# Smooth solutions to the Gauss image problem

Li Chen  Faculty of Mathematics and Statistics, Hubei Key Laboratory of
Applied Mathematics, Hubei University, Wuhan 430062, P.R. China
<EMAIL_ADDRESS>,  Di Wu  Faculty of Mathematics and Statistics, Hubei Key
Laboratory of Applied Mathematics, Hubei University, Wuhan 430062, P.R. China
<EMAIL_ADDRESS>and  Ni Xiang  Faculty of Mathematics and Statistics,
Hubei Key Laboratory of Applied Mathematics, Hubei University, Wuhan 430062,
P.R. China<EMAIL_ADDRESS>
###### Abstract.

In this paper we study the the Gauss image problem, which is a generalization
of the Aleksandrov problem in convex geometry. By considering a geometric flow
involving Gauss curvature and functions of normal vectors and radial vectors,
we obtain the existence of smooth solutions to this problem.

###### Key words and phrases:

Monge-Ampère equation, dual Orlicz-Minkowski problem, Gauss curvature flow,
Existence of solutions.

###### 2010 Mathematics Subject Classification:

35J96, 52A20, 53C44.

This research was supported by funds from Natural Science Foundation of China
No.11971157.

##  1\. Introduction

Let $K\subset\mathbb{R}^{n}$ be a convex body which contains the origin in its
interior and $x\in\partial K$ be a boundary point, then the normal cone of $K$
at $x$ is defined by

$\mathcal{N}(K,x)=\\{v\in\mathbb{S}^{n-1}:\langle y-x,v\rangle\leq
0\quad\text{for all}\quad y\in K\\},$

where $\langle y-x,v\rangle$ denotes the standard inner product of $y-x$ and
$v$ in $\mathbb{R}^{n}$. For $\omega\subset\mathbb{S}^{n-1}$, the radial Gauss
image of $\omega$ is defined by

$\alpha_{K}(\omega)=\bigcup_{x\in\rho_{K}(\omega)}\mathcal{N}(K,x)\subset\mathbb{S}^{n-1},$

where $\rho_{K}:\mathbb{S}^{n-1}\rightarrow\partial K$ is the radial function
of $\partial K$(see Section 2 for the definition). Recently, Boroczky, Lutwak,
Yang, Zhang and Zhao [4] proposed the Gauss image problem which link two given
submeasures via the radial Gauss image of a convex body:

The Gauss image problem. _Suppose $\lambda$ is a submeasure defined on the
Lebesgue measurable subsets of $\mathbb{S}^{n-1}$, and $\mu$ is a Borel
submeasure on $\mathbb{S}^{n-1}$. What are the necessary and sufficient
conditions, on $\lambda$ and $\mu$, so that there exists a convex body $K$
such that_

(1) $\lambda(\alpha_{K}(\cdot))=\mu$

_on the Borel subsets of $\mathbb{S}^{n-1}$? And if such a body exists, to
what extent is it unique?_

When $\lambda$ is spherical Lebesgue measure, the Gauss image problem is just
the classical Aleksandrov problem. It is necessary to contrast the Gauss image
problem with the various Minkowski problems and dual Minkowski problems that
have been extensively studied, see [7, 12, 15, 29, 32, 33, 38, 39, 40, 41, 42,
44, 48, 49] for the $L_{p}$-Minkowski problem, [6, 23, 25, 26, 36, 46, 47] for
the dual Minkowski problem, [5, 10, 11, 27, 28, 35, 43] for the $L_{p}$ dual
Minkowski problem, [3, 21, 24, 31] for the Orlicz Minkowski problem, [17, 19,
37] for the dual Orlicz Minkowski problem. In the Gauss image problem, a pair
of submeasures is given and it is asked if there exists a convex body
“linking” them via its radial Gauss image. However, in a Minkowski problem,
only one measure is given, and the question asks if this measure is a specific
geometric measure of a convex body.

To statement the solutions to the Gauss image problem in [4]. We introduce
some concepts (see [4] for details). If $\omega\subset\mathbb{S}^{n-1}$ is
contained in a closed hemisphere, then the polar set $\omega^{*}$ is defined
by

$\omega^{*}=\\{v\in\mathbb{S}^{n-1}:\langle u,v\rangle\leq 0\quad\text{for
all}\quad u\in\omega\\}.$

######  Definition 1.

Two Borel measures $\mu$ and $\lambda$ on $\mathbb{S}^{n-1}$ are called
Aleksandrov related if

(2)
$\lambda(\mathbb{S}^{n-1})=\mu(\mathbb{S}^{n-1})>\lambda(\omega^{*})+\mu(\omega)$

for any compact, spherically convex set $\omega\in\mathbb{S}^{n-1}$.

Note that $\lambda(\mathbb{S}^{n-1})=\mu(\mathbb{S}^{n-1})$ is obvious for a
solution to (1). The following existence result for solutions to the Gauss
image problem was proved in [4].

######  Theorem 1.

Suppose $\lambda,\mu$ are Borel measures on $\mathbb{S}^{n-1}$ and $\lambda$
is absolutely continuous. If $\mu$ and $\lambda$ are Aleksandrov related, then
there exists a body $K$ containing the origin in its interior such that
$\lambda(\alpha_{K}(\cdot))=\mu$.

Note that for the special case in which $\mu$ is a measure that has a density
with repect to the spherical Lebesgue measure, say $f$, and $\lambda$ is a
measure that has a density with repect to the spherical Lebesgue measure, say
$g$. In this case, $\mu$ and $\lambda$ on $\mathbb{S}^{n-1}$ are Aleksandrov
related if

(3)
$\int_{\mathbb{S}^{n-1}}f=\int_{\mathbb{S}^{n-1}}g>\int_{\omega}f+\int_{\omega^{*}}g$

for any compact, spherically convex set $\omega\in\mathbb{S}^{n-1}$. Moreover,
the geometric problem (1) is the equation of Monge-Ampère type

(4) $g\bigg{(}\frac{\nabla h+hx}{|\nabla h+hx|}\bigg{)}|\nabla
h+hx|^{-n}h\det(\nabla^{2}h+hI)=f\quad\text{ on }\quad\mathbb{S}^{n-1},$

where $h$ is the support function of the polar body $K^{*}$ of $K$ which is
defined as

$K^{*}=\\{x\in\mathbb{R}^{n}:\langle x,y\rangle\leq 1\quad\text{for all}\quad
y\in K\\}.$

Here $\nabla$ is the covariant derivative with respect to an orthonormal frame
on $\mathbb{S}^{n-1}$, $I$ is the unit matrix of order $n-1$, and $\nabla
h(x)+h(x)x$ is just the point on $\partial K^{*}$ whose outer unit normal
vector is $x\in\mathbb{S}^{n-1}$.

In this paper we study the existence of smooth solutions to the equation (4).
We obtain the following existence result.

######  Theorem 2.

Suppose that $f$ and $g$ are two positive smooth functions on
$\mathbb{S}^{n-1}$ . If $f$ and $g$ satisfy the condition (3), then there
exists a smooth solution to the equation (4).

The proof of Theorem 2 is inspired by [36], where the existence of smooth
solutions to the Aleksandrov and dual Minkowski problem was obtained by
studying a generalized Gauss curvature flow. In fact, the Gauss curvature flow
and its various generalizations have been extensively studied by many
scholars; see for example [1, 2, 8, 9, 10, 14, 16, 20, 22, 30, 37, 45] and the
references therein.

