# Interior estimates for the Monge-Ampère type
fourth order equations

Ling Wang and Bin $\text{Zhou}^{*}$ School of Mathematical Sciences, Peking
University, Beijing 100871, China<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

In this paper, we give several new approaches to study interior estimates for
a class of fourth order equations of Monge-Ampère type. First, we prove
interior estimates for the homogeneous equation in dimension two by using the
partial Legendre transform. As an application, we obtain a new proof of the
Bernstein theorem without using Caffarelli-Gutiérrez’s estimate, including the
Chern conjecture on affine maximal surfaces. For the inhomogeneous equation,
we also obtain a new proof in dimension two by an integral method relying on
the Monge-Ampère Sobolev inequality. This proof works even when the right hand
side is singular. In higher dimensions, we obtain the interior regularity in
terms of integral bounds on the second derivatives and the inverse of the
determinant.

###### Key words and phrases:

Monge-Ampère equation, linearized Monge-Ampère equation, fourth order
equation, partial Legendre transform, degenerate elliptic equations

###### 2020 Mathematics Subject Classification:

35J30, 35J96, 35B45, 35B65

*This research is partially supported by National Key R$\&$D Program of China SQ2020YFA0712800 and NSFC grants 11822101.

##  1\. Introduction

We study the regularity of the following fourth order equations of Monge-
Ampère type

(1.1) $U^{ij}w_{ij}=f,$

where $\\{U^{ij}\\}$ is the cofactor matrix of $D^{2}u$ of an unknown
uniformly convex function, and

(1.2) $w=\begin{cases}\ [\det D^{2}u]^{-(1-\theta)},&\theta\geq 0,\ \
\theta\neq 1,\\\\[4.0pt] \ \log\det D^{2}u,&\theta=1.\end{cases}$

When $\theta=\frac{1}{n+2}$, this is the affine mean curvature equation in
affine geometry [Ch]. When $\theta=0$, it is Abreu’s equation arising from the
problem of extremal metrics on toric manifolds in Kähler geometry [Ab], and is
equivalent to

$\sum_{i,j}\frac{\partial^{2}u^{ij}}{\partial x_{i}\partial x_{j}}=f,$

where $\\{u^{ij}\\}$ is the inverse matrix of $D^{2}u$. The regularity of
(1.1) has been extensively studied before, see [TW1, TW2, D, Z1, Z2, CHLS,
Le1, Le2, CW]. This equation is usually treated as a system of a Monge-Ampère
equation and a linearized Monge-Ampère equation. Therefore, in previous works,
its regularity relies heavily on Caffarelli-Gutiérrez’s deep result on the
interior regularity of the linearized Monge-Ampère equation [CG], which was
later extended by [LS, GN1, GN2] to the the boundary and to higher order
estimates. In this paper, we investigate the interior estimates of (1.1) by
several new approaches. We will mainly concentrate on the case
$\theta\in[0,1]$ due to the interesting geometric background.

We first consider the case of the homogeneous equation

(1.3) $U^{ij}w_{ij}=0,$

where $w$ is given by (1.2). We apply the partial Legendre transform to give a
new proof of the interior estimates of (1.3) in dimension two.

######  Theorem 1.1.

Assume $n=2$ and $\theta\in[0,1]$. Let $\Omega\subset\mathbb{R}^{2}$ be a
convex domain and let $u$ be a smooth convex solution to equation (1.3) on
$\Omega$ satisfying

(1.4) $0<\lambda\leq\det D^{2}u\leq\Lambda.$

Then for any $\Omega^{\prime}\Subset\Omega$, there exists a constant $C>0$
depending on $\sup_{\Omega}|u|$, $\lambda$, $\Lambda$, $\theta$ and
$dist(\Omega^{\prime},\partial\Omega)$, such that

$\|u\|_{C^{4,\alpha}(\Omega^{\prime})}\leq C.$

The partial Legendre transform for the fourth order equation was first used in
[LZ] where the authors deal with the second boundary value problem. After the
partial Legendre transform, equation (1.3) becomes a quasi-linear second order
equation (see (2.3)) for the determinant. The main ingredient in our proof is
an interior integral gradient estimate (Theorem 2.2). When
$\theta\in\left[0,\frac{1}{4}\right]$, condition (1.4) holds by the
determinant estimates and arguments of strict convexity [TW1, Z1]. By Theorem
1.1 and a rescaling argument as in [TW1, Theorem 2.1], we obtain a new proof
of the following Bernstein theorem [TW1, JL, Z1] without using Caffarelli-
Gutiérrez’s theory.

######  Theorem 1.2.

Assume $n=2$ and $0\leq\theta\leq\frac{1}{4}$. Let $u$ be an entire smooth
uniformly convex solution to (1.3) on $\mathbb{R}^{2}$. Then $u$ is a
quadratic polynomial.

In the case of the inhomogeneous equation and in higher dimensions, the
partial Legendre transform does not work. We will investigate the interior
regularity by an integral method motivated by De Giorgi-Nash-Moser’s theory.
Consider the inhomogeneous equation with general right hand side term

(1.5) $U^{ij}w_{ij}=f(x,u,Du,D^{2}u),$

where $w$ given by (1.2) with $\theta\in[0,1]$. This equation is introduced by
[Le4, Le5] in the study of convex functionals with a convexity constraint
related to the Rochet-Choné model for the monopolist problem in economics. It
is said to be singular since the right hand side term depends on $D^{2}u$. A
typical example considered in [Le4, Le5, LZ] is

(1.6) $f=\mbox{div}\,(|\nabla u|^{p-2}\nabla u)+f^{0}(x,u).$

Note that when $f\in L^{\infty}(\Omega)$, once we have the determinant
estimate (1.4), we can use Caffarelli-Gutiérrez’s theory to get the interior
regularity. The assumption on $f$ can be weakened to $f\in
L^{\frac{n}{2}+\varepsilon}(\Omega)$ by [LN]. Hence for $f$ defined as (1.6),
when $n=2$ and $p\geq 2$, we can obtain the interior regularity of $u$
directly by using interior $W^{2,1+\varepsilon}$-estimates of the Monge-Ampère
equations [DFS, S]. To settle the more singular case $1<p<2$, Le established
the interior estimate of the linearized Monge-Ampère equation with right hand
side term in divergence form in dimension two [Le3]. One of the main tools in
[Le3] are the Monge-Ampère Sobolev inequality (see Lemma 3.2). In this paper,
we will use the Monge-Ampère Sobolev inequality and the
$W^{2,1+\varepsilon}$-estimates for the Monge-Ampère equation directly in the
fourth order equation to obtain a $W^{2,p}$-estimate of the solution (Theorem
3.3). Then we can apply the regularity theory of second order elliptic
equation of divergence type to obtain a new proof for the interior estimates
of (1.5).

######  Theorem 1.3.

Assume $n=2$ and $\theta\in[0,1]$. Let $\Omega\subset\mathbb{R}^{2}$ be a
convex domain. Assume

$f=\mbox{div}\,(g)+h$

where $g:=\left(g^{1}(x),g^{2}(x)\right):\Omega\to\mathbb{R}^{2}$ is a bounded
vector function and $h\in L^{q}(\Omega)$ for some $q>\frac{n}{2}$. Suppose $u$
is a smooth convex solution to equation (1.5) on $\Omega$ satisfying (1.4).
Then for any $\Omega^{\prime}\Subset\Omega$, there exists a constant $C>0$
depending on $\sup_{\Omega}|u|$, $\lambda$, $\Lambda$, $\theta$,
$\|g\|_{L^{\infty}(\Omega)}$, $\|h\|_{L^{q}(\Omega)}$ and
$dist(\Omega^{\prime},\partial\Omega)$, such that

$\|u\|_{W^{4,q}(\Omega^{\prime})}\leq C.$

######  Remark 1.4.

(1) It is clear that the above theorem applies to the case (1.6) for any $p>1$
in dimension two. The higher dimensional case is still open.

(2) When $f\in W^{1,q}(\Omega)$ and $\theta=0$ or $1$, we give another new
proof in Section 3 inspired by [CC] for the complex setting. More precisely,
we get a $C^{2}$-estimate of $u$ in terms of the $W^{2,p}$-bound (Theorem
3.4), which makes (1.5) become a uniformly elliptic equation. Then the
classical theory of uniformly elliptic equations can be applied.

In higher dimensions, the interior regularity and the Bernstein theorem are
still widely open. In fact, according to the counterexample in [TW1] for the
affine maximal surface equation, there may be no interior estimates if no
further assumptions are made. More precisely, (1.4) may not hold. We give a
partial result by assuming integral bounds on the second derivatives and the
inverse of the determinant.

######  Theorem 1.5.

Let $\Omega\subset\mathbb{R}^{n}$ be a convex domain and let $u$ be a smooth
uniformly convex solution to equation (1.3) with $\theta\in[0,1]$ on $\Omega$.
Assume that $p,q>0$ satisfy $\frac{n}{p}+\frac{1}{q}<\frac{2}{n}$. Then for
any $\Omega^{\prime}\Subset\Omega$, there exists a constant $C$, depending
only on $p$, $q$, $\theta$, $\sup_{\Omega}|u|$,
$\|u\|_{W^{2,p}\left(\Omega\right)}$, $\left\|(\det
D^{2}u)^{-1}\right\|_{L^{q}\left(\Omega\right)}$ and
$dist(\Omega^{\prime},\partial\Omega)$, such that

$\|u\|_{C^{4,\alpha}(\Omega^{\prime})}\leq C.$

As an application, we obtain a Liouville type theorem in higher dimensions.

######  Corollary 1.6.

Let $u$ be an entire smooth uniformly convex solution to (1.1) with
$\theta\in[0,1]$ on $\mathbb{R}^{n}$. Suppose there are $p,q,C>0$ such that
$\frac{n}{p}+\frac{1}{q}<\frac{2}{n}$ and

(1.7) $\int_{B_{R}(0)}|D^{2}u|^{p}+(\det D^{2}u)^{-q}\,dx\leq
CR^{n}\,,\quad\forall\,R>0.$

Then $u$ is a quadratic polynomial.

The structure of the paper is as follows. In Section 2.1, we apply the partial
Legendre transform to (1.1) in dimension two to derive a new equation. The key
interior gradient estimate (Theorem 2.2) for the new equation is established
in Section 2.2. Then we prove Theorem 1.1 in Section 2.3 with this key
estimate. In Section 3, we first derive the $W^{2,p}$-estimate of $u$ (Theorem
3.3), and then we prove Theorem 1.3. Section 4 is devoted to some study on
interior regularity in higher dimensions.

Acknowledgments. The authors would like to thank Guohuan Qiu for some valuable
discussions.

##  2\. The homogeneous equation in dimension two

In this section, we present a new proof for the interior estimate for the
homogeneous equation without Caffarelli-Gutiérrez’s theory.

###  2.1. The new equation under partial Legendre transform

We first focus on the dimension two case. Write $u(x)=u(x_{1},x_{2})$. The
partial Legendre transform in the $x_{1}$-variable is

(2.1) $u^{\star}(\xi,\eta)=x_{1}u_{x_{1}}(x_{1},x_{2})-u(x_{1},x_{2}),$

where

$y=(\xi,\eta)=\mathcal{P}(x_{1},x_{2}):=(u_{x_{1}},x_{2})\in\mathcal{P}(\Omega)=:\Omega^{\star}.$

We have

$\frac{\partial(\xi,\eta)}{\partial(x_{1},x_{2})}=\begin{pmatrix}u_{x_{1}x_{1}}&\
\ u_{x_{1}x_{2}}\\\\[4.0pt] 0&\ \ 1\\\ \end{pmatrix},\quad\text{and }\
\frac{\partial(x_{1},x_{2})}{\partial(\xi,\eta)}=\begin{pmatrix}\frac{1}{u_{x_{1}x_{1}}}&\
\ -\frac{u_{x_{1}x_{2}}}{u_{x_{1}x_{1}}}\\\\[4.0pt] 0&\ \ 1\\\ \end{pmatrix}.$

Hence,

(2.2) $u^{\star}_{\xi}=x_{1},\ u^{\star}_{\xi\xi}=\frac{1}{u_{x_{1}x_{1}}},\
u^{\star}_{\eta}=-u_{x_{2}},\ u^{\star}_{\eta\eta}=-\frac{\det
D^{2}u}{u_{x_{1}x_{1}}},\
u^{\star}_{\xi\eta}=-\frac{u_{x_{1}x_{2}}}{u_{x_{1}x_{1}}}.$

The partial Legendre transform has been used widely in the study of the Monge-
Ampère equation [DS, F, GP, LS, Li]. Here we apply it to equation (1.3).

In order to derive the equation under the partial Legendre transform, we
consider the associated functionals of (1.3)

$A_{\theta}(u)=\begin{cases}\ \int_{\Omega}[\det
D^{2}u]^{\theta}\,dx,&\theta>0,\ \ \theta\neq 1,\\\\[4.0pt] \
\int_{\Omega}\log\det D^{2}u\,dx,&\theta=0,\\\\[4.0pt] \ \int_{\Omega}\det
D^{2}u\log\det D^{2}u\,dx,&\theta=1.\end{cases}$

The case of $\theta=0$ is essentially included in [LZ].

######  Proposition 2.1.

Let $u$ be a uniformly convex solution to (1.3) in $\Omega$. Then in
$\Omega^{\star}=\mathcal{P}(\Omega)$, its partial Legendre transform
$u^{\star}$ satisfies

(2.3)
$w^{\star}w^{\star}_{\xi\xi}+w^{\star}_{\eta\eta}+(\theta-1){w^{\star}_{\xi}}^{2}+\frac{\theta-2}{w^{\star}}{w^{\star}_{\eta}}^{2}=0,$

Here $w^{\star}=-\frac{u^{\star}_{\eta\eta}}{u^{\star}_{\xi\xi}}$.

