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

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