Let $M_{0}$ be a smooth, closed, uniformly convex hypersurface in
$\mathbb{R}^{n}$, which contains the origin in its interiors and given by a
smooth embedding $X_{0}:\mathbb{S}^{n-1}\rightarrow\mathbb{R}^{n}$. We
consider a family of closed hypersurfaces $\left\\{M_{t}\right\\}$ given by
$M_{t}=X(\mathbb{S}^{n-1},t)$, where
$X:\mathbb{S}^{n-1}\times[0,T)\rightarrow\mathbb{R}^{n}$ is a smooth map
satisfying the following initial value problem:

(5) $\left\\{\begin{aligned} \frac{\partial X}{\partial
t}(x,t)&=-\frac{f(\nu)}{g\Big{(}\frac{X}{|X|}\Big{)}}|X|^{n}\mathcal{K}\nu+X,\\\
X(x,0)&=X_{0}(x).\end{aligned}\right.$

Here $\nu$ is the unit outer normal vector of the hypersurface $M_{t}$ at the
point $X(x,t)$, $\mathcal{K}$ is the Gauss curvature of $M_{t}$ at $X(x,t)$,
and $T$ is the maximal time for which the solution exists. We obtain the long-
time existence and convergence of the flow (5).

######  Theorem 3.

Suppose $f$ and $g$ satisfy the assumptions of Theorem 2. Let $M_{0}$ be a
smooth, closed, uniformly convex hypersurface in $\mathbb{R}^{n}$, which
contains the origin in its interior. Then, the flow (5) has a unique smooth
solution, which exists for all time $t>0$. Moreover, when $t\to\infty$, a
subsequence of $M_{t}=X(\mathbb{S}^{n-1},t)$ converges in $C^{\infty}$ to a
smooth, closed, uniformly convex hypersurface, whose support function is a
smooth solution to the equation (4).

This paper is organized as follows. In section 2, we give some basic knowledge
about convex hypersurfaces and the flow (5). In section 3, more properties of
the flow (5) will be proved, based on which we can obtain the uniform lower
and upper bounds of support functions of $\left\\{M_{t}\right\\}$ via delicate
analyses. In the last section, the long-time existence and convergence of the
flow (5) will be proved, which completes the proofs of our theorems.

##  2\. Preliminaries

###  2.1. Basic properties of convex hypersurfaces

We first recall some basic properties of convex hypersurfaces in
$\mathbb{R}^{n}$; see [45] for details. Let $M$ be a smooth, closed, uniformly
convex hypersurface in $\mathbb{R}^{n}$ enclosing the origin. The support
function $h$ of $M$ is defined as

(6) $h(x):=\max_{y\in M}\langle
y,x\rangle,\quad\forall\,x\in\mathbb{S}^{n-1},$

where $\langle\cdot,\cdot\rangle$ is the standard inner product in
$\mathbb{R}^{n}$.

The convex hypersurface $M$ can be recovered by its support function $h$. In
fact, writing the Gauss map of $M$ as $\nu_{M}$, we parametrize $M$ by
$X:\mathbb{S}^{n-1}\to M$ which is given as

$X(x)=\nu_{M}^{-1}(x),\quad\forall\,x\in\mathbb{S}^{n-1}.$

Note that $x$ is the unit outer normal vector of $M$ at $X(x)$. On the other
hand, one can easily check that the maximum in the definition (6) is attained
at $y=\nu_{M}^{-1}(x)$, namely

(7) $h(x)=\langle x,X(x)\rangle,\quad\forall\,x\in\mathbb{S}^{n-1}.$

Let $e_{ij}$ be the standard metric of the unit sphere $\mathbb{S}^{n-1}$, and
$\nabla$ be the corresponding connection on $\mathbb{S}^{n-1}$. Then, it is
easy to check that

(8) $X(x)=\nabla h(x)+h(x)x,\quad\forall\,x\in\mathbb{S}^{n-1}.$

By differentiating (7) twice, the second fundamental form $A_{ij}$ of $M$ can
be also computed in terms of the support function:

(9) $A_{ij}=\nabla_{i}\nabla_{j}h+he_{ij},$

where $\nabla_{i}\nabla_{j}$ denotes the second order covariant derivative
with respect to $e_{ij}$. The induced metric matrix $g_{ij}$ of $M$ can be
derived by Weingarten’s formula:

(10) $e_{ij}=\langle\nabla_{i}x,\nabla_{j}x\rangle=A_{ik}A_{lj}g^{kl}.$

The principal radii of curvature are eigenvalues of the matrix
$b_{ij}=A^{ik}g_{jk}$. When considering a smooth local orthonormal frame on
$\mathbb{S}^{n-1}$, by virtue of (9) and (10), we have

(11) $b_{ij}=A_{ij}=\nabla_{i}\nabla_{j}h+h\delta_{ij}.$

In particular, the Gauss curvature of $M$ at $X(x)$ is given by

$\mathcal{K}(x)=[\det(\nabla_{i}\nabla_{j}h+h\delta_{ij})]^{-1}.$

We shall use $b^{ij}$ to denote the inverse matrix of $b_{ij}$.

The radial function $\rho$ of the convex hypersurface $M$ is defined as

$\rho(u):=\max\left\\{\lambda>0:\lambda u\in
M\right\\},\quad\forall\,u\in\mathbb{S}^{n-1}.$

Note that $\rho(u)u\in M$. The Gauss map $\nu_{M}$ can be computed as

$\nu_{M}(\rho(u)u)=\frac{\rho(u)u-\nabla\rho}{\sqrt{\rho^{2}+|\nabla\rho|^{2}}}.$

If we connect $u$ and $x$ through the following equality:

(12) $\rho(u)u=X(x)=\nabla h(x)+h(x)x=\overline{\nabla}h(x),$

where $\overline{\nabla}$ is the standard connection of $\mathbb{R}^{n}$, then
we have the following relations

(13) $x=\frac{\rho(u)u-\nabla\rho}{\sqrt{\rho^{2}+|\nabla\rho|^{2}}},\quad
u=\frac{\nabla h+h(x)x}{\sqrt{|\nabla h|^{2}+h^{2}}},$

and

(14) $\frac{h(x)}{\mathcal{K}(x)}dx=\rho^{n}(u)du.$

###  2.2. Geometric flow and its associated functional

Recalling the evolution equation of $X(x,t)$ in the geometric flow (5), and
using similar computations as in [45], we obtain the evolution equation of the
corresponding support function $h(x,t)$:

(15) $\frac{\partial h}{\partial
t}(x,t)=-\frac{f(x)\rho^{n}(u)}{g(u)}\mathcal{K}(x,t)+h(x,t)\ \text{ in }\
\mathbb{S}^{n-1}\times(0,T).$

Since $M_{t}$ can be recovered by $h(\cdot,t)$, the flow (15) is equivalent to
the original flow (5).

