# Star-shaped Centrosymmetric Curves under Gage’s Area-preserving Flow

Laiyuan Gao Shengliang Pan

Abstract It is proved that Gage’s area-preserving flow can evolve a
centrosymmetric star-shaped initial curve smoothly, make it convex in a finite
time and deform it into a circle as time tends to infinity.

Keywords curve shortening flow, area-preserving flow, star-shaped

Mathematics Subject Classification (2000) 35C44, 35K05, 53A04

##  1 Introduction

A regular closed embedded plane curve can be parameterized as
$X:S^{1}\rightarrow\mathbb{R}^{2},\varphi\mapsto(x(\varphi),y(\varphi))$. If
the total length of $X$ is $L$, $X$ is usually parameterized as
$X:[0,L]\rightarrow\mathbb{R}^{2},s\mapsto(x(s),y(s))$, where $s$ is the arc
length parameter. Denote by $T(s)$ the unit tangential vector at a point
$X(s)$ and by $N(s)=N_{in}(s)$ the inward unit normal vector such that every
ordered pair $(T(s),N(s))$ determines a positive orientation of the plane. The
curve $X$ is called star-shaped if there exists a point $O$ inside the region
bounded by $X$ such that

$\det(X(s),T(s))=\begin{vmatrix}x(s)&y(s)\\\
x^{\prime}(s)&y^{\prime}(s)\end{vmatrix}>0$

for every $s$, and the point $O$ is called a star center of the curve. If $X$
is a $C^{2}$ curve then its (relative) curvature is defined as

$\kappa(s):=\left\langle\frac{\partial T}{\partial
s}(s),~{}N(s)\right\rangle.$

In this paper, we will investigate the evolution behaviour of a star-shaped
centrosymmetric plane curve under Gage’s area-preserving flow (GAPF) [5], that
is to say, we will consider the following Cauchy problem

$\left\\{\begin{array}[]{l}\frac{\partial X}{\partial
t}(\varphi,t)=\left(\kappa-\frac{2\pi}{L}\right)N\ \ \ \text{in}\ \
S^{1}\times(0,\omega),\\\ X(\varphi,0)=X_{0}(\varphi)\ \ \ \ \ \ \text{on}\ \
S^{1},\end{array}\right.$ (1.1)

where
$X:S^{1}\times[0,\omega)\rightarrow\mathbb{R}^{2}((\varphi,t)\mapsto(x,y))$ is
a family of closed curves with $X_{0}$ being star-shaped and centrosymmetric
about the origin $O$ of the plane, $\kappa=\kappa(\varphi,t)$ the curvature
and $L=L(t)$ the length of $X(\cdot,t)$. In 1984, Gage ([5]) proved that the
evolving curve $X(\cdot,t)$ under this flow can be deformed into a circle if
the initial curve $X_{0}$ is convex. As Gage pointed out that there is a kind
of simple closed curves which may develop singularities in a finite time under
the flow (1.1), which is verified by Mayer’s numerical experiment (see [19]).
So, unlike Grayson’s theorem [10] for the curve shortening flow (CSF), there
is no convergence result of GAPF for generic simple closed initial curves. A
natural question is whether there is a class of non-convex initial curves
which may become convex and converge to a circle under the flow (1.1). To
guarantee such a convergence, referring to Gage’s example (see [5, 19]), we
should assume that the initial curve $X_{0}$ can not concave wildly. In this
paper, we consider $X_{0}$ to be star-shaped and centrosymmetric and obtain
the following main result:

######  Theorem 1.1.

Let $X_{0}$ be a smooth, embedded and star-shaped curve in the plane. If
$X_{0}$ is centrosymmetric then Gage’s area-preserving flow (1.1) with this
initial curve exists globally and makes the evolving curve convex in a finite
time and deforms it into a circle as time tends to infinity.

A flow is called global if the evolving curve is smooth for all
$t\in[0,+\infty)$. For an initial star-shaped curve in the plane, it is still
unknown whether Gage’s area-preserving flow exists globally or not (see Lemma
2.5 and Remark 3.4). If $X_{0}$ is only star-shaped, it seems that the flow
(1.1) may not preserve the star-shapedness of the evolving curve during the
evolution process. It causes essential difficulties to understand the
asymptotic behavior of $X(\cdot,t)$. The extra symmetric property of $X_{0}$
is inspired by Prof. Michael Gage [21] (also see the early work [6, 8] by Gage
and Li).

The proof of Theorem 1.1 is divided into two parts. In the first part, i.e.
the global existence of the flow, the star-shapedness of the evolving curve
plays an essential role, see Lemma 3.8 and Corollary 3.9. To show that
$X(\cdot,t)$ is star-shaped, $X_{0}$ needs to be of centrosymmetry. If $X_{0}$
is centrosymmetric with respect to the point $O$, the key idea in this part is
to show that the evolving curve never touches the point $O$ via a comparison
to the evolution behavior of the famous CSF, see Lemmas 3.1-3.7 for the
detaails. In the second part, i.e. the convergence of the evolving curve, some
ideas in Grayson’s papers [10, 11] are adopted, see Lemmas 4.2-4.3.

Once $X(\cdot,t)$ is proved to be star-shaped, the polar angle $\theta$ can be
used as a parameter of the evolving curve. In order to make $\theta$
independent of time, one can add a tangent component to the origin flow to get
a new one:

$\left\\{\begin{array}[]{l}\frac{\partial X}{\partial t}(\varphi,t)=\alpha
T+\left(\kappa-\frac{2\pi}{L}\right)N\ \ \ \text{in}\ \
S^{1}\times(0,\omega),\\\ X(\varphi,0)=X_{0}(\varphi)\ \ \ \ \ \ \text{on}\ \
S^{1}.\end{array}\right.$ (1.2)

The tangent component $\alpha(\varphi,t)T(\varphi,t)$ (not influence the shape
of $X$, see Proposition 1.1 on page 6 of [2]) will be determined in the next
section.

GAPF [5] has also been considered by Wang, Wo and Yang [23] if the initial
curve is closed and locally convex. They have studied some asymptotic
behaviours of the evolving curve, including the convergence for global flows
and some blow-up properties. For higher dimensional cases, one can refer to
Huisken’s volume-preserving flow of convex hypersurfaces [14] and its
generalization by Kim and Kwon [16] to the case of star-shaped hypersurfaces
with a so called $\rho$-reflection property.

This paper is organized as follows. In Section 2, some basic properties of the
flow (1.1) are obtained, including the short time existence and a property of
$X(\cdot,t)$ which implies its star-shapeness. In Section 3, it is proved that
the flow (1.1) exists on the time interval $[0,+\infty)$. And in the final
section, the proof of Theorem 1.1 is completed.

##  2 Preparation

Given a curve $X$ in the plane, its “support function” is defined by111$p$ is
usually called the support function of $X$ in convex geometry, see, for
example, [12, 13, 20].

$p=-\langle X,N\rangle.$

If the curve is expressed as $X(s)=(x(s),y(s))$, $s\in[0,L]$, where $L$ is the
length of $X$, then its unit tangent and normal vector fields can be written
as

$T(s)=(\dot{x}(s),\dot{y}(s)),\ \ \ \ \ \ \ N(s)=(-\dot{y}(s),\dot{x}(s)),$

where “$\cdot$” stands for derivative with respect to the arc length parameter
$s$. Since

$p(s)=x(s)\dot{y}(s)-\dot{x}(s)y(s)=\det(X(s),T(s)),$

$X$ is star-shaped with respect to the origin $O$ if and only if $p(s)>0$ for
all $s$.

Using the polar coordinate system $(r,\theta)$ for the plane, a smooth and
closed curve can be expressed as

$X(s)=r(\theta)P(\theta),$

where $\theta=\theta(s)$ and $P(\theta)=(\cos\theta,\sin\theta)$. Let
$Q(\theta)=(-\sin\theta,\cos\theta)$, then the unit tangential vector of $X$
can be given by

$T=\frac{dr}{ds}P+r\frac{d\theta}{ds}Q,$

and furthermore

$\det\langle X(s),T(s)\rangle=r^{2}(s)\dot{\theta}(s).$ (2.1)

If $X$ is closed and star-shaped then one can choose the origin $O$ of our
frame so that $r(s)>0$ and $\det\langle X,T\rangle>0$. The equation (2.1)
implies that one can use the polar angle $\theta$ to parameterize a star-
shaped plane curve.

Now let us deal with the flow (1.1) with a star-shaped initial value $X_{0}$.
We shall first derive some evolution equations and determine the tangential
component $\alpha T$ to make the polar angle $\theta$ independent of time $t$.
Then (1.1) can be reduced to a Cauchy problem of a single equation for the
radial function $r=r(\theta,t)$. After that, some basic properties of the flow
(1.1) will be explored in this section.

Let $g:=\sqrt{\langle\frac{\partial X}{\partial\varphi},\frac{\partial
X}{\partial\varphi}\rangle}$ be the metric of the evolving curve. Set
$\beta=\kappa-\frac{2\pi}{L}$. Under the flow (1.2), $g$ evolves according to

$\displaystyle\frac{\partial g}{\partial
t}=\frac{1}{g}\left\langle\frac{\partial}{\partial t}\frac{\partial
X}{\partial\varphi},\frac{\partial
X}{\partial\varphi}\right\rangle=g\left\langle\frac{\partial}{\partial
s}(\alpha T+\beta N),T\right\rangle=\left(\frac{\partial\alpha}{\partial
s}-\beta\kappa\right)g.$

The interchange of the operators $\partial/\partial s$ and $\partial/\partial
t$ is given by

$\displaystyle\frac{\partial}{\partial t}\frac{\partial}{\partial
s}=\frac{\partial}{\partial
t}\left(\frac{1}{g}\frac{\partial}{\partial\varphi}\right)=\frac{\partial}{\partial
s}\frac{\partial}{\partial t}-\left(\frac{\partial\alpha}{\partial
s}-\beta\kappa\right)\frac{\partial}{\partial s}.$

$T$ and $N$ evolve according to

$\displaystyle\frac{\partial T}{\partial t}=\frac{\partial}{\partial
t}\frac{\partial X}{\partial s}=\frac{\partial}{\partial s}\frac{\partial
X}{\partial t}-\left(\frac{\partial\alpha}{\partial
s}-\beta\kappa\right)T=\left(\alpha\kappa+\frac{\partial\beta}{\partial
s}\right)N,$ $\displaystyle\frac{\partial N}{\partial
t}=\left\langle\frac{\partial N}{\partial t},T\right\rangle
T+\left\langle\frac{\partial N}{\partial t},N\right\rangle
N=-\left(\alpha\kappa+\frac{\partial\beta}{\partial s}\right)T.$

If there is a family of star-shaped curves evolving under the flow (1.2) then
we can express the evolving curve as

$\displaystyle X(\theta,t)=r(\theta,t)P(\theta).$ (2.2)

Noticing that $\frac{\partial X}{\partial\theta}=\frac{\partial
r}{\partial\theta}P+rQ$, we obtain

$\displaystyle g=\left\|\frac{\partial
X}{\partial\theta}\right\|=\left(r^{2}+\left(\frac{\partial
r}{\partial\theta}\right)^{2}\right)^{1/2},~{}~{}T=\frac{\partial r}{\partial
s}P+\frac{r}{g}Q,~{}~{}N=-\frac{r}{g}P+\frac{\partial r}{\partial s}Q.$ (2.3)

Differentiating the right hand side of (2.2) and using (1.2) and (2.3), one
gets

$\displaystyle\frac{\partial r}{\partial t}P+r\frac{\partial\theta}{\partial
t}Q=\alpha T+\beta N=\left(\alpha\frac{\partial r}{\partial
s}-\frac{r\beta}{g}\right)P+\left(\frac{\alpha r}{g}+\beta\frac{\partial
r}{\partial s}\right)Q.$

Comparing the coefficients of both sides can yield the following evolution
equations:

$\displaystyle\frac{\partial r}{\partial t}=\alpha\frac{\partial r}{\partial
s}-\frac{r\beta}{g},\ \ \ \frac{\partial\theta}{\partial
t}=\frac{\alpha}{g}+\frac{\beta}{r}\frac{\partial r}{\partial s}.$ (2.4)

From now on, we choose

$\alpha=-\frac{\beta}{r}\frac{\partial r}{\partial
s}g=-\frac{\beta}{r}\frac{\partial r}{\partial\theta}$

so that $\frac{\partial\theta}{\partial t}\equiv 0$, the polar angle $\theta$
is independent of the time $t$. Since the curvature of the evolving curve is

$\kappa=\frac{1}{g^{3}}\left(-r\frac{\partial^{2}r}{\partial\theta^{2}}+2\left(\frac{\partial
r}{\partial\theta}\right)^{2}+r^{2}\right),$

one can immediately obtain the evolution equation of $r$,

$\displaystyle\frac{\partial r}{\partial
t}=\frac{1}{g^{2}}\frac{\partial^{2}r}{\partial\theta^{2}}-\frac{2}{rg^{2}}\left(\frac{\partial
r}{\partial\theta}\right)^{2}-\frac{r}{g^{2}}+\frac{2\pi g}{rL}.$ (2.5)

Now, if $r=r(\theta,t)>0$ is defined on $[0,2\pi]\times[0,\omega)$ and
satisfies the equation (2.5), then a family of curves
$\\{X=rP|t\in[0,\omega)\\}$ satisfies the flow (1.2). So we can reduce the
flow (1.2) to the equation (2.5) with initial value $r_{0}(\theta)>0$:

######  Lemma 2.1.

Suppose $X_{0}$ is star-shaped. The flow (1.2) is equivalent to the quasi-
linear parabolic equation (2.5) with a positive initial value $r_{0}$ in some
interval $[0,\omega)$.

The length of the curve can be calculated according to

$L(t)=\int_{0}^{2\pi}g(\theta,t)d\theta=\int_{0}^{2\pi}\sqrt{r^{2}+\left(\frac{\partial
r}{\partial\theta}\right)^{2}}d\theta.$

Let us define an operator $F$ from the space
$C^{2,\alpha}([0,2\pi]\times[0,\omega))$ to
$C^{\beta}([0,2\pi]\times[0,\omega))$, for $0<\beta<\alpha\leq 1$ according to

$F(r)=\frac{\partial r}{\partial
t}-\frac{1}{g^{2}}\frac{\partial^{2}r}{\partial\theta^{2}}+\frac{2}{rg^{2}}\left(\frac{\partial
r}{\partial\theta}\right)^{2}+\frac{r}{g^{2}}-\frac{2\pi g}{rL}.$

Since the Frechet derivative of $F$ at some point $r_{0}>0$ is

$DF(r_{0})f=\frac{\partial f}{\partial t}-\frac{1}{r_{0}^{2}+(\frac{\partial
r_{0}}{\partial\theta})^{2}}\frac{\partial^{2}f}{\partial\theta^{2}}+\text{lower
linear terms of}\ f,$

the equation (2.5) is uniformly parabolic near its initial value $r_{0}$. It
follows from the implicit function theorem of Banach spaces that the Cauchy
problem (2.5) has a unique solution in some small time interval (See Section
1.2 in [2]). Using Lemma 2.1 can give us the short time existence.

######  Lemma 2.2.

The flow (1.1) has a unique smooth solution in some time interval
$[0,\omega)$, where $\omega>0$.

Next, we shall derive some basic properties of the flow (1.1).

######  Lemma 2.3.

Under the flow (1.1), the area $A$ of the evolving curve is constant, that is,
$A(t)\equiv A_{0}$; and the length $L$ satisfies $\sqrt{4\pi A_{0}}\leq
L(t)\leq L_{0}$.

