Lemma $3.2$ - Mean curvature flow singularities for mean convex surfaces

Question

This is a lemma of the paper "Mean curvature flow singularities for mean convex surfaces" by Gerhard Huisken and Carlo Sinestrari (the paper is available here):

$\textbf{Lemma 3.2.}$ Suppose $(1 + \eta) H^2 \leq |A|^2 \leq c_0 H^2$ for some $\eta, c_0 > 0$ at some point of $\mathscr{M}_t$, then we also have

(i) $-2Z \geq \eta H^2|A|^2$;

(ii) $|H \nabla_i h_{kl} - \nabla_i H h_{kl}|^2 \geq \frac{\eta^2}{4n(n-1)^2c_0} H^2 |\nabla H^2|$

My doubt is concerning to item $(ii)$ and below is the argument given by the authors

We have (see [10, Lemma $2.3$ (ii)], reference [10] is available here)

$|H \ \nabla_i h_{kl} - \nabla_i H \ h_{kl}|^2 \geq \frac{1}{4} |\nabla_i H \ h_{kl} - \nabla_k H \ h_{il}|^2 = \frac{1}{2} (|A|^2 |\nabla H|^2 - |\nabla^i H h_{il}|^2).$

Let us denote with $\lambda_1, \cdots, \lambda_n$ the eigenvalues of $A$ in such a way that $\lambda_n$ is an eigenvalue with the largest modulus. Then we have $|\nabla^i H \ h_{il}|^2 \leq \lambda_n^2 |\nabla H|^2$ and

\begin{align*} |H \ \nabla_i h_{kl} - \nabla_i H \ h_{kl}|^2 &\geq \frac{1}{2} \sum\limits_{i=1}^{n-1} \lambda_i^2 |\nabla H|^2 = \sum\limits_{i=1}^{n-1} \lambda_i^2 \lambda_n^2 \frac{|\nabla H|^2}{2\lambda_n^2}\\ &\geq \sum\limits_{i=1}^{n-1} \sum\limits_{j=i+1}^n \lambda_i^2 \lambda_j^2 \frac{|\nabla H|^2}{2(n-1)|A|^2}\\ &\geq \left( \sum\limits_{i,j=1, \ i < j}^n \lambda_i \lambda_j \right)^2 \frac{|\nabla H|^2}{n(n-1)|A|^2}\\ &= \frac{(|A|^2 - H^2)^2}{4n(n-1)|A|^2} |\nabla H|^2 \geq \frac{\eta^2 H^2}{4n(n-1)c_0} |\nabla H|^2. \square \end{align*}

I would like to understand the following equality and inequalities:

a) $\frac{1}{4} |\nabla_i H \ h_{kl} - \nabla_k H \ h_{il}|^2 = \frac{1}{2} (|A|^2 |\nabla H|^2 - |\nabla^i H h_{il}|^2)$;

b) $|H \ \nabla_i h_{kl} - \nabla_i H \ h_{kl}|^2 \geq \frac{1}{2} \sum\limits_{i=1}^{n-1} \lambda_i^2 |\nabla H|^2$;

c) $\sum\limits_{i=1}^{n-1} \sum\limits_{j=i+1}^n \lambda_i^2 \lambda_j^2 \frac{|\nabla H|^2}{2(n-1)|A|^2} \geq \left( \sum\limits_{i,j=1, \ i < j}^n \lambda_i \lambda_j \right)^2 \frac{|\nabla H|^2}{n(n-1)|A|^2}$.

My thoughts:

$a$ and $b$) I just consider use normal coordinates, but this doesn't helps me because the right hand side in $a$ and $b$ would be zero.

$c$) I just try to prove that $\sum\limits_{i=1}^{n-1} \sum\limits_{j=i+1}^n \lambda_i^2 \lambda_j^2 \geq \left( \sum\limits_{i,j=1, \ i < j}^n \lambda_i \lambda_j \right)^2$, but I can't prove this because I don't know if all eigenvalues are non-negative. Indeed, I even don't know if the $H > 0$ because I didn't see the hypothesis that the hypersurfaces is mean convex on the paper until this lemma.

Thanks in advance!

Answer 1

(a) is just direct computations:

\begin{align*} \frac{1}{4} |\nabla_i H \ h_{kl} - \nabla_k H \ h_{il}|^2 &= \frac 14 \sum_{i,k,l} (\nabla _i H h_{kl} - \nabla _k H h_{il})^2 \\ &=\frac 14 \sum_{i,k,l}\bigg( (\nabla _i H h_{kl})^2 + (\nabla _k H h_{il})^2 - 2\nabla _i H h_{kl} \nabla _k H h_{il}\bigg)\\ &=\frac 12 \left(\sum_i (\nabla_iH)^2 \sum_{k,l} h_{kl}^2\right) -\frac 12 \sum_{i,k,l} \nabla _i H h_{kl} \nabla _k H h_{il} \\ &= \frac 12 |\nabla H|^2 |A|^2 -\frac 12 \sum_{i,k,l} \nabla _i H h_{kl} \nabla _k H h_{il} \end{align*}

For the remaining term, use (that at one point) $h_{il} = \lambda _i \delta_{il}$, so $$\sum_{i,k,l} \nabla _i H h_{kl} \nabla _k H h_{il} =\sum_i (\nabla_iH)^2 \lambda_i^2 = |\nabla^ i H h_{il}|^2$$

In the paper (b) is shown using (a). By (a) and that $|\nabla H h_{il}|^2 \le \lambda_n^2 |\nabla H|^2$,

\begin{align*} |H \ \nabla_i h_{kl} - \nabla_i H \ h_{kl}|^2 &\ge \frac{1}{2} (|A|^2 |\nabla H|^2 - |\nabla^i H h_{il}|^2)\\ &\ge \frac 12\bigg( \left(\sum_{i=1}^n \lambda_i^2\right) |\nabla H|^2 - \lambda_n^2 |\nabla H|^2 \bigg)\\ &= \frac 12 \left(\sum_{i=1}^{n-1} \lambda_i^2\right) |\nabla H|^2 \end{align*}

Lastly (c) has completely nothing to do with MCF. Note that

$$ \sum_{i=1}^{n-1} \sum_{j=i+1}^n = \sum_{i,j=1, i<j}^n.$$

Thus

\begin{align*} \left(\sum_{i,j=1, i<j}^n \lambda_i \lambda_j \right)^2 &= \left(\sum_{i=1}^{n-1} \sum_{j=i+1}^n \lambda_i \lambda_j\right)^2 \\ &\le \left(\sum_{i=1}^{n-1} \sum_{j=i+1}^n (\lambda_i \lambda_j)^2\right) \left(\sum_{i=1}^{n-1} \sum_{j=i+1}^n 1^2\right) \ \ \ \ \text{(Cauchy-Schwarz inequality)}\\ &= \frac{n(n-1)}{2} \sum_{i=1}^{n-1} \sum_{j=i+1}^n (\lambda_i \lambda_j)^2. \end{align*}