## A note on Huisken’s paper 1984

$\mathbf{Thm(Myers):}$ If a Riemannian manifold has $Ric\geq (n-1)K$, then $diam(M)\leq \frac{\pi}{K}$.

Suppose we know a function $h:M\to \mathbb{R}^+$, $M$ is compact and the gradient estimate

$|\nabla h|\leq \eta^2 h^2_{\max}$ on $M$      $(1)$

and $Ric_M\geq (n-1)c^2h^2$. Consider the relation $\displaystyle h_{\min}\geq (1-\eta)h_{\max}$.

$\mathbf{Proof:}$

Let $x$ is a point on $M$, $h$ achieves the maximum. Since $(1)$, then

$h(x)>(1-\eta)h_{\max}$ in $B=B(x,\eta^{-1}h^{-1}_{\max})$,

By the Myers thm, we get

$\displaystyle diam(B)\leq \frac{\pi}{(1-\eta)h_{\max}}$

But we know if $\eta$ is small enough, then $\displaystyle \eta^{-1}h^{-1}_{\max}>\frac{\pi}{(1-\eta)h_{\max}}$. This forces $diam(M)<\eta^{-1}h^{-1}_{\max}$, then

$\displaystyle h_{\min}\geq (1-\eta)h_{\max}$.

$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$ Gerhard Huisken. Flow by mean curvature of convex surfaces into spheres.1984. p258 lemma 8.6.

The parameter $\eta$ has relation with $M$ and $h_{\max}$. We can not apply Myers thm directly. Guo Bin told me this method.