Lecture 12. The distance associated to subelliptic diffusion operators

In this lecture we prove that most of the results that were proven for Laplace-Beltrami operators may actually be generalized to any locally subelliptic operator.
Let $L$ be a locally subelliptic diffusion operator defined on $\mathbb{R}^n$ . For every smooth functions $f,g: \mathbb{R} \rightarrow \mathbb{R}$ , we recall that the carre du champ operator is the symmetric first-order differential form defined by:

$\Gamma (f,g) =\frac{1}{2} \left( L(fg)-fLg-gLf \right).$
A straightforward computation shows that if

$L=\sum_{i,j=1}^n \sigma_{ij} (x) \frac{\partial^2}{ \partial x_i \partial x_j} +\sum_{i=1}^n b_i (x)\frac{\partial}{\partial x_i},$
then,

$\Gamma (f,g)=\sum_{i,j=1}^n \sigma_{ij} (x) \frac{\partial f}{\partial x_i} \frac{\partial g}{\partial x_j}.$
As a consequence, for every smooth function $f$ , $\Gamma(f) \ge 0.$

Definition: An absolutely continuous curve $\gamma: [0,T] \rightarrow \mathbb{R}^n$ is said to be subunit for the operator $L$ if for every smooth function $f : \mathbb{R}^n \to \mathbb{R}$ we have $\left| \frac{d}{dt} f ( \gamma(t) ) \right| \le \sqrt{ (\Gamma f) (\gamma(t)) }$ . We then define the subunit length of $\gamma$ as $\ell_s(\gamma) = T$ .

Given $x, y\in \mathbb{R}^n$ , we indicate with

$S(x,y) =\{\gamma:[0,T]\to \mathbb{R}^n \mid \gamma\ \text{is subunit for}\ \Gamma, \gamma(0) = x,\ \gamma(T) = y\}.$
In these lectures we always assume that

$S(x,y) \not= \emptyset,\ \ \ \ \text{for every}\ x, y\in \mathbb{R}^n.$
If $L$ is an elliptic operator or if $L$ is a sum of squares operator that satisfies Hormander’s condition, then this assumption is satisfied.

Under such assumption it is easy to verify that

$d(x,y) = \inf\{l_s(\gamma)\mid \gamma\in S(x,y)\},$
defines a true distance on $\mathbb{R}^n$ . This is the intrinsic distance associated to the subelliptic operator $L$ . A beautiful result by Fefferman and Phong relates the subellipticity of $L$ to the size of the balls for this metric:

Theorem: Let $x \in \mathbb{R}^n$ . There exist constants $\varepsilon >0$ , $r_0 >0$ and $C_1,C_2 >0$ such that for $0 \le r < r_0$ ,

$B(x,r) \subset B_d (x,C_1 r^\varepsilon) \subset B (x,C_2 r^\varepsilon) ,$
where $B_d$ denotes here the ball for the metric $d$ and $B$ the ball for the Euclidean metric on $\mathbb{R}^n$ .

A corollary of this result is that the topology induced by $d$ coincides with the Euclidean topology of $\mathbb{R}^n$ . The distance $d$ can also be computed using the following definition:

Proposition: For every $x,y \in \mathbb{R}^n$ ,
$d(x,y)=\sup \left\{ |f(x) -f(y) | , f \in C^\infty(\mathbb{R}^n) , \| \Gamma(f) \|_\infty \le 1 \right\},\ \ \ \ x,y \in \mathbb{R}^n.$

Proof: Let $x,y \in \mathbb{R}^n$ . We denote
$\delta (x,y)=\sup \{ | f(x)-f(y) |, f \in C_0^\infty(\mathbb{R}^n), \| \Gamma(f) \|_\infty \le 1 \}.$
Let $\gamma: [0,T] \to \mathbb{R}^n$ be a sub-unit curve such that
$\gamma(0)=x, \gamma(T)=x.$
We have $\left| \frac{d}{dt} f ( \gamma(t) ) \right| \le \sqrt{ (\Gamma f) (\gamma(t)) }$ , therefore, if $\Gamma(f) \le 1$ ,

$\left| f(y)-f(x) \right| \le T.$
As a consequence
$\delta (x,y) \le d(x,y).$

We now prove the converse inequality which is trickier. We already know that if $L$ is elliptic then $d(x,y)=\delta(x,y)$ . If $L$ is only subelliptic, we consider the sequence of operators $L_k=L+\frac{1}{k} \Delta$ where $\Delta$ is the usual Laplacian. We denote by $d_k$ the distance associated to $L_k$ . It is easy to see that $d_k$ increases with $k$ and that $d_k(x,y) \le d(x,y)$ . We can find a curve $\gamma_k:[0,1] \to \\mathbb{R}^n$ , such that $\gamma(0)=x,\gamma(1)=y$ and for every $f \in C^\infty(\mathbb{R}^n)$ ,

$\left| \frac{d}{dt} f(\gamma_k(t)) \right|^2 \le \left(d^2_k(x,y)+\frac{1}{k} \right)\left( \Gamma(f) (\gamma_k(t)) + \frac{1}{k} \Gamma_\Delta(f)(\gamma_k(t)) \right),$
where $\Gamma_\Delta$ is the carre du champ operator of $\Delta$ . Since $d_k \le d$ , we see that the sequence $\gamma_k$ is uniformly equicontinuous. As a consequence of the Arzela-Ascoli theorem, we deduce that there exists a subsequence which we continue to denote $\gamma_k$ that converges uniformly to a curve $\gamma:[0,1] \to \mathbb{R}^n$ , such that $\gamma(0)=x,\gamma(1)=y$ and for every $f \in C^\infty(\mathbb{R}^n)$ ,

$\left| \frac{d}{dt} f(\gamma (t)) \right|^2 \le \sup_k d^2_k(x,y) \Gamma(f) (\gamma_k(t)).$
By definition of $d$ , we deduce $d(x,y) \le \sup_k d_k(x,y)$ . As a consequence, we proved that $d(x,y)=\lim_{k \to \infty} d_k(x,y)$ . Since it is clear that

$d_k(x,y)= \sup \left\{ |f(x) -f(y) | , f \in C^\infty(\mathbb{R}^n) , \left\| \Gamma(f) + \frac{1}{k} \Gamma_\Delta(f) \right\|_\infty \le 1 \right\} \le \delta(x,y),$
we finally conclude that $d(x,y) \le \delta(x,y)$ , hence $d(x,y)=\delta(x,y)$ .
$\square$

A straightforward corollary of the previous proposition is the following useful result:

Corollary: If $f \in C^\infty(\mathbb{R}^n)$ satisfies $\Gamma(f)=0$ , then $f$ is constant.
The Hopf-Rinow theorem is still true with an identical proof in the case of subelliptic operators.

Theorem: The metric space $(\mathbb{R}^n,d)$ is complete (i.e. Cauchy sequences are convergent) if and only the compact sets are the closed and bounded sets.
Similarly, we also have the following key result:
Proposition: There exists an increasing sequence $h_n\in C^\infty_0(\mathbb{R}^n)$ , $0 \le h_n \le 1$ , such that $h_n\nearrow 1$ on $\mathbb{R}^n$ , and $\| \Gamma(h_n) \|_{\infty} \to 0$ , as $n\to \infty$ if and only if the metric space $(\mathbb{R}^n,d)$ is complete.