## Application of inverse mean curvature to a geometric inequality

In this note I am going to use inverse mean curvature flow to prove an inequality for star-shaped mean convex surface in the Euclidean space.

 Theorem 1 Let ${\Sigma}$ be a closed embedded smooth hypersurface in ${\mathbb R^n}$ which is star-shaped with respect to ${O}$ and has positive mean curvature ${H}$, then $\displaystyle n \mathrm{Vol}(\Omega)\le \frac 1{n-1}\int_\Sigma Hr^2d \mu. \ \ \ \ \ (1)$ Here ${\Omega}$ is the region enclosed by ${\Sigma }$ and ${r=|x|}$ is the distance from ${O}$. The equality holds if and only if ${\Sigma}$ is a sphere.

This result generalizes a result in my previous post. I would like to thank Prof. Pengzi Miao for ideas.

0.1. Inverse mean curvature flow

In this section we will fix a closed embedded smooth hypersurface ${\Sigma}$ in ${\mathbb R^n}$ given by the embedding ${X_0: \mathbb S^{n-1}\rightarrow \mathbb R^n}$ which satisfies the assumptions in Theorem 1. The corresponding inverse mean curvature flow is then defined by the following equation:

$\displaystyle \begin{cases} \frac{\partial X}{\partial t}= \frac 1H \nu\\ X(\cdot, 0)= X_0 \end{cases} \ \ \ \ \ (2)$

where ${\nu}$ is the unit outward normal of ${X(\cdot ,t)}$. The hypersurface parametrized by ${X(\cdot, t)}$ will be denoted by ${\Sigma_t}$ and the region enclosed by ${\Sigma_t}$ will be denoted by ${\Omega_t}$.

By the results of [Gerhardt] and [Urbas], if the embedding ${X_0}$ of ${\Sigma}$ satisfies the assumptions in Theorem 1, then there exists a unique solution to (2) for all ${t\in [0, \infty)}$. Moreover, the rescaled hypersurfaces ${\widetilde X= e^{-\frac t {n-1}}X}$ converge exponentially fast to a sphere in ${C^\infty}$ sense.

Recall that ${r=|x|}$. The key idea to prove Theorem 1 is to observe that the following quantity is monotone:

$\displaystyle Q (t) = e^{ - \frac{n-2}{n-1} t} \left[ n \mathrm{Vol}(\Omega_t) - \frac{1}{n-1} \int_{\Sigma_t} r^2 H d \mu \right].$

 Proposition 2 Let ${\Sigma_t}$ be the family of smooth surfaces evolving according to (2), then $\displaystyle H' =-\Delta (\frac 1H ) -\frac {|A|^2}H.$

Proof: We will use $\overline \nabla$ to denote the connection in $\mathbb R^n$. Let ${k=H^{-1}}$, then

$\displaystyle \begin{array}{rcl} g_{ij}'= \langle \overline \nabla _t \partial _i , \partial _j\rangle+ \langle \partial _i, \overline \nabla _t \partial _j\rangle\ = \langle \overline \nabla _ i \partial _t, \partial_j \rangle+ \langle \partial _i ,\overline \nabla _j \partial _t\rangle &=&k(\langle \overline \nabla _i \nu, \partial_j \rangle+\langle \partial_i ,\overline \nabla _j \nu\rangle)\\ &=& 2k h_{ij}. \end{array}$

Therefore ${\frac{\partial g^{ij}}{\partial t}= -2 g^{ik}g^{jl}\frac{\partial g_{kl}}{\partial t}= -2k h^{ij}}$. As ${\overline \nabla _t \nu}$ is tangential, by using normal coordinates we have

$\displaystyle \begin{array}{rcl} \frac{\partial h_{ij}}{\partial t} = -\overline \nabla_t \langle \nu, \overline \nabla _i \partial _j\rangle = - \langle \overline \nabla_t\nu, \overline \nabla _i \partial_j\rangle- \langle \nu, \overline \nabla_t\overline \nabla _i \partial _j\rangle &=& - \langle \nu, \overline \nabla _i \overline \nabla_t\partial _j\rangle\\ &=& - \langle \nu, \overline \nabla _i \overline \nabla_j\partial _t\rangle\\ &=& - \nabla _i\nabla _j k \langle \nu, \nu\rangle- k \langle \nu,\overline \nabla _i \overline \nabla _j \nu\rangle\\ &=& - \nabla _i\nabla _j k +k \langle \overline \nabla _i\nu, \overline \nabla _j \nu\rangle. \end{array}$

Therefore

$\displaystyle H' = -\frac 2H |A|^2 -\Delta (\frac 1H) +\frac 1H |A|^2=-\Delta (\frac 1H ) -\frac {|A|^2}H.$

$\Box$

 Proposition 3 Let ${\Sigma_t}$ be the family of smooth surfaces evolving according to (2), then $\displaystyle \left[e^{-\frac{n-2}{n-1}t} \left(n(n-1)\mathrm{Vol}(\Omega_t )- \int_{\Sigma_t} Hr^2\right)\right]'\ge 0.$

