Principal curvature and enclosed volume of surface

In 46 page of Huisken, Gerhard, The volume preserving mean curvature flow, J. Reine Angew. Math. 382, 35-48 (1987). ZBL0621.53007.

Assume $M\subset \mathbb R^{n+1}$ is a n-dim compact smooth convex manifold without boundary, if the volume enclosed by $M$ is constant, how to show the principal curvature of $M$ can't tend to infinity? In fact, I think it is wrong, for example, in the picture below, the max principle curvature tend to infinity. Besides, whether there are upper lower estimate of volume by the curvature of manifold?

• I don't think you are quoting the paper correctly. For starters, the $M$ in Huisken's paper is uniformly convex... – Willie Wong Apr 25 '17 at 3:37
• @WillieWong I have edit it. – lanse7pty Apr 25 '17 at 7:26

Maybe this will help you: write $$\mbox{Vol}(\Omega)=\int_{M}\int_{0}^{c(p)}\prod(1-t\lambda_i)dtdA,$$ where $\Omega\subset\mathbb{R}^{n+1}$ is a compact convex domain which $\partial\Omega=M$, $\lambda_i$ are the principal curvatures of $M$, $dA$ is the area of $M$, $c(p)=\max\left\{t\geq 0:\mbox{dist}(M,\mbox{exp}(t\nu_p)\right\}$ and $\nu_p$ is the upward unit normal of $M$ at $p$.
Then, by convexity, $c(p)\leq\lambda_{\mbox{max}}^{-1}$ and the result will follow.