###### Proof.

As

$\det D^{2}u=-\frac{u^{\star}_{\eta\eta}}{u^{\star}_{\xi\xi}},\ \
dxdy=u^{\star}_{\xi\xi}\,d\xi d\eta,$

we have

$\displaystyle A_{\theta}(u)$ $\displaystyle=$
$\displaystyle\int_{\Omega^{\star}}\left(-\frac{u^{\star}_{\eta\eta}}{u^{\star}_{\xi\xi}}\right)^{\theta}u^{\star}_{\xi\xi}\,d\xi
d\eta$ $\displaystyle=$
$\displaystyle\int_{\Omega^{\star}}(-u^{\star}_{\eta\eta})^{\theta}{u^{\star}_{\xi\xi}}^{1-\theta}\,d\xi
d\eta=:A_{\theta}^{\star}(u^{\star}),\ \theta\in(0,1);$ $\displaystyle
A_{0}(u)$ $\displaystyle=$
$\displaystyle\int_{\Omega^{\star}}\log\left(-\frac{u^{\star}_{\eta\eta}}{u^{\star}_{\xi\xi}}\right)u^{\star}_{\xi\xi}\,d\xi
d\eta=:A_{0}^{\star}(u^{\star});$ $\displaystyle A_{1}(u)$ $\displaystyle=$
$\displaystyle\int_{\Omega^{\star}}\left(-\frac{u^{\star}_{\eta\eta}}{u^{\star}_{\xi\xi}}\right)\log\left(-\frac{u^{\star}_{\eta\eta}}{u^{\star}_{\xi\xi}}\right)u^{\star}_{\xi\xi}\,d\xi
d\eta=:A_{1}^{\star}(u^{\star}).$

Since $u$ is maximal with respect to the functional $A_{\theta}$, $u^{\star}$
is maximal with respect to the functional $A_{\theta}^{\star}$. It suffices to
derive the Euler-Lagrange equation of $A_{\theta}^{\star}$. See [TW1, Z1] for
the case of the Legendre transform.

First, we consider $\theta\in(0,1)$. For $\varphi\in
C^{\infty}_{0}(\Omega^{\star})$, by integration by parts,

(2.4)
$\displaystyle\left.\frac{dA_{\theta}^{\star}(u^{\star}+t\varphi)}{dt}\right|_{t=0}$
$\displaystyle=$
$\displaystyle\int_{\Omega^{\star}}[-\theta(-u^{\star}_{\eta\eta})^{\theta-1}{u^{\star}_{\xi\xi}}^{1-\theta}\varphi_{\eta\eta}+(1-\theta)(-u^{\star}_{\eta\eta})^{\theta}(u^{\star}_{\xi\xi})^{-\theta}\varphi_{\xi\xi}]\,d\xi
d\eta$ $\displaystyle=$
$\displaystyle\int_{\Omega^{\star}}\\{-[\theta(-u^{\star}_{\eta\eta})^{\theta-1}{u^{\star}_{\xi\xi}}^{1-\theta}]_{\eta\eta}+[(1-\theta)(-u^{\star}_{\eta\eta})^{\theta}(u^{\star}_{\xi\xi})^{-\theta}]_{\xi\xi}\\}\varphi\,d\xi
d\eta.$

Denote $w^{\star}=-\frac{u^{\star}_{\eta\eta}}{u^{\star}_{\xi\xi}}$. Then the
equation, after the transformation, becomes

$-\theta({w^{\star}}^{\theta-1})_{\eta\eta}+(1-\theta)({w^{\star}}^{\theta})_{\xi\xi}=0.$

After simplification, this is (2.3). Similarly, for $\varphi\in
C^{\infty}_{0}(\Omega^{*})$,

$\displaystyle\left.\frac{dA_{0}^{\star}(u^{\star}+t\varphi)}{dt}\right|_{t=0}$
$\displaystyle=$
$\displaystyle\int_{\Omega^{\star}}-\frac{u^{\star}_{\xi\xi}}{u^{\star}_{\eta\eta}}\left(-\frac{\varphi_{\eta\eta}u^{\star}_{\xi\xi}-u^{\star}_{\eta\eta}\varphi_{\xi\xi}}{{u^{\star}_{\xi\xi}}^{2}}\right)u^{\star}_{\xi\xi}+\log\left(-\frac{u^{\star}_{\eta\eta}}{u^{\star}_{\xi\xi}}\right)\varphi_{\xi\xi}\,d\xi
d\eta$ $\displaystyle=$
$\displaystyle\int_{\Omega^{\star}}\frac{\varphi_{\eta\eta}u^{\star}_{\xi\xi}-u^{\star}_{\eta\eta}\varphi_{\xi\xi}}{u^{\star}_{\eta\eta}}+\log\left(-\frac{u^{\star}_{\eta\eta}}{u^{\star}_{\xi\xi}}\right)\varphi_{\xi\xi}\,d\xi
d\eta,$

and the equation, after the transformation, becomes

$-[(w^{\star})^{-1}]_{\eta\eta}+(\log w^{\star})_{\xi\xi}=0.$

After simplification, we obtain (2.3). Finally,

$\displaystyle\left.\frac{dA_{1}^{\star}(u^{\star}+t\varphi)}{dt}\right|_{t=0}$
$\displaystyle=$ $\displaystyle\int_{\Omega^{\star}}(1+\log
w^{\star})\left(-\frac{\varphi_{\eta\eta}u^{\star}_{\xi\xi}-u^{\star}_{\eta\eta}\varphi_{\xi\xi}}{{u^{\star}_{\xi\xi}}^{2}}\right)u^{\star}_{\xi\xi}+w^{\star}\log
w^{\star}\varphi_{\xi\xi}\,d\xi d\eta$ $\displaystyle=$
$\displaystyle-\int_{\Omega^{\star}}(1+\log
w^{\star})(w^{\star}\varphi_{\xi\xi}-\varphi_{\eta\eta})+w^{\star}\log
w^{\star}\varphi_{\xi\xi}\,d\xi d\eta.$

Then the equation after transformation is

$-w^{\star}_{\xi\xi}-(\log w^{\star})_{\eta\eta}=0,$

which is equivalent to (2.3). ∎

###  2.2. The interior gradient estimate of (2.3)

For simplicity, we change notations in this section and write (2.3) as

(2.5) $uu_{xx}+u_{yy}=(1-\theta)u_{x}^{2}+\frac{2-\theta}{u}u_{y}^{2}.$

It is easy to see that this is a quasi-linear equation with the right-hand
side depending on the gradient. We prove the following interior gradient
estimate.

######  Theorem 2.2.

Assume $u$ is a solution to (2.5) with $\theta\in[0,1]$ on $B_{R}:=B_{R}(0)$
and satisfies $0<\lambda\leq u\leq\Lambda$. Then there exist $\alpha,C>0$
depending on $\lambda$, $\Lambda$, $R$ and $\theta$, such that

(2.6) $\int_{B_{R}}|\nabla u|^{3}(R^{2}-x^{2}-y^{2})^{\alpha}\,dV\leq C.$

###### Proof.

Let $w=v\phi\eta$, where

$v=\sqrt{u_{x}^{2}+u_{y}^{2}+1},\ \eta=(R^{2}-x^{2}-y^{2})^{\alpha},\
\alpha>3$

and $\phi=\phi(u)$ is a positive function of $u$ to be determined later.

A direct calculation yields

(2.7) $\displaystyle uw_{xx}+w_{yy}$ $\displaystyle=$
$\displaystyle(uv_{xx}+v_{yy})\phi\eta+2[uv_{x}(\phi\eta)_{x}+v_{y}(\phi\eta)_{y}]+[u(\phi\eta)_{xx}+(\phi\eta)_{yy}]v.$

It is clear that

(2.8) $\displaystyle v_{x}$ $\displaystyle=$ $\displaystyle
v^{-1}(u_{x}u_{xx}+u_{y}u_{xy}),\quad v_{y}=v^{-1}(u_{x}u_{xy}+u_{y}u_{yy}),$
(2.9) $\displaystyle v_{xx}$ $\displaystyle=$
$\displaystyle-v^{-3}(u_{x}u_{xx}+u_{y}u_{xy})^{2}+v^{-1}(u_{xx}^{2}+u_{x}u_{xxx}+u_{xy}^{2}+u_{y}u_{xxy}),$
(2.10) $\displaystyle v_{yy}$ $\displaystyle=$
$\displaystyle-v^{-3}(u_{x}u_{xy}+u_{y}u_{yy})^{2}+v^{-1}(u_{xy}^{2}+u_{x}u_{xyy}+u_{yy}^{2}+u_{y}u_{yyy}).$

Differentiating the equation, we have

(2.11) $\displaystyle u_{x}u_{xx}+uu_{xxx}+u_{xyy}$ $\displaystyle=$
$\displaystyle
2(1-\theta)u_{x}u_{xx}-\frac{2-\theta}{u^{2}}u_{x}u_{y}^{2}+\frac{2(2-\theta)}{u}u_{y}u_{xy},$
(2.12) $\displaystyle u_{y}u_{xx}+uu_{xxy}+u_{yyy}$ $\displaystyle=$
$\displaystyle
2(1-\theta)u_{x}u_{xy}-\frac{2-\theta}{u^{2}}u_{y}u_{y}^{2}+\frac{2(2-\theta)}{u}u_{y}u_{yy}.$

By the Cauchy inequality, we have

(2.13) $(u_{x}u_{xx}+u_{y}u_{xy})^{2}\leq(v^{2}-1)(u_{xx}^{2}+u_{xy}^{2}),\ \
(u_{x}u_{xy}+u_{y}u_{yy})^{2}\leq(v^{2}-1)(u_{xy}^{2}+u_{yy}^{2}).$

Then by (2.8), (2.9), (2.10), (2.11), (2.12) and (2.13), the first term in
(2.7) satisfies

$\displaystyle uv_{xx}+v_{yy}$ $\displaystyle=$ $\displaystyle
u[v^{-1}(u^{2}_{xx}+u^{2}_{xy})-v^{-3}(u_{x}u_{xx}+u_{y}u_{xy})^{2}]+v^{-1}u(u_{x}u_{xxx}+u_{y}u_{xxy})$
$\displaystyle+[v^{-1}(u^{2}_{xy}+u^{2}_{yy})-v^{-3}(u_{x}u_{xy}+u_{y}u_{yy})^{2}]+v^{-1}(u_{x}u_{xyy}+u_{y}u_{yyy}))$
$\displaystyle\geq$ $\displaystyle
v^{-1}\left[2(1-\theta)u_{x}(u_{x}u_{xx}+u_{y}u_{xy})+\frac{2(2-\theta)}{u}u_{y}(u_{x}u_{xy}+u_{y}u_{yy})-(v^{2}-1)u_{xx}\right]$
$\displaystyle\
-v^{-1}\frac{2-\theta}{u^{2}}(u_{x}^{2}u_{y}^{2}+u_{y}^{4})+v^{-3}\left[u(u^{2}_{xx}+u^{2}_{xy})+u^{2}_{xy}+u^{2}_{yy}\right]$
$\displaystyle=$ $\displaystyle
2(1-\theta)u_{x}v_{x}+\frac{2(2-\theta)}{u}u_{y}v_{y}-\frac{2-\theta}{u^{2}}\frac{v^{2}-1}{v}u_{y}^{2}-\frac{v^{2}-1}{v}u_{xx}$
$\displaystyle\
+v^{-3}\left[u(u^{2}_{xx}+u^{2}_{xy})+u^{2}_{xy}+u^{2}_{yy}\right]$
$\displaystyle\geq$ $\displaystyle
2(1-\theta)u_{x}v_{x}+\frac{2(2-\theta)}{u}u_{y}v_{y}-\frac{2-\theta}{u^{2}}\frac{v^{2}-1}{v}u_{y}^{2}-vu_{xx}+v^{-1}u_{xx}+v^{-3}uu_{xx}^{2}$
$\displaystyle\geq$ $\displaystyle
2(1-\theta)u_{x}v_{x}+\frac{2(2-\theta)}{u}u_{y}v_{y}-\frac{2-\theta}{u^{2}}\frac{v^{2}-1}{v}u_{y}^{2}-vu_{xx}-\frac{v}{4u}-v^{-3}uu_{xx}^{2}+v^{-3}uu_{xx}^{2}$
$\displaystyle\geq$ $\displaystyle
2(1-\theta)u_{x}v_{x}+\frac{2(2-\theta)}{u}u_{y}v_{y}-\frac{2-\theta}{u^{2}}\frac{v^{2}-1}{v}u_{y}^{2}-vu_{xx}-Cv,$

where $C>0$ depends on $\lambda$. By integration by parts,

$\displaystyle\int_{B_{R}}(uv_{xx}+v_{yy})\phi\eta\,dV$ $\displaystyle\geq$
$\displaystyle-\int_{B_{R}}2(1-\theta)[u_{xx}v\phi\eta+u_{x}v(\phi\eta)_{x}]\,dV$
$\displaystyle-\int_{B_{R}}2(2-\theta)\left[\frac{u_{yy}}{u}v\phi\eta-\frac{u_{y}^{2}}{u^{2}}v\phi\eta+\frac{1}{u}u_{y}v(\phi\eta)_{y}\right]\,dV$
$\displaystyle-\int_{B_{R}}\left(\frac{2-\theta}{u^{2}}\frac{v^{2}-1}{v}u_{y}^{2}\phi\eta+vu_{xx}\phi\eta+C\phi\eta
v\right)\,dV$ $\displaystyle=$
$\displaystyle\int_{B_{R}}u_{xx}v\phi\eta\,dV-\int_{B_{R}}2(2-\theta)\frac{1}{u}\left[(1-\theta)u_{x}^{2}+\frac{2-\theta}{u}u_{y}^{2}\right]v\phi\eta\,dV$
$\displaystyle-\int_{B_{R}}\left[2(1-\theta)u_{x}v(\phi\eta)_{x}+2(2-\theta)\left(\frac{u_{y}v(\phi\eta)_{y}}{u}-\frac{u_{y}^{2}v\phi\eta}{u^{2}}\right)\right]\,dV$
$\displaystyle-\int_{B_{R}}\left[\frac{2-\theta}{u^{2}}\left(v-\frac{1}{v}\right)u_{y}^{2}\phi\eta+C\phi\eta
v\right]\,dV$ $\displaystyle=$
$\displaystyle\int_{B_{R}}u_{xx}v\phi\eta\,dV-\int_{B_{R}}\left[2(1-\theta)\phi^{\prime}+\frac{2(2-\theta)(1-\theta)}{u}\phi\right]u_{x}^{2}v\eta\,dV$
$\displaystyle-\int_{B_{R}}\left[2(2-\theta)\frac{\phi^{\prime}}{u}+(2-\theta)(3-2\theta)\frac{1}{u^{2}}\phi\right]u_{y}^{2}v\eta\,dV$
$\displaystyle-\int_{B_{R}}\left[2(1-\theta)u_{x}v\phi\eta_{x}+\frac{2(2-\theta)}{u}u_{y}v\phi\eta_{y}+C\phi\eta
v\right]\,dV+\int_{B_{R}}\frac{2-\theta}{u^{2}}v^{-1}u_{y}^{2}\phi\eta\,dV$
$\displaystyle\geq$
$\displaystyle\int_{B_{R}}u_{xx}v\phi\eta\,dV-\int_{B_{R}}\left[2(1-\theta)\phi^{\prime}+\frac{2(2-\theta)(1-\theta)}{u}\phi\right]u_{x}^{2}v\eta\,dV$
$\displaystyle-\int_{B_{R}}\left[2(2-\theta)\frac{\phi^{\prime}}{u}+(2-\theta)(3-2\theta)\frac{1}{u^{2}}\phi\right]u_{y}^{2}v\eta\,dV$
$\displaystyle-C\left[\int_{B_{R}}|\nabla
u|^{2}(R^{2}-x^{2}-y^{2})^{\alpha-1}\,dV+\int_{B_{R}}|\nabla
u|(R^{2}-x^{2}-y^{2})^{\alpha}\,dV+1\right].$

