is
called \erf{zero-norm} below, and is removed in Section 6. Under the normalization assumption, $f$ is called $\phi_\ep$ (see \erf{def-phi} below). Section 2 is devoted to the study of
the geometry of the curves $H_i=0$, $i=1,2$, and certain line segments
connecting points on those curves, which will be the level sets of the function $\psi$.
Section 3 introduces a condition (see \erf{cond-slope} below), which is
crucial to define the function $\psi$, with necessary properties, together with three
examples of cases of $H_i$, $i=1,2$, for which the critical condition \erf{cond-slope} is valid. Section 4 is devoted to building
the function $\psi$ and the study of its basic properties.
We define the function $\phi_\ep$ as the inf-convolution of the convex
$\psi^2+q_\ep$ in Section 5. In Section 6, the function $f$ is defined as the perturbation of
 $\phi_\ep$ by a linear function, which fits the general case, and we establish a comparison theorem for sub ans supersolutions of \erf{N-prob} as an application of the test function $f$. An existence result for bounded solutions of \erf{N-prob} is presented as well.

\section{Some geometric preliminaries}
We introduce a normalization assumption.
\beq\tag{B2}\label{zero-norm}
h_i(0)=0 \ \ \FOR i=1,2.
\eeq
This condition is clearly equivalent to the condition that $H_i(0)=0$ for $i=1,2$.
We \emph{assume} throughout this section that \erf{cond-H_i}--\erf{zero-norm} are satisfied. We come back to the general case without \erf{zero-norm} in Section 6.
We set
\begin{align*}
P&=\{x=(x_1,x_2)\in\R^2\mid x_1\geq h_1(x_2),\, x_2\geq h_2(x_1)\},
\\ Q&=\{x\in\R^2\mid x_1\leq h_1(x_2),\, x_2\geq h_2(x_1)\},
\\ R&=\{x\in\R^2\mid x_1\geq h_1(x_2),\, x_2\leq h_2(x_1)\},
\\ S&=\{x\in\R^2\mid x_1\leq h_1(x_2),\, x_2\leq h_2(x_1)\}.
\end{align*}
\begin{center}
\input two-d-regions.tex
\end{center}

We fix $v_0=(v_{01},v_{02}),v_1=(v_{11}.v_{12}),v_2=(v_{21},v_{22})\in\R^2$ such that
\begin{align}
\label{cond-v_i}& v_0\cdot e_1>0,\ v_0\cdot e_2>0,\quad
v_1\cdot (-e_1)>0,\ v_1\cdot e_2>0,\quad v_2\cdot e_1>0, \quad v_2\cdot (-e_2)>0,
\\\label{cond-v_1}&\einf_{t\geq 0}v_1\cdot(h_1'(t),1)>0,\quad \einf_{t\leq 0}v_1\cdot(-1,-h_2'(t))>0,
\\\label{cond-v_2}&\einf_{t\geq 0}v_2\cdot(1, h_2'(t))>0,\quad \einf_{t\leq 0}v_2\cdot (-h_1'(t),-1)>0.
\end{align}
Note that, since $h_1'(t)\geq 0$ and $h_2'(t)\geq 0$ a.e.,
\beq \label{cond-v_0}\einf_{t\geq 0}v_0\cdot (1,h_2'(t))\geq v_{01}>0,\quad \einf_{t\geq 0}v_0\cdot (h_1'(t),1)\geq v_{02}>0.\eeq
Since $0\leq h_i'(t)<1$ a.e. on $\R$ for $i=1,2$, it is easily seen that
\erf{cond-v_1} (resp., \erf{cond-v_2}) implies that
$v_1\cdot e_1<0$ and $v_1\cdot e_2>0$ (resp., $v_2\cdot e_1>0$ and $v_2\cdot e_2<0$).  Moreover, we deduce easily that \erf{cond-v_1} (resp., \erf{cond-v_2})
is equivalent to the condition that $v_1\cdot(\ga_1,1)>0$ and
$v_1\cdot (-1,-\ga_2^0)>0$ (resp., $v_2\cdot (1,\ga_2)>0$ and $v_2\cdot (-\ga_1^0,-1)>0$), where $\ga_i^0=\esup_{t\leq 0}h_i'(t)$ for $i=1,2$.

Let $\gl>0$. Let $x_{PQ}^\gl,\, x_{PR}^\gl$ be the points
where the line $v_0\cdot x=\gl$ intersects the curves $H_1=0$ and
$H_2=0$, respectively. Set $\gl_1=v_1\cdot x_{PQ}^\gl$ and
$\gl_2=v_2\cdot x_{PR}^\gl$.  Note that $\gl_1$ and $\gl_2$ are positive constants.
Let $x_{QS}^\gl,\, x_{RS}^\gl$ be the points in $(-\infty,0)^2$
where the lines $v_1\cdot x=\gl_1$ and $v_2\cdot x=\gl_2$ intersect,
respectively, the curves $H_2=0$ and $H_1=0$. The existence and uniqueness of
$(x_{PQ}^\gl,x_{PR}^\gl,x_{QS}^\gl,x_{RS}^\gl)$ is established in the next proposition.

\begin{center}\input two-d-points.tex\end{center}

\begin{proposition} \label{four-points} Under \erf{cond-H_i}--\erf{zero-norm},
for each $\gl>0$, there exists a unique $(x_{PQ}^\gl,x_{PR}^\gl,x_{QS}^\gl,x_{RS}^\gl)\in (\R^2)^4$ with the properties described above.
\end{proposition}

Let $I$ be an interval of $\R$ and $f\mid I\to\R$.
In this paper, we say that $f$ is \emph{uniformly increasing} on $I$
if $t\mapsto f(t)-\ep t$ is nondecreasing on $I$ for some constant $\ep>0$.
Also, we say that $f$ is \emph{uniformly decreasing} on $I$
if $t\mapsto f(t)+\ep t$ is nonincreasing on $I$ for some constant $\ep>0$.
It is easily seen that if $f\mid I \to\R$ is uniformly increasing (resp., decreasing) and
Lipschitz continuous on $I$, then $f\mid I\to f(I)$ has an inverse function
$f^{-1}$ which is uniformly increasing (resp., decreasing) and Lipschitz continuous on $f(I)$.

We introduce the functions $f_i$, $i=1,2$, on $[0,\infty)$,
$g_i$, $i=1,2$, on $[0,\infty)$, $p_i$, $i=1,2$, on $(-\infty,0]$,
$q_i$, $i=1,2$, on $(-\infty,0]$ by setting
\beq \label{fgpq} \left\{\ \bald
&f_1(t)=v_0\cdot(h_1(t),t), \ \ f_2(t)=v_0\cdot(t,h_2(t)), \ \
\\&g_1(t)=v_1\cdot (h_1(t),t),\ \ g_2(t)=v_2\cdot (t,h_2(t)), \ \
\\& p_1(t)=v_1\cdot (t,h_2(t)), \ \ p_2(t)=v_2\cdot (h_1(t),t),
\\& q_1(t)=m_\gam\cdot (t,h_2(t)), \ \ q_2(t)=n_\gam\cdot (h_1(t),t),
\eald \right.
\eeq
where $m_\gam:=(-1,-\gam)$ and $n_\gam:=(-\gam,-1)$, with $\gam$ being
any fixed positive constant. In the later sections, the constant $\gam$ will be fixed
as the one from \erf{cond-slope}.

\begin{lemma} \label{4 func} The functions $f_i$ and $g_i$, with $i=1,2$, are uniformly
increasing and Lipschitz continuous on $[0,\infty)$. They have inverse functions
$f_i^{-1}$ and $g_i^{-1}$, defined on $[0,\infty)$, and the inverse functions
$f_i^{-1},\, g_i^{-1}$ are uniformly increasing and Lipschitz continuous.
Similarly, the functions $p_i, q_i$, $i=1,2$,  are uniformly
decreasing and Lipschitz continuous on $(-\infty,0]$ and they have inverse functions
$p_i^{-1}$ and $q_i^{-1}$, defined on $[0,\infty)$, and the inverse functions
$p_i^{-1},\, q_i^{-1}$ are uniformly decreasing and Lipschitz continuous.
\end{lemma}

\bproof By \erf{cond-H_i}, the functions $h_i$ are Lipschitz continuous on $\R$ and hence, the functions $f_i,\,g_i,p_i,q_i$, $i=1,2$, are Lipschitz continuous. Note also that
$f_i(0)=g_i(0)=p_i(0)=q_i(0)=0$ for all $i=1,2$.

We have for a.e. $t\geq 0$,
\[\bald
&f_1'(t)=v_0\cdot (h_1'(t),1),\quad f_2'(t)=v_0\cdot (1,h_2'(t)), \quad
\\&g_1'(t)=v_1\cdot (h_1'(t),1), \quad   g_2'(t)=v_2\cdot(1,h_2'(t)).
\eald
\]
By \erf{cond-v_0}, \erf{cond-v_1} and \erf{cond-v_2}, we find that for some
constant $\ep_1>0$,
\[
f'_i(t)\geq \ep_1,\ g'_i(t)\geq \ep_1 \ \ \text{ for a.e. }t\geq 0 \text{ and all }i=1,2,
\]
which show that $f_i,\,g_i$, $i=1,2$, are uniformly increasing
and map onto $[0,\infty)$.
Now, since the functions $f_i,\,g_i\mid [0,\infty)\to[0,\infty)$, $i=1,2$, are
uniformly increasing, Lipschitz continuous, and surjective, we conclude as well
that their inverse functions exist and are uniformly increasing and Lipschitz continuous.

Next, note that for a.e. $t\leq 0$,
\[
m_\gam\cdot (1,h_2'(t))=-1-\gam h_2'(t)\leq -1 \ \ \AND \ \
n_\gam\cdot (h_1'(t),1)=-\gam h_1'(t)-1\leq -1.
\]
Using these,  \erf{cond-v_1}, and \erf{cond-v_2}, we easily find that for a.e. $t\leq 0$, all $i=1,2$, and some constant $\ep_2>0$,
\[
p_i'(t)\leq -\ep_2 \ \ \AND \ \ q_i'(t)\leq -1,
\]
which show that the functions $p_i, q_i$, $i=1,2$, are uniformly decreasing
and map onto $[0,\infty)$. Arguing as above, we conclude that the functions
$p_i, q_i\mid(-\infty,0]\to[0,\infty)$ are uniformly decreasing and Lipschitz continuous and have their inverse $p_i^{-1}, q_i^{-1}$ and that $p_i^{-1}, q_i^{-1}$ are
uniformly decreasing and Lipschitz continuous on $[0,\infty)$.  The proof is complete.
\eproof

In what follows, we use the functions $G_i\mid \R\to \R^2$, $i=1,2$, defined by
\[
G_1(t)=(h_1(t),t) \ \ \AND  \ \ G_2(t)=(t,h_2(t)).
\]

\bproof[Proof of Proposition \ref{four-points}]

{\sc The case of $x_{PQ}^\gl$. } The required conditions on $x_{PQ}^\gl\in\R^2$ are
that
\[
H_1(x_{PQ}^\gl)=0 \ \ \AND  \ \ v_0\cdot x_{PQ}^\gl=\gl.
\]

If we set $x_{PQ}^\gl=(s,t)$, then the conditions above are equivalent to that
\[
s=h_1(t) \ \ \AND \ \ v_{0}\cdot(s,t)=\gl,
\]
which can be stated as
\[
s=h_1(t) \ \ \AND \ \ v_{0}\cdot(h_1(t),t)=\gl.
\]
If $t\in\R$ satisfies the above conditions, then $t>0$. Indeed, if $t\leq 0$ were the case, then we would have $v_{0}\cdot(h_1(t),t)\leq 0$, which is impossible.
Hence, the unique existence of $x_{PQ}^\gl$ with the required properties is
equivalent to the unique existence of the solution $t>0$ of
$v_{0}\cdot(h_1(t),t)=\gl$, which can be written as  $f_1(t)=\gl$.   By Lemma \ref{4 func}, the solution of $f_1(t)=\gl$ is given uniquely by $t=f_1^{-1}(\gl)$.
Hence, $x_{PQ}^\gl$ is given uniquely as the point $(h_1\circ f_1^{-1}(\gl),f_1^{-1}(\gl))$. That is, $x_{PQ}^\gl=G_1\circ f_1^{-1}(\gl)$.

{\sc The case of $x_{PR}^\gl$. } Similarly to the above case,
the problem of finding $x_{PR}^\gl$ is stated as that of finding
a solution $t>0$ of $f_2(t)=\gl$, which can be solved uniquely
by $t=f_2^{-1}(\gl)$ and the point $x_{PR}^\gl$ is given by $G_2\circ f_2^{-1}(\gl)$.

{\sc The case of $x_{QS}^\gl$. } Set $\gl_1:= v_1\cdot x_{PQ}^\gl$ and $t=f_1^{-1}(\gl)$.  Since $t>0$, $x_{PQ}^\gl=G_1(t)$, and the function $g_1\mid s\mapsto v_1\cdot G_1(s)$ is strictly increasing on $[0,\infty)$ by Lemma \ref{4 func}, we find that
$\gl_1=g_1(t)>0$.
The conditions on $x_{QS}^\gl$ are now stated as
\[
H_2(x_{QS}^\gl)=0, \ \  v_1\cdot x_{QS}^\gl=\gl_1, \ \ \AND \ \
x_{QS}^\gl\in (-\infty,0)^2.
\]
The first condition above, together with the last one, can be stated as $x_{QS}^\gl=(s,h_2(s))$ for some $s<0$.  Hence,
finding a point $x_{QS}^\gl$ with the properties above is equivalent to finding
a solution $s<0$ of $p_1(s)=\gl_1$, which can be solved uniquely as
$s=p_1^{-1}(\gl_1)$. Thus, there is a unique point $x_{QS}^\gl$ having the required
properties. Moreover, we find the formula for $x_{QS}^\gl$:
\beq \label{x_QS}
x_{QS}^\gl=G_2\circ p_1^{-1}(\gl_1)=G_2\circ p_1^{-1}\circ g_1\circ f_1^{-1}(\gl).
\eeq

{\sc The case of $x_{RS}^\gl$. } We skip the detail, but we point out that $x_{RS}^\gl$ is given by the formula
\beq\label{x_RS}
x_{RS}^\gl=G_1\circ p_2^{-1}\circ g_2\circ f_2^{-1}(\gl).
\eeq
The proof is now complete.
\eproof

Using formulas \erf{fgpq}, we can define $f_i,g_i,p_i,q_i$, $i=1,2$, on $\R$.
The convexity property of functions $f_i, g_i, p_i, q_i$, $i=1,2$, are recorded
in the following.

\begin{lemma}\label{conv fgpq} The functions $f_i,\,-g_i,\,p_i,\,-q_i$, $i=1,2$,
are convex on $\R$.
\end{lemma}

\bproof The differentiation in the distributional sense twice yields
\[\bald
&f_1''=v_{01}h_1''\geq 0, \quad
f_2''=v_{02}h_2''\geq 0,\quad
g_1''=v_{11}h_1''\leq 0,\quad
g_2''=v_{22}h_2''\leq 0,\quad
\\& p_1''=v_{12}h_2''\geq 0, \quad
p_2''=v_{21}h_1''\geq 0,\quad
q_1''=-\gam h_2''\leq 0,\quad
q_2''=-\gam h_1''\leq 0,
\eald \]
which assures the convexity of the functions \
$f_1, \, f_2, \, -g_1,\, -g_2,\, p_1,\, p_2,\, -q_1,\,-q_2$\ on\ $\R$.
\eproof

\section{Additional assumption}

It is not clear for the authors if conditions \erf{cond-H_i}
are enough to build a smooth test function, by which one can
prove the comparison principle for solutions of \erf{N-prob}.
We here make another assumption on the functions $H_i$ in addition to
\erf{cond-H_i}--\erf{zero-norm},
as described below.

