Suppose $M$ is a uniformly convex hypersurface in $\mathbb{R}^{n+1}$ ($A\geq \alpha Hg$ for $\alpha>0$) undergoing mean curvature flow. $A$ denotes the second fundamental form and $H$ denotes the mean curvature.
The proof of Huisken's theorem uses a uniform bound on the following quantity. $$ f_{\sigma,\epsilon} = \left(|A - \frac1nHg|^2-\epsilon H^2\right)H^{2-\sigma} $$ for some $\sigma\in (0,1)$ and all $\epsilon>0$. This is done by computing the evolution of $f_{\sigma, \epsilon}$, estimating some of the terms, and then using the maximum principle.
I am confused about the following inequality in p.294 of Extrinsic Geometric Flows by Andrews et. al: $$ \gamma \frac{|\nabla A|^2}{H^2}H^\sigma\geq \gamma f_{\sigma,\epsilon}\frac{|\nabla A|^2}{H^2} \tag{1} $$ This implies that $H^\sigma\geq f_{\sigma,\epsilon}$ since $\gamma>0$ and $H>0$. However, using $|A|^2\leq H^2$, the best inequality I can get is \begin{align*} \frac{f_{\sigma,\epsilon}}{H^\sigma} = \frac{|A|^2}{H^2}+1-\frac{2}{n}-\epsilon\leq 2-\frac{2}{n}-\epsilon\,. \end{align*} In fact, using uniform convexity $A\geq \alpha Hg$, we have $$ \frac{f_{\sigma,\epsilon}}{H^\sigma} =\frac{|A|^2}{H^2}+1-\frac{2}{n}-\epsilon\geq \alpha + 1-\frac{2}{n}-\epsilon $$ so for large $\alpha$ we would have $f_{\sigma,\epsilon}\geq H^\sigma$. What am I missing here?
I have added an excerpt from the book below.
Context: one of the terms in the evolution equation for $f_{\sigma, \epsilon}$ is $$-2H^\sigma|\nabla\frac{A}{H}|^2$$ Then, in p.294 they show that if $A\geq \alpha Hg$ for $\alpha>0$, then
$$
|\nabla\frac{A}{H}|^2\geq \gamma\frac{|\nabla A|^2}{H^2}
$$
where $\gamma$ depends on $\alpha$ and $n$. Since $A\geq \alpha Hg$ is preserved under mean curvature flow, the above inequality holds for as long as the flow exists with the same constant $\gamma$. This leads to the left side of (1).
Edit: Huisken's original argument (Section 5 of this) makes sense to me. He considers the slightly different function $$ f_\sigma = (|A|^2 - H^2/n)H^{2-\sigma}\,. $$ Computing the evolution of $f_\sigma$, we have \begin{align} (\partial_t-\Delta) f_\sigma &= \frac{2(1-\sigma)}{H}\langle\nabla H,\nabla f_\sigma\rangle\\ &-\frac{2}{H^{4-\sigma}}|H\nabla A - A\otimes\nabla H|^2\\ &-\frac{\sigma(1-\sigma)}{H^{4-\sigma}}(|A|^2-H^2/n)|\nabla H|^2\\ &+\sigma|A|^2f_\sigma\,. \end{align} Using an earlier inequality (Lemma 2.3(ii)) and preservation of $A\geq \alpha Hg$, the following is true under the flow $$ |H\nabla A - A\otimes\nabla H|^2 \geq \frac12 \alpha^2 H^2|\nabla H|^2 $$ So the last three terms of $(\partial_t-\Delta) f_\sigma$ are bounded above By
\begin{align} &-\frac{\alpha^2 H^2}{H^{4-\sigma}}-\frac{\sigma(1-\sigma)}{H^{4-\sigma}}(|A|^2-H^2/n)|\nabla H|^2+\sigma|A|^2f_\sigma \\ &=-\frac{|\nabla H|^2}{H^{4-\sigma}}\left(\alpha^2 H^2+\sigma(1-\sigma)(|A|^2-H^2/n)\right)+\sigma|A|^2f_\sigma\\ &=-\frac{|\nabla H|^2}{H^{2-\sigma}}\left(\alpha^2+\sigma(1-\sigma)(|A|^2/H^2-1/n)\right)+\sigma|A|^2f_\sigma \end{align}
In the paper, this is simplified to $$ -\alpha^2\frac{|\nabla H|^2}{H^{2-\sigma}}+\sigma|A|^2f_\sigma $$ which makes sense since $$ \sigma(1-\sigma)(|A|^2/H^2-1/n) \geq 0 $$ so it is a good term.
Excerpt: (The book uses $II$ for the second fundamental form $A$ and $\mathring{II}$ for the trace free second fundamental form $A-Hg/n$)
