# Pogorelov type $C^{2}$ estimates for Sum Hessian equations and a rigidity
theorem

Yue Liu  Yue Liu, School of Mathematical Science, Jilin University, Changchun,
130012, Jilin Province, P.R. China<EMAIL_ADDRESS>and  Changyu
Ren  Changyu Ren, School of Mathematical Science, Jilin University, Changchun,
130012, Jilin Province, P.R. China<EMAIL_ADDRESS>
###### Abstract.

We mainly study Pogorelov type $C^{2}$ estimates for solutions to the
Dirichlet problem of Sum Hessian equations. We establish respectively
Pogorelov type $C^{2}$ estimates for $k$-convex solutions and admissible
solutions under some conditions. Furthermore, we apply such estimates to
obtain a rigidity theorem for $k$-convex solutions of Sum Hessian equations in
Euclidean space.

###### Key words and phrases:

Hessian equations, Pogorelov type $C^{2}$ estimate, Rigidity theorem.

###### 1991 Mathematics Subject Classification:

53C23, 35J60, 53C42

Research of the last author is supported by an NSFC Grant No. 11871243

##  1\. Introduction

In this paper, we consider the following Dirichlet problem of Sum Hessian
equations,

(1.1)
$\left\\{\begin{array}[]{lll}{{\sigma}_{k}}(D^{2}u)+\alpha{{\sigma}_{k-1}}(D^{2}u)=f(x,u,Du),&in&\Omega,\\\
u=0,&on&\partial{\Omega}.\end{array}\right.$

Here, $u$ is an unknown function defined on $\Omega$, $\alpha$ is a positive
constant. Denote by $Du$ and $D^{2}u$ the gradient and the Hessian of $u$. We
also require $f>0$ and smooth enough with respect to every variables. Let
$\sigma_{k}(D^{2}u)=\sigma_{k}(\lambda(D^{2}u))$ denotes the $k$-th elementary
symmetric function of the eigenvalues of the Hessian matrix $D^{2}u$. Namely,
for $\lambda=(\lambda_{1},\cdots,\lambda_{n})\in\mathbb{R}^{n}$,

$\sigma_{k}(\lambda)=\displaystyle\sum_{1\leq i_{1}<\cdots<i_{k}\leq
n}\lambda_{i_{1}}\cdots\lambda_{i_{k}}.$

As we all know, the following classic $k$-Hessian equation

(1.2) $\sigma_{k}(D^{2}u)=f(x,u,Du),~{}~{}~{}~{}~{}~{}~{}~{}x\in\Omega$

is an important research content in the field of fully nonlinear partial
differential equations and geometric analysis, which is closely related to
many important geometric problems, such as [1, 8, 9, 11, 17, 18, 26, 28, 33].
The equation (1.1) is a natural extension of (1.2), which is a Hessian type
equation formed by linear combination of the $k$ Hessian operators. These
Hessian type equations have been studied and applied in geometry. Harvey and
Lawson [21] considered the special Lagrangian equations which is one of this
category. Krylov [23] and Dong [14] also considered equations close to this
type and obtained the curvature estimates using the concavity of the
operators. In [20], Guan and Zhang studied the curvature estimates for
equations of such type with the right hand side not depending on gradient term
but with coefficients depending on the position of the hypersurfaces. The
geometrical problems in the hyperbolic space also reduce to equations of this
type [15].

An important problem in the study of Hessian equations (1.2) is how to obtain
$C^{2}$ estimates. For the relative work, we refer the readers to [7, 8, 13,
16, 17, 18, 19]. Especially when the right hand side function depends on the
gradient, how to obtain $C^{2}$ estimates for equations (1.2) is a
longstanding problem [17].

In this paper, we are interested in Pogorelov type $C^{2}$ estimates for the
Dirichlet problem (1.1). Pogorelov type $C^{2}$ estimates is a type of
interior $C^{2}$ estimates with boundary information. This type of $C^{2}$
estimates were established at first by Pogorelov [28] for Monge-Ampère
equations, which play an important role in studying the regularity for fully
nonlinear equations. For example, Pogorelov type $C^{2}$ estimates were
established in order to study the regularity for degenerate Monge-Ampère
equations [4, 31]. For $k$-Hessian equations (1.2), if the right hand side
function is $f=f(x,u)$, Chou and Wang [13] established Pogorelov $C^{2}$
estimates of $k$ convex solutions:

$(-u)^{1+\delta}\Delta u\leq C.$

Sheng-Ubas-Wang [32] established Pogorelov type curvature estimates for the
curvature equations of the graphic hypersurface including the Hessian
equations. If the right hand side function depends on the gradient, Li-Wang
and the last author [25] established Pogorelov type $C^{2}$ estimates for
$k+1$ convex solutions for equation (1.2):

$(-u)\Delta u\leq C.$

In particular, for $2$ convex solutions of the case $k=2$, the following
Pogorelov type $C^{2}$ estimates hold:

$(-u)^{\beta}\Delta u\leq C.$

Here, positive constants $\beta$ and $C$ depend on the domain $\Omega$, $f$,
$\displaystyle\sup_{\Omega}|u|$ and $\displaystyle\sup_{\Omega}|Du|$.

A natural question is whether Pogorelov type $C^{2}$ estimates are still valid
for the Dirichlet problem (1.1)?

First, we give the definition of $k$ convex solution [8]:

######  Definition 1.1.

For a domain $\Omega\subset\mathbb{R}^{n}$, a function $v\in C^{2}(\Omega)$ is
called $k$ convex if the eigenvalues
$\lambda(x)=(\lambda_{1}(x),\cdots,\lambda_{n}(x))$ of the Hessian
$\nabla^{2}v(x)$ is in $\Gamma_{k}$ for all $x\in\Omega$, where $\Gamma_{k}$
is the Garding’s cone

$\Gamma_{k}=\\{\lambda\in\mathbb{R}^{n}\ |\quad\sigma_{m}(\lambda)>0,\quad
m=1,\cdots,k\\}.$

The main results are as follows:

######  Theorem 1.2.

Let $\Omega$ be a bounded domain in $\mathbb{R}^{n}$, and let $u\in
C^{4}(\Omega)\cap C^{0,1}(\overline{\Omega})$ be a $k-1$ convex solution for
the Dirichlet problem (1.1), where $f(x,u,p)\in
C^{1,1}(\overline{\Omega}\times\mathbb{R}\times\mathbb{R}^{n})$ is a positive
function. Then we have

(a) For $k=2,k=3$, there are some positive constants $\beta$ and $C$, such
that

(1.3) $(-u)^{\beta}\Delta u\leq C.$

(b) For $1<k\leq n$, if $f^{\frac{1}{k}}(x,u,p)$ is a convex function with
respect to $p$, there are some positive constants $\delta$ and $C$, such that

(1.4) $(-u)^{1+\delta}\Delta u\leq C.$

Here, $\delta>0$ can be arbitrarily small, the positive constants $\beta$ and
$C$ depend on the domain $\Omega$, $\alpha$, $f$,
$\displaystyle\sup_{\Omega}|u|$ and $\displaystyle\sup_{\Omega}|Du|$.

######  Remark 1.3.

In [24], Li-Wang and the last author proved that Sum Hessian operator
$\sigma_{k}+\alpha\sigma_{k-1}$ is elliptic operator in the set

$\tilde{\Gamma}_{k}=\Gamma_{k-1}\cap\\{\lambda|\sigma_{k}(\lambda)+\alpha\sigma_{k-1}(\lambda)>0\\},$

i.e., $\tilde{\Gamma}_{k}$ is the admissible set of the equations (1.1). Here,
in addition to the requirements for $k-1$ convex solutions, the right hand
side function $f>0$ ensures the ellipticity of the equations (1.1).

######  Remark 1.4.

Similar to the result in [24], because $f$ depends on the gradient, the
constant $\beta$ is large in theorem 1.2, and we cannot improve $\beta$ to 1
or $1+\delta$.

######  Theorem 1.5.

Let $\Omega$ be a bounded domain in $\mathbb{R}^{n}$, and let $u\in
C^{4}(\Omega)\cap C^{0,1}(\overline{\Omega})$ be a $k$ convex solution for the
Dirichlet problem (1.1), where $f(x,u,p)\in
C^{1,1}(\overline{\Omega}\times\mathbb{R}\times\mathbb{R}^{n})$ is a positive
function. Then we have

$(-u)\Delta u\leq C.$

Here, positive constant $C$ depends on the domain $\Omega$, $\alpha$, $f$,
$\displaystyle\sup_{\Omega}|u|$ and $\displaystyle\sup_{\Omega}|Du|$.

An application of the interior $C^{2}$ estimates is to prove rigidity theorems
for equations. For the following $k$-Hessian equations in $n$-dimensional
Euclidean spaces,

(1.5) $\sigma_{k}(D^{2}u)=1,$

Chang and Yuan [11] proposed a problem that: Are the entire solutions of (1.5)
with lower bound only quadratic polynomials?

For $k=1$, equation (1.5) is a linear equation. It is a obvious result coming
from the Liouville property of the harmonic functions. For $k=n$, equation
(1.5) is Monge-Ampère equation, which is a well known theorem. The work of
Jörgens [22], Calabi [10], Pogorelov [27, 28] proved that every entire
strictly convex solution is a quadratic polynomial. Then, Cheng and Yau [12]
gave another more geometry proof. In 2003, Caffarelli and Li [6] extended the
theorem of Jörgens, Calabi and Pogorelov.

For general $1<k<n$, there are few related results. For $k=2$, Chang and Yuan
[11] have proved that, if

$D^{2}u\geq\delta-\sqrt{\dfrac{2n}{n-1}},$

for any $\delta>0$, then the entire solution of the equation (1.5) only has
quadratic polynomials. For general $k$, Bao-Chen-Guan-Ji [3] proved that
strictly entire convex solutions of equation (1.5) satisfying a quadratic
growth are quadratic polynomials. Here, the quadratic growth means that: there
are some positive constants $c,b$ and sufficiently large $R$, such that

(1.6) $u(x)\geq c|x|^{2}-b,|x|\geq R.$

[25] relaxed the condition of strictly convex solutions in [3] to $k+1$ convex
solutions, and proved that the entire $k+1$ convex solutions of the equations
(1.5) with quadratic growth are quadratic polynomials. In 2016, Warren [34]
proved that equation (1.5) has non-polynomial entire $k$ convex solutions for
$n\geq 2k-1$ by constructing examples.

In this paper, we apply Pogorelov type $C^{2}$ estimates to prove the rigidity
theorem of the following Sum Hessian equations in Euclidean space,

(1.7) $\sigma_{k}(D^{2}u)+\alpha\sigma_{k-1}(D^{2}u)=1.$

######  Theorem 1.6.

The entire $k$ convex solutions of the equations (1.7) defined in
$\mathbb{R}^{n}$ with quadratic growth (1.6) are quadratic polynomials.

At last, using the idea of Warren [34], we can give an example of non-
polynomial entire $k-1$ convex solution.

######  Example 1.7.

When $n=3$, the $1$-convex function

$u(x,y,t)=\dfrac{e^{4t}-1}{4}(x^{2}+y^{2})+\dfrac{1}{16}(\dfrac{7e^{-4t}}{4}-\dfrac{e^{4t}}{4}-4t^{2})$

solves

$\sigma_{2}(D^{2}u)+\sigma_{1}(D^{2}u)=1.$

The organization of our paper is as follows: In section $2$, we give the
preliminary knowledge. The proofs of Theorem 1.2 and Theorem 1.5 are
respectively given in section 3 and section 4\. In the last section, we prove
the rigidity theorem of the equations (1.7).