With the notation above, we introduce an additional condition on the functions $H_i$
and the choice of $v_i,\, i=0,1,2$.

\beq\tag{B3}\label{cond-slope} \left\{\ \begin{minipage}{0.8\textwidth}
There exists a constant $0<\gam<1$ such that for each $\gl>0$, the slope of the line passing through $x_{QS}^\gl$ and $x_{RS}^\gl$ ranges in the interval $[-1/\gam,-\gam]$.
\end{minipage}\right.
\eeq

In our presentation under the assumption \erf{cond-slope},
$m_\gam$ and $n_\gam$ are those with $\gam$ given by \erf{cond-slope}.

We give here three cases where \erf{cond-slope} holds.

\begin{example} \label{ex1}Assume that \erf{cond-H_i}--\erf{zero-norm} hold and that the functions $h_i$
are positively homogeneous of degree one, that is, for some constants $\ga_i^0
\in [0,\ga_i]$,
\[
h_i(t)= \bcases
\ga_i^0 t \ & \IF \ t\leq 0,\\
\ga_i t & \IF \ t\geq 0.
\ecases
\]
Then \erf{cond-slope} holds for some choice of $v_0,v_1,v_2\in\R^2$ satisfying \erf{cond-v_i},\erf{cond-v_1}, and \erf{cond-v_2}.
\end{example}

\bproof Consider the points
\[
\tilde x_{PQ}=(\ga_1,1),\quad \tilde x_{PR}=(1,\ga_2), \quad
\tilde x_{QS}=(-1,-\ga_2^0),\quad \tilde x_{RS}=(-\ga_1^0,-1),
\]
which are on the half lines $x_1=\ga_1x_2\ (x_2>0)$,
$x_2=\ga_2 x_1\ (x_1>0)$, $x_2=\ga_2^0 x_1 \ (x_1>0)$, and $x_1=\ga_1^0 x_2\ (x_2<0)$, respectively. The lines passing through the two points $\tilde x_{PQ}$
and $\tilde x_{PR}$, $\tilde x_{PQ}$ and $\tilde x_{QS}$, and $\tilde x_{PR}$ and $\tilde x_{RS}$, are given, respectively, by
\[
(1-\ga_2)x_1+(1-\ga_1)x_2=1-\ga_1\ga_2, \quad -(1+\ga_2^0)x_1+(1+\ga_1)x_2=1-\ga_1\ga_2^0,
\]
and
\[
(1+\ga_2)x_1-(1+\ga_1^0)x_2=1-\ga_1^0\ga_2.
\]
We set
\[
v_0=\Big(\fr{1-\ga_2}{1-\ga_1\ga_2},\fr{1-\ga_1}{1-\ga_1\ga_2}\Big),
\quad
v_1=\Big(-\fr{1+\ga_2^0}{1-\ga_1\ga_2^0},\fr{1+\ga_1}{1-\ga_1\ga_2^0}\Big),
\]
and
\[
v_2=\Big(\fr{1+\ga_2}{1-\ga_1^0\ga_2}, -\fr{1+\ga_1^0}{1-\ga_1^0\ga_2}\Big).
\]
It is now easy to check that \erf{cond-v_i}, \erf{cond-v_1}, and \erf{cond-v_2} hold
and that for each $\gl>0$,
\[
x_{PQ}^\gl=(\ga_1\gl,\gl),\quad x_{PR}^\gl=(\gl,\ga_2\gl),
\quad x_{QS}^\gl=(-\gl,-\ga_2^0\gl), \quad x_{RS}^\gl=(-\ga_1^0\gl,-\gl).
\]
Hence, the slope $s$ of the line passing through $x_{QS}^\gl$ and $x_{RS}^\gl$ is given by
\[
s=-\fr{1-\ga_2^0}{1-\ga_1^0},
\]
and, accordingly, if
\[
0<\gam\leq \min\Big\{\fr{1-\ga_2^0}{1-\ga_1^0},\fr{1-\ga_1^0}{1-\ga_2^0}\Big\},
\]
then $s\in[-1/\gam,-\gam]$. Thus, \erf{cond-slope} holds with the current choice
of $v_i$, $i=0,1,2$.
\eproof

A small perturbation of the $h_i$ in the example above keeps the validity of \erf{cond-slope}
as stated in the next example.

\begin{example}Assume that \erf{cond-H_i}--\erf{zero-norm} hold. For $i=1,2$, set
\[
\ga_i^-=\einf_{t\in\R}h_i'(t), \quad \ga_i^{0-}=\esup_{t\leq 0}h_i'(t) \ \ \AND \ \
\ga_i^{0+}=\einf_{t\geq 0}h_i'(t).
\]
Then there is a constant $\ep>0$ such that if, for $i=1,2$,
\[
\max\{\ga_i-\ga_i^{0+}, \ga_i^{0-}-\ga_i^-\}\leq \ep,
\] then \erf{cond-slope} holds.
\end{example}

\bproof  We start by noting that if $\ga_i^-=\ga_i^{0-}$ and $\ga_i^{0+}=\ga_i$
for $i=1,2$, then the situation is the same as Example \ref{ex1}, with
$\ga_i^0=\ga_i^{0-}$, $i=1,2$.

As in Example \ref{ex1}, we set \[
v_0=\Big(\fr{1-\ga_2}{1-\ga_1\ga_2},\fr{1-\ga_1}{1-\ga_1\ga_2}\Big),
\quad
v_1=\Big(-\fr{1+\ga_2^{0-}}{1-\ga_1\ga_2^{0-}},\fr{1+\ga_1}{1-\ga_1\ga_2^{0-}}\Big),
\]
and
\[
v_2=\Big(\fr{1+\ga_2}{1-\ga_1^{0-}\ga_2}, -\fr{1+\ga_1^{0-}}{1-\ga_1^{0-}\ga_2}\Big),
\]
and note that \erf{cond-v_i}, \erf{cond-v_1}, and \erf{cond-v_2} hold.

Fix any $\gl>0$. Writing $x_{PQ}^\gl=(x_1,x_2)$, we have
\[
(1-\ga_2)x_1+(1-\ga_1)x_2=(1-\ga_1\ga_2)\gl \ \ \ \AND \ \ \ x_1=h_1(x_2).
\]
Since $\ga_1^{0+}x_2\leq h_1(x_2)\leq \ga_1 x_2$, there is $\gb_1^+\in[\ga_1^{0+},\ga_1]$ such that $h_1(x_2)=\gb_1^+x_2$, and then,
\[
x_1=\fr{\gb_1^+(1-\ga_1\ga_2)\gl}{\gb_1^+(1-\ga_2)+1-\ga_1} \ \ \AND \ \ x_2=\fr{(1-\ga_1\ga_2)\gl}{\gb_1^+(1-\ga_2)+1-\ga_1}.
\]
Similarly, if $x_{PR}^\gl=(x_1,x_2)$, then we have for some $\gb_2^+\in[\ga_2^{0+},\ga_2]$,
\[h_2(x_1)=\gb_2^+ x_1, \ \ \
x_1=\fr{(1-\ga_1\ga_2)\gl}{1-\ga_2+\gb_2^+(1-\ga_1)} \ \  \AND \ \
x_2=\fr{\gb_2^+(1-\ga_1\ga_2)\gl}{1-\ga_2+\gb_2^+ (1-\ga_1)}.
\]
Setting
\[
\gth_1=\fr{1-\ga_1\ga_2}{1-\ga_1+\gb_1^+(1-\ga_2)} \ \ \AND \ \ \gth_2
=\fr{1-\ga_1\ga_2}{1-\ga_2+\gb_2^+(1-\ga_1)},
\]
we have
\[
x_{PQ}^\gl=(\gb_1^+\gth_1\gl,\gth_1\gl) \ \  \AND \ \ x_{PR}^\gl=(\gth_2 \gl,\gb_2^+\gth_2 \gl).
\]

If we set
\[
\mu:=\left(-(1+\ga_2^{0-}), 1+\ga_1\right)\cdot x_{PQ}^\gl \ \ \AND \ \ x_{QS}^\gl=:(x_1,x_2),
\]
then
\[
-(1+\ga_2^{0-})x_1+(1+\ga_1)x_2=\mu \ \ \AND \ \ x_2=h_2(x_1).
\]
Noting that $h_2(x_1)=\gb_2^- x_1$ for some $\gb_2^-\in[\ga_2^-,\ga_2^{0-}]$, we
find that
\[\begin{gathered}
\mu=\left(1+\ga_1 -\gb_1^+(1+\ga_2^{0-})\right)\gth_1\gl,
\\x_1=-\fr{\mu}{1+\ga_2^{0-}-\gb_2^-(1+\ga_1)} \ \ \AND
\ \ x_2=-\fr{\gb_2^-\mu}{1+\ga_2^{0-}-\gb_2^-(1+\ga_1)}.
\end{gathered}
\]
Similarly, we find that if we set
\[
\nu=\left(1+\ga_2-\gb_2^+(1+\ga_1^{0-}\right)\gth_2\gl,
\]
then for some $\gb_1^-\in[\ga_1^-,\ga_1^{0-}]$,
\[
x_{RS}^\gl=\Big(-\fr{\gb_1^-\nu}{1+\ga_1^{0-}-\gb_1^-(1+\ga_2)},-\fr{\nu}{
1+\ga_1^{0-}-\gb_1^-(1+\ga_2)}\Big).
\]

We note that, as $\ga_i^- \to \ga_i^{0-}$ and
$\ga_i^{0+}\to\ga_i$,
\[\bald
&\Big(-\fr{\mu}{1+\ga_2^{0-}-\gb_2^-(1+\ga_1)},-\fr{\gb_2^-\mu}{1+\ga_2^{0-}-\gb_2^-(1+\ga_1)}\Big)\ \to \ (-\gl,-\ga_2^{0-}\gl),
\\ &\Big(-\fr{\gb_1^-\nu}{1+\ga_1^{0-}-\gb_1^-(1+\ga_2)},-\fr{\nu}{
1+\ga_1^{0-}-\gb_1^-(1+\ga_2)}\Big) \ \to \ (-\ga_1^{0-}\gl, -\gl).
\eald
\]
We choose constants $\kappa_i$, $i=1,2$, so that
\[
0<\kappa_1<\min_{i=1,2}\ga_i^{0-}<\kappa_2<1.
\]
By the convergence above, we may select $\ep>0$ small enough so that if
\beq\label{ex2.1}
\max_{i=1,2}\max\{\ga_i^{0-}-\ga_i^-,\ga_i-\ga_i^{0+}\}\leq \ep,
\eeq
then
\[\bald
&-\fr{\mu}{1+\ga_2^{0-}-\gb_2^-(1+\ga_1)}, -\fr{\nu}{
1+\ga_1^{0-}-\gb_1^-(1+\ga_2)}\in[-2\gl,\, -\kappa_2\gl],
\\&
-\fr{\gb_2^-\mu}{1+\ga_2^{0-}-\gb_2^-(1+\ga_1)},
-\fr{\gb_1^-\nu}{1+\ga_1^{0-}-\gb_1^-(1+\ga_2)}\in [-\kappa_1\gl,\, 0].
\eald
\]
Thus, if \erf{ex2.1} is valid, then the slope $s$ of the line passing through
$x_{QS}^\gl$ and $x_{RS}^\gl$ satisfies
\[
s\in\Big[-\fr{2}{\kappa_2-\kappa_1},\,-\fr{\kappa_2-\kappa_1}{2}\Big],
\]
which shows that \erf{cond-slope} holds if \erf{ex2.1} is satisfied.
\eproof

\begin{example} Assume that \erf{cond-H_i}--\erf{zero-norm} hold and moreover that either
$h_1(t)=0$ for all $t\leq 0$ or $h_2(t)=0$ for all $t\leq 0$. Then \erf{cond-slope} holds.
\end{example}

\bproof  By symmetry, we may only consider the case when $h_2(t)=0$ for all
$t\leq 0$. Fix $v_0=(1,1)$ and $v_2=(1,-1)$, so that \erf{cond-v_i} for $v_0$ and $v_2$
and \erf{cond-v_2} hold.  Set
\[
\ga_1^{0}=\esup_{t\leq 0}h_1'(t),
\]
fix a constant $\ep_0>0$, and set
\[
\gb_0=\ep_0+\fr{\ga_1^0}{1-\ga_1^0}.
\]
Set
\[
\gb=\ga_1+\gb_0 (1+\ga_1) \ \ \AND \ \ v_1=(-1,\gb),
\]
and note that $\gb\geq \gb_0\geq \ep_0>0$
and that \erf{cond-v_i} for $v_1$ and \erf{cond-v_1} hold.

Fix any $\gl>0$ and let $x_{PQ}^\gl,x_{QS}^\gl, x_{PR}^\gl, x_{RS}^\gl$ denote the points
described above (see also Proposition \ref{four-points}). By solving
\[
v_0\cdot x=\gl,\quad x_2=h_2(x_1),
\]
and writing $\xi_1^\gl$ for the unique solution $t$ of $t+h_2(t)=\gl$,
we have
\[
x_{PR}^\gl=(\xi_1^\gl,h_2(\xi_1^\gl)) \ \ \AND \ \ \fr \gl {1+\ga_2}<\xi_1^\gl\leq \gl.
\]
Similarly, we have
\[
x_{PQ}^\gl=(h_1(\xi_2^\gl),\xi_2^\gl) \ \ \AND \ \ \fr{\gl}{1+\ga_1}<\xi_2^\gl\leq \gl,
\]
where $\xi_2^\gl$ is the unique solution $t$ of $t+h_1(t)=\gl$. Also, we have
\[
x_{QS}^\gl=(\eta_1^\gl,0) \ \ \AND \ \ x_{RS}^\gl=(h_1(\eta_2^\gl),\eta_2^\gl),
\]
where $\eta_1^\gl:=-v_1\cdot x_{PQ}^\gl$
and $\eta_2^\gl$ denotes the unique solution
$t$ of $t-h_1(t)=-v_2\cdot x_{PR}^\gl$.

