Whether we have $|\nabla^m \mathring A| \le C(m,n) |\nabla^m A|$?

Question

Consider a hypersurface $M^n\subset \mathbb R^{n+1}$, $A$ is the second fundamental form, $\mathring A= A-\frac{H}{n}g$ where $H$ is mean curvature and $g$ is induced metric. From Blow-up of the mean curvature at the first singular time of the mean curvature flow, it seems that they used $$ |\nabla^2 \mathring A| \le C(n)|\nabla^2 A|. $$ In fact, I don't know how to prove it. Besides, Whether we have $$ |\nabla^m \mathring A| \le C(m,n) |\nabla^m A| ~~? $$

Answer 1

Note that $$ \nabla^m \mathring A = \nabla^m A - \frac{1}{n} \nabla^m H\otimes g$$ since $\nabla g = 0$. Thus

$$\begin{split} |\nabla^m \mathring A| &\le |\nabla^m A| + \frac{1}{n} |\nabla^m H \otimes g|\\ &= |\nabla^m A| + \frac{1}{n} |\nabla ^m H| \cdot |g|\\ &=|\nabla^m A|+\frac{1}{\sqrt n} |\nabla ^m H|. \end{split}$$

Note $H = g^{ij} A_{ij} = \mathrm{tr}(g^{-1} \otimes A)$. Thus

$$\nabla^m H= \mathrm{tr} (g^{-1} \otimes \nabla^m A).$$

If we calculate at a point where $g_{ij} = \delta_{ij}$, then by Cauchy Schwarz,

$$\begin{split} |\nabla^m H|^2 &= \sum_{j_1, \cdots, j_m} \left(\sum_i \nabla^m_{j_1\cdots j_m} A_{ii}\right)^2 \\ &\le n\sum_{j_1, \cdots, j_m} \left(\sum_i (\nabla^m_{j_1\cdots j_m} A_{ii})^2\right)\\ &\le n\sum_{i,j, j_1, \cdots, j_m} \left( \nabla^m_{j_1\cdots j_m} A_{ij}\right)^2 \\ &= n|\nabla^m A|^2. \end{split}$$

So we have

$$|\nabla^m \mathring A| \le 2|\nabla ^m A|$$

and the constant $C(n,m)=2$ does not depend on $m,n$.