#  A Liouville theorem for the Neumann problem of Monge-Ampère equations

###### Abstract.

In this paper, we study the Neumann problem of Monge-Ampère equations in Semi-
space. For two dimensional case, we prove that its viscosity convex solutions
must be a quadratic polynomial. When the space dimension $n\geq 3$, we show
that the conclusion still holds if either the boundary value is zero or the
viscosity convex solutions restricted on some $n-2$ dimensional subspace is
bounded from above by a quadratic function.

This work was supported by NSFC 11771237.

Huaiyu Jian

Department of Mathematical Sciences, Tsinghua University

Beijing 100084, China

Xushan Tu

Department of Mathematical Sciences, Tsinghua University

Beijing 100084, China

AMS Mathematics Subject Classification: 35J60, 35B50.

Running head: Liouville theorem for Monge-Ampère equation

A Liouville theorem for the Neumann problem of Monge-Ampère equations

Huaiyu Jian Xushan Tu

##  1\.  Introduction

The Bernstein theorem for Monge-Ampère equation
$\operatorname{det}D^{2}u(x)=1$ in $\mathbb{R}^{n}$ concludes that its
solution must be a quadratic polynomial, which was proved in [14, 8, 21, 22,
9]. This type of theorems was extended to the Hessian equation[2], special
Lagrangian equation[25], affine maximal hypersurface equation[23] and more
general Monge-Ampére equation $\operatorname{det}D^{2}u(x)=f(x)$ [6, 7, 10,
18], and it is called Liouville theorem if additional assumption on the
solution is necessary, just as the Liouville theorem for harmonic functions.
The Liouville theorem for Monge-Ampère equations in semi-space
$\mathbb{R}_{+}^{n}$ with Dirichlet boundary value on
$\partial\mathbb{R}_{+}^{n}$ was studied in [17, 20].

In this paper, we study the Liouville theorem for the Neumann problem

(1.1) $\begin{cases}\operatorname{det}D^{2}u(x)=1&\text{ in
}\mathbb{R}_{+}^{n},\\\ D_{n}u(x^{\prime},0)=ax_{1}&\text{ on
}\partial\mathbb{R}_{+}^{n}.\end{cases}$

Here and below,
$\mathbb{R}_{+}^{n}:=\\{(x^{\prime},x_{n})\in\mathbb{R}^{n}:\;x_{n}>0\\}$,
$D_{n}u=D_{e_{n}}u=\frac{\partial u}{\partial x_{n}}$, $a\in R$ is a constant,
and $e_{n}$ is the standard unit vector on $x_{n}$-axis.

When $n\geq 3$, we need to assume that $u$ is quadratic growth in the
auxiliary subspace at the infinite, i.e.,

(1.2) $\lim_{z\in\mathbb{R}^{n-2},z\to\infty}\frac{u(0,z,0)}{|z|^{2}}<\infty.$

We use $\Theta$ to denote the family of functions satisfying (1.2) .

We will use the standard concept for (Alesandrov) generalized solutions and
viscosity solutions to the first equation of (1.1). See the book [12] for
example. Caffarelli[3] indicated that the generalized solution and viscosity
solution to the first equation of (1.1) are equivalent, which is proved in
detail in [12]. This fact will be used again and again. To describe the
Neumann boundary value condition in a weak sense, we choose the viscosity
solution.

We say that $u$ is a viscosity subsolution ( or supersolution) of problem
(1.1) if $u\in C(\overline{\mathbb{R}_{+}^{n}})$ is a convex function, and for
any $x_{0}\in\overline{\mathbb{R}_{+}^{n}}$ and any convex function
$\varphi\in C^{2}(\overline{\mathbb{R}_{+}^{n}})$, the condition that
$u-\varphi$ has a local maximun ( or minimum) at $x_{0}$, implies

$\begin{cases}\operatorname{det}D^{2}\varphi(x_{0})\geq(or\leq)1&if\;x_{0}\in\mathbb{R}_{+}^{n},\\\
D_{n}\varphi(x_{0})\geq(or\leq)\phi(x_{0})&if\;x_{0}\in\partial\mathbb{R}_{+}^{n}.\end{cases}$

If $u\in C(\overline{\mathbb{R}_{+}^{n}})$ is both viscosity subsolution and
viscosity supersolution, we call it a viscosity solution.

The main result of this paper is the following

######  Theorem 1.1.

Let $u$ be a viscosity solution to problem (1.1). If either $n=2$, or when
$n\geq 3$, $u\in\Theta$ or $a=0$, then $u$ is a quadratic polynomial.

In a future paper [15], we will use Theorem 1.1 and perturbation method to
prove the global Schauder estimate for the viscosity solution to the Neumann
problem of Monge-Ampère equation $\operatorname{det}D^{2}u(x)=f(x)$ in a
bounded convex domain. Such estimate was obtained in [19, 24] for Dirichlet
boundary value problem.

Observe that the condition (1.2) does not depend on the base point $0$, this
is because

$2u(a,b+z^{\prime\prime},0)\leq u(0,2z^{\prime\prime},0)+u(2a,2b,0)$

by the convexity, and (1.2) is invariant after subtracting any linear
functions. Therefore, under a suitable linear transformation, our theorem
actually applies to the following problem:

(1.3) $\begin{cases}\operatorname{det}D^{2}u(x)=c&\text{ in
}\mathbb{R}_{+}^{n},\\\ D_{\alpha}u(x^{\prime},0)=\beta\cdot
x^{\prime}+b&\text{ on }\partial\mathbb{R}_{+}^{n},\end{cases}$

where $c>0$ and $b\in R$ are constants, while $\alpha$ and $\beta$ are
constant vectors such that $\alpha\cdot e_{n}>0$ and $\beta\cdot e_{n}=0$. In
this situation, the assumption (1.2) changes to

(1.4) $\lim_{z^{\prime}\in
H^{\prime\prime},z^{\prime}\to\infty}\frac{u(z^{\prime},0)}{|z^{\prime}|^{2}}<\infty$

for some $n-2$ dimensional subspace $H^{\prime\prime}\subset R^{n}$ which is
not parallel to $\beta$. And Theorem 1.1 is turned to

######  Theorem 1.2.

Let $u\in C(\overline{\mathbb{R}_{+}^{n}})$ be a continuous, convex viscosity
solution to problem (1.3). Suppose that either $n=2$, or when $n\geq 3$, $u$
satisfies (1.4) or $\beta=0$, then $u$ is a quadratic polynomial.

Our start point is to consider the level set at boundary point. First, we need
to prove that $\nabla u(0)$ is well defined for a viscosity solution. Then the
main task is to study the geometry of the height section

(1.5) $S_{h}(0):=\\{x\in\overline{\mathbb{R}_{+}^{n}}|\,u(x)<u(0)+\nabla
u(0)\cdot x+h\\}$

by blowing down $u$ at the infinite point of auxiliary subplane and then
blowing up at origin. As a result, we will finally prove that $D_{n}u(x)$ is a
linear function. The case $n=2$ or $a=0$ is much more easy, which will be
proved in Remark 2.16 and Remark 8.2 respectively.

This paper is arranged as follows. In section 2, we will show the basic
properties for the viscosity solutions of problem (1.1) and list a few well-
known Lemmas to be used. In section 3, we will prove the strict convexity of
solution to (1.1) away from the boundary and then introduce the linearized
problem for $D_{n}u(x)$ to obtain a boundary estimate of the normal
derivatives of the solutions. In section 4, we will study a general global
strict convexity lemma and provide an example showing that the growth
assumption for the strict convexity is optimal. In section 5, we will prove
the height section of solution to (1.1) has a good shape. In section 6, we
will use the result for case $a=0$ to prove a stability result for small $a$.
In section 7, we will prove the global uniformly convexity of solution to
(1.1). In the last section, we will improve the famous $C^{1,\alpha}$ estimate
techniques[3, 5], so as to prove that the blow down-up limit has certain
$C^{2}$ regularity on a cone with vertex $0$, which implies that $u$ is a
quadratic polynomial.

##  2\. Preliminaries

We start at notations and definitions, then list some basic results. They will
be used throughout this paper.

A point in $\mathbb{R}^{n}$ is written as

$x=(x_{1},\cdots,x_{n-1},x_{n})=(x^{\prime},x_{n})$

and as $x=(x_{1},x^{\prime\prime},x_{n})$ if $n\geq 3$.

Let $\mathbb{R}_{+}^{n}=\\{x\in\mathbb{R}^{n}|\,x_{n}>0\\}$ be the upper half-
space. Then $\overline{\mathbb{R}_{+}^{n}}=\\{x\in\mathbb{R}^{n}|\,x_{n}\geq
0\\}$. We will take the convention

$R^{n-1}=\partial\mathbb{R}_{+}^{n}=\\{x\in\mathbb{R}^{n}|\,x_{n}=0\\},\ \
R^{n-2}=\\{x\in\mathbb{R}^{n}|\,x_{1}=x_{n}=0\\}$

from time to time.

Denote by $\mathcal{I}$ (or $\mathcal{I}^{\prime}$,
$\mathcal{I}^{\prime\prime}$) the identity matrices of size $n$ (or $n-1$,
$n-2$), $B_{r}(x)$ (or $B_{r}^{\prime}(x^{\prime})$,
$B_{r}^{\prime\prime}(x^{\prime\prime})$) the ball of radius $r$ centered at
$x$ (or $x^{\prime}$, $x^{\prime\prime}$) in $\mathbb{R}^{n}$ (or
$\mathbb{R}^{n-1}$, $\mathbb{R}^{n-2}$ respectively), $B_{r}^{+}(x)=\\{y\in
B_{r}(x):\;y_{n}>0\\}$ and $\mathcal{P}x=x^{\prime}$ is the projection mapping
onto the $\mathbb{R}^{n-1}$.

The oscillation function of a function $W$ over a domain $E$ is denoted by

$Osc_{E}W=\sup_{y,z\in E}|W(y)-W(z)|.$

Suppose that $u$ satisfies the assumption of Theorem 1.1. Since $u$ is
continuous up to boundary, we have

(2.1) $S_{1}(0)\supset B_{\kappa_{0}}^{+}(0)$

for some $\kappa_{0}>0$, where we have used notation (1.5) (Also see (2.4)
below).

When $n\geq 3$, the quadratic growth condition (1.2) will be replaced by

(2.2) $u(0,x^{\prime\prime},0)<\frac{K}{2}(|x^{\prime\prime}|^{2}+1),\forall
x^{\prime\prime}\in\mathbb{R}^{n-2}$

for a constant $K>0$. We will use $\sigma_{i}:[0,\infty)\to[0,\infty)$,
$(i=0,1,2)$, to denote some universal strictly increasing $C^{0}$ function
with $\sigma_{i}(0)=0$ that depends only on $(n,a,\kappa_{0},K)$.

$\lambda$ and $\Lambda$ will denote known positive constants. We always use
the convention that $c$ or $C$ denotes various positive constants depending
only on $n,a$,$\kappa_{0},K$, $\lambda$ and $\Lambda$,  which are often called
as universal constants.

If not specified, the $\operatorname{det}D^{2}u$ will be understood as the
general Monge-Ampère measure of convex functions[11, 12], which yields the
(Alexandrov) generalized solution to the equation
$\operatorname{det}D^{2}u=f$, and it is equivalent to the viscosity solution
in the domain where $f$ is continuous. If $u$ is a viscosity subsolution (or
supersolution) of the equation $\operatorname{det}D^{2}u=f$, we say that $u$
is a viscosity solution to the inequalities
$\operatorname{det}D^{2}u\geq(or\leq)f$, or $u$ solves these inequalities in
viscosity senses.

The sub-differential of $u$ at point $x_{0}$ is defined by

$\partial u(x_{0}):=\\{p|\,L:=u(x_{0})+p\cdot(x-x_{0})\text{ is a support
plane of }u\text{ at }x_{0}\\}.$

We define $\nabla u(x_{0})$ as a fixed element of $\partial u(x_{0})$ for
interior point. When $n=2$ or when $n\geq 3$ and the assumption (1.2) is
satisfied, we will show the viscosity solution to (1.1), $u$, is strictly
convex in $\mathbb{R}_{+}^{n}$. Then the classical interior estimate [3, 5]
means that $u\in C_{loc}^{\infty}(\mathbb{R}_{+}^{n})$ and $\nabla u$ is
uniquely determined in $\mathbb{R}_{+}^{n}$.

Take a point $y=(y^{\prime},0)\in\partial\mathbb{R}_{+}^{n}$. If $u$ is
convex, viscosity solution to problem(1.1), by the definition we have

(2.3) $\sup_{a\in\partial u(y)}\\{a\cdot e_{n}\\}\leq ay_{1},$

and by Lemma 3.2 in Section 3, one may define

$D_{n}u(y)=\varlimsup_{t\to 0^{+}}D_{n}u(y+te_{n}):=\lim_{t\to
0^{+}}\sup\\{p\cdot e_{n}:\;p\in\partial u(y+te_{n})\\}.$

And in Corollary 3.3, we will prove that there exists $p_{y}\in\partial u(y)$
such that

$p\cdot e_{n}=ay_{1}=D_{n}u(y).$

Hence, we define $\nabla u(y)$ as a fixed $p_{y}$ for any boundary point
$y\in\partial\mathbb{R}_{+}^{n}$.

In this way, the section of $u$ centered at
$y\in\overline{\mathbb{R}_{+}^{n}}$ with height $h>0$ can be defined by

(2.4)
$S_{h}^{u}(y):=\left\\{x\in\overline{\mathbb{R}_{+}^{n}}|\,u(x)<u(y)+\nabla
u(y)\cdot(x-y)+h\right\\},$

and its Neumann boundary  is defined by

(2.5) $G_{h}^{u}(y):=S_{h}^{u}(y)\cap\partial\mathbb{R}_{+}^{n}.$

Sometimes, the notations $S_{h}^{u}(0)$, $G_{h}^{u}(0)$ , $B_{r}(0)$ and
$B_{r}^{+}(0)$ will be shorten as $S_{h}$, $G_{h}$, $B_{r}$ and
$B_{r}^{+}$respectively.

When we are going to prove Theorem 1.1, we may assume

(2.6) $a\geq 0,\ \ u(0)=0,\ \ \nabla u(0)=0\;\text{ and }u\geq 0\ \ \text{in
}\overline{\mathbb{R}_{+}^{n}}$

by considering the function $u(x,-x_{n})$ and subtracting the linear function
$u(0)+\nabla u(0)\cdot x$.

######  Definition 2.1 (Good Shape).

We say that a section $S_{h}^{u}(0)$ is of good shape if it satisfies the
following two property:

$S_{h}^{u}(0)$ has finite density at $0$, i.e.,

$c\leq\frac{\operatorname{Vol}(S_{h}^{u}({0}))}{h^{\frac{n}{2}}}\leq C,$

and

$c\mathcal{P}S_{h}^{u}({0})\subset\mathcal{P}G_{h}^{u}(0)\cap(-\mathcal{P}G_{h}^{u}(0)).$

When blowing up at infinity, we will need to modify the class $\Theta$ to
$\Theta(\delta)$.

######  Definition 2.2 ($\Theta(\delta)$ Class).

Suppose that $n\geq 3$, $b\in R$ and $u$ satisfies

$D_{n}u(x^{\prime},0)=bx_{1},\forall x^{\prime}\in\mathbb{R}^{n-1}$

and

$u(0,z^{\prime\prime},0)\leq z^{\prime\prime
T}\mathcal{Q}z^{\prime\prime}+\delta,\forall
x^{\prime\prime}\in\mathbb{R}^{n-2}$

for some $(n-2)\times(n-2)$ symmetric matrix $\mathcal{Q}$ and $\delta>0$. We
say $u\in\Theta(\delta)$ if the pair $(b,\mathcal{Q})$ satisfies

$0\leq b\leq c\ \ \text{ and }\ \ b^{2}\operatorname{det}\mathcal{Q}\leq C.$

######  Definition 2.3.

We say that a closed set $E\in\mathbb{R}^{n}$ is balance about $x$ up to a
constant $\kappa$, if

$t(x-E)\subset E-x\;\text{ for }\;t\leq c\kappa.$

When $\kappa$ is universal, we say that $E\in\mathbb{R}^{n}$ is balance about
$x$.

The following lemmas will be used in our article.

Use $\operatorname{Vol}_{k}(E)$ to denote the $k-$dimensional Lebesgue measure
and denote $\operatorname{Vol}(E)=\operatorname{Vol}_{n}(E)$.