Here $C>0$ depends on $\lambda$, $\Lambda$, $R$ and $\theta$. For the third
term in (2.7)

$\displaystyle[u(\phi\eta)_{xx}+(\phi\eta)_{yy}]v$ $\displaystyle=$
$\displaystyle
v\left[\phi^{\prime\prime}(u)(uu^{2}_{x}+u_{y}^{2})\eta+\phi^{\prime}(u)(uu_{xx}+u_{yy})\eta+2\phi^{\prime}(u)(uu_{x}\eta_{x}+u_{y}\eta_{y})+\phi\cdot(u\eta_{xx}+\eta_{yy})\right]$
$\displaystyle=$ $\displaystyle
v\bigg{[}\phi^{\prime\prime}(u)(uu^{2}_{x}+u_{y}^{2})\eta+\phi^{\prime}(u)\left((1-\theta)u_{x}^{2}+\frac{2-\theta}{u}u_{y}^{2}\right)\eta+2\phi^{\prime}(u)(uu_{x}\eta_{x}+u_{y}\eta_{y})$
$\displaystyle\ +\phi\cdot(u\eta_{xx}+\eta_{yy})\bigg{]}$ $\displaystyle\leq$
$\displaystyle\left[(\phi^{\prime\prime}u+(1-\theta)\phi^{\prime})u^{2}_{x}+\left(\phi^{\prime\prime}+\phi^{\prime}\frac{2-\theta}{u}\right)u_{y}^{2}\right]v\eta$
$\displaystyle\ +C\left[|\nabla u|^{2}(R^{2}-x^{2}-y^{2})^{\alpha-1}+|\nabla
u|(R^{2}-x^{2}-y^{2})^{\alpha-2}\right].$

Here $C>0$ depends on $\lambda$, $\Lambda$, $R$ and $\theta$. Integrating by
parts, the second term in (2.7)

$2\int_{B_{R}}uv_{x}(\phi\eta)_{x}+v_{y}(\phi\eta)_{y}\,dV=-2\int_{B_{R}}u_{x}(\phi\eta)_{x}v\,dV-2\int_{B_{R}}[u(\phi\eta)_{xx}+(\phi\eta)_{yy}]v\,dV,$

and hence

(2.14) $\displaystyle\int_{B_{R}}uw_{xx}+w_{yy}\,dV$ $\displaystyle=$
$\displaystyle\int_{B_{R}}(uv_{xx}+v_{yy})\phi\eta\,dV-2\int_{B_{R}}u_{x}(\phi\eta)_{x}v\,dV-\int_{B_{R}}[u(\phi\eta)_{xx}+(\phi\eta)_{yy}]v\,dV$
$\displaystyle\geq$
$\displaystyle\int_{B_{R}}u_{xx}v\phi\eta\,dV+\int_{B_{R}}[(-5+3\theta)\phi^{\prime}-\frac{2(2-\theta)(1-\theta)}{u}\phi-\phi^{\prime\prime}u]u_{x}^{2}v\eta\,dV$
$\displaystyle+\int_{B_{R}}[-3(2-\theta)\frac{\phi^{\prime}}{u}-(2-\theta)(3-2\theta)\frac{\phi}{u^{2}}-\phi^{\prime\prime}]u_{y}^{2}v\eta\,dV$
$\displaystyle-C\left[\int_{B_{R}}|\nabla
u|^{2}(R^{2}-x^{2}-y^{2})^{\alpha-1}\,dV+\int_{B_{R}}|\nabla
u|(R^{2}-x^{2}-y^{2})^{\alpha-2}\,dV+1\right].$

Note that the left hand side term satisfies

(2.15) $\int_{B_{R}}uw_{xx}+w_{yy}\,dV=\int_{B_{R}}u_{xx}v\phi\eta\,dV.$

Now we choose

$\phi(u)=Au^{\theta-2}-\frac{u}{2\theta^{2}-9\theta+9}$

with

$A\geq\frac{\Lambda^{3-\theta}}{2\theta^{2}-9\theta+9}+1.$

Then it is clear that $\phi(u)>0$. Furthermore, since $\theta\in[0,1]$, we
have

$(-5+3\theta)\phi^{\prime}-\frac{2(2-\theta)(1-\theta)}{u}\phi-\phi^{\prime\prime}u=1$

and

$-3(2-\theta)\frac{\phi^{\prime}}{u}-(2-\theta)(3-2\theta)\frac{\phi}{u^{2}}-\phi^{\prime\prime}=\frac{2(2-\theta)(3-\theta)}{(2\theta^{2}-9\theta+9)u}\geq
C_{0}>0.$

Combining them with (2.14) and (2.15), we obtain

$\displaystyle\int_{B_{R}}|\nabla u|^{3}(R^{2}-x^{2}-y^{2})^{\alpha}\,dV$
$\displaystyle\leq$ $\displaystyle C_{1}\int_{B_{R}}|\nabla
u|^{2}(R^{2}-x^{2}-y^{2})^{\alpha-1}\,dV+C_{2}\int_{B_{R}}|\nabla
u|(R^{2}-x^{2}-y^{2})^{\alpha-2}\,dV+C_{3}$ $\displaystyle\leq$ $\displaystyle
C_{1}^{\prime}\left(\int_{B_{R}}|\nabla
u|^{3}(R^{2}-x^{2}-y^{2})^{\alpha}\,dV\right)^{\frac{2}{3}}+C_{2}^{\prime}\left(\int_{B_{R}}|\nabla
u|^{3}(R^{2}-x^{2}-y^{2})^{\alpha}\,dV\right)^{\frac{1}{3}}+C_{3}.$

Hence, (2.6) follows. ∎

###  2.3. Proof of Theorem 1.1

In order to use the partial Legendre transform, we first recall the modulus of
convexity. For a convex function on $\mathbb{R}^{n}$, the modulus of convexity
$m_{u}$ of $u$ is defined by

(2.16) $m_{u}(t)=\inf\\{u(x)-\ell_{z}(x):|x-z|>t\\},$

where $t>0$ and $\ell_{z}$ is the supporting function of $u$ at $z$. For a
strictly convex function, $m_{u}$ must be a positive function. A result of
Heinz [H] implies that in two dimensions, when $\det D^{2}u\geq\lambda>0$,
there exists a positive function $C(t)>0$ depending on $\lambda$ such that
$m_{u}(t)\geq C(t)>0$. Now for the partial Legendre transform (2.1), we
consider the mapping

(2.17) $(\xi,\eta)=\mathcal{P}(x,y)=(u_{x},y):B_{R}(0)\to\mathbb{R}^{2}.$

The following important property is revealed in [Li].

######  Lemma 2.3 ([Li, Lemma 2.1]).

There exists a constant $\delta>0$ depending on the modulus of convexity
$m_{u}$ defined in (2.16), such that
$B_{\delta}(0)\subset\mathcal{P}(B_{R}(0))$.

###### Proof of Theorem 1.1.

For any $p\in\Omega$, let $R=\frac{dist(p,\partial\Omega)}{2}$. Without loss
of generality, we assume $\mathcal{P}(p)=0$. It is clear that
$\sup_{B_{R}(p)}|Du|\leq C$ for some constant $C$ depending on $R$ and
$\sup_{\Omega}|u|$. By Lemma 2.3, there exists $\delta>0$ such that
$B_{\delta}(0)\subset\mathcal{P}(B_{R}(p))$. According to Proposition 2.1,
$u^{\star}$ satisfies (2.3) in $B_{\delta}(0)$ with

$0<\lambda\leq
w^{\star}=-\frac{u^{\star}_{\eta\eta}}{u^{\star}_{\xi\xi}}\leq\Lambda.$

By Theorem 2.2,

$\|w^{\star}\|_{W^{1,3}(B_{\frac{7\delta}{8}}(0))}\leq C.$

Note that $n=2$. By the Sobolev theorem, we have the $C^{\alpha}$ estimate of
$w^{\star}$. And by the interior $W^{2,p}$-estimate of the uniformly elliptic
equation (2.3), we have the estimate

$\|w^{\star}\|_{W^{2,\frac{3}{2}}(B_{\frac{3\delta}{4}}(0))}\leq C,$

which implies the $W^{1,6}$-estimate of $w^{\star}$. Again by the interior
$W^{2,p}$-estimate of the uniformly elliptic equation (2.3), we have the

$\|w^{\star}\|_{W^{2,3}(B_{\frac{\delta}{2}}(0))}\leq C,$

which implies the $C^{1,\alpha}$ estimate of $w^{\star}$. Then by the Schauder
estimate of (2.3), we have

$\|w^{\star}\|_{C^{2,\alpha}(B_{\frac{\delta}{4}}(0))}\leq C$

and all the higher order estimates of $u^{\star}$. Transforming back by the
partial Legendre transform, we obtain the lower bound of $u_{x_{1}x_{1}}$ by
(2.2). Since we can do partial Legendre transforms of $u$ in any direction, we
can obtain the lower bound for the smallest eigenvalue of $D^{2}u$, which
implies the boundedness of $D^{2}u$ by (1.4). Then we have all the higher
order estimates of $u$. ∎

##  3\. The inhomogeneous equations in dimension two

In this section, we will study the interior estimate for the inhomogeneous
equation (1.5).

We first recall the regularity theory of the second order elliptic equation in
divergence form

(3.1) $D_{j}(a_{ij}(x)D_{i}u)=D_{i}g^{i}+h\quad\text{in}\,\,\Omega,$

where $g=(g^{1},\cdots,g^{n})$ is a vector valued function and
$\\{a_{ij}(x)\\}$ satisfies

$0<\lambda(x)|\xi|^{2}\leq a_{ij}(x)\xi_{i}\xi_{j}\leq\Lambda(x)|\xi|^{2},\ \
\forall\ \xi\in\mathbb{R}^{n}\setminus\\{0\\}$

and can be discontinuous. When $\\{a_{ij}(x)\\}$ is uniformly elliptic, a
fundamental regularity theory was established by De Giorgi, Nash, Moser, etc.
In [MS, Tr], the classical De Giorgi-Nash-Moser’s theory was extended to
degenerate linear elliptic equation with $\lambda(x)^{-1}\in L^{p}(\Omega)$
and $\Lambda(x)\in L^{q}(\Omega)$ where $\frac{1}{p}+\frac{1}{q}<\frac{2}{n}$.
We denote by $S_{n}^{+}$ the set of $n\times n$ nonnegative definite matrices.

######  Theorem 3.1 ([Tr, Theorem 4.2]).

Let $\\{a_{ij}\\}:\Omega\to S_{n}^{+}$ be such that $\lambda^{-1}\in
L_{loc}^{p}(\Omega)$ for some $p>n$. Let $g\in
L^{\infty}(\Omega;\mathbb{R}^{n})$ and $h\in L^{q}(\Omega)$. Assume $p$ and
$q$ satisfy $\frac{1}{q}+\frac{1}{p}<\frac{2}{n}$. Suppose that $u$ is a
subsolution (supersolution) to (3.1) in $B_{R}=B_{R}(y)\subset\Omega$ and that
$u\leq 0(u\geq 0)$ on $\partial B_{R}$. Then

$\sup_{B_{R}}u(-u)\leq
C\|\lambda^{-1}\|_{L^{p}(B_{R})}\left(\|g\|_{L^{\infty}(B_{R})}R^{1-\frac{n}{p}}+\|h\|_{L^{q}(B_{R})}R^{2-\frac{n}{q}-\frac{n}{p}}\right).$

In [Lo], the author investigated the linearized Monge-Ampère equation with
right hand side $f=D_{i}g^{i}+h$, i.e. $a_{ij}(x)=U^{ij}$ in (3.1). Under the
stronger assumption that $\det D^{2}u$ is sufficiently close to a positive
constant, he can use the $W^{2,p}$-estimate of the Monge-Ampère equation and
Theorem 3.1 to obtain the interior regularity. It was later shown in [Le3]
that when $n=2$ we only need the determinant to be bounded from above and
below.

Now we turn to the fourth order equation (1.5). By the divergence free
property of $U^{ij}$, i.e., $\sum_{i}D_{i}U^{ij}=0$, we can rewrite equation
(1.5) in divergence form

(3.2) $D_{j}(U^{ij}D_{i}w)=D_{i}g^{i}+h.$

Note that in dimension two, it holds $\frac{\det D^{2}u}{\Delta u}I\leq
U^{ij}\leq\Delta uI$. In view of Theorem 3.1, it suffices to get the
$L^{p}$-bound of $\Delta u$ for sufficiently large $p$. In the following, we
will use an integral method directly in (1.5) to derive it. Instead of the
classical Sobolev inequality, we will need the following Monge-Ampère Sobolev
inequality.

######  Lemma 3.2 ([Le3, Proposition 2.6]).

Assume $n=2$. Let $u$ be a smooth, strictly convex function defined in a
neighborhood of a bounded domain $\Omega\subset\mathbb{R}^{2}$. Suppose $u$
satisfies

(3.3) $0<\lambda\leq\det D^{2}u\leq\Lambda.$

Then for any $\chi>2$ there exists a constant $C>0$, depending only on
$\lambda$, $\Lambda$ and $\chi$, such that

$\left(\int_{\Omega}|v|^{\chi}\,dx\right)^{\frac{1}{\chi}}\leq
C\left(\int_{\Omega}U^{ij}v_{i}v_{j}\,dx\right)^{\frac{1}{2}}\quad\text{for\,\,all}\,\,v\in
C_{0}^{\infty}(\Omega).$

The above inequality is the two dimensional counterpart of the Monge-Ampère
Sobolev inequality in higher dimensions derived by Tian and Wang [TW].

