## Lecture 1. Diffusion operators

Definition: A differential operator $L$ on $\mathbb{R}^n$, is called a diffusion operator if it can be written

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

where $b_i$ and $\sigma_{ij}$ are continuous functions on $\mathbb{R}^n$ and if for every $x \in \mathbb{R}^n$, the matrix $(\sigma_{ij}(x))_{1\le i,j\le n}$ is a symmetric and nonnegative matrix.

If for every $x \in \mathbb{R}^n$ the matrix $(\sigma_{ij}(x))_{1\le i,j\le n}$ is positive definite, then the operator $L$ is said to be elliptic. The first example of a diffusion operator is the Laplace operator on $\mathbb{R}^n$:

$\Delta=\sum_{i=1}^n \frac{\partial^2}{\partial x_i^2}.$

It is of course an elliptic operator.

One of the first property of diffusion operators is that they satisfy a maximum principle. Before we state this principle let us recall the following simple result from linear algebra.

Lemma. Let $A$ and $B$ be two symmetric and nonnegative matrices, then $\mathbf{tr} (AB) \ge 0.$

Proof:

Since $A$ is symmetric and non negative, there exists a symmetric and non negative matrix $S$ such that $S^2=A$. We have then

$\mathbf{tr} ( AB)=\mathbf{tr} ( S^2 B)=\mathbf{tr} ( S B S)=\mathbf{tr} ( S^* BS).$

The matrix $S^* BS$ is seen to be symmetric and nonnegative and thus has a non negative trace.$\square$

Proposition (Maximum principle for diffusion operators). Let $f:\mathbb{R}^n \rightarrow \mathbb{R}$ be a smooth function that attains a local minimum at $x$. If $L$ is a diffusion operator then $Lf(x) \ge 0$.

Proof:

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

and let $f:\mathbb{R}^n \rightarrow \mathbb{R}$ be a smooth function that attains a local minimum at $x$. We have

$Lf(x) =\sum_{i,j=1}^n \sigma_{ij} (x) \frac{\partial^2 f}{ \partial x_i \partial x_j} (x)$
$=\mathbf{tr} \left( \sigma (x) \mathbf{Hess } f (x) \right),$

where $\sigma(x)$ is the symmetric and non negative matrix with coefficients $\sigma_{ij}(x)$ and $\mathbf{Hess } f (x)$ is the Hessian matrix of $f$, that is the symmetric matrix with coefficients $\frac{\partial^2 f}{ \partial x_i \partial x_j} (x)$. Since $x$ is a local minimum of $f$, $\mathbf{Hess } f (x)$ is a non negative matrix. We can now use the previous lemma to get the expected result $\square$

It is remarkable that, together with the linearity, this maximum principle characterizes the diffusion operators:

Theorem. Let $\mathcal{C}^{\infty} (\mathbb{R}^n)$ be the set of smooth functions $\mathbb{R}^n \rightarrow \mathbb{R}$ and let $\mathcal{C} (\mathbb{R}^n)$ be the set of continuous functions $\mathbb{R}^n \rightarrow \mathbb{R}$. Let now $L: \mathcal{C}^{\infty} (\mathbb{R}^n) \rightarrow \mathcal{C} (\mathbb{R}^n)$ be an operator such that:

• $L$ is linear;
• If $f \in \mathcal{C}^{\infty} (\mathbb{R}^n)$ has a local minimum at $x$, $Lf (x) \ge 0$.

Then $L$ is a diffusion operator.

Proof:

Let us consider an operator $L$ that satisfies the two above properties. As a first observation, it is readily seen from the third point that $L$ transforms constant functions into the zero function. Let now $y \in \mathbb{R}^n$ be fixed in the following argument. We are going to show that if $g$ is a smooth function, then

$L( \| x-y \|^3 g) (y) =0.$

The idea will then be to use the Taylor expansion formula. For $\varepsilon >0$, the function

$x \rightarrow \| x-y \|^3 g(x) +\varepsilon \| x -y \|^2$

admits a local minimum at $y$, thus

$L( \| x-y \|^3 g) (y) \ge - \varepsilon L ( \| x -y \|^2)(y).$

By letting $\varepsilon \to 0$, we therefore obtain

$L( \| x-y \|^3 g) (y) \ge 0.$

By considering now the function

$x \rightarrow \| x-y \|^3 g(x) -\varepsilon \| x -y \|^2,$

we show in the very same way that

$L( \| x-y \|^3 g) (y) \le 0.$

As a conclusion

$L( \| x-y \|^3 g) (y) = 0.$

Let now $f$ be a smooth function. By the Taylor expansion formula, there exists a smooth function $g$ such that in a neighborhood of $y$

$f(x)$
$=f(y)+\sum_{i=1}^n (x_i -y_i) \frac{\partial f}{\partial x_i}(y)+\frac{1}{2} \sum_{i,j=1}^n (x_i-y_i)(x_j-y_j) \frac{\partial^2 f}{\partial x_i \partial x_j} (y) + \| x-y \|^3 g(x).$

By applying the operator $L$ to the previous equality, and by taking account the previous observations we obtain

$Lf(y)=\sum_{i=1}^n L(x_i -y_i)(y) \frac{\partial f}{\partial x_i}(y)+\frac{1}{2} \sum_{i,j=1}^n L((x_i-y_i)(x_j-y_j))(y) \frac{\partial^2 f}{\partial x_i \partial x_j} (y).$