Noting that $\eta_i^\gl<0$, $i=1,2$,
\[\bald
\eta_1^\gl&=-v_1\cdot x_{PR}^\gl =h_1(\xi_2^\gl)-\gb\xi_2^\gl,
\\ \eta_2^\gl&=h_1(\eta_2^\gl)-v_2\cdot x_{PR}^\gl=h_1(\eta_2^\gl)-\xi_1^\gl+h_2(\xi_1^\gl),
\eald
\]
we compute that
\[
\eta_1^\gl=h_1(\xi_2^\gl)-\gb \xi_2^\gl
\left\{
\bald
&\leq (\ga_1-\gb)\xi_2^\gl=-\gb_0 (1+\ga_1) \xi_2^\gl\leq -\gb_0\gl,
\\& \geq -\gb \xi_2^\gl\geq -\gb\gl,
\eald
\right.
\]
and
\[
\eta_2^\gl\leq -\xi_1^\gl+h_2(\xi_1^\gl) \leq -(1-\ga_2)\xi_1^\gl< -\fr{(1-\ga_2)\gl}{1+\ga_2}.
\]
Also, we have
\[
\eta_2^\gl\geq \ga_1^0\eta_2^\gl-\xi_1^\gl\geq \ga_1^0\eta_2^\gl -\gl,
\]
and hence,
\[
\eta_2^\gl\geq -\fr{\gl}{1-\ga_1^0}.
\]
Furthermore, we note that
\[
\eta_1^\gl-h_1(\eta_2^\gl) >\eta_1^\gl\geq -\gb \gl,
\]
and
\[
\eta_1^\gl-h_1(\eta_2^\gl)\leq \eta_1^\gl-\ga_1^0\eta_2^\gl
\leq-\gb_0\gl+\ga_1^0\fr{\gl}{1-\ga_1^0}
=(\ga_1^0-\gb_0(1-\ga_1^0))\fr{\gl}{1-\ga_1^0}=-\ep_0 \gl.
\]

If we denote by $s$ the slope of the line passing through $x_{QS}^\gl$ and $x_{RS}^\gl$, then
\[
s=\fr{-\eta_2^\gl}{\eta_1^\gl-h_1(\eta_2^\gl)}
\left\{\bald
&\leq \fr{-\eta_2^\gl}{-\gl\gb}\leq -\fr{1-\ga_2}{(1+\ga_2)\gb},
\\&\geq  \fr{-\eta_2^\gl}{-\ep_0\gl}\geq -\fr{1}{(1-\ga_1^0)\ep_0}.
\eald
\right.
\]
It is immediate to see that \erf{cond-slope} is satisfied with any positive
constant $\gam$ such that
\[
\gam\leq \min\Big\{\fr{1-\ga_2}{(1+\ga_2)\gb},(1-\ga_1^0)\ep_0\Big\}. \qedhere
\]
\eproof

 Fix any point $y=G_2(t)$, with $t<0$, on the curve $H_2=0$.
Let $s_0\geq 0$ be the slope of the line passing through the origin and the point $y$.
Let $G_1(s)$ be an arbitrary point on the curve $H_1=0$ and let $\gs(s)$ denote the slope of the line passing through the points $y$ and $G_1(s)$, which makes sense as far as $t\not=h_1(s)$. Since $h_1$ is nondecreasing,
$h_1(-\infty):=\lim_{s\to-\infty}h_1(s)\in[-\infty,0]$ makes sense. Since $h_1$ is convex and $h_1(0)=0$, if $h_1(-\infty)<t$, then there is a unique $\tau\in(-\infty,0)$ such that
$h_1(\tau)=t$. Similarly, if $h_1(-\infty)=t$, then there is a unique
$\tau\in[-\infty,0)$ such that $h_1(\tau)=t$ and $h_1(s)>t$ for every $s>\tau$.
If $h_1(-\infty)>t$, then we set $\tau=-\infty$.
We record the following simple observation.

\begin{lemma} \label{func-gs}With the notation introduced above, the function $\gs$
is increasing and continuous on $(\tau,0]$, $\gs(0)=s_0$, and $\lim_{s\to \tau+} \gs(s)=-\infty$. Furthermore, if $h_1(-\infty)<t$, then $\gs(s)>0$ on
$(-\infty,\tau)$ and if $h_1(-\infty)=t$ and $\tau>-\infty$, then for any $s\in(-\infty,\tau]$,  $h_1(s)=t$ and $\gs(s)$ does not make sense.
\end{lemma}

\bproof By the definition of $\gs$,
if $h_1(s)\not=t$, then
\[
\gs(s)=\fr{h_2(t)-s}{t-h_1(s)}.
\]
It is obvious that $\gs$ is continuous on $(\tau,0]$.
By the choice of $s_0$, we have $\gs(0)=s_0$.

Note that since $0\leq h_1'<1$ and $0\leq h_2'<1$ a.e. and $h_1(0)=h_2(0)=0$, we have
$\tau<t$ and $t<h_2(t)$.   Note also by the definition of $\tau$ that $t<h_1(s)$ for any $s\in(\tau,0]$.

To show the monotonicity of $\gs$ on $(\tau,0]$, we argue slightly in an informal way,
that is, we assume that $h_1\in C^2(\tau,0]$.
We compute that for any $s\in(\tau,0]$,
\[
\gs'(s)=\fr{-(t-h_1(s))+h_1'(s)(h_2(t)-s)}{(t-h_1(s))^2},
\]
and set
\[
\nu(s)=-(t-h_1(s))+h_1'(s)(h_2(t)-s).
\]
It follows that
\[
\nu'(s)=h_1''(s)(h_2(t)-s).
\]
Note that for any $s\in(\tau,h_2(t)]$,
\[
\nu(s)\geq h_1(s)-t>0.
\]
Observe that $\nu'(s)\leq 0$ for all $s\in[h_2(t),0]$, which implies that
$\nu$ is nonincreasing on $[h_2(t),0]$, and that
\[
\nu(0)=-t+h_1'(0)h_2(t)=h_1'(0)(h_2(t)-t)+t(h_1'(0)-1)>h_1'(0)(h_2(t)-t)\geq 0.
\]
Hence, we have $\nu(s)>0$ and $\gs'(s)>0$ on $(\tau,0]$.
Thus, $\gs$ is strictly increasing on $(\tau,0]$. The argument above can be justified
with a bit more elaboration, which we leave to the interested reader.

If $\tau>-\infty$, then it is obvious that
\[
\lim_{s\to\tau+}|\gs(s)|=\infty,
\]
and moreover, by the monotonicity of $\gs$,
\[
\lim_{s\to\tau+}\gs(s)=-\infty.
\]

If $h_1(-\infty)<t$, then we have $t=h_1(\tau)$ and
and for any $s\in(-\infty,\tau)$,
\[
t-h_1(s)>0 \ \ \AND \ \  h_2(t)-s >h_2(t)-\tau,
\]
which yields \ $\gs(s)>0$.

If $h_1(-\infty)=t$ and $\tau>-\infty$, then $h_1(s)=t$ for all $s\in(-\infty,\tau]$
due to the convexity and monotonicity of $h_1$, and hence, $\gs(s)$ does not make sense for $s\in(-\infty,\tau]$.
\eproof

\begin{lemma} \label{comp-slope} Let $\gl>0$ and let $t,s\in\R$ be such that $x_{QS}^\gl=G_2(t)$
and $x_{RS}^\gl=G_1(s)$. Set $\mu=m_\gam\cdot x_{QS}^\gl$ and $\nu=n_\gam\cdot x_{RS}^\gl$. Let $y_1=G_2(r_1)$ and $y_2=G_1(r_2)$ be the  unique points, respectively,  where the line $n_\gam\cdot x=\nu$ intersects
the curve $H_1=0$ and where the line $m_\gam\cdot x=\mu$ intersects
the curve $H_2=0$. Then, we have \ $r_1\leq t$ and $r_2\leq s$.
\end{lemma}

The unique existence of $y_i$, $i=1,2$, can be proved similarly to that of
$x_{PQ}^\gl,\,x_{PR}^\gl$, etc.

\bproof  We first remark that the lines $m_\gam\cdot x=\mu$
and $n_\gam\cdot x=\nu$ have slopes $-1/\gam$ and $-\gam$, respectively.
Since $x_{QS}^\gl=G_2(t), x_{RS}^\gl=G_1(s)\in(-\infty,0)^2$, we have
$t,s<0$ and $\mu, \nu>0$. Since $m_\gam\cdot G_2(r_1)=\mu$
and $n_\gam\cdot G_1(r_2)=\nu$, it is obvious that $r_1,r_2<0$.
As in Lemma \ref{func-gs}, let  $\gs(\xi)$ denote the slope of the line passing through the points $x_{QS}^\gl$ and $G_1(\xi)$ as far as the ``slope" makes sense.
We have $\gs(r_2)=-1/\gam$ and, by \erf{cond-slope},
$\gs(s)\in[-1/\gam,-\gam]$. The monotonicity of $\gs$ by Lemma
\ref{func-gs} assures that $r_2\leq s$. A parallel argument to the above yields
$r_1\leq t$.
\eproof

For later convenience, we remark that, with the notation of Lemma \ref{comp-slope} and \erf{fgpq},
we have
\beq\label{gl to mu}\left\{\ \bald
&t=f_1^{-1}(\gl), \quad x_{QS}^\gl=G_2\circ p_1^{-1}\circ g_1\circ f_1^{-1}(\gl), \quad
x_{RS}^\gl=G_1\circ p_2^{-1}\circ g_2\circ f_2^{-1}(\gl),
\\&  \mu=q_1\circ  p_1^{-1}\circ g_1\circ f_1^{-1}(\gl),\quad \nu
=q_2\circ p_2^{-1}\circ g_2\circ f_2^{-1}(\gl).
\eald \right.
\eeq

\section{Convex function $\psi$}

Our first step of constructing the test function $\phi_\ep$  in Theorem \ref{exist-phi}
is to build a nonnegative convex function $\psi$, the square of which is a good approximation of the convex conjugate function of $\phi_\ep$.

The idea of building $\psi$ is to let $\psi$ be a convex function that has a  linear
growth and that for each $\gl>0$, the level set $\{\psi=\gl\}$ includes the line segments $[x_{PQ}^\gl, x_{PR}^\gl]$,
$[x_{QS}^\gl,x_{PQ}^\gl]$, and $[x_{RS}^\gl,x_{PR}^\gl]$ (see Proposition \ref{four-points} for the definition of $x_{PQ}^\gl$, $x_{PR}^\gl$, etc.).
Here, $[x,y]$ denotes the set $\{(1-t)x+ty\mid 0\leq t\leq 1\}$ for $x,y\in\R^2$.
For notational
convenience, we write
\[
I_P^\gl=[x_{PQ}^\gl, x_{PR}^\gl],\quad I_Q^\gl= [x_{QS}^\gl,x_{PQ}^\gl], \ \ \AND \ \
I_R^\gl=[x_{RS}^\gl,x_{PR}^\gl].
\]

\begin{lemma}\label{I^gl} We have
\[
P\stm\{0\}=\bigcup_{\gl>0}I_P^\gl,\quad
Q\stm\{0\}=\bigcup_{\gl>0}I_Q^\gl, \ \ \AND \ \ R\stm\{0\}
=\bigcup_{\gl>0}I_R^\gl.
\]
\end{lemma}

\bproof Note first that
\[
\pl P=(\{x\in \R^2\mid H_1(x)=0\} \cup\{x\in\R^2 \mid H_2(x)=0\}) \cap\{x\in \R^2\mid v_0\cdot x\geq 0\}.
\]
Fix any $\gl>0$, note that
that
\[
\{(1-t)x_{PQ}^\gl+t x_{PR}^\gl\mid t\in\R\}=\{x\in\R^2\mid v_0\cdot x=\gl\},
\]
and, in particular,
\[
I_P^\gl\subset \{x\in\R^2\mid v_0\cdot x=\gl\}.
\]
Recall by Proposition \ref{four-points} that the points $x_{PQ}^\gl$  and $x_{PR}^\gl$ are the unique points where the line
$v_0\cdot x=\gl$ meets, respectively, the curves $H_1=0$ and $H_2=0$.
Set
\[
I=\{t\in\R\mid (1-t)x_{PQ}^\gl+t x_{PR}^\gl\in P\},
\]
and note that $I$ is a compact subset of $\R$ (since the intersection of the line $v_0\cdot x=\gl$ and $P$ is compact). In particular,  we have
$0,1\in I\subset [\min I,\max I]$
The uniqueness of the points $x_{PQ}^\gl$  and $x_{PR}^\gl$ shows that if $a\in \pl I$, then $(1-a)x_{PQ}^\gl+ax_{PR}^\gl\in \pl P$ and hence either $a=0$ or $a=1$.
It is clear that the line $v_0\cdot x=\gl$ intersects the interior of $P$, which implies
that $I$ has a nonempty interior.
We now deduce that $I=[0,1]$, which reads
\[
I_P^\gl=P\cap\{x\in \R^2\mid v_0\cdot x=\gl\}.
\]
Thus, we conclude that
\[
P\stm \{0\}=P\cap\{x\in\R^2\mid v_0\cdot x>0\}
=\bigcup_{\gl>0}P\cap \{x\in \R^2\mid v_0\cdot x=\gl\}
=\bigcup_{\gl>0}I_P^\gl.
\]

Let $\gl>0$ and note by Proposition \ref{four-points} (and its proof)
that $x_{PQ}^\gl$ and $x_{QS}^\gl\in (-\infty,0)^2$ are
the unique points where the line $v_1\cdot x=\gl_1$, with
$\gl_1=g_1\circ f_1^{-1}(\gl)$, meets the curves $H_1=0$ and $H_2=0$,
respectively.   Note also that $x_{PR}^\gl$ and $x_{RS}^\gl$ are the unique points where the line $v_2\cdot x=\gl_2$, with $\gl_2=g_2\circ f_2^{-1}(\gl)$,
meets the curves $H_2=0$ and $H_1=0$, respectively.
Arguing as above, we find that
\[
I_Q^\gl=Q\cap\{x\in \R^2\mid v_1\cdot x=\gl_1\} \ \ \AND \ \
I_R^\gl=R\cap\{x\mid v_2\cdot x=\gl_2\}.
\]
Since both $g_i\circ f_i^{-1}$, $i=1,2$, map $(0,\infty)$ onto itself, we conclude that
\[
Q\stm\{0\}=\bigcup_{\gl>0}I_Q^\gl \ \ \AND \ \ R\stm\{0\}
=\bigcup_{\gl>0}I_R^\gl. \qedhere
\]
\eproof

Define the functions $\psi_0$, $\psi_1$, and $\psi_2$ on $\R^2$ by
\beq\label{d-psi_i}
\psi_0(x)=(v_0\cdot x)_+, \quad
\psi_1(x)=f_1\circ g_1^{-1}((v_1\cdot x)_+), \quad \AND \quad
\psi_2(x)=f_2\circ g_2^{-1}((v_2\cdot x)_+),
\eeq
where $a_+$ denotes the positive part, $\max\{a,0\}$, of $a\in\R$.
It is clear that $\psi_0$ is convex on $\R^2$. By Lemma \ref{4 func}, we find that
the functions $f_i\circ g_i^{-1}$, $i=1,2$, are uniformly increasing and Lipschitz continuous on $[0,\infty)$. In particular, for some constants $0<\gd<C$,
\beq\label{psi_i l-growth}
\gd(v_i\cdot x)_+\leq \psi_i(x)\leq C(v_i\cdot x)_+ \ \ \FORALL x\in\R^2.
\eeq
By Lemma \ref{conv fgpq}, $f_i$, $i=1,2$, are convex and $g_i$, $i=1,2$, are
concave. Since $f_i, g_i$, $i=1,2$, are increasing, it follows that
$g_i^{-1}$, $i=1,2$, are convex on $[0,\infty)$ and moreover,
$f_i\circ g_i^{-1}$, $i=1,2$, are convex on $[0,\infty)$. Noting that for $i=1,2$,
$x\mapsto (v_i\cdot x)_+$ is convex on $\R^2$, we find that
$\psi_i$, $i=1,2$, are convex on $\R^2$.

