TITLE: Differential inequality regarding volume comparison
QUESTION [6 upvotes]: Let $(M,g)$ be a complete $n$-dimensional Riemannian manifold and let $p \in M$. Consider $(t,\Theta)$ , the geodesic spherical coordinates around $p$, where $t \in (0,\text{conj}_p(\Theta))$ and $\Theta$ is a unit vector in $T_pM$. Let $A_p(t,\Theta)$ be the density of the volume measure in these coordinates, i.e.
\begin{equation*}
     d\operatorname{Vol} = A_p(t,\Theta) dt  d\Theta
\end{equation*}
A well-known theorem of Gromov states that if $\operatorname{Ric}(M) \geqslant (n-1)\kappa$,  then the map
\begin{equation}
    t \mapsto \frac{{A}_p(t,\Theta)}{sn^{n-1}_{\kappa}(t)}
\end{equation}
is non-increasing in $t$. As usual, $sn_{\kappa}$ is given by
\begin{align*}
 sn_{\kappa}(t) = \begin{cases} 
\frac{\sin{\sqrt{k}t}}{\sqrt{k}} & k > 0\\
t & k = 0\\
\frac{\sinh{\sqrt{-k}t}}{\sqrt{-k}} & k < 0
\end{cases}
\end{align*}
Now I would like to prove a similar result when the sectional curvature of $M$ is bounded from above. That is, if $ \text{sec}(M) \leqslant \kappa$, then
\begin{equation*}
     \frac{d^2}{dt^2}\left(\frac{A_p(t,\Theta)}{sn^{n-2}_{\kappa}(t)}\right) + \kappa \left(\frac{A_p(t,\Theta)}{sn^{n-2}_{\kappa}(t)}\right) \geqslant 0
\end{equation*}
I'm trying to mimic the argument given by Gromov, letting $\varphi(t) = A_p(t,\Theta)^{\frac{1}{n-2}}$ and calculate that $(\log \varphi(t))' = \frac{1}{n-2}\text{tr}(\text{II}(t))$, where $\text{II}(t)$ is the second fundamental form of $\partial B(p,t)$ . But since we are not proving a statement about monotonicity, I don't know how I can get rid of the power $(n-2)$. Differentiating such expression directly seems intimidating and tedious, and I believe there's a shortcut to the problem since it is very similar to the estimate of the norm of Jacobi fields. Any insight of the problem will be appreciated.

$\textbf{Update}$:
The other day I thought I had a solution but after re-checked it I don't think that works. Here's what I did: Let $$\varphi(t) = \frac{A_p(t,\Theta)}{sn^{n-2}_{\kappa}(t)}$$ one can use the calculus trick $\varphi'(t) = (\log{\varphi(t)})'\varphi(t) $ to calculate that
\begin{align*}
    \varphi'(t) &= (\log \varphi(t))'\varphi(t)\\
                & = (\log A_{p}(t,\Theta) - (n-2)\log sn_{\kappa}(t))'\varphi(t)\\
                &= [\text{tr}(\text{II}(t))-(n-2)\frac{sn'_{\kappa}(t)}{sn_{\kappa}(t)}]\varphi(t)
\end{align*}
and that
\begin{align*}
  \varphi''(t) &= ((\log \varphi(t))'\varphi(t))'\\
               &= (\log \varphi(t))''\varphi(t) + [(\log \varphi(t))']^2\varphi(t)\\
& = \{[\text{tr}(\text{II}(t))-(n-2)\frac{sn'_{\kappa}(t)}{sn_{\kappa}(t)}]' + [\text{tr}(\text{II}(t))-(n-2)\frac{sn'_{\kappa}(t)}{sn_{\kappa}(t)}]^2 \} \varphi(t)
\end{align*}
Note that $\varphi(t)$ is non-negative when $t$ is small, so in order to show the original inequality is true, it suffices to show that
\begin{equation*}
[\text{tr}(\text{II}(t))-(n-2)\frac{sn'_{\kappa}(t)}{sn_{\kappa}(t)}]' + [\text{tr}(\text{II}(t))-(n-2)\frac{sn'_{\kappa}(t)}{sn_{\kappa}(t)}]^2  + \kappa \geqslant 0 \tag{$\star$} 
\end{equation*}
Recall the second comparison estimate (c.f. Peterson Page 145 Cor 2.4) states that if the sectional curvature is bounded above by $\kappa$, then each component of the second fundamental form can be bounded from below, i.e.
