Application of Sobolev inequality

Question

Sobolev inequality: For all Lipschitz functions $v$ on $M$ we have $$ \left( \int|v|^\frac{n}{n-1}\right)^{n-1/n} \le c(n)\left(\int _M |\nabla v| + \int _M H|v|\right) $$ where $H$ is mean curvature of $M$.

Now, if we have $$ \int_M |\nabla^m A|^p \le C(m,p) ~~~~\forall m,p \ge 0 $$ Then how to show $$ |\nabla^m A| \le C(m) ~~~~~~~ \forall m\ge 0~~~~? $$

This is the last step of proof of 8.3 Lemma of Flow by mean curvature of convex surfaces into spheres

Answer 1

I don't know how to use Sobolev inequality to prove it. But from the 4.1 Theorem of The volume preserving mean curvature flow, I found a way to prove it.

First we have $$ \max_{M_t} |A|^2 \le C ~~~~~\forall ~t\in[0,T) $$ from 7.1 Theorem of Flow by mean curvature of convex surfaces into spheres, we have $$ \partial_t |\nabla^m A|^2 = \Delta |\nabla^m A|^2-2|\nabla^{m+1} A|^2+\sum_{i+j+k=m} \nabla ^i A*\nabla ^j A*\nabla^kA*\nabla^m A \tag 1 $$ Now we prove it by induction on $m$. Suppose $|\nabla^n A|^2$ is uniformly bounded by $C_n$ for all $n\le m$. From (1), we have $$ \partial_t |\nabla^{m+1} A|^2 \le \Delta |\nabla^{m+1} A|^2 +C_{11}(|\nabla^{m+1} A|^2 +1) $$ choose $N\ge 2C_{11}$ and let $f=|\nabla^{m+1} A|^2 + N|\nabla^{m} A|^2 $, then we have $$ \partial_t f \le \Delta f -N|\nabla^{m+1}A|^2 +C_{12} $$ the $-N|\nabla^{m+1}A|^2$ is from $\partial_t (N|\nabla ^m A|^2)$, then we have $$ \partial_t f \le \Delta f -Nf +C_{13} $$ By Maximum principle, we have $$ f\le \frac{C_{13}}{N}+ e^{-Nt}(\max_{M_0} f- \frac{C_{13}}{N}) $$

so, we have proved Lemma 8.3 of Flow by mean curvature of convex surfaces into spheres.

In fact, I really want to know how to use Sobolev inequality to get it.

After read Sobolev imbedding theorem, I think it is a direct conclusion of Theorem 7.10 of Elliptic Partial Differential Equations of Second Order.

Theorem7.10 : $$ W_0^{1.p} \subset \left\{ \begin{aligned} &L^{np/(n-p)}(\Omega) &~~~p<n\\ &C^0(\overline \Omega) &~~~p>n \end{aligned}\right. $$ Firthermore, there exists a constant $C=C(n,p)$ such that, for any $u\in W_0^{1.p}$ $$ \begin{aligned} &||u||_{L^{np/(n-p)}} ~\le C ||Du||_{L^p} &~~~p<n \\ &\sup_\Omega |u| \le C|\Omega|^{1/n-1/p} ||Du||_{L^p} &~~~p>n \end{aligned} $$

Since $$ \int_M |\nabla^m A|^p \le C(m,p) ~~~~\forall m,p \ge 0 $$ and $M$ is compact, let $p>dim M$, we have $$ \nabla^{m-1}A\in W^{1,p}_0 $$ then $$ \max_M|\nabla^{m-1}A| \le C(m,p,M) $$