######  Lemma 2.4.

Given any convex set $E\in\mathbb{R}^{n}$. Suppose the line $l(t):=x+te_{n}$
intersects $E$ at the points $p,q$. Then

(2.7) $|p_{n}-q_{n}|\cdot\operatorname{Vol}_{n-1}(\mathcal{P}E)\leq
C\operatorname{Vol}(E).$

In addition, if $\mathcal{P}E$ is balance about $\mathcal{P}q$ up to a
constant $\kappa$, then we have the reverse inequality

(2.8) $|p_{n}-q_{n}|\cdot\operatorname{Vol}_{n-1}(\mathcal{P}E)\geq
c\kappa\operatorname{Vol}(E).$

###### Proof.

Consider the non-negative concave function defined on $\mathcal{P}E$

$L(x^{\prime})=\sup\\{(y-z)\cdot e_{n}|\,y,z\in E,\ \text{ such that
}\mathcal{P}y=\mathcal{P}z=x^{\prime}\\}.$

Then

$\operatorname{Vol}(E)\sim||L||_{L^{\infty}}\operatorname{Vol}_{n-1}(\mathcal{P}E),$

which yields (2.7). Now, suppose that $\mathcal{P}E$ is balance about
$\mathcal{P}q$ up to the constant $\kappa$. By concavity $L(\mathcal{P}q)\geq
c\kappa||L||_{L^{\infty}}$, and (2.8) follows. ∎

######  Lemma 2.5.

Let $u\in C(\overline{B_{1}^{+}(0)})$ be a convex function. Suppose that

$0\leq u\leq\sigma_{1}(|x|)\text{ in }B_{c}^{+}(0)$

and

$\varlimsup_{t\to
0^{+}}D_{n}u(x^{\prime},t)\geq-\sigma_{0}(|x^{\prime}|),\forall(x^{\prime},0)\in\overline{B_{c}^{+}(0)}.$

Then

(2.9) $|\nabla u(y)|\leq C\left(\inf_{|y|\leq t\leq
c}\frac{2\sigma_{1}(1+t)}{t}+\sigma_{0}(|y^{\prime}|)\right)\text{ in
}B_{c}^{+}(0).$

Here we do not require $\sigma_{i}(0)=0\;(i=0,1).$

###### Proof.

Given $y\in B_{c}^{+}(0)$, $c\leq\frac{1}{2}$, let $p\in\partial u(y)$. The
assumption and the convexity implies that

$p_{n}\geq-\sigma_{0}(|y^{\prime}|).$

Suppose that $|p^{\prime}|>0$. Note that for every constant $a\in[0,1]$, the
point $z=y+a|y|(\frac{p^{\prime}}{|p^{\prime}|},1)$ is in $B_{1}^{+}(0)$.
Hence, by the convexity we have

$a|y|(p_{n}+|p^{\prime}|)=p\cdot(z-y)\leq u(z)-u(y)\leq
2\sigma_{1}((1+a)|y|).$

Letting $t=a|y|$, we obtain the desired (2.9). We may ignore the estimation
for $y\in\partial\Omega$, because $\nabla u(y)$ in this paper is determined by
taking the internal limit. (See the above definition for $D_{n}u(y)$ and
Corollary 3.3). ∎

Similar to Lemma 2.5, we can prove

######  Lemma 2.6.

Write $x\in\mathbb{R}^{n}$ as $x=(x^{1},\cdots,x^{k})$,
$x^{j}\in\mathbb{R}^{a_{j}}$, $\sum_{j=1}^{k}a_{j}=n$. Let $u\in
C(\bar{B}_{r}(0))$ be a convex function, satisfying

$u(x)\leq u(0)+\sum_{i=1}^{k}\sigma_{i}(|x^{i}|).$

Then

$|\nabla_{x_{j}}u(0)|\leq C\inf_{0\leq t\leq cr}\frac{2\sigma_{j}(Ct)}{t}.$

######  Corollary 2.7.

Let $\epsilon>0$ be small constant, $u\in C(\overline{B_{1}^{+}(0)})$ and
$v\in C^{2}(\overline{B_{1}^{+}(0)})$ be two convex functions. Suppose that

$||u-v||_{L^{\infty}}\leq\epsilon$

and

$\varlimsup_{t\to 0^{+}}D_{n}u(x^{\prime},t)\geq
D_{n}v(x^{\prime},0)-\epsilon^{\frac{1}{2}},\forall(x,0)\overline{B_{1}^{+}(0)}.$

Then

(2.10) $|\nabla u(y)-\nabla v(y)|\leq C_{v}\epsilon^{\frac{1}{2}}\text{ in
}B_{c}^{+}(0),$

where $C_{v}$ is aconstant which may depends on $||v||_{C^{2}}$.

###### Proof.

Given point $y\in B_{c}^{+}(0)$. Consider the function

$U^{y}(x)=u(x)+2\epsilon-v(y)-\nabla v(y)\cdot(x-y).$

Then by the assumption we have

$0\leq U^{y}(x)\leq 2\epsilon+v(x)+2\epsilon-v(y)-\nabla v(y)\cdot(x-y)\leq
C_{v}|x-y|^{2}+4\epsilon,$

and

$\varlimsup_{t\to 0^{+}}D_{n}U^{y}(x^{\prime},t)\geq
D_{n}v(x^{\prime},0)-\epsilon^{\frac{1}{2}}-D_{n}v(y^{\prime},0)\geq-
C_{v}|x^{\prime}-y^{\prime}|-\epsilon^{\frac{1}{2}}.$

Regarding $y$ as origin and applying Lemmas 2.5 and 2.6 to $U^{y}(x)$, we get
the desired results. ∎

The following is well-known John’s Lemma. See [11, 12, 13] for example.

######  Lemma 2.8 (Jhon’s Lemma).

If $\Omega$ is a bounded convex set with non-empty interior in
$\mathbb{R}^{n}$ and $E$ is the ellipsoid of smallest volume containing
$\Omega,$ then after an affine transformation $T,$

$B_{c(n)}\subset T(\Omega)\subset B_{C(n)}:=T(E).$

######  Lemma 2.9.

If $\Omega$ is a bounded convex set in $\mathbb{R}^{n}$ and $u,v\in
C(\overline{\Omega})$ are convex functions in $\Omega$ satisfying

$\begin{cases}\operatorname{det}D^{2}v\leq\operatorname{det}D^{2}u&\text{ in
}\Omega\\\ u\leq v&\text{ on }\partial\Omega\\\ \end{cases},$

then $u\leq v$ on $\overline{\Omega}$.

###### Proof.

See Theorem 2.10 in [11]. ∎

######  Lemma 2.10 ( Comparison principle for mixed problem).

Supposed that $\Omega$ is a bounded convex set in $\mathbb{R}^{n}$,
$G_{1}\subset\partial\Omega$ and $G_{2}=\partial\Omega\backslash G_{1}$ are
piecewise $C^{1}$ smooth, and $\beta$ is inward vector fieled on
$\partial\Omega_{2}$. If $u,v$ are continuous, convex function defined on
$\overline{\Omega}$ such that either $u$ or $v$ is $C^{2}$ smooth, satisfying

(2.11)
$\begin{cases}\operatorname{det}D^{2}v<\operatorname{det}D^{2}u<\infty&\text{
in }\Omega\\\ u<v&\text{ on }G_{1}\\\ D_{\beta}v<D_{\beta}u&\text{ on
}G_{2}\end{cases}$

in the viscosity sense, then $u<v$ on $\overline{\Omega}$.

###### Proof.

On the contrary to the conclusion, we suppose that $w=u-v$ archives positive
maximum value $m$ at a point $x_{0}\in\overline{\Omega}$. According to the
assumption and the first two inequalities in (2.11) we can take
$x_{0}\in\partial\Omega_{2}$ such that $u(x_{0})-v(x_{0})=m$. This means that
$u-[v(x_{0})+\nabla v(x_{0})\cdot(x-x_{0})+m]$ archives the maximum at $x_{0}$
when $v$ is $C^{2}$ smooth, or $v-[u(x_{0})+\nabla u(x_{0})\cdot(x-x_{0})-m]$
archives the minimum at $x_{0}$ when $u$ is $C^{2}$ smooth, which contradicts
the third inequality in (2.6) by the definition of viscosity solution. ∎

######  Lemma 2.11.

If $\Omega$ is a bounded convex domain in $\mathbb{R}^{n}$ and $u$ is a convex
function satisfying

$\operatorname{det}D^{2}u>\lambda\text{ in }\Omega,$

then

(2.12) $\operatorname{Vol}(\Omega)\leq
C(\lambda)||u||_{L^{\infty}}^{\frac{n}{2}}.$

In particular,

(2.13) $\operatorname{Vol}(S_{h})\leq Ch^{\frac{n}{2}}.$

###### Proof.

(2.13) was proved in Corollary 3.2.5 in [12], and (2.12) can be proved as its
proof. ∎

######  Lemma 2.12.

Suppose that $\Omega$ is a convex domain in $\mathbb{R}^{n}$ such that
$B_{c}(0)\subset\Omega\subset B_{C}(0)$ and $u\in C(\overline{\Omega})$ is a
convex function. If

$\lambda\leq\operatorname{det}D^{2}u\leq\Lambda\text{ in }\Omega,\ u=0\text{
on }\partial\Omega,$

then

$u(x)\geq-C\operatorname{dist}(x,\partial\Omega)^{\frac{1}{n}}.$

###### Proof.

This is the Aleksandrove’s maximum principle. See Theorem 1.4.2 in [12]. ∎

######  Lemma 2.13.

Suppose that $B_{c}(0)\subset\Omega\subset B_{C}(0)$ and $u$ is a convex
function defined on convex domain $\bar{\Omega}$ such that

$\lambda\leq\operatorname{det}D^{2}u=f(x)\leq\Lambda\text{ in }\Omega,\
u=0\text{ on }\partial\Omega.$

If $||f||_{C^{k,\alpha}}(\Omega)\leq C$ for some $k\geq 0$ and
$\alpha\in(0,1)$, then

$||u||_{C^{k+2,\alpha}(\frac{1}{2}\Omega)}\leq C.$

###### Proof.

When $k=0$, this is the famous interior $C^{2,\alpha}$ estimate in [4]. Also
see [16]. Then the case of $k\geq 1$ follows from the standard regularity
result for linear elliptic equations. ∎

######  Lemma 2.14.

Let $u\in C(B_{1}(0))$ be a convex solution to

$0<\lambda\leq\operatorname{det}D^{2}u\leq\Lambda.$

For $x\in B_{1}(0)$ and $p\in\partial u(x)$, define $\Sigma:=\\{y\in
B_{1}(0)|\,u(y)=u(x)+p\cdot(y-x)\\}.$ Then, either $\Sigma=\\{x\\}$ or
$\Sigma$ has no extremal point inside $B_{1}(0)$.

###### Proof.

See [3]. ∎

The following result is already well-known. See[14, 8, 21, 22, 9] or the
books[11, 12].

######  Theorem 2.15.

Let $u\in C({\mathbb{R}^{n}})$ be a continuous, convex solution. If
$\operatorname{det}D^{2}u=1$ in $\mathbb{R}^{n}$, then $u$ is a quadratic
polynomial.

######  Remark 2.16.

The case $a=0$ of Theorem 1.1 can be obtained directly from Theorem 2.15. In
fact, Lemma 3.2 in next section allow us to reflect $u$, the solution to
problem (1.1), with respect to $\partial\mathbb{R}_{+}^{n}$. Then we get a
convex function $\bar{u}$. Corollary 3.3 also means that
$D_{n}\bar{u}(\partial\mathbb{R}_{+}^{n})\subset\partial\mathbb{R}_{+}^{n}.$
Thus $\bar{u}$ solves the equation

$\operatorname{det}D^{2}\bar{u}=1,\text{ in }\;\mathbb{R}^{n}.$

Hence $\bar{u}$ is a quadratic polynomial by Theorem 2.15, so is $u$.

##  3\. Interior strict convexity and normal derivative estimate

In this section, we first prove that a viscosity solution to problem (1.1) is
interior strict convexity, then give its normal derivative estimates.

######  Lemma 3.1.

Suppose $u$ is a viscosity solution to problem (1.1). Then the graph of $u$
can’t contains any ray, and $u$ is strictly convex away from the boundary
$\partial\mathbb{R}_{+}^{n}$.

###### Proof.

If the graph of $u$ contains a ray $l$ and take $y\in
l^{\prime}=\mathcal{P}l$, the function

$u^{y}(x)=u(x)-u(y)-\nabla u(y)\cdot(y-x)$

is bounded in the tubular domain
$E:=(l^{\prime}-y+B_{1}(y))\cap\overline{\mathbb{R}_{+}^{n}}$ and

$\operatorname{det}D^{2}u^{y}=1\text{ in }E,$

which contradicts Lemma 2.11 since $E$ has infinite volume.

Suppose $y\in\mathbb{R}_{+}^{n}$. Let

(3.1) $h_{y}^{u}:=\sup\\{h|\,S_{t}^{u}(y)\subset\mathbb{R}_{+}^{n},\ \forall
t<h\\}.$

Since $u$ is strictly convex in the interior of sections[3], it is enough for
the strict convexity of $u$ away from the boundary
$\partial\mathbb{R}_{+}^{n}$ to show $h_{y}^{u}>0$ for
$y\in\mathbb{R}_{+}^{n}$.

Otherwise, if $h_{y}=0$, the convex set

$\Gamma:=\\{x\in\mathbb{R}_{+}^{n}|\,u(x)=u(y)+\nabla u(y)\cdot(x-y)\\}$

should contain more than one point. Lemma 2.14 implies that the extremal point
of $\Gamma$ is not in interior. There are two cases: either $\Gamma$ contains
a line $l$ parallel to $\mathbb{R}^{n-1}$, or $\Gamma$ contains a ray $l$
starting from point $z=(z^{\prime},0)$ on $\partial\mathbb{R}_{+}^{n}$. Both
cases contradicts the previous fact that the graph of $u$ can’t contains any
ray. ∎

######  Lemma 3.2.

Suppose $u$ is a viscosity solution to problem (1.1) and
$y\in\partial\mathbb{R}_{+}^{n}$. Then $D_{n}u(y):=\overline{\lim_{t\to
0^{+}}}D_{n}u(y^{\prime},t)$ is well-defined. Moreover,

(3.2) $D_{n}u(y)=ay_{1}$

and

(3.3) $u(y+te_{n})\geq u(y)+tay_{1},\ \ \forall t>0.$

###### Proof.

Without loss of generality we may assume $y=0$. By Lemma 3.1 and the convexity
of $u$, $D_{n}(te_{n})$ are well-definend and continuous for $t>0$. Hence,

$D_{n}u(0):=\overline{\lim_{t\to 0^{+}}}D_{n}u(te_{n})=\lim_{t\to
0^{+}}D_{n}u(te_{n})$

is well-defined. Notice that $ay_{1}|_{y=0}=0$. Were if (3.2) false, then

$\lim_{t\to 0^{+}}D_{n}u(te_{n})<-2\epsilon$

for some $\epsilon>0$. Consider the tubular domain

$\Gamma=\\{(x^{\prime},0)+st_{0}e_{n}|\,x^{\prime}\in B^{\prime}_{r}(0),\
s\in(0,1)\\}$

and choose the function

$v(x)=u(0)+\frac{2\operatorname{Osc}_{B_{r}^{+}(0)}u}{r^{2}}|x^{\prime}|^{2}-\epsilon
x_{n},$

where $t_{0}$ and $r=r(\epsilon,t_{0},u)$ are positive and so small that

$D_{n}u(t_{0}e_{n})<-2\epsilon t_{0},\ \
2\operatorname{Osc}_{B_{r}^{+}(t_{0}e_{n})}u\leq\epsilon t_{0}\text{ and
}|a|r<\epsilon,$

which implies

$\sup_{y\in B_{r}(t_{0}e_{n})\cup\partial B_{r}^{+}(0)}(u(y)-v(y))<0.$

Notice that $v$ is linear along the $e_{n}$ direction and $u$ is convex, we
find that $v$ is strictly greater than $u$ on the boundary
$\partial\Gamma_{1}:=\overline{\partial\Gamma\setminus\mathbb{R}^{n-1}}$.
Then, $v(0)=u(0)$ means that $v+c_{0}$ will touch $u$ by above at some point
$z\in\overline{\Gamma}\setminus\partial\Gamma_{1}$ for a proper $c_{0}\in R$.
Moreover, $\operatorname{det}D^{2}v=0<1=\operatorname{det}D^{2}u$ implies
$z\notin\Gamma$. Therefore, $z\in\partial\Gamma\cap R^{n-1}$. However,

$D_{n}v(z)=-\epsilon<-|a|r\leq az_{1}=D_{n}u(z),$

which contradicts the definition of viscosity supersolution. Thus, (3.2) is
proved and (3.3) follows by a direct integral. ∎

######  Corollary 3.3.

If $u$ is a viscosity solution to problem (1.1) and
$y\in\partial\mathbb{R}_{+}^{n}$, then there exists an element $p\in\partial
u(y)$ such that $p\cdot e_{n}=D_{n}u(y)$.

###### Proof.

Without loss of generality we may assume $y=0$. Let $p_{t}=\nabla u(te_{n})$
and $l_{t}(x):=u(te_{n})+p_{t}\cdot(x-te_{n})$ be the support function of $u$
at $te_{n}$. Lemmas 3.2 and 2.5 implies that $p_{t}$ is uniformly bounded as
$t\to 0^{+}$. Then, as $t\to 0^{+}$, $\\{l_{t}\\}$ is pre-compact and contains
a subsequence that converges to some support function $l(x):=u(0)+p\cdot x$ of
$u$ at $0$. Hence $p\in\partial u(0)$ and $D_{n}u(te_{n})=p_{t}\cdot e_{n}$
monotonically decreases to $p\cdot e_{n}$. Furthermore, by (2.3) and (3.2) we
see that $p\cdot e_{n}=0=D_{n}u(0)$. ∎

######  Lemma 3.4.

Suppose $u$ is a viscosity solution to problem (1.1). Then $D_{n}u\in
C(\overline{\mathbb{R}_{+}^{n}})$ and $v(x):=D_{n}u(x)-D_{n}u(x^{\prime},0)$
is a viscosity solution to following linearized problem:

(3.4) $\begin{cases}U^{ij}D_{ij}v=0&\text{ in }{\mathbb{R}_{+}^{n}}\\\ v\geq
0&\text{ on }\overline{\mathbb{R}_{+}^{n}}\\\ v=0&\text{ on
}{\partial\mathbb{R}_{+}^{n}}\end{cases},$

where $U^{ij}$ is the cofactor of the Hessian $D^{2}u$.

###### Proof.

First, Lemmas $3.1$ and 2.13 means that $u\in
C_{loc}^{\infty}({\mathbb{R}_{+}^{n}})$. Then it follows from the proof of
Corollary 3.3 that the family of continuous function
$f_{t}(x^{\prime})=D_{n}u(x^{\prime},t)$ point-wisely decreases to function
$f_{0}(x^{\prime})=D_{n}u(x^{\prime},0)=ax_{1}$ monotonically as $t\to 0^{+}$.
Finally, the finite cover Lemma implies that this convergence is uniformly.
Therefore $D_{n}u$ is continuous up to boundary.

The linearized equation $U^{ij}D_{ij}v=0$ comes directly from the equation
$\operatorname{det}D^{2}u=1$. The other two in (3.4) follows from the
convexity of $u$ and the continuity of $D_{n}u$ at the boundary. ∎

Next, we introduce two functions, $w^{y,+}(x)$ and $w^{y,-}(x)$, which will be
used as the upper/below barriers of $v$ at $y$ to the linearized problem
(3.4).

Given $y\in G_{\frac{1}{2}}^{u}(0)$ and consider the functions

(3.5) $u^{y}(x)=u(x)-u(y)-\nabla u(y)\cdot(y-x)$

and

(3.6) $w^{y}(x)=u^{y}(x)-\frac{n}{2}x_{n}D_{n}u^{y}.$

Define

(3.7) $\begin{cases}w^{y,+}(x)=C_{1}[w^{y}(x)+C_{2}x_{n}]\\\
w^{y,-}(x)=c_{1}[x_{n}-c_{2}w^{y}(x)]\end{cases}.$

Since $\operatorname{det}D^{2}u=1$, we have $U^{ij}D_{ij}D_{n}u=0$ in
$\mathbb{R}_{+}^{n}$. Hence

(3.8)
$U^{ij}D_{ij}w^{y}=U^{ij}D_{ij}w^{0}=n\operatorname{det}D^{2}u-\frac{n}{2}(U^{in}D_{in}u+U^{ni}D_{ni}u)=0\;\text{in}\;\mathbb{R}_{+}^{n}.$

######  Lemma 3.5.

Suppose that $u$ is a viscosity solution to problem (1.1) and (2.6) is
satisfied. If

$B_{c}^{+}(0)\subset S_{1}(0)\subset B_{C}^{+}(0)$

and

(3.9) $||D_{n}u||_{L^{\infty}(S_{1}(0))}\leq C,$

then, $w^{0,+}(x)$ defined in (3.7) is the upper barrier of $v$ at $0$ to the
linearized problem (3.4) in $S_{1}(0)$. (See (3.11) below). This implies

(3.10) $D_{n}u(te_{n})\leq Ct\ \ \text{and}\ \ u(te_{n})\leq Ct^{2}$

for all $t\geq 0$ such that $te_{n}\in S_{1}(0)$.