######  Theorem 3.3.

Assume $n=2$ and $\theta\in[0,1]$. Let $g\in
L^{\infty}(\Omega;\mathbb{R}^{n})$ and $h\in L^{q}(\Omega)$. Let $u$ be a
uniformly convex smooth solution to equation (3.2) in $B_{R}$ satisfying
(3.3). Then for any $p\geq 1$, there exists a constant $C>0$ depending on
$\lambda$, $\Lambda$, $R$, $\theta$, $\|g\|_{L^{\infty}(B_{R})}$,
$\|h\|_{L^{q}(B_{R})}$ and $p$ such that

$\|\Delta u\|_{L^{p}(B_{R/2})}\leq C.$

###### Proof.

We first consider the case $\theta\in[0,1)$. Denote

$v=\Delta u,\,w=(\det D^{2}u)^{-(1-\theta)},\,\rho=R^{2}-|x|^{2}.$

We consider $z=v^{\beta}\varphi\rho^{\alpha}$, where $\alpha,\beta>0$ are
constants and $\varphi=\varphi(w)$ is a positive function to be determined.
Then

$\displaystyle z_{i}$ $\displaystyle=$ $\displaystyle\beta
v_{i}v^{\beta-1}\varphi\rho^{\alpha}+v^{\beta}(\varphi\rho^{\alpha})_{i},$
$\displaystyle z_{ij}$ $\displaystyle=$
$\displaystyle\beta(\beta-1)v_{i}v_{j}v^{\beta-2}\varphi\rho^{\alpha}+\beta
v_{ij}v^{\beta-1}\varphi\rho^{\alpha}+\beta
v_{i}v^{\beta-1}(\varphi\rho^{\alpha})_{j}+\beta
v_{j}v^{\beta-1}(\varphi\rho^{\alpha})_{i}+v^{\beta}(\varphi\rho^{\alpha})_{ij}.$

By integration by parts with (1.5) and choosing $\alpha>2$, we have

$\displaystyle 0$ $\displaystyle=$ $\displaystyle\int_{B_{R}}U^{ij}z_{ij}\,dx$
$\displaystyle=$
$\displaystyle\beta(\beta-1)\int_{B_{R}}U^{ij}v_{i}v_{j}\varphi\rho^{\alpha}v^{\beta-2}\,dx+\beta\int_{B_{R}}U^{ij}v_{ij}\varphi\rho^{\alpha}v^{\beta-1}\,dx+2\beta\int_{B_{R}}U^{ij}v_{i}w_{j}\varphi^{\prime}\rho^{\alpha}v^{\beta-1}\,dx$
$\displaystyle+2\alpha\beta\int_{B_{R}}U^{ij}v_{i}\rho_{j}\varphi\rho^{\alpha-1}v^{\beta-1}\,dx+\int_{B_{R}}U^{ij}w_{i}w_{j}\varphi^{\prime\prime}\rho^{\alpha}v^{\beta}\,dx+2\alpha\int_{B_{R}}U^{ij}w_{i}\rho_{j}\varphi^{\prime}\rho^{\alpha-1}v^{\beta}\,dx$
$\displaystyle+\int_{B_{R}}f\varphi^{\prime}\rho^{\alpha}v^{\beta}\,dx+\alpha\int_{B_{R}}U^{ij}\rho_{ij}\varphi\rho^{\alpha-1}v^{\beta}\,dx+\alpha(\alpha-1)\int_{B_{R}}U^{ij}\rho_{i}\rho_{j}\varphi\rho^{\alpha-2}v^{\beta}\,dx.$

Note that for any $m$,

$\displaystyle u^{ij}(u_{mm})_{ij}=u^{il}u^{kj}u_{mij}u_{mkl}+(\ln\det
D^{2}u)_{mm}=u^{il}u^{kj}u_{mij}u_{mkl}+\frac{1}{1-\theta}\frac{w_{m}^{2}}{w^{2}}-\frac{1}{1-\theta}\frac{w_{mm}}{w}.$

We have

$\displaystyle\int_{B_{R}}U^{ij}v_{ij}\varphi\rho^{\alpha}v^{\beta-1}$
$\displaystyle\,dx=\int_{B_{R}}\det
D^{2}uu^{il}u^{kj}u_{mij}u_{mkl}\varphi\rho^{\alpha}v^{\beta-1}\,dx$
$\displaystyle+\frac{1}{1-\theta}\int_{B_{R}}\det D^{2}u\frac{|\nabla
w|^{2}}{w^{2}}\varphi\rho^{\alpha}v^{\beta-1}\,dx-\frac{1}{1-\theta}\int_{B_{R}}\det
D^{2}u\frac{\triangle w}{w}\varphi\rho^{\alpha}v^{\beta-1}\,dx.$

By integration by parts, we have

$\displaystyle\int_{B_{R}}\det D^{2}u\frac{\triangle
w}{w}\varphi\rho^{\alpha}v^{\beta-1}\,dx=\int_{B_{R}}w^{-\frac{2-\theta}{1-\theta}}\triangle
w\varphi\rho^{\alpha}v^{\beta-1}\,dx$ $\displaystyle=$
$\displaystyle\frac{2-\theta}{1-\theta}\int_{B_{R}}w^{-\frac{3-2\theta}{1-\theta}}|\nabla
w|^{2}\varphi\rho^{\alpha}v^{\beta-1}\,dx-\int_{B_{R}}w^{-\frac{2-\theta}{1-\theta}}|\nabla
w|^{2}\varphi^{\prime}\rho^{\alpha}v^{\beta-1}\,dx$
$\displaystyle-\alpha\int_{B_{R}}w^{-\frac{2-\theta}{1-\theta}}w_{i}\rho_{i}\varphi\rho^{\alpha-1}v^{\beta-1}\,dx-(\beta-1)\int_{B_{R}}w^{-\frac{2-\theta}{1-\theta}}w_{i}v_{i}\varphi\rho^{\alpha}v^{\beta-2}\,dx.$

By changing coordinates at each point, say $x_{0}$, we can assume that
$D^{2}u(x_{0})$ is diagonal. Then

(3.6) $\displaystyle
u^{il}u^{kj}u_{mij}u_{mkl}=\frac{u_{mij}^{2}}{u_{ii}u_{jj}}\geq\frac{u_{mmi}^{2}}{u_{ii}u_{mm}}\geq\frac{v_{i}^{2}}{u_{ii}v}=\frac{u^{ij}v_{i}v_{j}}{v},$

where we used the Cauchy inequality in the second inequality. Hence by (3) and
(3.6), we have

(3.7) $\displaystyle\int_{B_{R}}U^{ij}v_{ij}\varphi\rho^{\alpha}v^{\beta-1}$
$\displaystyle\,dx\geq\int_{B_{R}}U^{ij}v_{i}v_{j}\varphi\rho^{\alpha}v^{\beta-2}\,dx+\frac{1}{1-\theta}\int_{B_{R}}w^{-\frac{2-\theta}{1-\theta}}(\varphi^{\prime}-\frac{\varphi
w^{-1}}{1-\theta})|\nabla w|^{2}\rho^{\alpha}v^{\beta-1}\,dx$
$\displaystyle+\frac{\alpha}{1-\theta}\int_{B_{R}}w^{-\frac{2-\theta}{1-\theta}}w_{i}\rho_{i}\varphi\rho^{\alpha-1}v^{\beta-1}\,dx+\frac{\beta-1}{1-\theta}\int_{B_{R}}w^{-\frac{2-\theta}{1-\theta}}w_{i}v_{i}\varphi\rho^{\alpha}v^{\beta-2}\,dx.$

Then putting (3.7) into (3) yields

$\displaystyle 0$ $\displaystyle\geq$
$\displaystyle\beta^{2}\int_{B_{R}}U^{ij}v_{i}v_{j}\varphi\rho^{\alpha}v^{\beta-2}\,dx+\frac{\beta}{1-\theta}\int_{B_{R}}w^{-\frac{2-\theta}{1-\theta}}(\varphi^{\prime}-\frac{\varphi
w^{-1}}{1-\theta})|\nabla w|^{2}\rho^{\alpha}v^{\beta-1}\,dx$
$\displaystyle+\frac{\alpha\beta}{1-\theta}\int_{B_{R}}w^{-\frac{2-\theta}{1-\theta}}w_{i}\rho_{i}\varphi\rho^{\alpha-1}v^{\beta-1}\,dx+\frac{\beta(\beta-1)}{1-\theta}\int_{B_{R}}w^{-\frac{2-\theta}{1-\theta}}w_{i}v_{i}\varphi\rho^{\alpha}v^{\beta-2}\,dx$
$\displaystyle+2\beta\int_{B_{R}}U^{ij}v_{i}w_{j}\varphi^{\prime}\rho^{\alpha}v^{\beta-1}\,dx+2\alpha\beta\int_{B_{R}}U^{ij}v_{i}\rho_{j}\varphi\rho^{\alpha-1}v^{\beta-1}\,dx$
$\displaystyle+\int_{B_{R}}U^{ij}w_{i}w_{j}\varphi^{\prime\prime}\rho^{\alpha}v^{\beta}\,dx+\int_{B_{R}}f\varphi^{\prime}\rho^{\alpha}v^{\beta}\,dx+2\alpha\int_{B_{R}}U^{ij}w_{i}\rho_{j}\varphi^{\prime}\rho^{\alpha-1}v^{\beta}\,dx$
$\displaystyle+\alpha\int_{B_{R}}U^{ij}\rho_{ij}\varphi\rho^{\alpha-1}v^{\beta}\,dx+\alpha(\alpha-1)\int_{B_{R}}U^{ij}\rho_{i}\rho_{j}\varphi\rho^{\alpha-2}v^{\beta}\,dx.$

Now we choose $\varphi(w)=e^{e^{Aw}}$, where $A>0$ is to be determined later.
Then we know that

(3.9) $\varphi^{\prime}(w)=Ae^{Aw}\varphi,\
\varphi^{\prime\prime}(w)=(A^{2}e^{Aw}+A^{2}e^{2Aw})\varphi.$

We will estimate the right hand side of (3) term by term.

By (3.9), the two terms of the third line in (3) satisfy

$\displaystyle\left|2\beta\int_{B_{R}}U^{ij}v_{i}w_{j}\varphi^{\prime}\rho^{\alpha}v^{\beta-1}\,dx\right|=\left|\int_{B_{R}}2\beta
Ae^{Aw}U^{ij}v_{i}w_{j}\varphi\rho^{\alpha}v^{\beta-1}\,dx\right|$
$\displaystyle\ \ \ \
\leq\int_{B_{R}}\beta^{2}\frac{e^{Aw}}{1/2+e^{Aw}}U^{ij}v_{i}v_{j}\varphi\rho^{\alpha}v^{\beta-2}\,dx+\int_{B_{R}}(\frac{A^{2}}{2}e^{Aw}+A^{2}e^{2Aw})U^{ij}w_{i}w_{j}\varphi\rho^{\alpha}v^{\beta}\,dx,$

and

$\displaystyle\left|2\alpha\beta\int_{B_{R}}U^{ij}v_{i}\rho_{j}\varphi\rho^{\alpha-1}v^{\beta-1}\,dx\right|$
$\displaystyle\leq$
$\displaystyle\int_{B_{R}}\frac{\beta^{2}}{8(1+2e^{Aw})}U^{ij}v_{i}v_{j}\varphi\rho^{\alpha}v^{\beta-2}\,dx$
$\displaystyle+\int_{B_{R}}8(1+2e^{Aw})\alpha^{2}U^{ij}\rho_{i}\rho_{j}\varphi\rho^{\alpha-2}v^{\beta}\,dx.$

Note that

$|\nabla\rho|^{2}\leq\frac{\triangle u}{\det D^{2}u}U^{ij}\rho_{i}\rho_{j},\ \
|\nabla v|^{2}\leq\frac{\triangle u}{\det D^{2}u}U^{ij}v_{i}v_{j}.$

The two terms of the second line in (3) satisfy

$\displaystyle\left|\frac{\alpha\beta}{1-\theta}\int_{B_{R}}w^{-\frac{2-\theta}{1-\theta}}w_{i}\rho_{i}\varphi\rho^{\alpha-1}v^{\beta-1}\,dx\right|$
$\displaystyle\leq$
$\displaystyle\int_{B_{R}}\frac{\beta}{(1-\theta)^{2}}w^{-\frac{3-2\theta}{1-\theta}}|\nabla
w|^{2}\varphi\rho^{\alpha}v^{\beta-1}\,dx+\int_{B_{R}}\frac{1}{4}\alpha^{2}\beta
w^{-\frac{1}{1-\theta}}|\nabla\rho|^{2}\varphi\rho^{\alpha-2}v^{\beta-1}\,dx$
$\displaystyle\leq$
$\displaystyle\int_{B_{R}}\frac{\beta}{(1-\theta)^{2}}w^{-\frac{3-2\theta}{1-\theta}}|\nabla
w|^{2}\varphi\rho^{\alpha}v^{\beta-1}\,dx+\int_{B_{R}}\frac{1}{4}\alpha^{2}\beta
U^{ij}\rho_{i}\rho_{j}\varphi\rho^{\alpha-2}v^{\beta}\,dx,$

and

$\displaystyle\left|\frac{\beta(\beta-1)}{1-\theta}\int_{B_{R}}w^{-\frac{2-\theta}{1-\theta}}w_{i}v_{i}\varphi\rho^{\alpha}v^{\beta-2}\,dx\right|$
$\displaystyle\leq$
$\displaystyle\int_{B_{R}}\frac{(\beta-1)^{2}}{3(1-\theta)^{2}}(1+2e^{Aw})w^{-\frac{3-2\theta}{1-\theta}}|\nabla
w|^{2}\varphi\rho^{\alpha}v^{\beta-1}\,dx+\int_{B_{R}}\frac{3\beta^{2}w^{-\frac{1}{1-\theta}}}{4(1+2e^{Aw})}|\nabla
v|^{2}\varphi\rho^{\alpha}v^{\beta-3}\,dx$ $\displaystyle\leq$
$\displaystyle\int_{B_{R}}\frac{(\beta-1)^{2}}{3(1-\theta)^{2}}(1+2e^{Aw})w^{-\frac{3-2\theta}{1-\theta}}|\nabla
w|^{2}\varphi\rho^{\alpha}v^{\beta-1}\,dx+\int_{B_{R}}\frac{3\beta^{2}}{4(1+2e^{Aw})}U^{ij}v_{i}v_{j}\varphi\rho^{\alpha}v^{\beta-2}\,dx.$