##  2\. Preliminary

In this section, we give the preliminary knowledge related to our theorems and
their proofs. To make it convenient, we denote

$S_{k}(\lambda):=\sigma_{k}(\lambda)+\alpha\sigma_{k-1}(\lambda).$

For the related notation of $\sigma_{k}$, we use the denotation in [24, 29].
Let $\lambda=(\lambda_{1},\cdots,\lambda_{n})\in\mathbb{R}^{n}$, we have

(i) $S_{k}^{pp}(\lambda):=\dfrac{\partial
S_{k}(\lambda)}{\partial\lambda_{p}}=\sigma_{k-1}(\lambda|p)+\alpha\sigma_{k-2}(\lambda|p)=S_{k-1}(\lambda|p)$,
$p=1,2,\cdots,n$;

(ii)
$S_{k}^{pp,qq}(\lambda):=\dfrac{\partial^{2}S_{k}(\lambda)}{\partial\lambda_{p}\partial\lambda_{q}}=S_{k-2}(\lambda|pq)$,
$p,q=1,2,\cdots,n$, and $S_{k}^{pp,pp}(\lambda)=0$;

(iii) $S_{k}(\lambda)=\lambda_{i}S_{k-1}(\lambda|i)+S_{k}(\lambda|i),$
$i=1,\cdots,n$;

(iv)
$\displaystyle\sum_{i=1}^{n}S_{k}(\lambda|i)=(n-k)S_{k}(\lambda)+\alpha\sigma_{k-1}(\lambda);$

(v)
$\displaystyle\sum_{i=1}^{n}\lambda_{i}S_{k-1}(\lambda|i)=kS_{k}(\lambda)-\alpha\sigma_{k-1}(\lambda).$

In the joint work with Li, Wang and the last author [24] proved that the
admissible solution set of Sum Hessian operator $S_{k}(\lambda)$ is

$\tilde{\Gamma}_{k}=\Gamma_{k-1}\cap\\{\lambda|S_{k}>0\\},$

and $S_{k}^{\frac{1}{k}}(\lambda)$ and
$(\dfrac{S_{k}}{S_{l}})^{\frac{1}{k-l}}$ are concave functions on
$\tilde{\Gamma}_{k}$. Using the method in [19], we can get the following
lemma.

######  Lemma 2.1.

Assume that $k>l$, for $\vartheta=\dfrac{1}{k-l}$, then for
$\lambda\in\tilde{\Gamma}_{k}$, we have

(2.1) $\displaystyle-$
$\displaystyle\frac{S_{k}^{pp,qq}}{S_{k}}u_{pph}u_{qqh}+\dfrac{S_{l}^{pp,qq}}{S_{l}}u_{pph}u_{qqh}$
$\displaystyle\geq\left(\dfrac{(S_{k})_{h}}{S_{k}}-\dfrac{(S_{l})_{h}}{S_{l}}\right)\left((\vartheta-1)\dfrac{(S_{k})_{h}}{S_{k}}-(\vartheta+1)\dfrac{(S_{l})_{h}}{S_{l}}\right).$

Furthermore, for sufficiently small $\delta>0$, we have

(2.2) $\displaystyle-$ $\displaystyle
S_{k}^{pp,qq}u_{pph}u_{qqh}+(1-\vartheta+\dfrac{\vartheta}{\delta})\dfrac{(S_{k})_{h}^{2}}{S_{k}}$
$\displaystyle\geq
S_{k}(\vartheta+1-\delta\vartheta)\left[\dfrac{(S_{l})_{h}}{S_{l}}\right]^{2}-\dfrac{S_{k}}{S_{l}}S_{l}^{pp,qq}u_{pph}u_{qqh}.$

The following Lemma comes from [2].

######  Lemma 2.2.

Denote by $Sym(n)$ the set of all $n\times n$ symmetric matrices. Let $F$ be a
$C^{2}$ symmetric function defined in some open subset $\Psi\subset Sym(n)$.
At any diagonal matrix $A\in\Psi$ with distinct eigenvalues,
let$\ddot{F}(B,B)$ be the second derivative of $C^{2}$ symmetric function $F$
in direction $B\in Sym(n)$, then

$\displaystyle\ddot{F}(B,B)=\sum_{j,k=1}^{n}{\ddot{f}}^{jk}B_{jj}B_{kk}+2\sum_{j<k}\frac{\dot{f}^{j}-\dot{f}^{k}}{{\kappa}_{j}-{\kappa}_{k}}B_{jk}^{2}.$

######  Lemma 2.3.

Assume $\lambda=(\lambda_{1},\cdots,\lambda_{n})\in\tilde{\Gamma}_{k}$. If
$S_{k+1}(\lambda)>0$, then $\lambda\in\tilde{\Gamma}_{k+1}$.

###### Proof.

By the definition of $\tilde{\Gamma}_{k+1}$, we only need to prove
$\sigma_{k}(\lambda)>0$. If $\sigma_{k}(\lambda)\leq 0$, since
$S_{k}(\lambda)>0,S_{k+1}(\lambda)>0$, then

$\alpha\sigma_{k-1}>|\sigma_{k}|,\quad\sigma_{k+1}>|\alpha\sigma_{k}|,$

namely, $\alpha\sigma_{k-1}\sigma_{k+1}>\alpha\sigma_{k}^{2}$, this
contradicts Newton’s inequality. ∎

######  Remark 2.4.

Using Lemma 2.3, we can define $\tilde{\Gamma}_{k}$ as

$\tilde{\Gamma}_{k}=\\{\lambda\in\mathbb{R}^{n}|S_{m}(\lambda)>0,\quad
m=1,2,\cdots,k\\}.$

######  Lemma 2.5.

Assume that $\lambda=(\lambda_{1},\cdots,\lambda_{n})\in\tilde{\Gamma}_{k}$,
$1\leq k\leq n$, $\lambda_{1}\geq\cdots\geq\lambda_{n}$, then

(a) For any $1\leq s<k$, we have

(2.3)
$S_{s}(\lambda)>\lambda_{1}\cdots\lambda_{s}+\alpha\lambda_{1}\cdots\lambda_{s-1};$

(b) For any $1\leq j\leq k-1$, there exists a positive constant $\theta$
depending on $n,k$, such that

(2.4) $S_{k}^{jj}(\lambda)\geq\dfrac{\theta S_{k}(\lambda)}{\lambda_{j}}.$

###### Proof.

(a) Using $\lambda\in\tilde{\Gamma}_{k}\subset\Gamma_{s}$, we have

$\lambda_{1}>0,~{}\lambda_{2}>0,\cdots,~{}\lambda_{s}>0,$

and

$S_{s}(\lambda|1)>0,~{}S_{s-1}(\lambda|12)>0,\cdots,~{}S_{1}(\lambda|12\cdots
s)>0.$

Using the above inequalities, we get

$\displaystyle S_{s}(\lambda)=$
$\displaystyle\lambda_{1}S_{s-1}(\lambda|1)+S_{s}(\lambda|1)$ $\displaystyle>$
$\displaystyle\lambda_{1}S_{s-1}(\lambda|1)=\lambda_{1}\lambda_{2}S_{s-2}(\lambda|12)+\lambda_{1}S_{s-1}(\lambda|12)$
$\displaystyle>$
$\displaystyle\lambda_{1}\lambda_{2}S_{s-2}(\lambda|12)=\cdots$
$\displaystyle>$
$\displaystyle\lambda_{1}\lambda_{2}\cdots\lambda_{s-1}(\lambda_{s}+\alpha).$

(b) If $S_{k}(\lambda|j)\leq 0$, obviously we have

$S_{k}^{jj}(\lambda)=\dfrac{S_{k}(\lambda)-S_{k}(\lambda|j)}{\lambda_{j}}\geq\dfrac{S_{k}(\lambda)}{\lambda_{j}}.$

If $S_{k}(\lambda|j)>0$, then $(\lambda|j)\in\tilde{\Gamma}_{k}$ by Lemma 2.3.
Using (2.3), we obtain

$S_{k}^{jj}(\lambda)>\dfrac{\lambda_{1}\cdots\lambda_{k}+\alpha\lambda_{1}\cdots\lambda_{k-1}}{\lambda_{j}}.$

On the other hand, $(\lambda|j)\in\tilde{\Gamma}_{k}$ implies $\lambda_{k}>0$
and

$\lambda_{k}+\cdots+\lambda_{n}+\alpha>0.$

So there exists some constant $\theta$ depending on $n,k$, such that

$\theta
S_{k}(\lambda)\leq\lambda_{1}\cdots\lambda_{k}+\alpha\lambda_{1}\cdots\lambda_{k-1}.$

Then we obtain (2.4). ∎

######  Lemma 2.6.

For $\lambda\in\tilde{\Gamma}_{k}$, we have
$S_{k}^{2}(\lambda)-S_{k-1}(\lambda)S_{k+1}(\lambda)\geq 0.$

###### Proof.

Since $\lambda\in\tilde{\Gamma}_{k}$, $S_{k}(\lambda)>0,S_{k-1}(\lambda)>0$.
If $S_{k+1}(\lambda)\leq 0$, the Lemma obviously holds.

If $S_{k+1}(\lambda)>0$, Lemma 2.3 implies
$\lambda\in\tilde{\Gamma}_{k+1}\subset\Gamma_{k}$. Thus

$\displaystyle S_{k}^{2}-S_{k-1}S_{k+1}=$
$\displaystyle(\sigma_{k}+\alpha\sigma_{k-1})^{2}-(\sigma_{k-1}+\alpha\sigma_{k-2})(\sigma_{k+1}+\alpha\sigma_{k})$
$\displaystyle=$
$\displaystyle\sigma_{k}^{2}+\alpha^{2}\sigma_{k-1}^{2}+\alpha\sigma_{k}\sigma_{k-1}-(\sigma_{k-1}\sigma_{k+1}+\alpha^{2}\sigma_{k-2}\sigma_{k}+\alpha\sigma_{k-2}\sigma_{k+1})$
$\displaystyle\geq$ $\displaystyle 0.$

In the above inequality, we used Newton’s inequality and the following
inequality which obtained from Newton’s inequality

$\displaystyle\sigma_{k}\sigma_{k-1}\geq$
$\displaystyle\sigma_{k-2}\sigma_{k+1}.$

∎

######  Lemma 2.7.

Assume that $\lambda=(\lambda_{1},\cdots,\lambda_{n})\in\Gamma_{k}$, $1\leq
k\leq n$, $\lambda_{1}\geq\cdots\geq\lambda_{n}$, and $S_{k}\leq N_{0}$ for
some positive constant $N_{0}>0$. Then

(a) There exists some positive constant $K_{0}$, such that

(2.5)
$\lambda_{k-1}\leq(\dfrac{N_{0}}{\alpha})^{\frac{1}{k-1}},~{}~{}~{}~{}~{}~{}\lambda_{n}>-K_{0};$

(b) Denote $\kappa_{i}=\lambda_{i}+K_{0}$. For any $1\leq i\leq n$, we have

(2.6)
$2\kappa_{1}^{k+2}S_{k}^{11}(\lambda)\geq\kappa_{i}^{k+2}S_{k}^{ii}(\lambda),$

if $\lambda_{1}$ is sufficiently large;

(c) For any $\varepsilon_{0}>0$, we have

(2.7) $S_{k}(\lambda)>(1-\varepsilon_{0})\lambda_{1}S_{k}^{11}(\lambda),$

if $\lambda_{1}$ is sufficiently large;

(d) For any $1\leq i\leq n$, there exists some positive constant $C_{0}$, such
that

(2.8) $C_{0}S_{k}(\lambda)\geq\lambda_{i}S_{k}^{ii}(\lambda).$

###### Proof.

(a) Since $\lambda\in\Gamma_{k}$, by Lemma 11 in [30], we have

$\displaystyle N_{0}\geq S_{k}=\sigma_{k}+\alpha\sigma_{k-1}>$
$\displaystyle\alpha\sigma_{k-1}\geq\alpha\lambda_{1}\cdots\lambda_{k-1}>\alpha\lambda_{k-1}^{k-1}.$

Thus, $\lambda_{k-1}\leq(\dfrac{N_{0}}{\alpha})^{\frac{1}{k-1}}$. Using
$\lambda_{k}\geq 0$, we have

$\lambda_{n}>-(n-k)\lambda_{k}\geq-(n-k)\lambda_{k-1}>-K_{0}.$

(b) From (2.5), we get $\kappa_{i}>0$. We divide into two cases to prove
(2.6).

Case 1: $\lambda_{i}\geq\lambda_{1}^{1/(k+2)}$. Since

(2.9) $\displaystyle 2\kappa_{1}^{2}S_{k}^{11}-\kappa_{i}^{2}S_{k}^{ii}=$
$\displaystyle\kappa_{1}^{2}S_{k}^{11}+\kappa_{1}^{2}[\lambda_{i}S_{k}^{11,ii}+S_{k-1}(\lambda|1i)]$
$\displaystyle-\kappa_{i}^{2}[\lambda_{1}S_{k}^{11,ii}+S_{k-1}(\lambda|1i)]$
$\displaystyle\geq$
$\displaystyle\kappa_{1}^{2}S_{k}^{11}+(\kappa_{1}^{2}-\kappa_{i}^{2})S_{k-1}(\lambda|1i)$
$\displaystyle\geq$
$\displaystyle(\kappa_{1}^{2}-\kappa_{i}^{2})[S_{k}^{11}+S_{k-1}(\lambda|1i)].$

Using $\lambda\in\Gamma_{k}$, we have

$S_{k-1}(\lambda|1i)=\sigma_{k-1}(\lambda|1i)+\alpha\sigma_{k-2}(\lambda|1i)\geq\sigma_{k-1}(\lambda|1i)>C_{n-2}^{k-1}\dfrac{\lambda_{2}\cdots\lambda_{k}\lambda_{n}}{\lambda_{i}},$

and

(2.10)
$S_{k}^{11}=\sigma_{k-1}(\lambda|1)+\alpha\sigma_{k-2}(\lambda|1)\geq\alpha\lambda_{2}\cdots\lambda_{k-1}.$

In view of (2.5), we have $|\lambda_{k}\lambda_{n}|<K_{0}^{2}$. Using the
above two formulas, we have

(2.11)
$S_{k}^{11}+S_{k-1}(\lambda|1i)>\lambda_{2}\cdots\lambda_{k-1}(\alpha+C_{n-2}^{k-1}\dfrac{\lambda_{k}\lambda_{n}}{\lambda_{i}})>0,$

if $\lambda_{1}$ is sufficiently large. Combining (2.11) with (2.9), we can
get (2.6).