###### Proof.

It follows from (3.9) that

$v(x)=D_{n}u(x)-D_{n}u(x^{\prime},0)\leq C,\ \ \forall x\in S_{1}(0)$

and

$u(x)-\frac{n}{2}x_{n}D_{n}u\geq 1-Cx_{n},\ \ \forall x\in\partial
S_{1}(0)\setminus G_{1}(0).$

In (3.7) we choose $C_{1}$ and $C_{2}$ such that they are larger than the $C$.
Then, $w^{0,+}(x)\geq C_{1}$ on $\partial S_{1}(0)\setminus G_{1}(0)$ by
(3.6). This, together with (3.5)-(3.8), we see that

(3.11) $\begin{cases}U^{ij}D_{ij}w^{0,+}=0&\text{ in }S_{1}(0)\\\ w^{0,+}\geq
v&\text{ on }\partial S_{1}(0)\setminus G_{1}(0)\\\ w^{0,+}\geq 0=v&\text{ on
}{G_{1}(0)}\end{cases}.$

Hence, the comparison principle implies that $v(x)\leq w^{0,+}(x)$ in
$S_{1}(0)$, which, together with the convexity, yields the desired (3.10). ∎

######  Lemma 3.6.

Under the assumption of Lemma 3.5, there exists a small $\bar{\sigma}>0$
depending on $\sigma_{0}$ such that if

(3.12) $v(x)\geq\sigma_{0}(x_{n})-\bar{\sigma}\text{ in }S_{1}(0),$

then $w^{0,-}(x)$ defined in (3.7) is the lower barrier of $v$ at $0$ to the
linearized problem (3.4) in $S_{1}(0)$ (see (3.14) below). This implies that

(3.13) $v(te_{n})\geq ct\ \ and\ \ u(te_{n})\geq ct^{2}$

for all $t\geq 0$ such that $te_{n}\in S_{1}(0)$.

###### Proof.

Denote $K=Cn||D_{n}u||_{L^{\infty}(S_{1}(0))}$. Assume that $\bar{\sigma}>0$
is small and satisfies

$\sigma_{0}^{-1}(2\bar{\sigma})\leq\frac{1}{2K}.$

Recall that $\sigma_{0}$ denotes a strictly increasing and continuous function
from $[0,+\infty)$ to $[0,+\infty)$ with $\sigma_{0}(0)=0$. Let
$c_{0}=\sigma_{0}^{-1}(2\bar{\sigma})$. By (3.12) and the convexity we have

$v(x)\geq\max\\{\sigma_{0}(x_{n})-\bar{\sigma},0\\}\geq\bar{\sigma}\chi_{\\{x_{n}\geq
c_{0}\\}},$

where $\chi_{E}$ denotes the characteristic function of the set $E$, i.e.,
$\chi_{E}(x)=1$ if $x\in E$, otherwise $\chi_{E}(x)=0$. Take
$c_{1}=\frac{\bar{\sigma}}{2c}$ and $c_{2}=2c_{0}$. We have

$w^{0}(x)\geq 1-Kx_{n},\ \ on\ \ \partial S_{1}(0)\setminus G_{1}(0).$

Hence on $\partial S_{1}(0)\setminus G_{1}(0)$, we see that

$\displaystyle w^{0,-}(x)$ $\displaystyle\leq c_{1}[x_{n}+c_{2}(Kx_{n}-1)]$
$\displaystyle\leq c_{1}[(2x_{n}-c_{2})\chi_{\\{x_{n}\leq
c_{0}\\}}+2x_{n}\chi_{\\{x_{n}\geq c_{0}\\}}]$ $\displaystyle\leq
2c_{1}C\chi_{\\{x_{n}\geq c_{0}\\}}$
$\displaystyle=\bar{\sigma}\chi_{\\{x_{n}\geq c_{0}\\}}\leq v(x).$

In conclusion, we have

(3.14) $\begin{cases}U^{ij}D_{ij}w^{0,-}=0&\text{ in }S_{1}(0)\\\ w^{0,-}\leq
v&\text{ on }\partial S_{1}(0)\setminus G_{1}(0)\\\ w^{0,-}\leq 0=v&\text{ on
}{G_{1}(0)}\end{cases}.$

The comparison principle implies that $w^{0,-}(x)\leq v(x)$ in $S_{1}(0)$. In
particular,

$v(te_{n})\geq w^{0,-}(te_{n})\geq c_{1}(t-Cc_{2}t)\geq\frac{c_{1}}{2}t$

when $c_{2}=2\sigma_{0}^{-1}(2\bar{\sigma})>0$ is small, which is equivalent
to $\bar{\sigma}>0$ is small enough. This proves (3.13). ∎

######  Corollary 3.7.

Under the assumption of Lemma 3.5, there exists a small $\rho>0$ such that if

(3.15)
$\sup_{B_{\rho}^{\prime}(0)}||u(y^{\prime},0)||\leq\sigma_{0}(\rho)\rho,$

then for any $y^{\prime}\in B_{\rho}^{\prime}(0)$ and $y=(y,0)$, the function
$w^{y,+}(x)$ defined in (3.7) is the upper barrier of $v$ at $y$ to the
linearized problem (3.4). This also implies that $u^{y}(y+te_{n})\leq Ct^{2}$
for small $t>0$ if $u(y)=0$ and $\nabla u(y)=0$.

###### Proof.

If $\rho$ is small, the assumption implies that

$\sup_{B_{\rho}^{\prime}(0)}||\nabla
u(y^{\prime},0||\leq\min\\{\sigma_{0}(\rho),C\rho\\}<<\frac{1}{2}.$

Then, the section

$S_{\frac{1}{2}}(y)\subset S_{1-C|\nabla u(y)|}(y)\subset S_{1}(0).$

Regarding $y$ as $0$ in Lemma 3.5, we see that this corollary is a direct
consequence of Lemma 3.5. ∎

Similarly, we have

######  Corollary 3.8.

Under the assumption of Lemma 3.6, there exists a small $\rho>0$ such that if
(3.15) holds, then for any $y^{\prime}\in B_{\rho}^{\prime}(0)$ and $y=(y,0)$,
the function $w^{y,-}(x)$ defined in (3.7) is the lower barrier of $v$ at $y$
to the linearized problem (3.4). This also implies $u^{y}(y+te_{n})\geq
ct^{2}$ for small $t>0$ if $u(y)=0$ and $\nabla u(y)=0$.

##  4\. Strict Convexity Lemma and Examples

Recalling that if $u\in C(\overline{B_{1}^{+}(0)})$ is a convex function and
satisfies

(4.1) $\lambda\leq\operatorname{det}D^{2}u\leq\Lambda\text{ in }B_{1}^{+}(0),$

we can naturally define

$D_{n}u(x^{\prime},0)=\varlimsup_{t\to 0+}D_{n}u(x^{\prime},t),\ \ \forall
x^{\prime}\in B^{\prime}_{1}(0).$

See Lemma 3.2 for the proof. In this section, we first study the strict
convexity for such functions. The we provide an example showing that the
growth assumption for the strict convexity is optimal.

######  Lemma 4.1.

Given positive constants $H,M_{1}$ and $M_{2}$. Suppose that $u\in
C(\overline{B_{1}^{+}(0)})$ is a convex function satisfying (4.1) and

$D_{n}u(x^{\prime},0)\geq-M_{2}\text{ in }B^{\prime}_{1}(0).$

There exists $\sigma=\sigma(H,M_{1},M_{2},\lambda,\Lambda,n)>0$ such that if
$u$ satisfies

$0\leq u(x_{1},x^{\prime\prime},0)\leq\sigma+M_{1}|x^{\prime\prime}|^{2}\text{
in }B_{1}^{\prime}(0),$

then

$\sup_{B_{1}^{+}(0)}u\geq H.$

###### Proof.

Assume by way of contradiction that there exists $\sigma_{k}\to 0^{+}$ and
convex functions $u_{k}$ defined on $B_{1}^{+}(0)\subset\mathbb{R}^{n}$,
satisfying

$0<\lambda\leq\operatorname{det}D^{2}u_{k}\leq\Lambda\text{ in }B_{1}^{+}(0),\
D_{n}u_{k}(x^{\prime},0)\geq-M_{2}\text{ in }B^{\prime}_{1}(0),$ $0\leq
u_{k}(x_{1},x^{\prime\prime},0)\leq\sigma_{k}+M_{1}|x^{\prime\prime}|^{2}\text{
in }B^{\prime}_{1}(0),$

and

$\sup_{B_{1}^{+}(0)}u_{k}\leq H.$

Applying Lemma 2.5 we see that $u_{k}$ is locally Lipschitz with

$||u_{k}||_{Lip(\overline{B_{4c}^{+}(0)})}\leq C(H+M_{2}).$

Hence,

$u_{k}(x)\leq\sigma_{k}+M_{1}|x^{\prime\prime}|^{2}+C(H+M_{2})|x_{n}|\text{ in
}B_{4c}^{+}(0).$

Let $h_{k}=\sigma_{k}^{\frac{1}{2}}$ and
$\mathcal{D}_{k}=\operatorname{diag}(1,h_{k}^{\frac{1}{2}}\mathcal{I}^{\prime\prime},h_{k})$.
Consider the functions

$v_{k}(x)=\frac{u_{k}(\mathcal{D}_{k}x)}{h_{k}},\
x\in\Omega_{k}:=\mathcal{D}_{k}^{-1}B_{4c}^{+}(0).$

Then $0<\lambda\leq\operatorname{det}D^{2}v_{k}\leq\Lambda$ in $\Omega_{k}$,

$v_{k}(x)\leq\sigma_{k}^{\frac{1}{2}}+M_{1}|x^{\prime\prime}|^{2}+C(H+M_{2})|x_{n}|\text{
in }\mathcal{D}_{k}^{-1}B_{1}^{+}(0)$

and

$D_{n}v_{k}(x^{\prime},0)\geq-M_{2}\text{ on
}\mathcal{D}_{k}^{-1}B^{\prime}_{1}(0).$

Similar to Lemma 2.5, for any given $R>0$, we can prove that

$||v_{k}||_{Lip(B_{R}(0)\cap\Omega_{k})}\leq C(H+M_{1}+M_{2})R^{2}.$

Therefore, $v_{k}$ is locally uniformly Lipschitz, and there is a subsequence
that locally uniformly converges to a convex function $v$ defined on
$\Omega:=[-4c,4c]\times\mathbb{R}^{n-1}$. Moreover, we have

(4.2) $0<\lambda\leq\operatorname{det}D^{2}v\leq\Lambda,$

and

(4.3) $v(x)\leq M_{1}|x^{\prime\prime}|^{2}+C(H+M_{2})|x_{n}|\text{ in
}B_{4c}^{+}(0),\ D_{n}v(x^{\prime},0)\geq-M_{2}\text{ on }B_{R}^{\prime}(0).$

Furthermore, (4.3) implies that

(4.4) $a(x_{1})=\lim_{t\to 0^{+}}\frac{v(x_{1},0,t)}{t}$

is well-defined, and

$-M_{2}\leq a(x_{1})\leq C(H+M_{2}).$

Again, we can consider the blow-up of $v$ along axis $e_{1}$ by taking

$\bar{h}_{k}=2^{k},\;\bar{\mathcal{D}}_{k}:=\operatorname{diag}(1,\bar{h}_{k}^{\frac{1}{2}}\mathcal{I}^{\prime\prime},\bar{h}_{k}),\;\bar{v}_{k}(x)=\frac{v(\bar{\mathcal{D}}_{k}x)}{\bar{h}_{k}},\;x\in\bar{\mathcal{D}}_{k}^{-1}\Omega.$

The limit $\bar{v}$ still satisfies (4.2), (4.3). And (4.4) gives

$\bar{v}(x_{1},0,x_{n})=a(x_{1})x_{n}.$

Obviously, $\nabla\bar{v}(e_{n})\cdot e_{n}=a(x_{1})$. The non-negative
function

$w(x)=\bar{v}(x)-\bar{v}(e_{n})-\nabla\bar{v}(e_{n})\cdot x,$

is zero on the line $te_{n}$. This is, $w$ is bounded on a convex domain with
infinite volume. However, $0<\lambda\leq\operatorname{det}D^{2}w\leq\Lambda$,
which contradicts with Lemma 2.11.

∎

######  Lemma 4.2.

Given positive constants $H$ and $M_{1}$. Suppose $u\in
C(\overline{B_{1}(0)})$ is a convex function satisfying

$\lambda\leq\operatorname{det}D^{2}u\leq\Lambda\text{ in }B_{1}(0).$

There exists $\sigma=\sigma(H,M_{1},M_{2},\lambda,\Lambda,n)>0$ such that if
$u$ satisfies

$0\leq u(x_{1},x^{\prime\prime},0)\leq\sigma+M_{1}|x^{\prime\prime}|^{2}\text{
on }B_{1}^{\prime}(0),$

then

$\sup_{B_{1}(0)}u\geq H.$

###### Proof.

Assume by way of contradiction that $\sup_{B_{1}(0)}u<H$. Then, the assumption
of Lemma 4.2 immediately implies that $u\in Lip(B_{c}(0))$ and
$|\nabla^{\prime}u(0)|\leq C(\sigma+\sigma^{\frac{1}{2}})$( See Lemma 2.6).
After subtracting the supporting function $u(0)+\nabla u(0)\cdot x$ at $0$,
$u$ satisfies the assumption of Lemma 4.1 with
$M_{2}=\sigma+\sigma^{\frac{1}{2}}$. Applying Lemma 4.1 we obtain a
contradiction, completing the proof. ∎

We point that the constant $\sigma$ in the growth assumption of Lemma 4.1 and
4.2 can not be allowed to be zero. Otherwise, the constant $H$ may tend to
$\infty$, yielding a contradiction. Moreover, the following example shows that
the growth exponent $2$ is optimal.

######  Example 4.3.

Suppose $n\geq 3$, constants $1<a,b<\infty$ and $\delta>0$ is small such that

$\frac{1}{a}=\frac{1}{2}+\frac{1}{n-2}\delta,\;1-\delta=\frac{1}{b}.$

Let

$E_{1}:=\\{x^{\prime}\in\mathbb{R}^{n-1}:\;|x^{\prime\prime}|^{a}\geq|x_{1}|^{b}\\},\
\
E_{2}:=\\{x^{\prime}\in\mathbb{R}^{n-1}:\;|x_{1}|^{b}\geq|x^{\prime\prime}|^{a}\\}.$

Consider the function

(4.5)
$W_{a,b}(x^{\prime})=\begin{cases}|x^{\prime\prime}|^{a}+|x^{\prime\prime}|^{a-\frac{2a}{b}}|x_{1}|^{2},&x\in
E_{1}\\\
\frac{2b+a-ab}{b}|x_{1}|^{b}+\frac{ab-a}{b}|x_{1}|^{b-\frac{2b}{a}}|x^{\prime\prime}|^{2},&x\in
E_{2}\end{cases}$

and

(4.6) $W(x)=(1+x_{n}^{2})W_{a,b}(x^{\prime}).$

We will show that there is a $\rho=\rho(\delta)>0$ such that $W$ is a convex
function in $R^{n-1}\times(-\rho,\rho))$ satisfying

(4.7) $c(\delta)\leq\operatorname{det}D^{2}W\leq C(\delta),\
D_{n}W(x^{\prime},0)=0\;\text{in}\;R^{n-1}\times(-\rho,\rho).$

###### Proof.

Notice that $W\in C^{1}(\mathbb{R}^{n})\cap
C^{2}(\mathbb{R}^{n}\setminus\\{x^{\prime}=0\\})$, $\nabla
W(\\{x^{\prime}=0\\})\subset\\{x^{\prime}=0\\}$ and the n-dim Hausdroff
measure of the set $\nabla W(\\{x^{\prime}=0\\})$ is zero. The second equation
in (4.7) is obvious. Hence, it is enough to show that $W$ is convex and
satisfies the first inequality in (4.7) in the sense of (Alexandrov)
generalized solution. For this purpose, we need to calculate
$\operatorname{det}D^{2}W$ at any point $x^{\prime}\neq 0$.

