We recall Topping's diameter estimates (Theorem 1.1 here):
(Topping): Let $M$ be an n-dimensional closed, connected manifold smoothly immersed in $\mathbb R^N$, where $N\ge n+1$. Then the intrinsic diameter and the mean curvature $H$ of $M$ are related by $$ \operatorname{diam}(M)\le C(n)\int_M |H|^{n-1}d\mu $$
Now, if we have $$ \max_{M} |\nabla H|\le C\epsilon^\alpha ~~~~\text{ and }~~~~~~ \max_M |H| \ge C\epsilon^{\frac{\alpha}{2} } $$ Then, by Topping Lemma, how to show $$ \min_M |H| \ge \frac{C}{2}\epsilon^{\frac{\alpha}{2} } $$ As my understand, we should have $$ \min_M |H|\ge C\epsilon^\frac{\alpha}{2} -diam(M)C\epsilon^\alpha $$ So it means $$ \int_M |H|^{n-1} \thicksim \epsilon^{-\frac{\alpha}{2}} $$ Obviously, it is not right. Where is my mistake ?
Last, this question is from the Lemma 3.4 of Blow-up of the mean curvature at the first singular time of the mean curvature flow. I am not sure a lot whether I understand the author, or I miss something, so ,I add the picture in below.
Topping's lemma together with the bound $|H| \le 4n\Lambda_0^2$ and $|M_t|\le |M_0|$ give
$$\text{diam} M_t \le C(n, \lambda_0, |M_0|). $$
Then
$$\min H \ge \eta - C\epsilon^\alpha = \eta - C \epsilon^{\alpha/2}\epsilon^{\alpha/2} = \eta - C\eta \epsilon^{\alpha/2}$$
(the constant might be changing, but it depends only on $n$, $\Lambda_0$ and $|M_0|$). Now choose $\epsilon$ small depending only on $C,\alpha$ so that $C\epsilon^{\alpha/2} <1/2$. Then you have
$$\min H \ge \eta - \frac 12 \eta = \frac 12 \eta$$
as claimed.