Lecture 5. Hormander’s type operators

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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