It is easy to see that

$D^{2}W=\left[\begin{array}[]{cc}(1+x_{n}^{2})D^{\prime
2}W_{a,b}&2x_{n}D^{\prime}W_{a,b}\\\
2x_{n}D^{\prime}W_{a,b}&2W_{a,b}\end{array}\right].$

So the convexity of $W$ and $\det D^{2}W$ is reduced to the study of the
matrix

$\mathcal{A}=\left[\begin{array}[]{cc}(1+x_{n}^{2})D^{\prime
2}W_{a,b}-2W_{a,b}^{-2}x_{n}^{2}D^{\prime}W_{a,b}\otimes
D^{\prime}W_{a,b}&0\\\ 0&2W_{a,b}\end{array}\right].$

First, we assume $x^{\prime}\in E_{1}$. Then

$|x^{\prime\prime}|^{a}\leq W_{a,b}\leq 2|x^{\prime\prime}|^{a},\ \ \forall
x^{\prime}\in E_{1}.$

Set the matrix functions

$\mathcal{K}_{1}^{\prime}(x^{\prime\prime})=\operatorname{diag}\\{|x^{\prime\prime}|^{a-\frac{2a}{b}},|x^{\prime\prime}|^{a-2}\mathcal{I}^{\prime\prime}\\},\
\
\mathcal{K}_{1}(x^{\prime\prime})=\operatorname{diag}\\{\mathcal{K}_{1}^{\prime},|x^{\prime\prime}|^{a}\\}.$

Then by the assumption on $a$ and $b$ we have

(4.8)
$\operatorname{det}\mathcal{K}_{1}=|x^{\prime\prime}|^{2a(\frac{n}{2}-\frac{1}{b}-\frac{n-2}{a})}\equiv
1.$

Observing that $C(a,b)$ can be written as $C(\delta)$, we have

$0\leq 2W_{a,b}^{-2}D^{\prime}W_{a,b}\otimes D^{\prime}W_{a,b}\leq
C(\delta)\mathcal{K}_{1}^{\prime},\ \ D^{\prime 2}W_{a,b}\leq
C(\delta)\mathcal{K}_{1}^{\prime},\ \ \forall x^{\prime}\in E_{1}.$

Hence, in $E_{1}$ we have

(4.9)
$C(\delta)\mathcal{K}_{1}\geq\mathcal{A}\geq\left[\begin{array}[]{cc}(1+x_{n}^{2})D^{\prime
2}W_{a,b}-x_{n}^{2}C(\delta)\mathcal{K}_{1}^{\prime}&0\\\
0&C(\delta)|x^{\prime\prime}|^{a}\end{array}\right].$

Since $|x_{n}|\leq\rho$ will be small, by (4.8)-(4.9) we can conclude the
convexity of $W$ and the first inequality in (4.7) after we prove

(4.10) $D^{\prime 2}W_{a,b}\geq\epsilon(\delta)\mathcal{K}_{1}^{\prime},\ \
\forall x^{\prime}\in E_{1}.$

Denote

$\mathcal{B}_{1}=\sqrt{\mathcal{K}_{1}^{\prime}},\
X=|x^{\prime\prime}|^{-1}x^{\prime\prime},\
y=|x_{1}||x^{\prime\prime}|^{-\frac{a}{b}}$

and

$m=a[\mathcal{I}^{\prime\prime}+(a-2)X\otimes
X]+(a-\frac{2a}{b})[\mathcal{I}^{\prime\prime}+(a-\frac{2a}{b}-2)X\otimes
X]y.$

By a direct calculation we have

(4.11) $\mathcal{B}_{1}^{-1}D^{\prime
2}W_{a,b}\mathcal{B}_{1}^{-1}=\left[\begin{array}[]{cc}2&2(a-\frac{2a}{b})yX\\\
2(a-\frac{2a}{b})yX^{T}&m\end{array}\right].$

Observe that

(4.12) $m-2(a-\frac{2a}{b})^{2}|y|^{2}X\otimes
X=f_{1}\mathcal{I}^{\prime\prime}-f_{2}X\otimes X$

where, $f_{1}=2a-\frac{2a}{b}$ and

$f_{2}=2(a-\frac{2a}{b})^{2}y^{2}-(a-\frac{2a}{b})(a-\frac{2a}{b}-2)y^{2}-a(a-2).$

Notice that

$\displaystyle 2(a-\frac{2a}{b})^{2}-(a-\frac{2a}{b})(a-\frac{2a}{b}-2)$
$\displaystyle=(a-\frac{2a}{b})(a-\frac{2a}{b}+2)$
$\displaystyle=a(a-\frac{2a}{b})(\frac{1}{n-2}-1)\delta$ $\displaystyle\geq
0.$

Obviously, $|y|\leq 1$ in $E_{1}$. Recall $\delta$ is small,
$a\geq\frac{2a}{b}-a\geq\frac{7}{4}>\frac{3}{2}$, thus we get

$f_{1}-f_{2}\geq(a-\frac{1}{2})^{2}-(\frac{2a}{b}-a-\frac{1}{2})^{2}\geq
c[a-(\frac{2a}{b}-a)]\geq c\delta a.$

Since

$\mathcal{I}^{\prime\prime}-X\otimes X\geq 0,$

we obtain

$m-2(a-\frac{2a}{b})^{2}|y|^{2}X\otimes X\geq
ca\delta\mathcal{I}^{\prime\prime}.$

Replacing this in (4.12) and using (4.10)-(4.11), we finally obtain (4.10).

Second, we assume $x\in E_{2}$. Then we have

$c(\delta)|x_{1}|^{b}\leq W_{a,b}\leq C(\delta)|x_{1}|^{b}.$

Denote

$\mathcal{K}_{2}^{\prime}=\operatorname{diag}\\{|x_{1}|^{b-2},|x_{1}|^{b-\frac{2b}{a}}\mathcal{I}^{\prime\prime}\\}\;\text{
and}\;\mathcal{K}_{2}=\operatorname{diag}\\{\mathcal{K}_{2}^{\prime},|x_{1}|^{b}\\}.$

Then

$\operatorname{det}\mathcal{K}_{2}=|x_{1}|^{2b(\frac{n}{2}-\frac{1}{b}-\frac{n-2}{a})}\equiv
1.$

Similar to the first case in $E_{1}$, it is enough to show that

(4.13) $D^{\prime 2}W_{a,b}\geq\epsilon(\delta)\mathcal{K}_{2}^{\prime}\ \
inE_{2}.$

Denote

$\mathcal{B}_{2}=\sqrt{\mathcal{K}_{2}^{\prime}},\
Y=|x^{\prime\prime}|^{-1}x^{\prime\prime},\ y\
=|x_{1}|^{-\frac{b}{a}}|x^{\prime\prime}|$

and

$d=(2-a\delta)b(b-1)+a\delta(b-\frac{2b}{a})(b-\frac{2b}{a}-1)y^{2}.$

By a direct calculation we obtain

(4.14) $\mathcal{B}_{2}^{-1}D^{\prime
2}W_{a,b}\mathcal{B}_{2}^{-1}=\left[\begin{array}[]{cc}d&2a\delta(b-\frac{2b}{a})yY\\\
2a\delta(b-\frac{2b}{a})yY&2a\delta\mathcal{I}^{\prime\prime}\end{array}\right].$

Write

$d-2a\delta(b-\frac{2b}{a})^{2}y^{2}|Y|^{2}=g_{1}-g_{2}y^{2},$

where

$\displaystyle g_{2}$
$\displaystyle=a\delta(b-\frac{2b}{a})(b-\frac{2b}{a}-1)-2a\delta(b-\frac{2b}{a})^{2}$
$\displaystyle=\delta b^{2}(2-a)(\frac{2}{a}+\frac{1}{b}-1)$
$\displaystyle\geq 0$

Obviously, $|y|\leq 1$ in $E_{1}$. If $\delta>0$ is small, we have

$\displaystyle d-2a\delta(b-\frac{2b}{a})^{2}y^{2}|Y|^{2}$ $\displaystyle\geq
b^{2}\delta[(2-a\delta)+a\frac{2\delta}{n-2}(1-\frac{2}{a}-\frac{1}{b})-2a(\frac{2\delta}{n-2})^{2}]$
$\displaystyle\geq cb^{2}\delta,$

which, together with (4.14), implies the desired (4.13). ∎

######  Remark 4.4.

Suppose that $R>0$, $W$ is given by (4.5)-(4.6), and consider the Dirichlet
problem

(4.15) $\begin{cases}\operatorname{det}D^{2}W_{R}^{+}=c(\delta)&\text{ in
}B_{R}^{\prime}(0)\times[-\rho,\rho]\\\ W_{R}^{+}=W&\text{ on
}\partial(B_{R}^{\prime}(0)\times[-\rho,\rho]).\end{cases}.$

It is known that there exists a convex solution $W_{R}^{+}$, which is still
symmetric about $x_{n}=0$ since so is $W$. This is to say

$D_{n}W_{R}^{+}(x^{\prime},0)=0.$

Comparison principle (Lemma 2.9) implies

$W_{R}^{+}\geq W\geq 0$

if $c(\delta)>0$ is small. The convexity means

$W_{R}^{+}(x^{\prime},t)\leq\max\\{W_{R}^{+}(x^{\prime},\rho),W_{R}^{+}(x^{\prime},-\rho)\\}\leq
4W(x^{\prime},0),\ \ x\in B_{R}^{\prime}(0),t\in[-\rho,\rho].$

and it follows that

$W_{R}^{+}(te_{n})=0.$

This mean $W_{R}^{+}$ has the same asymptotic behavior around $0$ as $W$.

##  5\. Good Shape Lemma

In this section, we assume that $u$ is a viscosity solution to problem (1.1),
satisfying (2.6) and the hypothesis of Theorem 1.1. We are going to prove that
its height section $S_{h}:=S_{h}^{u}(0)$ (defined by (2.4)) is of good shape
for large $h$, then we will introduce the normalization family
$(u_{h},S_{1}^{h})$ of $(u,S_{h})$.

At first, we mention here that the good shape property is invariant under any
linear transform of the form
$\mathcal{A}=\operatorname{diag}\\{\mathcal{A}^{\prime},a_{n}\\}$ with $c\leq
a_{n},\;\operatorname{det}\mathcal{A}^{\prime}\leq C$.

Let $p_{h}$ be the mass center of $S_{h}$, and consider the sliding
transformation

$\mathcal{A}_{h}^{c}x=x-\sum_{i=1}^{n-1}\frac{p_{h}\cdot e_{i}}{p_{h}\cdot
e_{n}}x_{n}e_{i}.$

Take $y_{h}\in\overline{\mathcal{P}(\mathcal{A}_{h}^{c}S_{h})}$ such that

$|y_{h}\cdot e_{1}|=\sup\\{x\cdot
e_{1}|\,x\in\mathcal{P}(\mathcal{A}_{h}^{c}S_{h})\\}$

and consider another sliding transformation

$\mathcal{A}_{h}x=x-\sum_{i=2}^{n-1}\frac{y_{h}\cdot e_{i}}{y_{h}\cdot
e_{1}}x_{1}e_{i}.$

Apply John’s lemma (Lemma 2.8) to $\\{x^{\prime\prime}\in
R^{n-2}|\,(x_{1},x^{\prime\prime},0)\in\mathcal{A}_{h}\mathcal{P}(\mathcal{A}_{h}^{c}S_{h})\\}$,
we can find a new coordinates $\\{e_{i}\\}_{i=1}^{n}$ which keeps $e_{1}$ and
$e_{n}$ (or only $e_{n}$ for $\mathcal{A}_{h}^{c}$) invariant such that

$\mathcal{D}_{h}B_{1}(q_{h})\subset\mathcal{A}_{h}\mathcal{A}_{h}^{c}S_{h}\subset\mathcal{D}_{h}B_{C(n)}(q_{h})\subset\mathcal{D}_{h}B_{2C(n)}^{+}(0),$

where $q_{h}=\mathcal{A}_{h}\mathcal{A}_{h}^{c}p_{h}$,
$\mathcal{D}_{h}:=\operatorname{diag}(d_{1}(h),\cdots,d_{n}(h))$ is the
diagonal matrix satisfying

$\Pi_{i=1}^{n}d_{i}(h)=\operatorname{det}\mathcal{D}_{h}=Vol(S_{h}).$

It follows from Lemma 2.11 that $\operatorname{Vol}(S_{h})\leq
Ch^{\frac{n}{2}}$. Denote

$\tau_{h}=\frac{(\operatorname{det}\mathcal{D}_{h})^{\frac{1}{n}}}{h^{\frac{1}{2}}},\;\bar{\mathcal{D}_{h}}=\tau_{h}^{-1}\mathcal{D}_{h}:=\operatorname{diag}(\bar{d}_{1}(h),\cdots,\bar{d}_{n}(h)).$

We first consider the rescaling $(u_{h},S_{1}^{h})$ of $(u,S_{h})$ given by

(5.1) ${u}_{h}(x)=\frac{u({\mathcal{A}}_{h}^{-1}\bar{\mathcal{D}_{h}}x)}{h},\
\ x\in S_{1}^{h}:=\bar{\mathcal{D}_{h}}^{-1}{\mathcal{A}}_{h}S_{h},$

then we have

(5.2) ${\mathcal{A}}_{h}^{*}B_{\tau_{h}}(x_{h})\subset
S_{1}^{h}\subset{\mathcal{A}}_{h}^{*}B_{C(n)\tau_{h}}(x_{h})\subset\\{x_{n}\leq
2\tau_{n}C(n)\\}$

for some $x_{h}$, where

$A_{h}^{*}=\bar{\mathcal{D}_{h}}^{-1}\mathcal{A}_{h}\mathcal{A}_{h}^{c}\mathcal{A}_{h}^{-1}\bar{\mathcal{D}_{h}}$

is still a sliding transformation of the form
$A_{h}^{*}x=x-\sum_{i=1}^{n-1}\beta_{h}^{i}x_{n}e_{i}$.

We use the notation $G_{1}^{h}:=S_{1}^{h}\cap\partial\mathbb{R}_{+}^{n}$
similar to (2.5). And we will simplify ${S}_{t}^{h},{G}_{t}^{h}$ by
${S}_{t},{G}_{t}$ when the dependence on $h$ and ${u}_{h}$ is self-evident
from the context. In Section 7, we will use the corresponding notations
$S_{t}^{R},G_{t}^{R}$ for the $(u_{R},S_{1}^{R})$ as we just defined
$S_{1}^{h},G_{1}^{h}$ for $(u,S_{1})$.

Now, the Neumann problem (1.1) becomes

(5.3) $\begin{cases}\operatorname{det}D^{2}{u}_{h}=1&\text{ in }{S}_{1}^{h}\\\
D_{n}{u}_{h}=a_{h}x_{1}&\text{ on }{G}_{1}^{h}\\\ {u}_{h}=1&\text{ on
}\partial{S}_{1}^{h}\setminus{G}_{1}^{h}\end{cases}$

where

(5.4) $a_{h}=\frac{\bar{d}_{1}(h)\bar{d}_{n}(h)}{h}a.$

Recall that if $n\geq 3,$ $u\in\Theta$ and (2.2) is satisfied. Let

$\delta_{h}=\frac{K}{h},\
\bar{\mathcal{Q}}_{h}=\frac{C\bar{\mathcal{D}_{h}}^{\prime\prime
T}I^{\prime\prime}\bar{\mathcal{D}_{h}}^{\prime\prime}}{h}.$

We have

(5.5) $u_{h}(0,z,0)\leq z^{T}\bar{\mathcal{Q}}_{h}z+\delta_{h},\;\forall z\in
R^{n-2}$

and

(5.6)
$a_{h}^{2}\operatorname{det}\mathcal{Q}_{h}=C^{n}\tau_{h}^{-2n}\frac{(\operatorname{det}\mathcal{D}_{h})^{2}}{h^{n}}a^{2}\leq
C.$

If $h\geq 2K$, (2.2) also implies $d_{i}(h)\geq ch^{\frac{1}{2}}$ for $2\leq
i\leq n-1$. More precisely,

$\Pi_{i=2}^{n-1}\bar{d}_{i}(h)\geq
C\tau_{h}^{-\frac{n-2}{2}}\operatorname{Vol}_{n-2}(S_{h}(0)\cap\mathbb{R}^{n-2})\geq
Ch^{\frac{n-2}{2}}.$

Therefore $\operatorname{det}\mathcal{Q}_{h}\geq c$ and $0\leq a_{h}\leq C$.
To conclude, we have obtained that

(5.7) $0\leq a_{h}\leq C,\ \ u_{h}\in\Theta(\delta_{h}),\ \ \forall h\geq 2K.$

The Good Shape Lemma is stated as follows.

######  Theorem 5.1 (Good Shape).

For large $h$, the normalization $(u_{h},S_{1}^{h})$ given by (5.1) satisfies

(5.8) $\displaystyle\ \mathcal{P}S_{1}^{h}(0)\subset
C\mathcal{P}G_{1}^{h}(0)\text{ and }S_{1}^{h}(0)\subset B_{C(n)\tau_{h}}(0),$
(5.9) $\displaystyle\ c\leq\operatorname{Vol}(S_{1}^{h}(0))\leq C\;\text{ and
}\;c\leq\tau_{h}\leq C,$ (5.10) $\displaystyle\
\mathcal{P}G_{1}^{h}(0)\subset-C\mathcal{P}G_{1}^{h}(0).$

We are going to prove the theorem by the coming three lemmas.

######  Lemma 5.2.

The normalization $(u_{h},S_{1}^{h})$ given by (5.1) satisfies (5.8) for large
$h$. More precisely,

$\mathcal{P}S_{1}^{h}(0)\subset(1+C|a_{h}|)\mathcal{P}G_{1}^{h}(0)\subset
C\mathcal{P}G_{1}^{h}(0).$

###### Proof.

If there exists an unit direction $e,e\bot e_{n}$ such that

$\frac{\sup\\{s|\,se\in\mathcal{P}S_{1}^{h}(0)\\}}{\sup\\{s|\,te\in\mathcal{P}G_{1}^{h}(0),\
\forall t\in(0,s)\\}}\geq M>1.$

We assume that the maximum values are attained at points $y$ and $z$
respectively:

$y=y_{e}e+y_{n}e_{n}\in\partial\overline{S_{1}^{h}(0)},\ \
z=z_{e}e\in\partial\overline{G_{1}^{h}(0)}.$

It is clear that

(5.11) $u(z)=u(y)=1,\ \ y_{e}\geq Mz_{e},\ \ 0<y_{n}\leq C,$

where we have used (5.2) for the last inequality.

Since the transforms $\mathcal{A}_{h}^{*}$ leaves $\mathbb{R}^{n-1}$
invariant, we have $|z|\leq C$. Let

$l_{y,0}=\\{sy,\ s\in(0,1)\\},$ $l_{z}^{e_{n}}=\\{z+te_{n},\
t\in[0,\infty)\\}.$

There exists a point $p=sy=z+ty_{n}e_{n}\in l_{y,0}\cap l_{z}^{e_{n}}$ with
$s,t\in(0,\frac{1}{M}]$. By convexity, (2.6) and (5.11) we have

$u(p)\leq su(y)+(1-s)u(0)\leq\frac{1}{M}.$

However, Lemma 3.2, (5.11), the convexity again, and the fact $|z|\leq C$
implies that

$u(p)\geq u(z)+ty_{n}D_{n}u(z)\geq 1+ta_{h}y_{n}z\cdot e_{1}\geq
1-Ct|a_{h}|y_{n}\geq 1-\frac{C|a_{h}|}{M}.$

These two inequalities mean that $M\leq 1+C|a_{h}|\leq C$ by (5.7), which
implies $S_{1}^{h}(0)\subset B_{C(n)\tau_{h}}(0)$. Combining this and (5.2) we
have

$S_{1}^{h}\subset\\{x_{n}\leq
C\tau_{h}\\}\cap\\{CG_{1}^{h}\times[0,\infty]\\}\subset B_{C\tau_{h}}(0).$

∎

######  Lemma 5.3.

The normalization $(u_{h},S_{1}^{h})$ given by (5.1) satisfies (5.9) for large
$h$.

###### Proof.

Since $\operatorname{det}\mathcal{D}_{h}=Vol(S_{h})\leq Ch^{\frac{n}{2}}$ by
Lemma 2.11, we see that $\tau_{h}\leq C$ by its definition. Notice that

$\tau_{h}^{n}\sim\operatorname{Vol}(S_{1}^{h}(0))=\tau_{h}^{n}\operatorname{\mathcal{A}_{h}}\sim\tau_{h}^{n}.$

It is enough for (5.9) to prove that $\tau_{h}\geq c$.