Case 2: $\lambda_{i}<\lambda_{1}^{1/(k+2)}$. In this case,
$2\kappa_{1}>\kappa_{i}^{k+2}$, using (2.4), we have

$\displaystyle\kappa_{1}^{k+1}S_{k}^{11}>\kappa_{1}^{k}\lambda_{1}S_{k}^{11}\geq\kappa_{1}^{k}c_{0}S_{k}\geq\kappa_{1}c_{0}S_{k}\lambda_{1}^{k-1}\geq
S_{k}^{ii},$

when $\lambda_{1}$ is sufficiently large. Then we get

$2\kappa_{1}^{k+2}S_{k}^{11}\geq\kappa_{i}^{k+2}S_{k}^{ii}.$

(c) Using (2.10) and

$\sigma_{k}(\lambda|1)>C_{n-1}^{k}\lambda_{2}\cdots\lambda_{k}\lambda_{n},$

we get

$\displaystyle S_{k}-(1-\varepsilon_{0})\lambda_{1}S_{k}^{11}=$
$\displaystyle\varepsilon_{0}\lambda_{1}S_{k}^{11}+S_{k}(\lambda|1)$
$\displaystyle>$
$\displaystyle\varepsilon_{0}\lambda_{1}S_{k}^{11}+\sigma_{k}(\lambda|1)$
$\displaystyle\geq$
$\displaystyle\lambda_{2}\cdots\lambda_{k-1}(\varepsilon_{0}\lambda_{1}\alpha+C_{n-1}^{k}\lambda_{k}\lambda_{n}).$

By (2.5), we know $|\lambda_{k}\lambda_{n}|<K_{0}^{2}$. Thus when
$\lambda_{1}$ is sufficiently large, the above formula is non-negative.

(d) Since $\lambda\in\Gamma_{k}$, we have

$S_{k}=\sigma_{k}+\alpha\sigma_{k-1}\geq\alpha\lambda_{1}\cdots\lambda_{k-1},$

and

$\sigma_{k}(\lambda|i)>C_{n}^{k}\lambda_{1}\cdots\lambda_{k-1}\lambda_{n},$

which implies

$\displaystyle C_{0}S_{k}-\lambda_{i}S_{k}^{ii}=$
$\displaystyle(C_{0}-1)S_{k}+S_{k}(\lambda|i)$ $\displaystyle=$
$\displaystyle(C_{0}-1)S_{k}+\sigma_{k}(\lambda|i)+\alpha\sigma_{k-1}(\lambda|i)$
$\displaystyle>$
$\displaystyle\lambda_{1}\cdots\lambda_{k-1}[(C_{0}-1)\alpha+C_{n}^{k}\lambda_{n}].$

By (2.5), when $C_{0}>1+\dfrac{K_{0}C_{n}^{k}}{\alpha}$, the above formula is
non-negative. ∎

##  3\. Pogorelov type $C^{2}$ estimates for $k-1$ convex solutions

In this section, we prove theorem 1.2. First, for the case of $k=2,3$, we
establish Pogorelov type $C^{2}$ estimates for $k-1$ convex solutions of the
Dirichlet problem (1.1), and then we give a proof for the case that
$f^{\frac{1}{k}}(x,u,p)$ is convex with respect to $p$.

Proof of (1.3): Let $\lambda_{1}=\lambda_{1}(x)$ denote the biggest eigenvalue
of $D^{2}u$. Consider the test function:

$\phi(x)=\lambda_{1}(-u)^{\beta}\exp\\{\dfrac{\varepsilon}{2}|Du|^{2}+\dfrac{a}{2}|x|^{2}\\},$

where $\beta,\varepsilon,a$ are undetermined positive constants. Using the
trick of Brendle-Choi-Daskalopoulos [5, 35], assume test function $\phi(x)$
obtains its maximum at point $x_{0}$. Define a function $\psi(x)$:

$\psi(x)=\phi(x_{0})(-u)^{-\beta}\exp\\{-\dfrac{\varepsilon}{2}|Du|^{2}-\dfrac{a}{2}|x|^{2}\\}.$

Since

$\displaystyle\phi(x_{0})=$
$\displaystyle\psi(x)(-u)^{\beta}\exp\\{\dfrac{\varepsilon}{2}|Du|^{2}+\dfrac{a}{2}|x|^{2}\\}$
$\displaystyle\geq$
$\displaystyle\phi(x)=\lambda_{1}(-u)^{\beta}\exp\\{\dfrac{\varepsilon}{2}|Du|^{2}+\dfrac{a}{2}|x|^{2}\\},$

we have $\psi(x)\geq\lambda_{1}(x)$ and $\psi(x_{0})=\lambda_{1}(x_{0})$. By
rotating the coordinate, we assume that $(u_{ij})$ is a diagonal matrix at
$x_{0}$, and $u_{ii}=\lambda_{i},~{}\lambda_{1}\geq\cdots\geq\lambda_{n}$.

Let $\mu$ denote the multiplicity of the biggest eigenvalue at $x_{0}$,
namely,
$\lambda_{1}=\cdots=\lambda_{\mu}>\lambda_{\mu+1}\geq\cdots\geq\lambda_{n}$.
Then from Brendle-Choi-Daskalopoulos result(Lemma 5 in [5]), we have the
followings exist at $x_{0}$:

$\psi=\lambda_{1}=u_{11},$ $u_{kji}=\psi_{i}\delta_{kj},\quad 1\leq
k,j\leq\mu,$ $\psi_{ii}\geq
u_{11ii}+2\displaystyle\sum_{j>\mu}\dfrac{1}{\lambda_{1}-\lambda_{j}}u_{1ji}^{2},$

Now we restrict all calculations at point $x_{0}$. Since the function

$\tilde{\phi}=\psi(x)(-u)^{\beta}\exp\\{\dfrac{\varepsilon}{2}|Du|^{2}+\dfrac{a}{2}|x|^{2}\\}=\phi(x_{0})$

has constant value, at $x_{0}$,
$(\ln\tilde{\phi})_{i}=0,\quad(\ln\tilde{\phi})_{ii}=0$, using the above
formulas of $\psi$, we obtain

(3.1) $\dfrac{\psi_{i}}{\psi}+\dfrac{\beta u_{i}}{u}+\varepsilon
u_{i}u_{ii}+ax_{i}=\dfrac{u_{11i}}{u_{11}}+\dfrac{\beta u_{i}}{u}+\varepsilon
u_{i}u_{ii}+ax_{i}=0,$

and

$\displaystyle 0=$
$\displaystyle\dfrac{\psi_{ii}}{\psi}-\dfrac{\psi_{i}^{2}}{\psi^{2}}+\dfrac{\beta
u_{ii}}{u}-\dfrac{\beta u_{i}^{2}}{u^{2}}+\displaystyle\sum_{s}\varepsilon
u_{s}u_{sii}+\varepsilon u_{ii}^{2}+a$ $\displaystyle\geq$
$\displaystyle\dfrac{\beta u_{ii}}{u}-\dfrac{\beta
u_{i}^{2}}{u^{2}}+\dfrac{u_{11ii}}{u_{11}}+2\displaystyle\sum_{j>\mu}\dfrac{u_{1ji}^{2}}{\lambda_{1}(\lambda_{1}-\lambda_{j})}-\dfrac{u_{11i}^{2}}{u_{11}^{2}}+\displaystyle\sum_{s}\varepsilon
u_{s}u_{sii}+\varepsilon u_{ii}^{2}+a.$

In the above inequality, contracting with $S_{k}^{ii}$, we have

(3.2) $\displaystyle 0\geq$ $\displaystyle\dfrac{\beta
S_{k}^{ii}u_{ii}}{u}-\dfrac{\beta
S_{k}^{ii}u_{i}^{2}}{u^{2}}+\dfrac{S_{k}^{ii}u_{11ii}}{u_{11}}+2\displaystyle\sum_{j>\mu}\dfrac{S_{k}^{ii}u_{1ji}^{2}}{\lambda_{1}(\lambda_{1}-\lambda_{j})}-\dfrac{S_{k}^{ii}u_{11i}^{2}}{u_{11}^{2}}$
$\displaystyle+\displaystyle\sum_{s}\varepsilon
u_{s}S_{k}^{ii}u_{sii}+\varepsilon
S_{k}^{ii}u_{ii}^{2}+a\displaystyle\sum_{i}S_{k}^{ii}.$

At $x_{0}$, differentiating equation (1.1) twice, we have

(3.3) $S_{k}^{ii}u_{iij}=f_{j}+f_{u}u_{j}+f_{p_{j}}u_{jj},$

and

(3.4) $S_{k}^{ii}u_{iijj}+S_{k}^{pq,rs}u_{pqj}u_{rsj}\geq-C-
Cu_{11}+f_{p_{j}p_{j}}u_{jj}^{2}+\displaystyle\sum_{s}f_{p_{s}}u_{sjj}.$

Substituting (3.4) into (3.2), we have

(3.5) $\displaystyle 0\geq$ $\displaystyle\dfrac{\beta
S_{k}^{ii}u_{ii}}{u}-\dfrac{\beta
S_{k}^{ii}u_{i}^{2}}{u^{2}}+\dfrac{1}{u_{11}}[-C-Cu_{11}+f_{p_{1}p_{1}}u_{11}^{2}+\displaystyle\sum_{s}f_{p_{s}}u_{s11}-K(S_{k})_{1}^{2}$
$\displaystyle+K(S_{k})_{1}^{2}-S_{k}^{pq,rs}u_{pq1}u_{rs1}]+2\displaystyle\sum_{j>\mu}\dfrac{S_{k}^{11}u_{11j}^{2}}{\lambda_{1}(\lambda_{1}-\lambda_{j})}-\dfrac{S_{k}^{ii}u_{11i}^{2}}{u_{11}^{2}}$
$\displaystyle+\displaystyle\sum_{s}\varepsilon
u_{s}S_{k}^{ii}u_{sii}+\varepsilon
S_{k}^{ii}u_{ii}^{2}+a\displaystyle\sum_{i}S_{k}^{ii},$

where $K$ is a positive constant which will be determined later.

Combining (3.1) with (3.3), we obtain

$\dfrac{1}{u_{11}}\displaystyle\sum_{s}f_{p_{s}}u_{s11}+\displaystyle\sum_{s}\varepsilon
u_{s}S_{k}^{ii}u_{sii}\geq-\displaystyle\sum_{s}\dfrac{\beta
u_{s}f_{p_{s}}}{u}-C.$

By Lemma 2.2,

$\displaystyle-S_{k}^{pq,rs}u_{pq1}u_{rs1}$
$\displaystyle=-S_{k}^{pp,qq}u_{pp1}u_{qq1}+\sum_{p\neq
q}S_{k}^{pp,qq}u_{pq1}^{2}$ $\displaystyle\geq-
S_{k}^{pp,qq}u_{pp1}u_{qq1}+2\sum_{j\neq 1}S_{k}^{11,jj}u_{11j}^{2}.$

Let $K=\dfrac{k-1}{kS_{k}}$. Using Lemma 2.1, we have

$K(S_{k})_{1}^{2}-S_{k}^{pp,qq}u_{pp1}u_{qq1}\geq 0.$

Because $\lambda(D^{2}u)\in\Gamma_{k-1}\subset\Gamma_{1}$, that is $\Delta
u=\sigma_{1}(D^{2}u)>0$, so we have $u<0$ at $x_{0}$. By (v) in section 2,

$\dfrac{\beta
S_{k}^{ii}u_{ii}}{u}=\dfrac{\beta}{u}(kS_{k}-\alpha\sigma_{k-1})>\dfrac{\beta
kS_{k}}{u}.$

Using the above inequalities, (3.5) becomes

(3.6) $\displaystyle-\dfrac{C}{u}\geq$ $\displaystyle-\dfrac{\beta
S_{k}^{ii}u_{i}^{2}}{u^{2}}+\dfrac{2}{u_{11}}\displaystyle\sum_{j\neq
1}S_{k}^{11,jj}u_{11j}^{2}+2\displaystyle\sum_{j>\mu}\dfrac{S_{k}^{11}u_{11j}^{2}}{\lambda_{1}(\lambda_{1}-\lambda_{j})}$
$\displaystyle-\dfrac{S_{k}^{ii}u_{11i}^{2}}{u_{11}^{2}}+\varepsilon
S_{k}^{ii}u_{ii}^{2}+a\displaystyle\sum_{i}S_{k}^{ii}+\Big{(}f_{p_{1}p_{1}}-\dfrac{(k-1)f_{p_{1}}^{2}}{kf}\Big{)}u_{11}-C.$

Next, we divide into two cases to deal with (3.6).