###### Proof.

This result is a direct consequence of the equations (1.18)-(1.19) in [2] and
the classical isoperimetric inequality. ∎

Under the flow (1.2), the “support function” is

$p=-\langle X,N\rangle=-\left\langle
rP,-\frac{r}{g}P+\frac{1}{g}\frac{\partial
r}{\partial\theta}Q\right\rangle=\frac{r^{2}}{g},$

and thus a closed curve is star-shaped if and only if $r>0$ and
$\left|\frac{\partial r}{\partial\theta}\right|$ is bounded everywhere. Since
$r$ and $\left|\frac{\partial r}{\partial\theta}\right|$ satisfy parabolic
equations, one can apply the comparison principle to bound these two
functions. For a continuous function $f=f(\theta,t)$, set

$f_{\min}(t)=\min\\{f(\theta,t)|\theta\in[0,2\pi]\\},~{}~{}f_{\max}(t)=\max\\{f(\theta,t)|\theta\in[0,2\pi]\\}.$

######  Lemma 2.4.

Given a star-shaped curve $X_{0}$, we choose a point $O$ in the plane such
that $r_{0}(\theta)>0$. If $r(\theta,t)\geq c>0$ for
$(\theta,t)\in[0,2\pi]\times[0,\omega)$ under the flow (1.2) then

$\displaystyle r(\theta,t)\leq\frac{L_{0}}{2},$ (2.6)
$\displaystyle\left|\frac{\partial r}{\partial\theta}(\theta,t)\right|\leq
C_{1},$ (2.7)

where $C_{1}=\max\left\\{\max\limits_{\theta}\left|\frac{\partial
r}{\partial\theta}(\theta,0)\right|,\frac{3L_{0}}{\pi}\right\\}$ is a constant
depending on the initial curve $X_{0}$.

###### Proof.

Fix a $t_{0}\in[0,\omega)$. Let $t\in[0,t_{0})$, if $r(\cdot,t)$ attains its
maximum value $r_{\max}(t)$ at $(\theta_{*},t)$, then

$\frac{\partial^{2}r}{\partial\theta^{2}}(\theta_{*},t)\leq
0,~{}~{}~{}\frac{\partial
r}{\partial\theta}(\theta_{*},t)=0,~{}~{}~{}g(\theta_{*},t)=r(\theta_{*},t).$

By (2.5),

$\displaystyle\frac{\partial r}{\partial
t}(\theta_{*},t)\leq\frac{2\pi}{L(t)}\leq\sqrt{\frac{\pi}{A_{0}}}.$

So the maximum principle implies that if $t\in[0,t_{0})$ then

$\displaystyle r(\theta,t)\leq r_{\max}(0)+\sqrt{\frac{\pi}{A_{0}}}t_{0}.$
(2.8)

Differentiating the equation (2.5), one gets

$\displaystyle\frac{\partial^{2}r}{\partial t\partial\theta}$ $\displaystyle=$
$\displaystyle\frac{1}{g^{2}}\frac{\partial^{3}r}{\partial\theta^{3}}-\frac{2}{g^{4}}\frac{\partial
r}{\partial\theta}\left(\frac{\partial^{2}r}{\partial\theta^{2}}\right)^{2}-\frac{4}{rg^{2}}\frac{\partial
r}{\partial\theta}\frac{\partial^{2}r}{\partial\theta^{2}}+\frac{2}{r^{2}g^{2}}\left(\frac{\partial
r}{\partial\theta}\right)^{3}$
$\displaystyle+\frac{4}{g^{4}}\left(\frac{\partial
r}{\partial\theta}\right)^{3}+\frac{4}{rg^{4}}\left(\frac{\partial
r}{\partial\theta}\right)^{3}\frac{\partial^{2}r}{\partial\theta^{2}}-\frac{1}{g^{2}}\frac{\partial
r}{\partial\theta}+\frac{2r^{2}}{g^{4}}\frac{\partial r}{\partial\theta}$
$\displaystyle+\frac{2\pi}{Lg}\frac{\partial
r}{\partial\theta}+\frac{2\pi}{rLg}\frac{\partial
r}{\partial\theta}\frac{\partial^{2}r}{\partial\theta^{2}}-\frac{2\pi
g}{r^{2}L}\frac{\partial r}{\partial\theta}.$

Let $w=(\frac{\partial r}{\partial\theta})^{2}$, it evolves according to

$\displaystyle\frac{1}{2}\frac{\partial w}{\partial t}$ $\displaystyle=$
$\displaystyle\frac{1}{2}\frac{1}{g^{2}}\frac{\partial^{2}w}{\partial\theta^{2}}-\frac{1}{g^{2}}\left(\frac{\partial^{2}r}{\partial\theta^{2}}\right)^{2}-\frac{2w}{g^{4}}\left(\frac{\partial^{2}r}{\partial\theta^{2}}\right)^{2}-\frac{2}{rg^{2}}\frac{\partial
r}{\partial\theta}\frac{\partial w}{\partial\theta}$
$\displaystyle+\frac{2w^{2}}{r^{2}g^{2}}+\frac{4w^{2}}{g^{4}}+\frac{2w}{rg^{4}}\frac{\partial
r}{\partial\theta}\frac{\partial
w}{\partial\theta}-\frac{w}{g^{2}}+\frac{2r^{2}w}{g^{4}}$
$\displaystyle+\frac{2\pi w}{Lg}+\frac{\pi}{rLg}\frac{\partial
r}{\partial\theta}\frac{\partial w}{\partial\theta}-\frac{2\pi gw}{r^{2}L}.$

If $w$ attains its maximum value $w_{\max}(t_{0})$ at $(\theta_{*},t_{0})$,
then $\frac{\partial
w}{\partial\theta}(\theta_{*},t_{0})=0,\frac{\partial^{2}w}{\partial\theta^{2}}(\theta_{*},t_{0})\leq
0,$ and thus at the point $(\theta_{*},t_{0})$, one obtains

$\displaystyle\frac{1}{2}\frac{\partial w}{\partial t}$ $\displaystyle\leq$
$\displaystyle\left(\frac{2w^{2}}{r^{2}g^{2}}+\frac{4w^{2}}{g^{4}}\right)+\left(-\frac{w}{g^{2}}+\frac{2r^{2}w}{g^{4}}\right)+\left(\frac{2\pi
w}{Lg}-\frac{2\pi gw}{r^{2}L}\right)$ $\displaystyle=$
$\displaystyle\frac{2w^{2}(3r^{2}+w)}{r^{2}g^{4}}+\frac{w}{g^{4}}(r^{2}-w)-\frac{2\pi
w^{2}}{r^{2}Lg}$ $\displaystyle=$ $\displaystyle\frac{2w^{2}(3Lr^{2}+wL)-2\pi
w^{2}(r^{2}+w)g}{r^{2}Lg^{4}}+\frac{w}{g^{4}}(r^{2}-w)$ $\displaystyle=$
$\displaystyle\frac{2w^{2}r^{2}(3L-\pi g)+2w^{3}(L-\pi
g)}{r^{2}Lg^{4}}+\frac{w}{g^{4}}(r^{2}-w).$

If
$w\geq\max\\{\left(r_{\max}(t)\right)^{2},\left(\frac{3L_{0}}{\pi}\right)^{2}\\}$
then $3L-\pi g\leq 0$ and $r^{2}-w\leq 0$. One gets $\frac{\partial
w}{\partial t}(\theta_{*},t_{0})\leq 0$. The maximum principle tells us

$\displaystyle\left|\frac{\partial
r}{\partial\theta}\right|\leq\max\left\\{\max_{\theta}\left|\frac{\partial
r}{\partial\theta}(\theta,0)\right|,~{}r_{\max}(t),~{}\frac{3L_{0}}{\pi}\right\\}.$
(2.9)

Combining (2.8) and (2.9), the support function $p=\frac{r^{2}}{g}$ is
positive on the time interval $[0,t_{0}]$. The evolving curve $X(\cdot,t)$ is
star-shaped with respect to $O$ for $t\in[0,t_{0}]$. Since $r$ is the distance
from $O$ to $X(\theta,t)$, one obtains

$\displaystyle r(\theta,t)\leq\frac{L(t)}{2}\leq\frac{L_{0}}{2}$

which gives us (2.6) and enables us to revise the estimate (2.9) as

$\displaystyle\left|\frac{\partial
r}{\partial\theta}\right|\leq\max\left\\{\max_{\theta}\left|\frac{\partial
r}{\partial\theta}(\theta,0)\right|,~{}\frac{L_{0}}{2},~{}\frac{3L_{0}}{\pi}\right\\}=\max\left\\{\max_{\theta}\left|\frac{\partial
r}{\partial\theta}(\theta,0)\right|,~{}\frac{3L_{0}}{\pi}\right\\}.$ (2.10)

So we have done. ∎

It follows from Lemma 2.4 that the support function of the evolving curve is
positive everywhere once a positive lower bound of $r$ is given. So the flow
(1.1) preserves star-shapedness of the evolving curve under the condition of
Lemma 2.4. Now one can conclude that:

######  Lemma 2.5.

If the flow (1.1) does not blow up in the time interval $[0,\omega)$ and
$r_{\min}(t)>0$ for $t\in[0,\omega)$ then the point $O$ is the star center of
every evolving curve $X(\cdot,t)$.

##  3 Global existence

In this section, it is shown that Gage’s area-preserving flow (1.1) exists in
the time interval $[0,+\infty)$ if the initial smooth curve $X_{0}$ is
centrosymmetric and star-shaped with respect to the origin $O$.

###  3.1 Star-shaped curves under the CSF

The popular curve shortening flow with a smooth, closed and embedded initial
curve $X_{0}$ is defined by

$\left\\{\begin{array}[]{l}\frac{\partial Y}{\partial
t}(\varphi,t)=\widetilde{\kappa}(\varphi,t)\widetilde{N}(\varphi,t)\ \ \
\text{in}\ \ S^{1}\times(0,\omega),\\\ Y(\varphi,0)=X_{0}(\varphi)\ \ \ \ \ \
\text{on}\ \ S^{1},\end{array}\right.$ (3.1)

where $\widetilde{\kappa}(\varphi,t)$ is the relative curvature with respect
to the Frenet frame $\\{\widetilde{T},\widetilde{N}\\}$. Grayson theorem [10]
asserts that the evolving curve $Y(\cdot,t)$ is smooth, preserves its
embeddedness and becomes convex on the time interval
$\left[0,\frac{A_{0}}{2\pi}\right)$, where $A_{0}$ is the area bounded by
$X_{0}$. Gage-Hamilton theorem [3, 4, 7] says that a closed convex curve
$X_{0}$ evolving under the CSF (3.1) becomes asymptotically circular as
$t\rightarrow\frac{A_{0}}{2\pi}$.

If $X_{0}$ is star-shaped with respect to the origin $O$, it follows from the
continuity of the evolving curve that there exists $t_{0}>0$ such that
$Y(\cdot,t)$ is star-shaped with respect to $O$ for $t\in[0,t_{0})$. So one
can add a proper tangent component to the flow (3.1)

$\left\\{\begin{array}[]{l}\frac{\partial Y}{\partial
t}(\varphi,t)=\widetilde{\alpha}\widetilde{T}+\widetilde{\kappa}\widetilde{N}\
\ \ \text{in}\ \ S^{1}\times(0,\omega),\\\ Y(\varphi,0)=X_{0}(\varphi)\ \ \ \
\ \ \text{on}\ \ S^{1}\end{array}\right.$ (3.2)

to make the polar angle $\theta$ of $Y(\cdot,t)$ independent of time. Now let
us parameterize $Y(\cdot,t)$ by $\theta$ and set
$Y(\theta,t)=\rho(\theta,t)P(\theta)$ where
$P(\theta)=(\cos\theta,\sin\theta)$. As is well known that the solution to the
flow (3.1) differs from that of the flow (3.2) by a reparametrization and a
Euclidean translation. The evolution equation of the radial function
$\rho(\theta,t)$ is

$\displaystyle\frac{\partial\rho}{\partial
t}=\frac{1}{(g_{\rho})^{2}}\frac{\partial^{2}\rho}{\partial\theta^{2}}-\frac{2}{\rho(g_{\rho})^{2}}\left(\frac{\partial\rho}{\partial\theta}\right)^{2}-\frac{\rho}{(g_{\rho})^{2}},$
(3.3)

where
$g_{\rho}=\sqrt{\rho^{2}+\left(\frac{\partial\rho}{\partial\theta}\right)^{2}}$
is the metric of $Y(\theta,t)$.

######  Lemma 3.1.

Suppose the initial smooth curve $X_{0}$ is star-shaped and centrosymmetric
with respect to $O$. Under the CSF (3.2), if the evolving curve $Y(\cdot,t)$
is star-shaped with respect to $O$ then it is centrosymmetric with respect to
$O$.

###### Proof.

For $t_{*}\in\left[0,\frac{A_{0}}{2\pi}\right)$, the Gage-Hamilton-Grayson
theorem tells us that there exist constants $M_{i}(t_{*})>0$ such that

$\displaystyle\left|\frac{\partial^{i}\rho}{\partial\theta^{i}}(\theta,t)\right|\leq
M_{i}(t_{*}),\ \ \ \ (\theta,t)\in[0,2\pi]\times[0,t_{*}],\ \ i=1,2,\cdots.$
(3.4)

Define $\widetilde{\rho}(\theta,t)=\rho(\theta+\pi,t)$ and
$\varphi_{1}(\theta,t)=\widetilde{\rho}(\theta,t)-\rho(\theta,t)$. Since
$X_{0}$ is centrosymmetric, $\varphi_{1}(\theta,0)\equiv 0$. By the equation
(3.3), $\varphi_{1}(\theta,t)$ evolves according to

$\displaystyle\frac{\partial\varphi_{1}}{\partial t}$ $\displaystyle=$
$\displaystyle\frac{1}{(g_{\widetilde{\rho}})^{2}}\frac{\partial^{2}\varphi_{1}}{\partial\theta^{2}}+\frac{1}{(g_{\widetilde{\rho}})^{2}}\frac{\partial^{2}\rho}{\partial\theta^{2}}-\frac{1}{(g_{\rho})^{2}}\frac{\partial^{2}\rho}{\partial\theta^{2}}+\frac{2}{\rho(g_{\rho})^{2}}\left[\left(\frac{\partial\rho}{\partial\theta}\right)^{2}-\left(\frac{\partial\widetilde{\rho}}{\partial\theta}\right)^{2}\right]$
(3.5)
$\displaystyle+\frac{2}{\rho(g_{\rho})^{2}}\left(\frac{\partial\widetilde{\rho}}{\partial\theta}\right)^{2}-\frac{2}{\widetilde{\rho}(g_{\widetilde{\rho}})^{2}}\left(\frac{\partial\widetilde{\rho}}{\partial\theta}\right)^{2}+\frac{\rho-\widetilde{\rho}}{(g_{\rho})^{2}}+\frac{\widetilde{\rho}}{(g_{\rho})^{2}}-\frac{\widetilde{\rho}}{(g_{\widetilde{\rho}})^{2}}$
$\displaystyle=$
$\displaystyle\frac{1}{(g_{\widetilde{\rho}})^{2}}\frac{\partial^{2}\varphi_{1}}{\partial\theta^{2}}-\frac{1}{(g_{\rho})^{2}(g_{\widetilde{\rho}})^{2}}\frac{\partial^{2}\rho}{\partial\theta^{2}}\left(\frac{\partial\rho}{\partial\theta}+\frac{\partial\widetilde{\rho}}{\partial\theta}\right)\frac{\partial\varphi_{1}}{\partial\theta}-\frac{2}{\rho(g_{\rho})^{2}}\left(\frac{\partial\rho}{\partial\theta}+\frac{\partial\widetilde{\rho}}{\partial\theta}\right)\frac{\partial\varphi_{1}}{\partial\theta}$
$\displaystyle+\frac{2}{\rho(g_{\rho})^{2}(g_{\widetilde{\rho}})^{2}}\left(\frac{\partial\widetilde{\rho}}{\partial\theta}\right)^{2}\left(\frac{\partial\rho}{\partial\theta}+\frac{\partial\widetilde{\rho}}{\partial\theta}\right)\frac{\partial\varphi_{1}}{\partial\theta}+\frac{\widetilde{\rho}}{(g_{\rho})^{2}(g_{\widetilde{\rho}})^{2}}\left(\frac{\partial\rho}{\partial\theta}+\frac{\partial\widetilde{\rho}}{\partial\theta}\right)\frac{\partial\varphi_{1}}{\partial\theta}$
$\displaystyle-\frac{\widetilde{\rho}+\rho}{(g_{\rho})^{2}(g_{\widetilde{\rho}})^{2}}\frac{\partial^{2}\rho}{\partial\theta^{2}}\varphi_{1}+\frac{2(\widetilde{\rho}+\rho)}{\rho(g_{\rho})^{2}(g_{\widetilde{\rho}})^{2}}\left(\frac{\partial\widetilde{\rho}}{\partial\theta}\right)^{2}\varphi_{1}+\frac{2}{\rho\widetilde{\rho}(g_{\widetilde{\rho}})^{2}}\left(\frac{\partial\widetilde{\rho}}{\partial\theta}\right)^{2}\varphi_{1}$
$\displaystyle-\frac{\varphi_{1}}{(g_{\rho})^{2}}+\frac{\widetilde{\rho}(\widetilde{\rho}+\rho)}{(g_{\rho})^{2}(g_{\widetilde{\rho}})^{2}}\varphi_{1}.$

The equation (3.5) is linear with smooth coefficients (see (3.4)) and zero
initial value. By the uniqueness of the solution to linear parabolic
equations, one has

$\displaystyle\varphi_{1}(\theta,t)\equiv 0.$ (3.6)

So the evolving curve $Y(\cdot,t)$ is centrosymmetric with respect to $O$ for
$t\in[0,t_{*}]$. The proof is completed by the arbitrary choice of $t_{*}$. ∎

######  Lemma 3.2.

Let the initial smooth curve $X_{0}$ be star-shaped and centrosymmetric with
respect to $O$, then under the CSF (3.2), the evolving curve $Y(\cdot,t)$ is
star-shaped.