Consider the function $\upsilon$ defined by

$\upsilon(x)=\tau_{h}^{-1}\left(\frac{|x|^{2}}{8C_{n}^{2}}+\frac{x_{n}}{2C_{n}}\right).$

If $\tau_{h}>0$ is small, by (5.3), (5.7) and (5.8) we have

$\upsilon\leq\tau_{h}^{-1}\left(\frac{\tau_{h}^{2}}{8}+\frac{\tau_{h}}{2}\right)<1\leq
u_{h}\;\text{ on }\partial S_{1}^{h}\setminus G_{1}^{h},$
$\operatorname{det}D^{2}\upsilon\geq
c\tau_{h}^{-n}>1=\operatorname{det}D^{2}u_{h}\;\text{ in }S_{1}^{h},$

and

$D_{n}v\geq\frac{1}{2C_{n}\tau_{h}}>C\geq D_{n}u_{h}\;\text{ on }G_{1}^{h}.$

By comparison principle (Lemma 2.10) $v<u_{h}$ in $S_{1}^{h}$ , which
contradicts $v(0)=0=u_{h}(0)$. Therefore $\tau_{h}\geq c$.
∎

######  Lemma 5.4.

The normalization $(u_{h},S_{1}^{h})$ given by (5.1) satisfies (5.10) for
large $h$.

###### Proof.

Lemmas 5.2 and 5.3 mean that $\tau_{h}$ is between two positive universal
constants, and

(5.12) $\mathcal{P}S_{1}^{h}\subset C\mathcal{P}G_{1}^{h}\text{ and
}B_{c}(x_{h})\subset S_{1}^{h}\subset B_{C}(0)$

for some $x_{h}\in R_{+}^{n}$. Therefore, we only need to show
$\mathcal{P}G_{1}^{h}$ is balance about $0$, this is

(5.13) $\mathcal{P}{G}_{1}^{h}\supset B_{c}^{\prime}(0).$

Let $M\geq 4n^{4n}C^{4n}$ is large universal, $\theta=M^{-\frac{1}{n+1}}$,
$m=M^{-1}$ and $\delta=\frac{1}{M^{3}}$ is small enough. On the contrary to
(5.13), there would exist a
$p=(p_{1},p^{\prime\prime})\in\partial\mathcal{P}{G}_{1}^{h}$ such that

(5.14)
$|p|=\operatorname{dist}(0,p)=\operatorname{dist}(0,\partial\mathcal{P}{G}_{1}^{h})\leq\delta.$

Without loss of generality, we may assume that $p_{1}\geq 0$. There are two
cases:

Case 1: $|a_{h}|\geq CM^{-1}$. Then we have

$\max\\{d_{2}(h),\cdots,d_{n-1}(h)\\}\leq CMh^{\frac{1}{2}},$

which means that

$B_{cM^{-1}}^{\prime\prime}(0)\subset\mathcal{P}{S}_{1}^{h}(0)\subset
B_{C}^{\prime}(0).$

By convexity and (5.14), we get

$\sup\\{x_{1}|x\in{G}_{1}^{h}\\}\leq CM\delta\leq\frac{C}{M}.$

Therefore, (5.7) and the Neumann condition in (5.3) implies

$|D_{n}u_{h}(x^{\prime},0)|\leq\frac{C}{M},\text{ on }x_{1}\geq-\frac{1}{M}.$

Case 2: $|a_{h}|\leq CM^{-1}$. Then

$|D_{n}u_{h}(x^{\prime},0)|\leq\frac{C}{M},\text{ on }{G}_{1}^{h}.$

Next, we consider a new coordinate with zero as based point, axis
${e}_{1}=p/|p|$ and $e_{i}(i\geq 2)$ unchanged. For simplicity, we still use
the notation $(e_{1},e_{2},\cdots,e_{n})$.

In both cases, consider the domain $E=\\{x\in
S_{1}^{h}|x_{1}>-\frac{1}{M}\\}$. Then $\partial E=E_{1}\cup E_{2}\cup E_{3}$,
where

$\begin{array}[]{l}E_{1}=\overline{E}\cap\\{x_{1}=-\frac{1}{M}\\},\\\
E_{2}=\overline{E}\cap G_{1}^{h}\cap\\{x_{1}>-\frac{1}{M}\\},\\\
E_{3}=\overline{E}\cap(\partial S_{1}^{h}\setminus G_{1}^{h}).\end{array}$

For the function

$\upsilon(x)=\left[Q(Mx_{1})+\theta\sum_{i=2}^{n}Q(\frac{x_{i}}{C})\right]$

where $Q(t)=\frac{1}{4nC^{2}}(\frac{t^{2}}{2}+2t),$ we claim that

(5.15)
$\begin{cases}\operatorname{det}D^{2}{u_{h}}<\operatorname{det}D^{2}\upsilon&\text{
in }E\\\ D_{-x}{u_{h}}<D_{-x}\upsilon&\text{ on }E_{1}\\\
D_{n}{u_{h}}<D_{n}\upsilon&\text{ on }E_{2}\\\ \upsilon<{u_{h}}&\text{ on
}E_{3}\\\ \end{cases}.$

By comparison principle (Lemma 2.10) we have $\upsilon<u_{h}$ in $E$. This
would contradict $\upsilon(0)=u_{h}(0)$ and hence (5.13) is proved.

Finally, let us verify (5.15). First, by convexity and (2.6) we see that

$D_{-x}{u_{h}}=-D{u_{h}}\cdot x\leq u_{h}(0)-u_{h}(x)\leq 0$

holds in viscosity sense. Applying this, the results of Case 1 and 2, and the
largeness of $M$, we have:

in $E$, $\operatorname{det}D^{2}\upsilon\geq
CM^{2}\theta^{n-1}>1=\operatorname{det}D^{2}{u_{h}}$;

on $E_{1}$, $D_{-x}\upsilon\geq c-\theta\sum_{i=2}^{n-1}C>0\geq D_{-x}u_{h}$;

on $E_{2}$, $D_{n}\upsilon\geq c\theta>\frac{C}{M}\geq D_{n}u_{h}$;

on $E_{3}$, $\upsilon\leq\frac{3}{4}<1={u_{h}}$. ∎

In this way, we have proved Theorem 5.1. Since $c\leq\tau_{h}\leq C$, we can
still take

$\mathcal{D}_{h}=\operatorname{diag}\\{d_{1}(h),\cdots,d_{n}(h)\\}$

in (5.1) instead of $\bar{\mathcal{D}}_{h}$. This yields the definition of
normalization family.

######  Definition 5.5.

The normalization family $(u_{h},S_{1}^{h})$ of $(u,S_{h})$ is defined as

(5.16) ${u}_{h}(x)=\frac{u({\mathcal{D}_{h}}{\mathcal{A}}_{h}^{-1}x)}{h},\
x\in S_{1}^{h}:={\mathcal{A}}_{h}{\mathcal{D}_{h}}^{-1}S_{h}.$

By (2.1), (5.3)-(5.10), we have the following properties for large $h$:

(5.17) $B_{c}^{+}(0)\subset S_{1}^{h}\subset B_{C}(0),$ (5.18)
$\begin{cases}\operatorname{det}D^{2}{u}_{h}=1&\text{ in }{S}_{1}^{h}\\\
D_{n}{u}_{h}=a_{h}x_{1}&\text{ on }{G}_{1}^{h}\\\ {u}_{h}=1&\text{ on
}\partial{S}_{1}^{h}\setminus{G}_{1}^{h}\end{cases}$

where

(5.19) $a_{h}=\frac{{d}_{1}(h){d}_{n}(h)}{h}a,$

(5.20) $u_{h}(0,z,0)\leq z^{T}\mathcal{Q}_{h}z+\delta_{h}$

where

(5.21) $\delta_{h}=\frac{K}{h},\
\bar{\mathcal{Q}}=\frac{C{\mathcal{D}_{h}}^{\prime\prime
T}I^{\prime\prime}{\mathcal{D}_{h}}^{\prime\prime}}{h},$

and

(5.22) $0\leq a_{h}\leq C\text{ and
}a_{h}^{2}\operatorname{det}\mathcal{Q}_{h}\leq C.$

Moreover, Theorem 5.1 still holds for the normalization family
$(u_{h},S_{1}^{h})$.

##  6\. Stationary Theorem

In this section, we study the normalization family $(u_{h},S_{1}^{h})$ given
by (5.16) in the case $a_{h}$ is small. For simplify we omit $h$ and write it
by $(u,S_{1})$. This leads us to study the standard model

(6.1) $\begin{cases}\operatorname{det}D^{2}{u}=1&\text{ in }{S}_{1}\\\
D_{n}{u}=ax_{1}&\text{ on }{G}_{1}\\\ {u}=1&\text{ on
}\partial{S}_{1}\setminus{G}_{1}\end{cases}$

and

(6.2) $u(0)=0,\ \nabla u(0)=0,\ u\geq 0\;in\;S_{1},\ B_{c}^{+}(0)\subset
S_{1}\subset B_{C}(0).$

For every $a\in R$, set

(6.3)
$Q_{a}(x):=\frac{1}{2}\left[\sqrt{1+a^{2}}(x_{1}^{2}+x_{n}^{2})+2ax_{1}x_{n}+\sum_{i=2}^{n-1}x_{i}^{2}\right].$

Obviously, $\operatorname{det}D^{2}Q_{a}(x)\equiv 1$ and
$D_{n}Q_{a}(x^{\prime},0)=ax_{1}$.

We will prove the following

######  Theorem 6.1.

Suppose that $u\in(\overline{S_{1}})$ is a viscosity solution to problem (6.1)
and (6.2) is satisfied. There exists universal constants $b_{0}>0$, $c_{0}>0$
such that if

(6.4)
$\begin{split}||u-Q_{a}||_{L^{\infty}(S_{1}^{Q_{a}}(0))}+||D_{n}u-D_{n}Q_{a}||_{L^{\infty}(S_{1}^{Q_{a}}(0))}\leq
b_{0},\end{split}$

then $u\in C^{2,\alpha}(\overline{B_{c_{0}}^{+}(0)})$ for some $\alpha>0$ and

$||u||_{C^{2,\alpha}(\overline{B_{c_{0}}^{+}(0)})}\leq C.$

With this theorem, we will also show

######  Theorem 6.2.

Suppose that $u\in(\overline{S_{1}})$ is a viscosity solution to problem (6.1)
and (6.2) is satisfied. There exists universal constants $\delta_{0}>0$,
$c_{0}>0$ such if $a\in(-\delta_{0},\delta_{0})$, then then $u\in
C^{2,\alpha}(\overline{B_{c_{0}}^{+}(0)})$ for some $\alpha>0$ and

(6.5) $\begin{split}||u||_{C^{2,\alpha}(\overline{B_{c_{0}}^{+}(0)})}\leq
C.\end{split}$

Denote

$\displaystyle\mathcal{F}_{a}^{*}(x):=$
$\displaystyle\\{Q|\,Q(x)=\sum_{i,j=1}^{n}a^{ij}x_{i}x_{j}\text{ is a
quadratic polynomial,}$ $\displaystyle\text{ and }Q\text{ solves the first and
second equations of problem }\eqref{6.1}\\}$

and

$\mathcal{F}_{a}(x):=\\{l(x^{\prime})+Q(x)|\,Q\in\mathcal{F}_{a}^{*},l(x^{\prime})\text{
is a linear function },l(0)=0\\}.$

Theorem 6.1 relies on the following Lemma.

######  Lemma 6.3.

Let $u\in C(\overline{S_{1}(0)})$ be a viscosity solution to problem (6.1),
and satisfy (6.2). For any small constant $\mu>0$, there exists a small
$\epsilon_{0}=\epsilon_{0}(\mu)>0$ such that if

(6.6) $||u-P||_{L^{\infty}(S_{1})}\leq\epsilon,\ \
||D_{n}u-D_{n}Q||_{L^{\infty}(S_{1})}\leq C\epsilon$

for some $\epsilon\in(0,\epsilon_{0})$ and
$P(x)=l(x^{\prime})+Q(x)\in\mathcal{F}_{a}$, then

(6.7) $|Dl|\leq C\epsilon^{\frac{1}{2}},$

and

(6.8) $|u-P^{0}|\leq C\epsilon\mu^{\frac{3}{2}},\ \ |D_{n}u-D_{n}Q^{0}|\leq
C\epsilon\mu\text{ in }S_{\mu}^{Q^{0}}$

for some $P^{0}=l^{0}+Q^{0}\in\mathcal{F}_{a}$. Moreover,

(6.9) $|D^{2}Q-D^{2}Q^{0}|\leq 2\epsilon\text{ in }S_{\mu}^{Q^{0}}.$

###### Proof.

By virtue of the fact $l(0)=0$, (6.6) gives $|Dl|\leq C\epsilon$. One can
choose $x=(x^{\prime},0)\in G_{1}$ such that

$-l(x^{\prime})=\frac{1}{n}|Dl||x^{\prime}|\,\text{ and
}\;|x^{\prime}|\sim\epsilon^{\frac{1}{2}}.$

It follows from (6.2) and (6.6) that

$\displaystyle\frac{1}{n}|Dl||x^{\prime}|=-l(x^{\prime})$ $\displaystyle\leq
u(x^{\prime},0)-l(x^{\prime})$
$\displaystyle\leq|u(x,0)-l(x^{\prime})-Q(x^{\prime},0))|+|Q(x^{\prime},0)|$
$\displaystyle\leq\epsilon+C|x^{\prime}|^{2}\leq C|x^{\prime}|^{2}.$

Therefore $|Dl|\leq C\epsilon^{\frac{1}{2}}$ and (6.7) holds.

In following discussion, we will replace $u$ by $u-l(x^{\prime})$ and need
only to consider the case $P=Q$ in (6.6). (6.2) and (6.6) implies that
$c\mathcal{I}\leq D^{2}Q\leq C\mathcal{I}$. For simplicity, we further assume
$Q=Q_{a}$ (see (6.3)). This is because we can always find a new coordinate
with the aid of the sliding transform
$\mathcal{B}x:=x-2D^{\prime\prime}Q^{-1}D_{1i}Qx_{1}e_{i}$, under which $Q$
may be transformed to $Q_{a}$.

Let

(6.10) ${u}_{0}:=u-P=u-Q,\ v_{\varepsilon}:=\frac{{u}_{0}}{\varepsilon},\ \
\varepsilon>0.$

In fact, we may assume $||u-Q||_{L^{\infty}(S_{1})}>0$ and take
$\varepsilon=||u-Q||_{L^{\infty}(S_{1})}$. We will show the system
$(v_{\varepsilon},D_{n}v_{\varepsilon})$ could be well approximated by
$(w,D_{n}w)$, a solution of some linearized problem. More precisely, let
$F=S_{\frac{1}{2}}^{Q}(0)$ and

$Lw:=\sqrt{1+a^{2}}(w_{11}+w_{nn})+aw_{1n}+\Delta^{\prime\prime}w,$

where $\Delta^{\prime\prime}:=\sum_{i=2}^{n-1}\partial_{x_{i}}^{2}$. We will
show that as if $\varepsilon\to 0^{+}$, the corresponding
$(v_{\varepsilon},D_{n}v_{\varepsilon})$ will have a subsequence converging
uniformly to $(w,D_{n}w)$, a solution of the problem:

(6.11) $Lw=0\text{ in }F,\ w(0)=0,\ D_{n}w=0\text{ on }\\{x_{n}=0\\}.$

Step 1. We first consider the convergence on compact subset $E=\\{x_{n}\geq
C\epsilon^{\frac{1}{2}}\\}\cap F$, where $\epsilon>0$ is sufficiently small.
For any point $y$ such that $B_{C\rho}(y)\subset E$ for some
$\rho>>c\epsilon^{\frac{1}{2}}$. Then $B_{c\rho}(y)\subset
S_{\rho^{2}}^{Q}(y)\subset B_{C\rho}(y)$ by the fact $Q=Q_{a}$. It follows
from (6.1) that

(6.12)
$0=\operatorname{det}D^{2}{u}-\operatorname{det}D^{2}Q=Tr(AD^{2}({u}-Q))=\varepsilon
Tr(AD^{2}v_{\varepsilon}),$

where

$A=[A_{ij}]_{n\times n}:=\int_{0}^{1}cof((1-t)D^{2}Q+tD^{2}{u})dt,$

and $cofM$ denotes the cofactor matrix of $M$.

Since (6.6) gives us $|u-Q|\leq\epsilon$ in $S_{1}$, we see that

$B_{2c\rho}(y)\subset S_{\rho^{2}/2}^{u}(y)\subset B_{C\rho}(y).$

Hence, the classical interior estimate (Lemma 2.13 and its proof in [3]) means
that

$||{u}||_{C^{k}(B_{c\rho}(y))}\leq C\rho^{2-k}\text{ and }c\mathcal{I}\leq
D^{2}{u}\leq C\mathcal{I}\text{ in }B_{c\rho}(y).$

Therefore, the operator $\bar{L}f:=A_{ij}f_{ij}$ is uniformly elliptic with
smooth coefficients. By (6.12), we obtain

(6.13) $||{v}_{\varepsilon}||_{C^{k}(B_{c\rho}(y))}\leq C\rho^{-k},$

which means

$|D^{2}{u}-D^{2}Q|\leq C\varepsilon\rho^{-2}\text{ and }|A-D^{2}Q|\leq
C\varepsilon\rho^{-2}\;\text{in}\;B_{c\rho}(y)).$

Let $\varepsilon\to 0^{+}$, we have $(v_{\varepsilon},D_{n}{v}_{\varepsilon})$
converges to $(w,D_{n}w)$ in $B_{c\rho}(y)$. By (6.10), we see that

$w(0)=0,\ \ Lw=0\;\text{ in }\;B_{c\rho}(y).$

Step 2. We gives the uniform control of
$(v_{\varepsilon},D_{n}v_{\varepsilon})$ near $\\{x_{n}=0\\}$, so as to show

(6.14) $|D_{n}v_{\varepsilon}(x)|\leq C|x_{n}|\text{ in }S_{\frac{1}{4}}(0)$

and

(6.15) $\operatorname{Osc}_{B_{r}(x)}v_{\varepsilon}\leq
C\max\\{r^{\frac{2}{3}},\epsilon\\}\text{ for }x\in G_{\frac{1}{4}}.$

Given $y\in G_{\frac{1}{2}}(0)$. It follows from (6.6) and Corollary 2.7 that

$|\nabla u-\nabla P|\leq
C\epsilon^{\frac{1}{2}}\;\text{in}\;S_{\frac{1}{2}}(y)$

if $\epsilon$ is small enough. Hence, the function $u^{y}(x)$ defined in (3.5)
still satisfies

$u^{y}(x)\geq c\text{ on }\partial S_{\frac{1}{4}}(y).$

As in Lemma 3.4, $V:=D_{n}{v}_{\varepsilon}$ satisfies

(6.16) $\begin{cases}U^{ij}D_{ij}(\pm V)=0&\text{ in }S_{\frac{1}{4}}(0)\\\
|V|\leq C&\text{ on }\partial S_{\frac{1}{4}}(0)\setminus
G_{\frac{1}{4}}(0)\\\ \pm V=0&\text{ on }{G_{\frac{1}{4}}(0)}\end{cases}.$

Notice that the second in (6.16) is equivalent to $V\leq C$ or $-V\leq C$,
which comes from (6.6). Similar to Corollary 3.7, we see that the function
$w^{y,+}(x)=C_{1}[w^{y}(x)+C_{2}x_{n}]$ defined in (3.7) is a supersolution
for $V$ and $-V$ in(6.16) respectively.