Case A: $\lambda_{n}\geq-\dfrac{\lambda_{1}}{2}$. For index $i>\mu$, using
(3.1), we have

(3.7) $-\dfrac{\beta u_{i}^{2}}{u^{2}}\geq-\dfrac{2u_{11i}^{2}}{\beta
u_{11}^{2}}-\dfrac{4}{\beta}(\varepsilon
u_{i}u_{ii})^{2}-\dfrac{4}{\beta}(ax_{i})^{2}.$

For any $j>\mu$ and $\beta>6$, we have

(3.8) $\displaystyle\dfrac{2}{u_{11}}$ $\displaystyle
S_{k}^{11,jj}u_{11j}^{2}+2\dfrac{S_{k}^{11}}{\lambda_{1}(\lambda_{1}-\lambda_{j})}u_{11j}^{2}-\dfrac{2+\beta}{\beta}\dfrac{S_{k}^{jj}}{u_{11}^{2}}u_{11j}^{2}$
$\displaystyle=$
$\displaystyle\dfrac{(\beta-2)\lambda_{1}+(\beta+2)\lambda_{j}}{\beta(\lambda_{1}-\lambda_{j})}\dfrac{S_{k}^{jj}u_{11j}^{2}}{\lambda_{1}^{2}}\geq
0.$

Substituting (3.7), (3.8) into (3.6), we get

(3.9) $\displaystyle-\dfrac{C}{u}\geq$
$\displaystyle-\displaystyle\sum_{i=1}^{\mu}\dfrac{\beta
S_{k}^{ii}u_{i}^{2}}{u^{2}}-\displaystyle\sum_{i=1}^{\mu}\dfrac{S_{k}^{ii}u_{11i}^{2}}{u_{11}^{2}}+\varepsilon
S_{k}^{ii}u_{ii}^{2}-\dfrac{4}{\beta}\displaystyle\sum_{i>\mu}S_{k}^{ii}(\varepsilon
u_{i}u_{ii})^{2}$
$\displaystyle-\dfrac{4}{\beta}\displaystyle\sum_{i>\mu}S_{k}^{ii}(ax_{i})^{2}+a\displaystyle\sum_{i}S_{k}^{ii}+\Big{(}f_{p_{1}p_{1}}-\dfrac{(k-1)f_{p_{1}}^{2}}{kf}\Big{)}u_{11}-C.$

For index $i\leq\mu$, using (3.1), we have

(3.10) $-\dfrac{u_{11i}^{2}}{u_{11}^{2}}\geq-2(\dfrac{\beta
u_{i}}{u})^{2}-4(\varepsilon u_{i}u_{ii})^{2}-4(ax_{i})^{2}.$

Substituting (3.10) into (3.9), we get

(3.11) $\displaystyle-\dfrac{C}{u}$
$\displaystyle+\displaystyle\sum_{i=1}^{\mu}\dfrac{(\beta+2\beta^{2})S_{k}^{ii}u_{i}^{2}}{u^{2}}$
$\displaystyle\geq\varepsilon
S_{k}^{ii}u_{ii}^{2}-4\displaystyle\sum_{i=1}^{\mu}S_{k}^{ii}(\varepsilon
u_{i}u_{ii})^{2}-4\displaystyle\sum_{i=1}^{\mu}S_{k}^{ii}(ax_{i})^{2}-\dfrac{4}{\beta}\displaystyle\sum_{i>\mu}S_{k}^{ii}(\varepsilon
u_{i}u_{ii})^{2}$
$\displaystyle-\dfrac{4}{\beta}\displaystyle\sum_{i>\mu}S_{k}^{ii}(ax_{i})^{2}+a\displaystyle\sum_{i}S_{k}^{ii}+\Big{(}f_{p_{1}p_{1}}-\dfrac{(k-1)f_{p_{1}}^{2}}{kf}\Big{)}u_{11}-C.$

We choose $\varepsilon$, such that

$\varepsilon>8\varepsilon^{2}\displaystyle\max_{\Omega}|Du|^{2}.$

Note that $i\leq\mu$, $u_{ii}=u_{11}=\lambda_{1}$, thus for the above selected
$\varepsilon$,

(3.12) $\dfrac{3}{4}\varepsilon
S_{k}^{ii}u_{ii}^{2}-4\displaystyle\sum_{i=1}^{\mu}S_{k}^{ii}(\varepsilon
u_{i}u_{ii})^{2}-4\displaystyle\sum_{i=1}^{\mu}S_{k}^{ii}(ax_{i})^{2}-\dfrac{4}{\beta}\displaystyle\sum_{i>\mu}S_{k}^{ii}(\varepsilon
u_{i}u_{ii})^{2}>0,$

when $\lambda_{1}$ is sufficiently large. By (iv) in section 2, for $k=2,3$,
when $\lambda\in\tilde{\Gamma}_{k}$,

(3.13)
$\displaystyle\sum_{i=1}^{n}S_{2}^{ii}=(n-1)\sigma_{1}+n\alpha>(n-1)\sigma_{1},$
$\displaystyle\sum_{i=1}^{n}S_{3}^{ii}=(n-2)\sigma_{2}+(n-1)\alpha\sigma_{1}>(n-1)\alpha\sigma_{1}.$

Note that, when $\lambda\in\tilde{\Gamma}_{2}$,

$S_{2}^{11}=\lambda_{2}+\cdots+\lambda_{n}+\alpha>0,$

which implies $\sigma_{1}>\lambda_{1}-\alpha$. Taking $a$ sufficiently large,
we have

(3.14)
$\dfrac{a}{3}\displaystyle\sum_{i=1}^{n}S_{k}^{ii}+\Big{(}f_{p_{1}p_{1}}-\dfrac{(k-1)f_{p_{1}}^{2}}{kf}\Big{)}u_{11}-C>0.$

Choose $\beta>a^{2}$, such that

(3.15)
$-\dfrac{4}{\beta}\displaystyle\sum_{i>\mu}S_{k}^{ii}(ax_{i})^{2}+\dfrac{a}{3}\displaystyle\sum_{i>\mu}S_{k}^{ii}>0.$

Substituting (3.12), (3.14), (3.15) into (3.11), we get

$\displaystyle-\dfrac{C}{u}+\displaystyle\sum_{i=1}^{\mu}\dfrac{(\beta+2\beta^{2})S_{k}^{ii}u_{i}^{2}}{u^{2}}\geq$
$\displaystyle\dfrac{\varepsilon}{4}\displaystyle\sum_{i=1}^{\mu}S_{k}^{ii}u_{ii}^{2}+\dfrac{a}{3}\displaystyle\sum_{i}S_{k}^{ii}.$

When

$\displaystyle-\dfrac{C}{u}\geq\displaystyle\sum_{i=1}^{\mu}\dfrac{(\beta+2\beta^{2})S_{k}^{ii}u_{i}^{2}}{u^{2}},$

we get

$\displaystyle-\dfrac{2C}{u}\geq$
$\displaystyle\dfrac{a}{3}\displaystyle\sum_{i}S_{k}^{ii}.$

Using (3.13), we obtain (1.3).

When

$\displaystyle-\dfrac{C}{u}<\displaystyle\sum_{i=1}^{\mu}\dfrac{(\beta+2\beta^{2})S_{k}^{ii}u_{i}^{2}}{u^{2}},$

we get

$\displaystyle
2\displaystyle\sum_{i=1}^{\mu}\dfrac{(\beta+2\beta^{2})S_{k}^{ii}u_{i}^{2}}{u^{2}}\geq$
$\displaystyle\dfrac{\varepsilon}{4}\displaystyle\sum_{i=1}^{\mu}S_{k}^{ii}u_{ii}^{2}.$

Since $u_{11}=\cdots=u_{\mu\mu}$, then $S_{k}^{11}=\cdots=S_{k}^{\mu\mu}$, and
then we get

$(-u)^{2}u_{11}^{2}\leq C$

by the above inequality.

Case B: $\lambda_{n}<-\dfrac{\lambda_{1}}{2}$. Using (3.6) and (3.10), we get

(3.16)
$\displaystyle-\dfrac{C}{u}+\dfrac{(\beta+2\beta^{2})S_{k}^{ii}u_{i}^{2}}{u^{2}}\geq$
$\displaystyle\varepsilon S_{k}^{ii}u_{ii}^{2}-4S_{k}^{ii}(\varepsilon
u_{i}u_{ii})^{2}-4S_{k}^{ii}(ax_{i})^{2}$
$\displaystyle+a\displaystyle\sum_{i}S_{k}^{ii}+\Big{(}f_{p_{1}p_{1}}-\dfrac{(k-1)f_{p_{1}}^{2}}{kf}\Big{)}u_{11}-C.$

Note that $S_{k}^{nn}\geq\cdots\geq S_{k}^{11}$ and
$\lambda_{n}^{2}>\dfrac{\lambda_{1}^{2}}{4}$. We choose
$\varepsilon>8\varepsilon^{2}\displaystyle\max_{\Omega}|Du|^{2}$, then

(3.17) $\displaystyle\dfrac{3}{4}\varepsilon
S_{k}^{ii}u_{ii}^{2}-4S_{k}^{ii}(\varepsilon
u_{i}u_{ii})^{2}-4S_{k}^{ii}(ax_{i})^{2}>0,$

if $\lambda_{1}$ is sufficiently large. Let $a$ be sufficiently large
satisfying (3.14). Substituting (3.14), (3.17) into (3.16), we get

$-\dfrac{C}{u}+\dfrac{(\beta+2\beta^{2})S_{k}^{ii}u_{i}^{2}}{u^{2}}\geq\dfrac{\varepsilon}{4}S_{k}^{ii}u_{ii}^{2}+\dfrac{2a}{3}\displaystyle\sum_{i}S_{k}^{ii}.$

Using again (3.13), we obtain (1.3).

Proof of (1.4): Let $\lambda_{1}=\lambda_{1}(x)$ denote the biggest eigenvalue
of $D^{2}u$. We consider the following test function:

$\phi(x)=\lambda_{1}(-u)^{\beta}\exp\\{\dfrac{\varepsilon}{2}|Du|^{2}\\},$

where $\beta=1+\delta^{\prime}=1+2\delta$. We may assume that
$0<\delta<\dfrac{1}{3}$.

Suppose that function $\phi$ achieves its maximum value in $\Omega$ at some
point $x_{0}$, and similar to the process of obtaining (3.1) and (3.6), let
$a=0$ and use the convexity of $f^{\frac{1}{k}}(x,u,p)$, we obtain

(3.18) $\dfrac{\beta u_{i}}{u}+\dfrac{u_{11i}}{u_{11}}+\varepsilon
u_{i}u_{ii}=0,$ (3.19) $\displaystyle-\dfrac{C}{u}\geq$
$\displaystyle-\dfrac{\beta
S_{k}^{ii}u_{i}^{2}}{u^{2}}+\dfrac{2}{u_{11}}\displaystyle\sum_{j\neq
1}S_{k}^{11,jj}u_{11j}^{2}+2\displaystyle\sum_{j>\mu}\dfrac{S_{k}^{11}u_{11j}^{2}}{\lambda_{1}(\lambda_{1}-\lambda_{j})}$
$\displaystyle-\dfrac{S_{k}^{ii}u_{11i}^{2}}{u_{11}^{2}}+\varepsilon
S_{k}^{ii}u_{ii}^{2}-C.$

Next, we divide into two cases to deal with (3.19).