By denoting now,

$b_i(y)= L(x_i -y_i)(y),$

and

$\sigma_{ij} (y)=\frac{1}{2} L((x_i-y_i)(x_j-y_j))(y) ,$

we reach the conclusion

$Lf(y)=\sum_{i,j=1}^n \sigma_{ij} (y) \frac{\partial^2 f}{ \partial x_i \partial x_j}(y) +\sum_{i=1}^n b_i (y)\frac{\partial f}{\partial x_i}(y).$

The matrix, $(\sigma_{ij}(y))_{1\le i,j \le n}$ is seen to be non negative, because for every $\lambda \in \mathbb{R}^n$,

$\sum_{i,j=1}^n \lambda_i \lambda_j \sigma_{ij} (y)$
$= \frac{1}{2}\sum_{i,j=1}^n \lambda_i \lambda_j L((x_i-y_i)(x_j-y_j))(y)$
$=\frac{1}{2} L( \langle \lambda , x-y \rangle^2)(y).$

Now, the function $x \rightarrow \langle \lambda , x-y \rangle^2$ is seen to attain a local minimum at $y$, so that from the maximum principle

$L( \langle \lambda , x-y \rangle^2)(y) \ge 0.$

Finally, since $L$ transforms smooth into continuous functions, the functions $b_i$‘s and $\sigma_{ij}$‘s are seen to be continuous $\square$

Exercise. Let $L: \mathcal{C}^{\infty} (\mathbb{R}^n) \rightarrow \mathcal{C} (\mathbb{R}^n)$ be a linear operator such that:

• $L$ is a local operator, that is if $f=g$ on a neighborhood of $x$ then $Lf(x)=Lg(x)$;
• If $f \in \mathcal{C}^{\infty} (\mathbb{R}^n)$ has a global maximum at $x$ with $f(x)\ge 0$ then $Lf (x) \le 0$.

Show that for $f \in \mathcal{C}^{\infty} (\mathbb{R}^n)$ and $x \in \mathbb{R}^n$,

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

where $b_i$, $c$ and $\sigma_{ij}$ are continuous functions on $\mathbb{R}^n$ such that for every $x \in \mathbb{R}^n$, $c(x) \ge 0$ and the matrix $(\sigma_{ij}(x))_{1\le i,j\le n}$ is symmetric and nonnegative.

The previous theorem is actually a special case of a beautiful theorem that is due to Courrège that classifies the operators satisfying the positive global maximum principle. We mention this theorem without proof because the result will not be needed in the following. A complete proof may be found in the original article by Courrège.

We denote by $\mathcal{C}_c^{\infty} (\mathbb{R}^n)$ the space of smooth and compactly supported functions $\mathbb{R}^n \rightarrow \mathbb{R}$. A linear operator $A:\mathcal{C}_c^{\infty} (\mathbb{R}^n) \rightarrow \mathcal{C} (\mathbb{R}^n)$ is said to satisfy the positive maximum principle if for every function $f \in \mathcal{C}_c^{\infty}(\mathbb{R}^n)$ that has a global maximum at $x$ with $f(x)\ge 0$ then $Af (x) \le 0$.

In the following statement $\mathcal{B}(\mathbb{R}^n)$ denotes the set of Borel sets on $\mathbb{R}^n$ and a kernel $\mu$ on $\mathbb{R}^n \times \mathcal{B}(\mathbb{R}^n)$ is a family $\left\{ \mu(x,\cdot), x \in \mathbb{R}^n\right\}$ of Borel measures.

Theorem (Courrège’s theorem) Let $A:\mathcal{C}_c^{\infty} (\mathbb{R}^n) \rightarrow \mathcal{C} (\mathbb{R}^n)$ be a linear operator. Then $A$ satisfies the positive maximum principle if and only if there exist functions $(\sigma_{ij}(x))_{1\le i,j\le n}$, $b_i$, $c:\mathbb{R}^n \rightarrow \mathbb{R}$ and a kernel $\mu$ on $\mathbb{R}^n \times \mathcal{B}(\mathbb{R}^n)$ such that for every $f \in \mathcal{C}_c^{\infty} (\mathbb{R}^n)$ and $x \in \mathbb{R}^n$,

$Af(x) = \sum_{i,j=1}^n \sigma_{ij} (x) \frac{\partial^2 f}{ \partial x_i \partial x_j} +\sum_{i=1}^n b_i (x)\frac{\partial f}{\partial x_i} -c(x)f(x)$
$+\int_{\mathbb{R}^n} \left( u(y) -\chi(y-x)u(x)-\sum_{j=1}^n\frac{\partial u}{\partial x_j}(x)\chi(y-x)(y_j-x_j) \right)\mu(x,dy),$

where $\chi \in \mathcal{C}_c^{\infty} (\mathbb{R}^n)$, with $0 \le \chi \le 1$ takes the constant value 1 on the ball $\mathbf{B}(0,1)$. In addition, for every $x \in \mathbb{R}^n$, $c(x) \ge 0$ and the matrix $(\sigma_{ij}(x))_{1\le i,j\le n}$ is a symmetric and nonnegative matrix. The functions $b_j$‘s and $c$ are continuous. Moreover for every $y \in \mathbb{R}^n$, the function $x \to \sum_{i,j} \sigma_{ij}(x) y_i y_j$ is upper semicontinuous.

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### 2 Responses to Lecture 1. Diffusion operators

1. luc says:

Bonjour,

1. In the proof of the first theorem, you write “satisfies the three above properties”, aren’t there just two in the list ?

2. Concerning second condition in the exercise ” ….. $Lf(x) \geq 0$ ” , I am wondering if it should be replaced by $Lf(x) \leq 0$, which is consistent with the statement of Th\’eor\eme 0.1 of the paper by courr\ege.

Best, luc

2. Merci ! I made the corrections.