# How to evolve a star-shaped mean convex set to a strictly mean convex set?

I have trouble in going through a proof in a paper for quite a while. To simplify notation and make it readable to a larger audience, let me just present the simplest case:

Let $\Omega$ be a bounded smooth star-shaped domain in $\mathbb{R}^3$ with $H \ge 0$ everywhere on $\partial \Omega$. Here $H$ is the mean curvature defined on $\partial \Omega$. The authors claim one can easily perturb $\Omega$ in the following way:

There exist smooth star-shaped domain $\Omega_i$ with mean curvature $H_i>0$ everywhere on $\partial \Omega_i$, such that $\Omega_i \rightarrow \Omega$ in $L^1$, and $A(\partial \Omega_i) \rightarrow A(\partial \Omega)$ as $i \rightarrow \infty$, where $A$ is the surface area function.

The authors didn't give a proof. Is it obvious? I just can't see it. Can anyone here give me some hint? Like how to perturb a mean convex domain to get strictly mean convex condition?

My try is just consider $\partial \Omega$ as a graph $(x, u(x))$ and write down the mean curvature formula, but it's complicated and I've no idea how to perturb $u$ to get mean curvature strictly positive. The other way I tried is to choose $S^2$ as reference manifold, but it's still very complicated. I got stuck...

By the way, I've seen this argument in many places without proof. For example, people always claim one can perturb convex set to smooth strictly convex set, so forth. I've no idea how to do that without a proof. Anyway my question here is mainly about perturbing mean convex set.

Represent $\partial \Omega$ as an embedding $F:\mathbb S^2 \to \mathbb R^3$ and consider the mean curvature flow $\{F_t :\mathbb S^2 \to \mathbb R^3: t\in [0,T)\}$, which is a family of embeddings so that

$$\frac{\partial F}{\partial t} = H_t v_t,.$$

where $v_t$ is the inward unit normal of the embedding $F_t$. Since $\mathbb S^2$ is compact, the solution exist for short time $T$. The evolution of $H$ is given by

$$\frac{\partial H}{\partial t} = \Delta H + |A|^2 H.$$

The strong maximum principle implies that $H>0$ when $t>0$.

• Thank you very much! I still have a question. I just searched some references about mean curvature flows, and I learned from math.nyu.edu/~bkleiner/mean_convex_flow.pdf that the $H \ge 0$ can be preserved, but I didn't find result about $H>0$ in a short time. Could you show me a good reference about this? – student Nov 12 '16 at 8:51
• This is really a consequence of (parabolic) strong maximum principle for $\partial_t - L f = 0$, where $Lf =\Delta f + |A|^2 f$. For the maximum principle you might see " Second Order Parabolic Differential Equations" by Gary M. Lieberman. The statement I used is stated in "Mean curvature flow and isoperimetric inequalities", proposition 5.2. @student – Arctic Char Nov 12 '16 at 9:18
• Why is it still strar-shaped? The tangent plane at some point of the surface might pass through the origin, so in principle $F_t$ might fail to be star-shaped for all small $t$. – Anton Petrunin Nov 12 '16 at 18:19
• @Anton Petrunin, I think one can argue like this. Star-shaped domain can be rescaled a little bit so that it becomes a strictly star-shaped domain, that is, we can find a point $O$ such that all the tangent planes on the surface are strictly away from $O$. Then by running mean curvature flow, since the solution is smooth, the new tangent planes are also away from $O$. – student Nov 12 '16 at 19:41
• @Anton Petrunin, how about considering the parallel surface? Then it becomes strictly star-shaped. Also parallel surface preserves mean curvature. – student Nov 12 '16 at 20:14