###### Proof.

Suppose that there is a time $t_{*}\in[0,\frac{A_{0}}{2\pi})$ such that
$Y(\cdot,t)$ is star-shaped with respect to $O$ for $t\in[0,t_{*})$ but
$Y(\cdot,t_{*})$ is not so.One can claim that

$\displaystyle\rho(\theta,t)>0,\ \ \ \ (\theta,t)\in[0,2\pi]\times[0,t_{*}),$
(3.7)

and

$\displaystyle\rho_{\min}(t_{*}):=\min\\{\rho(\theta,t_{*})|\theta\in[0,2\pi]\\}=0.$
(3.8)

In fact, $Y(\cdot,t)$ is star-shaped with respect to $O$ for $t\in[0,t_{*})$
so (3.7) holds. If (3.8) does not hold then, for some $\delta>0$, one obtains

$\displaystyle\rho_{\min}(t_{*})\geq\delta.$ (3.9)

By the evolution equation (3.3), the quantity
$v=\frac{1}{2}\left(\frac{\partial\rho}{\partial\theta}\right)^{2}$ evolves
according to

$\displaystyle\frac{\partial v}{\partial t}$ $\displaystyle=$
$\displaystyle\frac{1}{(g_{\rho})^{2}}\frac{\partial^{2}v}{\partial\theta^{2}}-\frac{1}{(g_{\rho})^{2}}\left(\frac{\partial^{2}\rho}{\partial\theta^{2}}\right)^{2}-\frac{4v}{(g_{\rho})^{4}}\left(\frac{\partial^{2}\rho}{\partial\theta^{2}}\right)^{2}-\frac{4}{\rho(g_{\rho})^{2}}\frac{\partial\rho}{\partial\theta}\frac{\partial
v}{\partial\theta}$
$\displaystyle+\frac{8v^{2}}{\rho^{2}(g_{\rho})^{2}}+\frac{16v^{2}}{(g_{\rho})^{4}}+\frac{8v}{\rho(g_{\rho})^{4}}\frac{\partial\rho}{\partial\theta}\frac{\partial
v}{\partial\theta}-\frac{2v}{(g_{\rho})^{2}}+\frac{4\rho^{2}v}{(g_{\rho})^{4}}.$

At the point $(\theta_{*},t)$ where $v(\theta,t)$ attains $v_{\max}(t)$, one
gets

$\displaystyle\frac{\partial^{2}v}{\partial\theta^{2}}(\theta_{*},t)\leq
0,~{}~{}~{}~{}~{}~{}\frac{\partial v}{\partial\theta}(\theta_{*},t)=0,$

and thus

$\displaystyle\frac{\partial v}{\partial t}(\theta_{*},t)$ $\displaystyle\leq$
$\displaystyle\frac{8v^{2}}{\rho^{2}(g_{\rho})^{2}}(\theta_{*},t)+\frac{16v^{2}}{(g_{\rho})^{4}}(\theta_{*},t)-\frac{2v}{(g_{\rho})^{2}}(\theta_{*},t)+\frac{4\rho^{2}v}{(g_{\rho})^{4}}(\theta_{*},t)$
$\displaystyle\leq$
$\displaystyle\frac{4v}{\rho^{2}}(\theta_{*},t)+4+0+\frac{1}{2}\leq\frac{4v}{\delta^{2}}+\frac{9}{2},$

where $t\in[0,t_{*})$. It follows from the maximum principle that

$\displaystyle
v_{\max}(t)\leq\left(v_{\max}(0)+\frac{9}{8}\delta^{2}\right)e^{\frac{4}{\delta^{2}}t_{*}},$

i.e.,

$\displaystyle\left|\frac{\partial\rho}{\partial\theta}\right|_{\max}(t)\leq\sqrt{\left(2v_{\max}(0)+\frac{9}{4}\delta^{2}\right)e^{\frac{4}{\delta^{2}}t_{*}}}.$
(3.10)

By (3.9) and (3.10), the support function of
$Y(\cdot,t)\left(\mbox{i.e.},p_{Y}(\theta,t_{*})=\frac{\rho^{2}(\theta,t_{*})}{g_{\rho}(\theta,t_{*})}\right)$
has a positive lower bound. So $Y(\cdot,t)$ is star-shaped with respect to $O$
for $t\in[0,t_{*}+\varepsilon)$. This is a contradiction to the assumption
that $Y(\cdot,t_{*})$ is not star-shaped with respect to $O$. Therefore, the
claim (3.8) holds and there exists a $\theta_{*}\in[0,2\pi]$ such that

$\displaystyle\rho(\theta_{*},t_{*})=0.$ (3.11)

By the choice of $t_{*}\in(0,\frac{A_{0}}{2\pi})$, the evolving curve
$Y(\cdot,t_{*})$ does not blow up. By Lemma 3.1, $Y(\cdot,t)$ is
centrosymmetric with respect to $O$. So

$\displaystyle\rho(\theta_{*}+\pi,t_{*})=0.$ (3.12)

The equations (3.11) and (3.12) contradict to Huisken’s monotonic formula [15]
or the fact that the evolving curve keeps embedded before it shrinks to a
point (see Corollary 3.2.4 of [7]). Therefore, the equation (3.8) can not
hold. For every $t\in\left[0,\frac{A_{0}}{2\pi}\right)$, the evolving curve
$Y(\cdot,t)$ is star-shaped with respect to $O$. ∎

As a direct corollary of Lemma 3.1 and Lemma 3.2, we have

######  Corollary 3.3.

Suppose the initial smooth curve $X_{0}$ is star-shaped and centrosymmetric
with respect to $O$. Under the CSF (3.2), the evolving curve $Y(\cdot,t)$
shrinks to the point $O$ as $t\rightarrow\frac{A_{0}}{2\pi}$.

In 2010, Mantegazza in his note [18] showed that a star-shaped (with respect
to $O$) initial curve remains so under the CSF till the point $O$ is contained
in the open region bounded by the evolving curve.

######  Remark 3.4.

There exist some smooth, closed, star-shaped but not embedded curves which do
not always preserve the star-shapedness of evolving curves under the CSF.
Figure 1 presents such a curve $X_{0}$ which is of positive curvature
everywhere and is star-shaped with respect to $O$, but the two small loops
shrink so that they do not intersect each other after some seconds under the
CSF. This makes the evolving curve no longer star shaped. This example also
indicates that the embeddedness of the initial curve in Theorem 1.1 is so
important that it can not be omitted.

Figure 1: An Immersed Star-shaped Curve $X_{0}$.

###  3.2 Star-shapedness of the evolving curve under the flow (1.1)

It is proved in this subsection that Gage’s area-preserving flow (1.1)
preserves the star-shapedness of the evolving curve if the initial smooth
curve $X_{0}$ is both centrosymmetric and star-shaped with respect to the
origin $O$.

######  Lemma 3.5.

Suppose the initial smooth curve $X_{0}$ is star-shaped and centrosymmetric
with respect to $O$ and Gage’s area-preserving flow (1.1) exists on the time
interval $[0,\omega)$. If the evolving curve $X(\cdot,t)$ under (1.1) is star-
shaped with respect to $O$ for all $t\in[0,\omega)$ then it is centrosymmetric
with respect to $O$.

###### Proof.

For every $t_{*}\in[0,\omega)$, the radial function $r(\theta,t)$ evolves
according to the equation (2.5) under Gage’s area-preserving flow (1.1). By
assumption, $X(\cdot,0)$ is star-shaped with respect to $O$, there exists a
constant $c=c(t_{*})>0$ such that for all
$(\theta,t)\in[0,2\pi]\times[0,t_{*})$,

$\displaystyle r(\theta,t)\geq c(t_{*}).$ (3.13)

Lemma 2.4 tells us that (2.6) and (2.7) hold for $t\in[0,t_{*}]$. By the
classical theory of parabolic equations [17], there exist constants
$C_{i}(t_{*})>0$ such that

$\displaystyle\left|\frac{\partial^{i}r}{\partial\theta^{i}}(\theta,t)\right|\leq
C_{i}(t_{*}),\ \ \ \ (\theta,t)\in[0,2\pi]\times[0,t_{*}],\ \ \ i=1,2,\cdots.$
(3.14)

Set $\widetilde{r}(\theta,t):=r(\theta+\pi,t)$. By (2.5), the function
$\varphi_{2}(\theta,t):=\widetilde{r}(\theta,t)-r(\theta,t)$ satisfies a
linear, uniformly parabolic equation which has smooth coefficients (similar to
the equation (3.5)) and involves a non-local quantity $L=L(t)$ of the curve,

$\displaystyle\frac{\partial\varphi_{2}}{\partial t}$ $\displaystyle=$
$\displaystyle\frac{1}{(g_{\widetilde{r}})^{2}}\frac{\partial^{2}\varphi_{2}}{\partial\theta^{2}}-\frac{1}{g^{2}(g_{\widetilde{r}})^{2}}\frac{\partial^{2}r}{\partial\theta^{2}}\left(\frac{\partial
r}{\partial\theta}+\frac{\partial\widetilde{r}}{\partial\theta}\right)\frac{\partial\varphi_{2}}{\partial\theta}-\frac{2}{rg^{2}}\left(\frac{\partial
r}{\partial\theta}+\frac{\partial\widetilde{r}}{\partial\theta}\right)\frac{\partial\varphi_{2}}{\partial\theta}$
$\displaystyle+\frac{2}{rg^{2}(g_{\widetilde{r}})^{2}}\left(\frac{\partial\widetilde{r}}{\partial\theta}\right)^{2}\left(\frac{\partial
r}{\partial\theta}+\frac{\partial\widetilde{r}}{\partial\theta}\right)\frac{\partial\varphi_{2}}{\partial\theta}+\frac{\widetilde{r}}{g^{2}(g_{\widetilde{r}})^{2}}\left(\frac{\partial
r}{\partial\theta}+\frac{\partial\widetilde{r}}{\partial\theta}\right)\frac{\partial\varphi_{2}}{\partial\theta}$
$\displaystyle-\frac{\widetilde{r}+r}{g^{2}(g_{\widetilde{r}})^{2}}\frac{\partial^{2}r}{\partial\theta^{2}}\varphi_{2}+\frac{2(\widetilde{r}+r)}{rg^{2}(g_{\widetilde{r}})^{2}}\left(\frac{\partial\widetilde{r}}{\partial\theta}\right)^{2}\varphi_{2}+\frac{2}{r\widetilde{r}(g_{\widetilde{r}})^{2}}\left(\frac{\partial\widetilde{r}}{\partial\theta}\right)^{2}\varphi_{2}$
$\displaystyle-\frac{\varphi_{2}}{(g_{r})^{2}}+\frac{\widetilde{r}(\widetilde{r}+r)}{g^{2}(g_{\widetilde{r}})^{2}}\varphi_{2}+\frac{2\pi(\widetilde{r}+r)}{\widetilde{r}L(g_{\widetilde{r}}+g)}\varphi_{2}$
$\displaystyle+\frac{2\pi}{\widetilde{r}L(g+g_{\widetilde{r}})}\left(\frac{\partial\widetilde{r}}{\partial\theta}+\frac{\partial
r}{\partial\theta}\right)\frac{\partial\varphi_{2}}{\partial\theta}-\frac{2\pi
g}{\widetilde{r}rL}\varphi_{2},$

where
$g_{\widetilde{r}}:=\sqrt{\widetilde{r}^{2}+\left(\frac{\partial\widetilde{r}}{\partial\theta}\right)^{2}}$.
Since $\varphi_{2}(\theta,0)\equiv 0$, one can obtain
$\varphi_{2}(\theta,t)\equiv 0,$that is to say, the evolving curve
$X(\cdot,t)$ is symmetric with respect to $O$. ∎

######  Corollary 3.6.

Suppose the initial smooth curve $X_{0}$ is star-shaped and centrosymmetric
with respect to $O$. If the evolving curve $X(\cdot,t)$ under Gage’s area-
preserving flow (1.1) is star-shaped then $O$ is one of its star centers.

###### Proof.

Suppose Gage’s area-preserving flow (1.1) with initial $X_{0}$ exists on a
time interval $[0,\omega)$. By continuity, there exists a small
$t_{0}\in[0,\omega)$ such that $X(\cdot,t)$ is star-shaped with respect to $O$
if $t<t_{0}$. Lemma 3.5 implies that $X(\cdot,t)$ is symmetric with respect to
$O$ if $t<t_{0}$.

Suppose there exists a $t_{*}\in(0,\omega)$ such that (i) $X(\cdot,t_{*})$ is
star-shaped with respect to some point but not the origin $O$ and (ii)
$X(\cdot,t)$ is star-shaped with respect to $O$ for all $t\in[0,t_{*})$. By
Lemma 3.5 and the continuity of $X(\cdot,t)$, $X(\cdot,t_{*})$ is
centrosymmetric with respect to $O$. So one can conclude that

$\displaystyle r_{\min}(t_{*})>0.$

Otherwise, $X(\cdot,t_{*})$ is not star-shaped with respect to any point
because it is centrosymmetric with respect to $O$.

Let $P$ be a star center of $X(\cdot,t_{*})$, then the symmetry of the curve
implies that $-P$ is also a star center (see [22]). Since the set of all star
centers of $X(\cdot,t_{*})$ is convex, $O$ is one of its star centers, which
leads to a contradiction. ∎

Next we shall show that Gage’s area-preserving flow (1.1) preserves star-
shapedness of the evolving curve.

######  Lemma 3.7.

Suppose the initial smooth curve $X_{0}$ is star-shaped and centrosymmetric
with respect to $O$ and Gage’s area-preserving flow (1.1) exists on a time
interval $[0,\omega)$. Then the evolving curve $X(\cdot,t)$ is star-shaped
with respect to $O$ for all $t\in[0,\omega)$.

###### Proof.

Suppose there is a $t_{*}\in(0,\omega)$ such that $X(\cdot,t)$ is star-shaped
for $t\in[0,t_{*})$ but $X(\cdot,t_{*})$ is not so. It follows from Corollary
3.6 that $X(\cdot,t)$ is star-shaped with respect to $O$ for all
$t\in[0,t_{*})$. By Lemma 3.5, $X(\cdot,t)$ is centrosymmetric with respect to
$O$ on the same time interval $[0,t_{*})$. Let
$\varepsilon_{0}=\min\left\\{\frac{t_{*}}{2},\frac{A_{0}}{4\pi}\right\\}$,
then $X(\cdot,t_{*}-\varepsilon_{0})$ is a centrosymmetric and star-shaped
curve with respect to $O$.

Let $X(\cdot,t_{*}-\varepsilon_{0})$ evolve according to the CSF, then we
obtain a family of smooth curves $Y(\cdot,t)$ for
$t\in\left[t_{*}-\varepsilon_{0},t_{*}-\varepsilon_{0}+\frac{A_{0}}{2\pi}\right).$
Write $Y(\theta,t)=\rho(\theta,t)P(\theta)$, where $\theta$ is the polar angle
independent of $t$. By Lemma 3.1, Lemma 3.2 and Corollary 3.3, $Y(\cdot,t)$ is
star-shaped and centrosymmetric with respect to $O$ and it shrinks to $O$ as
$t$ tends to the time $\left(t_{*}-\varepsilon_{0}+\frac{A_{0}}{2\pi}\right)$.
Therefore, there exists a $\delta=\delta(t_{*})>0$ such that

$\displaystyle\rho(\theta,t)\geq\delta(t_{*})$ (3.15)

for all
$(\theta,t)\in[0,2\pi]\times\left[t_{*}-\varepsilon_{0},t_{*}+\frac{3A_{0}}{8\pi}\right]$.