Hence we get

$|D_{n}v_{\varepsilon}(x)|=|V(x)|\leq
w^{(x^{\prime},0),+}(x^{\prime},x_{n})\leq Cx_{n},\text{ in
}S_{\frac{1}{4}}(0).$

This roves (6.14).

To prove (6.15), we observe that (6.14) implies

(6.17) $|v_{\varepsilon}(x^{\prime},0)-v_{\varepsilon}(x^{\prime},x_{n})|\leq
Cx_{n}^{2}\text{ in }S_{\frac{1}{4}}(0).$

It is enough to estimate $|v_{\varepsilon}(p)-v_{\varepsilon}(q)|$ for
$p=(p^{\prime},0),q=(q^{\prime},0)$ in $G_{\frac{1}{2}}(0)$. Suppose that

$|p^{\prime}-q^{\prime}|=r,\
\,\rho=C\max\\{r^{\frac{1}{3}},\epsilon^{\frac{1}{2}}\\},\ \,A=p+\rho e_{n},\
\ B=q+\rho e_{n}.$

As $\rho\geq C\epsilon^{\frac{1}{2}}$, (6.13) implies

$|v_{\varepsilon}(A)-v_{\varepsilon}(B)|\leq\frac{r}{\rho},$

which, together with (6.17), gives

$|v_{\varepsilon}(p)-v_{\varepsilon}(q)|\leq C\rho^{2}+\frac{r}{\rho}\leq
C\max\\{r^{\frac{2}{3}},\epsilon\\}.$

In summary, we have proved (6.14) and (6.15). By this and the result of Step 1
we conclude that as $\varepsilon\to 0^{+}$,
$(v_{\varepsilon},D_{n}v_{\varepsilon})$ has a subsequence converging to a
solution $w$ of problem (6.11) in $S_{\frac{1}{4}}(0)$ uniformly.

Step 3. The classical regular results then shows the function $w\in
C_{loc}^{3}$. Since $w(0)=0$, $D_{n}w(0)=0$ and $D^{\prime}D_{n}w(0)=0$ by
(6.11), there exists a linear function

$l^{0}(x)=\sum_{i=1}^{n-1}a_{i}x_{i}$

and a quadratic function

$R(x)=\sum_{1\leq i,j\leq n-1}a_{ij}x_{i}x_{j}+b_{n}x_{n}^{2}$

such that

$|w(x)-l^{0}(x)-R(x)|\leq C|x|^{3}\text{ and }|D_{n}w(x)-2b_{n}x_{n}|\leq
C|x|^{2}.$

Here, $a_{i},a_{ij},b_{n}$ are bounded by a universal constant, and

$\sqrt{1+a^{2}}(a_{11}+b_{n})+\sum_{i=2}^{n-1}a_{ii}=0$

by (6.11). As $\varepsilon\to 0^{+}$, the uniform convergence means

$|v_{\varepsilon}(x)-l^{0}(x)-R(x)|\leq\sigma+C|x|^{3}\text{ and
}|D_{n}v_{\varepsilon}(x)-D_{n}R|\leq\sigma+C|x|^{2},$

where $\sigma=\sigma(\varepsilon)\to 0^{+}$. If we let
$\mu\geq\sigma^{\frac{1}{3}}+\varepsilon^{\frac{1}{3}}$, then

(6.18) $|u-\left[Q+\varepsilon(l^{0}+R)\right]|\leq
C\varepsilon\mu^{\frac{3}{2}}\text{ in }S_{\mu}^{Q},$

and

(6.19) $|D_{n}u-(D_{n}Q+\varepsilon D_{n}R)|\leq C\varepsilon\mu\text{ in
}S_{\mu}^{Q}.$

Step 4. Let

$f(t):=\operatorname{det}\left[D^{2}Q+\epsilon D^{2}R+t\mathcal{I}\right].$

Notice that $|f(0)-1|\leq C\epsilon^{2}$, $f^{\prime}(t)\sim C$. The equation
$f(t)=1$ is solvable. The solution $t_{0}$ satisfies

(6.20) $|t_{0}|\leq C\epsilon^{2}.$

Let

(6.21) $Q^{0}(x)=Q(x)+\epsilon R(x)+\frac{t_{0}}{2}|x|^{2},\ P^{0}=\epsilon
l^{0}+Q^{0}.$

Then

$\operatorname{det}D^{2}P^{0}=1,\ D_{n}P^{0}=D_{n}Q=ax_{1}\text{ on
}\\{x_{n}=0\\},$

which means that $P^{0}\in\mathcal{F}_{a}$.

Finally, we combine (6.18), (6.19), (6.20) and (6.21) to obtain

$|{u}(x)-P^{0}(x)|\leq C\epsilon^{2}\mu+C\epsilon\mu^{\frac{3}{2}}\leq
C\epsilon\mu^{\frac{3}{2}}\text{ in }S_{\mu}^{Q^{0}},$

and

$|D_{n}{u}(x)-D_{n}P^{0}(x)|\leq
C\epsilon^{2}\mu^{\frac{1}{2}}+C\epsilon\mu\leq C\epsilon\mu\text{ in
}S_{\mu}^{Q^{0}}.$

Noticing that $S_{\mu/2}^{Q_{0}}\subset S_{\mu}^{Q}$ and replacing $\mu$ with
$\mu/2$, we have proved the desired (6.8), while (6.9) follows directly from
(6.21). ∎

Proof of Theorem 6.1:  Let $C,\mu,$ and $\epsilon_{0}(\mu)$ are the same as in
Lemma 6.3. Choose $0<\alpha<1$ and $0<\mu\leq(4C)^{\frac{2}{\alpha-1}}$ . Let

$C_{\infty}=\max\\{\Pi_{j=0}^{\infty}(1+4C^{2}\mu^{\frac{j\alpha}{2}}),e^{4C}\\}.$

Observing that

$\log
C+\Sigma_{j=0}^{k}\log(1+4CC_{\infty}^{2}\epsilon_{0}\mu^{\frac{k\alpha}{2}-2})\leq
2C+\epsilon_{0}C(\mu)C_{\infty}^{3}\leq\log C_{\infty},$

we can assume that $\epsilon_{0}$ is sufficiently small such that the sequence

$C_{0}=C,\
C_{k}=(1+4CC_{\infty}^{2}\epsilon_{0}\mu^{\frac{k\alpha}{2}-2})C_{k-1},k=1,2,3,\cdots$

is bounded by $C_{\infty}$.

We will prove by mathematical induction that there exists a sequence of
constants $a_{k}$, transformations

$\mathcal{M}_{k}=\operatorname{diag}\\{\mathcal{M}_{k}^{\prime},\mathcal{M}_{k,nn}\\},\
\det{\mathcal{M}_{k}}=1$

(where $\mathcal{M}_{k}^{\prime}$ is a $n-1$ square matrix), and linear
functions $l_{k}=\sum_{i=1}^{n-1}a_{i}^{k}x_{i}$ such that at height
$h_{k}:=\mu^{k}$, the normalization solution $(u_{k},\Omega_{k})$ of
$(u,\Omega)$ given by

(6.22) $u_{k}(x)=\frac{u(\mathcal{T}_{k}x)}{h_{k}},\
x\in\Omega_{k}:=\mathcal{T}_{k}^{-1}S_{h_{k}}(0),\
\mathcal{T}_{k}=h_{k}^{\frac{1}{2}}\mathcal{M}_{k}$

satisfying problem (6.1) and

(6.23)
$||u_{k}-l_{k}-Q_{a_{k}}||_{L^{\infty}(S_{1}^{P_{k}})}+||D_{n}u_{k}-D_{n}Q_{a_{k}}||_{L^{\infty}(S_{1}^{P_{k}})}\leq\epsilon_{k}:=\epsilon_{0}\mu^{\frac{{k\alpha}}{2}},$

for $P_{k}=l_{k}+Q_{k}\in\mathcal{F}_{a}$, where

$Q_{k}(x):=\frac{1}{2}\left[\sqrt{1+a_{k}^{2}}(x_{1}^{2}+x_{2}^{2})+2a_{k}x_{1}x_{2}+\sum_{i=2}^{n-1}x_{i}^{2}\right].$

Moreover,

(6.24) $\begin{cases}&|b_{k-1}|\leq
C(C_{\infty})\epsilon_{k-1}^{\frac{1}{2}},\ \ |a_{k}|\leq C_{k},\ \
|{\mathcal{M}}_{k}{\mathcal{M}}_{k-1}^{-1}-\mathcal{I}|\leq
C_{k}\epsilon_{k-1}\\\ &C_{k}^{-1}B_{1}(0)\cap\Omega_{k}\subset
S_{1}^{u_{k}}(0)\subset C_{k}B_{1}(0).\end{cases}.$

The case $k=0$ is the assumption of Theorem 6.1. If the case of $k$ holds we
will prove it for the case $k+1$. We check for $\epsilon_{k}$ whether $u_{k}$
satisfies the conditions in Lemma 6.3. By induction hypothesis (6.23), we have

$|u_{k}-P_{k}|\leq\epsilon_{k}\text{ in }\Omega_{k}\cap S_{1}^{P_{k}}.$

Apply lemma 6.3 to $u_{k}$, we obtain the quadratic function
$\bar{P}_{k+1}=\bar{l}_{k+1}+\bar{Q}_{k+1}\in\mathcal{F}_{a}$ satisfies

(6.25) $\begin{cases}&\operatorname{dist}_{\mu}(u_{k},\bar{P}_{k+1},0)\leq
C{\epsilon}_{k}\mu^{\frac{3}{2}}\leq\epsilon_{k+1}\mu,\\\ &|b_{k}|\leq
C\epsilon_{k}^{\frac{1}{2}}\;\text{ and }\;|D^{2}Q_{k}-D^{2}\bar{Q}_{k+1}|\leq
C{\epsilon}_{k}.\end{cases}.$

Then there exists $a_{k}$ and positive matrix
$\mathcal{B}_{k+1}=\operatorname{diag}\\{\mathcal{B}_{k}^{\prime},\mathcal{B}_{k,nn}\\}$
satisfies

$\mathcal{B}_{k+1}^{T}D^{2}Q_{k}\mathcal{B}_{k+1}=D^{2}Q_{k+1}$

and

$|a_{k+1}-a_{k}|\leq CC_{k}\epsilon_{k}\text{ and
}|\mathcal{B}_{k+1}-\mathcal{I}|\leq CC_{k}{\epsilon}_{k}.$

In particular, $|a_{k+1}|\leq C_{k+1}$. Take
$\mathcal{M}_{k+1}=\mathcal{M}_{k}\mathcal{B}_{k+1}$,
$\mathcal{T}_{k+1}=\mu^{\frac{1}{2}}\mathcal{T}_{k}\mathcal{B}_{k+1}$ and let

$u_{k+1}(x)=\frac{u({\mathcal{T}}_{k+1}x)}{h_{k+1}},\
S_{1}^{u_{k+1}}(0)=\mathcal{T}_{k+1}^{-1}S_{h_{k+1}}(0),\
h_{k+1}=\mu^{{(k+1)\alpha}}.$

Then we have

$(1-CC_{\infty}^{2}\mu^{-1}\epsilon_{k})C_{k}^{-1}B_{1}(0)\cap\Omega_{k+1}\subset
S_{1}^{u_{k+1}}(0)\subset(1+CC_{\infty}^{2}\mu^{-1}\epsilon_{k})C_{k}B_{1}(0),$

which is

$C_{k+1}^{-1}B_{1}(0)\cap\Omega_{k+1}\subset S_{1}^{u_{k+1}}(0)\subset
C_{k+1}B_{1}(0).$

Using the above facts and (6.25), one may verify by direct calculations that
(6.23)-(6.24) holds for $k+1$.

It follows that $|a_{k}|\leq C_{k}$, $|\mathcal{M}_{k}|\leq C_{k}$ converge
geometrically to $a_{\infty}$, $\mathcal{M}_{\infty}$ and

$|\mathcal{M}_{\infty}\mathcal{M}_{k}^{-1}-\mathcal{I}|\leq
C_{k}\epsilon_{k}\leq C_{\infty}h_{k}^{\frac{{\alpha}}{2}}.$

We replace each $\mathcal{M}_{k}$ by $\mathcal{M}_{\infty}$ and $Q_{k}$ by
$Q_{\infty}$, the inductive hypothesis still holds with a large universal
constant. This is

$|u(x)-h_{k}^{\frac{1}{2}}l_{k}(\mathcal{M}_{\infty}^{-1}x)-Q_{\infty}(\mathcal{M}_{\infty}^{-1}x)|\leq
C_{1}C_{\infty}^{2}h_{k}^{1+\frac{{\alpha}}{2}}\text{ in
}B_{ch_{k}^{\frac{1}{2}}}(0).$

Recalling that $|Dl_{k}|\leq C\epsilon_{k-1}^{\frac{1}{2}}$, we obtain

$|u(x)-Q_{\infty}(\mathcal{M}_{\infty}^{-1}x)|\leq
C_{1}C_{\infty}^{2}h_{k}^{1+\frac{{\alpha}}{2}}\text{ in
}B_{ch_{k}^{\frac{1}{2}}}(0),$

which means

(6.26) $|u(x)-Q_{\infty}(\mathcal{M}_{\infty}^{-1}x)|\leq
C_{1}C_{\infty}^{2}|x|^{2+\alpha}.$

The above induction argument and Lemma 6.3 could be applies to any point $y\in
G_{c_{0}}(0)$. As a result, we may obtain that

$|u(y+x)-Q^{y}(x)|\leq C|x|^{2+{\alpha}}$

for a quadratic polynomial depending on the based point $y$. In fact, consider
$p=y+\theta e_{n}$ with $\theta>0$ is small enough. The height section
$S_{c_{1}\theta^{2}}^{Q^{y}}(p)$ is contained in $S_{1}$ with

$B_{c\theta^{2}}(p)\subset S_{c_{1}\theta^{2}}^{Q^{y}}(p)\subset
B_{C\theta^{2}}(p).$

By (6.26) we have

$\begin{cases}\operatorname{det}D^{2}u(x)=1&\text{ in
}S_{c_{1}\theta^{2}}^{Q^{y}}(p)\\\ |u-Q^{y}(x)|\leq
C|\theta|^{2+\alpha}&\text{ on }\partial
S_{c_{1}\theta^{2}}^{Q^{y}}(p)\end{cases}.$

By interior $C^{2,\alpha}$ theory and its perturbation method (See [3] for the
details), $||u||_{C^{2,\alpha}(B_{c\theta^{2}}(p))}\leq C$. Since the constant
$C$ is independent of $y$ and $\theta$, $u\in
C^{2,\alpha}(\overline{B_{c_{0}}^{+}(0)})$ for a small $c_{0}>0$.

Proof of Theorem 6.2:  Choose $\delta_{0}=b_{0}^{6n}$ and $c_{0}=cb_{0}$,
where $\alpha,\mu,$ and $b_{0}$ are the same as in the proof of Theorem 6.1.
Take $a\in(-\delta_{0},\delta_{0})$. We are going to construct a quadratic
approximation of $u$ near $0$ as (6.4), and applying Theorem 6.1 to complete
the proof of Theorem 6.2.

Lemma 5.2 implies

$\mathcal{P}S_{1}(0)\subset(1+C\delta_{0})G_{1}(0).$

For $(x^{\prime},0)\in PS_{1}(0)$, let $g(x^{\prime})$ be the maximum/minimum
value such that $(x^{\prime},g(x^{\prime}))\in S_{1}(0)$. We consider the
convex set

$E=\\{(x^{\prime},x_{n})|(x^{\prime},0)\in PS_{1}(0),0\leq x_{n}\leq
g(x^{\prime})\\},$

and

$S=\\{(x^{\prime},\pm x_{n})|(x^{\prime},x_{n})\in E\\}.$

Then

$\partial S_{1}\setminus G_{1}\subset
E_{C\delta_{0}}=\\{x\in\mathbb{R}^{n}|\operatorname{dist}(x,\partial E)\leq
C\delta_{0}\\}.$

By the standard theory(see [11, 12] for example) we see that there is a convex
function $v$ symmetric about the hyperplane $\\{x_{n}=0\\}$ solving

$\begin{cases}\operatorname{det}D^{2}v=1&\text{ in }S\\\ v=1&\text{ on
}\partial S\\\ \end{cases}.$

Moreover, by (6.2) and Lemma 2.13, $v$ is smooth in the interior and Lemma
2.12 tells us

$v(x)\geq 1-C\operatorname{dist}(x,\partial S)^{\frac{1}{n}}.$

Thus, $v$ is also a solution to the following Neumann problem

(6.27) $\begin{cases}\operatorname{det}D^{2}v=1&\text{ in }S_{1}\\\
1-b_{0}^{6}\leq v\leq 1&\text{ on }\partial S_{1}\setminus G_{1}\\\
D_{n}v=0&\text{ on }G_{1}\\\ \end{cases}.$

Since $|a|<\delta_{0}$ and $b_{0}>0$ is small, then $v^{\pm}:=v\pm
2b_{0}^{6}(2C-x_{n})$ must be supersolution/subsolution of $u$ to (6.1), which
implies

(6.28) $||v-u||_{L^{\infty}(S_{1})}\leq Cb_{0}^{6}$

by the comparison principle (Lemma 2.10). Let $w=u-v$, by Corollary 2.7 we
have

(6.29) $|\nabla w|\leq Cb_{0}^{3}\text{ in }B_{c}^{+}(0).$

By (6.27) we have $D_{in}v=0$ for $i\neq n$ in $G_{1}$. Let

$\mathcal{M}=D_{ij}v(0),\ \ R(x)=x^{T}\mathcal{M}x,\ \
\mathcal{M}=\mathcal{B}^{2},$

where $\mathcal{B}=(b_{ij})_{n\times n}$ may be chosen such that it is bounded
symmetric matrix with $b_{in}=0$ for $i\neq n$ because of the same property of
$\mathcal{M}$. Consider the diagonal matrices $\mathcal{B}_{1}$ defined by

$\mathcal{B}_{1}^{2}=[\frac{1}{2}diag\\{\sqrt{1+a^{2}},I^{\prime\prime},\sqrt{1+a^{2}}\\}]^{-1}$

and the function

$\tilde{u}(x)=\frac{u(b_{0}(\mathcal{B}\mathcal{B}^{-1})x)}{b_{0}^{2}},\\\
Q(x)=\frac{Q_{a}(b_{0}(\mathcal{B}\mathcal{B}_{1})^{-1}x)}{b_{0}^{2}},\ x\in
B_{C}(0),$

where $Q_{a}$ is the same as in (6.3). Letting $E_{0}=B_{Cb_{0}}(0)$ and
noticing $v\in C^{3}(B_{c}(0))$, we see that

$\displaystyle||\tilde{u}-Q||_{L^{\infty}(E_{0})}$
$\displaystyle\leq\frac{1}{b_{0}^{2}}||u-R||_{L^{\infty}(E_{0})}+C|a|$
$\displaystyle\leq\frac{1}{b_{0}^{2}}||u-v||_{L^{\infty}(E_{0})}+Cb_{0}+C|a|$
$\displaystyle\leq Cb_{0}$

and

$\displaystyle||D_{n}\tilde{u}-D_{n}Q||_{L^{\infty}(E_{0})}$
$\displaystyle\leq\frac{1}{b_{0}}||D_{n}u-x_{n}D_{nn}v||_{L^{\infty}(E_{0})}+C|a|$
$\displaystyle\leq\frac{1}{b_{0}}||D_{n}u-D_{n}v||_{L^{\infty}(E_{0})}+Cb_{0}+C|a|$
$\displaystyle\leq Cb_{0}.$

By Theorem 6.1, we have $\tilde{u}\in C^{2,\alpha}(\overline{B_{c}^{+}(0)})$,
therefore $u\in C^{2,\alpha}(\overline{B_{c_{0}}^{+}(0)})$ for $c_{0}=cb_{0}$.