Case A: $\lambda_{n}\geq-\dfrac{\delta\lambda_{1}}{3}$. For index $i>\mu$,
using (3.18), we have

(3.20) $-\dfrac{\beta
u_{i}^{2}}{u^{2}}\geq-\dfrac{(1+\delta)u_{11i}^{2}}{\beta
u_{11}^{2}}-\dfrac{1+1/\delta}{\beta}(\varepsilon u_{i}u_{ii})^{2}.$

For any $j>\mu$, we have

(3.21) $\displaystyle\dfrac{2}{u_{11}}$ $\displaystyle
S_{k}^{11,jj}u_{11j}^{2}+2\dfrac{S_{k}^{11}}{\lambda_{1}(\lambda_{1}-\lambda_{j})}u_{11j}^{2}-\dfrac{1+\delta+\beta}{\beta}\dfrac{S_{k}^{jj}}{u_{11}^{2}}u_{11j}^{2}$
$\displaystyle=\dfrac{\delta\lambda_{1}+(2+3\delta)\lambda_{j}}{\beta(\lambda_{1}-\lambda_{j})}\dfrac{S_{k}^{jj}u_{11j}^{2}}{\lambda_{1}^{2}}\geq
0.$

Substituting (3.20), (3.21) into (3.19), we obtain

(3.22) $\displaystyle-\dfrac{C}{u}\geq$
$\displaystyle-\displaystyle\sum_{i=1}^{\mu}\dfrac{\beta
S_{k}^{ii}u_{i}^{2}}{u^{2}}-\displaystyle\sum_{i=1}^{\mu}\dfrac{S_{k}^{ii}u_{11i}^{2}}{u_{11}^{2}}+\varepsilon
S_{k}^{ii}u_{ii}^{2}-\dfrac{1+1/\delta}{\beta}\displaystyle\sum_{i>\mu}S_{k}^{ii}(\varepsilon
u_{i}u_{ii})^{2}-C.$

For index $i\leq\mu$, using (3.18), we have

(3.23) $-\dfrac{u_{11i}^{2}}{u_{11}^{2}}\geq-2(\dfrac{\beta
u_{i}}{u})^{2}-2(\varepsilon u_{i}u_{ii})^{2}.$

Substituting (3.23) into (3.22), we obtain

(3.24) $\displaystyle-$
$\displaystyle\dfrac{C}{u}+\displaystyle\sum_{i=1}^{\mu}\dfrac{(\beta+2\beta^{2})S_{k}^{ii}u_{i}^{2}}{u^{2}}$
$\displaystyle\geq\varepsilon
S_{k}^{ii}u_{ii}^{2}-2\displaystyle\sum_{i=1}^{\mu}S_{k}^{ii}(\varepsilon
u_{i}u_{ii})^{2}-\dfrac{1+1/\delta}{\beta}\displaystyle\sum_{i>\mu}S_{k}^{ii}(\varepsilon
u_{i}u_{ii})^{2}-C.$

We take $\varepsilon$ sufficiently small, such that

$\varepsilon>\max\\{4,\dfrac{2+2/\delta}{\beta}\\}\varepsilon^{2}\displaystyle\max_{\Omega}|Du|^{2}.$

Then we have

(3.25) $\dfrac{1}{2}\varepsilon
S_{k}^{ii}u_{ii}^{2}-2\displaystyle\sum_{i=1}^{\mu}S_{k}^{ii}(\varepsilon
u_{i}u_{ii})^{2}-\dfrac{1+1/\delta}{\beta}\displaystyle\sum_{i>\mu}S_{k}^{ii}(\varepsilon
u_{i}u_{ii})^{2}>0.$

Using (3.24), (3.25) and $\lambda_{1}S_{k}^{11}\geq\theta S_{k}$, we obtain

$\displaystyle-\dfrac{C}{u}+\displaystyle\sum_{i=1}^{\mu}\dfrac{(\beta+2\beta^{2})S_{k}^{ii}u_{i}^{2}}{u^{2}}\geq$
$\displaystyle\dfrac{\varepsilon}{3}S_{k}^{ii}u_{ii}^{2}\geq\dfrac{\varepsilon}{6}\displaystyle\sum_{i=1}^{\mu}S_{k}^{ii}u_{ii}^{2}+\dfrac{\varepsilon\theta
f}{6}u_{11}.$

Hence, we get (1.4).

Case B: $\lambda_{n}<-\dfrac{\delta\lambda_{1}}{3}$, then
$\lambda_{n}^{2}>\dfrac{\delta^{2}\lambda_{1}^{2}}{9}$. Using (3.19) and
(3.23), we obtain

(3.26)
$\displaystyle-\dfrac{C}{u}+\dfrac{(\beta+2\beta^{2})S_{k}^{ii}u_{i}^{2}}{u^{2}}\geq$
$\displaystyle\varepsilon S_{k}^{ii}u_{ii}^{2}-2S_{k}^{ii}(\varepsilon
u_{i}u_{ii})^{2}-C.$

We choose $\varepsilon>4\varepsilon^{2}\displaystyle\max_{\Omega}|Du|^{2}$. It
then follows that

$-\dfrac{C}{u}+\dfrac{(\beta+2\beta^{2})S_{k}^{ii}u_{i}^{2}}{u^{2}}\geq\dfrac{\varepsilon}{3}S_{k}^{ii}u_{ii}^{2}.$

Hence, we complete the proof of Theorem 1.2.

##  4\. Pogorelov type $C^{2}$ estimates for $k$-convex solutions

In this section, we will prove Theorem 1.5. Similar to the method in [25], we
consider the following test function.

$\phi=m\log(-u)+\log P_{m}+\dfrac{mN}{2}|Du|^{2},$

where

$P_{m}=\displaystyle\sum_{j}\kappa_{j}^{m},\kappa_{j}=\lambda_{j}+K_{0},$

$m$ and $N$ are some sufficiently large undetermined constants. The
$\lambda_{1},\lambda_{2},\cdots,\lambda_{n}$ are eigenvalues of the Hessian
matrix $D^{2}u$. Let
$K_{0}=n\Big{(}\dfrac{\sup_{\Omega}f}{\alpha}\Big{)}^{\frac{1}{k-1}}$. By
(2.5), $\kappa_{1},\kappa_{2},\cdots,\kappa_{n}$ are nonnegative. Suppose that
function $\phi$ achieves its maximum value in $\Omega$ at some point $x_{0}$.
By rotating the coordinates, we assume that $(u_{ij})$ is a diagonal matrix at
$x_{0}$, and $\kappa_{1}\geq\kappa_{2}\geq\cdots\geq\kappa_{n}$.

Differentiating test function twice and using Lemma 2.2, at $x_{0}$, we have

(4.1)
$\dfrac{\displaystyle\sum_{j}\kappa_{j}^{m-1}u_{jji}}{P_{m}}+Nu_{i}u_{ii}+\dfrac{u_{i}}{u}=0,$

and

(4.2) $\displaystyle 0\geq$
$\displaystyle\dfrac{1}{P_{m}}\Big{[}\displaystyle\sum_{j}\kappa_{j}^{m-1}u_{jjii}+(m-1)\displaystyle\sum_{j}\kappa_{j}^{m-2}u_{jji}^{2}+\displaystyle\sum_{p\neq
q}\dfrac{\kappa_{p}^{m-1}-\kappa_{q}^{m-1}}{\kappa_{p}-\kappa_{q}}u_{pqi}^{2}\Big{]}$
$\displaystyle-\dfrac{m}{P_{m}^{2}}\Big{(}\displaystyle\sum_{j}\kappa_{j}^{m-1}u_{jji}\Big{)}^{2}+\displaystyle\sum_{s}Nu_{s}u_{sii}+Nu_{ii}^{2}+\dfrac{u_{ii}}{u}-\dfrac{u_{i}^{2}}{u^{2}}.$

At $x_{0}$, differentiating the equation (1.1) twice, we have

(4.3) $S_{k}^{ii}u_{iij}=f_{j}+f_{u}u_{j}+f_{p_{j}}u_{jj}$

and

(4.4) $S_{k}^{ii}u_{iijj}+S_{k}^{pq,rs}u_{pqj}u_{rsj}\geq-C-
Cu_{11}^{2}+\displaystyle\sum_{s}f_{p_{s}}u_{sjj}.$

Here, $C$ is a constant depending on $f$, the diameter of the domain $\Omega$,
$\displaystyle\sup_{\Omega}|u|$ and $\displaystyle\sup_{\Omega}|Du|$.
Contacting $S_{k}^{ii}$ in both sides of (4.2), using (4.4), we have

(4.5) $\displaystyle 0\geq$
$\displaystyle\dfrac{1}{P_{m}}\Big{[}\displaystyle\sum_{j}\kappa_{j}^{m-1}\Big{(}-C-Cu_{11}^{2}+\displaystyle\sum_{s}f_{p_{s}}u_{sjj}-K(S_{k})_{j}^{2}+K(S_{k})_{j}^{2}$
$\displaystyle-
S_{k}^{pq,rs}u_{pqj}u_{rsj}\Big{)}+(m-1)S_{k}^{ii}\displaystyle\sum_{j}\kappa_{j}^{m-2}u_{jji}^{2}+S_{k}^{ii}\displaystyle\sum_{p\neq
q}\dfrac{\kappa_{p}^{m-1}-\kappa_{q}^{m-1}}{\kappa_{p}-\kappa_{q}}u_{pqi}^{2}\Big{]}$
$\displaystyle-\dfrac{mS_{k}^{ii}}{P_{m}^{2}}\Big{(}\displaystyle\sum_{j}\kappa_{j}^{m-1}u_{jji}\Big{)}^{2}+\displaystyle\sum_{s}Nu_{s}S_{k}^{ii}u_{sii}+NS_{k}^{ii}u_{ii}^{2}+\dfrac{S_{k}^{ii}u_{ii}}{u}-\dfrac{S_{k}^{ii}u_{i}^{2}}{u^{2}}.$

Using (4.1) and (4.3), we have

(4.6)
$\displaystyle\dfrac{1}{P_{m}}\displaystyle\sum_{j}\displaystyle\sum_{s}\kappa_{j}^{m-1}f_{p_{s}}u_{sjj}+\displaystyle\sum_{s}Nu_{s}S_{k}^{ii}u_{sii}\geq-\displaystyle\sum_{s}f_{p_{s}}\dfrac{u_{s}}{u}-C.$

By Lemma 2.2,

(4.7) $\displaystyle-S_{k}^{pq,rs}u_{pqj}u_{rsj}=$ $\displaystyle-
S_{k}^{pp,qq}u_{ppj}u_{qqj}+\displaystyle\sum_{p\neq
q}S_{k}^{pp,qq}u_{pqj}^{2}$ $\displaystyle\geq$ $\displaystyle-
S_{k}^{pp,qq}u_{ppj}u_{qqj}+2\sum_{j\neq i}S_{k}^{jj,ii}u_{jji}^{2}.$

For the given index $1\leq i,j\leq n$, we denote

$\displaystyle
A_{i}=\dfrac{\kappa_{i}^{m-1}}{P_{m}}\Big{[}K(S_{k})_{i}^{2}-\displaystyle\sum_{p,q}S_{k}^{pp,qq}u_{ppi}u_{qqi}\Big{]},~{}~{}B_{i}=\dfrac{2\kappa_{j}^{m-1}}{P_{m}}\displaystyle\sum_{j}S_{k}^{jj,ii}u_{jji}^{2},$
$\displaystyle
C_{i}=\dfrac{m-1}{P_{m}}S_{k}^{ii}\displaystyle\sum_{j}\kappa_{j}^{m-2}u_{jji}^{2},\qquad\qquad\qquad
D_{i}=\dfrac{2S_{k}^{jj}}{P_{m}}\displaystyle\sum_{j\neq
i}\dfrac{\kappa_{j}^{m-1}-\kappa_{i}^{m-1}}{\kappa_{j}-\kappa_{i}}u_{jji}^{2},$
$\displaystyle
E_{i}=\dfrac{mS_{k}^{ii}}{P_{m}^{2}}\Big{(}\displaystyle\sum_{j}\kappa_{j}^{m-1}u_{jji}\Big{)}^{2}.$

Using (4.6), (4.7) and the above definitions, (4.5) becomes

(4.8) $\displaystyle 0\geq$
$\displaystyle\displaystyle\sum_{i}(A_{i}+B_{i}+C_{i}+D_{i}-E_{i})+NS_{k}^{ii}u_{ii}^{2}+\dfrac{S_{k}^{ii}u_{ii}}{u}-\dfrac{S_{k}^{ii}u_{i}^{2}}{u^{2}}$
$\displaystyle-\displaystyle\sum_{s}f_{p_{s}}\dfrac{u_{s}}{u}-C(K)u_{11}.$

Let’s deal with the third derivatives. We divide two cases $i\neq 1$ and
$i=1$. The proof is the same as in [25].

######  Lemma 4.1.

For any $i\neq 1$, we have

$A_{i}+B_{i}+C_{i}+D_{i}-(1+\dfrac{1}{m})E_{i}\geq 0,$

for sufficiently large $m$ and $\lambda_{1}$.