Proof: Let ${f= r^2}$, then

$\displaystyle \overline \nabla ^2 f = f''g=2 g.$

We have

$\displaystyle f'= \langle x,x\rangle'= 2\langle x, \frac \nu H\rangle.$

So

$\displaystyle \int_\Sigma f'H d\mu = 2 \int_\Sigma \langle x,\nu\rangle= 2 \int_\Omega \overline {\mathrm{div}}(x)=2n \mathrm{Vol}(\Omega).$

By proposition 2

$\displaystyle \int_\Sigma f H' = \int_\Sigma - \frac{\Delta f }{H}- \frac{f|A|^2 }H.$

Consider

$\displaystyle \Delta f = \overline \Delta f - H \frac{\partial f }{\partial \nu}- \overline \nabla ^2 f (\nu,\nu)= 2n-2 -H\frac{\partial f}{\partial \nu}.$

So

$\displaystyle \begin{array}{rcl} \int_\Sigma f H' & = &\int_\Sigma -\frac {(2n-2)}H + \frac{\partial f}{\partial \nu} - \frac{ f |A|^2 }H\\ &= &-(2n-2)\int_\Sigma \frac {1}H +2n \mathrm{Vol}(\Omega)-\int_\Sigma\frac{ f |A|^2 }H. \end{array}$

As ${\frac{\partial }{\partial t}d\mu= d\mu}$, we have

$\displaystyle \begin{array}{rcl} (\int_\Sigma fH)' &=& \int_\Sigma f'H + fH' + fH\\ &=& 4n \mathrm{Vol}(\Omega ) -2(n-1) \int_\Sigma \frac 1H - \int_\Sigma \frac{f|A|^2} H + \int_\Sigma fH\\ &\le & 4n \mathrm{Vol}(\Omega ) -2(n-1) \int_\Sigma \frac 1H - \frac{1 }{n-1}\int_\Sigma fH + \int_\Sigma fH\\ &\le & 4n \mathrm{Vol}(\Omega ) -2(n-1) \int_\Sigma \frac 1H +\frac{n-2 }{n-1}\int_\Sigma fH\\ &\le & 4n \mathrm{Vol}(\Omega ) -2n \mathrm{Vol}(\Omega)+\frac{n-2 }{n-1}\int_\Sigma fH\\ &=& 2n \mathrm{Vol}(\Omega ) +\frac{n-2 }{n-1}\int_\Sigma fH. \end{array}$

On the other hand, by [Ros] Theorem 1, we have

$\displaystyle \frac{d}{dt}\mathrm{Vol}(\Omega _t) = \int_\Sigma \frac 1H \ge \frac n{n-1}\mathrm{Vol}(\Omega_t).$

We deduce that

$\displaystyle \left(n(n-1)\mathrm{Vol}(\Omega _t) - \int_\Sigma fH\right)' \ge \frac{n-2}{n-1}\left(n(n-1) \mathrm{Vol(\Omega_t)}- \int_\Sigma fH\right).$

So

$\displaystyle \left[e^{-\frac{n-2}{n-1}t} \left(n(n-1)\mathrm{Vol}(\Omega_t )- \int_{\Sigma_t} Hr^2\right)\right]'\ge 0.$

$\Box$

 Proposition 4 Let ${Q(t)=e^{-\frac{n-2}{n-1}t} \left(n(n-1)\mathrm{Vol}(\Omega_t )- \int_{\Sigma_t} Hr^2\right)}$, then ${\lim_{t\rightarrow \infty}Q(t)\le 0}$.

Proof: First of all, by Proposition 3, ${Q(t)}$ is increasing. In particular, ${\lim_{t\rightarrow \infty}Q(t)}$ exists (which can be ${\infty}$).

On the other hand, by the result of [Urbas], the rescaled surface ${\widetilde \Sigma_t}$ given by ${\widetilde x= e^{-\frac t{n-1}}x}$ converges exponentially fast in ${C^\infty}$ sense to a sphere. In particular, ${\widetilde \Sigma_t}$, and hence ${\Sigma_t}$, must be convex (i.e. second fundamental form being positive-definite) for large enough ${t}$. By Lemma 3.4 of this paper, for such ${t}$, we have

$\displaystyle n(n-1)\mathrm{Vol}(\Omega _t) \le \int_{\Sigma_t} Hr^2. \ \ \ \ \ (3)$

We conclude that ${\lim_{t\rightarrow \infty}Q(t)<\infty}$, but then by (3) again, we must have ${\lim_{t\rightarrow \infty }Q(t)\le 0}$. $\Box$

0.2. Proof of the main result

Proof: The proof of Theorem 1 is now straightforward. Indeed the (1) holds if and only if ${Q(0)\le 0}$, but this follows from the monotonicity property of ${Q(t)}$ by Proposition 3 and Proposition 4.

If the equality case holds, then clearly ${Q(t)}$ must be constant and so by Proposition 3 ${\Sigma_t}$ must be a sphere for all ${t}$, in particular ${\Sigma}$ must be a sphere. The converse is easy to verify.

$\Box$