Lecture 19. Volume doubling property

In this Lecture we consider a complete and $n$ -dimensional Riemannian manifold $(\mathbb{M},g)$ with non negative Ricci curvature. Our goal is to prove the following fundamental result, which is known as the volume doubling property.

Theorem: There exists a constant $C=C(n) > 0$ such that for every $x\in \mathbb{M}$ and every $r > 0$ one has $\mu(B(x,2r))\le C \mu (B(x,r)).$

Actually by suitably adapting the arguments given in this Lecture, the previous result can be extended to the case of negative Ricci curvature as follows:

Theorem: Assume $\mathbf{Ric} \ge -K$ with $K \ge 0$ . There exist positive constants $C_1=C_1(n,K), C_2=C_2(n,K)$ such that for every $x\in \mathbb{M}$ and every $r > 0$ one has
$\mu(B(x,2r))\le C_1e^{KC_2 r^2} \mu (B(x,r)).$

For simplicity, we show the arguments in the case $K=0$ and let the reader work out the arguments in the case $K \neq 0$ .

This result can be obtained from geometric methods as a consequence of the Bishop-Gromov comparison theorem. The proof we give instead only relies on the previous methods and has the advantage to generalize to a much larger class of operators than Laplace-Beltrami on Riemannian manifolds.

The key heat kernel estimate that leads to the doubling property is the following uniform and scale invariant lower bound on the heat kernel measure of balls.

Theorem: There exist an absolute constant $K > 0$ , and $A > 0$ , depending only on $n$ , such that
$P_{Ar^2}(\mathbf 1_{B(x,r)})(x) \ge K, \ \ \ \ \ x\in \mathbb{M}, r > 0.$

Proof: We first recall the following result that was proved in a previous Lecture: Let $a \in C^1([0,T],[0,\infty))$ and $\gamma \in C((0,T),\mathbb{R})$ . Given $f \ge 0$ , which is bounded and such that $\sqrt{f}$ is Lipschitz, we have
$a(T) P_T \left( f \Gamma (\ln f) \right) -a(0)(P_{T} f) \Gamma (\ln P_{T}f)$
$\ge \int_0^T \left(a'+2\rho a -\frac{4a\gamma}{n} \right)\Phi (s) ds +\left(\frac{4}{n}\int_0^T a\gamma ds\right)LP_{T} f -\left(\frac{2 }{n}\int_0^T a\gamma^2ds\right)P_T f.$

We choose $a(t)=\tau+T-t$ ,and $\gamma(t)=-\frac{n}{4(\tau+T-t)}$ where $\tau > 0$ will later be optimized. Noting that we presently have
$a' = 1,\ \ \ a\gamma = - \frac{n}{4},\ \ \ \ a\gamma^2 = \frac{n^2}{16(\tau + T - t)^2},$
we obtain the inequality
$\tau P_T(f \Gamma(\ln f)) -(T+\tau) P_T f \Gamma(\ln P_T f) \ge -T L P_T f -\frac{n}{8} \ln \left( 1+\frac{T}{\tau}\right) P_T f$
In what follows we consider a bounded function $f$ on $\mathbb{M}$ such that $\Gamma(f) \le 1$ almost everywhere on $\mathbb{M}$ . For any $\lambda \in \mathbb R$ we consider the function $\psi$ defined by
$\psi(\lambda,t) = \frac{1}{\lambda} \log P_t(e^{\lambda f}), \ \ \ \text{or alternatively}\ \ \ P_t(e^{\lambda f}) = e^{\lambda \psi}.$
Notice that Jensen’s inequality gives $\lambda \psi \ge \lambda P_t f,$ and so we have $P_t f \le \psi.$ We now apply the previous inequality to the function $e^{\lambda f}$ , obtaining
$\lambda^2 \tau P_T\left(e^{\lambda f} \Gamma(f)\right) - \lambda^2 (T+\tau) e^{\lambda \psi} \Gamma(\psi) \ge - T L P_T(e^{\lambda f}) - \frac{n}{8} e^{\lambda \psi} \ln\left(1+\frac{T}{\tau}\right).$
Keeping in mind that $\Gamma(f) \le 1$ , we see that $P_T(e^{\lambda f} \Gamma(f)) \le e^{\lambda \psi}.$ Using this observation in combination with the fact that
$L \left(P_t (e^{\lambda f})\right) = \frac{\partial}{\partial t} \left(P_t (e^{\lambda f})\right) = \frac{\partial e^{\lambda \psi}}{\partial t} = \lambda e^{\lambda \psi} \frac{\partial \psi}{\partial t} ,$
and switching notation from $T$ to $t$ , we infer
$\lambda^2 \tau \ge \lambda^2 (t+\tau) e^{\lambda \psi} \Gamma(\psi)- \lambda t \frac{\partial \psi}{\partial t} - \frac{n}{8} \ln\left(1+\frac{t}{\tau}\right).$
The latter inequality finally gives
$\frac{\partial \psi}{\partial t} \ge - \frac{\lambda}{t}\left(\tau + \frac{n}{8\lambda^2} \ln\left(1+\frac{t}{\tau}\right)\right)\ge - \frac{\lambda}{t}\left(\tau + \frac{nt}{8\lambda^2\tau} \right).$
We now optimize the right-hand side of the inequality with respect to $\tau$ . We notice explicitly that the maximum value of the right-hand side is attained at $\tau_0 = \sqrt{\frac{nt}{8 \lambda^2} }$ . We find therefore
$\frac{\partial \psi}{\partial t} \ge -\sqrt{\frac{n}{2t} }$
We now integrate the inequality between $s$ and $t$ , obtaining
$\psi(\lambda,s) \le \psi(\lambda,t) +\sqrt{ \frac{n}{2}} \int_{s}^t \frac{d\tau}{\sqrt{\tau}}.$
We infer then
$P_s(\lambda f) \le \lambda \psi(\lambda,t) + \lambda \sqrt{ 2nt}.$
Letting $s\to 0^+$ we conclude
$\lambda f \le \lambda \psi(\lambda,t) + \lambda \sqrt{ 2nt}.$