###### Proof.

First, let $l=1$ in the formula (2.2) of Lemma 2.1, then we have $A_{i}\geq 0$
for sufficiently large $K$.

Next, it is the same as (3.13) in [25]. Using

$\displaystyle\kappa_{j}S_{k}^{jj,ii}+S_{k}^{jj}$
$\displaystyle=(\lambda_{j}+K_{0})S_{k}^{jj,ii}+S_{k}^{jj}$
$\displaystyle=K_{0}S_{k}^{jj,ii}+S_{k}^{ii}-S_{k-1}(\lambda|ij)+\lambda_{i}S_{k-2}(\lambda|ij)+S_{k-1}(\lambda|ij)$
$\displaystyle=(\lambda_{i}+K_{0})S_{k}^{ii,jj}+S_{k}^{ii}\geq S_{k}^{ii},$

and Cauchy-Schwarz inequalities we obtain

(4.9) $\displaystyle P_{m}^{2}$
$\displaystyle\Big{[}B_{i}+C_{i}+D_{i}-(1+\dfrac{1}{m})E_{i}\Big{]}$
$\displaystyle\geq\displaystyle\sum_{j\neq
i}\Big{[}(m+1)\kappa_{i}^{m}\kappa_{j}^{m-2}S_{k}^{ii}+2\kappa_{1}^{m}S_{k}^{jj}\displaystyle\sum_{l=0}^{m-3}\kappa_{i}^{m-2-l}\kappa_{j}^{l}\Big{]}u_{jji}^{2}$
$\displaystyle+\Big{[}(m-1)(P_{m}-\kappa_{i}^{m})-2\kappa_{i}^{m}\Big{]}S_{k}^{ii}\kappa_{i}^{m-2}u_{iii}^{2}$
$\displaystyle-2(m+1)S_{k}^{ii}\kappa_{i}^{m-1}u_{iii}\displaystyle\sum_{j\neq
i}\kappa_{j}^{m-1}u_{jji}.$

By (2.6), we have

$2\kappa_{1}^{m}S_{k}^{jj}\displaystyle\sum_{l=0}^{m-3}\kappa_{i}^{m-2-l}\kappa_{j}^{l}\geq
7S_{k}^{ii}\kappa_{j}^{m-2}\kappa_{i}^{m}$

for $m\geq\displaystyle\max\\{k+2,10\\}$. Using the above formula and (4.9),

$\displaystyle P_{m}^{2}\Big{[}$ $\displaystyle
B_{i}+C_{i}+D_{i}-(1+\dfrac{1}{m})E_{i}\Big{]}$ $\displaystyle\geq$
$\displaystyle\displaystyle\sum_{j\neq
i}(m+8)\kappa_{i}^{m}\kappa_{j}^{m-2}S_{k}^{ii}u_{jji}^{2}+[(m-1)(P_{m}-\kappa_{i}^{m})-2\kappa_{i}^{m}]S_{k}^{ii}\kappa_{i}^{m-2}u_{iii}^{2}$
$\displaystyle-2(m+1)S_{k}^{ii}\kappa_{i}^{m-1}u_{iii}\displaystyle\sum_{j\neq
i}\kappa_{j}^{m-1}u_{jji}$ $\displaystyle\geq$
$\displaystyle(m+8)\kappa_{i}^{m}\kappa_{1}^{m-2}S_{k}^{ii}u_{11i}^{2}+[(m-1)\kappa_{1}^{m}-2\kappa_{i}^{m}]S_{k}^{ii}\kappa_{i}^{m-2}u_{iii}^{2}$
$\displaystyle-2(m+1)S_{k}^{ii}\kappa_{i}^{m-1}u_{iii}\kappa_{1}^{m-1}u_{11i}$
$\displaystyle\geq$
$\displaystyle(m+8)\kappa_{i}^{m}\kappa_{1}^{m-2}S_{k}^{ii}u_{11i}^{2}+(m-3)\kappa_{1}^{m}S_{k}^{ii}\kappa_{i}^{m-2}u_{iii}^{2}$
$\displaystyle-2(m+1)S_{k}^{ii}\kappa_{i}^{m-1}u_{iii}\kappa_{1}^{m-1}u_{11i}$
$\displaystyle\geq$ $\displaystyle 0.$

In the second inequality we have used, for $m\geq 10$

$\displaystyle\sum_{j\neq
i,1}(m+8)\kappa_{i}^{m}\kappa_{j}^{m-2}S_{k}^{ii}u_{jji}^{2}+(m-1)\displaystyle\sum_{j\neq
i,1}\kappa_{j}^{m}\kappa_{i}^{m-2}S_{k}^{ii}u_{iii}^{2}$
$-2(m+1)S_{k}^{ii}\kappa_{i}^{m-1}u_{iii}\displaystyle\sum_{j\neq
i,1}\kappa_{j}^{m-1}u_{jji}\geq 0.$

In the forth inequality we have used

$(m+8)(m-3)\geq(m+1)^{2}.$

Take $m=\max\\{k+2,10\\}$. Then we obtain Lemma 4.1. ∎

######  Lemma 4.2.

For $\mu=1,\cdots,k-2$, if there exists some positive constant $\delta\leq 1$,
such that $\dfrac{\lambda_{\mu}}{\lambda_{1}}\geq\delta$, then there exist two
sufficiently small positive constants $\eta,\delta^{\prime}$ depending on
$\delta$, such that, if
$\dfrac{\lambda_{\mu+1}}{\lambda_{1}}\leq\delta^{\prime}$, we have

$A_{1}+B_{1}+C_{1}+D_{1}-(1+\dfrac{\eta}{m})E_{1}\geq 0.$

###### Proof.

First, it is the same as (3.19) in [25]. For $m\geq 5$, we have

(4.10) $\displaystyle P_{m}^{2}\Big{[}$ $\displaystyle
B_{1}+C_{1}+D_{1}-(1+\dfrac{\eta}{m})E_{1}\Big{]}$ $\displaystyle\geq$
$\displaystyle-(1+\eta)S_{k}^{11}\kappa_{1}^{2m-2}u_{111}^{2}+2P_{m}\kappa_{1}^{m-2}\displaystyle\sum_{j\neq
1}S_{k}^{jj}u_{jj1}^{2}.$

Next, we will replace $-(1+\eta)S_{k}^{11}\kappa_{1}^{2m-2}u_{111}^{2}$ from
$A_{1}$. We divide two cases $\mu=1$ and $2\leq\mu\leq k-2$.

For $\mu=1$, since $S_{1}=\sigma_{1}+\alpha$ and $S_{1}^{aa,bb}=0$, by Lemma
2.1, for sufficiently large $K$, we have

(4.11) $\displaystyle A_{1}$
$\displaystyle=\dfrac{\kappa_{1}^{m-1}}{P_{m}}\Big{[}K(S_{k})_{1}^{2}-\displaystyle\sum_{p,q}S_{k}^{pp,qq}u_{pp1}u_{qq1}\Big{]}$
$\displaystyle\geq\dfrac{\kappa_{1}^{m-1}}{P_{m}}S_{k}(1+\dfrac{\vartheta}{2})\dfrac{(S_{1})_{1}^{2}}{S_{1}^{2}}.$

Using $S_{1}^{aa}=1$, we have

(4.12) $\displaystyle(1+\dfrac{\vartheta}{2})(S_{1})_{1}^{2}=$
$\displaystyle(1+\dfrac{\vartheta}{2})\Big{(}\displaystyle\sum_{a}u_{aa1}\Big{)}^{2}$
$\displaystyle=$
$\displaystyle(1+\dfrac{\vartheta}{2})u_{111}^{2}+2(1+\dfrac{\vartheta}{2})\displaystyle\sum_{a\neq
1}u_{aa1}u_{111}+(1+\dfrac{\vartheta}{2})\Big{(}\displaystyle\sum_{a\neq
1}u_{aa1}\Big{)}^{2}$ $\displaystyle\geq$
$\displaystyle(1+\dfrac{\vartheta}{4})u_{111}^{2}-C_{\vartheta}\displaystyle\sum_{a\neq
1}u_{aa1}^{2}.$

Combining (4.11), (4.12) and (2.7), we obtain

(4.13) $\displaystyle P_{m}^{2}A_{1}$
$\displaystyle\geq\dfrac{P_{m}\kappa_{1}^{m-1}S_{k}}{S_{1}^{2}}(1+\dfrac{\vartheta}{4})u_{111}^{2}-\dfrac{P_{m}\kappa_{1}^{m-1}C_{\vartheta}}{S_{1}^{2}}\displaystyle\sum_{a\neq
1}u_{aa1}^{2}$
$\displaystyle\geq\dfrac{P_{m}\kappa_{1}^{m-2}S_{k}^{11}(1-\varepsilon_{0})}{(1+\displaystyle\sum_{j\neq
1}\dfrac{\lambda_{j}}{\lambda_{1}}+\dfrac{\alpha}{\lambda_{1}})^{2}}(1+\dfrac{\vartheta}{4})u_{111}^{2}-\dfrac{P_{m}\kappa_{1}^{m-1}C_{\vartheta}}{S_{1}^{2}}\displaystyle\sum_{a\neq
1}u_{aa1}^{2}$
$\displaystyle\geq(1+\eta)\kappa_{1}^{2m-2}S_{k}^{11}u_{111}^{2}-P_{m}\kappa_{1}^{m-3}C_{\vartheta}\displaystyle\sum_{a\neq
1}u_{aa1}^{2}.$

The last inequality comes from choosing positive constants
$\delta^{\prime},\varepsilon_{0},\eta$ such that

$\dfrac{\lambda_{j}}{\lambda_{1}}\leq\delta^{\prime},\quad(1+\dfrac{\vartheta}{4})(1-\varepsilon_{0})\geq(1+\eta)(1+n\delta^{\prime})^{2}.$

For $2\leq\mu\leq k-2$, by Lemma 2.1, we have

(4.14) $\displaystyle A_{1}\geq$
$\displaystyle\dfrac{\kappa_{1}^{m-1}}{P_{m}}\Big{[}S_{k}\dfrac{(S_{\mu})_{1}^{2}}{S_{\mu}^{2}}-\dfrac{S_{k}}{S_{\mu}}S_{\mu}^{pp,qq}u_{pp1}u_{qq1}\Big{]}$
$\displaystyle=$
$\displaystyle\dfrac{\kappa_{1}^{m-1}S_{k}}{P_{m}S_{\mu}^{2}}\Big{[}\displaystyle\sum_{a}(S_{\mu}^{aa}u_{aa1})^{2}+\displaystyle\sum_{a\neq
b}(S_{\mu}^{aa}S_{\mu}^{bb}-S_{\mu}S_{\mu}^{aa,bb})u_{aa1}u_{bb1}\Big{]}.$

Next, we split $\displaystyle\sum_{a\neq
b}(S_{\mu}^{aa}S_{\mu}^{bb}-S_{\mu}S_{\mu}^{aa,bb})u_{aa1}u_{bb1}$ into three
terms to deal with:

$\displaystyle\sum_{a\neq b}=\displaystyle\sum_{a\neq
b;a,b\leq\mu}+2\displaystyle\sum_{a\leq\mu,b>\mu}+\displaystyle\sum_{a\neq
b;a,b>\mu}:=T_{1}+T_{2}+T_{3}$

First, let’s deal with $T_{1}$. By Lemma 2.6, for $a\neq b$, we have

(4.15) $\displaystyle
S_{\mu}^{aa}S_{\mu}^{bb}-S_{\mu}S_{\mu}^{aa,bb}=S_{\mu-1}^{2}(\lambda|ab)-S_{\mu}(\lambda|ab)S_{\mu-2}(\lambda|ab)\geq
0.$

Note that $\lambda\in\Gamma_{k},\mu\leq k-2$, which implies
$S_{\mu}(\lambda|ab)>0$. So (4.15) implies

(4.16) $\displaystyle T_{1}=$ $\displaystyle\displaystyle\sum_{a\neq
b;a,b\leq\mu}(S_{\mu}^{aa}S_{\mu}^{bb}-S_{\mu}S_{\mu}^{aa,bb})u_{aa1}u_{bb1}$
$\displaystyle\geq$ $\displaystyle-\displaystyle\sum_{a\neq
b;a,b\leq\mu}\Big{[}S_{\mu-1}^{2}(\lambda|ab)-S_{\mu}(\lambda|ab)S_{\mu-2}(\lambda|ab)\Big{]}u_{aa1}^{2}$
$\displaystyle\geq$ $\displaystyle-\displaystyle\sum_{a\neq
b;a,b\leq\mu}S_{\mu-1}^{2}(\lambda|ab)u_{aa1}^{2}.$

