A volume comparison result

Let ${(M^{n+1},g)}$ be a complete Riemannian manifold with non-negative Ricci curvature. Suppose we have a geodesic ball ${B}$ in ${(M,g)}$ whose boundary has the same surface area as a Euclidean ball (not necessarily of the same radius), we want to compare the volumes of these two balls.

Note that the usual comparison theorem says that the volume of a geodesic ball is bounded from above (often by the volume of a ball in a space form with the same radius) given that ${\mathrm{Ric}}$ is bounded from below. In this note, we investigate a dual problem: instead of comparing the volumes of balls with the same radius, we compare the balls with the same surface area. Naturally, the main ingredient is a comparison theorem of Bishop.

We use the following notations. Denote the geodesic ball and geodesic sphere in ${(M,g)}$ centered at ${p}$ with radius ${r}$ by ${B(p,r)}$ and ${S(p,r)}$ respectively. Let ${p\in M}$ be fixed and define ${A(r)=\mathrm{Area}(S(p,r))}$ and ${V(r)= \mathrm{Vol}(B(p,r))}$. Similarly, we define ${\overline A(r)=\mathrm{Area}(\mathbb{S}(r))}$ and ${\overline V(r)=\mathrm{Vol}(\mathbb{B}(r))}$, where ${\mathbb{S}(r)=\{x\in \mathbb{R}^{n+1}: |x|=r\}}$ and ${\mathbb{B}(r)=\{x\in \mathbb{R}^{n+1}: |x|\le r\}}$. Let ${\omega_n=\overline A(1)}$, we define

$\displaystyle \overline r(t)=\left(\frac{t}{\omega_n}\right)^{\frac{1}{n}},$

so that the Euclidean ball of radius ${\overline r(A)}$ has surface area ${A}$.

 Theorem 1 Let ${(M^{n+1},g)}$ be a complete Riemannian manifold with ${\mathrm{Ric}\ge 0}$. Suppose ${B}$ is a geodesic ball such that its boundary has area ${A_0}$, then ${\mathrm{Vol}(B)\ge \overline V(\overline r(A_0))}$. The equality holds if and only if ${B}$ is isometric to a Euclidean ball.

From now on, we fix a geodesic ball ${B=B(p, r_0)}$ in ${(M,g)}$ and let ${A_0}$ be the area of ${\partial B}$. Let ${r_1=\min \{r\le r_0: A(r)=A_0\}}$. Note that ${A(r)\le A_0}$ for ${r\le r_1}$. Obviously, to prove Theorem 1, it suffices to show ${V(r_1)\ge \overline V(\overline r(A_0))}$.

For ${0\le t\le A_0}$, we define

$\displaystyle r(t)=\max\{r\in [0,r_1]: A(r)=t\}.$

It is easy to see that ${r(t)}$ is increasing (and is upper semi-continuous). Indeed, if ${t_1, then by the continuity of ${A(r)}$, there exists ${r(t_1) such that ${A(r')=t_2}$. Therefore ${r(t_2)>r(t_1)}$.

We also have the simple properties that ${A(r(t))=t \textrm{ and }r(A(\rho))\ge \rho.}$

 Lemma 2 There exists a non-decreasing function ${\tau(t)\ge1}$ such that for ${0\le t\le A_0}$, $\displaystyle r(t)= \tau(t)\overline r (t).$

Proof: By dividing ${\overline A}$ and ${A}$ by the same constant, we can assume ${\overline A(r)= r^n}$, and regard ${\overline A}$ and ${A}$ as functions of ${r^n}$ (i.e. ${\overline A=\mathrm{id}}$). In the same way we normalize ${r}$ and ${\overline r}$ so that ${\overline r^n(t)=t}$. By a comparison theorem of Bishop (cf. [BC] p.256 Cor. 3 or [SY] Theorem 1.2), there exists a non-increasing function ${s=s(r)\le 1}$ such that

$\displaystyle \begin{array}{rl} A(r^n)=s(r)r^n. \end{array}$

Let ${t\le A_0}$ be fixed and substitute ${r^n=r^n(t)}$ in the above equation,

$\displaystyle t=A(r^n(t))= s(r(t))r^n(t).$

So

$\displaystyle r^n (t)=\frac{1}{s(r(t))}t=\frac{1}{s(r(t))}\overline r^n(t).$

Clearly ${\frac{1}{s(r(t))}\ge1 }$ is non-decreasing (as ${r(t)}$ is increasing). The result follows. $\Box$

As a consequence, we have:

 Lemma 3 The function ${r-\overline r}$ is non-decreasing for ${0\le t\le A_0}$. In particular, ${ r(A_0)-r(t)\ge\overline r(A_0)-\overline r(t)}$.

Proof of Theorem 1: Recall that ${B=B(p,r_0)}$, ${A_0=\mathrm{Area}(\partial B)}$, ${r_1=\min \{r\le r_0: A(r)=A_0\}}$, and that ${r(t)=\max\{r\in [0,r_1]: A(r)=t\}}$ for ${t\le A_0}$. By the layercake integral formula and Lemma 3,

$\displaystyle \begin{array}{rl} \mathrm{Vol}(B)\ge V(r_1) =\int_0^{r_1}A(r)dr =&\int_0^{A_0}|\{r\in [0,r_1]: A(r)\ge t\}|dt\\ \ge&\int_0^{A_0}(r(A_0)-r(t))dt\\\ \ge&\int_0^{A_0}(\overline r(A_0)-\overline r(t))dt\\ =&\int_0^{\overline r(A_0)}\overline A(r)dr\\ =&\overline V(\overline r(A_0)). \end{array}$

By the equality case of the comparison theorem of Bishop, ${\mathrm{Vol}(B)=\overline V(\overline r(A))}$ if and only if ${B}$ is isometric to a Euclidean ball. $\Box$

Question: Can Theorem 1 be generalized to a Riemannian manifold with smooth boundary?