Denote the radial function of $M_{t}$ by $\rho(u,t)$. For any $t$, let $u$ and
$x$ be related through the following equality:

$\rho(u,t)u=\overline{\nabla}h(x,t)=\nabla h(x,t)+h(x,t)x.$

Therefore, $x$ can be expressed as $x=x(u,t)$. By a direction computation (see
[10]), we have

(16)
$\frac{1}{\rho(u,t)}\partial_{t}\rho(u,t)=\frac{1}{h(x,t)}\partial_{t}h(x,t).$

Now by virtue of (15) and (16), we obtain the evolution equation of
$\rho(u,t)$:

(17) $\frac{\partial\rho}{\partial
t}(u,t)=-\frac{f(x)\rho^{n+1}(u,t)}{g(u)h(x,t)}\mathcal{K}(x,t)+\rho(u,t)\
\text{ in }\ \mathbb{S}^{n-1}\times(0,T),$

where $x=x(u,t)$ is the unit outer normal vector of $M_{t}$ at the point
$\rho(u,t)u$.

Consider the following functional:

(18) $J(t)=\int_{\mathbb{S}^{n-1}}f(x)\log
h(x,t)\mathop{}\\!\mathrm{d}x-\int_{\mathbb{S}^{n-1}}g(u)\log\rho(u,t)\mathop{}\\!\mathrm{d}u,\quad
t\geq 0,$

which will turn out to be monotonic along the flow (15).

######  Lemma 1.

$J(t)$ is non-increasing along the flow (15). Namely $\frac{d}{dt}J(t)\leq 0$,
and the equality holds if and only if $M_{t}$ satisfies the elliptic equation
(4).

###### Proof.

Using (14) and (16), we have

$\displaystyle\frac{d}{dt}J(t)$ $\displaystyle=$
$\displaystyle\int_{\mathbb{S}^{n-1}}\frac{f\partial_{t}h}{h}dx-\int_{\mathbb{S}^{n-1}}\frac{g\partial_{t}\rho}{\rho}du$
$\displaystyle=$
$\displaystyle\int_{\mathbb{S}^{n-1}}\frac{\partial_{t}h}{h}(f-\frac{gh}{\mathcal{K}\rho^{n}})dx$
$\displaystyle=$
$\displaystyle\int_{\mathbb{S}^{n-1}}\frac{1}{h}(f-\frac{gh}{\mathcal{K}\rho^{n}})(h-\frac{f\mathcal{K}\rho^{n}}{g})dx$
$\displaystyle=$
$\displaystyle-\int_{\mathbb{S}^{n-1}}\frac{1}{gh\mathcal{K}\rho^{n}}(gh-f\mathcal{K}\rho^{n})^{2}dx$
$\displaystyle\leq$ $\displaystyle 0.$

Clearly $\frac{d}{dt}J(t)=0$ if and only if

$gh=f\mathcal{K}\rho^{n}.$

Namely $M_{t}$ satisfies (4). This completes the proof. ∎

######  Lemma 2.

Assume that

$\int_{\mathbb{S}^{n-1}}f(x)dx=\int_{\mathbb{S}^{n-1}}g(u)du,$

then the log-volume of $M_{t}$

(20)
$V_{g}(M_{t})=\int_{\mathbb{S}^{n-1}}g(u)\log\rho(u,t)\mathop{}\\!\mathrm{d}u,$

remain unchanged under the flow (15).

###### Proof.

Using (14) and (17), we have

$\displaystyle\frac{d}{dt}V_{g}(M_{t})$ $\displaystyle=$
$\displaystyle\int_{\mathbb{S}^{n-1}}\frac{g\partial_{t}\rho}{\rho}du$
$\displaystyle=$
$\displaystyle\int_{\mathbb{S}^{n-1}}\frac{g}{\rho}(\rho-\frac{f\rho^{n+1}}{gh}\mathcal{K})du$
$\displaystyle=$
$\displaystyle\int_{\mathbb{S}^{n-1}}g(u)du-\int_{\mathbb{S}^{n-1}}f(x)\frac{\rho^{n}\mathcal{K}}{h}du$
$\displaystyle=$
$\displaystyle\int_{\mathbb{S}^{n-1}}g(u)du-\int_{\mathbb{S}^{n-1}}f(x)dx$
$\displaystyle=$ $\displaystyle 0.$

So, we complete the proof. ∎

##  3\. Uniform bounds of support functions

In this section, we will derive uniformly positive lower and upper bounds of
support functions along the flow (5). Our idea comes from the proof of Lemma
3.2 in [36].

######  Lemma 3.

Suppose $f$ and $g$ satisfy the assumptions in Theorem 2. Let $X(\cdot,t)$ be
a strictly convex solution to the flow (5) which encloses the origin for
$t\in[0,T)$, then exists a positive constant $C$ depending only on $M_{0}$,
$f$ and $g$, such that for every $t\in[0,T)$,

(21) $1/C\leq h(\cdot,t)\leq C\quad\text{ on }\ \mathbb{S}^{n-1},$

and

(22) $1/C\leq\rho(\cdot,t)\leq C\quad\text{ on }\ \mathbb{S}^{n-1}.$

###### Proof.

Since $f$ and $g$ satisfy the condition (3), the measure $\mu$ with the
density $f$, and the measure $\lambda$ with the density $g$ are Aleksandrov
related. Then, using Theorem 1, there exists a body $N^{*}$ containing the
origin in its interior satisfies the equation (1). Let $N$ be the polar dual
of $N^{*}$. Choosing the constants $s_{1}>s_{0}>0$ such that

$N_{0}=s_{0}N\subset K_{0}\subset s_{1}N=N_{1}.$

Let $r_{0}$ and $r_{1}$ be respectively the radial functions of $N_{0}$ and
$N_{1}$. Clearly, $sN$ is a stationary solution to (5) in the generalised
sense. We firstly prove that

$K_{t}\subset N_{1}$

for all $t>0$ by a contradiction. Otherwise, there exists a first time
$t_{0}>0$ such that

$\sup_{u\in\mathbb{S}^{n-1}}\frac{\rho(u,t_{0})}{r_{1}(u)}=1.$

Set

$P=M_{t_{0}}\cap N_{1},$

which can be a point or a closed set. Clearly, the unit normal vector of
$M_{t_{0}}$ coincide with that of $N_{1}$ for any $p\in P$. Namely,
$\nu_{M_{t_{0}}}(p)=\nu_{N_{1}}(p)$ for any $p\in P$. Moreover, replacing
$r_{1}$ by $(1+a)r_{1}$ for a small constant $a$, we may assume that

$\frac{\partial}{\partial t}\rho(u,t)>0\quad\text{on}\quad P\times\\{t_{0}\\}$

and also in a neighbourhood of $P\times\\{t_{0}\\}$. There exists sufficiently
small constants $\epsilon,\delta>0$ such that

$\frac{\partial}{\partial t}\rho(u,t)>\delta$

for all $u\in
E=\\{\xi\in\mathbb{S}^{n-1}:\rho(\xi,t)>(1-\epsilon)r_{1}(\xi)\\}$. Since
$\nu_{M_{t_{0}}}(p)=\nu_{N_{1}}(p)$ for any $p\in P$, making $\epsilon$ small
again, we have $\nu_{N_{1}}(u)\approx\nu_{M_{t_{0}}}(u)$ for $u\in E$. Thus,
using the equation (17), the Gauss curvature of $M_{t_{0}}$ satisfies