##  7\. Universal Strict Convexity

Throughout this section, we assume that $u$ is a viscosity solution to problem
(1.1), satisfying (1.2) (or (2.2)), (2.1) and (2.6). By (2.1) and Theorem 5.1
we may assume

(7.1) $B_{c}^{+}(0)\subset S_{1}(0)\subset B_{C}(0).$

We will study the geometry of the section $S_{R}(0)$ for large $R$. In
particular, we will show the uniformly strict convexity for the normalization
family $(u_{R},S_{1}^{R})$ defined by (5.16).

Recall the notations from the beginning in Section 5 to (5.1). When we fix a
$R$, we will use the notation $\mathcal{A}_{R}$, $d_{i}(R)$ and
$\mathcal{D}_{R}$ instead of $\mathcal{A}_{h}$, $d_{i}(h)$ and
$\mathcal{D}_{h}$. We define the section $S_{t}^{R}(0)$ of $(u_{R},S_{1}^{R})$
as we defined $S_{1}^{h}(0)$ of $(u,S_{h})$ by (5.1) in Section 5. Set

$d_{i}^{R}(t)=\frac{d_{i}(tR)}{d_{i}(R)},\
\mathcal{D}_{t}^{R}=\mathcal{D}_{tR}(\mathcal{D}_{R})^{-1}\text{ and
}{\mathcal{A}}_{t}^{R}=(D_{R})^{-1}\mathcal{A}_{tR}(\mathcal{A}_{R})^{-1}D_{R},$

The normalization of $(u_{R},S_{1}^{R})$ at height $t$ is essentially the same
as $(u_{tR},S_{1}^{tR})$:

${u}_{tR}(x):=\frac{u_{R}((\mathcal{A}_{t}^{R})^{-1}\mathcal{D}_{t}^{R}x)}{t},\
x\in S_{1}^{tR}:=(\mathcal{D}_{t}^{R})^{-1}\mathcal{A}_{t}^{R}S_{1}^{R}.$

Here is the main result of this section.

######  Theorem 7.1.

Given large constants $R_{0}>0$ and $C_{0}>1$. There exists a small constant
$\eta>0$ such that for any $R\geq R_{0}$,

(7.2) $C_{0}\eta S_{R}(0)\subset S_{\eta R}(0)\subset(1-\eta)S_{R}(0).$

Moreover, there exist $\bar{\delta}=\delta_{R_{0}}$ and $\gamma\in(0,1)$ such
that

(7.3) $c|x|^{\frac{1+\gamma}{\gamma}}-C\delta_{R}/\bar{\delta}\leq
u_{R}(x)\leq C|x|^{1+\gamma}+C\delta_{R}/\bar{\delta},\ \ \forall x\in
S_{1}^{R},$

where $\delta_{R}$ denotes a constant depending on $R$ and tends to $0$ as
$R\to\infty$.

######  Claim 7.2.

(7.2) is a scaling invariant property. Using iterative techniques, one can see
that (7.2) and (7.3) are equivalent. More precisely, regardless of the term
$C\delta_{R}/\bar{\delta}$, the left side of (7.2) and the right side of (7.3)
are equivalent to

$u_{R}(x)\leq\sigma_{1}(|x|)|x|\text{ for all large }R,$

and the right side of (7.2) and the left side of (7.3) are equivalent to

$u_{R}(x)\geq\sigma_{0}(|x|)\text{ for all large }R,$

where $\sigma_{0},\sigma_{1}$ are some universal strictly increasing $C^{0}$
function with $\sigma_{i}(0)=0$ $(i=0,1)$.

The right side of (7.2) implies $S_{\eta^{k}}^{\eta^{-k}R}\subset
B_{(1-\eta)^{k}C}(0)$. Thus, if we need to obtain the global type estimate of
$u_{R}$ in $S_{1}^{R}$, we only need to prove the corresponding local estimate
of $u_{\eta^{-k}R}$ in $S_{\eta^{k}}^{\eta^{-k}R}$ for some universal large
$k$. Especially, we can obtain that $D_{n}u_{R}$ is uniformly bounded in
$S_{1}^{R}(0)$.

By Lemma 2.5 and the Neumann boundary value condition, the right side of (7.3)
provides a universal $C^{1,\gamma}$ module for $u_{R}$ at $0$.

To prove Theorem 7.1, in addition to the notations $\delta_{0},c_{0}$ in
Theorem 6.2 and $a_{R}$ in (5.19) we need the following lemma.

######  Lemma 7.3.

If $R\geq R_{0}$ is large, then $a_{R}\geq c\delta_{0}a$ and

(7.4) $cR^{\frac{1}{2}}\mathcal{I}^{\prime\prime}\leq D_{R}^{\prime\prime}\leq
CR^{\frac{1}{2}}\mathcal{I}^{\prime\prime}.$

###### Proof.

Assume by way of contradiction that $a_{R}\leq c\delta_{0}a\leq\delta_{0}$.
Then Theorem 6.2, together with (2.6) and (7.1), implies that the
normalization solution $u_{R}$ is $C^{2,\alpha}$ near $0$. This is

$B_{c}^{+}(0)\subset t^{-\frac{1}{2}}S_{t}^{R}(0)\subset
B_{C}^{+}(0),\;\forall t\in(0,cc_{0}^{2}).$

We recall (7.1) and
$S_{1}(0)=\mathcal{D}_{R}^{-1}\mathcal{A}_{R}S_{\frac{1}{R}}^{R}(0).$ If
$R\geq Cc_{0}^{-2}$ is large, we can take $t=\frac{1}{R}$ to obtain

$c\mathcal{I}\leq\mathcal{D}_{R}^{-1}\mathcal{A}_{R}R^{\frac{1}{2}}\leq
C\mathcal{I}.$

This gives $d_{i}(R)\leq CR^{\frac{1}{2}}$ for $2\leq i\leq n-1$. Hence,

$a_{R}=\frac{d_{1}(R)d_{n}(R)}{R}a\geq Ca,$

which is a contradiction.

Thus $a_{R}\geq c\delta_{0}a$, which implies $\Pi_{i=2}^{n-2}d_{i}(R)\leq
C_{1}\delta_{0}^{-1}R^{\frac{n-2}{2}}$. Noticing that $d_{i}(R)\geq
cR^{\frac{1}{2}}$ for $2\leq i\leq n-1$ (see (2.2)), we see that (7.4) holds
for $C=(C_{1}\delta_{0}^{-1})^{\frac{1}{n-2}}$. ∎

######  Remark 7.4.

By Lemma 7.3 we could improve (2.2) to

$c|z^{\prime\prime}|^{2}-\bar{\delta}^{-1}\leq u(0,z^{\prime\prime},0)\leq
C|z^{\prime\prime}|^{2},$

which implies

(7.5) $c|z^{\prime\prime}|^{2}-\delta_{R}/\bar{\delta}\leq
u_{R}(0,z^{\prime\prime},0)\leq C|z^{\prime\prime}|^{2}\text{ for
}R\geq\bar{\delta}.$

At this moment, the above also holds when $a=0$.

Proof of Theorem 7.1: Due to Claim 7.2, it is sufficient to prove (7.3), which
will be completed by four steps. According to Theorem 5.1, we only need to
prove (7.3) in normal direction and on tangent plane.

Step 1. Let $E_{R}=S_{1}^{R}\setminus S_{\delta_{R}/\bar{\delta}}^{R}$, we
first prove that

(7.6)
$\begin{split}u_{R}(x)\geq\sigma_{0}(\max\\{|x_{1}|,|x_{n}|\\})-C\delta_{R}\text{
on }E_{R}\end{split}$

for some universal module $\sigma_{0}$.

Assume $q=(q^{\prime},0)=(q_{1},q^{\prime\prime},0)\in\mathcal{P}E_{R}$ with
$q_{1}\neq 0$, let

$\mathcal{B}x=x-\sum_{i=2}^{n-1}\frac{q_{i}}{q_{1}}x_{1}e_{i}$

and consider the positive function

$\upsilon(x)=\frac{u(\frac{q}{2}+\frac{q_{1}\mathcal{B}^{-1}x}{8C})}{q_{1}^{2}},\
x\in B_{1}^{+}(0).$

By the convexity, (7.5) and the assumption on $u$, we have

$0\leq\upsilon(x)\leq\frac{C}{q_{1}^{2}}$

and

$\upsilon(x_{1},x^{\prime\prime},0)\leq\frac{u(q)}{q_{1}^{2}}+C(|x^{\prime\prime}|^{2}+\frac{\delta_{R}}{q_{1}^{2}})\leq
C|x^{\prime\prime}|^{2}+\frac{u(q)+C\delta_{R}}{q_{1}^{2}}.$

The Neumann boundary value condition guarantees

$D_{n}\upsilon(x_{1},x^{\prime\prime},0)\geq-C.$

Regarding $\frac{u(q_{n})+C\delta_{R}}{q_{1}^{2}}$ as small constant and
$q_{1}^{-1}$ as universal constant, we use Lemma 4.1 to obtain

$\begin{split}\frac{u(q)+C\delta_{R}}{q_{1}^{2}}\geq\sigma(\frac{C}{q_{1}^{2}}).\end{split}$

Letting $\sigma_{0}(t)=t^{2}{\sigma}(C/t^{2})$, we get

(7.7) $u(q^{\prime},0)\geq\hat{\sigma}_{0}(q_{1})-C\delta_{R},\ \
\forall(q^{\prime},0)\in\mathcal{P}E_{R}.$

Next, assume $q=q_{n}e_{n}$ and consider the positive function

$v(x)=\frac{u(\frac{3q}{4}+\frac{q_{n}x}{8C})}{q_{n}^{2}},\ x\in B_{1}(0).$

Similarly, we have

$0\leq v(x)\leq\frac{C}{q_{n}^{2}}\text{ and }v(0,x^{\prime\prime},x_{n})\leq
C|x^{\prime\prime}|^{2}+\frac{u(q)+C\delta_{R}}{q_{n}^{2}}.$

Using Lemma 4.2, we get

(7.8) $u(q_{n}e_{n})\geq\sigma_{0}(q_{n})-C\delta_{R}$

for $\sigma_{0}(t)=t^{2}{\sigma}(C/t^{2})$.

Since $S_{t}$ is of good shape for $t\geq\delta_{R}/\bar{\delta}$, combining
(7.7) and (7.8), we have proved the desired (7.6).

Step 2. We first choose a large $R_{0}$ such that $\bar{\delta}R_{0}^{-1}$ is
small enough, and define $\delta_{R}=(\bar{\delta}R^{-1}$. Then
$\delta_{R}\leq\delta R_{0}$ for $R\geq R_{0}$.

For simplicity, we only consider the point $x\in E:=S_{1}^{R}(0)\setminus
S_{\delta_{R}}^{R}(0)$ and assume $t\geq\delta_{R}$ in the rest of this Step.
By iteration, (7.6) means that there exists universal $\gamma\in(0,1)$ such
that

$u_{R}(x)\geq c\max\\{|x_{1}|,|x_{n}|\\}^{\frac{1+\gamma}{\gamma}},$

which is

$\max\\{d_{1}^{R}(t),d_{n}^{R}(t)\\}\leq Ct^{\frac{\gamma}{1+\gamma}}.$

Recalling (7.4) and $\Pi_{i=1}^{n}d_{i}^{R}(t)=t^{\frac{n}{2}}$, we have

$\min\\{d_{1}^{R}(t),d_{n}^{R}(t)\\}\geq ct^{\frac{1}{1+\gamma}}.$

This gives (7.3) in the normal direction.

Step 3. We now have

(7.9) $ct^{\frac{1}{1+\gamma}}\leq d_{1}^{R}(t)\leq
Ct^{\frac{\gamma}{1+\gamma}},$

whose right side gives

$u_{R}(x^{\prime},0)\geq c|x_{1}|^{\frac{1+\gamma}{\gamma}}.$

Noticing (7.5) and that $u_{R}$ is locally Lipschitz, we get

(7.10)
$u_{R}(x^{\prime},0)\geq\max\\{c|x^{\prime\prime}|^{2}-\frac{C\delta_{R}}{\delta}|x_{1}|,0\\}.$

Let
$E_{1}=\\{x^{\prime}\mathcal{P}(E):\;|x_{1}|\leq\frac{c^{2}}{4}|x^{\prime\prime}|^{2}\\}$
and $E_{2}=\mathcal{P}(E)\setminus E_{1}$. The above two inequalities implies
that

$\displaystyle u_{R}(x^{\prime},0)$
$\displaystyle\geq\max\\{c|x^{\prime\prime}|^{2}-C|x_{1}|,0\\}\chi_{E_{1}}+c|x_{1}|^{\frac{1+\gamma}{\gamma}}\chi_{E_{2}}$
$\displaystyle\geq\frac{c}{2}|x^{\prime\prime}|^{2}\chi_{E_{1}}+c|x_{1}|^{\frac{1+\gamma}{\gamma}}\chi_{E_{2}}$
$\displaystyle\geq c|x^{\prime}|^{\frac{2(1+\gamma)}{\gamma}},$

Which proves the right side of (7.3) on the tangent plane.

Step 4. Given constant $M$, we claim that there exist universal $\varepsilon$,
such that

$\sup\\{s|\,se_{1}\in G_{t}^{R}(0)\\}\geq M\varepsilon$

for some $t\in[\varepsilon,\frac{1}{2}]$. Then This claim, (7.5) and (5.8)
implies that

$\varepsilon^{-1}G_{\varepsilon}^{R}(0)\supset t^{-1}G_{t}^{R}(0)\supset
cMB_{1}(0)\supset cM\mathcal{P}(S_{1}^{R}(0)),$

which proves the left side of (7.3) on the tangent plane.

On the contrary to this claim, by the balance property of $G_{t}^{R}(0)$ we
get

$u_{R}(x_{1},0)\geq M^{-1}|x_{1}|\text{ for }x_{1}\geq CM\varepsilon.$

For simplicity, we may require the point $x\in
S_{1}^{R}(0)\setminus(S_{CM\varepsilon}^{R}(0)\cap\\{|x_{1}|\leq
2CM\varepsilon\\})$ and assume constant $t\geq CM\varepsilon$ in the rest of
this Step. Then by the convexity and (7.5) we have

$u_{R}(x_{1},x^{\prime\prime},0)\geq
2u_{R}(2x_{1},0,0)-u_{R}(0,-x^{\prime\prime},0)\geq
2M^{-1}|x_{1}|-C|x^{\prime\prime}|^{2}.$

This together with (7.10) implies that

$G_{t}^{R}(0)\subset\\{c(|x^{\prime\prime}|^{2}-t)\leq|x_{1}|\leq
CM(|x^{\prime\prime}|^{2}+t)\\}.$

Since $0\in G_{t}^{R}(0)$, the above relation then implies that

$G_{t}^{R}(0)\subset\\{|x_{1}|\leq CM^{4}t\\}.$

Thus, $d_{1}^{R}(t)\leq CM^{4}t$. Taking $t=\varepsilon$ is small enough, this
contradicts (7.9). In this way, we completes the proof of Theorem 7.1.

##  8\. Liouville Theorem

In this section, we will modify the techniques developed from the classical
$C^{1,\alpha}$ estimate [5] to prove Theorem 1.1. We always suppose that $u$
is a viscosity solution to problem (1.1) and the hypothesis of Theorem 1.1 is
satisfied. As we have said, we may assume (2.6). Recall the normalization
family $(u_{R},S_{1}^{R})$ defined by (5.16), which satisfies (5.17)-(5.22).

Consider the function

(8.1) $v_{R}(x)=D_{n}u_{R}(x)-a_{R}x_{1},$

and let

(8.2) $\begin{split}&m_{R}=\inf\\{\frac{v_{R}(x)}{x_{n}}|\,x\in
S_{1}^{R}(0)\\},\\\ &M_{R}=\sup\\{\frac{v_{R}(x)}{x_{n}}|\,x\in
S_{1}^{R}(0)\\},\\\ &\omega_{R}=\frac{M_{R}}{m_{R}}-1.\end{split}$

Our goal is to prove that

(8.3) $c\leq m_{R}\leq M_{R}\leq C$

and $\omega_{R}$ is strictly increasing in $R$. For this purpose, we need the
auxiliary function like those from (3.5) to (3.7). Given $y\in
G_{\frac{1}{2}}^{R}(0)$, let

(8.4) $u^{R,y}(x)=u_{R}(x)-u_{R}(y)-\nabla u_{R}(y)\cdot(y-x)$

and define the functions

(8.5)
$\begin{split}&w^{R,y}(x)=[u^{R,y}(x)-\frac{n}{2}x_{n}D_{n}u^{R,y}(x)],\\\
&w^{R,y,+}(x)=C_{1}[w^{R,y}(x)+C_{2}x_{n}],\\\
&w^{R,y,-}(x)=c_{1}[x_{n}-c_{2}w^{R,y}(x)].\end{split}$

In order to show that $w^{R,y,\pm}$ are supersolution/subsolution to the
following linearized problem (8.17)for $v_{R}$, we need the following two
lemmas.