Let $\gl>0$. We next observe that
\beq\label{psi_012}
\psi_0(x)=\gl \ \ \ON I_P^\gl,
\quad
\psi_1(x)=\gl \ \ \ON I_Q^\gl,
 \ \ \AND \ \
\psi_2(x)=\gl \ \ \ON I_R^\gl.
\eeq
By definition, we have $v_0\cdot x_{PQ}^\gl=v_0\cdot x_{PR}^\gl=\gl$.
In particular, if $x\in I_P$, we have $v_0\cdot x=\gl$ and
$\psi_0(x)=\gl$.
If $x_{PQ}^\gl=G_1(t)$ and $x_{PR}^\gl=G_2(s)$ for some $t,s\in\R$, then
$t,s>0$ and $\gl=f_1(t)=f_2(s)$. Setting $\gl_1:=v_1\cdot x_{PQ}^\gl
=v_1\cdot G_1(t)$ and
$\gl_2:=v_2\cdot x_{PR}^\gl=v_2\cdot G_2(s)$, by the definition of $g_i$, we have $g_1(t)=\gl_1$ and $g_2(s)=\gl_2$, which can be stated as $\gl_1=g_1^{-1}(t)$
and $\gl_2=g_2^{-1}(s)$. Hence, if $x\in I_Q^\gl$, then
$v_1\cdot x=\gl_1$ and $\psi_1(x)=f_1\circ g_1^{-1}(\gl_1)=f_1(t)=\gl$.
Similarly, if $x\in I_{R}^\gl$, then $\psi_2(x)
=f_2\circ g_2^{-1}(\gl_2)=f_2(s)=\gl$.

Set
\[
\tilde \psi(x)=\max_{i=1,2,3}\psi_i(x)
\ \ \FOR x\in\R^2.
\]
Obviously, $\tilde \psi$ is a nonnegative convex function on $\R^2$.
We define $\psi\mid \R^2\to \R$ by
\beq\label{def-psi}
\psi(x)=\max\{\phi(x)\mid \phi \text{ convex on }\R^2,\ \phi\leq \tilde \psi \ON P\cup Q\cup R\}.
\eeq
It is clear that $\psi$ is a convex function on $\R^2$ and that
$\tilde\psi(x)\leq \psi(x)$ for all $x\in\R^2$ and $\psi(x)= \tilde \psi(x)$ for all $x\in P\cup Q\cup R$.

\begin{lemma} \label{psi=psi_012} We have
\[
\psi=\psi_0 \ \ON \ P,\quad \psi=\psi_1 \ \ON \ Q,\ \ \AND \ \  \psi=\psi_2 \ \ON \ R.
\]
\end{lemma}

\bproof Note that $\psi=\tilde \psi$ on $P\cup Q\cup R$.
Let $\gl>0$ and let $y_1^\gl, \, y_2^\gl$ be the unique points where the line $v_0\cdot x=\gl$ intersects the lines $x_1=0$ (i.e., the $x_2$-axis)
and $x_2=0$ (the $x_1$-axis), respectively.
Consider the convex functions $\psi_i$, $i=0,1,2$,
on the line segment $I_0:=[y_1^\gl,y_2^\gl]$. By the definition of $\psi_0$, we have
$\psi_0(x)=\gl$ on $I_0$.
Since $v_{22}<0$, we have $v_2\cdot y_1^\gl<0$ and $\psi_2(y_1^\gl)=
f_2\circ g_2^{-1}(0)=0$. Similarly, we have $\psi_1(y_2^\gl)=0$.
Note that $I_P^\gl\subset I_0$ and that $\psi_1(x_{PQ}^\gl)=\gl$
and $\psi_2(x_{PR}^\gl)=\gl$. By the convexity of $\psi_i$, $i=0,1,2$, we infer that
$\psi_1(x)\leq \psi_0(x)$ on $[x_{PQ}^\gl,y_2^\gl]$,
$\psi_2(x)\leq \psi_0(x)$ on $[y_1^\gl,x_{PR}^\gl]$, and
$\psi(x)=\tilde\psi(x)=\psi_0(x)$ on $I_P^\gl$.

Set $\gl_1:=v_1\cdot x_{PQ}^\gl$ and note that $\gl_1=g_1\circ f_1^{-1}(\gl)>0$.
As we have observed above, we have  $\psi_1(x_{PQ}^\gl)=\psi_0(x_{PQ}^\gl)=\gl\geq \psi_2(x_{PQ}^\gl)$.  Since $x_{QS}^\gl\in(-\infty,0)^2$, we have
$v_0\cdot x_{QS}^\gl<0$ and $\psi_0(x_{QS}^\gl)=0$.
We claim that $v_2\cdot x_{QS}^\gl<0$. Indeed, since $v_{22}<0$, if we define
$g_2$ on $\R$ by $g_2(t)=v_2\cdot G_2(t)$, then $g_2$ is concave on $\R$ by Lemma \ref{conv fgpq},
$g_2(0)=0$, and $g_2(t)>0$ for all $t>0$ by Lemma \ref{4 func}.
Hence, choosing $r<0$ so that $x_{QS}^\gl=G_2(r)$, we have
\[
0=g_2(0)\geq \tfrac 12(g_2(r)+g_2(-r)) >\tfrac 12 g_2(r),
\]
which assures that $v_2\cdot x_{QS}^\gl=g_2(r)<0$.
This shows that $\psi_2(x_{QS}^\gl)=0$.
It now follows that $\psi_0(x)\leq \psi_1(x)$,
$\psi_2(x)\leq \psi_1(x)$,
and $\psi(x)=\psi_1(x)=\gl$ on  $I_Q^\gl$.

An argument parallel to the above shows that
$\max\{\psi_0(x),\psi_1(x)\}\leq \psi_2(x)=\psi(x)=\gl$ on
$I_R^\gl$.

Thus, noting that $\psi(0)=\psi_i(0)=0$ for all $i=0,1,2$, we conclude by Lemma \ref{I^gl} that
\[
\psi=\psi_0 \ \ \ON P,\quad
\psi=\psi_1 \ \ \ON Q,\ \ \AND \ \
\psi=\psi_2 \ \ \ON  R.  \qedhere
\]
\eproof

\begin{lemma} \label{psi l-growth}There exist constants $0<\gd\leq C<\infty$ such that
\[
\gd|x|\leq \psi(x)\leq C|x| \ \ \FORALL x\in\R^2.
\]
\end{lemma}

\bproof By \erf{psi_i l-growth}, there are constants $0<\gd_0<C_0<\infty$ such that
\beq\label{psi_i l-growth2}
\gd_0(v_i\cdot x)_+\leq \psi_i(x)\leq C_0(v_i\cdot x)_+ \ \ \FORALL x\in \R^2,\, i=0,1,2.
\eeq

We claim that for some constant $\gd_1>0$,
\beq \label{lb-PQR}
\max_{i=0,1,2}v_i\cdot x\geq \gd_1|x| \ \ \FORALL x\in P\cup Q\cup R.
\eeq
Observe first that since $v_{01}>0$ and $v_{02}>0$, we have for any $x\in[0,\infty)^2$,
\[
v_0\cdot x\geq \min_{j=1,2}v_{0j}(x_1+x_2)\geq \min_{j=1,2}|v_{0j}||x|.
\]
Next,  since $v_{11}<0$ and $v_{12}>0$,
we have for any $x\in (-\infty,0]\tim[0,\infty)$,
\[
v_1\cdot x=|v_{11}||x_1|+|v_{12}||x_2|\geq \min_{j=1,2}|v_{1j}||x|.
\]
Similarly, we have for any $x\in[0,\infty)\tim(-\infty,0]$,
\[
v_2\cdot x\geq \min_{j=1,2}|v_{2j}||x|.
\]
Hence, we have
\
\max_{i=0,1,2}v_i\cdot x\geq \min_{i=0,1,2,\, j=1,2}|v_{ij}||x|
\ \ \FORALL x\in [0,\infty)\tim \R \, \cup\,  \R\tim [0,\infty).
\]

To conclude the validity of \erf{lb-PQR}, it remains to check the case where
either $x\in Q\cap(-\infty,0]^2$ or $x\in R\cap(-\infty,0]^2$.
Let $x\in Q\cap(-\infty,0]^2$. Since $v_{12}>0$ and $x_2\geq h_2(x_1)$, we have
\[
v_1\cdot x \geq v_1\cdot G_2(x_1)=p_1(x_1).
\]
Hence, by the uniform monotonicity of $p_1$, a consequence of  Lemma \ref{4 func},
we find that for some constant $\ep_1>0$,
\[
v_1\cdot x\geq p_1(x_1)\geq \ep_1|x_1|.
\]
By \erf{cond-H_i}, the function $h_2$ has the Lipschitz constant $\ga_2$ and $h_2(0)=0$. Accordingly, we have
\[
|x|^2\leq x_1^2+h_2(x_1)^2\leq (1+\ga_2^2)|x_1|^2,
\]
and hence,
\[
v_1\cdot x\geq \fr{\ep_1}{|(1,\ga_2)|}|x|
\]
Similarly, if $x\in R\cap(-\infty,0]^2$, then we have
\[
v_2\cdot x \geq v_2\cdot G_1(x_2)=p_2(x_2)\geq \ep_2|x_2|,
\]
where $\ep_2$ is a positive constant,
and moreover,
\[
v_2\cdot x\geq \fr{\ep_2}{|(\ga_1,1)|}|x|.
\]
Now, setting
\[
\gd_1=\min\left\{\min_{i=0,1,2,\,j=1,2}|v_{ij}|, \fr{\ep_1}{|(1,\ga_2)|},\fr{\ep_2}{|(\ga_1,1)|}\right\},
\]
we find that \erf{lb-PQR} holds. Combining this with \erf{psi_i l-growth2}, we have
\[
\psi(x)=\tilde\psi(x)\geq \gd_0\gd_1 |x| \ \ \FORALL x\in P\cup Q\cup R.
\]
Since the function $x\mapsto \gd_0\gd_1|x|$ is convex on $\R^2$, we conclude that
\[
\psi(x)\geq \gd_0\gd_1|x| \ \ \FORALL x\in\R^2.
\]

From \erf{psi_i l-growth2}, we find immediately that
\[
\tilde\psi(x)\leq C_0\max_{i=0,1,2}|v_i||x| \ \ \FORALL x\in\R^2,
\]
which implies that
\[
\psi(x)\leq C_0\max_{i=0,1,2}|v_i||x| \ \ \FORALL x\in P\cup Q\cup R.
\]
Since $\psi$ is convex on $\R^2$ and $\pl (-\infty,0)^2\subset Q\cup R$, if $x\in(-\infty,0)^2$, then $x=(1-t)(|x_1|+|x_2|)(-e_1)+t(|x_1|+|x_2|)(-e_2)$, where $t=|x_2|/(|x_1|+|x_2|)$,  and
\[\bald
\psi(x)&\leq (1-t)\psi((|x_1|+|x_2|)(-e_1))+t\psi((|x_1|+|x_2|)(-e_2))
\\&\leq C_0\max_{i=0,1,2}|v_i| (|x_1|+|x_2|)
\leq \sqrt 2 C_0\max_{i=0,1,2}|v_i| |x|.
\eald\]
Thus, we find that
\[
\psi(x)\leq \sqrt 2 C_0\max_{i=0,1,2}|v_i||x| \ \ \FORALL x\in\R^2.
\]
This completes the proof.
\eproof

We are interested in the sublevel sets $\{\psi\leq \gl\}$ of $\psi$, with $\gl>0$, and
we introduce functions $\chi_i$, $i=1,2$, by
\[\bald
\chi_1(x)&=f_1\circ g_1^{-1}\circ p_1\circ q_1^{-1}((m_\gam\cdot x)_+),
\\ \chi_2(x)&=f_2\circ g_2^{-1}\circ p_2\circ q_2^{-1}((n_\gam\cdot x)_+).
\eald
\]
For $i=1,2$, by Lemmas \ref{4 func} and \ref{conv fgpq}, the functions
$f_i,\,g_i$ are increasing, $p_i,\,q_i$ are decreasing, $f_i,\,p_i$ are
convex, and $g_i,\,q_i$ are concave on their respective domains of definition. Hence,
$q_i^{-1}$ is concave on $[0,\infty)$, which implies that $p_i\circ q_i^{-1}$ is
convex (and increasing) on $[0,\infty)$. Moreover, since $f_i,\,g_i^{-1}$ are increasing and convex on
$[0,\infty)$, we find that $f_i\circ g_i^{-1}\circ p_i\circ q_i^{-1}$ is
convex and increasing on $[0,\infty)$. Hence, the functions $\chi_i$, $i=1,2$, are convex on $\R^2$.

\begin{lemma} \label{chi-psi}We have $\,\chi_1=\psi\,$ on the curve $H_2=0$ in $S$, $\,\chi_2=\psi\,$
on the curve $H_1=0$ in $S$, and $\,\chi_i\leq \psi\,$ on $\R^2$ for $i=1,2$.
\end{lemma}

Remark that the curve $H_2=0$ (resp., $H_1=0$) in $S$ is the set
\[\begin{gathered}
\{x\in (-\infty,0]^2\mid H_2=0\}=\{G_2(t)\mid t\leq 0\}=
\{0\}\cup \{x_{QS}^\gl\mid \gl>0\}
\\(\text{resp., } \quad
\{x\in (-\infty,0]^2\mid H_1=0\}=\{G_1(t)\mid t\leq 0\}=
\{0\}\cup \{x_{RS}^\gl\mid \gl>0\}).
\end{gathered}
\]

\bproof Note first that $\psi(0)=\chi_i(0)=0$ for $i=1,2$.

Let $\gl>0$. Note by \erf{gl to mu} that
\[
x_{QS}^\gl=G_2\circ p_1^{-1}\circ g_1\circ f_1^{-1}(\gl) \ \ \AND \ \
x_{RS}^\gl=G_1\circ p_2^{-1}\circ g_2\circ f_2^{-1}(\gl).
\]
From the above, we have
\[
m_\gam\cdot x_{QS}^\gl=q_1\circ p_1^{-1}\circ g_1\circ f_1^{-1}(\gl)
\ \ \AND \ \ n_\gam\cdot x_{RS}^\gl=q_2\circ p_2^{-1}\circ g_2\circ f_2^{-1}(\gl).
\]
Since \ $
x_{QS}^\gl, \,x_{RS}^\gl, \,m_\gam,\, n_\gam\in(-\infty,0)^2$,
we have \ $
m_\gam\cdot x_{QS}^\gl>0\ \AND \ n_\gam\cdot x_{RS}^\gl>0$.
Hence, by the definition of $\chi_i$, we find that
\[
\chi_1(x_{QS}^\gl)=\gl \ \ \AND \ \ \chi_2(x_{RS}^\gl)=\gl.
\]
Moreover, Lemmas \ref{I^gl} and  \ref{psi=psi_012} ensure that
$\,\chi_1=\psi_1=\psi\,$ on the curve $\,H_2=0\,$ in $S$
and $\,\chi_2=\psi_2=\psi\,$ on the curve $\,H_1=0\,$ in $S$.

We fix any $x\in P\cup Q\cup R$ and show that $\chi_i(x)\leq \psi(x)$ for $i=1,2$.
Once this is done, since $\chi_i$ are convex on $\R^2$, we conclude that
$\chi_i\leq \psi$ on $\R^2$ for $i=1,2$ and finish the proof.

We may henceforth assume that $x\not=0$.
Choose $\gl>0$ so that $\psi(x)=\gl$, and
note that $x\in I_P^\gl \cup I_Q^\gl \cup I_R^\gl$.

Consider the case $x\in I_P^\gl$. Since $x\in(0,\infty)^2$ and $m_\gam, \, n_\gam\in (-\infty,0)^2$,  we have $m_\gam\cdot x<0$, $n_\gam\cdot x<0$, and hence,
$\chi_i(x)=0\leq \psi(x)$ for $i=1,2$.

Recall the monotonicity of $\chi_i$: for any
$\xi,\eta\in\R^2$, if $m_\gam\cdot \xi\leq m_\gam\cdot \eta$, then
$\chi_1(\xi)\leq \chi_1(\eta)$ and, if  $n_\gam\cdot \xi\leq n_\gam\cdot \eta$, then
$\chi_2(\xi)\leq \chi_2(\eta)$.