Under Gage’s area-preserving flow, the radial function $r(\theta,t)$ of
$X(\theta,t)$ satisfies the equation (2.5) for
$t\in[t_{*}-\varepsilon_{0},t_{*})$. Set
$\varphi_{3}(\theta,t)=r(\theta,t)-\rho(\theta,t)$, where
$(\theta,t)\in[0,2\pi]\times[t_{*}-\varepsilon_{0},t_{*})$. By (2.5) and
(3.3), $\varphi_{3}$ evolves according to

$\displaystyle\frac{\partial\varphi_{3}}{\partial t}$ $\displaystyle=$
$\displaystyle\frac{1}{g^{2}}\frac{\partial^{2}\varphi_{3}}{\partial\theta^{2}}-\frac{1}{(g_{\rho})^{2}g^{2}}\frac{\partial^{2}\rho}{\partial\theta^{2}}\left(\frac{\partial\rho}{\partial\theta}+\frac{\partial
r}{\partial\theta}\right)\frac{\partial\varphi_{3}}{\partial\theta}-\frac{2}{\rho(g_{\rho})^{2}}\left(\frac{\partial\rho}{\partial\theta}+\frac{\partial
r}{\partial\theta}\right)\frac{\partial\varphi_{3}}{\partial\theta}$ (3.16)
$\displaystyle+\frac{2}{\rho(g_{\rho})^{2}g^{2}}\left(\frac{\partial\rho}{\partial\theta}\right)^{2}\left(\frac{\partial\rho}{\partial\theta}+\frac{\partial
r}{\partial\theta}\right)\frac{\partial\varphi_{3}}{\partial\theta}+\frac{r}{(g_{\rho})^{2}g^{2}}\left(\frac{\partial\rho}{\partial\theta}+\frac{\partial
r}{\partial\theta}\right)\frac{\partial\varphi_{3}}{\partial\theta}$
$\displaystyle-\frac{r+\rho}{(g_{\rho})^{2}g^{2}}\frac{\partial^{2}\rho}{\partial\theta^{2}}\varphi_{3}+\frac{2(r+\rho)}{\rho(g_{\rho})^{2}g^{2}}\left(\frac{\partial\rho}{\partial\theta}\right)^{2}\varphi_{3}+\frac{2}{\rho
rg^{2}}\left(\frac{\partial\rho}{\partial\theta}\right)^{2}\varphi_{3}$
$\displaystyle-\frac{\varphi_{3}}{g^{2}}+\frac{r(r+\rho)}{(g_{\rho})^{2}g^{2}}\varphi_{3}+\frac{2\pi
g}{rL}.$

Noticing that $\frac{2\pi g}{rL}>0$ and (3.16) is a linear parabolic equation
with smooth and bounded coefficients, the maximum principle implies that there
exists a constant $\widetilde{c}>0$ such that for
$t\in[t_{*}-\varepsilon_{0},t_{*})$,

$\displaystyle\varphi_{3}(\theta,t)\geq\min\\{\varphi_{3}(\theta,t_{*}-\varepsilon_{0})|\theta\in[0,2\pi]\\}e^{-\widetilde{c}t}.$

Since $\varphi_{3}(\theta,t_{*}-\varepsilon_{0})\equiv 0$, we have
$\varphi_{3}(\theta,t)\geq 0$. Thus

$\displaystyle r(\theta,t_{*})\geq\rho(\theta,t_{*})\geq\delta(t_{*})>0.$
(3.17)

By Lemma 3.5 and Corollary 3.6, $X(\cdot,t_{*})$ is star-shaped, which
contradicts with the assumption. ∎

###  3.3 Extending Gage’s area-preserving flow

Now let us extend Gage’s area-preserving flow (1.1). Once the curvature of the
evolving curve is bounded on the time interval $[0,T_{*})$ for any $T_{*}>0$,
one can prove the smoothness of $X(\cdot,t)$ and the flow (1.1) can be
extended globally.

Suppose the initial smooth curve $X_{0}$ is star-shaped and centrosymmetric
with respect to $O$ and Gage’s area-preserving flow (1.1) with initial $X_{0}$
exists on the time interval $[0,T_{*})$, where $T_{*}>0$ is a finite number.
By Lemma 3.5, Corollary 3.6 and Lemma 3.7, the evolving curve $X(\cdot,t)$ is
star-shaped and centrosymmetric with respect to $O$. Let
$\varepsilon_{0}=\min\\{\frac{T_{*}}{2},\frac{A_{0}}{4\pi}\\}$. $X(\cdot,t)$
is star-shaped with respect to $O$ for $t\in[0,T_{*})$, so there exists a
$\delta_{1}=\delta_{1}(T_{*})>0$ such that $r(\theta,t)\geq\delta_{1}$ holds
for $t\in[0,T_{*}-\varepsilon_{0}]$. By the proof of Lemma 3.7, there exists a
$\delta_{2}=\delta_{2}(T_{*})>0$ such that $r(\theta,t)\geq\delta_{2}$ for
$t\in[T_{*}-\varepsilon_{0},T_{*}+\frac{A_{0}}{8\pi}]$. Let
$\delta=\delta(T_{*})=\min\\{\delta_{1},\delta_{2}\\}>0$, then for all
$(\theta,t)\in[0,2\pi]\times[0,T_{*}+\frac{A_{0}}{8\pi})$, one can obtain

$\displaystyle r(\theta,t)\geq\delta>0,$ (3.18)

which together with Lemma 2.4 and Lemma 2.5 tells us that if
$t\in\left[0,T_{*}+\frac{A_{0}}{8\pi}\right)$, then the evolving curve
$X(\cdot,t)$ is star-shaped with respect to $O$. So there exists an
$h=h(T_{*})>0$ such that the “support function” with respect to $O$ satisfies

$\displaystyle p(\theta,t)\geq 2h(T_{*})>0,\ \ \ \
(\theta,t)\in[0,2\pi]\times\left[0,T_{*}+\frac{A_{0}}{8\pi}\right).$ (3.19)

Under the flow (1.1), the curvature evolves according to

$\displaystyle\frac{\partial\kappa}{\partial
t}=\frac{\partial^{2}\kappa}{\partial
s^{2}}+\kappa^{3}-\frac{2\pi}{L}\kappa^{2},$ (3.20)

where $s$ stands for the arc length parameter. Define

$\displaystyle\varphi_{4}(s,t)=\frac{\kappa(s,t)}{p(s,t)-h(T_{*})},\ \ \ \ \ \
t\in[0,T_{*}),$

we have

$\displaystyle\frac{\partial\varphi_{4}}{\partial
s}=\frac{1}{p-h}\frac{\partial\kappa}{\partial
s}-\frac{\kappa}{(p-h)^{2}}\frac{\partial p}{\partial s},$
$\displaystyle\frac{\partial^{2}\varphi_{4}}{\partial
s^{2}}=\frac{1}{p-h}\frac{\partial^{2}\kappa}{\partial
s^{2}}-\frac{2}{(p-h)^{2}}\frac{\partial\kappa}{\partial s}\frac{\partial
p}{\partial s}-\frac{\kappa}{(p-h)^{2}}\frac{\partial^{2}p}{\partial
s^{2}}+\frac{2\kappa}{(p-h)^{3}}\left(\frac{\partial p}{\partial
s}\right)^{2}.$

Noticing that the support function evolves according to

$\displaystyle\frac{\partial p}{\partial t}$ $\displaystyle=$
$\displaystyle-\frac{\partial}{\partial t}\langle
X,N\rangle=-\left\langle\left(\kappa-\frac{2\pi}{L}\right)N,N\right\rangle+\left\langle
X,\frac{\partial\kappa}{\partial s}T\right\rangle$ (3.21) $\displaystyle=$
$\displaystyle\frac{2\pi}{L}-\kappa+\langle
X,T\rangle\frac{\partial\kappa}{\partial s},$

the evolution equation of $\varphi_{4}$ can be given by

$\displaystyle\frac{\partial\varphi_{4}}{\partial t}$ $\displaystyle=$
$\displaystyle\frac{\partial^{2}\varphi_{4}}{\partial
s^{2}}+\frac{2}{p-h}\frac{\partial p}{\partial
s}\frac{\partial\varphi_{4}}{\partial s}-\frac{\kappa}{p-h}\langle
X,T\rangle\frac{\partial\varphi_{4}}{\partial
s}+\frac{\kappa}{(p-h)^{2}}\frac{\partial^{2}p}{\partial s^{2}}$ (3.22)
$\displaystyle+(p-h)^{2}(\varphi_{4})^{3}-\frac{2\pi}{L}(p-h)(\varphi_{4})^{2}+(\varphi_{4})^{2}-\frac{2\pi}{L}\frac{\varphi_{4}}{p-h}$
$\displaystyle-\frac{\kappa^{2}}{(p-h)^{3}}\langle X,T\rangle\frac{\partial
p}{\partial s}.$

Using

$\displaystyle\frac{\partial p}{\partial s}=\kappa\langle
X,T\rangle,~{}~{}~{}~{}\frac{\partial^{2}p}{\partial
s^{2}}=\frac{\partial\kappa}{\partial s}\langle
X,T\rangle+\kappa-\kappa^{2}p,$

we can compute that

$\displaystyle\frac{\kappa}{(p-h)^{2}}\frac{\partial^{2}p}{\partial
s^{2}}-\frac{\kappa^{2}}{(p-h)^{3}}\langle X,T\rangle\frac{\partial
p}{\partial s}=\varphi_{4}\langle
X,T\rangle\frac{\partial\varphi_{4}}{\partial
s}+(\varphi_{4})^{2}-p(p-h)(\varphi_{4})^{3}.$ (3.23)

Substituting (3.23) into the evolution equation of $\varphi_{4}$ can give us

$\displaystyle\frac{\partial\varphi_{4}}{\partial t}$ $\displaystyle=$
$\displaystyle\frac{\partial^{2}\varphi_{4}}{\partial
s^{2}}+\frac{2}{p-h}\frac{\partial p}{\partial
s}\frac{\partial\varphi_{4}}{\partial
s}-h(p-h)(\varphi_{4})^{3}-\frac{2\pi}{L}(p-h)(\varphi_{4})^{2}$ (3.24)
$\displaystyle+2(\varphi_{4})^{2}-\frac{2\pi}{L}\frac{\varphi_{4}}{p-h}.$

As is well known that the arc length parameter $s$ depends on both space
parameter and time $t$, one can not apply the maximum principle directly. In
Section 2, the parameter $\varphi$ (independent of $t$) is used to parametrize
$X(\cdot,t)$. Now, let $g=\frac{\partial s}{\partial\varphi}$ and suppose that
for fixed $t$, $\varphi_{4}(\cdot,t)$ attains its maximum
$(\varphi_{4})_{\max}(t)$ at the point $(\varphi_{*},t)$. Denoted by
$s_{*}=s(\varphi_{*})$, then

$\frac{\partial\varphi_{4}}{\partial
s}(s_{*},t)=\left(\frac{\partial\varphi_{4}}{\partial\varphi}\frac{1}{g}\right)(s(\varphi_{*}),t)=0$

and

$\frac{\partial^{2}\varphi_{4}}{\partial
s^{2}}(s_{*},t)=\left(\frac{\partial^{2}\varphi_{4}}{\partial\varphi^{2}}\frac{1}{g^{2}}\right)(s(\varphi_{*}),t)-\left(\frac{\partial\varphi_{4}}{\partial\varphi}\frac{\partial
g}{\partial\varphi}\frac{1}{g^{3}}\right)(s(\varphi_{*}),t)\leq 0.$

By (3.24),

$\displaystyle\frac{\partial\varphi_{4}}{\partial t}(s_{*},t)$
$\displaystyle\leq$
$\displaystyle-h(p-h)(\varphi_{4}(s_{*},t))^{3}-(p-h)\frac{2\pi}{L}(\varphi_{4}(s_{*},t))^{2}$
$\displaystyle+2(\varphi_{4}(s_{*},t))^{2}-\frac{2\pi}{L}\frac{\varphi_{4}(s_{*},t)}{p-h}$
$\displaystyle<$
$\displaystyle-h(p-h)(\varphi_{4}(s_{*},t))^{3}+2(\varphi_{4}(s_{*},t))^{2}$
$\displaystyle<$
$\displaystyle-h^{2}(\varphi_{4}(s_{*},t))^{3}+2(\varphi_{4}(s_{*},t))^{2}.$

Once $\varphi_{4}(s_{*},t)$ is greater than $\frac{2}{h^{2}}$, then
$\frac{\partial\varphi_{4}}{\partial t}(s_{*},t)<0$. By the maximum principle,
we get

$\displaystyle\varphi_{4}(s,t)\leq\max\left\\{(\varphi_{4})_{\max}(0),~{}~{}\frac{2}{h^{2}(T_{*})}\right\\}$
(3.25)

for $(s,t)\in[0,L]\times\left[0,T_{*}+\frac{A_{0}}{8\pi}\right]$. Since the
“support function” satisfies $p(s,t)\leq\frac{L(t)}{2}\leq\frac{L(0)}{2}$, we
have

$\displaystyle\kappa(s,t)\leq\left(\frac{L(0)}{2}-h(T_{*})\right)\cdot\max\left\\{(\varphi_{4})_{\max}(0),~{}~{}\frac{2}{h^{2}(T_{*})}\right\\}.$
(3.26)

Similarly, if $\varphi_{4}$ attains $(\varphi_{4})_{\min}(t)$ at a point
$(s_{*},t)$ then $\frac{\partial\varphi_{4}}{\partial
s}(s_{*},t)=0,~{}~{}\frac{\partial^{2}\varphi_{4}}{\partial
s^{2}}(s_{*},t)\geq 0$. Once $\varphi_{4}(s_{*},t)\leq-\frac{2\pi}{\sqrt{4\pi
A_{0}}h(T_{*})}$, we have

$\displaystyle\frac{\partial\varphi_{4}}{\partial t}(s_{*},t)$
$\displaystyle\geq$
$\displaystyle-h(p-h)(\varphi_{4}(s_{*},t))^{3}-(p-h)\frac{2\pi}{L(t)}(\varphi_{4}(s_{*},t))^{2}$
$\displaystyle+2(\varphi_{4}(s_{*},t))^{2}-\frac{2\pi}{L(t)}\frac{\varphi_{4}(s_{*},t)}{p-h}$
$\displaystyle\geq$
$\displaystyle-\left(h\varphi_{4}(s_{*},t)+\frac{2\pi}{L(t)}\right)(p-h)(\varphi_{4}(s_{*},t))^{2}>0,$

and

$\varphi_{4}(s,t)\geq\min\left\\{(\varphi_{4})_{\min}(0),~{}-\frac{2\pi}{\sqrt{4\pi
A_{0}}h(T_{*})}\right\\}$

for $t\in[0,L]\times\left[0,T_{*}+\frac{A_{0}}{8\pi}\right]$. Therefore we can
get

$\displaystyle\kappa(s,t)\geq\left(\frac{L(0)}{2}-h(T_{*})\right)\cdot\min\left\\{(\varphi_{4})_{\min}(0),~{}-\frac{2\pi}{\sqrt{4\pi
A_{0}}h(T_{*})}\right\\}.$ (3.27)

######  Lemma 3.8.

Let the initial smooth curve $X_{0}$ be star-shaped and centrosymmetric with
respect to $O$ and Gage’s area-preserving flow (1.1) exist on a time interval
$[0,T_{*})$ for some $T_{*}>0$, then the curvature of $X(\cdot,t)$ is bounded
on the time interval $[0,T_{*}]$ uniformly.

As a consequence of Lemma 3.8, one can extend Gage’s area-preserving flow
(1.1) globally.

######  Corollary 3.9.

Gage’s area-preserving flow (1.1) exists on the time interval $[0,+\infty)$ if
the initial smooth curve $X_{0}$ is star-shaped and centrosymmetric with
respect to $O$.