$\displaystyle\mathcal{K}(M_{t_{0}})$ $\displaystyle<$
$\displaystyle\frac{(\rho(u,t_{0})-\delta)g(u)}{f(\nu_{M_{t_{0}}})\rho^{n+1}(u,t_{0})}$
$\displaystyle<$
$\displaystyle\frac{(r_{1}(u,t_{0})-\delta)g(u)}{f(\nu_{N_{1}})(r_{1}(u,t_{0})-\epsilon)^{n+1}}$
$\displaystyle<$
$\displaystyle\frac{1}{(1-\epsilon)^{n}}\frac{g(u)}{f(\nu_{N_{1}})(r_{1}(u,t_{0}))^{n}}$
$\displaystyle<$ $\displaystyle\mathcal{K}((1-\epsilon)N_{1}).$

Namely, the Gauss curvature of $M_{t_{0}}$ is strictly small than that of
$(1-\epsilon)N_{1}$ for all $\xi\in E$. Applying the comparison principle for
generalised solutions to the elliptic Monge-Ampère equation (see Theorem 1.4.6
in [18]) to the functions $\rho(u,t_{0})$ and $(1-\epsilon)r_{1}(u)$, we reach
a contradiction. Similarly, we can prove that $N_{0}\subset M_{t}$ for all
$t>0$. ∎

From Theorems 2 and 3 in [4], we know that if $f$ and $g$ are even functions
satisfying

(23) $\int_{\mathbb{S}^{n-1}}f(x)dx=\int_{\mathbb{S}^{n-1}}g(u)du,$

then $f$ and $g$ satisfy the condition (3). In this case, if $M_{0}$ is
origin-symmetric, we can give a proof of Lemma 3 without using Theorem 1.

######  Lemma 4.

Suppose that $M_{0}$ is origin-symmetric, $f$ and $g$ are two smooth, positive
even functions satisfying the condition (23), then the conclusions in Lemma 3
hold true.

###### Proof.

Note that $M_{0}$ is origin-symmetric, $f$ and $g$ are even functions, thus
$M_{t}$ is origin-symmetric and $h(x,t)$ is an even function. For fixed
$t\in[0,T)$, assume $h(x,t)$ attains its maximum at $x_{t}$. Since $h(x,t)$ is
an even function, we have by the definition of the support function (6)

(24) $h(x,t)\geq|\langle x,x_{t}\rangle|h(x_{t},t).$

By Lemmas 1 and 2,

$\displaystyle\int_{\mathbb{S}^{n-1}}f(x)\log h(x,0)dx$ $\displaystyle\geq$
$\displaystyle\int_{\mathbb{S}^{n-1}}f(x)\log h(x,t)dx$ $\displaystyle\geq$
$\displaystyle\int_{\mathbb{S}^{n-1}}f(x)\log|\langle
x,x_{t}\rangle|h(x_{t},t)dx$ $\displaystyle\geq$ $\displaystyle C\log
h(x_{t},t)-C,$

which implies that

$\max_{\mathbb{S}^{n-1}\times[0,T)}h(x,t)\leq C$

for some positive constant $C$.

The positive lower bound of $h$ will be proved by contradiction. Let
$\\{t_{k}\\}\subset[0,T)$ be a sequence such that

$\min_{\mathbb{S}^{n-1}}h(\cdot,t_{k})\rightarrow 0\quad\text{as}\quad
k\rightarrow\infty.$

Let $K_{t}$ be the convex body enclosed by $M_{t}$. By Blaschke selection
theorem, there is a sequence in $\\{K_{t_{k}}\\}$, which is still denoted by
$\\{K_{t_{k}}\\}$, such that

$K_{t_{k}}\rightarrow\widetilde{K}\quad\text{as}\quad k\rightarrow+\infty.$

Since $K_{t_{k}}$ is an origin-symmetric convex body, $\widetilde{K}$ is also
origin-symmetric. Then

$\displaystyle\min_{\mathbb{S}^{n-1}}h_{\widetilde{K}}=\lim_{k\rightarrow+\infty}\min_{\mathbb{S}^{n-1}}h_{K_{t_{k}}}=0.$

It follows that $\widetilde{K}$ is contained in a hyperplane in
$\mathbb{R}^{n}$. Then

$\rho_{\widetilde{K}}=0,\quad\text{a.e. in $\mathbb{S}^{n-1}$}.$

Using Lemma 2, we have for any $\epsilon>0$

$\displaystyle V_{g}(M_{0})$ $\displaystyle=$ $\displaystyle V_{g}(M_{t_{k}})$
$\displaystyle\leq$
$\displaystyle\lim_{k\rightarrow+\infty}\int_{\mathbb{S}^{n-1}}g(u)\log[\rho(u,t_{k})+\epsilon]du$
$\displaystyle=$ $\displaystyle\int_{\mathbb{S}^{n-1}}g(u)\log\epsilon du$
$\displaystyle=$ $\displaystyle
C\log\epsilon\rightarrow-\infty\quad\text{as}\quad\epsilon\rightarrow 0,$

which is a contradiction. Then

$\min_{\mathbb{S}^{n-1}\times[0,T)}h(x,t)\geq C$

for some positive constant $C$. So we complete the proof. ∎

Due to the convexity of $M_{t}$, Lemma 3 also implies the gradient estimates
of $h(\cdot,t)$ and $\rho(\cdot,t)$.

######  Lemma 5.

Let $X(\cdot,t)$ be a strictly convex solution to the flow (5) which encloses
the origin for $t\in[0,T)$, then we have

$\displaystyle|\nabla h(x,t)|\leq
C,\quad\forall(x,t)\in\mathbb{S}^{n-1}\times[0,T),$
$\displaystyle|\nabla\rho(u,t)|\leq
C,\quad\forall(u,t)\in\mathbb{S}^{n-1}\times[0,T),$

where $C$ is a positive constant depending only on the constant in Lemma 3.

###### Proof.

By virtue of (12), we have

(25) $\rho^{2}=|\nabla h|^{2}+h^{2}\geq|\nabla h|^{2}.$

By (7), (12) and (13), we have

(26)
$h=\frac{\rho^{2}}{\sqrt{\rho^{2}+|\nabla\rho|^{2}}}\leq\frac{\rho^{2}}{|\nabla\rho|}.$

Using the two inequalities (25) and (26), the estimates of this lemma now
follows directly from Lemma 3. ∎

##  4\. Uniform bounds for principal curvatures

In this section, we continue to establish uniform upper and lower bounds for
principal curvatures. These estimates can be obtained by considering proper
auxiliary functions, see [10, 13, 36, 37] for similar techniques. First, we
need the following lemma.

######  Lemma 6.

Given two positive constants $r<R$. For the function

$G(y)=\frac{|y|^{n}}{g(\frac{y}{|y|})},\quad y=(y^{1},...,y^{n})\in
A(r,R)=\\{y\in\mathbb{R}^{n}:r<|y|<R\\},$

we have

$\displaystyle\left\|G\right\|_{C^{k}(A(r,R))}\leq
C_{k}\left\|g\right\|_{C^{k}(\mathbb{S}^{n-1})},$

where $k=0,1,2$, and $C_{k}$ is a positive constant depending only on $n,r,R$,
$\left\|g\right\|_{C^{k}(\mathbb{S}^{n-1})}$, and
$\min\limits_{\mathbb{S}^{n-1}}g$.