Now, assume that $x\in I_Q^\gl$.
Note that $x=x_{QS}^\gl+\tau (x_{PQ}^\gl-x_{QS}^\gl)$ for some $\tau\in[0,\,1]$
and that $x_{PQ}^\gl-x_{QS}^\gl\in (0,\infty)^2$, and hence, $m_\gam\cdot(x_{PQ}^\gl-x_{QS}^\gl)<0$,\ $n_\gam\cdot (x_{PQ}^\gl-x_{QS}^\gl)<0$.
Moreover, we have
\[\bald
m_\gam \cdot x\leq m_\gam\cdot x_{QS}^\gl \ \ \ \AND \ \ \
n_\gam\cdot x\leq n_\gam\cdot x_{QS}^\gl.
\eald\]
Similarly, we find that if $x\in I_R^\gl$, then
\[m_\gam\cdot x\leq m_\gam\cdot x_{RS}^\gl, \quad
n_\gam\cdot x\leq n_\gam\cdot x_{RS}^\gl.
\]
Hence, if $x\in I_Q^\gl$, then
\[
\chi_1(x)\leq \chi_1(x_{QS}^\gl)=\gl=\psi(x),
\]
and, if $x\in I_R^\gl$, then
\[
\chi_2(x)\leq \chi_2(x_{RS}^\gl)=\gl=\psi(x).
\]

By Lemma \ref{comp-slope}, there are points $y_1=G_2(r_1)$ and $y_2=G_1(r_2)$,
with $r_i<0$ for $i=1,2$,  such that $m_\gam\cdot x_{QS}^\gl=m_\gam\cdot y_2$,
$n_\gam\cdot x_{RS}^\gl=n_\gam\cdot y_1$. Moreover, the lemma says that if we write
$x_{QS}^\gl=G_2(t)$ and $x_{RS}^\gl=G_1(s)$, then
$r_1\leq t<0$ and $r_2\leq s<0$.  The last two inequalities imply that
\[
x_{QS}^\gl-y_1,\, x_{RS}^\gl-y_2\in[0,\infty)^2,
\]
and hence,
\[
n_\gam\cdot (x_{QS}^\gl-y_1)\leq 0 \quad\AND\quad
m_\gam\cdot(x_{RS}^\gl-y_2)\leq 0.
\]
Consequently, if $x\in I_Q^\gl$, then $n_\gam\cdot x\leq n_\gam\cdot y_1$ and
\[\bald
\chi_2(x)& \leq \chi_2(y_1)= f_2\circ g_2^{-1}\circ p_2\circ q_2^{-1}(n_\gam \cdot y_1)
= f_2\circ g_2^{-1}\circ p_2\circ q_2^{-1}(n_\gam \cdot x_{RS}^\gl)
\\&=\chi_2(x_{RS}^\gl)=\gl=\psi(x),
\eald
\]
and  if $x\in I_R^\gl$, then $m_\gam\cdot x\leq m_\gam\cdot y_2$ and
\[\bald
\chi_1(x)&\leq \chi_1(y_2)
=f_1\circ g_1^{-1}\circ p_1\circ q_1^{-1}(m_\gam \cdot x_{QS}^\gl)
= \chi_1(x_{QS}^\gl) =\gl=\psi(x).
\eald
\]
The proof is complete.
\eproof

We remark that we may choose $\gam>0$ as small as we wish in
\erf{cond-slope}.

\begin{lemma} \label{normal}Let $\gl>0$ and $L_\gl:=\{x\in\R^2 \mid \psi(x)\leq \gl\}$. Let $y\in\pl L_\gl$.  We select $\gam>0$ sufficiently small. Then,
\[ \bald
 & N(y,L_\gl)\subset \cone\{v_1, n_\gam\} \ \ & \IF \ y_1<h_1(y_2),
\\&
N(y,L_\gl)\subset \cone\{v_2, m_\gam\} \ \ &\IF \ y_2<h_2(y_1),
\\&
N(y,L_\gl)\subset \cone\{v_0,v_2\} \ \ &\IF \ y_1>h_1(y_2),
\\&
N(y,L_\gl)\subset \cone\{v_0,v_1\} \ \ &\IF \ y_2>h_2(y_1).
\eald
\]
\end{lemma}

We formulate a simple lemma for the proof of the lemma above.

\begin{lemma}\label{cone} Let $v,w,p,q\in\R^2$. Assume that
\[
v\cdot p=w\cdot q=0, \quad v\cdot q<0, \  \ \AND \ \ w\cdot p<0.
\]
Set \ $
C=\{x\in\R^2\mid v\cdot x\leq 0,\ w\cdot x\leq 0\}$ \ and \ $C^*=\{\xi\in\R^2
\mid \xi\cdot x\leq 0 \ \FORALL x\in C\}$.
Then,
\[
C=\cone\{p,q\} \quad\AND\quad C^*=\cone\{v,w\}.
\]
\end{lemma}

\bproof It is easily seen that the vectors $p$ and $q$ are linearly independent.
For any $x=sp+tq$, with $s,t\in\R$, we have
\[
v\cdot x=t v\cdot q \ \ \AND \ \ w\cdot x=s w\cdot p,
\]
and, since $v\cdot q<0$ and $w\cdot p<0$, we find that
$x\in C$ if and only if $s\geq 0$ and $t\geq 0$. Hence, we have
\[
C=\cone\{p,q\}.
\]

Next, note first that $v,w\in C^*$ and $C^*$ is a closed convex cone. Hence,
we have $\cone\{v,w\}\subset C^*$. To show that $\cone\{v,w\}=C^*$,
we suppose to the contrary that $C^*\stm \cone\{v,w\}\not=\emptyset$. Select
$u\in C^*\stm\cone\{v,w\}$. By the Hahn-Banach theorem, there is $x_0\in\R^2$
such that $v\cdot x_0\leq 0$, $w\cdot x_0\leq 0$, and $u\cdot x_0>0$, which
can be stated that $x_0\in C$ and $u\cdot x_0>0$. These contradict
that $u\in C^*$.
\eproof

\bproof[Proof of Lemma \ref{normal}]   We consider only
the case when $y_1<h_1(y_2)$. The other cases are treated similarly, whose
details are left to the reader.

Set \
$
C=\{x\in\R^2\mid v_1\cdot x\leq 0,\ n_\gam\cdot x\leq 0\},
$
which is a closed convex cone with vertex at the origin.
We prove that
\beq\label{inc-NC*}
N(y,L_\gl)\subset C^*,
\eeq
where \ $
C^*=\{\xi\in\R^2\mid \xi\cdot x\leq 0 \ \FORALL x\in C\}.$
Once this is done, thanks to
Lemma \ref{cone}, we conclude the proof.

To check \erf{inc-NC*}, we set $p=x_{PQ}^\gl$, $q=x_{RS}^\gl$,
and $D=\{x\in\R^2\mid v_1\cdot(x-p)\leq 0,\ n_\gam\cdot (x-q)\leq 0\}$,
and note by the definition of $L_\gl$ that \ $
L_\gl\subset D$ \ and \ $p,q\in\pl L_\gl.$

Since $v_1$ and $n_\gam$ are linearly independent, there is a unique solution
$x\in\R^2$ of the system \ $v_1\cdot(x-p)=n_\gam\cdot (x-q)=0$.
Let $r$ denote the unique solution $x$ of the above system. Note that
\[\bald
D&=\{x\in\R^2\mid v_1\cdot(x-r)\leq 0,\ n_\gam\cdot (x-r)\leq 0\}
=r+C.
\eald
\]

The proof of Lemma
\ref{4 func} easily extends to show that
the functions
\[
t\mapsto v_1\cdot G_1(t) \ \ \AND \ \ t\mapsto n_\gam\cdot G_1(t)
\]
are, respectively, uniformly increasing and decreasing on $\R$.
The characterization of the points $p,q$ due to Proposition \ref{four-points}
shows that $v_1\cdot q<0<v_1\cdot p$ and $n_\gam\cdot p <0<n_\gam\cdot q$.
Also, since $v_1\cdot e_1<0$,
$n_\gam\cdot e_1<0$,  and $y_1<h_1(y_2)$ by assumption, we have
\[\bald
v_1\cdot p\geq v_1\cdot y>v_1\cdot G_1(y_2) \ \ \AND \ \
n_\gam\cdot q\geq n_\gam\cdot y>n_\gam\cdot G_1(y_2),
\eald \]
which assures that $q_2<y_2<p_2$.

Noting that
\beq\label{vnpq}
v_1\cdot (p-r)=0>v_1\cdot (q-r) \ \ \AND \ \ n_\gam\cdot(q-r)=0>n_\gam\cdot(p-r),
\eeq
we invoke Lemma \ref{cone}, to find that
\beq\label{cones}
C=\cone\{p-r,q-r\} \ \ \AND \ \ C^*=\cone\{v_1,n_\gam\}.
\eeq

Since $y-r\in C$, there are unique $s,t\in[0,\infty)$ such that
$y-r=s(p-r)+t(q-r)$.
We claim that $s+t<1$. Indeed,
if we write $e_1=\gs (p-r)+\tau(q-r)$,
with $\gs,\tau\in\R$, we find by \erf{vnpq} that $\gs>0$ and $\tau>0$.
Since $q_2<y_2<p_2$, we have $G_1(y_2)\in L_\gl$.
Note that $G_1(y_2)=y+(h_1(y_2)-y_1)e_1$. There exists $\gth>0$
such that $y+\gth e_1\in [q,p]$, which reads that the point
\[
y-r+\gth e_1=(s+\gth \gs)(p-r)+(t+\gth\tau)(q-r)
\]
lies in the interval $[q-r,\,p-r]$. This yields
\[
s+\gth\gs+ t+\gth\tau=1,
\]
which ensures that $s+t<1$.

Now, let $\xi\in N(y,L_\gl)$, which implies that
\[
\xi\cdot(q-y)\leq 0 \ \ \AND \ \ \xi\cdot (p-y)\leq 0.
\]
Since $y-r=s(p-r)+t(q-r)$, we have
\[\bald
& 0\leq \xi\cdot(y-p)=\xi\cdot(r-p)+s\xi\cdot(p-r)+t\xi\cdot(q-r),
\\& 0\leq \xi\cdot(y-q)=\xi\cdot(r-q)+s\xi\cdot(p-r)+t\xi\cdot(q-r).
\eald
\]
Combining these with the inequalities $s+t<1,\,s\geq 0,\,t\geq 0$, we infer that
\[
\xi\cdot (p-r)\leq 0 \ \ \AND \ \ \xi\cdot(q-r)\leq 0,
\]
which imply together with the first identity of \erf{cones} that $\xi\in C^*$.
The proof is complete.
\eproof

\begin{lemma} \label{psi^2} For any $x\in\R^2$,
 \[
\pl \psi^2(x)=2\psi(x)\pl \psi(x).
\]
\end{lemma}

\bproof  If $p\in \pl\psi(x)$, then, as $\R^2\ni y\to 0$,
\[\bald
\psi^2(x+y)-\psi^2(x)&=(\psi(x+y)+\psi(x))(\psi(x+y)-\psi(x))
\\&\geq (\psi(x+y)+\psi(x)) p\cdot y
=2\psi(x)p\cdot y+O(|y|^2),
\eald\]
which assures that $2\psi(x)p\in \pl \psi^2(x)$. If $p\in \pl\psi^2(x)$ and $x\not=0$, then, as $y\to 0$,
\[\bald
(\psi(x+y)+\psi(x))(\psi(x+y)-\psi(x))=\psi^2(x+y)-\psi^2(x)
\geq p\cdot y,
\eald\]
and
\[\bald
\psi(x+y)-\psi(x)
\geq \fr{p\cdot y}{\psi(x+y)+\psi(x)}
=\fr{p}{2\psi(x)}\cdot y+O(|y|^2),
\eald\]
which reads that $p/(2\psi(x))\in \pl \psi(x)$. On the other hand, we clearly  have
$\pl \psi^2(0)=\{0\}=2\psi(0)\pl \psi(0)$.
\eproof

\begin{lemma}\label{bounds-Dpsi} For some constants $0<\gd\leq C<\infty$,
\[
\gd\leq |p|\leq C \ \ \FORALL p\in \pl \psi(x) \ \AND \ x\in\R^2\stm\{0\}.
\]
\end{lemma}

\bproof According to Lemma \ref{psi l-growth}, there exist constants
$0<\gd_0\leq C_0<\infty$ such that
\[
\gd_0|x|\leq \psi(x)\leq C_0|x| \ \ \FORALL \ x\in\R^2.
\]
Let $x\in\R^2\stm\{0\}$ and $p\in\pl \psi(x)$.  For any $y\in\R^2$,
we have
\beq\label{bounds-Dpsi.1}
\psi(x+y)-\psi(x)\geq p\cdot y.
\eeq
Put $y=-x$, to obtain
\[
\gd_0|x|\leq \psi(x)\leq p\cdot x\leq |p||x|,
\]
and hence, $|p|\geq\gd_0$. Next, set $y=|x|p/|p|$, to find that
\[
|p||x|\leq \psi\left(x+\tfrac{|x|p}{|p|}\right)\leq 2C_0|x|,
\]
which yields that $|p|\leq 2C_0$. Thus, we have \ $\gd_0\leq|p|\leq 2C_0$.
\eproof

\section{$C^{1,1}$ test function}

In the previous section, we have constructed a convex function $\psi$ on $\R^2$ such that for some constants $0<\gd\leq C<\infty$,
\beq \label{deltaC1}
\gd|x|\leq \psi(x)\leq C|x| \ \ \FORALL \ x\in \R^2,
\eeq
and
\beq\label{deltaC2}
\gd\leq |p|\leq C \ \ \FORALL \ p\in \pl\psi(x),\, x\in\R^2\stm\{0\}.
\eeq
(See Lemmas \ref{psi l-growth} and \ref{bounds-Dpsi} for the estimates above.)
More importantly, thanks to Lemma \ref{normal}, if we set for $\gl>0$,
\[
L_\gl:=\{p\in\R^2\mid \psi(p)\leq\gl\},
\]
then for any $q\in \pl L_\gl$, we have
\beq \label{normal2} N(q,L_\gl)\subset \bcases
\cone\{v_1,n_\gam\}  \ & \IF\ q_1<h_1(q_2),\\
\cone\{v_2,m_\gam\} & \IF\ q_2<h_2(q_1),\\
\cone\{v_0,v_2\} & \IF\ q_1>h_1(q_2), \\
\cone\{v_0,v_1\} & \IF\ q_2>h_2(q_1).
\ecases
\eeq