###### Proof.

Let $T_{*}$ be a positive number which can be chosen arbitrarily. By Lemma
2.4, both $r$ and $\frac{\partial r}{\partial\theta}$ are bounded uniformly on
the time interval $[0,T_{*}]$. By Lemmas (3.1)-(3.8),
$\left|\frac{\partial^{2}r}{\partial\theta^{2}}\right|$ is also bounded
uniformly on the same interval $[0,T_{*}]$. All the higher derivatives
$\frac{\partial^{i}r}{\partial\theta^{i}}(i\geq 3)$ satisfy linear parabolic
equations, they also have uniform bounds. Therefore, the evolving curve is
smooth on any finite time interval, and the flow (1.1) with initial $X_{0}$
exists globally. ∎

##  4 Convergence and a convexity theorem

In this section, it is shown that Gage’s area preserving flow can deform every
smooth, closed and embedded curve into a convex one if the flow exists on the
time interval $[0,+\infty)$.

Since we have proved that the flow (1.1) with inial curve $X_{0}$ being
smooth, centrosymmetric and star-shaped does not blow up at any finite time in
Section 3, the main result in this paper can be obtained by the following
theorem immediately.

######  Theorem 4.1.

Given a smooth, embedded and closed initial curve $X_{0}$, if the flow (1.1)
does not blow up in the time interval $[0,+\infty)$, then the evolving curve
converges to a finite circle as time goes to infinity.

In order to prove this result, we need do some preparations. From now on, we
use the subindex to stand for partial derivatives, such as
$\kappa_{t}=\frac{\partial\kappa}{\partial
t},\kappa_{s}=\frac{\partial\kappa}{\partial
s},\kappa_{ss}=\frac{\partial^{2}\kappa}{\partial s^{2}},\cdots$. The length
of the evolving curve is decreasing during the evolution process and bounded
from below by $\sqrt{4\pi A_{0}}$. Following the parlance from [10], the time
derivative of the length must approach zero at an $\varepsilon$-dense set of
sufficiently large time. Now one can use this fact in the following special
case. Denote by $I_{n}$ the time interval $[n-\frac{1}{10},n+\frac{1}{10}]$,
where $n$ is a natural number. For a positive number $\varepsilon$, all of the
intervals which satisfy that
$\inf_{I_{n}}\int_{0}^{L}\kappa^{2}ds-\frac{4\pi^{2}}{L}\geq\varepsilon$
consists of a finite set. So there exists an $N>0$ such that whenever $n>N$ we
have $t_{n}\in I_{n}$ satisfying

$\displaystyle\int_{0}^{L}\kappa(\cdot,t_{n})^{2}ds-\frac{4\pi^{2}}{L(t_{n})}\leq\varepsilon.$
(4.1)

######  Lemma 4.2.

The $L^{2}$ norm of the difference $(\kappa-\frac{2\pi}{L})$ converges to zero
as $t\rightarrow\infty$.

###### Proof.

Integrating the evolution equation of $L$ can give us

$L(t)-L(0)=-\int_{0}^{t}\int_{0}^{L}\left(\kappa-\frac{2\pi}{L}\right)^{2}dsdt.$

Letting $t$ tend to infinity can yield

$\displaystyle\int_{0}^{\infty}\int_{0}^{L}\left(\kappa-\frac{2\pi}{L}\right)^{2}dsdt<L(0).$
(4.2)

We consider the time derivative of the integral
$\int_{0}^{L}\left(\kappa-\frac{2\pi}{L}\right)^{2}ds$:

$\displaystyle\frac{d}{dt}\int_{0}^{L}\left(\kappa-\frac{2\pi}{L}\right)^{2}ds$
$\displaystyle=$
$\displaystyle-2\int_{0}^{L}(\kappa_{s})^{2}ds+\int_{0}^{L}\kappa^{2}\left(\kappa-\frac{2\pi}{L}\right)^{2}ds$
(4.3)
$\displaystyle+\frac{2\pi}{L}\int_{0}^{L}\kappa\left(\kappa-\frac{2\pi}{L}\right)^{2}ds.$

One only needs to deal with the case that the evolving curve is not convex. In
this case, $\inf_{s}\kappa^{2}=0$. Otherwise, the convex curve $X(\cdot,t)$
will converge to a finite circle (see [3]) and the proof is completed. Since

$\sup_{s}\kappa^{2}\leq L\int_{0}^{L}(\kappa_{s})^{2}ds,$

one obtains

$-\int_{0}^{L}(\kappa_{s})^{2}ds\leq-\frac{1}{L}\sup_{s}\kappa^{2}.$

By the Cauchy-Schwartz inequality,

$\displaystyle\left|\int_{0}^{L}\kappa\left(\kappa-\frac{2\pi}{L}\right)^{2}ds\right|$
$\displaystyle\leq$
$\displaystyle\sqrt{\int_{0}^{L}\kappa^{2}\left(\kappa-\frac{2\pi}{L}\right)^{2}ds}\cdot\sqrt{\int_{0}^{L}\left(\kappa-\frac{2\pi}{L}\right)^{2}ds}$
$\displaystyle\leq$
$\displaystyle\sqrt{\int_{0}^{L}\sup_{s}\kappa^{4}ds}\cdot\sqrt{\int_{0}^{L}\left(\kappa-\frac{2\pi}{L}\right)^{2}ds}.$
$\displaystyle=$
$\displaystyle\sup_{s}\kappa^{2}\cdot\sqrt{L(t)}\cdot\sqrt{\int_{0}^{L}\left(\kappa-\frac{2\pi}{L}\right)^{2}ds}.$

Taking this estimate into the equation (4.3) can yield

$\displaystyle\frac{d}{dt}\int_{0}^{L}\left(\kappa-\frac{2\pi}{L}\right)^{2}ds$
$\displaystyle\leq$
$\displaystyle-\frac{2}{L(t)}\sup_{s}\kappa^{2}+\sup_{s}\kappa^{2}\int_{0}^{L}\left(\kappa-\frac{2\pi}{L}\right)^{2}ds$
$\displaystyle+\sup_{s}\kappa^{2}\cdot\frac{2\pi}{\sqrt{L(t)}}\cdot\sqrt{\int_{0}^{L}\left(\kappa-\frac{2\pi}{L}\right)^{2}ds}$
$\displaystyle\leq$
$\displaystyle\sup_{s}\kappa^{2}\left(-\frac{2}{L_{0}}+\int_{0}^{L}\left(\kappa-\frac{2\pi}{L}\right)^{2}ds+\frac{2\pi}{\sqrt{L_{\infty}}}\sqrt{\int_{0}^{L}\left(\kappa-\frac{2\pi}{L}\right)^{2}ds}\right),$

where $L_{\infty}=\sqrt{4\pi A}$. Choose $\varepsilon>0$ in (4.1) small enough
such that

$-\frac{2}{L_{0}}+\frac{3}{2}\varepsilon+\frac{2\pi}{\sqrt{L_{\infty}}}\sqrt{\frac{3\varepsilon}{2}}\leq
0.$

We now claim that
$\int_{0}^{L}(\kappa-\frac{2\pi}{L})^{2}ds<\frac{3\varepsilon}{2}$.

Firstly, assume that at time $t=t_{n}$, we have

$\int_{0}^{L}(\kappa-\frac{2\pi}{L})^{2}ds\leq\varepsilon.$

Secondly, for $t>t_{n}$, suppose there exists a
$t_{*}\in\left(t_{n},t_{n}+\frac{11}{10}\right]$ such that

$\displaystyle\int_{0}^{L}\left(\kappa(\cdot,t_{*})-\frac{2\pi}{L(t_{*})}\right)^{2}ds=\frac{3\varepsilon}{2}$

and

$\displaystyle\int_{0}^{L}\left(\kappa-\frac{2\pi}{L}\right)^{2}ds<\frac{3\varepsilon}{2}$

for all $t\in[t_{n},t_{*})$. In the time interval $(t_{n},t_{*})$, we have

$\displaystyle\frac{d}{dt}\int_{0}^{L}\left(\kappa-\frac{2\pi}{L}\right)^{2}ds$
$\displaystyle\leq$
$\displaystyle\sup_{s}\kappa^{2}\left(-\frac{2}{L_{0}}+\frac{3}{2}\varepsilon+\frac{2\pi}{\sqrt{L_{\infty}}}\sqrt{\frac{3\varepsilon}{2}}\right)\leq
0.$

Thus

$\displaystyle\int_{0}^{L}\left(\kappa(\cdot,t_{*})-\frac{2\pi}{L(t_{*})}\right)^{2}ds$
$\displaystyle\leq$
$\displaystyle\int_{0}^{L}\left(\kappa(\cdot,t_{n})-\frac{2\pi}{L(t_{n})}\right)^{2}ds<\frac{3\varepsilon}{2},$

which contradicts with the definition of $t_{*}$. For any $\varepsilon>0$,
there exists an $N>0$ such that $t>N$ implies that
$\int_{0}^{L}(\kappa-\frac{2\pi}{L})^{2}ds<\frac{3}{2}\varepsilon$. That is to
say, one has the limit

$\lim_{t\rightarrow\infty}\int_{0}^{L}\left(\kappa-\frac{2\pi}{L}\right)^{2}ds=0.$

∎

The integral of $\int_{0}^{L}(\kappa-\frac{2\pi}{L})^{2}ds$ in the equation
(4.3) is bounded in the time interval $[0,\infty)$. Next, the $L^{2}$-norm of
$(\kappa-\frac{2\pi}{L})_{s}$ will be estimated.

######  Lemma 4.3.

Under the assumption of Theorem 4.1, we have the limit

$\lim_{t\rightarrow\infty}\int_{0}^{L}(\kappa_{s})^{2}ds=0.$

###### Proof.

Denote $u=\kappa-\frac{2\pi}{L}$. It is easy to calculate that

$u_{t}=u_{ss}+\left(u+\frac{2\pi}{L}\right)^{2}u-\frac{2\pi}{L^{2}}\int_{0}^{L}u^{2}ds$

and

$u_{st}=u_{sss}+3\left(u+\frac{2\pi}{L}\right)uu_{s}+\left(u+\frac{2\pi}{L}\right)^{2}u_{s}.$

By using these equations one can get

$\displaystyle\frac{d}{dt}\int_{0}^{L}(u_{s})^{2}ds$ $\displaystyle=$
$\displaystyle-2\int_{0}^{L}(u_{ss})^{2}ds+7\int_{0}^{L}u^{2}(u_{s})^{2}ds$
(4.4)
$\displaystyle+\frac{18\pi}{L}\int_{0}^{L}u(u_{s})^{2}ds+\frac{8\pi^{2}}{L^{2}}\int_{0}^{L}(u_{s})^{2}ds.$

If $\int_{0}^{L}(u_{s})^{2}ds>C\int_{0}^{L}u^{2}ds$ for some positive $C$ then
from

$\int_{0}^{L}(u_{s})^{2}ds=-\int_{0}^{L}uu_{ss}ds\leq\sqrt{\frac{1}{C}\int_{0}^{L}(u_{s})^{2}ds\int_{0}^{L}(u_{ss})^{2}ds}$

it follows that

$\displaystyle\int_{0}^{L}(u_{s})^{2}ds\leq\frac{1}{C}\int_{0}^{L}(u_{ss})^{2}ds.$
(4.5)

We now estimate the terms in the right hand side of the equation (4.4).

$\displaystyle\int_{0}^{L}u^{2}(u_{s})^{2}ds$ $\displaystyle\leq$
$\displaystyle\int_{0}^{L}u^{2}ds\cdot\sup_{s}(u_{s})^{2}\leq\int_{0}^{L}u^{2}ds\left(\int_{0}^{L}|u_{ss}|ds\right)^{2}$
(4.6) $\displaystyle\leq$
$\displaystyle\int_{0}^{L}u^{2}dsL\int_{0}^{L}(u_{ss})^{2}ds.$

Using (4.5)-(4.6) can yield

$\displaystyle\int_{0}^{L}u(u_{s})^{2}ds$ $\displaystyle\leq$
$\displaystyle\frac{1}{2}\int_{0}^{L}u^{2}(u_{s})^{2}ds+\frac{1}{2}\int_{0}^{L}(u_{s})^{2}ds$
(4.7) $\displaystyle\leq$
$\displaystyle\frac{L}{2}\int_{0}^{L}u^{2}ds\int_{0}^{L}(u_{ss})^{2}ds+\frac{1}{2C}\int_{0}^{L}(u_{ss})^{2}ds.$

Taking the estimates (4.5)-(4.7) into the equation (4.4), one obtains

$\displaystyle\frac{d}{dt}\int_{0}^{L}(u_{s})^{2}ds$ $\displaystyle=$
$\displaystyle\left(-2+(7L+9\pi)\int_{0}^{L}u^{2}ds+\frac{9\pi}{CL}+\frac{8\pi^{2}}{CL^{2}}\right)\int_{0}^{L}(u_{ss})^{2}ds$
$\displaystyle\leq$
$\displaystyle\left(-2+(7L_{0}+9\pi)\int_{0}^{L}u^{2}ds+\frac{9\pi}{CL_{\infty}}+\frac{8\pi^{2}}{CL_{\infty}^{2}}\right)\int_{0}^{L}(u_{ss})^{2}ds.$

Lemma 4.2 tells us that there exists a positive $t_{0}$ such that for
$t>t_{0}$,

$(7L_{0}+9\pi)\int_{0}^{L}u^{2}ds\leq\frac{1}{2}.$

If $C\geq\max\\{\frac{36\pi}{L_{\infty}},\frac{32\pi^{2}}{L_{\infty}^{2}}\\}$
then $\frac{9\pi}{CL_{\infty}}+\frac{8\pi^{2}}{CL_{\infty}^{2}}\leq 1/2$ and
furthermore

$\displaystyle\frac{d}{dt}\int_{0}^{L}(u_{s})^{2}ds\leq-\int_{0}^{L}(u_{ss})^{2}ds$

if $t>t_{0}$. For a $C^{1}$ and $2\pi$-periodic function $f$ defined on
$\mathbb{R}$ such that $\int_{0}^{2\pi}fdx=0$, Wirtinger’s inequality tells us

$\int_{0}^{2\pi}(f^{\prime})^{2}dx\geq\int_{0}^{2\pi}f^{2}dx.$

Let $s=\frac{L}{2\pi}\phi$, then

$u_{s}=u_{\phi}\frac{2\pi}{L}\ \ \ \mbox{and}\ \ \ \
u_{ss}=u_{\phi\phi}\left(\frac{2\pi}{L}\right)^{2}.$

By Wirtinger’s inequality, we get

$\displaystyle\int_{0}^{L}(u_{ss})^{2}ds=\left(\frac{2\pi}{L}\right)^{3}\int_{0}^{2\pi}(u_{\phi\phi})^{2}d\phi\geq\left(\frac{2\pi}{L}\right)^{3}\int_{0}^{2\pi}(u_{\phi})^{2}d\phi=\left(\frac{2\pi}{L}\right)^{2}\int_{0}^{L}(u_{s})^{2}ds.$

If $C\geq\max\\{\frac{36\pi}{L_{\infty}},\frac{32\pi^{2}}{L_{\infty}^{2}}\\}$,
then

$\frac{d}{dt}\int_{0}^{L}(u_{s})^{2}ds\leq-\left(\frac{2\pi}{L_{0}}\right)^{2}\int_{0}^{L}(u_{s})^{2}ds,\
\ \ t>t_{0}.$

Thus, for $t>t_{0}$,

$\displaystyle\int_{0}^{L}(u_{s})^{2}ds\leq\int_{0}^{L}(u_{s}(\cdot,t_{0}))^{2}ds\cdot\exp\left[-\left(\frac{2\pi}{L_{0}}\right)^{2}t\right].$
(4.8)

For large time, the quantity $\int_{0}^{L}(u_{s})^{2}ds$ either decays
exponentially or is bounded,

$\int_{0}^{L}(u_{s})^{2}ds\leq\left(1+\max\left\\{\frac{36\pi}{L_{\infty}},\frac{32\pi^{2}}{L_{\infty}^{2}}\right\\}\right)\int_{0}^{L}u^{2}ds.$

In either event, it decreases to zero. ∎

Now we go back to the proof of Theorem 4.1. It follows from Sobolev’s
inequality that $|\kappa-\frac{2\pi}{L}|$ tends to 0 as $t\rightarrow\infty$.
Thus there is a time $T_{0}>0$ such that $\kappa>0$ for all $t>T_{0}$. The
evolving curve becomes a convex one. By the result in Gage’s original paper
[5] (see also the note [1]), we know that the curvature of the evolving curve
converges to $\sqrt{\frac{\pi}{A_{0}}}$ in the $C^{\infty}$ metric.