###### Proof.

Denote $\partial_{i}=\frac{\partial}{\partial y^{i}}$ for $1\leq i\leq n$. It
is clearly see that

$\displaystyle\overline{\nabla}_{i}|y|^{n}=n|y|^{n-2}y^{i}\quad\text{and}\quad\overline{\nabla}_{i}g=\frac{1}{|y|^{3}}\langle\overline{\nabla}g,|y|^{2}\partial_{i}-yy^{i}\rangle.$

Thus,

$\displaystyle\overline{\nabla}_{i}G(y)$ $\displaystyle=$
$\displaystyle\frac{n|y|^{n-2}y^{i}}{g}-\frac{|y|^{n-3}\langle\overline{\nabla}g,|y|^{2}\partial_{i}-yy^{i}\rangle}{g^{2}}.$

It follows consequently

$\displaystyle\left\|G\right\|_{C^{1}(A(r,R))}\leq
C_{1}\left\|g\right\|_{C^{1}(\mathbb{S}^{n})}.$

Moreover, we have

$\displaystyle\overline{\nabla}_{j}\overline{\nabla}_{i}|y|^{n}=n(n-2)|y|^{n-4}y^{i}y^{j}+n|y|^{n-2}\delta_{ij}$

and

$\displaystyle\overline{\nabla}_{j}\overline{\nabla}_{i}g$ $\displaystyle=$
$\displaystyle\frac{1}{|y|^{3}}\overline{\nabla}^{2}g\Big{(}|y|^{2}\partial_{i}-yy^{i},|y|^{2}\partial_{i}-yy^{i}\Big{)}+\frac{1}{|y|^{3}}\Big{\langle}\overline{\nabla}g,2y^{j}\partial_{i}-y^{i}\partial_{j}-y\delta_{ij}\Big{\rangle}$
$\displaystyle-\frac{3y^{j}}{|y|^{5}}\langle\overline{\nabla}g,|y|^{2}\partial_{i}-yy^{i}\rangle.$

Note that

$\displaystyle\overline{\nabla}_{j}\overline{\nabla}_{i}G$ $\displaystyle=$
$\displaystyle\frac{1}{g}\overline{\nabla}_{j}\overline{\nabla}_{i}|y|^{n}-|y|^{n}\frac{\overline{\nabla}_{j}\overline{\nabla}_{i}g}{g^{2}}+2|y|^{n}\frac{\overline{\nabla}_{i}g\overline{\nabla}_{j}g}{g^{3}}-2\overline{\nabla}_{i}(|y|^{n})\frac{\overline{\nabla}_{j}g}{g}.$

Thus, we get

$\displaystyle\left\|G\right\|_{C^{2}(A(r,R))}\leq
C_{2}\left\|g\right\|_{C^{2}(\mathbb{S}^{n})}$

in view of

$\displaystyle\overline{\nabla}^{2}g(e_{i},e_{j})=\nabla^{2}g(e_{i},e_{j})-\langle\overline{\nabla}g,\frac{y}{|y|}\rangle\delta_{ij}$

for a local orthonormal frame $\\{e_{1},\cdots,e_{n-1}\\}$ on
$\mathbb{S}^{n-1}$. So, our proof is completed. ∎

If we have proved Lemma 6, we can derive the uniform upper bound of the Gauss
curvature of $M_{t}$ and the uniform lower bound for principal curvatures by
similar calculations which have been done in [13]. In the rest of this
section, we take a local orthonormal frame $\\{e_{1},\cdots,e_{n-1}\\}$ on
$\mathbb{S}^{n-1}$ such that the standard metric on $\mathbb{S}^{n-1}$ is
$\\{\delta_{ij}\\}$. And double indices always mean to sum from $1$ to $n-1$.

######  Lemma 7.

Let $X(\cdot,t)$ be a strictly convex solution to the flow (5) which encloses
the origin for $t\in[0,T)$, then we have

$\mathcal{K}(x,t)\leq C,\quad\forall(x,t)\in\mathbb{S}^{n-1}\times[0,T),$

where $C$ is a positive constant depending only on the constants in Lemmas 3
and 5.

###### Proof.

Set

$Q(x,t)=\frac{-\partial_{t}h(x,t)+h(x,t)}{h(x,t)-\varepsilon_{0}}=\frac{f(x)\rho^{n}(u)}{(h-\varepsilon_{0})g(u)}\mathcal{K}(x,t),$

where

$\varepsilon_{0}=\frac{1}{2}\,\inf_{\mathbb{S}^{n-1}\times[0,T)}h(x,t)$

and the second equality follows from (15). For each $t\in[0,T)$, assume
$Q(\cdot,t)$ attains its maximum at some point $x_{t}\in\mathbb{S}^{n-1}$. At
$(x_{t},t)$, we can obtain

(27)
$0=Q_{i}=\frac{-\partial_{t}h_{i}+h_{i}}{h-\varepsilon_{0}}+\frac{\partial_{t}h-h}{(h-\varepsilon_{0})^{2}}h_{i},$

and

(28) $0\geq
Q_{ij}=\frac{-\partial_{t}h_{ij}+h_{ij}}{h-\varepsilon_{0}}+\frac{(\partial_{t}h-h)h_{ij}}{(h-\varepsilon_{0})^{2}},$

where (27) is used in (28). Recall that $b_{ij}=h_{ij}+h\delta_{ij}$, and
$b^{ij}$ is its inverse matrix. Using the inequality (28), it yields

$\begin{split}\partial_{t}b_{ij}&=\partial_{t}h_{ij}+\partial_{t}h\delta_{ij}\\\
&\geq
h_{ij}+\frac{\partial_{t}h-h}{h-\varepsilon_{0}}h_{ij}+\partial_{t}h\delta_{ij}\\\
&=b_{ij}-Q(b_{ij}-\varepsilon_{0}\delta_{ij}).\end{split}$

Noticing that $\mathcal{K}=1/\det(b_{ij})$, we have

(29)
$\begin{split}\partial_{t}\mathcal{K}&=-\mathcal{K}b^{ji}\partial_{t}b_{ij}\\\
&\leq-\mathcal{K}b^{ji}[b_{ij}-Q(b_{ij}-\varepsilon_{0}\delta_{ij})]\\\
&=-\mathcal{K}\bigl{[}(n-1)(1-Q)+Q\varepsilon_{0}\operatorname{tr}(b^{ij})\bigr{]}.\end{split}$

Note that

(30) $Q(x,t)=\frac{f(x)\rho^{n}(u)}{(h-\varepsilon_{0})g(u)}\mathcal{K}(x,t),$

we derive by Lemma 3

(31) $\frac{1}{C_{1}}Q(x,t)\leq\mathcal{K}(x,t)\leq C_{1}Q(x,t),$

where $C_{1}$ is a positive constant depending only on the constant $C$ in
Lemma 3, and the upper and lower bounds of $f,g$ on $\mathbb{S}^{n-1}$.
Combining Lemma 3 and the inequalities (29) and (31), we have

(32)
$\begin{split}\partial_{t}\mathcal{K}&\leq(n-1)\mathcal{K}Q-(n-1)\varepsilon_{0}Q\mathcal{K}^{\frac{n}{n-1}}\\\
&\leq C_{2}Q^{2}-C_{3}Q^{\frac{2n-1}{n-1}},\end{split}$

where the inequality

$\frac{1}{n-1}\operatorname{tr}(b^{ij})\geq\det(b^{ij})^{\frac{1}{n-1}}=\mathcal{K}^{\frac{1}{n-1}}$

is used. Here $C_{2}$, $C_{3}$ are positive constants depending only on $n$,
$\varepsilon_{0}$ and $C_{1}$. By the definition of $Q$ and (27),