By (3.9), the last term of the fourth line in (3) satisfies

$\displaystyle\left|2\alpha\int_{B_{R}}U^{ij}w_{i}\rho_{j}\varphi^{\prime}\rho^{\alpha-1}v^{\beta}\,dx\right|\leq\int_{B_{R}}\frac{A^{2}}{2}e^{Aw}U^{ij}w_{i}w_{j}\varphi\rho^{\alpha}v^{\beta}\,dx$
$\displaystyle+$
$\displaystyle\int_{B_{R}}2\alpha^{2}e^{Aw}U^{ij}\rho_{i}\rho_{j}\varphi\rho^{\alpha-2}v^{\beta}\,dx.$

Note that $f=\mbox{div}\,(g)+h$, where $g$ is a bounded vector field. We
denote $|g|=\sqrt{\sum_{i}(g^{i})^{2}}$. Then by integration by parts, we have

$\displaystyle\left|\int_{B_{R}}f\varphi^{\prime}\rho^{\alpha}v^{\beta}\,dx\right|$
$\displaystyle=$
$\displaystyle\left|\int_{B_{R}}h\varphi^{\prime}\rho^{\alpha}v^{\beta}\,dx-\int_{B_{R}}g^{i}w_{i}\varphi^{\prime\prime}\rho^{\alpha}v^{\beta}\,dx-\alpha\int_{B_{R}}g^{i}\rho_{i}\varphi^{\prime}\rho^{\alpha-1}v^{\beta}\,dx-\beta\int_{B_{R}}g^{i}v_{i}\varphi^{\prime}\rho^{\alpha}v^{\beta-1}\,dx\right|$
$\displaystyle\leq$
$\displaystyle\int_{B_{R}}\frac{1}{2}w^{-\frac{3-2\theta}{1-\theta}}|\nabla
w|^{2}\varphi\rho^{\alpha}v^{\beta-1}\,dx+\int_{B_{R}}\frac{1}{2}(A^{2}e^{Aw}+A^{2}e^{2Aw})^{2}w^{\frac{3-2\theta}{1-\theta}}|g|^{2}\varphi\rho^{\alpha}v^{\beta+1}\,dx$
$\displaystyle+\int_{B_{R}}\frac{\alpha}{2}w^{-\frac{1}{1-\theta}}|\nabla\rho|^{2}Ae^{Aw}\varphi\rho^{\alpha-2}v^{\beta-1}\,dx+\int_{B_{R}}\frac{\alpha}{2}w^{\frac{1}{1-\theta}}|g|^{2}Ae^{Aw}\varphi\rho^{\alpha}v^{\beta+1}\,dx+\int_{B_{R}}|h|\varphi^{\prime}\rho^{\alpha}v^{\beta}\,dx$
$\displaystyle+\int_{B_{R}}\frac{\beta^{2}w^{-\frac{1}{1-\theta}}}{16(1+2e^{Aw})}|\nabla
v|^{2}\varphi\rho^{\alpha}v^{\beta-3}\,dx+\int_{B_{R}}4(1+2e^{Aw})A^{2}e^{2Aw}w^{\frac{1}{1-\theta}}|g|^{2}\varphi\rho^{\alpha}v^{\beta+1}\,dx$
$\displaystyle\leq$
$\displaystyle\int_{B_{R}}\frac{1}{2}w^{-\frac{3-2\theta}{1-\theta}}|\nabla
w|^{2}\varphi\rho^{\alpha}v^{\beta-1}\,dx+\int_{B_{R}}\frac{1}{2}(A^{2}e^{Aw}+A^{2}e^{2Aw})^{2}w^{\frac{3-2\theta}{1-\theta}}|g|^{2}\varphi\rho^{\alpha}v^{\beta+1}\,dx$
$\displaystyle+\int_{B_{R}}\frac{\alpha}{2}U^{ij}\rho_{i}\rho_{j}Ae^{Aw}\varphi\rho^{\alpha-2}v^{\beta}\,dx+\int_{B_{R}}\frac{\alpha}{2}w^{\frac{1}{1-\theta}}|g|^{2}Ae^{Aw}\varphi\rho^{\alpha}v^{\beta+1}\,dx+\int_{B_{R}}|h|\varphi^{\prime}\rho^{\alpha}v^{\beta}\,dx$
$\displaystyle+\int_{B_{R}}\frac{\beta^{2}}{16(1+2e^{Aw})}U^{ij}v_{i}v_{j}\varphi\rho^{\alpha}v^{\beta-2}\,dx+\int_{B_{R}}4(1+2e^{Aw})A^{2}e^{2Aw}w^{\frac{1}{1-\theta}}|g|^{2}\varphi\rho^{\alpha}v^{\beta+1}\,dx.$

Hence, (3) reduces to

$\displaystyle 0$ $\displaystyle\geq$
$\displaystyle\int_{B_{R}}\frac{\beta^{2}}{16(1+2e^{Aw})}U^{ij}v_{i}v_{j}\varphi\rho^{\alpha}v^{\beta-2}\,dx+\alpha\int_{B_{R}}U^{ij}\rho_{ij}\varphi\rho^{\alpha-1}v^{\beta}\,dx-\int_{B_{R}}|h|\varphi^{\prime}\rho^{\alpha}v^{\beta}\,dx$
$\displaystyle+\int_{B_{R}}\left(\frac{\beta}{1-\theta}Awe^{Aw}-\frac{(\beta-1)^{2}}{3(1-\theta)^{2}}(1+2e^{Aw})-\frac{2\beta}{(1-\theta)^{2}}-\frac{1}{2}\right)w^{-\frac{3-2\theta}{1-\theta}}|\nabla
w|^{2}\varphi\rho^{\alpha}v^{\beta-1}\,dx$
$\displaystyle+\int_{B_{R}}\left(\alpha(\alpha-1)-(8+18e^{Aw}+\frac{1}{4}\beta)\alpha^{2}-\frac{\alpha}{2}Ae^{Aw}\right)U^{ij}\rho_{i}\rho_{j}\varphi\rho^{\alpha-2}v^{\beta}\,dx$
$\displaystyle-\int_{B_{R}}\frac{1}{2}(A^{2}e^{Aw}+A^{2}e^{2Aw})^{2}w^{\frac{3-2\theta}{1-\theta}}|g|^{2}\varphi\rho^{\alpha}v^{\beta+1}\,dx-\int_{B_{R}}\frac{\alpha}{2}w^{\frac{1}{1-\theta}}|g|^{2}Ae^{Aw}\varphi\rho^{\alpha}v^{\beta+1}\,dx$
$\displaystyle-\int_{B_{R}}4(1+2e^{Aw})A^{2}e^{2Aw}w^{\frac{1}{1-\theta}}|g|^{2}\varphi\rho^{\alpha}v^{\beta+1}\,dx.$

Now we choose $A$ sufficiently large such that

$\displaystyle\frac{\beta}{1-\theta}Awe^{Aw}-\frac{(\beta-1)^{2}}{3(1-\theta)^{2}}(1+2e^{Aw})-\frac{2\beta}{(1-\theta)^{2}}-\frac{1}{2}>0,$

i.e.,

$\displaystyle(3\beta(1-\theta)Aw-2(\beta-1)^{2})e^{Aw}-(\beta-1)^{2}-6\beta-\frac{3(1-\theta)^{2}}{2}>0.$

Note that

$U^{ij}\rho_{i}\rho_{j}\leq
v^{n-1}|\nabla\rho|^{2}\quad\text{and}\quad|U^{ij}\rho_{ij}|\leq 2nv^{n-1}.$

Then we have

$\displaystyle\int_{B_{R}}U^{ij}v_{i}v_{j}\rho^{\alpha}v^{\beta-2}\,dx\leq
C\int_{B_{R}}\rho^{\alpha}v^{\beta+1}\,dx+C\int_{B_{R}}\rho^{\alpha-2}v^{\beta+n-1}\,dx+C\int_{B_{R}}|h|\rho^{\alpha}v^{\beta}\,dx.$

Since

$\displaystyle
U^{ij}(v^{\frac{\beta}{2}}\rho^{\frac{\alpha}{2}})_{i}(v^{\frac{\beta}{2}}\rho^{\frac{\alpha}{2}})_{j}$
$\displaystyle=$
$\displaystyle\frac{\beta^{2}}{4}U^{ij}v_{i}v_{j}\rho^{\alpha}v^{\beta-2}+\frac{\alpha^{2}}{4}U^{ij}\rho_{i}\rho_{j}\rho^{\alpha-2}v^{\beta}+\frac{\alpha\beta}{2}U^{ij}v_{i}\rho_{j}\rho^{\alpha-1}v^{\beta-1}$
$\displaystyle\leq$
$\displaystyle\frac{\beta^{2}}{2}U^{ij}v_{i}v_{j}\rho^{\alpha}v^{\beta-2}+\frac{\alpha^{2}}{2}U^{ij}\rho_{i}\rho_{j}\rho^{\alpha-2}v^{\beta},$

we have

$\displaystyle\int_{B_{R}}U^{ij}(v^{\frac{\beta}{2}}\rho^{\frac{\alpha}{2}})_{i}(v^{\frac{\beta}{2}}\rho^{\frac{\alpha}{2}})_{j}\,dx$
$\displaystyle\leq$ $\displaystyle
C\int_{B_{R}}\rho^{\alpha}v^{\beta+1}\,dx+C\int_{B_{R}}\rho^{\alpha-2}v^{\beta+n-1}\,dx+C\int_{B_{R}}|h|\rho^{\alpha}v^{\beta}\,dx$
$\displaystyle\leq$ $\displaystyle
C\int_{B_{R}}v^{\beta+n-1}\,dx+C\left(\int_{B_{R}}|h|^{q}\,dx\right)^{\frac{1}{q}}\left(\int_{B_{R}}(v^{\frac{\beta}{2}}\rho^{\frac{\alpha}{2}})^{\frac{2q}{q-1}}\,dx\right)^{1-\frac{1}{q}}.$

Choosing $\chi\geq\frac{2q}{q-1}$ and using Lemma 3.2 with $n=2$, we have

(3.10)
$\displaystyle\left[\int_{B_{R}}(v^{\frac{\beta}{2}}\rho^{\frac{\alpha}{2}})^{\chi}\,dx\right]^{\frac{2}{\chi}}\leq
C\int_{B_{R}}v^{\beta+1}\,dx+C.$

By the interior $W^{2,1+\varepsilon}$-estimates for Monge-Ampère equation
[DFS, S] with (3.3), we know that there is a small $\varepsilon_{0}>0$, that
depends only on $\lambda$ and $\Lambda$ such that
$\|v\|_{L^{1+\varepsilon_{0}}}\leq C(\lambda,\Lambda)$. Then in (3.10), we
choose $\beta=\varepsilon_{0}$ and $\chi=\frac{2p}{\varepsilon_{0}}$ for
$p\in(\frac{q}{q-1}\varepsilon_{0},+\infty)$ to get $\|\Delta
u\|_{L^{p}(B_{R/2})}\leq C$.

For $\theta=1$, we know that $w=\ln\det D^{2}u$, and we can obtain the
$L^{p}$-bound of $\Delta u$ following the same method used above.
Alternatively, we can apply tha Legendre transform to $u$ for the case
$\theta=1$ to get the estimate of $D^{2}u$ by the strictly convexity of $u$
with condition (3.3). ∎

Now we establish the higher estimates of $u$ in terms of the
$W^{2,p}$-estimate. By chaining together a sequence of balls, in a standard
fashion, we know $\Delta u\in L^{p}_{loc}(\Omega)$ for any fixed
$p\in[1,+\infty)$. Then by Theorem 3.1 and the same arguments as [Lo,
Proposition 6.1] or [Le3, Theorem 1.3], we get the Hölder continuity of $w$
and all the higher order estimates of $u$.

As we mentioned in the introduction, there is another approach to establish
the $C^{2}$-estimate from $W^{2,p}$ when $f\in W^{1,q}(B_{1})$, which holds in
any dimension.

