Lecture 23. The isoperimetric inequality

In this Lecture, we study in further details the connection between volume growth of metric balls, heat kernel upper bounds and the $L^1$ Sobolev inequality. As we shall see, on a manifold with non negative Ricci curvature, all these properties are equivalent one to each other and equivalent to the isoperimetric inequality as well. We start with some preliminaries about geometric measure theory on Riemannian manifolds.

Let $(\mathbb{M},g)$ be a complete and non compact Riemannian manifold.

In what follows, given an open set $\Omega \subset \mathbb{M}$ we will indicate with $\mathcal F(\Omega)$ the set of $C^1$ vector fields $V$ ‘s, on $\Omega$ such that $\| V \|_\infty \le 1$ .

Given a function $f\in L^1_{loc}(\Omega)$ we define the total variation of $f$ in $\Omega$ as
$\text{Var} (f;\Omega) = \underset{\phi\in \mathcal{F}(\Omega)}{\sup} \int_\Omega f \mathbf{div} \phi d\mu.$
The space $BV (\Omega) = \{f\in L^1(\Omega)\mid \text{Var}(f;\Omega)<\infty\},$ endowed with the norm
$||f||_{BV(\Omega)} = ||f||_{L^1(\mathbb{M})} + \text{Var} (f;\Omega),$
is a Banach space. It is well-known that $W^{1,1}(\Omega) = \{f\in L^1(\Omega)\mid \| \nabla f \| \in L^1(\Omega )\}$ is a strict subspace of $BV(\Omega)$ . It is important to note that when $f\in W^{1,1}(\Omega)$ , then $f\in BV(\Omega)$ , and one has in fact $\text{Var}(f;\Omega) = ||\sqrt{\Gamma(f)}||_{L^1(\Omega)}.$ Given a measurable set $E\subset \mathbb{M}$ we say that it has finite perimeter in $\Omega$ if $\mathbf 1_E\in BV(\Omega)$ . In such case the horizontal perimeter of $E$ relative to $\Omega$ is by definition
$P(E;\Omega) = \text{Var}(\mathbf 1_E;\Omega).$
We say that a measurable set $E\subset \mathbb{M}$ is a Caccioppoli set if $P(E;\Omega) < \infty$ for any $\Omega \subset \mathbb{M}$ . For instance, if $O$ is an open relatively compact set in $\mathbb{M}$ whose boundary $E$ is $n-1$ dimensional sub manifold of $\mathbb{M}$ , then it is a Caccioppoli set and $P(E;\mathbb{M})=\mu_{n-1} (E)$ where $\mu_{n-1}$ is the Riemannian measure on $E$ . We will need the following approximation result.

Proposition: Let $f\in BV(\Omega)$ , then there exists a sequence $\{f_n\}_{n\in \mathbb N}$ of functions in $C^\infty(\Omega)$ such that:

(i) $||f_n - f||_{L^1(\Omega)} \to 0$ ;

(ii) $\int_\Omega \sqrt{\Gamma(f_n)} d\mu \to \text{Var}(f;\Omega)$ .

If $\Omega = \mathbb{M}$ , then the sequence $\{f_n\}_{n\in \mathbb N}$ can be taken in $C^\infty_0(\mathbb{M})$ .

Our main result of the Lecture is the following result.

Theorem: Let $n > 1$ . Let us assume that $\mathbf{Ric} \ge 0$ . then the following assertions are equivalent:

(1) There exists a constant $C_1 > 0$ such that for every $x \in \mathbb{M}$ , $r \ge 0$ ,
$\mu (B(x,r)) \ge C_1 r^n.$

(2) There exists a constant $C_2 > 0$ such that for $x \in \mathbb{M}$ , $t > 0$ ,
$p(x,x,t) \le \frac{C_2}{t^{\frac{n}{2}}}.$

(3) There exists a constant $C_3 > 0$ such that for every Caccioppoli set $E\subset \mathbb{M}$ one has
$\mu(E)^{\frac{n-1}{n}} \le C_3 P(E;\mathbb{M}).$

(4) With the same constant $C_3 > 0$ as in (3), for every $f \in BV(\mathbb{M})$ one has
$\left( \int_\mathbb{M} |f|^{\frac{n}{n-1}} d\mu\right)^{\frac{n-1}{n}} \le C_3 \emph{Var}(f;\mathbb{M})$ .