(33) $\partial_{t}h_{i}=(1-Q)h_{i}.$

Thus, we have

$\displaystyle\partial_{t}(\nabla h+hx)$ $\displaystyle=$
$\displaystyle\partial_{t}(h_{i}e_{i}+hx)$ $\displaystyle=$
$\displaystyle\partial_{t}h_{i}e_{i}+(\partial_{t}h)x$ $\displaystyle=$
$\displaystyle(1-Q)h_{i}e_{i}+(h-(h-\varepsilon_{0})Q)x$ $\displaystyle=$
$\displaystyle(1-Q)(\nabla h+hx)+\varepsilon_{0}Qx.$

So, we can say that

$\displaystyle\partial_{t}G(\nabla h+hx)$ $\displaystyle=$
$\displaystyle\langle\overline{\nabla}G,\partial_{t}(\nabla h+hx)\rangle$
$\displaystyle=$ $\displaystyle(1-Q)\langle\overline{\nabla}G,\nabla
h+hx\rangle+\varepsilon_{0}Q\langle\overline{\nabla}G,x\rangle$
$\displaystyle\leq$ $\displaystyle(1-Q)|\overline{\nabla}G|\cdot|\nabla
h+hx|+\varepsilon_{0}Q|\overline{\nabla}G|$ $\displaystyle\leq$ $\displaystyle
C_{4}(1-Q)|\nabla h+hx|+C_{4}\varepsilon_{0}Q,$

where we know by Lemma 6 that $C_{4}$ is a positive constant depending on $n$,
the constant $C$ in Lemmas 3 and 5,
$\left\|g\right\|_{C^{1}(\mathbb{S}^{n-1})}$ and
$\min\limits_{\mathbb{S}^{n-1}}g$.

Thus,

(36) $\begin{split}&\frac{\partial}{\partial t}\Bigl{[}\frac{G(\nabla
h+hx)}{h(x,t)-\varepsilon_{0}}\Bigr{]}\\\
=&\frac{\partial_{t}G}{h-\varepsilon_{0}}-\frac{G\partial_{t}h}{(h-\varepsilon_{0})^{2}}\\\
\leq&\frac{(1-Q)|\overline{\nabla}G|\cdot|\overline{\nabla}h|+\varepsilon_{0}Q|\overline{\nabla}G|}{h-\varepsilon_{0}}+\frac{[(1-Q)h+\varepsilon_{0}Q]G}{(h-\varepsilon_{0})^{2}}\\\
\leq&C_{5}+C_{5}Q,\end{split}$

where $C_{5}$ is a positive constant depending only on $\varepsilon_{0}$,
$C_{4}$ and the constant $C$ in Lemmas 3 and 5.

By virtue of (30), (32) and (36), we have at $(x_{t},t)$

$\displaystyle\partial_{t}Q$ $\displaystyle=$ $\displaystyle
f\partial_{t}[\frac{G}{h-\varepsilon_{0}}]\mathcal{K}+f\frac{G}{h-\varepsilon_{0}}\partial_{t}\mathcal{K}$
$\displaystyle\leq$ $\displaystyle\max_{\mathbb{S}^{n-1}}f\cdot
C_{5}(1+Q)C_{1}Q+\frac{Q}{\mathcal{K}}(C_{2}Q^{2}-C_{3}Q^{\frac{2n-1}{n-1}})$
$\displaystyle\leq$ $\displaystyle\max_{\mathbb{S}^{n-1}}f\cdot
C_{1}C_{5}Q(1+Q)+C_{1}C_{2}Q^{2}-C_{1}^{-1}C_{3}Q^{\frac{2n-1}{n-1}})$
$\displaystyle\leq$ $\displaystyle CQ+CQ^{2}-CQ^{\frac{2n-1}{n-1}},$

where we have used (31) to obtain the third inequality and $C$ is a positive
constant depending only on $\max\limits_{\mathbb{S}^{n-1}}f$, $C_{1}$,
$C_{2}$, $C_{3}$ and $C_{5}$. Thus, whenever $Q(x_{t},t)$ is greater than some
constant which is independent of $t$, we have

$\partial_{t}Q<0,$

which implies that $Q$ has an uniform upper bound. By (31), $\mathcal{K}$ has
a uniform upper bound. ∎

######  Lemma 8.

Let $X(\cdot,t)$ be a strictly convex solution to the flow (5) which encloses
the origin for $t\in[0,T)$, then for the principal curvatures
$\kappa_{i}(x,t)$ of $M_{t}$, we have

$\kappa_{i}(x,t)\geq
C,\quad\forall(x,t)\in\mathbb{S}^{n-1}\times[0,T),\quad\forall
i=1,\cdots,n-1,$

where $C$ is a positive constant depending only on the constant in Lemmas 3
and 5.

###### Proof.

Set

$\widetilde{\Lambda}(x,t)=\log\lambda_{\max}(b_{ij})-A\log h+B|\nabla
h|^{2},\quad\forall(x,t)\in\mathbb{S}^{n-1}\times[0,T),$

where $b_{ij}=h_{ij}+h\delta_{ij}$ as before, $\lambda_{\max}(b_{ij})$ denotes
the maximal eigenvalue of the matrix $(b_{ij})$, and $A$ and $B$ are positive
constants to be chosen later.

For any fixed $T^{\prime}\in(0,T)$, assume
$\max_{\mathbb{S}^{n-1}\times[0,T^{\prime}]}\widetilde{\Lambda}(x,t)$ is
attained at some point $(x_{0},t_{0})\in\mathbb{S}^{n-1}\times[0,T^{\prime}]$.
By choosing a suitable orthonormal frame, we may assume

$\\{b_{ij}(x_{0},t_{0})\\}\quad\text{is diagonal
and}\quad\lambda_{\max}(b_{ij})(x_{0},t_{0})=b_{11}(x_{0},t_{0}).$

Thus, the new function defined on $\mathbb{S}^{n-1}\times[0,T^{\prime}]$

$\Lambda(x,t)=\log b_{11}-A\log h+B|\nabla h|^{2}$

also attains its maximum at $(x_{0},t_{0})$. Thus, we have at $(x_{0},t_{0})$

(38) $0=\Lambda_{i}=b^{11}b_{11;i}-A\frac{h_{i}}{h}+2B\sum_{k}h_{k}h_{ki},$

and

(39)
$\begin{split}0\geq\Lambda_{ij}&=b^{11}b_{11;ij}-(b^{11})^{2}b_{11;i}b_{11;j}\\\
&\hskip
11.00008pt-A\Bigl{(}\frac{h_{ij}}{h}-\frac{h_{i}h_{j}}{h^{2}}\Bigr{)}+2B\sum_{k}(h_{kj}h_{ki}+h_{k}h_{kij}),\end{split}$

where $(b^{ij})$ is the inverse of the matrix $(b_{ij})$. Without loss of
generality, we can assume $t_{0}>0$. Then, we get at $(x_{0},t_{0})$