######  Theorem 3.4.

Let $u$ be a uniformly convex smooth solution to equation (1.5) in $B_{1}$
with $\theta=0$ or $1$. Assume that $f\in W^{1,q}(B_{1})$ for some
$q>\frac{n}{2}$. Suppose $u$ satisfies (3.3). Assume that $\|\Delta
u\|_{L^{p_{n}}(B_{1})}$ is bounded for some $p_{n}>\frac{n(n-1)}{2}$. Then
there exists a constant $C>0$ depending on $\lambda$, $\Lambda$, $\|\Delta
u\|_{L^{p_{n}}(B_{1})}$ and $\|f\|_{W^{1,q}(B_{1})}$, such that

$\sup_{B_{1/2}}\Delta u\leq C.$

###### Proof.

The proof is inspired by [CC]. We consider the case $\theta=1$. The case
$\theta=0$ then follows by using the Legendre transform. Denote
$v=u^{kl}w_{k}w_{l}$, where $\\{u^{kl}\\}$ is the inverse matrix of $D^{2}u$.
By direct calculations,

$\displaystyle v_{i}$ $\displaystyle=$
$\displaystyle-u^{ks}u^{tl}u_{sti}w_{k}w_{l}+2u^{kl}w_{ik}w_{l},$
$\displaystyle v_{ij}$ $\displaystyle=$ $\displaystyle
u^{kp}u^{rs}u^{tl}u_{prj}u_{sti}w_{k}w_{l}+u^{ks}u^{lp}u^{rt}u_{prj}u_{sti}w_{k}w_{l}-u^{ks}u^{tl}u_{stij}w_{k}w_{l}$
$\displaystyle-2u^{ks}u^{tl}u_{sti}w_{jk}w_{l}-2u^{ks}u^{tl}u_{stj}w_{ik}w_{l}+2u^{kl}w_{ijk}w_{l}+2u^{kl}w_{ik}w_{jl}.$

For any $x_{0}\in B_{1}$, we can choose a coordinate transformation so that
$u_{ij}(x_{0})=u_{ii}(x_{0})\delta_{ij}$. Then

(3.11) $\displaystyle u^{ij}v_{ij}(x_{0})$ $\displaystyle=$ $\displaystyle
2u^{ii}u^{kk}u^{rr}u^{ll}u_{kri}u_{lri}w_{k}w_{l}-u^{ii}u^{kk}u^{ll}u_{iikl}w_{k}w_{l}-4u^{ii}u^{kk}u^{ll}u_{kli}w_{ik}w_{l}$
$\displaystyle+2u^{ii}u^{kk}w_{iik}w_{k}+2u^{ii}u^{kk}w_{ik}^{2}.$

Differentiating $\ln\det D^{2}u=w$ twice, we have

(3.12) $\displaystyle u^{ij}u_{ijkl}-u^{is}u^{tj}u_{ijk}u_{stl}=(\ln\det
D^{2}u)_{kl}=w_{kl}.$

Note that (1.5) can be written as $u^{ij}w_{ij}=e^{-w}f$. Differentiating the
equation respect to $x_{k}$-direction directly yields

(3.13)
$\displaystyle-u^{ip}u^{rj}u_{prk}w_{ij}+u^{ij}w_{ijk}=e^{-w}(f_{k}-w_{k}f).$

Inserting (3.12) and (3.13) into (3.11), we have

(3.14) $\displaystyle u^{ij}v_{ij}(x_{0})$ $\displaystyle=$ $\displaystyle
u^{kk}u^{ll}u^{ii}u^{jj}u_{kji}u_{lji}w_{k}w_{l}-2u^{ii}u^{kk}u^{ll}u_{kli}w_{ik}w_{l}+u^{ii}u^{kk}w_{ik}^{2}$
$\displaystyle+u^{ii}u^{kk}w_{ik}w_{ik}-u^{kk}u^{ll}w_{kl}w_{k}w_{l}+2u^{kk}e^{-w}(f_{k}-w_{k}f)w_{k}$
$\displaystyle=$ $\displaystyle
u^{ii}u^{kk}\left(\sum_{l}u^{ll}u_{ikl}w_{l}\right)^{2}-2u^{ii}u^{kk}\left(\sum_{l}u^{ll}u_{ikl}w_{l}\right)w_{ik}+u^{ii}u^{kk}w_{ik}^{2}$
$\displaystyle+u^{ii}u^{kk}w_{ik}^{2}-u^{kk}u^{ll}w_{kl}w_{k}w_{l}+2u^{kk}e^{-w}(f_{k}-w_{k}f)w_{k}$
$\displaystyle=$ $\displaystyle
u^{ii}u^{kk}|w_{ik}-\sum_{l}u^{ll}u_{ikl}w_{l}|^{2}+u^{ii}u^{kk}w_{ik}^{2}-u^{kk}u^{ll}w_{kl}w_{k}w_{l}+2u^{kk}e^{-w}(f_{k}-w_{k}f)w_{k}.$

Next, we compute

(3.15) $\displaystyle u^{ij}(e^{\frac{1}{2}w}v)_{ij}(x_{0})$ $\displaystyle=$
$\displaystyle\frac{1}{4}e^{\frac{1}{2}w}u^{ii}w_{i}^{2}u^{kk}w_{k}^{2}+\frac{1}{2}e^{\frac{1}{2}w}u^{ii}w_{ii}v+e^{\frac{1}{2}w}u^{ii}w_{i}v_{i}+e^{\frac{1}{2}w}u^{ii}v_{ii}.$

Note that

(3.16) $\displaystyle u^{ii}w_{i}v_{i}$ $\displaystyle=$
$\displaystyle-u^{ii}u^{kk}u^{ll}u_{ikl}w_{k}w_{l}w_{i}+2u^{ii}u^{kk}w_{ki}w_{k}w_{i}$
$\displaystyle=$ $\displaystyle
u^{ii}u^{kk}w_{k}w_{i}(w_{ki}-\sum_{l}u^{ll}u_{ikl}w_{l})+u^{ii}u^{kk}w_{ki}w_{k}w_{i}.$

Combining (3.14), (3.15) and (3.16), we have

(3.17) $\displaystyle u^{ij}(e^{\frac{1}{2}w}v)_{ij}(x_{0})$ $\displaystyle=$
$\displaystyle
u^{ii}u^{kk}|w_{ik}-\sum_{l}u^{ll}u_{ikl}w_{l}+\frac{1}{2}w_{i}w_{k}|^{2}e^{\frac{1}{2}w}+u^{ii}u^{kk}w_{ik}^{2}e^{\frac{1}{2}w}$
$\displaystyle-\frac{3}{2}e^{-\frac{1}{2}w}fv+2e^{-\frac{1}{2}w}u^{kk}f_{k}w_{k}.$

By (3.12), we have

(3.18) $\displaystyle u^{ij}(\Delta
u)_{ij}(x_{0})=u^{ii}u^{jj}u_{ijk}^{2}+\Delta w.$

Let $z=e^{\frac{1}{2}w}v+\Delta u$. Note that

$\Delta
w=\sum_{k}w_{kk}\leq\frac{1}{2}(u^{kk})^{2}w_{kk}^{2}e^{\frac{1}{2}w}+\frac{1}{2}u_{kk}^{2}e^{-\frac{1}{2}w}\leq\frac{1}{2}(u^{kk})^{2}w_{kk}^{2}e^{\frac{1}{2}w}+\frac{1}{2}(\Delta
u)^{2}e^{-\frac{1}{2}w}.$

Then by (3.17) and (3.18), we have

(3.19) $\displaystyle u^{ij}z_{ij}(x_{0})$ $\displaystyle\geq$
$\displaystyle-C|f|v-C|f_{k}|v-C|f_{k}|+u^{kk}u^{ll}w_{kl}^{2}e^{\frac{1}{2}w}+\Delta
w$ $\displaystyle\geq$
$\displaystyle-C(|f|+|f_{k}|)v-C|f_{k}|+(u^{kk})^{2}w_{kk}^{2}e^{\frac{1}{2}w}-\frac{1}{2}(u^{kk})^{2}w_{kk}^{2}e^{\frac{1}{2}w}-\frac{1}{2}(\Delta
u)^{2}e^{-\frac{1}{2}w}$ $\displaystyle\geq$
$\displaystyle-C(|f|+|f_{k}|+\Delta u)z,$

where $C=C(\lambda,\Lambda)$. In the last inequality we used $z\geq\Delta
u\geq n(\det D^{2}u)^{\frac{1}{n}}\geq n\lambda^{\frac{1}{n}}$. Since (3.19)
is valid at every point in $B_{1}$, by (3.3), we have the following inequality

(3.20) $D_{j}(U^{ij}D_{i}z)\geq-gz-C\Delta uz\quad\text{in}\,\,B_{1},$

where $g=C(|f|+|f_{k}|)\in L^{q}(B_{1})$.

Next, we drive the upper bound of $z$ by integration and iteration. Let
$\eta\in C_{0}^{\infty}(B_{1})$ be a cutoff function. Multiplying (3.20) by
$\varphi=\eta^{2}z^{\beta-1}$ with $\beta\geq 2$ and by integration by parts,
we have

$\displaystyle(\beta-1)\int_{B_{1}}U^{ij}z_{i}z_{j}\eta^{2}z^{\beta-2}\,dx$
$\displaystyle\leq$ $\displaystyle-2\int_{B_{1}}U^{ij}z_{i}\eta_{j}\eta
z^{\beta-1}\,dx+C\int_{B_{1}}\Delta
u\cdot\eta^{2}z^{\beta}\,dx+\int_{B_{1}}g\eta^{2}z^{\beta}\,dx$
$\displaystyle\leq$
$\displaystyle\frac{\beta-1}{2}\int_{B_{1}}U^{ij}z_{i}z_{j}\eta^{2}z^{\beta-2}\,dx+\frac{2}{\beta-1}\int_{B_{1}}U^{ij}\eta_{i}\eta_{j}z^{\beta}\,dx$
$\displaystyle+C\int_{B_{1}}\Delta
u\cdot\eta^{2}z^{\beta}\,dx+\int_{B_{1}}g\eta^{2}z^{\beta}\,dx,$

which implies

$\displaystyle\int_{B_{1}}U^{ij}(\eta z^{\frac{\beta}{2}})_{i}(\eta
z^{\frac{\beta}{2}})_{j}\,dx\leq
C\beta\left(\int_{B_{1}}U^{ij}\eta_{i}\eta_{j}z^{\beta}\,dx+\int_{B_{1}}\Delta
u\cdot\eta^{2}z^{\beta}\,dx+\int_{B_{1}}g\eta^{2}z^{\beta}\,dx\right).$

Then by the Monge-Ampère Sobolev inequality [TW] and $(U^{ij})\leq(\Delta
u)^{n-1}I$, we have

(3.21) $\displaystyle\|\eta z^{\frac{\beta}{2}}\|_{L^{p}(B_{1})}^{2}\leq
C_{0}\beta\left(\int_{B_{1}}(\Delta
u)^{n-1}|D\eta|^{2}z^{\beta}\,dx+\int_{B_{1}}\Delta
u\cdot\eta^{2}z^{\beta}\,dx+\int_{B_{1}}g\eta^{2}z^{\beta}\,dx\right),$

where $p=2^{*}=\frac{2n}{n-2}$ for $n>2$ and $p>2$ for $n=2$. Then by Hölder’s
inequality

$\int_{B_{1}}g\eta^{2}z^{\beta}\,dx\leq\left(\int_{B_{1}}g^{q}\,dx\right)^{\frac{1}{q}}\left(\int_{B_{1}}|\eta
z^{\frac{\beta}{2}}|^{\frac{2q}{q-1}}\right)^{1-\frac{1}{q}}.$

Since $q>\frac{n}{2}$, we have $2<\frac{2q}{q-1}<2^{*}$. By interpolation
inequality, we obtain

$\|\eta z^{\frac{\beta}{2}}\|_{L^{\frac{2q}{q-1}}(B_{1})}\leq\varepsilon\|\eta
z^{\frac{\beta}{2}}\|_{L^{p}(B_{1})}+C(n,q)\varepsilon^{-\frac{n}{2q-n}}\|\eta
z^{\frac{\beta}{2}}\|_{L^{2}(B_{1})}.$

Then we choose
$\varepsilon=\left(4C_{0}\|g\|_{L^{q}(B_{1})}\beta+1\right)^{-\frac{1}{2}}$.
By (3.21),

(3.22) $\displaystyle\|\eta z^{\frac{\beta}{2}}\|_{L^{p}(B_{1})}^{2}\leq
C\beta^{\alpha}\left(\int_{B_{1}}(\Delta
u)^{n-1}|D\eta|^{2}z^{\beta}\,dx+\int_{B_{1}}\Delta
u\cdot\eta^{2}z^{\beta}\,dx+\int_{B_{1}}\eta^{2}z^{\beta}\,dx\right),$

where $\alpha=\frac{2q}{2q-n}$. Then by Hölder’s inequality, we have

(3.23) $\displaystyle\int_{B_{1}}(\Delta u)^{n-1}|D\eta|^{2}z^{\beta}\,dx$
$\displaystyle\leq$ $\displaystyle\|\Delta
u\|_{L^{p_{n}}(B_{1})}\cdot\||D\eta|z^{\frac{\beta}{2}}\|_{L^{\frac{2p_{n}}{p_{n}-n+1}}(B_{1})}^{2},$
(3.24) $\displaystyle\int_{B_{1}}\Delta u\cdot\eta^{2}z^{\beta}\,dx$
$\displaystyle\leq$ $\displaystyle\|\Delta u\|_{L^{p_{n}}(B_{1})}\cdot\|\eta
z^{\frac{\beta}{2}}\|_{L^{\frac{2p_{n}}{p_{n}-1}}(B_{1})}^{2}.$

Combining (3.22), (3.23) and (3.24), we get

$\displaystyle\|\eta z^{\frac{\beta}{2}}\|_{L^{p}(B_{1})}$ $\displaystyle\leq$
$\displaystyle
C\beta^{\alpha/2}\left(\||D\eta|z^{\frac{\beta}{2}}\|_{L^{\frac{2p_{n}}{p_{n}-n+1}}(B_{1})}+\|\eta
z^{\frac{\beta}{2}}\|_{L^{\frac{2p_{n}}{p_{n}-1}}(B_{1})}+\|\eta
z^{\frac{\beta}{2}}\|_{L^{2}(B_{1})}\right)$ $\displaystyle\leq$
$\displaystyle
C\beta^{\alpha/2}\left(\||D\eta|z^{\frac{\beta}{2}}\|_{L^{\frac{2p_{n}}{p_{n}-n+1}}(B_{1})}+\|\eta
z^{\frac{\beta}{2}}\|_{L^{\frac{2p_{n}}{p_{n}-n+1}}(B_{1})}\right).$

Now for any $0<r<R\leq 1$, we choose a cutoff function $\eta\in
C_{0}^{\infty}(B_{R})$ such that

$0\leq\eta\leq 1,\quad\eta\equiv
1\,\,\text{in}\,\,B_{r}\quad\text{and}\quad|D\eta|\leq\frac{2}{R-r}.$

Then we obtain

$\|z^{\frac{\beta}{2}}\|_{L^{p}(B_{r})}\leq\frac{C\beta^{\frac{\alpha}{2}}}{R-r}\|z^{\frac{\beta}{2}}\|_{L^{\frac{2p_{n}}{p_{n}-n+1}}(B_{R})}.$

By the assumption $p_{n}>\frac{n(n-1)}{2}$, we know
$p>\frac{2p_{n}}{p_{n}-n+1}$. Denote $\chi=p\frac{p_{n}-n+1}{2p_{n}}>1$. Then

(3.25) $\displaystyle\|z\|_{L^{\frac{\beta
p_{n}}{p_{n}-n+1}\chi}(B_{r})}\leq\frac{C^{\frac{2}{\beta}}\beta^{\frac{\alpha}{\beta}}}{(R-r)^{\frac{2}{\beta}}}\|z\|_{L^{\frac{\beta
p_{n}}{p_{n}-n+1}}(B_{R})}.$

