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?




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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.



How to evolve a star-shaped mean convex set to a strictly 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 $t>0$. 