(40)
$\begin{split}0\leq\partial_{t}\Lambda&=b^{11}(\partial_{t}h_{11}+\partial_{t}h)-A\frac{\partial_{t}h}{h}+2B\sum_{k}h_{k}\partial_{t}h_{k}.\end{split}$

From the equation (15), we know

(41) $\log(h-\partial_{t}h)=\log\mathcal{K}(x,t)+\log f(x)G(\nabla h+hx).$

Set

$w(x,t)=\log\Big{[}f(x)G(\nabla h+hx)\Big{]},$

where

$G(\nabla h+hx)=\frac{|\nabla h+hx|^{n}}{g\bigg{(}\frac{\nabla h+hx}{|\nabla
h+hx|}\bigg{)}}.$

Differentiating (41) gives

(42) $\frac{h_{k}-\partial_{t}h_{k}}{h-\partial_{t}h}=-b^{ji}b_{ij;k}+w_{k},$

and

(43)
$\frac{h_{11}-\partial_{t}h_{11}}{h-\partial_{t}h}=\frac{(h_{1}-\partial_{t}h_{1})^{2}}{(h-\partial_{t}h)^{2}}-b^{ii}b_{ii;11}+b^{ii}b^{jj}(b_{ij;1})^{2}+w_{11}.$

Multiplying both sides of (43) by $-b^{11}$, it yields

(44)
$\begin{split}\frac{b^{11}\partial_{t}h_{11}-b^{11}h_{11}}{h-\partial_{t}h}&\leq
b^{11}b^{ii}b_{ii;11}-b^{11}b^{ii}b^{jj}(b_{ij;1})^{2}-b^{11}w_{11}\\\ &\leq
b^{11}b^{ii}b_{11;ii}-b^{11}b^{ii}b^{11}(b_{i1;1})^{2}-\sum_{i}b^{ii}\\\
&\hskip 11.00008pt+b^{11}(n-1-w_{11}),\end{split}$

where we use the Ricci identity $b_{ii;11}=b_{11;ii}-b_{11}+b_{ii}$. We know
that $b^{ij}\Lambda_{ij}\leq 0$ from (39) which implies at $(x_{0},t_{0})$

$\begin{split}b^{11}b^{ii}b_{11;ii}-(b^{11})^{2}b^{ii}(b_{11;i})^{2}\leq
Ab^{ii}\Bigl{(}\frac{h_{ii}}{h}-\frac{h_{i}^{2}}{h^{2}}\Bigr{)}-2B\sum_{k}b^{ii}(h_{ki}^{2}+h_{k}h_{kii}).\end{split}$

Thus,

(45) $\begin{split}&b^{11}b^{ii}b_{11;ii}-(b^{11})^{2}b^{ii}(b_{11;i})^{2}\\\
\leq&\frac{(n-1)A}{h}-A\sum_{i}b^{ii}-\frac{Ab^{ii}h_{i}^{2}}{h^{2}}+4(n-1)Bh-2B\sum_{i}b_{ii}-2Bh^{2}\sum_{i}b^{ii}\\\
&-2Bb^{ii}h_{k}b_{ii;k}+2Bb^{ii}h_{i}^{2},\end{split}$

where we use the following equalities

$\displaystyle b^{ii}h_{ii}=b^{ii}(b_{ii}-h)=n-1-h\sum_{i}b^{ii},$
$\displaystyle\sum_{k}b^{ii}h_{ki}^{2}=b^{ii}h_{ii}^{2}=b^{ii}(b_{ii}^{2}-2hb_{ii}+h^{2})=-2(n-1)h+\sum_{i}b_{ii}+h^{2}\sum_{i}b^{ii},$
$\displaystyle\sum_{k}b^{ii}h_{k}h_{kii}=\sum_{k}b^{ii}h_{k}(b_{ki;i}-h_{i}\delta_{ki})=\sum_{k}b^{ii}h_{k}b_{ii;k}-b^{ii}h_{i}^{2}.$

Here the fact that $b_{ij;k}$ is symmetric in all indices is used to get the
third equality above. Inserting the inequality (45) into (44), we obtain that

(46)
$\begin{split}\frac{b^{11}\partial_{t}h_{11}-b^{11}h_{11}}{h-\partial_{t}h}&\leq\frac{(n-1)A}{h}-(A+2Bh^{2}+1)\sum_{i}b^{ii}-\frac{A-2Bh^{2}}{h^{2}}b^{ii}h_{i}^{2}\\\
&\hskip 11.99998pt+4(n-1)Bh-2B\sum_{i}b_{ii}\\\ &\hskip
11.99998pt-2B\sum_{k}b^{ii}h_{k}b_{ii;k}+b^{11}(n-1-w_{11}).\end{split}$

Using (42), we get

(47) $\frac{2B\sum_{k}h_{k}\partial_{t}h_{k}}{h-\partial_{t}h}=\frac{2B|\nabla
h|^{2}}{h-\partial_{t}h}+2B\sum_{k}b^{ii}h_{k}b_{ii;k}-2B\langle\nabla
h,\nabla w\rangle.$

Now dividing (40) by $h-\partial_{t}h$ gives

$\begin{split}0&\leq\frac{b^{11}(\partial_{t}h_{11}-h_{11}+b_{11}-h+\partial_{t}h)}{h-\partial_{t}h}-\frac{A\partial_{t}h}{h(h-\partial_{t}h)}+\frac{2B\sum_{k}h_{k}\partial_{t}h_{k}}{h-\partial_{t}h}\\\
&=\frac{b^{11}(\partial_{t}h_{11}-h_{11})}{h-\partial_{t}h}+\frac{2B\sum_{k}h_{k}\partial_{t}h_{k}}{h-\partial_{t}h}-b^{11}+\frac{A}{h}-\frac{A-1}{h-\partial_{t}h},\end{split}$

which together with (46) and (47) implies that

$\begin{split}0&\leq\frac{nA}{h}-(A+2Bh^{2}+1)\sum_{i}b^{ii}-\frac{A-2Bh^{2}}{h^{2}}\sum_{i}b^{ii}h_{i}^{2}\\\
&\hskip 11.99998pt+4(n-1)Bh-2B\sum_{i}b_{ii}+b^{11}(n-2-w_{11})\\\ &\hskip
11.99998pt-2B\langle\nabla h,\nabla w\rangle-\frac{A-1-2B|\nabla
h|^{2}}{h-\partial_{t}h}.\end{split}$

Now we choose $A=n+2BC^{2}$, where $C$ is the constant in Lemma 3, we can
obtain

(48) $(A-n+3)\sum_{i}b^{ii}+2B\sum_{i}b_{ii}\leq
C_{1}(A+B)-b^{11}w_{11}-2B\langle\nabla h,\nabla w\rangle,$

where $C_{1}$ is a positive constant depending only on $n$ and the constant
$C$ in Lemma 3.