We iterate (3.25) to get the desired estimate. Set

$\beta_{i}=2\chi^{i}\quad\text{and}\quad R_{i}=r+\frac{R-r}{2^{i}},\
i=0,1,2,\cdots,$

i.e.,

$\beta_{i}=\chi\beta_{i-1}\ \text{and}\ R_{i-1}-R_{i}=\frac{R-r}{2^{i}},\
i=1,2,\cdots.$

By (3.25),

$\|z\|_{L^{\frac{2p_{n}}{p_{n}-n+1}\chi^{i+1}}(B_{R_{i+1}})}\leq
C^{\sum_{j=0}^{i}\frac{2}{\beta_{j}}}\cdot\prod_{j=0}^{i}\beta_{j}^{\frac{\alpha}{\beta_{j}}}\cdot
4^{\sum_{j=0}^{i}\frac{j}{\beta_{j}}}\frac{1}{(R-r)^{\sum_{j=0}^{i}\frac{2}{\beta_{j}}}}\cdot\|z\|_{L^{\frac{2p_{n}}{p_{n}-n+1}}(B_{R})}.$

Letting $i\to\infty$, by Young’s inequality, we have

$\displaystyle\|z\|_{L^{\infty}(B_{r})}$
$\displaystyle\leq\frac{C}{(R-r)^{\frac{\chi}{\chi-1}}}\|z\|_{L^{\frac{2p_{n}}{p_{n}-n+1}}(B_{R})}$
$\displaystyle=\frac{C}{(R-r)^{\frac{\chi}{\chi-1}}}\|z\|_{L^{1}(B_{R})}^{\frac{p_{n}-n+1}{2p_{n}}}\cdot\|z\|_{L^{\infty}(B_{R})}^{\frac{p_{n}+n-1}{2p_{n}}}$
$\displaystyle\leq\frac{1}{2}\|z\|_{L^{\infty}(B_{R})}+\frac{C}{(R-r)^{\frac{\chi}{\chi-1}\cdot\frac{2p_{n}}{p_{n}-n+1}}}\|z\|_{L^{1}(B_{R})}.$

Set $f(t)=\|z\|_{L^{\infty}(B_{t})}$ for $t\in(0,1].$ Then for any
$0<r<R\leq\overline{R}<1$,

$f(r)\leq\frac{1}{2}f(R)+\frac{C}{(R-r)^{\frac{\chi}{\chi-1}\cdot\frac{2p_{n}}{p_{n}-n+1}}}\|z\|_{L^{1}(B_{\overline{R}})}.$

We apply Lemma 3.5 below to get

(3.26) $\displaystyle
f(r)\leq\frac{C}{(R-r)^{\frac{\chi}{\chi-1}\cdot\frac{2p_{n}}{p_{n}-n+1}}}\|z\|_{L^{1}(B_{\overline{R}})}.$

It remains to show $\|z\|_{L^{1}(B_{\overline{R}})}\leq C$. It is clear that
$\Delta u\in L^{1}(B_{\overline{R}})$, hence it is suffices to estimate the
integral of $v$. Let $\eta\in C_{0}^{\infty}(B_{1})$ be a cutoff function such
that $\eta\equiv 1$ in ${B_{\overline{R}}}$. Multiplying (1.5) by
$\varphi=\eta^{2}w$ and integrating by parts, we have

$\int_{B_{1}}U^{ij}w_{i}w_{j}\eta^{2}\,dx+2\int_{B_{1}}U^{ij}w_{i}\eta_{j}\eta
w\,dx=\int_{B_{1}}fw\eta^{2}\,dx.$

Then by the Cauchy inequality, we get

$\displaystyle\frac{1}{2}\int_{B_{1}}U^{ij}w_{i}w_{j}\eta^{2}\,dx$
$\displaystyle\leq$ $\displaystyle
2\int_{B_{1}}U^{ij}\eta_{i}\eta_{j}w^{2}\,dx+\int_{B_{1}}|f|w\eta^{2}\,dx$
$\displaystyle\leq$ $\displaystyle 2\int_{B_{1}}(\Delta
u)^{n-1}|D\eta|^{2}w^{2}\,dx+\int_{B_{1}}|f|w\eta^{2}\,dx.$

Hence $\int_{B_{\overline{R}}}u^{kl}w_{k}w_{l}\,dx\leq C$ follows by $\Delta
u\in L^{p_{n}}(B_{1})$, $f\in L^{q}(B_{1})$. Then we complete the proof by
choosing $r=\frac{1}{2}$ and $R=\overline{R}$ in (3.26). ∎

######  Lemma 3.5 ([HL, Lemma 4.3]).

Let $f(t)\geq 0$ be bounded in $[\tau_{0},\tau_{1}]$ with $\tau_{0}\geq 0$.
Suppose for $\tau_{0}\leq t<s\leq\tau_{1}$ we have

$f(t)\leq\beta f(s)+\frac{A}{(s-t)^{\alpha}}+B$

for some $\beta\in[0,1)$. Then for any $\tau_{0}\leq t<s\leq\tau_{1}$ there
holds

$f(t)\leq C(\alpha,\beta)\left\\{\frac{A}{(s-t)^{\alpha}}+B\right\\}.$

######  Remark 3.6.

Note that for $n=2$, we have already got an $L^{p}$-bound of $\Delta u$ in
Theorem 3.3 for all $p\in[1,+\infty)$. However, it is still unknown how to
obtain $L^{p}$-bound of $\Delta u$ when $n\geq 3$ by the integral method used
in Theorem 3.3. We also do not know how to obtain higher estimates for other
exponents $\theta$ by the above method.

##  4\. An interior estimate in higher dimensions

In this section, we will prove Theorem 1.5. By [CG, Ca], it suffices to get
the interior estimates on the upper and lower bound of $\det D^{2}u$. Note
that there exists a constant $c_{n}>0$ such that

(4.1) $\frac{\det D^{2}u}{c_{n}\cdot\Delta u}I\leq U^{ij}\leq c_{n}(\Delta
u)^{n-1}I.$

In view of (3.2) and (4.1), we first consider the following degenerate linear
elliptic equation

(4.2) $-D_{j}(a_{ij}(x)D_{i}u)+c(x)u=f(x)\quad\text{in}\,\,\Omega.$

######  Lemma 4.1.

Assume $\\{a_{ij}(x)\\}$ satisfies

(4.3) $\frac{d(x)}{\lambda(x)}\leq
a_{ij}(x)\leq\lambda(x)^{n-1}\quad\text{in}\,\,\,B_{1},$

where $\lambda(x)\in L^{p}(B_{1})$, $d^{-1}\in L^{q}(B_{1})$. Assume
$c(x),f(x)\in L^{p_{0}}(B_{1})$. Let $u\in W^{1,p}(B_{1})$ be a subsolution in
the following sense:

(4.4)
$\int_{B_{1}}a_{ij}D_{i}uD_{j}\varphi+cu\varphi\,dx\leq\int_{B_{1}}f\varphi\,dx\quad\text{for\,\,any}\,\,\varphi\in
W^{1,p}_{0}(B_{1})\text{\ and}\,\,\varphi\geq 0.$

Suppose $p$, $q$ and $p_{0}$ satisfy $\frac{n}{p}+\frac{1}{q}<\frac{2}{n}$ and
$p_{0}\geq\frac{p}{n-1}$. Then

$\sup_{B_{1/2}}u\leq
C\left(\|u^{+}\|_{L^{1}(B_{1})}+\|f\|_{L^{p_{0}}(B_{1})}\right),$

where $C$ depends only on $p$, $\|\lambda\|_{L^{p}(B_{1})}$,
$\|d^{-1}\|_{L^{q}(B_{1})}$ and $\|c\|_{L^{p_{0}}(B_{1})}$.

###### Proof.

For some $k>0$ and $m>0$, set $\overline{u}=u^{+}+k$ and

$\overline{u}_{m}=\left\\{\begin{array}[]{lc}\overline{u},&\ u<m,\\\ k+m,&\
u\geq m.\end{array}\right.$

Let $\eta\in C^{\infty}_{0}(B_{1})$. We choose a test function
$\varphi=\eta^{2}(\overline{u}_{m}^{\beta}\overline{u}-k^{\beta+1})\in
W^{1,p}_{0}(B_{1})$ for some $\beta\geq 0$ to be determined later.
Substituting $\varphi$ into (4.4), we have

$\displaystyle\beta\int_{B_{1}}a_{ij}D_{i}\overline{u}D_{j}\overline{u}_{m}\overline{u}_{m}^{\beta-1}\overline{u}\eta^{2}\,dx+\int_{B_{1}}a_{ij}D_{i}\overline{u}D_{j}\overline{u}\,\overline{u}_{m}^{\beta}\eta^{2}\,dx$
$\displaystyle\ \ \ \ \ \ \
\leq-2\int_{B_{1}}a_{ij}D_{i}\overline{u}D_{j}\eta\overline{u}_{m}^{\beta}\overline{u}\eta\,dx+\int_{B_{1}}\left(|c|\eta^{2}\overline{u}_{m}^{\beta}\overline{u}^{2}+|f|\eta^{2}\overline{u}_{m}^{\beta}\overline{u}\right)\,dx.$

Note that $D\overline{u}=D\overline{u}_{m}$ in $\\{u<m\\}$ and
$D\overline{u}_{m}=0$ in $\\{u\geq m\\}$. By the Cauchy inequality, we have

$\displaystyle\beta$
$\displaystyle\int_{B_{1}}a_{ij}D_{i}\overline{u}_{m}D_{j}\overline{u}_{m}\overline{u}_{m}^{\beta-1}\overline{u}\eta^{2}\,dx+\int_{B_{1}}a_{ij}D_{i}\overline{u}D_{j}\overline{u}\,\overline{u}_{m}^{\beta}\eta^{2}\,dx$
$\displaystyle\leq\frac{1}{2}\int_{B_{1}}a_{ij}D_{i}\overline{u}D_{j}\overline{u}\,\overline{u}_{m}^{\beta}\eta^{2}\,dx+4\int_{B_{1}}a_{ij}D_{i}\eta
D_{j}\eta\overline{u}_{m}^{\beta}\overline{u}^{2}\,dx+\int_{B_{1}}\left(|c|\eta^{2}\overline{u}_{m}^{\beta}\overline{u}^{2}+|f|\eta^{2}\overline{u}_{m}^{\beta}\overline{u}\right)\,dx.$

Then by (4.3)

(4.5)
$\displaystyle\beta\int_{B_{1}}\frac{d}{\lambda(x)}|D\overline{u}_{m}|^{2}\overline{u}_{m}^{\beta}\eta^{2}\,dx+\frac{1}{2}\int_{B_{1}}\frac{d}{\lambda(x)}|D\overline{u}|^{2}\overline{u}_{m}^{\beta}\eta^{2}\,dx$
$\displaystyle\leq$
$\displaystyle\beta\int_{B_{1}}a_{ij}D_{i}\overline{u}_{m}D_{j}\overline{u}_{m}\overline{u}_{m}^{\beta-1}\overline{u}\eta^{2}\,dx+\frac{1}{2}\int_{B_{1}}a_{ij}D_{i}\overline{u}D_{j}\overline{u}\,\overline{u}_{m}^{\beta}\eta^{2}\,dx$
$\displaystyle\leq$ $\displaystyle
4\int_{B_{1}}\lambda(x)^{n-1}|D\eta|^{2}\overline{u}_{m}^{\beta}\overline{u}^{2}\,dx+\int_{B_{1}}c_{0}\eta^{2}\overline{u}_{m}^{\beta}\overline{u}^{2}\,dx,$

where $c_{0}=|c|+\frac{|f|}{k}$. Choose $k=\|f\|_{L^{p_{0}}(B_{1})}$ if $f$ is
not identically 0. Otherwise choose arbitrary $k>0$ and let $k\to 0^{+}$. Let
$w=\overline{u}_{m}^{\frac{\beta}{2}}\overline{u}$. There holds

$|Dw|^{2}=\left|\frac{\beta}{2}\overline{u}_{m}^{\frac{\beta}{2}-1}\overline{u}_{m}D\overline{u}_{m}+\overline{u}_{m}^{\frac{\beta}{2}}D\overline{u}\right|^{2}\leq(1+\beta)(\beta\overline{u}_{m}^{\beta}|D\overline{u}_{m}|^{2}+\overline{u}_{m}^{\beta}|D\overline{u}|^{2}).$

Therefore by (4.5)

$\int_{B_{1}}\frac{d}{\lambda(x)}|Dw|^{2}\eta^{2}\,dx\leq
8(1+\beta)\int_{B_{1}}\lambda(x)^{n-1}|D\eta|^{2}w^{2}\,dx+2(1+\beta)\int_{B_{1}}c_{0}w^{2}\eta^{2}\,dx.$

By $|D(w\eta)|^{2}\leq 2|Dw|^{2}\eta^{2}+2|D\eta|^{2}w^{2}$, we have

(4.6) $\displaystyle\int_{B_{1}}\frac{d}{\lambda(x)}|D(w\eta)|^{2}\,dx$
$\displaystyle\leq$
$\displaystyle\int_{B_{1}}\left(\frac{2d}{\lambda(x)}+16(1+\beta)\lambda(x)^{n-1}\right)|D\eta|^{2}w^{2}\,dx+4(1+\beta)\int_{B_{1}}c_{0}w^{2}\eta^{2}\,dx$
$\displaystyle\leq$ $\displaystyle
18(1+\beta)\int_{B_{1}}\lambda(x)^{n-1}|D\eta|^{2}w^{2}\,dx+4(1+\beta)\int_{B_{1}}c_{0}w^{2}\eta^{2}\,dx.$

Next, we deal with the $\lambda(x)$ in the above estimate. By the assumptions
$\lambda\in L^{p}(B_{1})$ and $d^{-1}\in L^{q}(B_{1})$, we have $\lambda
d^{-1}\in L^{\frac{pq}{p+q}}(B_{1})$. By Hölder’s inequality, we have