\begin{equation*}
 (\text{II}(t))_{_{2 \leqslant \alpha, \beta \leqslant n}} \geqslant \frac{\text{sn}'_{k}(t)}{\text{sn}_{k}(t)}
\end{equation*}
which gives us
\begin{equation}
    \text{tr}(\text{II}(t)) \geqslant (n-1) \frac{\text{sn}'_{K}(t)}{\text{sn}_{K}(t)} \tag{$\star \star$} 
\end{equation}
and
\begin{equation*}
     \text{tr}(\text{II}(t))-(n-2)\frac{\text{sn}'_{\kappa}(t)}{\text{sn}_{\kappa}(t)} \geqslant \frac{\text{sn}'_{\kappa}(t)}{\text{sn}_{\kappa}(t)}
\end{equation*}
I thought plug this expression back in $\star$ will solve the problem, since
\begin{equation*}
(\frac{\text{sn}'_{\kappa}(t)}{\text{sn}_{\kappa}(t)})' + (\frac{\text{sn}'_{\kappa}(t)}{\text{sn}_{\kappa}(t)})^2 + \kappa = 0
\end{equation*}
But I made a fatal mistake by declaring that $f \geqslant g$ implies $f' \geqslant g'$! It seems like I'm back to square one, though I got a relatively nice expression. Any insight or help would be appreciated.

REPLY [2 votes]: So my professor gave me an idea of how to solve this problem. After we get
\begin{equation*}
\underbrace{[\text{tr}(\text{II}(t))-(n-2)\frac{\text{sn}'_{\kappa}(t)}{\text{sn}_{\kappa}(t)}]'}_{\text{Part A}} + \underbrace{[\text{tr}(\text{II}(t))-(n-2)\frac{\text{sn}'_{\kappa}(t)}{\text{sn}_{\kappa}(t)}]^2}_{\text{Part B}} + \kappa \geqslant 0  \tag{$\star$}
\end{equation*}
We can use Riccati's equation to rewrite
\begin{align*}
\text{Part A} = &[\text{tr}(\text{II}(t))-(n-2)\frac{\text{sn}'_{\kappa}(t)}{\text{sn}_{\kappa}(t)}]' \\
\geqslant & -\text{tr}(\text{II}(t)^2)-(n-1)\kappa -  (n-2)[-(\frac{\text{sn}'_{\kappa}(t)}{\text{sn}_{\kappa}(t)})^{2} -\kappa]\\
=& -\text{tr}(\text{II}(t)^2) + (n-2)(\frac{\text{sn}'_{\kappa}(t)}{\text{sn}_{\kappa}(t)})^{2}-\kappa 
\end{align*}
And after expanding out $\text{Part B}$, $\star$ becomes
\begin{align*}
&\underbrace{[\text{tr}(\text{II}(t))-(n-2)\frac{\text{sn}'_{\kappa}(t)}{\text{sn}_{\kappa}(t)}]'}_{\text{Part A}} + \underbrace{[\text{tr}(\text{II}(t))-(n-2)\frac{\text{sn}'_{\kappa}(t)}{\text{sn}_{\kappa}(t)}]^2}_{\text{Part B}} + \kappa \\ \geqslant & \text{tr}(\text{II}(t))^2-\text{tr}(\text{II}(t)^2) -2(n-2)\text{tr}(\text{II}(t))\frac{\text{sn}'_{\kappa}(t)}{\text{sn}_{\kappa}(t)}' + (n-1)(n-2)(\frac{\text{sn}'_{\kappa}(t)}{\text{sn}_{\kappa}(t)})^2\\
\geqslant & \text{tr}(\text{II}(t))^2-\text{tr}(\text{II}(t)^2) -2(n-2)\text{tr}(\text{II}(t))\frac{\text{sn}'_{\kappa}(t)}{\text{sn}_{\kappa}(t)}' + (\frac{\text{sn}'_{\kappa}(t)}{\text{sn}_{\kappa}(t)})^2\\
= &\sum_{1,i\neq j,n-1}\lambda_{i}(t)\lambda_j(t) - [\lambda_{i}(t)+\lambda_{j}(t)]\frac{\text{sn}'_{\kappa}(t)}{\text{sn}_{\kappa}(t)}'  + (\frac{\text{sn}'_{\kappa}(t)}{\text{sn}_{\kappa}(t)})^2\\
=& \sum_{1,i\neq j,n-1}(\lambda_i(t)-\frac{\text{sn}'_{\kappa}(t)}{\text{sn}_{\kappa}(t)}')(\lambda_j(t)-\frac{\text{sn}'_{\kappa}(t)}{\text{sn}_{\kappa}(t)}')\\
\geqslant  & 0
\end{align*}
where $\lambda_{i}(t), i=1,\dots,n-1$ are the eigenvalues of $\text{II}(t)$.  The last inequality follows from Hessian comparison, which is indicated in Corollary 2.4 in Petersen's book.