######  Lemma 8.1.

Suppose that $u$ is a viscosity solution to the first equation of problem
(1.1) and (2.6) is satisfied. Assume that

(8.6) $B_{c}^{+}(0)\subset S_{1}^{u}(0)\subset
B_{C}^{+}(0)\;\text{and}\;||u||_{Lip(B_{C}(0))}\leq C.$

Define $h_{x}^{u}=\sup\\{h|\,S_{t}^{u}(x)\subset B_{1}^{+}(0),\ \forall
t<h\\}$. There exists small universal constants $\delta>0$, $\rho>0$ such that
if

(8.7) $c|x|^{\frac{1+\gamma}{\gamma}}-\delta\leq u(x)\leq
C|x|^{1+\gamma}+\delta,\ \ \forall x\in S_{1}^{u}(0),$

then

$h_{x}^{u}\geq\sigma_{0}(x_{n})\text{ in }B_{\rho}^{+}(0).$

###### Proof.

If the result were false, we could find a constant $\tau>0$, a sequence of
convex viscosity solutions $u_{k}$ satisfying (8.6)-(8.7) and points $p_{k}\in
B_{\rho}(0)$ such that

$p_{k}\cdot e_{n}\geq\tau,\ \ \text{and}\;h_{p_{k}}^{u_{k}}\to 0.$

Letting $k\to\infty$, we get a limit $v$ which still satisfies our assumption
and $p_{k}$ converges to a point $p$ such that $h_{p}^{v}=0$. Let
$\Sigma:=S_{0}^{v}(p)$. Then $\Sigma$ is not a single point set. (8.7) implies
that $\Sigma\subset B_{\delta_{1}}(0)$ for some constant $\delta_{1}>0$
depending on $\delta,\rho$. Hence we can find an interior extremal point

$q\in\\{y|\,y\in\Sigma,y_{n}=\sup_{x\in\Sigma}x_{n}\\},$

which contradicts Lemma 2.14. ∎

######  Lemma 8.2.

For large $R\geq R_{0}$, one has

(8.8) $\sigma_{0}(x_{n})\leq v_{R}(x)\leq C\text{ in }S_{1}^{R}(0).$

###### Proof.

As we have discussed in Claim 7.2, we need only to prove (8.8) for $y\in
S_{\sigma_{0}(|c_{0}|)}^{R}(0)\subset B_{c_{0}}(0)$ for a small universal
$c_{0}$. According to Lemma 2.5, the Neumann boundary value condition and
(7.3) provides a $C^{1}$ module for $u_{R}$ at $0$. This is,

(8.9) $|\nabla u(x)|\leq C|x|^{\gamma},\;\text{for}\;x\;\text{near}\;0,$

which implies the right side of (8.8).

As in Lemma 8.1 we denote

(8.10) $h_{y}^{R}=\sup\\{h|\,S_{t}^{R}(y)\subset\mathbb{R}_{+}^{n},\ \forall
t<h\\}.$

We will find a universal $C^{0}$ module $\sigma_{0}$ such that the following
three inequalities hold in $S_{1}^{R}(0)$:

(8.11) $h_{x}^{R}\geq\sigma_{0}(x_{n}),$ (8.12)
$\sigma_{0}(x_{n})\mathcal{I}\leq
D^{2}u_{R}(x)\leq\frac{1}{\sigma_{0}(x_{n})}\mathcal{I},$

and

(8.13) $v_{R}(x)\geq\sigma_{0}(x_{n}).$

Again, it is sufficient to prove (8.10)-(8.12) for $x$ around $0$. Suppose
$c_{0}>0$ is small and $y\in S_{c_{0}}^{R}(0)$, then (7.3) and (8.9) means

$|\nabla u_{R}(y)|+u_{R}(y)\leq c_{1}:=c_{0}+Cc_{0}^{\gamma}$

and $c_{1}$ is still small. Thus, the section $S_{h_{y}^{R}}^{R}(y)\cap
B_{C}(0)$ is contained in $S_{c_{1}}^{R}(0)\subset S_{\frac{1}{2}}^{R}(0)$ and
will touch $\partial\mathbb{R}_{+}^{n}$ at some point $z=(z^{\prime},0)\in
G_{c_{1}}^{R}(0)$.

Let $l_{zy}$ be the line segment connecting $z$ and $y$. Extend $l_{zy}$ such
that it intersects $\partial B_{{c}}^{+}(0)$ at point $q$. Set

$w(x)=u_{R}(x)-u_{R}(y)-\nabla u_{R}(y)\cdot(x-y).$

Then

(8.14) $w(x)\leq h_{y}^{R},\ x\in l_{yz}.$

Furthermore, it follows from Theorem 7.1 and Lemma 8.1 that

(8.15) $h_{y}^{R}\geq\sigma_{0}(y_{n})$

for $\sigma_{0}(t):=ct^{\frac{2}{n}}\sigma(ct^{-\frac{2}{n}})$.

The classical estimate result (see Lemma 2.11) gives
$\operatorname{Vol}(_{h_{y}^{R}}^{R}(y))\sim(h_{y}^{R})^{\frac{n}{2}}$. Notice
that $S_{h_{y}^{R}}^{R}\subset B_{C}^{+}(0)$, thus $S_{h_{y}^{R}}^{R}\supset
B^{+}_{c(h_{y}^{R})^{\frac{1}{2}}}(0)$ and the classical interior estimate
(See Lemma 2.13 and its proof) then gives

(8.16) $\hat{\sigma}_{0}(y_{n})\mathcal{I}\leq
D^{2}u_{R}(y)\leq\frac{1}{\hat{\sigma}_{0}(y_{n})}\mathcal{I}$

for
$\hat{\sigma}_{0}(y_{n})=\min\\{[\sigma_{0}(y_{n})]^{n-2},\sigma_{0}(y_{n})\\}$.

Now, up to a bounded linear transform which maps in $S_{\sigma_{0}}^{R}(0)$ to
$S_{1}^{\sigma_{0}R}$, (8.15)-(8.16) and the statement in Claim 7.2 implies
that (8.11) and (8.12) in $S_{1}^{R}(0)$.

Finally, let us prove the (8.13). Similarly as Lemma 3.4, the non-negative
function $v_{R}(x)$ in (8.1) solves the linearized problem:

(8.17) $\begin{cases}U_{R}^{ij}D_{ij}v_{R}=0&\text{ in }S_{1}^{R}(0)\\\ 0\leq
v_{R}\leq C&\text{ in }S_{1}^{R}(0)\\\ v_{R}=0&\text{ on
}{G_{1}^{R}(0)}\end{cases}.$

Notice that we also have $v_{R}(ce_{n})\geq c$ by the convexity. Due to
(8.12), the first equation in (8.17) is uniformly elliptic with norm of the
elliptic coefficients depending on $\sigma_{0}(y_{n})$ in the domain
$E(y_{n})=S_{1}^{R}(0)\cap\\{x_{n}\geq\frac{y_{n}}{2}\\}$. We apply the
classical Harnack inequality to obtain

(8.18) $v_{R}(y^{\prime},y_{n})\geq
C(||U_{R}^{ij}||_{E(y_{n})},||U_{R}^{ij,-1}||_{E(y_{n})},\frac{y_{n}}{2})v_{R}(ce_{n})\geq\sigma_{0}(y_{n})$

in $S_{\frac{1}{2}}\cap\\{x_{n}\geq\frac{y_{n}}{2}\\}$ for some new
$\sigma_{0}$. Thus, the proof is completed.

∎

Proof of Theorem 1.1: It follows from Lemma 8.2 that $v_{R}(x)$ satisfies the
following problem:

(8.19) $\begin{cases}U_{R}^{ij}D_{ij}v_{R}=0&\text{ in }S_{1}^{R}(0)\\\
\sigma_{0}(x_{n})\leq v_{R}\leq C&\text{ on }\partial S_{1}^{R}(0)\setminus
G_{1}^{R}(0)\\\ v_{R}=0&\text{ on }{G_{1}^{R}(0)}\end{cases}.$

Then as Corollary 3.7 and 3.8 we see that the function $w^{R,y,\pm}(x)$ in
(8.5) are supersolution/subsolution of problem (8.19) in $S_{c}^{R}(0)$, and
$cx_{n}\leq v_{R}(x)\leq Cx_{n}$ in $S_{c}^{R}(0)$. Using the definition
(8.2), we get

$c\leq m_{R}\leq M_{R}\leq C,$

which is the desired (8.3).

Next, we are going to prove

(8.20) $\omega_{\rho R}\leq(1-\theta)\omega_{R}$

for some universal $\rho>0$ and $\theta\in(0,1)$.

Let

$\begin{split}&m_{R,t}=\inf\\{\frac{v_{R}(x)}{x_{n}}|\,x\in
S_{t}^{R}(0)\\},\\\ &M_{R,t}=\sup\\{\frac{v_{R}(x)}{x_{n}}|\,x\in
S_{t}^{R}(0)\\},\\\ &\omega_{R,t}=\frac{M_{R,t}}{m_{R,t}}-1.\end{split}$

It is clear that for all $t\in(0,1)$,

$m_{R,1}\leq m_{R,t}\leq M_{R,t}\leq M_{R,1}$

and

$\omega_{R,t}=\omega_{Rt}.$

To prove (8.20), it is sufficient to prove

(8.21) $\omega_{R,\rho}\leq(1-\theta)\omega_{R,1}$

for some universal $\rho>0$ and $\theta\in(0,1)$. Assume $\omega_{R,1}>0$ and
consider the two non-negative functions:

(8.22)
$\begin{split}H_{1}(x)=\frac{v_{R}(x)-m_{R,1}x_{n}}{M_{R,1}-m_{R,1}},\\\
H_{2}(x)=\frac{M_{R,1}x_{n}-v_{R}(x)}{M_{R,1}-m_{R,1}}.\end{split}$

It follows from (8.19) that each $H_{i}$ $(i=1,2)$ solves

(8.23) $\begin{cases}U_{R}^{ij}D_{ij}H_{i}=0&\text{ in }S_{1}^{R}(0)\\\ 0\leq
H_{i}\leq 1&\text{ in }S_{1}^{R}(0)\\\ H_{i}=0&\text{ on
}{G_{1}^{R}(0)}\end{cases}.$

While $H_{1}+H_{2}=x_{n}$, we may assume $H_{1}(ce_{n})\geq\frac{c}{2}$
without loss of generality.

Since (8.12) means that the first equation in (8.23) is universal elliptic
away from $G_{1}^{R}(0)$, as (8.18) we have

(8.24) $\begin{split}H_{1}(x)\geq\sigma_{0}(x_{n})H_{1}(ce_{n})\text{ in
}B_{c/2}^{+}(0).\end{split}$

Again, for any $y=(y^{\prime},0)\in G_{c_{0}}^{R}(0)$ with $c_{0}>0$ small
enough, we could find a universal constant $\bar{c}>0$ such that
$B_{c(\bar{c})/2}^{+}(0)\subset S_{\bar{c}}^{R}(y)\subset B_{c/2}^{+}(0)$.

With some universal constant $C_{1},C_{2},c_{1},c_{2}$ depending on $\bar{c}$
and $\sigma_{0}(x_{n})$, the function $w^{R,y,\pm}(x)$ in (8.5) is still
supersolution/subsolution of $H_{1}$ to the following problem:

(8.25) $\begin{cases}U_{R}^{ij}D_{ij}H_{1}=0&\text{ in }S_{\bar{c}}^{R}(y)\\\
\sigma_{0}(x_{n})\leq H_{1}\leq 1&\text{ on }\partial
S_{\bar{c}}^{R}(y)\setminus G_{\bar{c}}^{R}(y)\\\ H_{1}=0&\text{ on
}G_{\bar{c}}^{R}(y)\end{cases}.$

In conclusion, we have

(8.26) $H_{1}(y+te_{n})\geq w^{R,y,-}(y+te_{n})\geq\frac{c_{1}}{2}t\;\text{
for }\;t\leq t_{0}.$

On the other hand, it follows from (7.2) (or (7.3)) that

(8.27) $S_{\rho}^{R}(0)\subset G_{c_{0}}^{R}(0)\times[0,t_{0})\;\text{
for}\;\rho=c\min\\{c_{0}^{\frac{(1+\gamma)^{2}}{\gamma}},t_{0}\\}.$

Combining (8.26) and (8.27), we have

$\frac{v_{R}(x)-m_{R,1}x_{n}}{M_{R,1}-m_{R,1}}\geq\theta x_{n}\;\text{ in
}\;S_{\rho}^{R}(0)$

for $\theta=\frac{cc_{1}}{2}$. That is

$m_{R,\rho}\geq m_{R,1}+\theta(M_{R,1}-m_{R,1})\;\text{ in
}\;S_{\rho}^{R}(0),$

which yields the desired (8.21). In this way, (8.20) has been proved.

By (8.3) and (8.20) we see that

$\omega_{R}=\omega_{\rho^{k}\rho^{-k}R}\leq(1-\theta)^{k}\omega_{R\rho^{-k}}\leq
C(1-\theta)^{k}.$

Letting $k\to 0$, we get $\omega_{R}=0$ for any $R$. Hence,
$v_{R}(x)=D_{n}u_{R}-a_{R}x_{1}$ is a linear function in $S_{R}(0)$. Then
letting $R\to\infty$, we get the limit function
$\hat{v}(x):=D_{n}u(x)-ax_{1}=cx_{n}$ for some $c>0$.

Lemma 3.1 means that $u$ is strict convex and smooth away from
$\partial\mathbb{R}_{+}^{n}$. Thus

$u(x)=f(x^{\prime})+ax_{1}x_{n}+cx_{n}^{2}$

for some smooth function $f$ in $\mathbb{R}^{n-1}$. Let

$g(x^{\prime})=f(x^{\prime})-\frac{a^{2}}{4c}x_{1}^{2}.$

Since $D^{2}u(x^{\prime},1)>0$, we have $D^{2}g(x^{\prime})>0$. Therefore, $g$
is a convex, smooth function solving

$\operatorname{det}D^{2}g=1\;\text{ in }\;\mathbb{R}^{n-1}.$

The classical results (Theorem 2.15) shows that $g$ is quadratic polynomial of
$x^{\prime}$, which implies that $u$ is a quadratic polynomial of $x$. The
proof of Theorem 1.1 is completed.

######  Remark 8.3.

In the case $n=2$, there are two other proofs for Theorem 1.1, which depends
on the strict convex Lemma 4.1 and 4.2. With the two lemmas, one can show the
solution is $C^{1}$ up to boundary and is of super-linearity growth at
infinite. This fact, together with partial Legendre transform and the
Liouville theorem for harmonic functions, will gives us the first proof. As
for the second proof, we need to use Wang’s function[26] to construct
supersolution/subsolution and so prove that the $u$ is $C^{1,1}$ up to
boundary, then by blowing up at infinity to obtain $D^{2}u=c\mathcal{I}$.

As the end of this paper, we give the two proofs in more details.

Proof 1. We write a point in $\mathbb{R}^{2}$ as $(x,y)$ and use the classical
partial Legendre transform in the $x$-variable. For any $p\in R$, define

$u^{*}\left(p,y\right):=\sup_{x\in R}\\{px-u\left(x,y\right)\\}.$

The uniformly strict convexity of $u$ at boundary and the smoothness of $u$ at
interior also implies the superlinearity at infinite, which guarantees the
uniqueness and existence of $x=X\left(p,y\right)$ such that the above supremum
is attained.

By a calculation the two dimensional case of problem (1.1),

$\operatorname{det}D^{2}u=1\;in\;\mathbb{R}_{+}^{2},\ \
D_{2}u\left(x,0\right)=ax\;on\;R$

is turned to the problem for $u^{*}$:

$\begin{cases}\Delta u^{*}=1&\text{ in }\mathbb{R}_{+}^{2}\\\
D_{v^{*}}u^{*}=0&\text{ on }\mathbb{R}\end{cases},$

where $v^{*}=e_{2}+ae_{1}$. Differentiating $u^{*}$ with respect to $v^{*}$
and letting $V=D_{v^{*}}u^{*}$ we obtain

$\begin{cases}\Delta V=0&\text{ in }\mathbb{R}_{+}^{2}\\\ V=0&\text{ on
}\mathbb{R}\end{cases}.$

Note that

$V=D_{v^{*}}u^{*}\left(p,y\right)=aX\left(p,y\right)-u_{2}\left(X,y\right)=u_{2}\left(X,0\right)-u_{2}\left(X,y\right)\leq
0.$

We can use the Liuoville theorem for harmonic functions to see that the only
solutions to above problem are

$V(x,y)=D_{v^{*}}u^{*}\left(x,y\right)\equiv A$

for some constant $A$. Then

$u^{*}\left(p,y\right)=g\left(p-ay\right)+Ay$

for some function $g:[0,\infty)\to R$. Recalling $\Delta
u^{*}=1,D_{v^{*}}u^{*}=0$, we have

$g\left(0\right)=0\ and\ \left(1+a^{2}\right)g^{\prime\prime}=1,$

which implies that $u^{*}$ is a quadratic polynomial, and so is the $u$.

Proof 2.  As we have said, when $n=2$, one could use strict convexity Lemma
4.1 and 4.2 to show that the solution $u$ to problem (1.1) is uniformly
$C^{1,\alpha}$ up to boundary[5].

We consider Wang’s singular function with bounded positive Monge-Ampépre
measure for large $a>>2$:

$W_{a,b}\left(p,q\right)=\begin{cases}p^{a}+\frac{a^{2}-1}{a(a-2)}p^{a-\frac{2a}{b}}q^{2},\
|p|^{a}\geq|q|^{b}\\\
\frac{4a-5}{2(a-2)}q^{b}+\frac{1}{2a}q^{b-\frac{2b}{a}}p^{2},\
|q|^{b}\geq|p|^{a}\end{cases}.$

Let

$W\left(x_{1},x_{2}\right)=CaW_{a,b}\left(\tau x_{1},x_{2}/\tau\right),$

where $\tau>0$ is small and satisfies $a\tau^{a}\leq c$. One can use this
function to construct a supersolution $(W^{+})$/subsolution $(W^{-})$at
boundary point:

$\begin{cases}w^{+}\left(x\right)=u\left(x\right)-W\left(x_{1},x_{2}\right)+Mx_{2}\\\
w^{-}\left(x\right)=\epsilon_{1}\left(W\left(x_{1},x_{2}\right)-u\left(x\right)+\epsilon_{2}x_{2}\right)\end{cases}.$

This will give the uniform estimate $c\leq D_{nn}u\leq C$ on the boundary.
Therefore $u$ is $C^{1,1}$ and then in $C^{2,\alpha}$. After blowing up at
infinite, this together with the superlinearity will imply
$D^{2}u=c\mathcal{I}$ and the proof is completed.

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