In the section 4 of the paper [9] by the first author and Zhang, the evolving
curve of GAPF is proved to converge to a fixed limiting circle not escaping to
infinity or oscillating indefinitely as $t\rightarrow+\infty$. So the proof of
Theorem 1.1 can be completed by combining Corollary 3.9 and Theorem 4.1.

Acknowledgments Laiyuan Gao is supported by the National Natural Science
Foundation of China (No.11801230). Shengliang Pan is supported by the National
Natural Science Foundation of China (No.12071347). The authors are very
grateful to Professor Michael Gage for his inspiration on the flow (1.1). The
first author thanks the Department of Mathematics of UC San Diego for
providing a fantastic research atmosphere because a part of this paper was
finished when he visited Professor Lei Ni.

## References

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Laiyuan Gao

School of Mathematics and Statistics, Jiangsu Normal University.

No.101, Shanghai Road, Xuzhou City, Jiangsu Province, China.

Email<EMAIL_ADDRESS>

Shengliang Pan

School of Mathematical Sciences, Tongji University.

No.1239, Siping Road, Shanghai, China.

Email<EMAIL_ADDRESS>


# Trudinger-Moser inequalities on the entire Heisenberg group

Yunyan Yang<EMAIL_ADDRESS>Department of Mathematics, Renmin
University of China, Beijing 100872, P. R. China

###### Abstract

Continuing our previous work (Cohn, Lam, Lu, Yang, Nonlinear Analysis (2011),
doi: 10.1016 /j.na.2011.09.053), we obtain a class of Trudinger-Moser
inequalities on the entire Heisenberg group, which indicate what the best
constants are. All the existing proofs of similar inequalities on unbounded
domain of the Euclidean space or the Heisenberg group are based on
rearrangement argument. In this note, we propose a new approach to solve this
problem. Specifically we get the global Trudinger-Moser inequality by gluing
local estimates with the help of cut-off functions. Our method still works for
similar problems when the Heisenberg group is replaced by the Eclidean space
or complete noncompact Riemannian manifolds.

###### keywords:

Trudinger-Moser inequality, singular Trudinger-Moser inequality, Adams
inequality

###### MSC:

46E35

††journal: ***

##  1 Introduction

Let $\mathbb{H}^{n}=\mathbb{R}^{2n}\times\mathbb{R}$ be the Heisenberg group
whose group action is defined by

$(x,y,t)\circ({x}^{\prime},{y}^{\prime},{t}^{\prime})=(x+{x}^{\prime},y+{y}^{\prime},t+{t}^{\prime}+2(\langle
y,x^{\prime}\rangle-\langle x,{y}^{\prime}\rangle)),$ (1.1)

where $x,y,x^{\prime},y^{\prime}\in\mathbb{R}^{n}$,
$t,t^{\prime}\in\mathbb{R}$, and $\langle\cdot,\cdot\rangle$ denotes the
standard inner product in $\mathbb{R}^{n}$. Let us denote the parabolic
dilation in $\mathbb{R}^{2n}\times\mathbb{R}$ by $\delta_{\lambda}$, namely,
$\delta_{\lambda}(\xi)=(\lambda x,\lambda y,\lambda^{2}t)$ for any
$\xi=(x,y,t)\in\mathbb{R}^{2n}\times\mathbb{R}$. The Jacobian determinant of
$\delta_{\lambda}$ is $\lambda^{Q}$, where $Q=2n+2$ is the homogeneous
dimension of $\mathbb{H}^{n}$. The following norm

$|\xi|_{h}=\left[\left(\sum_{i=1}^{n}(x_{i}^{2}+y_{i}^{2})\right)^{2}+t^{2}\right]^{\frac{1}{4}}$
(1.2)

is homogeneous of degree one with respect to the dilation $\delta_{\lambda}$.
The associated distance between two points $\xi$ and $\eta$ of
$\mathbb{H}^{n}$ is defined accordingly by

$d_{h}(\xi,\eta)=|\eta^{-1}\circ\xi|_{h},$ (1.3)

where $\eta^{-1}$ denotes the inverse of $\eta$ with respect to the group
action, i.e. $\eta^{-1}=-\eta$. Obviously $d_{h}(\cdot,\cdot)$ is symmetric.
The open ball of radius $r$ centered at $\xi$ is

$B_{h}(\xi,r)=\\{\eta\in\mathbb{H}^{n}:d_{h}(\eta,\xi)<r\\}.$

It is important to note that (see for example Stein [11], Section 5 of Chapter
VIII)

$|B_{h}(\xi,r)|=|B_{h}(0,r)|=|B_{h}(0,1)|r^{Q},$ (1.4)

where $|\cdot|$ denotes the Lebesgue measure. The Lie algebra of
$\mathbb{H}^{n}$ is generated by the left-invariant vector fields

$T=\frac{\partial}{\partial t},\,\,X_{i}=\frac{\partial}{\partial
x_{i}}+2y_{i}\frac{\partial}{\partial t},\,\,Y_{i}=\frac{\partial}{\partial
y_{i}}-2x_{i}\frac{\partial}{\partial t},\,i=1,\cdots,n.$ (1.5)

These generators satisfy the non-commutative formula
$[X_{i},Y_{i}]=-4\delta_{ij}T$. Denote by $|\nabla_{\mathbb{H}^{n}}u|$ the
norm of the sub-elliptic gradient of a smooth function
$u:\mathbb{H}^{n}\rightarrow\mathbb{R}$:

$|\nabla_{\mathbb{H}^{n}}u|=\left(\sum_{i=1}^{n}\left((X_{i}u)^{2}+(Y_{i}u)^{2}\right)\right)^{{1}/{2}}.$

Let $\Omega$ be an open set in $\mathbb{H}^{n}$. We use $W_{0}^{1,p}(\Omega)$
to denote the completion of $C_{0}^{\infty}(\Omega)$ under the norm

$\|u\|_{W_{0}^{1,p}(\Omega)}=\left(\int_{\Omega}\left(|\nabla_{\mathbb{H}^{n}}u|^{p}+|u|^{p}\right)d\xi\right)^{{1}/{p}}.$
(1.6)

In [4], Cohn-Lu proved a Trudinger-Moser inequality on bounded smooth domains
in the Hesenberg group $\mathbb{H}^{n}$. Precisely, there exists some constant
$C_{n}$ depending only on $n$ such that for all bounded smooth domain
$\Omega\subset\mathbb{H}^{n}$, if $u\in W_{0}^{1,Q}(\Omega)$ satisfies
$\|\nabla_{\mathbb{H}^{n}}u\|_{L^{Q}(\Omega)}\leq 1$, then

$\int_{\Omega}e^{\alpha_{Q}|u|^{Q^{\prime}}}d\xi\leq C_{n}|\Omega|,$ (1.7)

where $Q^{\prime}=Q/(Q-1)$, $\alpha_{Q}=Q\sigma_{Q}^{1/(Q-1)}$,
$\sigma_{Q}=\Gamma(\frac{1}{2})\Gamma(n+\frac{1}{2})\omega_{2n-1}/n!$,
$\omega_{2n-1}$ is the surface area of the unit sphere in $\mathbb{R}^{2n}$.
Furthermore, the integrals of all $u\in W_{0}^{1,Q}(\Omega)$ satisfying
$\|\nabla_{\mathbb{H}^{n}}u\|_{L^{Q}(\Omega)}\leq 1$ are not uniformly bounded
if $\alpha_{Q}$ is replaced by any larger number. Recently, Cohn, Lam, Lu and
the author [3] obtained a Trudinger-Moser inequality on the Heisenberg group
$\mathbb{H}^{n}$. Note that $W^{1,Q}(\mathbb{H}^{n})$ is the completion of
$C_{0}^{\infty}(\mathbb{H}^{n})$ under the norm (1.6) with $\Omega$ replaced
by $\mathbb{H}^{n}$. We have the following:

Theorem A ([3]). There exists some constant
$\alpha^{\ast}:0<\alpha^{\ast}\leq\alpha_{Q}$ such that for any pair $\beta$
and $\alpha$ satisfying $0\leq\beta<Q$, $0<\alpha\leq\alpha^{\ast}$, and
$\frac{\alpha}{\alpha^{\ast}}+\frac{\beta}{Q}\leq 1$, there holds

$\sup_{\|u\|_{W^{1,Q}(\mathbb{H}^{n})}\leq
1}\int_{\mathbb{H}^{n}}\frac{1}{|\xi|_{h}^{\beta}}\left\\{e^{\alpha|u|^{Q\,^{\prime}}}-\sum_{k=0}^{Q-2}\frac{\alpha^{k}|u|^{kQ\,^{\prime}}}{k!}\right\\}d\xi<\infty.$
(1.8)

When $\frac{\alpha}{\alpha^{\ast}}+\frac{\beta}{Q}>1$, the integral in
$(\ref{TMH})$ is still finite for any $u\in W^{1,Q}(\mathbb{H}^{n})$, but the
supremum is infinite if further $\frac{\alpha}{\alpha_{Q}}+\frac{\beta}{Q}>1$.

Theorem A is an analogue of (Adimurthi-Yang [1], Theorem 1.1). Earlier works
on this topic (Trudinger-Moser inequalities on unbounded domain of
$\mathbb{R}^{n}$) were done by Cao [2], Panda [9], do Ó [5], Ruf [10], Li-Ruf
[8] and others. The proof of Theorem A is based on symmetrization argument,
radial lemma and the Young inequality. Note that $\alpha^{*}$ in Theorem A is
not explicitly known. A natural question is what the best constant $\alpha$
for (1.8) is. Denote an equivalent norm in $W^{1,Q}(\mathbb{H}^{n})$ by

$\|u\|_{1,\tau}=\left(\int_{\mathbb{H}^{n}}(|\nabla_{\mathbb{H}^{n}}u|^{Q}+\tau|u|^{Q})d\xi\right)^{\frac{1}{Q}}$
(1.9)

for any fixed number $\tau>0$. Our main result is the following:

Theorem 1.1. Let $\tau$ be any positive real number. Let $Q$, $Q^{\prime}$ and
$\alpha_{Q}$ be as in (1.7). For any $\beta:0\leq\beta<Q$ and
$\alpha:0<\alpha<\alpha_{Q}(1-\beta/Q)$, there holds

$\sup_{\|u\|_{1,\tau}\leq
1}\int_{\mathbb{H}^{n}}\frac{1}{|\xi|_{h}^{\beta}}\left\\{e^{\alpha|u|^{Q\,^{\prime}}}-\sum_{k=0}^{Q-2}\frac{\alpha^{k}|u|^{kQ\,^{\prime}}}{k!}\right\\}d\xi<\infty.$
(1.10)

When $\alpha>\alpha_{Q}(1-\beta/Q)$, the above integral is still finite for
any $u\in W^{1,Q}(\mathbb{H}^{n})$, but the supremum is infinite.

Clearly Theorem 1.1 implies that the best constant for the inequality (1.10)
is $\alpha_{Q}(1-\beta/Q)$. But we do not know whether or not (1.10) still
holds when $\alpha=\alpha_{Q}(1-\beta/Q)$. Even so, $(\ref{Ttau})$ gives more
information than (1.8).

According to the author’s knowledge, the existing proofs of Trudinger-Moser
inequalities for unbounded domains are all based on the rearrangement theory
[6]. It is not known that whether or not this technique can be successfully
applied to the Heisenberg group case.

To prove Theorem 1.1, we propose a new approach. The idea can be described as
follows. Firstly, using (1.7), we derive a local Trudinger-Moser inequality,
namely, for any fixed $r>0$ and all $\xi_{0}\in\mathbb{H}^{n}$, there exists
some constant $C$ depending only on $n$, $r$ and $\beta$ such that

$\int_{B_{h}(\xi_{0},r)}\frac{1}{|\xi|_{h}^{\beta}}\left\\{e^{\alpha|u|^{Q\,^{\prime}}}-\sum_{k=0}^{Q-2}\frac{\alpha^{k}|u|^{kQ\,^{\prime}}}{k!}\right\\}d\xi\leq
C\int_{B_{h}(\xi_{0},r)}|\nabla_{\mathbb{H}^{n}}u|^{Q}d\xi$ (1.11)

provided that $0\leq\alpha<\alpha_{Q}(1-\beta/Q)$ and
$\int_{B_{h}(\xi_{0},r)}|\nabla_{\mathbb{H}^{n}}u|^{Q}d\xi\leq 1$. Secondly,
fixing sufficiently large $r>0$, we select a specific sequence of Heisenberg
balls $\\{B_{h}(\xi_{i},r)\\}_{i=1}^{\infty}$ to cover the Heisenberg group
$\mathbb{H}^{n}$. Then we choose appropriate cut-off function $\phi_{i}$ on
each $B_{h}(\xi_{i},r)$. Finally, we obtain (1.10) by gluing all local
estimates (1.11) for $\phi_{i}u$. We remark that our method still works for
similar problems when the Heisenberg group is replaced by the Eclidean space
or complete noncompact Riemannian manifolds. In the Eclidean space case,
$\tau$ can also be arbitrary in (1.10). But in the manifold case, the choice
of $\tau$ may depend on the geometric structure (see [13], Theorem 2.3). As an
easy consequence of Theorem 1.1 (in fact a special case $\beta=0$), the
following corollary holds.

Corollary 1.2. Let $Q=2n+2$. For any $q\geq Q$, $W^{1,Q}(\mathbb{H}^{n})$ is
continuously embedded in $L^{q}(\mathbb{H}^{n})$.

The remaining part of this note is organized as follows. In section 2, we
prove a covering lemma for $\mathbb{H}^{n}$; Cut-off functions are selected
for the subsequent analysis in section 3; The proof of Theorem 1.1 is
completed in section 4.

##  2 A covering lemma for the Heisenberg group

In this section, we will use a sequence of Heisenberg balls with the same
radius to cover the entire Heisenberg group $\mathbb{H}^{n}$. We require these
balls to satisfy the following properties: $(i)$ For any
$\xi\in\mathbb{H}^{n}$, $\xi$ belongs to at most $N$ balls for some constant
integer $N$ which is independent of the base point $\xi$; $(ii)$ If the radius
of those balls becomes appropriately smaller, then they are disjoint.

Firstly, we need to understand the Heisenberg distance between two points of
the Heisenberg group $\mathbb{H}^{n}$. The following two properties are more
or less standard. We prefer to present them by our own way.

Proposition 2.1. Let $\xi$ and $\eta$ be two points of $\mathbb{H}^{n}$. There
holds

$|\eta^{-1}\circ\xi|_{h}\leq 3(|\xi|_{h}+|\eta|_{h}),$

where $|\cdot|_{h}$ is the homogeneous norm defined by (1.2).

Proof. Write $\xi=(x,y,t)$, $\eta=(x^{\prime},y^{\prime},t^{\prime})$. Then
(1.1) gives

$\eta^{-1}\circ\xi=(x-x^{\prime},y-y^{\prime},t-t^{\prime}-2(\langle
y,x^{\prime}\rangle-\langle x,y^{\prime}\rangle)).$

Since
$(|x-x^{\prime}|^{2}+|y-y^{\prime}|^{2})^{1/2}\leq(|x|^{2}+|y|^{2})^{1/2}+(|x^{\prime}|^{2}+|y^{\prime}|^{2})^{1/2}$
and

$\left|2\left(\langle y,x^{\prime}\rangle-\langle
x,y^{\prime}\rangle\right)\right|\leq|x|^{2}+|y|^{2}+|x^{\prime}|^{2}+|y^{\prime}|^{2},$

we have by using the inequality $\sqrt{a+b}\leq\sqrt{a}+\sqrt{b}$ ($a\geq 0$,
$b\geq 0$) repeatedly

$\displaystyle|\eta^{-1}\circ\xi|_{h}$ $\displaystyle=$
$\displaystyle\left[\left(\sum_{i=1}^{n}\left((x_{i}-x_{i}^{\prime})^{2}+(y_{i}-y_{i}^{\prime})^{2}\right)\right)^{2}+(t-t^{\prime}-2\left(\langle
y,x^{\prime}\rangle-\langle
x,y^{\prime}\rangle)\right)^{2}\right]^{\frac{1}{4}}$ $\displaystyle\leq$
$\displaystyle\left(\sum_{i=1}^{n}\left((x_{i}-x_{i}^{\prime})^{2}+(y_{i}-y_{i}^{\prime})^{2}\right)\right)^{\frac{1}{2}}+\left|t-t^{\prime}-2\left(\langle
y,x^{\prime}\rangle-\langle x,y^{\prime}\rangle\right)\right|^{\frac{1}{2}}$
$\displaystyle\leq$ $\displaystyle
2\left(\sum_{i=1}^{n}\left(x_{i}^{2}+y_{i}^{2}\right)\right)^{\frac{1}{2}}+2\left(\sum_{i=1}^{n}\left({x_{i}^{\prime}}^{2}+{y_{i}^{\prime}}^{2}\right)\right)^{\frac{1}{2}}+|t|^{\frac{1}{2}}+|t^{\prime}|^{\frac{1}{2}}$
$\displaystyle\leq$ $\displaystyle 3(|\xi|_{h}+|\eta|_{h}).$

$\hfill\Box$

Proposition 2.2. Let $\xi$, $\eta$, $\zeta$ be arbitrary points of
$\mathbb{H}^{n}$. Then we have

$d_{h}(\xi,\eta)\leq 3\left(d_{h}(\xi,\zeta)+d_{h}(\zeta,\eta)\right),$

where $d_{h}(\cdot,\cdot)$ is the distance function defined by (1.3).