For $\ep>0$, we set
\[
\eta_\ep(p)=\psi^2(p)+\ep|p|^2,
\]
and
\beq \label{def-phi}
\phi_\ep(x)=\max_{p\in\R^2}(x\cdot p-\eta_\ep(p)).
\eeq
Note that $\phi_\ep$ is convex on $\R^2$,
\[\phi_\ep\in C^{1,1}(\R^2), \ \ \AND \ \
\gd'|x|^2\leq \phi_\ep(x)\leq C'|x|^2 \ \ \FORALL \ x\in\R^2,
\]
where $0<\gd'\leq C'<\infty$ are constants.
Indeed, by the definition of $\phi_\ep$, it is clear that $\phi_\ep$ is convex on $\R^2$. Writing
\beq \label{inf-conv}
\phi_\ep(x)=\tfrac{1}{4\ep}|x|^2-\min_{p\in\R^2}\left(\psi^2(p)
+\ep\left|p-\tfrac{1}{2\ep}x\right|^2\right),
\eeq
we find that $\phi_\ep$ is semi-concave on $\R^2$. Hence, $\phi_\ep\in C^{1,1}(\R^2)$.  (Notice that the function $x\mapsto -\phi_\ep(x)+|x|^2/(4\ep)$ is the inf-convolution of the function $\psi^2$.) Finally, note that, since
\[
(\ep+\gd^2)|p|^2\leq \eta_\ep(p)\leq (\ep+C^2)|p|^2,
\]
we have
\[
\phi_\ep(x)
\bcases
\leq \max_{p\in\R^2}\left(x\cdot p-(\ep+C^2)|p|^2\right)= \tfrac{1}{4(\ep+C^2)}|x|^2& \\
\geq \max_{p\in\R^2}\left(x\cdot p-(\ep+\gd^2)|x|^2\right)=\tfrac{1}{4(\ep+\gd^2)|x|^2},&
\ecases
\]
and conclude that for all $x\in\R^2$,
\[
\tfrac{1}{4(\ep+C^2)}|x|^2\leq \phi_\ep(x)\leq \tfrac{1}{4(\ep+\gd)}|x|^2.
\]

\begin{theorem} \label{exist-phi} Let $\ep>0$ be sufficiently small.
The function $\phi_\ep$ defined by \erf{def-phi} satisfies:
\beq \label{cond-phi_ep}
H_1(D\phi_\ep(x)) \bcases
\geq 0 & \IF \ x_1\leq 0,\\
\leq 0 &\IF\ x_1\geq 0,
\ecases  \ \ \ \AND \ \ \
H_2(D\phi_\ep(x))\bcases
\geq 0 & \IF \ x_2\leq 0,\\
\leq 0 & \IF \ x_2\geq 0.
\ecases
\eeq
\end{theorem}

A possible choice of the constant $\ep>0$ in the theorem above is
\beq \label{cond-ep}
0<\ep<\fr{\ep_0 \gd^2}{\max\{|v_0|,|v_1|,|v_2|,|m_\gam|,|n_\gam|\}},
\eeq
where $\gd$ is the constant appearing in \erf{deltaC1} and \erf{deltaC2}
and
\beq\label{ep_0}
\ep_0=\min\{v_0\cdot e_1,v_0\cdot e_2,
-v_1\cdot e_1, v_1\cdot e_2, v_2\cdot e_1, -v_2\cdot e_2, -m_\gam\cdot e_2,-n_\gam \cdot e_1\}.
\eeq
Since $m_\gam\cdot e_1=-\gam=n_\gam\cdot e_2<0$, we find by \erf{cond-v_i} that
$\ep_0>0$.

\bproof We assume henceforth that $\ep$ satisfies \erf{cond-ep}, and prove that
\erf{cond-phi_ep} holds.
As before, we set $ L_\gl=\{q\in\R^2\mid \psi(q)\leq\gl\}$ for $\gl>0$.
Fix any $x\in\R^2$. Assume first $x_1\leq 0$ and show that
$H_1(D\phi_\ep(x))\geq 0$. We argue by contradiction. Set $p=D\phi_\ep(x)$ and suppose that
$
H_1(p)=-p_1+h(p_2)<0.
$

We have $p_1>h_1(p_2)$ and $p\not=0$.
By the convex duality, using  Lemma \ref{psi^2}, we have
\[
x\in \pl \eta_\ep(p)=2\psi(p)\pl\psi(p)+2\ep p,
\]
that is,  $x-2\ep p\in 2\psi(p)\pl\psi(p)$.
Set
$\gl=\psi(p), $
and note that $p\in \pl L_\gl$, $x-2\ep p\in N(p,L_\gl)$, and
$|x-2\ep p|\geq 2\gl\gd\geq 2\gd^2|p|$ by \erf{deltaC1} and \erf{deltaC2}.
Since $p_1>h_1(p_2)$, we have, by \erf{normal2},
\[
N(p,L_\gl)\subset \cone\{v_0,v_2\}.
\]
Accordingly, there are $t,s\geq 0$, with $t+s>0$, such that
$x-2\ep p=tv_0+sv_2$.  We have
\[
2\gd^2|p|\leq |x-2\ep p|\leq \max\{|v_0|,|v_2|\}(t+s),
\]
and also,  by \erf{ep_0},
\[
x_1\geq e_1\cdot (x-2\ep p)-2\ep |p| \geq \ep_0(t+s)-2\ep|p|.
\]
Hence,
\[
\max\{|v_0|,|v_2|\}x_1\geq \max\{|v_0|,|v_2|\}(\ep_0(t+s)-2\ep |p|)
 \geq 2\left(\ep_0\gd^2-\ep \max\{|v_0|,|v_2|\} \right)|p|.
\]
This, together with \erf{cond-ep}, gives a contradiction that $x_1>0$.

Next, we assume that $x_1\geq 0$ and show that  $H_1(D\phi_\ep(x))\leq 0$.
The proof in this case is similar to the above, and we argue by contradiction.
Suppose that $H_1(p)> 0$, that is, $p_1<h_1(p_2)$, where $p=D\phi_\ep(x)$.
We have $x-2\ep p\in 2\gl \pl\psi(p)$, where $\gl=\psi(p)$, and by \erf{normal2},
\[
\pl\psi(p)\subset N(p,L_\gl)\subset \cone\{v_1,n_\gam\}.
\]
We choose $t,s\geq 0$, with $t+s>0$, so that $x-2\ep p=t v_1+s n_\gam$ and note that
\[
2\gd^2|p|\leq |x-2\ep p|\leq \max\{|v_1|,|n_\gam|\}(t+s),
\]
and, by \erf{ep_0},
\[
x_1\leq e_1\cdot(x-2\ep p)+2\ep|p|
\leq -\ep_0(t+s)+2\ep|p|.
\]
Consequently, we have
\[
 \max\{|v_1|,|n_\gam|\} x_1\leq -2\left(\ep_0\gd^2 -\ep\max\{|v_1|,|n_\gam|\} \right)|p|,
\]
which yields a contradiction.

Now, we assume that $x_2\leq 0$. Similarly to the above,
we set $p=D\phi_\ep(x)$ and  suppose to the contrary that $H_2(p)< 0$, that is, $p_2>h_2(p_1)$.
We have $x-2\ep p\in 2\gl \pl\psi(p)$, with $\gl=\psi(p)$, and by \erf{normal2},
\[
x-2\ep p\in N(p,L_\gl)\subset \cone\{v_0,v_1\}.
\]
We choose $t,s\geq 0$, with $t+s>0$, so that $x-2\ep p=t v_0+s v_1$ and compute as before that
\[
2\gd^2|p|\leq |x-2\ep p|\leq \max\{|v_0|,|v_1|\}(t+s),
\]
and
\[
x_2\geq e_2\cdot(x-2\ep p)-2\ep|p|
\geq \ep_0(t+s)-2\ep|p|.
\]
Combining these two, we have
\[
 \max\{|v_0|,|v_1|\} x_2\geq 2\left(\ep_0\gd^2 -\ep\max\{|v_0|,|v_1|\} \right)|p|.
\]
This yields a contradiction, which shows that $H_2(p)\geq 0$.

Finally, we assume that $x_2\geq 0$ and show that  $H_2(p)\leq 0$, where $p=D\phi_\ep(x)$.  As always, we argue by contradiction and, hence, suppose that $H_2(p)> 0$, that is, $p_2<h_2(p_1)$.
We have $x-2\ep p\in 2\gl \pl\psi(p)$, where $\gl=\psi(p)$, and by \erf{normal2},
\[
x-2\ep p\in  N(p,L_\gl)\subset \cone\{v_2,m_\gam\}.
\]
We choose $t,s\geq 0$, with $t+s>0$, so that $x-2\ep p=t v_2+s m_\gam$ and note that
\[
2\gd^2|p|\leq |x-2\ep p|\leq \max\{|v_2|,|m_\gam|\}(t+s),
\]
and
\[
x_2\leq e_2\cdot(x-2\ep p)+2\ep|p|
\leq -\ep_0(t+s)+2\ep|p|.
\]
Hence, we have
\[
 \max\{|v_2|,|m_\gam|\} x_2\leq -2\left(\ep_0\gd^2 -\ep\max\{|v_2|,|m_\gam|\} \right)|p|,
\]
and get a contradiction.
\eproof

\def\th{\tilde h}

\section{Comparison principle} \label{comparison}

As an application of Theorem \ref{exist-phi},
we establish a comparison theorem for \erf{N-prob} in this section.

Let $\gO:=(0,\infty)^2$, as in the previous sections, and we reformulate Theorem
\ref{exist-phi} by removing the normalization \erf{zero-norm}.

Under the assumption \erf{cond-H_i}, the functions $h_i$ are both Lipschitz continuous on $\R$, with Lipschitz bound $\ga_0:=\max\{\ga_1,\ga_2\}\in[0,1)$, and, if we define the function $h\mid \R^2\to\R^2$ by $h(x)=(h_1(x_2),h_2(x_1))$, then
$h$ is Lipschitz continuous, with Lipschitz bound $\ga_0$. By the standard fixed point theorem, there exists a unique fixed point $a=(a_1,a_2)\in\R^2$ of the mapping $h$.
In particular, we have
\beq \label{fixed point} h_1(a_2)=a_1\quad\AND\quad h_2(a_1)=a_2.
\eeq
We define the functions
$\tH_i\mid \R^2\to\R$, $\tilde h_i\mid \R\to\R$, $i=1,2$, by
\beq\label{tilde H}
\tH_i(p)=H_i(p+a),\quad \th_1(t)=-a_1+h_1(t+a_2), \quad \th_2(t)=-a_2+h_2(t+a_1),
\eeq
and note that for all $p\in\R^2$ and $i=1,2$,
\[
\tH_1(p)=-p_1+\th_1(p_2), \quad \tH_2(p)=-p_2+\th_2(p_1),
\quad \th_i(0)=0.
\]
It is easily checked that \erf{cond-H_i} holds with $\tH_i$ replacing $H_i$.

\begin{corollary} \label{exist-f}  Assume \erf{cond-H_i}. Let $a\in\R^2$ be the point such that \erf{fixed point} holds. Define $\tH_i$ and $\th_i$, $i=1,2$,
by \erf{tilde H}, and assume that \erf{cond-slope} is satisfied for $\tH_i$
in place of $H_i$, $i=1,2$, and a choice of vectors $v_i$, $i=0,1,2$.
Then, there exists a convex function $f\in C^{1,1}(\R^2)$ such that
\beq \label{f1}
H_1(Df(x))
\bcases
\geq 0\ \ &\IF\ x_1\leq 0, \\
\leq 0 &\IF\ x_1\geq 0,
\ecases
\quad\AND \quad
H_2(Df(x))
\bcases
\geq 0 \ \ &\IF\ x_2\leq 0, \\
\leq 0 &\IF\ x_2\geq 0,
\ecases
\eeq
and for some positive constants $\gd, C$,
\beq \label{f2}
\gd|x|^2-C\leq f(x)\leq C(|x|^2+1) \ \ \ \FORALL x\in\R^2.
\eeq
\end{corollary}

\bproof We apply Theorem \ref{exist-phi}, to choose a function $\phi\in C^{1,1}(\R^2)$ which satisfies the conditions \erf{f1}, with $H_i$ replaced by $\tH_i$, and
\erf{f2} for some constants $0<\gd<C<\infty$.
We set $f(x)=\phi(s)+a\cdot x$ for $x\in\R^2$, observe that for all $x\in\R^2$ and $i=1,2$,
\[
H_i(Df(x))=H_i(D\phi(x)+a)=\tH_i(D\phi(x)),
\]
and conclude that $f$ satisfies \erf{f1}. It is clear that $f$ satisfies \erf{f2}.
\eproof

As an application of Corollary \ref{exist-f}, we present a comparison theorem for bounded sub and supersolutions
of \erf{N-prob}.

We write $\bS^n$ for the space of real $n\tim n$ symmetric matrices and say that
a function $\go: [0,\infty)\to[0,\infty)$ is a modulus if it is continuous
and nondecreasing on $[0,\infty)$. We denote by $I_n$ the unit matrix in $\bS^n$.
For $X,Y\in\bS^n$, the inequality $X\leq Y$ means that $Y-X$ is positive  semidefinite.  The choice of norm for $X\in\bS^n$ here is
$|X|=\max\{ |X\xi|\mid \xi\in\R^n, \,|\xi|=1\}$.

Let $F$ be a continuous, real-valued function on $\bS^2\tim\R^2\tim\R\tim\ol\gO$.
Assume in the spirit of \cite{CIL1992} that $F$ satisfies the following three conditions
\erf{F1}--\erf{F3}.
\renewcommand{\theenumi}{F\arabic{enumi}}
\renewcommand{\labelenumi}{(\theenumi)}

\begin{enumerate}
\item
\label{F1}
There is a constant $\mu>0$ such that for each $(X,p,x)\in\bS^2\tim\R^2\tim\gO$, the function \ $u\to F(X,p,u,x)-\mu u$\ is nondecreasing on $\R$.
\item
\label{F2}
There is a constant $L>0$ such that for every $(X,p),(Y,q)\in \bS^2\tim \R^2$ and
$(u,x)\in\R\tim\gO$,
\[
|F(X,p,u,x)-F(Y,q,u,x)|\leq L(|X-Y|+|p-q|).
\]
\item \label{F3}
There is a modulus $\go$ such that for any $X,Y\in \bS^2, x,y\in\gO$, $(p,u)\in\R^2\tim\R$,
and $\ga>0$, if
\[
-\ga \bmat I_2 & 0\\ 0&I_2\emat \leq \bmat X&0\\0&-Y\emat
\leq \ga\bmat I_2&-I_2\\
-I_2&I_2\emat,
\]
then
\[
F(X,p,u,x)-F(Y,p,u,y)\geq -\go(\ga|x-y|^2+|x-y|(|p|+1)).
\]
\end{enumerate}

Note (see \cite{CIL1992}) that if $F\in C(\bS^2\tim\R^2\tim\R\tim\gO)$ satisfies \erf{F3}, then $F$ is degenerate elliptic. The degenerate ellipticity means here that
if $X,Y\in\bS^2$ and $X\leq Y$, then $F(X,p,u,x)\geq F(Y,p,u,x)$ for all $(p,u,x)\in
\R^2\tim\R\tim\gO$.

We remark that a large class of elliptic operators arising as the dynamic programming
equations (Hamilton-Jacobi-Bellman or Hamilton-Jacobi-Isaacs equations) in stochastic optimal control or differential games satisfy \erf{F1}--\erf{F3}.

\begin{theorem}\label{comp.thm} Let $F\in C(\bS^2\tim\R^2\tim\R\tim\gO,\R)$ satisfy \erf{F1}--\erf{F3}. Let $H_i\in C(\R^2,\R)$, $i=1,2$, satisfy \erf{cond-H_i}.
Let $a\in\R^2$ satisfy \erf{fixed point} and define $\tH_i$, $i=1,2$, by \erf{tilde H}.
Assume that \erf{cond-slope} holds for $\tH_i$, $i=1,2$, and some $v_i\in\R^2$, $i=0,1,2$, satisfying \erf{cond-v_i}--\erf{cond-v_2}. Let $v,w\mid\ol\gO\to \R$
be, respectively, a lower semicontinuous viscosity subsolution and
upper semicontinuous viscosity supersolution of \erf{N-prob}. Moreover, assume that
$v,w$ are bounded on $\ol\gO$. Then, $v\leq w$ on $\ol\gO$.
\end{theorem}

We state an existence theorem of a (continuous) viscosity solution
of \erf{N-prob}, where the boundedness of the function $x\mapsto F(0,0,0,x)$
assures the existence of a bounded solution.

\begin{corollary}\label{exist} In addition to the hypotheses of Theorem \ref{comp.thm},
assume that the function $\gO\ni x\mapsto F(0,0,0,x)$ is bounded. Then, there exists a unique viscosity solution $u\in C(\ol\gO)$ of \erf{N-prob}.
\end{corollary}

\bproof The uniqueness assertion is an obvious consequence of
Theorem \ref{comp.thm}.