A direct calculation gives

$\displaystyle\nabla_{i}G(\nabla
h+hx)=\langle\overline{\nabla}G,\overline{\nabla}_{i}\overline{\nabla}h\rangle=\langle\overline{\nabla}G,e_{k}\rangle
b_{ik},$

and

$\displaystyle\nabla_{i}\nabla_{j}G(\nabla h+hx)$ $\displaystyle=$
$\displaystyle\overline{\nabla}^{2}G\Big{(}\overline{\nabla}_{j}(\overline{\nabla}h)\rangle,\overline{\nabla}_{i}(\overline{\nabla}h)\Big{)}+\Big{\langle}\overline{\nabla}G,\overline{\nabla}_{i}\overline{\nabla}_{j}(\overline{\nabla}h)\Big{\rangle}$
$\displaystyle=$
$\displaystyle\overline{\nabla}^{2}G\Big{(}\sum_{k}b_{ik}e_{k},\sum_{l}b_{jl}e_{l}\Big{)}-\langle\overline{\nabla}G,x\rangle
b_{ij}+\sum_{k}\langle\overline{\nabla}G,e_{k}\rangle b_{ij;k}.$

Thus, we have

$\begin{split}-2B\langle\nabla h,\nabla
w\rangle&=-2B\sum_{k}h_{k}\left(\frac{f_{k}}{f}+\frac{\sum_{l}\langle\overline{\nabla}G,e_{l}\rangle
b_{kl}}{G}\right)\\\ &\leq
C_{2}B+\sum_{k}\frac{2Bh_{k}\langle\overline{\nabla}G,e_{k}\rangle
b_{kk}}{G},\end{split}$

where $C_{2}$ is a positive constant depending only on the constants $C$ in
Lemma 3, $\left\|f\right\|_{C^{1}(\mathbb{S}^{n-1})}$ and
$\min\limits_{\mathbb{S}^{n-1}}f$. Moreover, we have by Lemma 6

$\displaystyle-w_{11}$ $\displaystyle=$
$\displaystyle\frac{f_{1}^{2}}{f^{2}}-\frac{f_{11}}{f}+\frac{[\langle\overline{\nabla}G,e_{1}\rangle
b_{11}]^{2}}{G^{2}}$
$\displaystyle-\frac{1}{G}\Big{[}\overline{\nabla}^{2}G(e_{1},e_{1})b^{2}_{11}-\langle\overline{\nabla}G,x\rangle
b_{11}+\langle\overline{\nabla}G,e_{k}\rangle b_{11,k}\Big{]}$
$\displaystyle\leq$ $\displaystyle
C_{3}(1+b_{11}+b_{11}^{2})+\frac{\langle\overline{\nabla}G,e_{k}\rangle
b_{11,k}}{G},$

where $C_{3}$ is a positive constant depending only on the constants $C$ in
Lemmas 3 and 5, $\left\|f\right\|_{C^{2}(\mathbb{S}^{n-1})}$,
$\left\|g\right\|_{C^{2}(\mathbb{S}^{n-1})}$,
$\min\limits_{\mathbb{S}^{n-1}}f$ and $\min\limits_{\mathbb{S}^{n-1}}g$.
Therefore, combining the two inequalities above, we have

(49) $\begin{split}&-b^{11}w_{11}-2B\langle\nabla h,\nabla w\rangle\\\
\leq&C_{3}(b^{11}+1+b_{11})+C_{2}B+\sum_{k}\frac{\langle\overline{\nabla}G,e_{k}\rangle}{G}(b^{11}b_{11;k}+2Bh_{k}h_{kk})\\\
=&C_{3}(b^{11}+1+b_{11})+C_{2}B+\sum_{k}\frac{\langle\overline{\nabla}G,e_{k}\rangle}{G}\cdot\frac{Ah_{k}}{h}\\\
\leq&C_{3}(b^{11}+1+b_{11})+C_{2}B+C_{4}A,\end{split}$

where we have used the equality (38), and $C_{4}$ is a positive constant
depending only on the constants $C$ in Lemma 3 and $n$. Inserting (49) into
(48), we have

$(A-n)\sum_{i}b^{ii}+2B\sum_{i}b_{ii}\leq(C_{1}+C_{4})A+(C_{1}+C_{2})B+C_{3}(b^{11}+1+b_{11}),$

which together with $A=n+2BC^{2}$ implies that

$\begin{split}&(2BC^{2}-C_{3})\sum_{i}b^{ii}+(2B-C_{3})\sum_{i}b_{ii}\\\
\leq&(C_{1}+C_{4})(n+2BC^{2})+(C_{1}+C_{2})B+C_{3}.\end{split}$

If we choose
$B=\max\left\\{\frac{C_{3}+1}{2},\frac{C_{3}+1}{2C^{2}}\right\\}$, we see that
$b_{11}(x_{0},t_{0})$ is bounded from above by a positive constant depending
only on $n$, $C_{1}$, $C_{2}$, $C_{3}$ and $C_{4}$. Using Lemmas 3,
$\widetilde{\Lambda}(x_{0},t_{0})$ is bounded from above by a positive
constant depending only on $n$, $C$, $C_{1}$, $C_{2}$, $C_{3}$ and $C_{4}$.
Thus, we prove the conclusion of this lemma, by noticing that $T^{\prime}$ can
be any number in $(0,T)$. ∎

##  5\. Existence of solutions

In this section, we will complete the proof of Theorem 3. Combining Lemma 7
and Lemma 8, we see that the principal curvatures of $M_{t}$ has uniform
positive upper and lower bounds. This together with Lemmas 3 and 5 implies
that the evolution equation (15) is uniformly parabolic on any finite time
interval. Thus, using Krylov- Safonov estimates [34] and Schauder estimates of
the parabolic equations, we can say that the smooth solution of (15) exists
for all time. And by these estimates again, a subsequence of $M_{t}$ converges
in $C^{\infty}$ to a positive, smooth, uniformly convex hypersurface
$M_{\infty}$ in $\mathbb{R}^{n}$. Now to complete the proof of Theorem 3, it
remains to check the support function of $M_{\infty}$ satisfies Eq. (4).

Let $\tilde{h}$ be the support function of $M_{\infty}$. We need to prove that
$\tilde{h}$ is a solution to the following equation

(50) $g\bigg{(}\frac{\nabla h+hx}{|\nabla h+hx|}\bigg{)}|\nabla
h+hx|^{-n}h\det(\nabla^{2}h+hI)=f\text{ on }\mathbb{S}^{n-1}.$

By Lemma 1, $J^{\prime}(t)\leq 0$ for any $t>0$. Since

$\int_{0}^{t}[-J^{\prime}(t)]\mathop{}\\!\mathrm{d}t=J(0)-J(t)\leq
C\int_{\mathbb{S}^{n-1}}fdx,$

we get

$\int_{0}^{\infty}[-J^{\prime}(t)]\mathop{}\\!\mathrm{d}t\leq
C\int_{\mathbb{S}^{n-1}}fdx.$

Thus, there exists a subsequence of times $t_{j}\to\infty$ such that

$-J^{\prime}(t_{j})\to 0\text{ as }t_{j}\to\infty.$

Thus, we obtain using (2.2)

$\int_{\mathbb{S}^{n-1}}\frac{1}{g\tilde{h}\widetilde{\mathcal{K}}\tilde{\rho}^{n}}(g\tilde{h}-f\widetilde{\mathcal{K}}\tilde{\rho}^{n})^{2}dx=0,$

where $\widetilde{\mathcal{K}}$ is the Gauss curvature of $M_{\infty}$. It
implies that

$g(u)\frac{\tilde{h}(x)}{\widetilde{\mathcal{K}}\tilde{\rho}^{n}(u)}=f(x)\quad\mbox{on}\quad\mathbb{S}^{n-1},$

which means $\tilde{h}$ is a solution to the equation (50).

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