At this point we let $B = B(x,r) = \{x\in \mathbb{M}\mid d(y,x) < r\}$ , and consider the function $f(y) = - d(y,x)$ . Since we clearly have $e^{\lambda f} \le e^{-\lambda r} \mathbf 1_{B^c} + \mathbf 1_B,$ it follows that for every $t > 0$ one has $e^{\lambda \psi(\lambda,t)(x)} = P_t(e^{\lambda f})(x) \le e^{-\lambda r} + P_t(\mathbf 1_B)(x).$ This gives the lower bound $P_t(\mathbf 1_B)(x) \ge e^{\lambda \psi(\lambda,t)(x)} - e^{-\lambda r}$ . To estimate the first term in the right-hand side of the latter inequality, we use the previous estimate which gives $P_{t}(\mathbf 1_B)(x) \ge e^{-\lambda \sqrt{2nt}} - e^{-\lambda r}.$ To make use of this estimate, we now choose $\lambda = \frac{1}{r}$ , $t = Ar^2$ , obtaining $P_{Ar^2}(\mathbf 1_B)(x) \ge e^{-A\sqrt{2n}} - e^{-1}$ . The conclusion follows then easily $\square$

We now turn to the proof of the volume doubling property. We first recall the following basic result which is a straightforward consequence of the Li Yau inequality.

Proposition: Let $p(x,y,t)$ be the heat kernel on $\mathbb{M}$ . For every $x,y, z\in \mathbb{M}$ and every $0 < s < t < \infty$ one has
$p(x,y,s) \le p(x,z,t) \left(\frac{t}{s}\right)^{\frac{n}{2}} \exp\left( \frac{d(y,z)^2}{4(t-s)} \right).$

We are now in position to prove the doubling.

From the semigroup property and the symmetry of the heat kernel we have for any $y\in \mathbb{M}$ and $t > 0$ $p(y,y,2t) = \int_\mathbb{M} p(y,z,t)^2 d\mu(z).$ Consider now a function $h\in C^\infty_0(\mathbb{M})$ such that $0\le h\le 1$ , $h\equiv 1$ on $B(x,\sqrt{t}/2)$ and $h\equiv 0$ outside $B(x,\sqrt t)$ . We thus have
$P_t h(y) = \int_\mathbb{M} p(y,z,t) h(z) d\mu(z) \le \left(\int_{B(x,\sqrt t)} p(y,z,t)^2 d\mu(z)\right)^{\frac{1}{2}} \left(\int_\mathbb{M} h(z)^2 d\mu(z)\right)^{\frac{1}{2}}$
$\le p(y,y,2t)^{\frac{1}{2}} \mu(B(x,\sqrt t))^{\frac{1}{2}}.$
If we take $y=x$ , and $t =r^2$ , we obtain
$P_{r^2} \left(\mathbf 1_{B(x,r)}\right)(x)^2 \le P_{r^2} h(x)^2 \leq p(x,x,2r^2)\ \mu(B(x,r)).$
At this point we use the crucial previous theorem, which gives for some $0 < A = A(n) < 1$
$P_{Ar^2}(\mathbf 1_{B(x,r)})(x) \ge K, \ \ \ \ \ x\in \mathbb{M}, r > 0.$ Combining the latter inequality with the Harnack inequality, we obtain the following on-diagonal lower bound
$p(x,x,2r^2) \ge \frac{K^*}{\mu(B(x,r))},\ \ \ \ \ x\in \mathbb{M},\ r> 0.$
Applying the Harnack inequality for every $y\in B(x,\sqrt t)$ we find
$p(x,x,t) \le C(n) p(x,y,2t).$
Integration over $B(x,\sqrt t)$ gives
$p(x,x,t)\mu(B(x,\sqrt t)) \le C(n) \int_{B(x,\sqrt t)}p(x,y,2t)d\mu(y) \le C(n),$ where we have used $P_t1\le 1$ . Letting $t = r^2$ , we obtain from this the on-diagonal upper bound $p(x,x,r^2) \le \frac{C(n)}{\mu(B(x,r))}$ . We finally obtain
$\mu(B(x,2r)) \le \frac{C}{p(x,x,4r^2)} \le \frac{C^*}{p(x,x,2r^2)} \le C^{**} \mu(B(x,r)),$
where we have used once more the Harnack inequality,
which gives $\frac{p(x,x,2r^2)}{p(x,x,4r^2)}\le C$ .