Choose a nonnegative function $\eta\in C^2([0,\infty))$ such that
\[
\eta'(0)=-1 \quad \AND\quad\max\{\eta(t), |\eta'(t)|,|\eta''(t)|\}\leq 1 \ \ \FORALL t\geq 0,
\]
set $\psi(x)=\eta(x_1)+\eta(x_2)$ for $x\in\ol\gO$, and observe that for any
$x\in\ol\gO$ and constant $C>0$, if $x_1=0$,
\[
H_1(CD\psi(x))\geq -C\eta'(0)+h_1(C\eta'(x_2)\leq h_1(0)+C(1-\ga_1),
\]
and
\[
H_1(-CD\psi(x))\leq C\eta'(0)+h_1(-C\eta'(x_2))\leq h_1(0)-C(1-\ga_1).
\]
If $x_2=0$, then
\[
H_2(CD\psi(x))\geq -C\eta'(0)+h_2(C\eta'(x_1)\geq h_2(0)+C(1-\ga_2),
\]
and
\[
H_2(-CD\psi(x))\leq C\eta'(0)+h_2(-C\eta'(x_2))\leq h_2(0)-C(1-\ga_2).
\]
For instance, we may choose the function $\eta(t)=e^{-t}$.

Choose a constant $C_0>0$ so that $|F(0,0,0,x)|\leq C_0$ for all $x\in\gO$.
Let $C_1,C_2$ be positive constants to be fixed later, and set
\[
v(x)=-C_1\psi(x)-C_2 \ \ \ \AND \ \ \ w(x)=C_1\psi(x)+C_2\quad\FOR x\in\ol\gO.
\]
Compute that for any $x\in\ol\gO$,
\[\bald
F(D^2v(x),Dv(x),v(x),x)
&\leq F(0,0,v(x),x)-\mu v(x)+\mu v(x)
\\&\quad +LC_1(|\diag\{\eta''(x_1),\eta''(x_2)\}|+2)
\\&\leq C_0-\mu C_2+3LC_1,
\eald\]
if $x_1=0$, then
\[
H_1(Dv(x))=H_1(-C_1D\psi(x))\leq h_1(0)-C_1(1-\ga_1),
\]
and if $x_2=0$, then
\[
H_2(Dv(x))\leq h_2(0)-C_1(1-\ga_2).
\]
Similarly, we have for any $x\in\ol\gO$,
\[
F(D^2w(x),Dw(x),w(x),x)
\geq F(0,0,0,x)+\mu w(x)-3LC_1
\geq -C_0+\mu C_2-3LC_1,
\]
if $x_1=0$, then
\[
H_1(Dw(x))\geq h_1(0)+C_1(1-\ga_1),
\]
and if $x_2=0$, then
\[
H_2(Dw(x))\leq h_2(0)+C_1(1-\ga_2).
\]
We fix $C_1$ so that
\[
|h_i(0)|\leq C_1(1-\ga_i) \ \ \ \FOR i=1,2,
\]
and then $C_2$ so that
\[
\mu C_2\geq C_0+3LC_1,
\]
to find that $v$ and $w$ are, respectively, a classical (and hence, viscosity)
subsolution and supersolution of \erf{N-prob}.
If we set
\[
u(x)=\sup\{\phi(x)\mid v\leq \phi\leq w \ON \ol\gO,\ \phi \text{ viscosity subsolution of } \erf{N-prob}\} \ \ \FOR x\in\ol\gO,
\]
then, according to the Perron method (see \cite{CIL1992}),
$u$ is an upper semicontinuous subsolution of \erf{N-prob}
and the lower semicontinuous envelope $u_*$ of $u$ is a supersolution of \erf{N-prob}.
Theorem \ref{comp.thm} assures that $u\leq u_*$ on $\ol\gO$, which implies that
$u\in C(\ol\gO)$. Since $v\leq u\leq w$ on $\ol\gO$, it is clear that $u$ is bounded on $\ol\gO$.
\eproof

\bproof[Proof of Theorem \ref{comp.thm}] We fix a nonnegative function $\eta\in C^2([0,\infty))$ so that
\[
\eta'(0)=-1, \quad \max\{|\eta'(t)|,|\eta''(t)|\}\leq 1 \ \ \FORALL t\geq 0, \ \ \AND \ \
\lim_{t\to\infty} \eta(t)=+\infty.
\]
Let $\ep>0$, $C_1>0$, and set for $x\in\ol\gO$.
\[
\psi(x)=\eta(x_1)+\eta(x_2), \quad
\tilde v(x)=v(x)-\ep(\psi(x)+C_1) \ \ \AND \ \ \tilde w(x)=w(x)+\ep(\psi(x)+C_1).
\]

We first observe that if $C_1$ is sufficiently large, then, for any $\ep>0$, $\tilde v$ and $\tilde w$ are a ``strict'' subsolution and supersolution of \erf{N-prob}, respectively.
To check this, we compute slightly in an informal manner that for $x\in\gO$,
\[
\bald
|D^2\psi(x)|&= |\diag\{\eta''(x_1),\eta''(x_2)\}|\leq 2,
\quad |D\eta(x)|\leq 1,
\\F(D^2\tilde v,D\tilde v,\tilde v,x)
&\leq F(D^2 v,Dv,\tilde v,x)-\mu \tilde v+\mu \tilde v+3\ep\,L
\\& \leq F(D^2 v,Dv,v,x)+3\ep\,L-\mu \ep C_1\leq -\ep(\mu C_1-3L),
\eald
\]
for $x\in \pl_1\gO$,
\[
H_1(D\tilde v(x))=H_1(Dv(x)-\ep D\psi(x))\leq -\ep(1-\ga_1),
\]
and for $x\in\pl_2\gO$,
\[
H_2(D\tilde v(x))=H_2(Dv(x)-\ep D\psi(x))\leq -\ep(1-\ga_1).
\]
Similarly, if $x\in\gO$,
\[
\bald
F(D^2\tilde w,D\tilde w,\tilde w,x)
\geq F(D^2 w,Dw,w,x)-3\ep\,L+\mu \ep C_1\geq \ep(\mu C_1-3L),
\eald
\]
if $x\in \pl_1\gO$,
\[
H_1(D\tilde w(x))=H_1(Dw(x)+\ep D\psi(x))\geq \ep (1-\ga_2),
\]
and if $x\in\pl_2\gO$,
\[
H_2(D\tilde w(x))=H_2(Dw(x)+\ep D\psi(x))\geq \ep (1-\ga_2).
\]

Fixing  $C_1$ by $\mu C_1=3L+1$ and setting $\rho:=\min_{i=1,2}(1-\ga_i)$,
we easily find that $\tilde v$ and $\tilde w$ are respectively a subsolution of
\beq\label{strict-sub}
\left\{\bald
&F(D^2u,Du,u,x)=-\ep\ \ \IN \gO,
\\&H_1(Du)=-\ep\rho \ \ \ON \pl_1\gO,\quad H_2(Du)=-\ep\rho \ \ \ON \pl_2\gO,
\eald \right.
\eeq
and supersolution of
\beq\label{strict-super}
\left\{\bald
&F(D^2u,Du,u,x)=\ep \ \ \IN \gO,
\\&H_1(Du)=\ep\rho \ \ \ON \pl_1\gO,\quad H_2(Du)=\ep\rho \ \ \ON \pl_2\gO,
\eald \right.
\eeq
Note also that
\beq\label{at-infty}
\lim_{|x|\to \infty}\tilde v(x)=-\infty \ \ \AND \ \
\lim_{|x|\to \infty}\tilde w(x)=+\infty.
\eeq

It is enough to prove that $\max_{\ol\gO}(\tilde v-\tilde w)\leq 0$
 for all $\ep>0$.

To do this, we argue by contradiction, and suppose that
for some $\ep>0$,
\beq\label{positive}
\max_{\ol\gO}(\tilde v-\tilde w)>0.
\eeq
We fix $\ep>0$ so that \erf{positive} holds.
Let $f$ be the function from Corollary \ref{exist-f}.
Let $\ga>0$ and consider the function
\[
\tilde v(x)-\tilde w(y)-\ga^{-1} f(\ga(x-y))
\]
on $\ol\gO\tim\ol\gO$. In view of \erf{at-infty} and \erf{f2}, this function takes a maximum
at some $(x_\ga,y_\ga)\in{\ol\gO\,}^2$. Moreover, the set
$\{(x_\ga,y_\ga)\mid \ga>0\}$ is bounded.

Since $f$ is a $C^{1,1}$ function, there is a constant $B>0$ such that $\xi\mapsto
f(\xi)-B|\xi|^2/2$ is concave on $\R^2$, and hence,
\beq\label{semi-cv}
f(x)-f(\xi)\leq Df(\xi)\cdot (x-\xi)+\tfrac{B}{2}|x-\xi|^2 \ \ \FORALL x,\xi\in\R^2.
\eeq
Consequently,  we have for any $x,y\in{\ol\gO\,}^2$,
\[\bald
\tilde v(x)-\tilde w(y) &
-\tilde v(x_\ga)-\tilde w(y_\ga)
\leq \ga^{-1}(f(\ga(x-y))-f(\ga(x_\ga-y_\ga))
\\& \leq Df(\ga(x_\ga-y_\ga))\cdot (x-y-((x_\ga-y_\ga))
+\tfrac{B \ga}{2}|x-y-(x_\ga-y_\ga)|^2,
\eald\]
which implies that if we write $\,p_\ga=Df(x_\ga-y_\ga)\,$ and
\[
\psi_\ga(x,y)=p_\ga \cdot (x-y-(x_\ga-y_\ga))
+\tfrac{B \ga}{2}|x-y-(x_\ga-y_\ga)|^2 \ \ \FOR x,y\in\R^2,
\]
then the function $(x,y)\mapsto \tilde v(x)-\tilde w(y) -\psi_\ga(x,y)$ has a maximum
at $(x_\ga,y_\ga)$ on ${\ol\gO\,}^2$. Noting that
\[
D^2\psi_\ga(x,y)=B\ga \bmat I_2&-I_2 \\ -I_2 & I_2\emat,
\]
we deduce thanks to \cite[Theorem 3.2]{CIL1992} that
there are $X_\ga, Y_\ga\in\bS^2$ such that
\begin{gather}\label{semi-max1}
(p_\ga, X_\ga)\in\ol{J}^{2,+}\tilde v(x_\ga),\quad
(p_\ga,Y_\ga)\in \ol{J}^{2,-}\tilde w(y_\ga),
\\ \label{semi-max2}
-3B \ga\bmat I_2 &0 \\ 0& I_2\emat
\leq \bmat X_\ga & 0 \\ 0 & -Y_\ga\emat
\leq 3B \ga\bmat I_2&-I_2 \\ -I_2& I_2\emat,
\end{gather}
where $\ol{J}^{2,+}$ and $\ol {J}^{2,-}$
are the ``closures'' of the second-order superjet and subjet, respectively (see \cite{CIL1992} for the definition of $\ol{J}^{2,\pm}$).
It follows from \erf{semi-max1}, \erf{strict-sub}, and \erf{strict-super} that
\begin{gather}\label{st-visco1}
F(X_\ga, p_\ga, \tilde v(x_\ga),x_\ga)\leq -\ep \ \ \IF x_\ga\in\gO, \\
\label{st-visco2}
F(Y_\ga, p_\ga, \tilde w(y_\ga),y_\ga)\leq -\ep \ \ \IF y_\ga\in\gO,\\
\label{st-visco3}
\min\{H_i(p_\ga),\,F(X_\ga, p_\ga, \tilde v(x_\ga),x_\ga)\}\leq -\ep\rho \ \ \IF x_\ga\in\pl_i\gO,\,i=1,2, \\
\label{st-visco4}
\min\{H_i(p_\ga),\,F(Y_\ga, p_\ga, \tilde w(y_\ga),y_\ga)\}\leq -\ep\rho \ \ \IF y_\ga\in\pl_i\gO,\,i=1,2.
\end{gather}

It is a standard observation (see, e.g.,  \cite{CIL1992}) that for some
$x_0\in\ol\gO$, as $\ga \to \infty$ along a sequence $\{\ga_j\}$,
\[
(x_\ga,y_\ga) \to (x_0,x_0),  \quad
\tilde v(x_\ga) \to \tilde v(x_0), \quad
\tilde w(y_\ga)\to \tilde w(x_0), \quad \AND \quad
\ga^{-1}f(\ga(x_\ga-y_\ga)) \to 0.
\]
In particular, \erf{f2} ensures that, as $\ga=\ga_j$ and $j\to\infty$,
\beq\label{est-(x-y)}
\ga|x_{\ga}-y_\ga|^2 \ \to \ 0.
\eeq
Furthermore, plugging $x=\xi+q|\xi|$, with $|q|=1$, and $\xi=\ga(x_\ga-y_\ga)$ into \erf{semi-cv} yields
\[
\ga|x_\ga-y_\ga||p_\ga|\leq f(\ga(x_\ga-y_\ga))+B\ga^2|x_\ga-y_\ga|^2,
\]
which, combined with \erf{est-(x-y)} and the previous observations, shows that, as $\ga=\ga_j$ and $j\to\infty$,
\beq \label{est-p}
\ga^{-1}|p_\ga||x_\ga-y_\ga| \,\to\, 0.
\eeq