Proof:

That (1) $\rightarrow$ (2) follows immediately from the Li-Yau upper Gaussian bound.

The proof that (2) $\rightarrow$ (3) is not straightforward, it relies on the Li-Yau inequality. Let $f \in C_0(\mathbb{M})$ with $f\ge 0$ . By Li-Yau inequality, we obtain
$\Gamma (P_t f) - P_tf \frac{\partial P_tf}{\partial t} \le \frac{d}{2t}(P_tf)^2.$
This gives in particular, ,
$\left(\frac{\partial P_tf}{\partial t}\right)^- \le \frac{d}{2t} P_tf,$
where we have denoted $a^+ = \sup\{a,0\}$ , $a^- = \sup\{-a,0\}$ . Since $\int_\mathbb{M} \frac{\partial P_tf}{\partial t} d\mu=0$ , we deduce
$||\frac{\partial P_tf}{\partial t}||_{L^1(\mathbb{M})} \le \frac{d}{t} ||f||_{L^1(\mathbb{M})},\ \ \ \ t > 0.$
By duality, we deduce that for every $f \in C^\infty_0(\mathbb{M})$ , $f\ge 0$ ,
$\|\frac{\partial P_tf}{\partial t} \|_{L^\infty(\mathbb{M})} \le \frac{d}{t} \| f\|_{L^\infty(\mathbb{M})}.$
Once we have this crucial information we can return to the Li-Yau inequality and infer
$\Gamma (P_t f) \le \frac{1}{t} \frac{3d}{2} \| f\|^2_{L^\infty(\mathbb{M})}.$
Thus,
$\| \sqrt{\Gamma (P_t f)} \|_{L^\infty(\mathbb{M})} \le \sqrt{\frac{3d}{2t} }\| f\|_{L^\infty(\mathbb{M})}.$
Applying this inequality to $g \in C_0^\infty(\mathbb{M})$ , with $g\ge 0$ and $||g||_{L^\infty(\mathbb{M})}\le 1$ , if $f \in C_0^1(\mathbb{M})$ we have
$\int_\mathbb{M}g(f-P_tf) d\mu = \int_0^t \int_\mathbb{M} g \frac{\partial P_sf}{\partial s} d\mu ds$
$= \int_0^t \int_\mathbb{M} g L P_sf d\mu ds = \int_0^t \int_\mathbb{M} L g P_sf d\mu ds$
$= \int_0^t \int_\mathbb{M} P_sLg f d\mu ds = \int_0^t \int_\mathbb{M} L P_sg f d\mu ds$
$= - \int_0^t \int_\mathbb{M} \Gamma(P_sg,f) d\mu ds$
$\le \int_0^t \| \sqrt{\Gamma(P_sg)} \|_{L^\infty(\mathbb{M})}\int_\mathbb{M} \sqrt{\Gamma(f)} d\mu ds \le \sqrt{6d} \sqrt{t} \int_\mathbb{M} \sqrt{\Gamma(f)} d\mu.$
We thus obtain the following basic inequality: for $f \in C_0^1(\mathbb{M})$ ,
$\|P_tf - f\|_{L^1(\mathbb{M})} \le \sqrt{6d}\ \ \sqrt{t}\ \| \sqrt{\Gamma(f)} \|_{L^1(\mathbb{M})}.$
Suppose now that $E\subset \mathbb{M}$ is a bounded Caccioppoli set. But then, $\mathbf 1_E\in BV(\Omega)$ , for any bounded open set $\Omega \supset E$ . It is easy to see that $\text{Var}(\mathbf 1_E;\Omega) = \text{Var}(\mathbf 1_E;\mathbb{M})$ , and therefore $\mathbf 1_E\in BV(\mathbb{M})$ . There exists a sequence $\{f_n\}_{n\in \mathbb N}$ in $C^\infty_0(\mathbb{M})$ satisfying (i) and
(ii) above. Applying the previous inequality to $f_n$ we obtain
