Lecture 5. Hormander’s type operators

Posted on September 20, 2016 by Fabrice Baudoin

For geometric purposes, it is often very useful to use the language of vector fields to study diffusion operators.
Let \mathcal{O} \subset \mathbb{R}^n be a non-empty open set. A
smooth vector field  V on \mathcal{O} is a smooth map

\begin{array}{llll} V: & \mathcal{O} & \rightarrow & \mathbb{R}^{n} \\ & x & \rightarrow & (v_{1}(x),...,v_{n}(x)). \end{array}

We will often regard a vector field V as a differential operator acting on the space of smooth functions f: \mathcal{O} \rightarrow \mathbb{R} as follows:

(Vf) (x)=\sum_{i=1}^n v_i (x) \frac{\partial f}{\partial x_i}.
By the chain rule, we note that V is a derivation, that is an operator on
C^{\infty} (\mathcal{O} ), linear over \mathbb{R}, satisfying for f,g \in C^{\infty} (\mathcal{O} ),

V(fg)=(Vf)g +f (Vg).
Conversely, it easily seen that any derivation on C^{\infty} (\mathcal{O} ) defines a vector field on \mathcal{O} (Pick x_0 \in \mathcal{O}, observe that if g is a smooth function such that g(x_0)=0 then V((x-x_0)g)(x_0)=0 and then, to compute Vf(x_0) use a Taylor expansion around x_0).

With these notations, it is readily checked that if V_0,V_1,\cdots, V_d are smooth vector fields on \mathbb{R}^n, then the second order differential operator

L=V_0+\sum_{i=1}^d V_i^2
is a diffusion operator. Here V_i^2 has to be understood as the operator $V_i$ composed with itself. Diffusion operators that may be written under the previous form are called Hormander’s type diffusion operators. It is easily seen that a Hormander’s type diffusion operator is elliptic in \mathbb{R}^n if and only if for every x \in \mathbb{R}^n, the vectors V_1(x),\cdots,V_d(x) form a basis of \mathbb{R}^n.
Proposition:  Let

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},

be a diffusion operator on \mathbb{R}^n such that the \sigma_{ij}‘s and the b_i‘s are smooth functions. Let us assume that for every x \in \mathbb{R}^n, the rank of the matrix (\sigma_{ij}(x))_{1 \le i,j \le n} is constant equal to d. Then, there exist smooth vector fields V_0,V_1,\cdots, V_d on \mathbb{R}^n such that V_1,\cdots, V_d are linearly independent and

L=V_0+\sum_{i=1}^d V_i^2.

 

Proof:  We first assume d=n.  Since the matrix (\sigma_{ij}(x))_{1 \le i,j \le n} is symmetric and positive, it admits a unique symmetric and positive square root v(x)=(v_{ij}(x))_{1 \le i,j \le n}. Let us assume for a moment that the v_{ij}‘s are smooth functions, in that case by denoting

V_i=\sum_{j=1}^n v_{ij} \frac{\partial}{\partial x_j},
the vector fields V_1,\cdots,V_n are linearly independent and it is readily seen that the differential operator

L-\sum_{i=1}^n V_i^2
is actually a first order differential operator and thus a vector field. We therefore are let to prove that the v_{ij}‘s are smooth functions.  Let \mathcal{O} be a bounded non empty set of \mathbb{R}^n and \Gamma be any contour in the half plane \mathbf{Re} (z) >0 that contains all the eigenvalues of \sigma(x), $x \in \mathcal{O}$. We claim that

v(x)=\frac{1}{2 \pi} \int_{\Gamma} \sqrt{z} (\sigma(x)-z\mathbf{I}_n)^{-1} dz, \quad x \in \mathcal{O}.
Indeed, if \Gamma' is another contour in the half plane \mathbf{Re} (z) >0 whose interior contains \Gamma, as a straightforward application of the Fubini’s theorem and Cauchy’s formula we have

\begin{array}{ll} & \frac{1}{2 \pi} \int_{\Gamma} \sqrt{z} (\sigma(x)-z\mathbf{I}_n)^{-1} dz \times \frac{1}{2 \pi} \int_{\Gamma'} \sqrt{z} (\sigma(x)-z\mathbf{I}_n)^{-1} dz \\ =&\frac{1}{4 \pi^2} \int_{\Gamma'}\int_{\Gamma} \sqrt{z z'} (\sigma(x)-z\mathbf{I}_n)^{-1}(\sigma(x)-z'\mathbf{I}_n)^{-1} dz dz' \\ = & \frac{1}{4 \pi^2} \int_{\Gamma'}\int_{\Gamma} \sqrt{z z'} \frac{ (\sigma(x)-z\mathbf{I}_n)^{-1}-(\sigma(x)-z'\mathbf{I}_n)^{-1}}{z-z'} dz dz' \\ = & - \frac{1}{4 \pi^2} \int_{\Gamma}\int_{\Gamma'} \sqrt{z z'} \frac{ (\sigma(x)-z\mathbf{I}_n)^{-1}}{z'-z} dz' dz \\ =& - \frac{1}{4 \pi^2} \int_{\Gamma} (\sigma(x)-z\mathbf{I}_n)^{-1} \left( \int_{\Gamma'}\frac{ \sqrt{z z'} }{z'-z} dz' \right) dz \\ =& \frac{1}{2 i \pi} \int_{\Gamma} (\sigma(x)-z\mathbf{I}_n)^{-1} z dz \\ =& \frac{1}{2 i \pi} \int_{\Gamma} (\sigma(x)-z\mathbf{I}_n)^{-1} \sigma(x) dz. \end{array}
In the last expression above, we may modify \Gamma into a circle \Gamma_R =\{ z, | z|=R \}. Then by choosing R big enough (R > \sup_{x \in \mathcal{O}} \sqrt{ \| \sigma (x) \|}), and expanding  \int_{\Gamma} (\sigma(x)-z\mathbf{I}_n)^{-1} dz in powers of z, we see that

\lim_{R \to +\infty} \frac{1}{2 i \pi} \int_{\Gamma_R} (\sigma(x)-z\mathbf{I}_n)^{-1} dz= \mathbf{Id} .
As a conclusion,

\left( \frac{1}{2 \pi} \int_{\Gamma} \sqrt{z} (\sigma(x)-z\mathbf{I}_n)^{-1} dz \right)^2 = \sigma(x),
so that, as we claimed it,

v(x)=\frac{1}{2 \pi} \int_{\Gamma} \sqrt{z} (\sigma(x)-z\mathbf{I}_n)^{-1} dz, \quad x \in \mathcal{O}.
This expression of the square root of \sigma clearly shows that the v_{ij}‘s are smooth functions. By putting things together, we therefore proved the proposition in the case d=n.

Let us now turn to the case d < n. By smoothly choosing an orthonormal basis of \mathbb{R}^n which is adapted to the orthogonal decomposition

\mathbb{R}^n=\mathbf{Ker} (\sigma(x) ) \oplus \mathbf{Ker} (\sigma(x) )^\perp,
we get a decomposition

\sigma(x)=M(x)\left( \begin{array}{ll} \mathcal{V}(x) & 0 \\ 0 & 0 \end{array} \right)M(x)^{-1},
where M(x) is an orthogonal matrix with smooth coefficients and where $\mathcal{V}(x)$ is a d \times d symmetric and positive matrix with smooth coefficients. We may now apply the first part of the proof to the matrix \mathcal{V}(x) and easily conclude \square

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