How to show the inequation by using Hessian comparison.

Question

In the below picture ,how to show the inequation 1?

In fact,I'm not familiar with Hessian comparison.So, hope a detail answer , Thanks very much.

The below picture is form 194th page of here ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Answer 1

The Hessian comparison theorem is:

If the sectional curvatures of a manifold are bounded below by $M$, then the distance function $s(x) = d(p,x)$ satisfies $\nabla^2 s \le \nabla^2_M s_M$, where $s_M$ is the corresponding distance function on the space of constant curvature $M$.

[see e.g. ON THE DISTRIBUTIONAL HESSIAN OF THE DISTANCE FUNCTION by Mantegazza, Mascellani & Uraltsev.]

We know $|R_{ijkl}|<M$ and thus our sectional curvatures $K(e_i,e_j) = R_{ijij}$ satisfy $K > -M$; so we can compare to the hyperbolic space with curvature $-M$. In polar coordinates this space has metric $$g_M = ds^2 + \frac1M \sinh^2(\sqrt M s) d\Omega^2$$ where $d \Omega^2$ is the round metric on the unit $(n-1)$-sphere. The Hessian of the distance function is (see e.g. Petersen Chapter 2.3) $$\nabla^2s_M = \frac1{\sqrt M} \sinh(\sqrt M s) \cosh(\sqrt M s) d\Omega^2 = \sqrt M \coth(\sqrt M s) (g_M - ds^2).$$

If we throw away the $-ds^2$ we get the estimate $$|\nabla^2 s_M| \le \sqrt M \coth(\sqrt M s)|g_M|.$$

Since $|g_M| = \sqrt{g^{ij}g_{ij}} = \sqrt n$ depends only on the dimension $n$, the theorem (along with an upper bound for $\coth$ you can try proving) gives us the estimate $$|\nabla^2 s| \le |\nabla^2 s_M| \le \sqrt {nM} \coth \sqrt M s \le \sqrt n \left(\sqrt M + \frac1s\right).$$

It looks like the authors have a slightly better estimate here with $\sqrt M$ instead of $\sqrt {nM}$ - I'm not sure whether the mistake is mine or theirs. (Taking the $-ds^2$ in to account only improves the $\sqrt n$ to $\sqrt{n-1}$.) It doesn't matter anyway, since we can still choose a constant $C_3$ dependent only on dimension that makes it work.

Answer 2

I just using this Hessian comparison,which I am familiar with.