$\|P_tf_n - f_n\|_{L^1(\mathbb{M})} \le \sqrt{6d} \ \sqrt{t}\ \| \sqrt{\Gamma(f_n)} \|_{L^1(\mathbb{M})} = \sqrt{6d} \sqrt{t}\ Var(f_n,\mathbb{M})$ .
Letting $n\to \infty$ in this inequality, we conclude
$\|P_t \mathbf 1_E - \mathbf 1_E\|_{L^1(\mathbb{M})} \le \sqrt{6d} \sqrt{t}\ Var(\mathbf 1_E,\mathbb{M}) = \sqrt{6d} \ \sqrt{t}\ P(E;\mathbb{M})$
Observe now that, using $P_t 1 = 1$ , we have
$||P_t \mathbf 1_E - \mathbf 1_E||_{L^1(\mathbb{M})} = 2\left(\mu(E) - \int_E P_t \mathbf 1_E d\mu\right).$
On the other hand,
$\int_E P_t \mathbf 1_E d\mu = \int_\mathbb{M} \left(P_{t/2}\mathbf 1_\mathbb{M}\right)^2 d\mu.$
We thus obtain
$||P_t \mathbf 1_E - \mathbf 1_E||_{L^1(\mathbb{M})} = 2 \left(\mu(E) - \int_\mathbb{M} \left(P_{t/2}\mathbf 1_E\right)^2 d\mu\right).$
We now observe that the assumption (1) implies
$p(x,x,t) \le \frac{C_4}{t^{n/2}},\ \ \ x\in \mathbb{M}, t > 0.$
This gives
$\int_\mathbb{M} (P_{t/2} \mathbf 1_E)^2 d\mu \le \left(\int_E \left(\int_\mathbb{M} p(x,y,t/2)^2 d\mu(y)\right)^{\frac{1}{2}}d\mu(x)\right)^2$
$= \left(\int_E p(x,x,t)^{\frac{1}{2}}d\mu(x)\right)^2 \le \frac{C_4}{t^{n/2}} \mu(E)^2.$
Combining these equations we reach the conclusion
$\mu(E) \le \frac{\sqrt{6d}}{2} \ \sqrt{t}\ P(E;\mathbb{M}) + \frac{C_4}{t^{n/2}} \mu(E)^2$
Now the absolute minimum of the function $g(t) = A t^\alpha + B t^{-\beta}$ , $t > 0$ , where $A, B, \alpha, \beta > 0$ , is given by
$g_{\min} = \left[\left(\frac{\alpha}{\beta}\right)^{\frac{\beta}{\alpha + \beta}} + \left(\frac{\beta} {\alpha}\right)^{\frac{\alpha}{\alpha + \beta}}\right] A^{\frac{\beta}{\alpha + \beta}} B^{\frac{\alpha}{\alpha + \beta}}$
Applying this observation with $\alpha = \frac{1}{2}, \beta = \frac{n}{2}$ , we conclude
$\mu(E)^{\frac{n-1}{n}} \le C_3 P(E,\mathbb{M}).$
The fact that 3) implies 4) is classical geometric measure theory. It relies on the co-area formula that we recall: For every $f,g \in C_0^\infty(\mathbb{M})$ ,
$\int_\mathbb{M} g \| \nabla f \| d\mu=\int_{-\infty}^{+\infty} \left( \int_{f(x)=t} g(x) d\mu_{n-1}(x) \right) dt.$
Let now $f \in C_0^\infty(\mathbb{M})$ . We have
$f(x)=\int_0^{+\infty} \mathbf{1}_{f(x) > t} (t) dt.$
By using Minkowski inequality, we get then
$\| f \|_{\frac{n}{n-1}} \le \int_0^\infty \| \mathbf{1}_{f(\cdot) > t} \|_{\frac{n}{n-1}}dt$
$\le \int_0^\infty \mu ( f > t )^{\frac{n}{n-1}}dt$
$\le C_3 \int_0^\infty \mu_{n-1} ( f=t) dt =C_3 \int_\mathbb{M} \sqrt{\Gamma(f)} d\mu$