Proof. Note that $|\gamma^{-1}|_{h}=|\gamma|_{h}$ for all
$\gamma\in\mathbb{H}^{n}$. It follows from Proposition 2.1 that

$\displaystyle d_{h}(\xi,\eta)$ $\displaystyle=$
$\displaystyle|\eta^{-1}\circ\xi|_{h}$ $\displaystyle=$
$\displaystyle|\eta^{-1}\circ\zeta\circ\zeta^{-1}\circ\xi|_{h}$
$\displaystyle\leq$ $\displaystyle
3(|\eta^{-1}\circ\zeta|_{h}+|\zeta^{-1}\circ\xi|_{h})$ $\displaystyle=$
$\displaystyle 3\left(d_{h}(\zeta,\eta)+d_{h}(\xi,\zeta)\right).$

This gives the desired result. $\hfill\Box$

Secondly, by adapting an argument of (Hebey [7], Lemma 1.6), we obtain the
following useful covering lemma.

Lemma 2.3. Let $\rho>0$ be given. There exists a sequence $(\xi_{i})$ of
points of $\mathbb{H}^{n}$ such that for any $r\geq\rho$:
$(i)$ $\cup_{i}B_{h}(\xi_{i},\rho)=\mathbb{H}^{n}$ and for any $i\not=j$,
$B_{h}(\xi_{i},\rho/6)\cap B_{h}(\xi_{j},\rho/6)=\varnothing$;
$(ii)$ for any $\xi\in\mathbb{H}^{n}$, $\xi$ belongs to at most
$[(24r/\rho)^{Q}]$ balls $B_{h}(\xi_{i},r)$, where $[(24r/\rho)^{Q}]$ denotes
the integral part of $(24r/\rho)^{Q}$.

Proof. Firstly, we claim that there exists a sequence $(\xi_{i})$ of points of
$\mathbb{H}^{n}$ such that

$\cup_{i}B_{h}(\xi_{i},\rho)=\mathbb{H}^{n}\,\,{\rm and}\,\,\forall
i\not=j,B_{h}(\xi_{i},\rho/6)\cap B_{h}(\xi_{j},\rho/6)=\varnothing.$ (2.1)

To see this, we set

$X_{\rho}=\left\\{{\rm sequence}\,\,(\xi_{i})_{i\in
I}:\xi_{i}\in\mathbb{H}^{n},I\,\,{\rm is\,\,countable\,\,and}\,\,\forall
i\not=j,d_{h}(\xi_{i},\xi_{j})\geq\rho\right\\}.$

Then $X_{\rho}$ is partially ordered by inclusion and every element in
$X_{\rho}$ has an upper bound in the sense of inclusion. Hence, by Zorn’s
lemma, $X_{\rho}$ contains a maximal element $(\xi_{i})_{i\in I}$. On one
hand, if $\cup_{i}B_{h}(\xi_{i},\rho)\not=\mathbb{H}^{n}$, then there exists a
point $\xi\in\mathbb{H}^{n}$ such that $d_{h}(\xi_{i},\xi)\geq\rho$ for all
$i\in I$. This contradicts the maximality of $(\xi_{i})_{i\in I}$. Hence
$\cup_{i}B_{h}(\xi_{i},\rho)=\mathbb{H}^{n}$. On the other hand, if
$B_{h}(\xi_{i},\rho/6)\cap B_{h}(\xi_{j},\rho/6)\not=\varnothing$ for some
$i\not=j$, then we can take some $\eta\in B_{h}(\xi_{i},\rho/6)\cap
B_{h}(\xi_{j},\rho/6)$. It follows from Proposition 2.2 that

$\displaystyle d_{h}(\xi_{i},\xi_{j})$ $\displaystyle\leq$ $\displaystyle
3\left(d_{h}(\xi_{i},\eta)+d_{h}(\eta,\xi_{j})\right)$ $\displaystyle<$
$\displaystyle 3\left(\frac{\rho}{6}+\frac{\rho}{6}\right)=\rho.$

This contradicts the fact that $d_{h}(\xi_{i},\xi_{j})\geq\rho$ for any
$i\not=j$. Thus our claim (2.1) holds.

Assume $(\xi_{i})$ satisfies (2.1). For any fixed $r>0$ and
$\xi\in\mathbb{H}^{n}$ we set

$I_{r}(\xi)=\left\\{i\in I:\xi\in B_{h}(\xi_{i},r)\right\\}.$

By (1.4) and Proposition 2.2, we have for $r\geq\rho$

$\displaystyle|B_{h}(\xi,r)|$ $\displaystyle=$ $\displaystyle
4^{-Q}|B_{h}(\xi,4r)|$ $\displaystyle\geq$ $\displaystyle 4^{-Q}\sum_{i\in
I_{r}(\xi)}|B_{h}(\xi_{i},\rho/6)|$ $\displaystyle=$ $\displaystyle
4^{-Q}\,\,{\rm Card}\,\,I_{r}(\xi)\,\,({\rho}/{6})^{Q}|B_{h}(0,1)|,$

where ${\rm Card}\,\,I_{r}(\xi)$ denotes the cardinality of the set
$I_{r}(\xi)$. As a consequence, for $r\geq\rho$ there holds

${\rm Card}\,\,I_{r}(\xi)\leq(24r/\rho)^{Q}.$

This completes the proof of the lemma. $\hfill\Box$

##  3 Cut-off functions on Heisenberg balls

In this section, we will construct cut-off functions on Heisenberg balls. To
do this, we first estimate the gradient of the distance function as follows.

Lemma 3.1. Let $\xi_{0}$ be any fixed point of $\mathbb{H}^{n}$. Define a
function $\rho(\xi)=d_{h}(\xi,\xi_{0})$. Then we have
$|\nabla_{\mathbb{H}^{n}}\rho(\xi)|\leq 1$ for any $\xi\not=\xi_{0}$.

Proof. Write $\xi=(x_{1},\cdots,x_{n},y_{1},\cdots,y_{n},t)$ and
$\xi_{0}=(x_{01},\cdots,x_{0n},y_{01},\cdots,y_{0n},t_{0})$. For any
$\xi\not=\xi_{0}$, we set

$E=\sum_{i=1}^{n}\left((x_{i}-x_{0i})^{2}+(y_{i}-y_{0i})^{2}\right),\quad
F=t-t_{0}-2\sum_{i=1}^{n}(x_{i}y_{0i}-y_{i}x_{0i}).$

Then by (1.1) and (1.3),

$\rho(\xi)=|\xi_{0}^{-1}\circ\xi|_{h}=\left(E^{2}+F^{2}\right)^{1/4}.$

We calculate

$\displaystyle\frac{\partial}{\partial
x_{i}}\rho=\rho^{-3}\left((x_{i}-x_{0i})E-y_{0i}F\right),\quad
2y_{i}\frac{\partial}{\partial t}\rho=\rho^{-3}y_{i}F,$

and then by (1.5),

$X_{i}\rho=\frac{\partial}{\partial x_{i}}\rho+2y_{i}\frac{\partial}{\partial
t}\rho=\rho^{-3}\left((x_{i}-x_{0i})E+(y_{i}-y_{0i})F\right).$

Similarly we have

$\frac{\partial}{\partial
y_{i}}\rho=\rho^{-3}\left((y_{i}-y_{0i})E+x_{0i}F\right)$

and thus by (1.5),

$Y_{i}\rho=\frac{\partial}{\partial y_{i}}\rho-2x_{i}\frac{\partial}{\partial
t}\rho=\rho^{-3}\left((y_{i}-y_{0i})E+(x_{0i}-x_{i})F\right).$

It follows that

$(X_{i}\rho)^{2}+(Y_{i}\rho)^{2}=\rho^{-6}\left((y_{i}-y_{0i})^{2}+(x_{i}-x_{0i})^{2}\right)(E^{2}+F^{2}).$

Note that $E^{2}+F^{2}=\rho^{4}$. We obtain

$\displaystyle|\nabla_{\mathbb{H}^{n}}\rho|$ $\displaystyle=$
$\displaystyle\left(\sum_{i=1}^{n}\left((X_{i}\rho)^{2}+(Y_{i}\rho)^{2}\right)\right)^{1/2}$
$\displaystyle=$ $\displaystyle\rho^{-3}E^{1/2}(E^{2}+F^{2})^{1/2}$
$\displaystyle=$ $\displaystyle\rho^{-1}E^{1/2}\leq 1.$

This completes the proof of the lemma. $\hfill\Box$

Now we construct cut-off functions. Let $\phi:\mathbb{R}\rightarrow\mathbb{R}$
be a smooth function such that $0\leq\phi\leq 1$, $\phi\equiv 1$ on the
interval $[-1,1]$, $\phi\equiv 0$ on $(-\infty,-2)\cup(2,\infty)$, and
$|\phi^{\prime}(t)|\leq 2$ for all $t\in\mathbb{R}$. Let $r>0$ be given.
Define a function on $\mathbb{H}^{n}$ by

$\phi_{0}(\xi)=\phi\left(\frac{d_{h}(\xi,\xi_{0})}{r}\right).$ (3.1)

Then $\phi_{0}$ is a cut-off function supported on the Heisenberg ball
$B_{h}(\xi_{0},2r)$. The estimate of the gradient of $\phi_{0}$ is very
important for the subsequent analysis. Precisely we have the following:

Lemma 3.2. For any fixed $r>0$ and $\xi_{0}\in\mathbb{H}^{n}$, let $\phi_{0}$
be defined by (3.1). Then $\phi_{0}$ is supported in $B_{h}(\xi_{0},2r)$,
$0\leq\phi_{0}\leq 1$, $\phi_{0}\equiv 1$ on $B_{h}(\xi_{0},r)$, and
$|\nabla_{\mathbb{H}^{n}}\phi_{0}(\xi)|\leq 2/r$ for all
$\xi\in\mathbb{H}^{n}$.

Proof. We only need to explain the last assertion, namely
$|\nabla_{\mathbb{H}^{n}}\phi_{0}(\xi)|\leq 2/r$ for all
$\xi\in\mathbb{H}^{n}$. Since $\phi_{0}\equiv 1$ on $B_{h}(\xi_{0},r)$, we
have $\nabla_{\mathbb{H}^{n}}\phi_{0}\equiv 0$ on $B_{h}(\xi_{0},r)$,
particularly $\nabla_{\mathbb{H}^{n}}\phi_{0}(0)=0$. For $\xi\not=\xi_{0}$, a
simple calculation shows

$\nabla_{\mathbb{H}^{n}}\phi_{0}(\xi)=\frac{1}{r}\phi^{\prime}\nabla_{\mathbb{H}^{n}}d_{h}(\xi,\xi_{0}).$

This together with Lemma 3.1 and $|\phi^{\prime}|\leq 2$ concludes the last
assertion. $\hfill\Box$

##  4 Proof of Theorem 1.1

In this section, we will prove Theorem 1.1. For simplicity, we define a smooth
function $\zeta:\mathbb{N}\times\mathbb{R}\rightarrow\mathbb{R}$ by

$\zeta(m,s)=e^{s}-\sum_{k=0}^{m-2}\frac{s^{k}}{k!},\quad\forall m\geq 2.$
(4.1)

As we promised in the introduction, we first derive a local Trudinger-Moser
inequality for the Heisenberg group $\mathbb{H}^{n}$ by using (1.7). Let $Q$,
$Q^{\prime}$ and $\alpha_{Q}$ be given by (1.7). Then we have the following:

Lemma 4.1. Let $r>0$ be given and $\xi_{0}$ be any point of $\mathbb{H}^{n}$.
If $0\leq\beta<Q$, $0\leq\alpha\leq\alpha_{Q}(1-\beta/Q)$, and $w\in
W_{0}^{1,Q}(B_{h}(\xi_{0},r))$ satisfies
$\int_{B_{h}(\xi_{0},r)}|\nabla_{\mathbb{H}^{n}}w|^{Q}d\xi\leq 1$, then there
exists some constant $C$ depending only on $n$, $r$ and $\beta$ such that

$\int_{B_{h}(\xi_{0},r)}\frac{1}{|\xi|_{h}^{\beta}}\zeta(Q,\alpha|w|^{Q^{\prime}})d\xi\leq
C\int_{B_{h}(\xi_{0},r)}|\nabla_{\mathbb{H}^{n}}w|^{Q}d\xi.$ (4.2)

Proof. Using Proposition 2.2, we have that

$|\xi_{0}|_{h}\leq
3(d_{h}(\xi,\xi_{0})+|\xi|_{h}),\quad\forall\xi\in\mathbb{H}.$

If $|\xi_{0}|_{h}>6r$, then for any $\xi\in B_{h}(\xi_{0},r)$ there holds

$|\xi|_{h}\geq\frac{|\xi_{0}|_{h}}{3}-d_{h}(\xi,\xi_{0})>r.$ (4.3)

Let
$\widetilde{w}=w/\|\nabla_{\mathbb{H}^{n}}w\|_{L^{Q}({B}_{h}(\xi_{0},r))}$.
Since $\|\nabla_{\mathbb{H}^{n}}w\|_{L^{Q}({B}_{h}(\xi_{0},r))}\leq 1$ and
$0\leq\alpha\leq\alpha_{Q}(1-\beta/Q)$, we have

$\displaystyle\zeta\left(Q,\alpha|w|^{Q^{\prime}}\right)$ $\displaystyle=$
$\displaystyle\sum_{k=Q-1}^{\infty}\frac{\alpha^{k}|w|^{Q^{\prime}k}}{k!}{}$
(4.4) $\displaystyle=$
$\displaystyle\sum_{k=Q-1}^{\infty}\frac{\alpha^{k}\|\nabla_{\mathbb{H}^{n}}w\|_{L^{Q}({B}_{h}(\xi_{0},r))}^{Q^{\prime}k}|\widetilde{w}|^{Q^{\prime}k}}{k!}{}$
$\displaystyle\leq$
$\displaystyle\|\nabla_{\mathbb{H}^{n}}w\|_{L^{Q}({B}_{h}(\xi_{0},r))}^{Q}\zeta\left(Q,\alpha|\widetilde{w}|^{Q^{\prime}}\right).$

By (1.4) and (1.7),

$\int_{B_{h}(\xi_{0},r)}\zeta\left(Q,\alpha_{Q}|\widetilde{w}|^{Q^{\prime}}\right)d\xi\leq
C_{n}r^{Q}|B_{h}(0,1)|,$

where $C_{n}$ is given by (1.7). Hence when $|\xi_{0}|_{h}>6r$ and
$0\leq\alpha\leq\alpha_{Q}(1-\beta/Q)$, we have by using (4.3) and (4.4),

$\displaystyle\int_{B_{h}(\xi_{0},r)}\frac{1}{|\xi|_{h}^{\beta}}\zeta\left(Q,\alpha|w|^{Q^{\prime}}\right)d\xi$
$\displaystyle\leq$ $\displaystyle
r^{-\beta}\int_{B_{h}(\xi_{0},r)}\zeta\left(Q,\alpha|w|^{Q^{\prime}}\right)d\xi$
$\displaystyle\leq$ $\displaystyle
C_{n}r^{Q-\beta}|B_{h}(0,1)|\int_{B_{h}(\xi_{0},r)}|\nabla_{\mathbb{H}^{n}}w|^{Q}d\xi.$