By Lemma 2.5, for any $a,b\leq\mu$,

(4.17)
$S_{\mu}^{aa}=\sigma_{\mu-1}(\lambda|a)+\alpha\sigma_{\mu-2}(\lambda|a)\geq\dfrac{\lambda_{1}\cdots\lambda_{\mu}}{\lambda_{a}},$
(4.18) $S_{\mu-1}(\lambda|ab)\leq
C(\dfrac{\lambda_{1}\cdots\lambda_{\mu+1}}{\lambda_{a}\lambda_{b}}+\alpha\dfrac{\lambda_{1}\cdots\lambda_{\mu}}{\lambda_{a}\lambda_{b}})\leq
C(\dfrac{\alpha+\lambda_{\mu+1}}{\lambda_{b}})S_{\mu}^{aa}.$

Using (4.16), (4.17), (4.18), we obtain

(4.19) $\displaystyle T_{1}\geq$ $\displaystyle-\displaystyle\sum_{a\neq
b;a,b\leq\mu}C_{1}\dfrac{(\alpha+\lambda_{\mu+1})^{2}}{\lambda_{b}^{2}}(S_{\mu}^{aa}u_{aa1})^{2}$
$\displaystyle\geq$
$\displaystyle-\dfrac{C_{1}}{\delta^{2}}\dfrac{(\alpha+\delta^{\prime}\lambda_{1})^{2}}{\lambda_{1}^{2}}\displaystyle\sum_{a\leq\mu}(S_{\mu}^{aa}u_{aa1})^{2}$
$\displaystyle\geq$
$\displaystyle-\varepsilon\displaystyle\sum_{a\leq\mu}(S_{\mu}^{aa}u_{aa1})^{2}.$

Here, we choose positive constants $\delta^{\prime},\varepsilon$ sufficiently
small and $\lambda_{1}$ sufficiently large, satisfying

$\dfrac{4C_{1}\delta^{\prime
2}}{\delta^{2}}<\varepsilon,\quad\alpha<\delta^{\prime}\lambda_{1}.$

Next, let’s deal with $T_{2}$ and $T_{3}$. Using Cauchy-Schwarz inequalities
we get

(4.20) $\displaystyle T_{2}\geq$
$\displaystyle-2\displaystyle\sum_{a\leq\mu;b>\mu}S_{\mu}^{aa}S_{\mu}^{bb}|u_{aa1}u_{bb1}|$
$\displaystyle\geq$
$\displaystyle-\varepsilon\displaystyle\sum_{a\leq\mu;b>\mu}(S_{\mu}^{aa}u_{aa1})^{2}-\dfrac{1}{\varepsilon}\displaystyle\sum_{a\leq\mu;b>\mu}(S_{\mu}^{bb}u_{bb1})^{2},$
(4.21) $\displaystyle T_{3}\geq$ $\displaystyle-\displaystyle\sum_{a\neq
b;a,b>\mu}S_{\mu}^{aa}S_{\mu}^{bb}|u_{aa1}u_{bb1}|\geq-\displaystyle\sum_{a\neq
b;a,b>\mu}(S_{\mu}^{aa}u_{aa1})^{2}.$

Hence, combining (4.14), (4.19), (4.20) and (4.21), we obtain

(4.22)
$A_{1}\geq\dfrac{\kappa_{1}^{m-1}S_{k}}{P_{m}S_{\mu}^{2}}\Big{[}(1-2\varepsilon)\displaystyle\sum_{a\leq\mu}(S_{\mu}^{aa}u_{aa1})^{2}-C_{\varepsilon}\displaystyle\sum_{a>\mu}(S_{\mu}^{aa}u_{aa1})^{2}\Big{]}.$

For $a>\mu$, we have

(4.23) $S_{\mu}^{aa}\leq C_{2}\lambda_{1}\cdots\lambda_{\mu-1},\quad
S_{\mu}\geq\lambda_{1}\cdots\lambda_{\mu}.$

So

(4.24)
$-\dfrac{1}{S_{\mu}^{2}}\displaystyle\sum_{a>\mu}(S_{\mu}^{aa}u_{aa1})^{2}\geq-\dfrac{C_{2}^{2}}{\lambda_{\mu}^{2}}\displaystyle\sum_{a>\mu}u_{aa1}^{2}\geq-\dfrac{C_{3}}{\kappa_{1}^{2}\delta^{2}}\displaystyle\sum_{a>\mu}u_{aa1}^{2}.$

For $a\leq\mu$, we have

(4.25) $S_{\mu}(\lambda|a)\leq
C_{4}(\dfrac{\lambda_{1}\cdots\lambda_{\mu+1}}{\lambda_{a}}+\alpha\dfrac{\lambda_{1}\cdots\lambda_{\mu}}{\lambda_{a}}).$

Using (4.23), (4.25) and (2.7), we have

(4.26)
$\displaystyle\dfrac{S_{k}}{S_{\mu}^{2}}\displaystyle\sum_{a\leq\mu}(S_{\mu}^{aa}u_{aa1})^{2}\geq$
$\displaystyle\dfrac{\lambda_{1}S_{k}^{11}(1-\varepsilon_{0})}{S_{\mu}^{2}}\displaystyle\sum_{a\leq\mu}(S_{\mu}^{aa}u_{aa1})^{2}$
$\displaystyle\geq$
$\displaystyle\dfrac{S_{k}^{11}(1-\varepsilon_{0})}{\lambda_{1}}\displaystyle\sum_{a\leq\mu}\Big{(}\dfrac{\lambda_{a}S_{\mu}^{aa}}{S_{\mu}}\Big{)}^{2}u_{aa1}^{2}$
$\displaystyle\geq$
$\displaystyle\dfrac{S_{k}^{11}(1-\varepsilon_{0})}{\kappa_{1}}\displaystyle\sum_{a\leq\mu}\Big{(}1-\dfrac{S_{\mu}(\lambda|a)}{S_{\mu}}\Big{)}^{2}u_{aa1}^{2}$
$\displaystyle\geq$
$\displaystyle\dfrac{S_{k}^{11}(1-\varepsilon_{0})}{\kappa_{1}}\Big{(}1-\dfrac{C_{5}(\alpha+\delta^{\prime}\lambda_{1})}{\delta\lambda_{1}}\Big{)}^{2}\displaystyle\sum_{a\leq\mu}u_{aa1}^{2}$
$\displaystyle\geq$
$\displaystyle\dfrac{S_{k}^{11}(1-\varepsilon_{0})}{\kappa_{1}}\Big{(}1-\dfrac{C_{6}\delta^{\prime}}{\delta}\Big{)}^{2}\displaystyle\sum_{a\leq\mu}u_{aa1}^{2}.$

Combining (4.22), (4.24), (4.26) and $P_{m}>(1+\delta^{m})\kappa_{1}^{m}$, we
obtain

(4.27) $\displaystyle P_{m}^{2}A_{1}\geq$
$\displaystyle(1+\delta^{m})\kappa_{1}^{2m-2}S_{k}^{11}(1-\varepsilon_{0})(1-2\varepsilon)\Big{(}1-\dfrac{C_{6}\delta^{\prime}}{\delta}\Big{)}^{2}\displaystyle\sum_{a\leq\mu}u_{aa1}^{2}$
$\displaystyle-\dfrac{P_{m}\kappa_{1}^{m-3}C_{\varepsilon}^{\prime}S_{k}}{\delta^{2}}\displaystyle\sum_{a>\mu}u_{aa1}^{2}$
$\displaystyle\geq$
$\displaystyle(1+\eta)\kappa_{1}^{2m-2}S_{k}^{11}\displaystyle\sum_{a\leq\mu}u_{aa1}^{2}-\dfrac{P_{m}\kappa_{1}^{m-3}C_{\varepsilon}^{\prime}S_{k}}{\delta^{2}}\displaystyle\sum_{a>\mu}u_{aa1}^{2}.$

Here, the last inequality comes from choosing positive constants
$\varepsilon_{0},\varepsilon,\delta^{\prime},\eta$ sufficiently small such
that

$\dfrac{C_{6}\delta^{\prime}}{\delta}\leq
2\varepsilon,\quad(1-\varepsilon_{0})(1-2\varepsilon)^{3}(1+\delta^{m})\geq
1+\eta.$

Using (4.10), (4.13) or (4.27), we have

(4.28) $\displaystyle P_{m}^{2}\Big{[}$ $\displaystyle
A_{1}+B_{1}+C_{1}+D_{1}-(1+\dfrac{\eta}{m})E_{1}\Big{]}$ $\displaystyle\geq$
$\displaystyle 2P_{m}\kappa_{1}^{m-2}\displaystyle\sum_{j\neq
1}S_{k}^{jj}u_{jj1}^{2}-\dfrac{P_{m}\kappa_{1}^{m-3}C_{\varepsilon}^{\prime}S_{k}}{\delta^{2}}\displaystyle\sum_{j>\mu}u_{jj1}^{2}.$

Now, for $k-1\geq j>\mu$, by (2.4), we have

$\displaystyle\kappa_{1}S_{k}^{jj}$
$\displaystyle=\dfrac{\kappa_{1}}{\lambda_{j}}\lambda_{j}S_{k}^{jj}\geq\dfrac{\kappa_{1}}{\lambda_{j}}\theta
S_{k}\geq\dfrac{\theta S_{k}}{\delta^{\prime}}.$

For $j\geq k$, we have

$\kappa_{1}S_{k}^{jj}\geq\kappa_{1}S_{k}^{k-1,k-1}\geq\dfrac{\kappa_{1}}{\lambda_{k-1}}\theta
S_{k}\geq\dfrac{\theta S_{k}}{\delta^{\prime}}.$

Hence, choose $\delta^{\prime}$ small enough such that

$\delta^{\prime}<\dfrac{\theta\delta^{2}}{C_{\varepsilon}^{\prime}},$

then (4.28) is nonnegative. We complete the proof of the lemma. ∎

Hence, a direct corollary of Lemma 4.2 is the following.

######  Corollary 4.3.

There exist two finite sequences of positive numbers
$\\{\delta_{i}\\}_{i=1}^{k-1}$ and $\\{\eta_{i}\\}_{i=1}^{k-1}$, such that, if
the following inequality holds for some index $1\leq r\leq k-2$,

$\dfrac{\lambda_{r}}{\lambda_{1}}\geq\delta_{r},\quad\dfrac{\lambda_{r+1}}{\lambda_{1}}\leq\delta_{r+1},$

then, for sufficiently large $K$ and $\lambda_{1}$, we have

(4.29) $A_{1}+B_{1}+C_{1}+D_{1}-(1+\dfrac{\eta_{r}}{m})E_{1}\geq 0.$

Now, we can prove Theorem 1.5.

By Corollary 4.3, there exists some sequence $\\{\delta_{i}\\}_{i=1}^{k-1}$.
We consider the following two cases.

Case 1: If $r=k-1$,$\lambda_{k-1}\geq\delta_{k-1}\lambda_{1}$, obviously we
have

$f=\sigma_{k}+\alpha\sigma_{k-1}>\alpha\lambda_{1}\cdots\lambda_{k-1}\geq\delta_{k-1}^{k-2}\lambda_{1}^{k-1},$

which implies $\lambda_{1}\leq C$. Hence, we complete the proof of Theorem
1.5.

Case 2: There exists some index $1\leq r\leq k-2$, such that

$\dfrac{\lambda_{r}}{\lambda_{1}}\geq\delta_{r},\quad\dfrac{\lambda_{r+1}}{\lambda_{1}}\leq\delta_{r+1}.$

Using Lemma 4.1 and Corollary 4.3, we have

(4.30)
$\displaystyle\sum_{i}(A_{i}+B_{i}+C_{i}+D_{i})-E_{1}-(1+\dfrac{1}{m})\displaystyle\sum_{i=2}^{n}E_{i}\geq
0.$

Combining (4.8) with (4.30), we have

(4.31) $\displaystyle 0\geq$
$\displaystyle\displaystyle\sum_{i=2}^{n}\dfrac{S_{k}^{ii}}{P_{m}^{2}}(\displaystyle\sum_{j}\kappa_{j}^{m-1}u_{jji})^{2}+Nu_{ii}^{2}S_{k}^{ii}+\dfrac{S_{k}^{ii}u_{ii}}{u}-\dfrac{S_{k}^{ii}u_{i}^{2}}{u^{2}}-\displaystyle\sum_{s}f_{p_{s}}\dfrac{u_{s}}{u}-Cu_{11}.$

By (4.1), for any fixed $i\geq 2$, we have

$\displaystyle-\dfrac{S_{k}^{ii}u_{i}^{2}}{u^{2}}=$
$\displaystyle-\dfrac{S_{k}^{ii}}{P_{m}^{2}}(\displaystyle\sum_{j}\kappa_{j}^{m-1}u_{jji})^{2}+S_{k}^{ii}N^{2}u_{i}^{2}u_{ii}^{2}+\dfrac{2NS_{k}^{ii}u_{i}^{2}u_{ii}}{u}$
$\displaystyle\geq$
$\displaystyle-\dfrac{S_{k}^{ii}}{P_{m}^{2}}(\displaystyle\sum_{j}\kappa_{j}^{m-1}u_{jji})^{2}+\dfrac{2NC_{0}S_{k}u_{i}^{2}}{u}.$

In the above inequality, we have used (2.8). Using the above formula and (v)
in section 2, (4.31) becomes

(4.32) $\displaystyle-\dfrac{C}{u}\geq$ $\displaystyle
Nu_{ii}^{2}S_{k}^{ii}-\dfrac{u_{1}^{2}S_{k}^{11}}{u^{2}}-Cu_{11}.$

Since $\lambda\in\Gamma_{k}$, we know $u_{11}S_{k}^{11}\geq\theta S_{k}$ by
(2.4). Choose $N$ sufficiently large in the above formula, such that

$\dfrac{N}{2}u_{ii}^{2}S_{k}^{ii}-Cu_{11}>0,$

we obtain

$-\dfrac{C}{u}+\dfrac{CS_{k}^{11}}{u^{2}}\geq\dfrac{N}{4}S_{k}^{11}\lambda_{1}^{2}+\dfrac{N\theta
S_{k}}{4}\lambda_{1}.$

Hence, we complete the proof of Theorem 1.5.