Finally, we show that $(4) \rightarrow (1)$ . In what follows we let $\nu = n/(n-1)$ . Let $p,q\in (0,\infty)$ and $0 < \theta\le 1$ be such that
$\frac{1}{p} = \frac{\theta}{\nu} + \frac{1-\theta}{q}.$
Holder inequality, combined with assumption (4), gives for any $f\in Lip_d(\mathbb{M})$ with compact support
$||f||_{L^p(\mathbb{M})} \le ||f||^\theta_{L^{\nu}(\mathbb{M})} ||f||^{1-\theta}_{L^q(\mathbb{M})}\le \left(C_3 ||\sqrt{\Gamma(f)}||_{L^1(\mathbb{M})}\right)^\theta ||f||^{1-\theta}_{L^q(\mathbb{M})}.$
For any $x\in \mathbb{M}$ and $r > 0$ we now let $f(y) = (r-d(y,x))^+$ . Clearly such $f\in Lip_d(\mathbb{M})$ and supp $\ f = \overline B(x,r)$ . Since with this choice $||\sqrt{\Gamma(f)}||_{L^1(\mathbb{M})}^\theta \le \mu(B(x,r))^{\theta}$ , the above inequality implies
$\frac{r}{2} \mu(B(x,\frac{r}{2})^{\frac{1}{p}} \le r^{1-\theta} \left(C_3 \mu(B(x,r)\right)^\theta \mu(B(x,r))^{\frac{1-\theta}{q}},$
which, noting that $\frac{1-\theta}{q} + \theta = \frac{n+\theta p}{pn}$ , we can rewrite as follows
$\mu(B(x,r)) \ge \left(\frac{1}{2C_3^\theta}\right)^{pa} \mu(B(x,\frac{r}{2}))^a r^{\theta p a},$
where we have let $a = \frac{n}{n+\theta p}$ . Notice that $0 < a < 1$ .
Iterating the latter inequality we find
$\mu(B(x,r)) \ge \left(\frac{1}{2C_3^\theta}\right)^{p\sum_{j=1}^k a^j} r^{\theta p \sum_{j=1}^k a^j} 2^{-\theta p \sum_{j=1}^k (j-1)a^j}\mu(B(x,\frac{r}{2^k}))^{a^k},\ \ \ k\in \mathbb N.$
From the doubling property for any $x\in \mathbb{M}$ there exist constants $C, R > 0$ such that with $Q = \log_2 C$ one has
$\mu(B(x,tr)) \ge C^{-1} t^{Q} \mu(B(x,r)),\ \ \ 0\le t\le 1, 0<r\le R.$
This estimate implies that
$\underset{k\to \infty}{\liminf}\ \mu(B(x,\frac{r}{2^k}))^{a^k}\ge 1,\ \ \ x\in \mathbb{M}, r > 0.$
Since on the other hand $\sum_{j=1}^\infty a^j = \frac{n}{\theta p}$ , and $\sum_{j=1}^\infty (j-1) a^j = \frac{n^2}{\theta^2p^2}$ , we conclude that
$\mu(B(x,r)) \ge \left(2^{-\frac{1}{\theta}(1+\frac{n}{p})} C_3^{-1}\right)^n r^n,\ \ \ x\in \mathbb{M}, r > 0.$
This establishes (1), thus completing the proof $\square$