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.

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