A sharp quantitative isoperimetric inequality in hyperbolic n-space

Abstract

In this paper we prove a quantitative version of the classical isoperimetric inequality in the hyperbolic space $\mathbb {H}^n$. The constant only depends on the dimension and an upper bound for the volume of the set.

Appendix: Elementary facts in hyperbolic space

Here we give the proofs of certain facts we used in the course of the slicing lemma. We use the notation from the proof of Lemma 3.3 without any further explanation.

Lemma 7.1

The function

$$\begin{aligned} (0,\infty )\ni y \mapsto \frac{y\varphi '(y)}{\varphi (y)} \end{aligned}$$

is strictly increasing.

Proof

Because of the strict monotonicity of $\mathbf {v}(r)$ it is equivalent to show that

$$\begin{aligned} h(r):= \frac{\mathbf {v}(r)\varphi '(\mathbf {v}(r))}{\varphi (\mathbf {v}(r))} \end{aligned}$$

is increasing in $r>0$. Using the identities

$$\begin{aligned} \varphi (\mathbf {v}(r))=\mathbf {p}(r)=n\omega _n\sinh ^{n-1}r \quad \text{ and }\quad \varphi '(\mathbf {v}(r))=(n-1)\coth r \end{aligned}$$

(cf. (3.17)), we compute that $h'(r)>0$ is equivalent to

$$\begin{aligned} \mathbf {v}(r)\sinh ^{n-1}r(n\cosh ^2r-\sinh ^2r) < \mathbf {v}'(r)\cosh r\sinh ^nr, \end{aligned}$$

which, taking into account that $\mathbf {v}'(r)=\mathbf {p}(r)$, is equivalent to (2.10). $\square$

Lemma 7.2

The expression

$$\begin{aligned} \frac{\mathbf {p}(r)}{\psi _r(s)} \end{aligned}$$

is increasing in $r>0$ for every $s\in (0,1)$.

Proof

We begin by calculating

$$\begin{aligned} \frac{\psi _r(s)}{\mathbf {p}(r)} = \frac{\varphi (s\mathbf {v}(r))}{\varphi (\mathbf {v}(r))} + \frac{\varphi ((1-s)\mathbf {v}(r))}{\varphi (\mathbf {v}(r))} -1. \end{aligned}$$

It therefore suffices to prove that the function

$$\begin{aligned} f(y):=\frac{\varphi (sy)}{\varphi (y)},\quad y>0, \end{aligned}$$

is decreasing for every $s\in [0,1]$. A straightforward calculation yields that $f'(y)<0$ is equivalent to

$$\begin{aligned} \frac{sy\varphi '(sy)}{\varphi (sy)} < \frac{y\varphi '(y)}{\varphi (y)}. \end{aligned}$$

(6.25)

But Lemma 7.1 implies that the right-hand side is increasing in y. Since $s\in (0,1)$ and $y>0$, we infer the asserted estimate (6.25) and thereby complete the proof of the lemma. $\square$

Lemma 7.3

For any $r>0$ we have

$$\begin{aligned} \int _0^1\frac{1}{\psi _r(s)}\,ds <\infty . \end{aligned}$$

Proof

We first note that it is enough to prove the integrability of $1/\psi _r$ in a neighborhood of the singular points 0 and 1. Therefore, we consider $s\in [0,\frac{1}{2}]$. By the mean value theorem there exist $\xi _1\in [0,s]$ and $\xi _2\in [1-s,1]$ such that there holds:

$$\begin{aligned} \psi _r(s)&= \varphi (s\mathbf {v}(r)) + \varphi ((1-s)\mathbf {v}(r)) - \varphi (\mathbf {v}(r)) \\&= s\bigg [\frac{d}{ds}\Big |_{s=\xi _1}\varphi (s\mathbf {v}(r)) - \frac{d}{ds}\Big |_{s=\xi _2}\varphi (s\mathbf {v}(r))\bigg ] \\&= s\mathbf {v}(r)\bigg [\frac{n-1}{\tanh (\mathbf v^{-1}(\xi _1\mathbf {v}(r)))} - \frac{n-1}{\tanh (\mathbf v^{-1}(\xi _2\mathbf {v}(r)))}\bigg ] \\&\ge s\mathbf {v}(r)\bigg [\frac{n-1}{\tanh (\mathbf v^{-1}(s\mathbf {v}(r)))} - \frac{n-1}{\tanh (\mathbf v^{-1}((1-s)\mathbf {v}(r)))}\bigg ] \\&= \frac{(n-1)s\mathbf {v}(r)}{\tanh (\mathbf v^{-1}(s\mathbf {v}(r)))} \bigg [1-\frac{\tanh (\mathbf v^{-1}(s\mathbf {v}(r)))}{\tanh (\mathbf v^{-1}((1-s)\mathbf {v}(r)))}\bigg ]. \end{aligned}$$

We now choose $s_o\in [0,\frac{1}{2}]$ in dependence of r small enough to have

$$\begin{aligned} \tanh (\mathbf v^{-1}(s_o\mathbf {v}(r))) \le \tfrac{1}{2}\tanh (\mathbf v^{-1}((1-s_o)\mathbf {v}(r))). \end{aligned}$$

Then, for $s\in (0,s_o]$ we find that

$$\begin{aligned} \psi _r(s)&\ge \frac{(n-1)s\mathbf {v}(r)}{\tanh (\mathbf v^{-1}(s\mathbf {v}(r)))} \bigg [1-\frac{\tanh (\mathbf v^{-1}(s_o\mathbf {v}(r)))}{\tanh (\mathbf v^{-1}((1-s_o)\mathbf {v}(r)))}\bigg ] \nonumber \\&\ge \frac{(n-1)s\mathbf {v}(r)}{2\tanh (\mathbf v^{-1}(s\mathbf {v}(r)))} \ge \frac{(n-1)s\mathbf {v}(r)}{2\sinh (\mathbf v^{-1}(s\mathbf {v}(r)))}. \end{aligned}$$

(6.26)

Next, we note that for $0\le \sigma \le 1$ we have $\cosh \sigma \le 2$ and therefore we have

$$\begin{aligned} \sinh ^{n-1}\sigma \cosh \sigma \le 2\sinh ^{n-1}\sigma . \end{aligned}$$

Integrating both sides with respect to $\sigma$ over (0, t), we obtain for $0\le t\le 1$ that

$$\begin{aligned} \sinh ^{n}t \le 2n\int _0^t\sinh ^{n-1}\sigma \,d\sigma = \tfrac{2}{\omega _n}\mathbf {v}(t). \end{aligned}$$

Assuming that $\mathbf v^{-1}(s\mathbf {v}(r))\le 1$, an assumption which can be imposed after possibly reducing the value of $s_o$, we can use the preceding estimate in (6.26) to infer that for all $s\in (0,s_o]$

$$\begin{aligned} \frac{1}{\psi _r(s)} \le \frac{2^{\frac{n+1}{n}} \mathbf v(\mathbf v^{-1}(s\mathbf {v}(r)))^{\frac{1}{n}}}{(n-1)\omega _n^{\frac{1}{n}}s\mathbf {v}(r)} = \frac{2^{\frac{n+1}{n}}}{(n-1)\omega _n^{\frac{1}{n}}(s\mathbf {v}(r))^{\frac{n-1}{n}}} = c(n,r)\, s^{-\frac{n-1}{n}}. \end{aligned}$$

But this ensures that the integral $\int _0^{s_o} 1/\psi _r \,ds$ is finite and by symmetry we also have that $\int _{1-s_o}^1 1/\psi _r \,ds = \int _0^{s_o} 1/\psi _r \,ds<\infty$. This finishes the proof of the lemma. $\square$