(4.7)
$\displaystyle\left\|D(w\eta)\right\|_{L^{\frac{2pq}{pq+p+q}}(B_{1})}^{2}$
$\displaystyle\leq$ $\displaystyle\|\lambda
d^{-1}\|_{L^{\frac{pq}{p+q}}(B_{1})}\cdot\int_{B_{1}}\frac{1}{\lambda(x)d^{-1}}\left|D(w\eta)\right|^{2}\,dx,$
(4.8) $\displaystyle\int_{B_{1}}\lambda(x)^{n-1}|D\eta|^{2}w^{2}\,dx$
$\displaystyle\leq$
$\displaystyle\|\lambda\|_{L^{p}(B_{1})}\cdot\|wD\eta\|_{L^{\frac{2p}{p-n+1}}(B_{1})}^{2},$
(4.9) $\displaystyle\int_{B_{1}}c_{0}w^{2}\eta^{2}\,dx$ $\displaystyle\leq$
$\displaystyle\|c_{0}\|_{L^{p_{0}}(B_{1})}\cdot\left(\int_{B_{1}}(\eta
w)^{\frac{2{p_{0}}}{{p_{0}}-1}}\,dx\right)^{1-\frac{1}{{p_{0}}}}.$

Combining (4.6)-(4.9), we get

$\displaystyle\|D(w\eta)\|_{L^{\frac{2pq}{pq+p+q}}(B_{1})}$
$\displaystyle\leq$ $\displaystyle
C(1+\beta)^{\frac{1}{2}}\left(\|wD\eta\|_{L^{\frac{2p}{p-n+1}}(B_{1})}+\|w\eta\|_{L^{\frac{2{p_{0}}}{{p_{0}}-1}}(B_{1})}\right)$
$\displaystyle\leq$ $\displaystyle
C(1+\beta)^{\frac{1}{2}}\left(\|wD\eta\|_{L^{\frac{2p}{p-n+1}}(B_{1})}+\|w\eta\|_{L^{\frac{2p}{p-n+1}}(B_{1})}\right)$

for $C$ depending on $\|\lambda\|_{L^{p}(B_{1})}$,
$\|d^{-1}\|_{L^{q}(B_{1})}$, $\|c\|_{L^{p_{0}}(B_{1})}$. Here we used
$p_{0}\geq\frac{n-1}{p}$. By the Sobolev inequality

$\|w\eta\|_{L^{\alpha}(B_{1})}\leq
C(1+\beta)^{\frac{1}{2}}\left(\|wD\eta\|_{L^{\frac{2p}{p-n+1}}(B_{1})}+\|w\eta\|_{L^{\frac{2p}{p-n+1}}(B_{1})}\right),$

where

(4.10) $\frac{1}{\alpha}=\frac{pq+p+q}{2pq}-\frac{1}{n}.$

Now for any $0<r<R\leq 1$, we choose a cutoff function $\eta\in
C_{0}^{\infty}(B_{R})$ such that

$0\leq\eta\leq 1,\quad\eta\equiv
1\,\,\text{in}\,\,B_{r}\quad\text{and}\quad|D\eta|\leq\frac{2}{R-r}.$

Then we obtain

$\|w\|_{L^{\alpha}(B_{r})}\leq\frac{C(1+\beta)^{\frac{1}{2}}}{R-r}\|w\|_{L^{\frac{2p}{p-n+1}}(B_{R})}.$

By (4.10) and the assumption $\frac{n}{p}+\frac{1}{q}<\frac{2}{n}$, we have
$\alpha>\frac{2p}{p-n+1}$. We can do the iteration as follows.

Recalling the definition of $w$, we have

$\|\overline{u}_{m}^{\frac{\beta}{2}}\overline{u}\|_{L^{\alpha}(B_{r})}\leq\frac{C(1+\beta)^{\frac{1}{2}}}{R-r}\|\overline{u}_{m}^{\frac{\beta}{2}}\overline{u}\|_{L^{\frac{2p}{p-n+1}}(B_{R})}.$

Set $\gamma=\beta+2\geq 2$. By $\overline{u}_{m}\leq\overline{u}$, we obtain

$\|\overline{u}_{m}^{\frac{\gamma}{2}}\|_{L^{\alpha}(B_{r})}\leq\frac{C\gamma^{\frac{1}{2}}}{R-r}\|\overline{u}^{\frac{\gamma}{2}}\|_{L^{\frac{2p}{p-n+1}}(B_{R})}.$

Letting $m\to\infty$, we get

$\|\overline{u}^{\frac{\gamma}{2}}\|_{L^{\alpha}(B_{r})}\leq\frac{C\gamma^{\frac{1}{2}}}{R-r}\|\overline{u}^{\frac{\gamma}{2}}\|_{L^{\frac{2p}{p-n+1}}(B_{R})},$

i.e.,

$\|\overline{u}\|_{L^{{\frac{\gamma}{2}}\alpha}(B_{r})}\leq\frac{(C\gamma)^{\frac{1}{\gamma}}}{(R-r)^{\frac{2}{\gamma}}}\|\overline{u}\|_{L^{{\frac{\gamma}{2}}\frac{2p}{p-n+1}}(B_{R})}.$

Denote $\chi=\alpha\cdot\frac{p-n+1}{2p}>1$. Then

(4.11)
$\|\overline{u}\|_{L^{{\frac{{\gamma}p}{p-n+1}\chi}}(B_{r})}\leq\frac{(C\gamma)^{\frac{1}{\gamma}}}{(R-r)^{\frac{2}{\gamma}}}\|\overline{u}\|_{L^{{\frac{{\gamma}p}{p-n+1}}}(B_{R})}.$

We iterate (4.11) to get the desired estimate. Set

$\gamma_{i}=2\chi^{i}\quad\text{and}\quad R_{i}=r+\frac{R-r}{2^{i}},\
i=0,1,2,\cdots,$

i.e.,

$\gamma_{i}=\chi\gamma_{i-1}\ \text{and}\ R_{i-1}-R_{i}=\frac{R-r}{2^{i}},\
i=1,2,\cdots.$

By (4.11),

$\|\overline{u}\|_{L^{\frac{2p}{p-n+1}\chi^{i+1}}(B_{R_{i+1}})}\leq
C^{\sum_{j=0}^{i}\frac{1}{\gamma_{j}}}\cdot\prod_{j=0}^{i}\gamma_{j}^{\frac{1}{\gamma_{j}}}\cdot
4^{\sum_{j=0}^{i}\frac{j}{\gamma_{j}}}\frac{1}{(R-r)^{\sum_{j=0}^{i}\frac{2}{\gamma_{j}}}}\cdot\|\overline{u}\|_{L^{\frac{2p}{p-n+1}}(B_{R})}.$

Letting $i\to\infty$, by Young’s inequality, we have

$\displaystyle\|\overline{u}\|_{L^{\infty}(B_{r})}$
$\displaystyle\leq\frac{C}{(R-r)^{\frac{\chi}{\chi-1}}}\|\overline{u}\|_{L^{\frac{2p}{p-n+1}}(B_{R})}$
$\displaystyle=\frac{C}{(R-r)^{\frac{\chi}{\chi-1}}}\|\overline{u}\|_{L^{1}(B_{R})}^{\frac{p-n+1}{2p}}\cdot\|\overline{u}\|_{L^{\infty}(B_{R})}^{\frac{p+n-1}{2p}}$
$\displaystyle\leq\frac{1}{2}\|\overline{u}\|_{L^{\infty}(B_{R})}+\frac{C}{(R-r)^{\frac{\chi}{\chi-1}\cdot\frac{2p}{p-n+1}}}\|\overline{u}\|_{L^{1}(B_{R})}.$

Set $f(t)=\|\overline{u}\|_{L^{\infty}(B_{t})}$ for $t\in(0,1].$ Then for any
$0<r<R\leq 1$

$f(r)\leq\frac{1}{2}f(R)+\frac{C}{(R-r)^{\frac{\chi}{\chi-1}\cdot\frac{2p}{p-n+1}}}\|\overline{u}\|_{L^{1}(B_{1})}.$

We apply Lemma 3.5 to get

$f(r)\leq\frac{C}{(R-r)^{\frac{\chi}{\chi-1}\cdot\frac{2p}{p-n+1}}}\|\overline{u}\|_{L^{1}(B_{1})}.$

The lemma follows by choosing $r=\frac{1}{2}$ and $R=1$. ∎

Now we can use Lemma 4.1 to obtain the interior estimates for (1.1). For
simplicity, we only consider the homogeneous equation.

###### Proof of Theorem 1.5.

Denote $a_{ij}=U^{ij}$, $\lambda=\Delta u$ and $d=\det D^{2}u$. Firstly, we
consider the case $0\leq\theta<1$. We apply Lemma 4.1 to equation (1.3), which
yields

$\sup_{B_{1/2}}w\leq C\|w\|_{L^{1}(B_{1})}\leq
C\|w\|_{L^{\frac{q}{1-\theta}}(B_{1})}.$

Since $w=(\det D^{2}u)^{-(1-\theta)}\in L^{\frac{q}{1-\theta}}(B_{1}),$ we
know $\sup_{B_{1/2}}\\{(\det D^{2}u)^{-1}\\}\leq C$. For the upper bound of
the determinant, we set $w=\frac{1}{v}$. A direct calculation yields

$w_{i}=-\frac{v_{i}}{v^{2}},\ \
w_{ij}=\frac{2v_{i}v_{j}}{v^{3}}-\frac{v_{ij}}{v^{2}}.$

Then $v$ satisfies

$0=U^{ij}w_{ij}=\frac{2}{v^{3}}U^{ij}v_{i}v_{j}-\frac{1}{v^{2}}U^{ij}v_{ij}\geq-\frac{1}{v^{2}}U^{ij}v_{ij},$

i.e.,

$-D_{i}(U^{ij}D_{j}v)\leq 0\quad\text{in}\,\,B_{1}.$

Similarly, we use Lemma 4.1 to obtain

$\sup_{B_{1/2}}v\leq C\|v\|_{L^{1}(B_{1})}\leq
C\|v\|_{L^{\frac{p}{n(1-\theta)}}(B_{1})},$

which implies $\sup_{B_{1/2}}\det D^{2}u\leq C$.

Next, we consider the case $\theta=1$. Write $w=\log d$, where $d\in
L^{\frac{p}{n}}(B_{1})$. Then we have

$w_{i}=\frac{d_{i}}{d},\ \ w_{ij}=-\frac{d_{i}d_{j}}{d^{2}}+\frac{d_{ij}}{d},$

which yields

$0=U^{ij}w_{ij}=-\frac{1}{d^{2}}U^{ij}d_{i}d_{j}+\frac{1}{d}U^{ij}d_{ij}\leq\frac{1}{d}U^{ij}d_{ij},$

i.e.,

$-D_{i}(U^{ij}D_{j}d)\leq 0\quad\text{in}\,\,B_{1}.$

Similarly, write $w=-\log z$ with $z=(\det D^{2}u)^{-1}\in L^{q}(B_{1})$. We
have

$w_{i}=-\frac{z_{i}}{z},\ \ w_{ij}=\frac{z_{i}z_{j}}{z^{2}}-\frac{z_{ij}}{z},$

which means

$0=U^{ij}w_{ij}=\frac{1}{z^{2}}U^{ij}z_{i}z_{j}-\frac{1}{z}U^{ij}z_{ij}\geq-\frac{1}{z}U^{ij}z_{ij},$

i.e.,

$-D_{i}(U^{ij}D_{j}z)\leq 0\quad\text{in}\,\,B_{1}.$

Then we use Lemma 4.1 to obtain the bounds of $\det D^{2}u$.

Once we have the determinant estimates, all the interior estimates follow. ∎

######  Remark 4.2.

To get the upper bound of $\det D^{2}u$, we only need $u\in W^{2,p}(B_{1})$
with $p>\frac{n^{2}}{2}$. In fact, by taking $\det
D^{2}u=v^{\frac{1}{1-\theta}}\ (0\leq\theta<1)$, where $v$ defined as the
proof of Theorem 1.5, condition (4.3) in Lemma 4.1 becomes

$\frac{v^{\frac{1}{1-\theta}}}{\lambda(x)}\leq a_{ij}(x)\leq\lambda(x)^{n-1}$

Then the upper bound can be obtained by similar arguments as in Lemma 4.1
without assumption on $(\det D^{2}u)^{-1}$.

######  Remark 4.3.

We can also consider the inhomogeneous equation

(4.12) $D_{j}(U^{ij}D_{i}w)=f\quad\,\,\text{in}\,\,B_{1}$

where $f\in L^{{p_{0}}}(B_{1}).$ For the lower bound of $\det D^{2}u$, if we
assume ${p_{0}}\geq\frac{p}{n-1}$, we can apply Lemma 4.1 to (4.12) directly
for all $\theta\in[0,1)$. However, for the upper bound of $\det D^{2}u$,
$v=\frac{1}{w}$ satisfies

(4.13) $-D_{j}(U^{ij}D_{i}v)\leq fv^{2}\quad\,\,\text{in}\,\,B_{1}.$

Even we assume ${p_{0}}=\infty$, we only know $fv\in
L^{\frac{p}{n(1-\theta)}}(B_{1})$. Then we can only apply Lemma 4.1 to the
equation (4.13) to get the upper bound of $\det D^{2}u$ when
$\frac{1}{n}\leq\theta<1$. Hence, we have all the higher interior estimates
for $\theta\in[\frac{1}{n},1)$. Note that the case $\theta=0$ (Abreu’s
equation) and $\theta=\frac{1}{n+2}$ (affine mean curvature equation) are not
included.

Finally, we prove a Liouville type theorem.

###### Proof of Corollary 1.6.

For $u$ in any $B_{R}\subset\mathbb{R}^{n}$, define
$u_{R}(x)=\frac{1}{R^{2}}u(Rx)$. Tthen we know that $U_{R}^{ij}(w_{R})_{ij}=0$
in $B_{1}.$ Since $u$ satisfies (1.7), we know $u_{R}$ satisfies

$\int_{B_{1}}|D^{2}u_{R}|^{p}+(\det D^{2}u_{R})^{-q}\,dx\leq C.$

Applying Theorem 1.5 to $u_{R}$, we find that there exists a $C>0$ independent
of $R$, such that $\|u_{R}\|_{C^{4,\alpha}(B_{1/2})}\leq C$. In particular, we
know that $\|D^{3}u_{R}\|_{L^{\infty}(B_{1/2})}\leq C$. Hence

$\|D^{3}u(Rx)\|_{L^{\infty}(B_{\frac{1}{2}})}\leq\frac{C}{R},$

i.e.,

$\|D^{3}u(x)\|_{L^{\infty}(B_{\frac{R}{2}})}\leq\frac{C}{R}.$

Let $R\to+\infty$, we have $D^{3}u\equiv 0,$ which means $u$ is a quadratic
function. ∎

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