In the following we assume $|\xi_{0}|_{h}\leq 6r$. If $\xi\in
B_{h}(\xi_{0},r)$, then Proposition 2.2 implies that

$|\xi|_{h}\leq 3(d_{h}(\xi,\xi_{0})+|\xi_{0}|_{h})<21r.$

Hölder’s inequality together with (1.7) implies that there exits some constant
$\widetilde{C}$ depending only on $n$, $r$ and $\beta$ such that

$\int_{B_{h}(\xi_{0},r)}\frac{1}{|\xi|_{h}^{\beta}}\zeta\left(Q,\alpha|\widetilde{w}|^{Q^{\prime}}\right)d\xi\leq\int_{|\xi|_{h}\leq
21r}\frac{1}{|\xi|_{h}^{\beta}}\zeta\left(Q,\alpha|\widetilde{w}|^{Q^{\prime}}\right)d\xi\leq\widetilde{C}.$

It then follows from (4.4) that

$\displaystyle\int_{B_{h}(\xi_{0},r)}\frac{1}{|\xi|_{h}^{\beta}}\zeta\left(Q,\alpha|w|^{Q^{\prime}}\right)d\xi$
$\displaystyle\leq$
$\displaystyle\|\nabla_{\mathbb{H}^{n}}w\|_{L^{Q}({B}_{h}(\xi_{0},r))}^{Q}\int_{B_{h}(\xi_{0},r)}\frac{1}{|\xi|_{h}^{\beta}}\zeta\left(Q,\alpha|\widetilde{w}|^{Q^{\prime}}\right)d\xi$
$\displaystyle\leq$
$\displaystyle\widetilde{C}\int_{B_{h}(\xi_{0},r)}|\nabla_{\mathbb{H}^{n}}w|^{Q}d\xi.$

Hence (4.2) holds. $\hfill\Box$

Proof of Theorem 1.1. Firstly, we prove (1.10). Let $\tau>0$ and
$\alpha:0\leq\alpha<\alpha_{Q}(1-\beta/Q)$ be fixed. Since
$C_{0}^{\infty}(\mathbb{H}^{n})$ is dense in $W^{1,Q}(\mathbb{H}^{n})$ under
the norm (1.9), it suffices to prove (1.10) for all $u\in
C_{0}^{\infty}(\mathbb{H}^{n})$ with

$\int_{\mathbb{H}^{n}}(|\nabla_{\mathbb{H}^{n}}u|^{Q}+\tau|u|^{Q})d\xi\leq 1.$
(4.5)

Assume $u\in C^{\infty}(\mathbb{H}^{n})$ satisfies (4.5). Let $r>0$ be a
sufficiently large number to be determined later. By Lemma 2.3, there exists a
sequence $(\xi_{i})$ of points of $\mathbb{H}^{n}$ such that

$\cup_{i}B_{h}(\xi_{i},r)=\mathbb{H}^{n}\,\,{\rm and}\,\,\forall
i\not=j,\,\,B_{h}(\xi_{i},r/6)\cap B_{h}(\xi_{j},r/6)=\varnothing,$ (4.6)

and for any $\xi\in\mathbb{H}^{n}$,

$\xi\,\,{\rm belongs\,\,to\,\,at\,\,most}\,\,48^{Q}\,\,{\rm
balls}\,\,B_{h}(\xi_{i},2r).$ (4.7)

Let $\phi$ be a smooth function given by (3.1). For each $\xi_{i}$, we set

$\phi_{i}(\xi)=\phi\left(\frac{d_{h}(\xi,\xi_{i})}{r}\right),\quad\forall\xi\in\mathbb{H}^{n}.$

It follows from Lemma 3.2 that $0\leq\phi_{i}\leq 1$, $\phi_{i}\equiv 1$ on
$B_{h}(\xi_{i},r)$, $\phi_{i}\equiv 0$ outside $B_{h}(\xi_{i},2r)$, and

$|\nabla_{\mathbb{H}^{n}}\phi_{i}(\xi)|\leq\frac{2}{r},\,\,\forall\xi\in\mathbb{H}^{n}.$
(4.8)

Clearly $\phi_{i}^{2}u\in W_{0}^{1,Q}\left(B_{h}(\xi_{i},2r)\right)$. Since
$u$ satisfies (4.5), we have that

$\int_{\mathbb{H}^{n}}|\nabla_{\mathbb{H}^{n}}u|^{Q}d\xi\leq 1,\,\,{\rm
and}\,\,\int_{\mathbb{H}^{n}}|u|^{Q}d\xi\leq\frac{1}{\tau}.$

Minkowski inequality together with (4.8) and $0\leq\phi_{i}\leq 1$ leads to

$\displaystyle\left(\int_{B_{h}(\xi_{i},2r)}|\nabla_{\mathbb{H}^{n}}(\phi_{i}^{2}u)|^{Q}d\xi\right)^{1/Q}$
$\displaystyle\leq$
$\displaystyle\left(\int_{B_{h}(\xi_{i},2r)}\phi_{i}^{2Q}|\nabla_{\mathbb{H}^{n}}u|^{Q}d\xi\right)^{1/Q}+\left(\int_{B_{h}(\xi_{i},2r)}|\nabla_{\mathbb{H}^{n}}\phi_{i}^{2}|^{Q}|u|^{Q}d\xi\right)^{1/Q}{}$
(4.9) $\displaystyle\leq$
$\displaystyle\left(\int_{B_{h}(\xi_{i},2r)}|\nabla_{\mathbb{H}^{n}}u|^{Q}d\xi\right)^{1/Q}+\frac{4}{r}\left(\int_{B_{h}(\xi_{i},2r)}|u|^{Q}d\xi\right)^{1/Q}{}$
$\displaystyle\leq$ $\displaystyle 1+\frac{4}{\tau r}.$

Define $\widetilde{u}_{i}={\phi_{i}^{2}u}/(1+\frac{4}{\tau r})$. Then
$\widetilde{u}_{i}\in W_{0}^{1,Q}(B_{h}(\xi_{i},2r))$ and
$\int_{B_{h}(\xi_{i},2r)}|\nabla_{\mathbb{H}^{n}}\widetilde{u}_{i}|^{Q}d\xi\leq
1$. Since $\alpha<\alpha_{Q}(1-\beta/Q)$, we can select $r$ sufficiently large
such that

$\alpha\left(1+\frac{4}{\tau r}\right)^{Q^{\prime}}<\alpha_{Q}(1-\beta/Q).$

This together with Lemma 4.1 implies that there exists some constant $C$
depending only on $n$, $r$ and $\beta$ such that

$\displaystyle\int_{B_{h}(\xi_{i},2r)}\frac{1}{|\xi|_{h}^{\beta}}\zeta\left(Q,\alpha|\phi_{i}^{2}u|^{Q^{\prime}}\right)d\xi$
$\displaystyle=$
$\displaystyle\int_{B_{h}(\xi_{i},2r)}\frac{1}{|\xi|_{h}^{\beta}}\zeta\left(Q,\alpha\left(1+\frac{4}{\tau
r}\right)^{Q^{\prime}}|\widetilde{u}_{i}|^{Q^{\prime}}\right)d\xi{}$ (4.10)
$\displaystyle\leq$ $\displaystyle
C\int_{B_{h}(\xi_{i},2r)}|\nabla_{\mathbb{H}^{n}}\widetilde{u}_{i}|^{Q}d\xi.$
$\displaystyle\leq$ $\displaystyle
C\int_{B_{h}(\xi_{i},2r)}|\nabla_{\mathbb{H}^{n}}(\phi_{i}^{2}u)|^{Q}d\xi.$

Combining (4.6) and (4.10), we obtain

$\displaystyle\int_{\mathbb{H}^{n}}\frac{1}{|\xi|_{h}^{\beta}}\zeta\left(Q,\alpha|u|^{Q^{\prime}}\right)d\xi$
$\displaystyle\leq$
$\displaystyle\sum_{i}\int_{B_{h}(\xi_{i},r)}\frac{1}{|\xi|_{h}^{\beta}}\zeta\left(Q,\alpha|\phi_{i}^{2}u|^{Q^{\prime}}\right)d\xi{}$
(4.11) $\displaystyle\leq$
$\displaystyle\sum_{i}\int_{B_{h}(\xi_{i},2r)}\frac{1}{|\xi|_{h}^{\beta}}\zeta\left(Q,\alpha|\phi_{i}^{2}u|^{Q^{\prime}}\right)d\xi$
$\displaystyle\leq$ $\displaystyle
C\sum_{i}\int_{\mathbb{H}^{n}}|\nabla(\phi_{i}^{2}u)|^{Q}d\xi.$

Using the inequality $|a+b|^{Q}\leq 2^{Q}|a|^{Q}+2^{Q}|b|^{Q}$, $\forall
a,b\in\mathbb{R}$, $0\leq\phi_{i}\leq 1$ and (4.8), we get

$\displaystyle\int_{\mathbb{H}^{n}}|\nabla_{\mathbb{H}^{n}}(\phi_{i}^{2}u)|^{Q}d\xi$
$\displaystyle\leq$ $\displaystyle
2^{Q}\int_{\mathbb{H}^{n}}\left(\phi_{i}^{2Q}|\nabla_{\mathbb{H}^{n}}u|^{Q}+|\nabla_{\mathbb{H}^{n}}\phi_{i}^{2}|^{Q}|u|^{Q}\right)d\xi{}$
$\displaystyle\leq$ $\displaystyle
2^{Q}\int_{\mathbb{H}^{n}}\phi_{i}|\nabla_{\mathbb{H}^{n}}u|^{Q}d\xi+\left(\frac{8}{r}\right)^{Q}\int_{\mathbb{H}^{n}}\phi_{i}|u|^{Q}d\xi.$

In view of (4.7), it then follows that

$\displaystyle\sum_{i}\int_{\mathbb{H}^{n}}|\nabla_{\mathbb{H}^{n}}(\phi_{i}^{2}u)|^{Q}d\xi$
$\displaystyle\leq$ $\displaystyle
2^{Q}\sum_{i}\int_{\mathbb{H}^{n}}\phi_{i}|\nabla_{\mathbb{H}^{n}}u|^{Q}d\xi+\left(\frac{8}{r}\right)^{Q}\sum_{i}\int_{\mathbb{H}^{n}}\phi_{i}|u|^{Q}d\xi$
$\displaystyle\leq$ $\displaystyle
96^{Q}\int_{\mathbb{H}^{n}}|\nabla_{\mathbb{H}^{n}}u|^{Q}d\xi+\left(\frac{384}{r}\right)^{Q}\int_{\mathbb{H}^{n}}|u|^{Q}d\xi.$

This together with (4.11) implies

$\int_{\mathbb{H}^{n}}\frac{1}{|\xi|^{\beta}}\zeta\left(Q,\alpha|u|^{Q^{\prime}}\right)d\xi\leq\widetilde{C}$

for some constant $\widetilde{C}$ depending only on $C$, $Q$, and $r$. Hence
we conclude (1.10).

Secondly, we prove that for any fixed $\beta:0\leq\beta<Q$, $\alpha>0$, and
$u\in W^{1,Q}(\mathbb{H}^{n})$, there holds

$\int_{\mathbb{H}^{n}}\frac{1}{|\xi|_{h}^{\beta}}\zeta\left(Q,\alpha|u|^{Q^{\prime}}\right)d\xi<\infty.$
(4.12)

Since $C_{0}^{\infty}(\mathbb{H}^{n})$ is dense in $W^{1,Q}(\mathbb{H}^{n})$,
we can take some $u_{0}\in C_{0}^{\infty}(\mathbb{H}^{n})$ such that
$\|u-u_{0}\|_{W^{1,Q}(\mathbb{H}^{n})}<\epsilon$, where $\epsilon>0$ is a
small number to be determined later. Set

$w=\frac{u-u_{0}}{\|u-u_{0}\|_{W^{1,Q}(\mathbb{H}^{n})}}.$

Then $\|w\|_{W^{1,Q}(\mathbb{H}^{n})}=1$. We divide the proof of (4.12) into
two cases:

Case $1$. $\beta=0$.

Recall (4.1). By ([12], Lemma 2.2), $\zeta(Q,t)$ is convex with respect to
$t$. Since
$|a+b|^{\gamma}\leq(1+\delta)|a|^{\gamma}+C(\delta,\gamma)|b|^{\gamma}$,
$\forall a,b\in\mathbb{R},\gamma\geq 1,\delta>0$, for some constant
$C(\delta,\gamma)$ depending only on $\delta$ and $\gamma$, we obtain

$\displaystyle\int_{\mathbb{H}^{n}}\zeta\left(Q,\alpha|u|^{Q^{\prime}}\right)d\xi$
$\displaystyle=$
$\displaystyle\int_{\mathbb{H}^{n}}\zeta\left(Q,\alpha|u-u_{0}+u_{0}|^{Q^{\prime}}\right)d\xi$
$\displaystyle\leq$
$\displaystyle\int_{\mathbb{H}^{n}}\zeta\left(Q,\alpha(1+\delta)|u-u_{0}|^{Q^{\prime}}+\alpha
C(\delta,Q^{\prime})|u_{0}|^{Q^{\prime}}\right)d\xi$ $\displaystyle\leq$
$\displaystyle\frac{1}{\mu}\int_{\mathbb{H}^{n}}\zeta\left(Q,\mu\alpha(1+\delta)|u-u_{0}|^{Q^{\prime}})\right)d\xi+\frac{1}{\nu}\int_{\mathbb{H}^{n}}\zeta\left(Q,\nu\alpha
C(\delta,Q^{\prime})|u_{0}|^{Q^{\prime}}\right)d\xi$ $\displaystyle\leq$
$\displaystyle\frac{1}{\mu}\int_{\mathbb{H}^{n}}\zeta\left(Q,\mu\alpha(1+\delta)\epsilon^{Q^{\prime}}|w|^{Q^{\prime}})\right)d\xi+\frac{1}{\nu}\int_{\mathbb{H}^{n}}\zeta\left(Q,\nu\alpha
C(\delta,Q^{\prime})|u_{0}|^{Q^{\prime}}\right)d\xi,$

where ${1}/{\mu}+1/\nu=1$, $\mu>1$, $\nu>1$. Now we choose $\epsilon>0$
sufficiently small such that
$\mu\alpha(1+\delta)\epsilon^{Q^{\prime}}<\alpha_{Q}$. By (1.10), there holds

$\int_{\mathbb{H}^{n}}\zeta\left(Q,\mu\alpha(1+\delta)\epsilon^{Q^{\prime}}|w|^{Q^{\prime}})\right)d\xi\leq
C_{1}$

for some constant $C_{1}$ depending only on $n$ and $\tau$. In addition, since
$u_{0}\in C_{0}^{\infty}(\mathbb{H}^{n})$, it is obvious that

$\int_{\mathbb{H}^{n}}\zeta\left(Q,\nu\alpha
C(\delta,Q^{\prime})|u_{0}|^{Q^{\prime}}\right)d\xi<\infty.$

Therefore, we have

$\int_{\mathbb{H}^{n}}\zeta\left(Q,\alpha|u|^{Q^{\prime}}\right)d\xi<\infty.$

Case $2$. $0<\beta<Q$.

Note that

$\displaystyle\int_{\mathbb{H}^{n}}\frac{1}{|\xi|_{h}^{\beta}}\zeta\left(Q,\alpha|u|^{Q^{\prime}}\right)d\xi\leq\int_{|\xi|_{h}\leq
1}\frac{1}{|\xi|_{h}^{\beta}}\zeta\left(Q,\alpha|u|^{Q^{\prime}}\right)d\xi+\int_{\mathbb{H}^{n}}\zeta\left(Q,\alpha|u|^{Q^{\prime}}\right)d\xi.$

This together with Hölder’s inequality and Case $1$ implies (4.12).

Finally, we confirm that for any $\alpha>\alpha_{Q}(1-\beta/Q)$, there holds

$\sup_{\|u\|_{1,\tau}\leq
1}\int_{\mathbb{H}^{n}}\frac{1}{|\xi|_{h}^{\beta}}\zeta\left(Q,\alpha|u|^{Q^{\prime}}\right)d\xi=\infty.$

This is based on calculations of related integrals of the Moser function
sequence. We omit the details but refer the reader to [3]. $\hfill\Box$

Acknowledgements. This work was partly supported by the NSFC 11171347 and the
NCET program 2008-2011.

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