Focusing on $\ga=\ga_j$, with large $j$, we may assume that
$\tilde v(x_\ga)>\tilde w(y_\ga)$.
Observe by the choice of $f$ that if $x_\ga\in\pl_1\gO$, then
$(x_\ga-y_\ga)\cdot e_1=-y_\ga\cdot e_1\leq 0$ and, by \erf{f1},
\[H_1(p_\ga)=H_1(Df(\ga(x_\ga-y_\ga))\geq 0.
\]
Similarly, if $x_\ga\in\pl_2\gO$, then $(x_\ga-y_\ga)\cdot e_2= -y_\ga\cdot e_2\leq 0$ and $H_2(p_\ga)=H_2(Df(x_\ga-y_\ga))\geq 0$.
Also, if $y_\ga\in\pl_1\gO$, then
$(x_\ga-y_\ga)\cdot e_1=\ga x_\ga\cdot e_1\geq 0$ and $H_1(p_\ga)
=H_1(Df(x_\ga-y_\ga))\leq 0$.
If $y_\ga\in\pl_2\gO$, then $(x_\ga-y_\ga) \cdot e_2 =x_\ga\cdot e_2\geq 0$
and $H_2(p_\ga)=H_2(Df(x_\ga-y_\ga))\leq 0$.
Taking into account these sign properties, we obtain by \erf{st-visco1}--\erf{st-visco4},
\[\bald
 &F(X_\ga,p_\ga,\tilde v(x_\ga),x_\ga)\leq -\ep\rho,
\\&F(Y_\ga,p_\ga,\tilde w(y_\ga),y_\ga)\geq \ep\rho,
\eald
\]
Since $u\mapsto F(X,p,u,x)$ is increasing on $\R$ by \erf{F1}, we have
\[
F(X_\ga,p_\ga,\tilde w(y_\ga),x_\ga)\leq -\ep\rho,
\]
and
\[
-2\ep\rho\geq F(X_\ga,p_\ga,\tilde w(y_\ga),x_\ga)-F(Y_\ga,p_\ga,\tilde w(y_\ga),y_\ga),
\]
which yields, together with \erf{F3},
\[
-2\ep\rho\geq -\go(3B\ga|x_\ga-y_\ga|^2+|x_\ga-y_\ga|(|p_\ga|+1)).
\]
Recalling \erf{est-(x-y)} and \erf{est-p} and sending $j\to\infty$ for $\ga=\ga_j$, we
get a contradiction, $\go(0)=0\leq -2\ep\rho$. The proof is complete.
\eproof

\begin{bibdiv}
\begin{biblist}
\bib{B1999}{article}{
   author={Barles, Guy},
   title={Nonlinear Neumann boundary conditions for quasilinear degenerate
   elliptic equations and applications},
   journal={J. Differential Equations},
   volume={154},
   date={1999},
   number={1},
   pages={191--224},
   issn={0022-0396},
   review={\MR{1685618}},
   doi={10.1006/jdeq.1998.3568},
}
\bib{BISW2017}{article}{
   author={Biswas, Anup},
   author={Ishii, Hitoshi},
   author={Saha, Subhamay},
   author={Wang, Lin},
   title={On viscosity solution of HJB equations with state constraints and
   reflection control},
   journal={SIAM J. Control Optim.},
   volume={55},
   date={2017},
   number={1},
   pages={365--396},
   issn={0363-0129},
   review={\MR{3606413}},
   doi={10.1137/15M103830X},
}
\bib{CIL1992}{article}{
   author={Crandall, Michael G.},
   author={Ishii, Hitoshi},
   author={Lions, Pierre-Louis},
   title={User's guide to viscosity solutions of second order partial
   differential equations},
   journal={Bull. Amer. Math. Soc. (N.S.)},
   volume={27},
   date={1992},
   number={1},
   pages={1--67},
   issn={0273-0979},
   review={\MR{1118699}},
   doi={10.1090/S0273-0979-1992-00266-5},
}
\bib{DI1991}{article}{
   author={Dupuis, Paul},
   author={Ishii, Hitoshi},
   title={On oblique derivative problems for fully nonlinear second-order
   elliptic PDEs on domains with corners},
   journal={Hokkaido Math. J.},
   volume={20},
   date={1991},
   number={1},
   pages={135--164},
   issn={0385-4035},
   review={\MR{1096165}},
   doi={10.14492/hokmj/1381413798},
}

\bib{I1991}{article}{
   author={Ishii, Hitoshi},
   title={Fully nonlinear oblique derivative problems for nonlinear
   second-order elliptic PDEs},
   journal={Duke Math. J.},
   volume={62},
   date={1991},
   number={3},
   pages={633--661},
   issn={0012-7094},
   review={\MR{1104812}},
   doi={10.1215/S0012-7094-91-06228-9},
}

\end{biblist}
\end{bibdiv}
\bye

redpajama_set_name: RedPajamaArXiv


text:
Monday, November 9, 2015 - 7:00pm
Green Street Arts Center
On Monday night, November 9, at 7:00PM, Connecticut Heritage Productions, will join theaters, libraries, and colleges nationwide in the simultaneous staged readings of The Other Woman, a full-length, one-act play adapted from five essays appearing in the bestselling anthology The Other Woman: Twenty-one Wives, Lovers, and Others Talk Openly About Sex, Deception, Love, and Betrayal. Co-producers Cynthia Comsky and Victoria Zackheim (anthology editor and playwright) have engaged the participation of artistic directors across the country, including experimental as well as traditional theaters, schools and libraries reflecting the diversity of our population in the United States.
Connecticut Heritage Productions will be highlighted in a local/regional/national media campaign designed to draw attention to the range of venues for creativity around the country, and to forge a connection with our own community.
The Other Woman tells the stories of five women with very different viewpoints and experiences around infidelity. Authors whose work appears in the play are Caroline Leavitt, Aviva Layton, Mary Jo Eustace, Connie May Fowler, and Maxinne Rhea Leighton. The play's first reading took place at the Players Club in Manhattan, with parts read by Kathleen Chalfant, Ellen McLaughlin, Penny Fuller, Winslow Corbett, and Connie May Fowler.
Connecticut Heritage Productions production will be directed by Teresa Langston with roles read by Susan Smith Thom, Melissa McLean, Barbara Gallow, Elizabeth Reynolds and Amy Bentley. The event will be held at the Green Street Center for the Arts in Middletown, CT.

redpajama_set_name: RedPajamaCommonCrawl


text:
Items Engine Co. No. 21 was organized Dec. 21st, 1872, at No. 13 Taylor St.; [A group of colored policemen]; Lieutenant William Middleton; Officer Albert C. Blue.
Engine Co. No. 21 was organized Dec. 21st, 1872, at No. 13 Taylor St.; [A group of colored policemen]; Lieutenant William Middleton; Officer Albert C. Blue.
Schomburg Center for Research in Black Culture, Jean Blackwell Hutson Research and Reference Division
Taitt, John
Souvenir of Negro progress: Chicago, 1779-1925.
Taitt, John (Author)
Place: Chicago
Publisher: De Saible Assoc.
Shelf locator: Sc 977.3-T (Taitt, J. Souvenir of Negro progress)
RLIN/OCLC: NYPG92-B21085
Universal Unique Identifier (UUID): ffbc2b30-c6cf-012f-3bcc-3c075448cc4b
Schomburg Center for Research in Black Culture, Jean Blackwell Hutson Research and Reference Division, The New York Public Library. "Engine Co. No. 21 was organized Dec. 21st, 1872, at No. 13 Taylor St.; [A group of colored policemen]; Lieutenant William Middleton; Officer Albert C. Blue." The New York Public Library Digital Collections. 1925. http://digitalcollections.nypl.org/items/510d47de-4d74-a3d9-e040-e00a18064a99
Schomburg Center for Research in Black Culture, Jean Blackwell Hutson Research and Reference Division, The New York Public Library. "Engine Co. No. 21 was organized Dec. 21st, 1872, at No. 13 Taylor St.; [A group of colored policemen]; Lieutenant William Middleton; Officer Albert C. Blue." New York Public Library Digital Collections. Accessed July 18, 2019. http://digitalcollections.nypl.org/items/510d47de-4d74-a3d9-e040-e00a18064a99
Schomburg Center for Research in Black Culture, Jean Blackwell Hutson Research and Reference Division, The New York Public Library. (1925). Engine Co. No. 21 was organized Dec. 21st, 1872, at No. 13 Taylor St.; [A group of colored policemen]; Lieutenant William Middleton; Officer Albert C. Blue. Retrieved from http://digitalcollections.nypl.org/items/510d47de-4d74-a3d9-e040-e00a18064a99
<ref name=NYPL>{{cite web | url=http://digitalcollections.nypl.org/items/510d47de-4d74-a3d9-e040-e00a18064a99 | title= (still image) Engine Co. No. 21 was organized Dec. 21st, 1872, at No. 13 Taylor St.; [A group of colored policemen]; Lieutenant William Middleton; Officer Albert C. Blue., (1925) }} |author=Digital Collections, The New York Public Library |accessdate=July 18, 2019 |publisher=The New York Public Library, Astor, Lennox, and Tilden Foundation}}</ref>


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text:
Oklahoma addiction treatment facilities report trouble with insurance coverage
Published: Sun, December 31, 2017 5:00 AM
Pregnant woman in hospital gown rubs belly
Oklahoma City — When patients check into Eagle Ridge Institute, CEO Michael Hanes knows his staff may have as little as 10 days to try to change a life's downward trajectory.
Eagle Ridge, in Oklahoma City, treats pregnant women and mothers with addictions. Many have extensive histories of trauma that contribute to their drug use, Hanes said, but insurance companies sometimes only authorize a few days of inpatient treatment — assuming they agree to it at all. In those cases, the providers don't dig deeply into a patients' histories, because they won't have the time to help patients process the memories that come up, he said.
Patients' brains need to begin to heal from the effects of drugs before they can work with providers to address any mental health conditions or socioeconomic problems they face, Hanes said. The process can take 90 days or longer, he said, and not all insurers are willing to allow that much time.
Insurers "are unwilling to treat it as a chronic condition," he said. "They see it like, you get an infection, you get an antibiotic and you're cured."
Federal law requires parity in insurance benefits for mental health and addiction treatment, meaning companies can't set tougher restrictions for those types of care than for physical health services. For example, insurers can't set limits on the number of times a patient can visit a mental health provider, if they don't have the same limits for medical providers. They also aren't allowed to charge patients higher out-of-pocket costs for behavioral health care, which includes both mental health and addiction.
In practice, however, it may not always work out that way. Some insurers still have higher out-of-pocket costs for substance use disorder treatment than for medical care, Hanes said — which would appear to be a clear parity violation. Other disparities, however, are harder to pin down.
A report by Milliman, a consulting firm, found that in 2015 primary care providers in Oklahoma were paid about 18 percent more than behavioral health providers. The gap narrowed when the analysts compared two specific services that both types of providers could offer, but primary care providers still earned about 8 percent more. The gap was wider when behavioral health providers were compared to specialists. Disparities were even larger in many other states.
Lower rates could also contribute to other disparities, the report suggested, because providers tend not to join plans that don't offer competitive rates. In Oklahoma, people who needed inpatient behavioral care were four times more likely to have to go out-of-network than people who needed hospital treatment for a medical issue in 2015. Insurance companies set up networks of providers who have agreed to accept certain rates, and steer patients toward those. If patients go to an out-of-network provider, they generally end up paying more.
Oklahomans who needed outpatient behavioral health care had an even greater disparity: they were eight times as likely to use out-of-network providers than people seeking similar medical care. The report didn't consider other types of restrictions, like how much patients must pay out-of-pocket for medications, and whether they are required to try cheaper drugs first.
So are the disparities a smoking gun that Oklahoma insurance plans are breaking the law? It's not so simple. For one thing, it's possible that other factors, like patients' reluctance to change therapists after beginning treatment, could explain part of the difference. There also could be legitimate explanations for paying some providers more than others.
The Oklahoma Insurance Department typically gets three or four complaints alleging parity violations each year, said Mike Rhoads, deputy commissioner of life and health insurance. In almost all cases, patients went out of network and were left with large bills — a difficult situation, but not a parity violation, he said.
The department also compiles market conduct reports, where staff examine a sample of an insurer's cases for parity violations and other problems, Rhoads said. The parity portion looks at whether insurers pay less for an initial behavioral health visit than for a primary care visit, and insurers typically pass muster, he said.
The comparison doesn't look at other services, because of the difficulty of finding matches. It also doesn't compare out-of-pocket costs or how many in-network providers a company has, though the department will look bring it up with companies if staff believe the network is inadequate, Rhoads said. Overall, he thinks Oklahoma insurers are following federal and state law.
"We've got some carriers in Oklahoma that get it," he said.
Even state insurance regulators can't always tell if an insurance plan meets parity, however, according to a report compiled jointly by the National Center on Addiction and Substance Abuse, the Legal Action Center, the Partnership for Drug-Free Kids and the Treatment Research Institute. That makes it difficult for average patients to know if they have a case, let alone what they need to do to resolve the problem, the report said.
"Individuals do not have access to data necessary to show a violation that occurs at the systems level," the report said.
Insurers point to efforts
Cigna, Aetna, Blue Cross Blue Shield of Oklahoma and United Healthcare all emphasized that they cover "medically necessary" services for substance use disorder.
Only GlobalHealth, in an email from Chief Operations Officer J. David Thompson, said it sets limits on inpatient and outpatient treatment days.
Blue Cross also said some of its grandfathered plans still could include limits of 30 days for inpatient care and 20 days for outpatient care, but that most of its plans don't include limits. Grandfathered plans were in place before the Affordable Care Act passed.
A few insurers say they've taken steps to make it easier to get treatment.
Rick Watson, senior medical director at Cigna for Oklahoma and north Texas, said the company reduced the paperwork needed to authorize treatment with methadone or buprenorphine. The two drugs are used to replace opioids without causing a high, so patients don't go through the withdrawal symptoms that could lead them to relapse.
Mark Friedlander, chief medical officer for Aetna Behavioral Health, said the company lists both medications as "preventative," so customers don't need prior authorization and may have lower co-pays.
Representatives for United Healthcare said the company also doesn't require prior authorization for some addiction medications, including buprenorphine and naloxone, which can reverse an opioid overdose.
Blue Cross said it removed prior authorization requirements for medication-assisted treatment in February this year.
"We continue to provide our members with access to high-value, evidence-based health care treatments and encourage our members to speak with their doctors about the best treatment options available for their specific needs," Blue Cross said in a statement.
Cigna also offers specialized case managers for people who appear to be at risk of a substance use disorder because they have multiple prescribers or take high doses, Watson said. The case manager can lay out options for treatment, if needed, he said. Customers aren't required to participate, though their employers may give them incentives to work with a case manager.
"We are not only reactive when someone is in need, but try to be proactive as well," he said. "We want to get them back on the path to wellness."
Aetna encourages patients to work with a case manager to come up with a treatment plan, Friedlander said. The hope is that reaching patients earlier will allow them to find options that meet their needs well, rather than whatever happens to be available in an emergency, he said.
"Many times people don't seek treatment until there's a crisis," he said.
Questions about need
One reason for disparities may be that insurers are still confused, because the parity rules are relatively new, said Ky Humble, development director at The Recovery Center in Oklahoma City. The first parity law passed in 2008, and federal officials only began enforcing finalized regulations in January 2014. Before the Affordable Care Act expanded the 2008 parity law to cover more plans, most insurers wouldn't cover medications to assist with detoxification under any circumstances, he said.
Another problem is that it's difficult to tell whether legal restrictions, like requiring providers to show that services are medically necessary for a patient, are being used appropriately.
Insurance companies don't always agree with their assessment of a client's needs, such as whether a client needs medication to help with detox, Humble said. One company may agree to cover a medication, while another might think a client in a similar situation doesn't need it, he said.
"Different insurance companies require different levels of need," he said.
Medical necessity criteria are based on the severity of patients' symptoms and their levels of functioning, which makes them somewhat subjective, Hanes said.
"A lot of times the criteria for medical necessity can be very high and somewhat obscure," he said. "Each reviewer with an insurance company may have their own interpretation."
The problem may go even deeper than differences in interpretation, however. Medical criteria sometimes aren't a good way to evaluate the needs of people with substance use disorders, said Pam Music, assistant to the business manager at Valley Hope Cushing. She said staff have to get on the phone with insurers nearly every day to try to extend a patient's stay beyond 14 days.
Patients may only have medical symptoms that need around-the-clock monitoring for the first few days of detox, Music said, but they may not be ready to stay away from the drugs while living in the community — and may not even have access to outpatient services if they live in rural areas, she said.
"After the first few days or week, people aren't going to have the tremors," she said. "Anymore, (insurers) are looking at us to triage, get (patients) stable and send them out."
Not paying for enough treatment for people who need to be in an inpatient program can set patients up to cycle in and out of treatment as they get out and relapse — which ultimately may cost insurers more in the long run, Music said. Or worse, patients may never return to treatment because they overdose after returning to drugs like opioids, she said.
"They're playing with people's lives, and that's something I can't tolerate," Music said. "We're losing way too many people."
Mental Health Act
To read more about the law, go to cms.gov and search for "Mental Health Parity and Addiction Equity Act."
CommentsOklahoma addiction treatment facilities report trouble with insurance coverage


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