##  5\. A rigidity theorem for $k$ convex solutions

In this section, we prove Theorem 1.6. First, we have the following lemma.

######  Lemma 5.1.

For the following Dirichlet problem of Sum Hessian equations,

(5.1) $\left\\{\begin{array}[]{lll}S_{k}(D^{2}u)=f(x),&in&\Omega,\\\
u=0,&on&\partial{\Omega}.\end{array}\right.$

Assume that $f>0$ is a smooth function in some domain $\Omega$, $u$ is a $k$
convex solution of the above problem. Then there exists some sufficiently
large constant $\beta>0$, such that

(5.2) $(-u)^{\beta}\Delta u\leq C.$

Here, the constants $C$ and $\beta$ only depend on $n,k$ and the diameters of
the domain $\Omega$.

###### Proof.

Obviously, for sufficiently large $a$ and $b$, the function
$w=\dfrac{a}{2}|x|^{2}-b$ can control $u$ by comparison principal [8], namely

$w\leq u\leq 0.$

Here $a$ and $b$ depend on the diameter of the domain $\Omega$. Hence, in the
following proof, the constant $\beta$, $C$ in (5.2) can contain
$\displaystyle\sup_{\Omega}|u|$.

Since $u$ is $k$ convex solution, by (2.5), there is some constant $K_{0}>0$,
such that $D^{2}u+K_{0}I\geq 0$. We consider the test function,

$\phi=m\beta\log(-u)+\log P_{m}+\dfrac{m}{2}|x|^{2},$

where

$P_{m}=\displaystyle\sum_{j}\kappa_{j}^{m},\quad\kappa_{j}=\lambda_{j}+K_{0}.$

Suppose $\phi$ achieves its maximum value at $x_{0}$ in $\Omega$. By rotating
the coordinate, we assume that $(u_{ij})$ is a diagonal matrix at $x_{0}$, so
$u_{ii}=\lambda_{i}$ and $\kappa_{1}\geq\kappa_{2}\geq\cdots\geq\kappa_{n}$.

Differentiating test function twice at $x_{0}$, we have

(5.3)
$\dfrac{\displaystyle\sum_{j}\kappa_{j}^{m-1}u_{jji}}{P_{m}}+x_{i}+\dfrac{\beta
u_{i}}{u}=0$

and

(5.4) $\displaystyle 0\geq$
$\displaystyle\dfrac{1}{P_{m}}\Big{[}\displaystyle\sum_{j}\kappa_{j}^{m-1}u_{jjii}+(m-1)\displaystyle\sum_{j}\kappa_{j}^{m-2}u_{jji}^{2}+\displaystyle\sum_{p\neq
q}\dfrac{\kappa_{p}^{m-1}-\kappa_{q}^{m-1}}{\kappa_{p}-\kappa_{q}}u_{pqi}^{2}\Big{]}$
$\displaystyle-\dfrac{m}{P_{m}^{2}}\Big{(}\displaystyle\sum_{j}\kappa_{j}^{m-1}u_{jji}\Big{)}^{2}+\dfrac{\beta
u_{ii}}{u}-\dfrac{\beta u_{i}^{2}}{u^{2}}+1.$

At $x_{0}$, differentiating the equation (5.1) twice, we have

(5.5) $S_{k}^{ii}u_{iij}=f_{j}$

and

(5.6) $S_{k}^{ii}u_{iijj}+S_{k}^{pq,rs}u_{pqj}u_{rsj}=f_{jj}.$

Using (5.3), we have

(5.7) $-\dfrac{\beta
u_{i}^{2}}{u^{2}}\geq-\dfrac{2}{\beta}\dfrac{\Big{(}\displaystyle\sum_{j}\kappa_{j}^{m-1}u_{jji}\Big{)}^{2}}{P_{m}^{2}}-\dfrac{2x_{i}^{2}}{\beta}.$

Contacting $S_{k}^{ii}$ in both sides of (5.4), using (5.6) and (5.7), we
obtain

(5.8) $\displaystyle 0\geq$
$\displaystyle\dfrac{1}{P_{m}}\Big{[}\displaystyle\sum_{j}\kappa_{j}^{m-1}(f_{jj}-S_{k}^{pq,rs}u_{pqj}u_{rsj})+(m-1)S_{k}^{ii}\displaystyle\sum_{j}\kappa_{j}^{m-2}u_{jji}^{2}$
$\displaystyle+S_{k}^{ii}\displaystyle\sum_{p\neq
q}\dfrac{\kappa_{p}^{m-1}-\kappa_{q}^{m-1}}{\kappa_{p}-\kappa_{q}}u_{pqi}^{2}\Big{]}-\dfrac{(m+\frac{2}{\beta})S_{k}^{ii}}{P_{m}^{2}}\Big{(}\displaystyle\sum_{j}\kappa_{j}^{m-1}u_{jji}\Big{)}^{2}$
$\displaystyle+\dfrac{\beta
S_{k}^{ii}u_{ii}}{u}-\dfrac{2S_{k}^{ii}x_{i}^{2}}{\beta}+\displaystyle\sum_{i}S_{k}^{ii}.$

Using the definitions of $A_{i},B_{i},C_{i},D_{i},E_{i}$ and (5.8), we have

(5.9) $\displaystyle 0\geq$
$\displaystyle\displaystyle\sum_{i=1}^{n}\Big{[}A_{i}+B_{i}+C_{i}+D_{i}-(1+\dfrac{2}{m\beta})E_{i}\Big{]}+\dfrac{\beta
S_{k}^{ii}u_{ii}}{u}$
$\displaystyle-\dfrac{2S_{k}^{ii}x_{i}^{2}}{\beta}+\displaystyle\sum_{i}S_{k}^{ii}-\dfrac{C}{\kappa_{1}}.$

For sequence $\\{\delta_{i}\\}_{i=1}^{k-1}$ and $\\{\eta_{i}\\}_{i=1}^{k-1}$
appearing in Corollary 4.3, we choose $\beta$ sufficiently large such that

$\dfrac{2}{\beta}<\min\\{\dfrac{1}{2},\eta_{1},\cdots,\eta_{k-1}\\}.$

If $r=k-1$, $\lambda_{k-1}\geq\delta_{k-1}\lambda_{1}$, obviously we have

$f=\sigma_{k}+\alpha\sigma_{k-1}>\alpha\lambda_{1}\cdots\lambda_{k-1}\geq\delta_{k-1}^{k-2}\lambda_{1}^{k-1},$

which implies $\lambda_{1}\leq C$. Thus, we obtain Lemma 5.1.

If there exists some index $1\leq r\leq k-2$, such that

$\dfrac{\lambda_{r}}{\lambda_{1}}\geq\delta_{r},\quad\dfrac{\lambda_{r+1}}{\lambda_{1}}\leq\delta_{r+1},$

then (5.9) becomes

(5.10) $0\geq\dfrac{\beta
S_{k}^{ii}u_{ii}}{u}-\dfrac{2S_{k}^{ii}x_{i}^{2}}{\beta}+\displaystyle\sum_{i}S_{k}^{ii}-\dfrac{C}{\kappa_{1}}$

by Lemma 4.1 and Corollary 4.3. We choose $\beta$ sufficiently large, such
that

$-\dfrac{2S_{k}^{ii}x_{i}^{2}}{\beta}+\dfrac{1}{2}\displaystyle\sum_{i}S_{k}^{ii}>0.$

By Newton-Maclaurin inequality, we have

$\sigma_{k-1}\geq\sigma_{1}^{\frac{1}{k-1}}\sigma_{k}^{\frac{k-2}{k-1}},~{}~{}~{}\sigma_{k-2}\geq\sigma_{1}^{\frac{1}{k-2}}\sigma_{k-1}^{\frac{k-3}{k-2}},$

which implies

$\displaystyle\sum_{i}S_{k}^{ii}=(n-k+1)\sigma_{k-1}+\alpha(n-k+2)\sigma_{k-2}\geq
c_{0}\sigma_{1}^{\frac{1}{k-1}}.$

By (v) in section 2,

$\dfrac{S_{k}^{ii}u_{ii}}{u}=\dfrac{kS_{k}-\alpha\sigma_{k-1}}{u}>\dfrac{kS_{k}}{u}.$

Substituting the above inequalities into (5.10), we obtain

$-\dfrac{\beta
C}{u}\geq\dfrac{1}{2}\displaystyle\sum_{i}S_{k}^{ii}-\dfrac{C}{\kappa_{1}}\geq\dfrac{c_{0}}{4}\sigma_{1}^{\frac{1}{k-1}}.$

Thus, we obtain Lemma 5.1. ∎

Proof of Theorem 1.3: The proof is classical as in [33, 25]. For the
convenience of the readers, the complete proof is given here. Suppose $u$ is
an entire solution of the equation (1.7). For arbitrary positive constant
$R>1$, we consider the set

$\Omega_{R}=\\{y\in R^{n};u(Ry)\leq R^{2}\\}.$

Let

$v(y)=\dfrac{u(Ry)-R^{2}}{R^{2}}.$

We consider the following Dirichlet problem

(5.11)
$\left\\{\begin{array}[]{lll}{{\sigma}_{k}}(D^{2}v)+\alpha{{\sigma}_{k-1}}(D^{2}v)=1,&in&\Omega_{R},\\\
v=0,&on&\partial{\Omega_{R}}.\end{array}\right.$

By Lemma 5.1, we have the following estimates

(5.12) $(-v)^{\beta}\Delta v\leq C.$

Here the constants $C$ and $\beta$ depend on $k$ and the diameter of the
domain $\Omega_{R}$. Now using the quadratic growth condition appears in
Theorem 1.6, we have

$c|Ry|^{2}-b\leq u(Ry)\leq R^{2}.$

Namely

$|y|^{2}\leq\dfrac{1+b}{c}.$

Hence, $\Omega_{R}$ is bounded. Thus, the constant $C$ and $\beta$ become two
absolutely constants. Consider the domain

$\Omega_{R}^{\prime}=\\{y;u(Ry)\leq\dfrac{R^{2}}{2}\\}\subset\Omega_{R}.$

In $\Omega_{R}^{\prime}$, we have

$v(y)\leq-\dfrac{1}{2}.$

Hence, (5.12) implies that in $\Omega_{R}^{\prime}$, we have

$\Delta v\leq 2^{\beta}C.$

Note that

$\nabla_{y}^{2}v=\nabla_{x}^{2}u.$

Thus, using the above two formulas, in
$\Omega_{R}^{\prime}=\\{x;u(x)\leq\dfrac{R^{2}}{2}\\}$, we have

$\Delta u\leq C,$

where $C$ is an absolutely constant. Since $R$ is arbitrary, we have the above
inequality in whole $R^{n}$. Using Evan-Krylov theory [10], we have

$|D^{2}u|_{C^{\alpha}(B_{R})}\leq
C\dfrac{|D^{2}u|_{C^{0}(B_{R})}}{R^{\alpha}}\leq\dfrac{C}{R^{\alpha}}.$

Hence, when $R\rightarrow+\infty$, we obtain Theorem 1.3.

## Acknowledgments:

The last author wish to thank Professor Pengfei Guan for his valuable
suggestions and comments. We wish to thank the referees for useful comments
and suggestions.

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