# A note on the generator of some Levy process

In this below, we will discuss some connections on the several representations on the generator of Levy process.

For simplicity, we only discuss one dimensional Levy measure. Throughout this discussion, we assume ${\nu}$ is a Levy measure on ${\mathbb R}$, i.e. a non-negative measure satisfying

$\displaystyle \nu(\{0\}) = 0; \ \int_{\mathbb R} (1 \wedge x^{2}) \nu(dx) <\infty.$

Let’s consider an operator ${\mathcal I}$ given by

$\displaystyle \mathcal I \phi(x) = \int_{\mathbb R} [ \phi(x+y) - \phi(x) - \phi'(x) y I_{B_{1}} (y)] \nu (dy). \ \ \ \ \ (1)$

The above (1) is often given as the generator of a purely non-gaussian Levy processes in the probability literature. Another operator ${\mathcal L}$ used in [Caffarelli and Silvestre 09] is

$\displaystyle \mathcal L \phi(x) = PV \int_{\mathbb R} (\phi(x+y) - \phi(x)) \nu(dy) := \lim_{\epsilon \rightarrow 0^{+}} \int_{|y|> \epsilon} (\phi(x+y) - \phi(x)) \nu(dy). \ \ \ \ \ (2)$

The main goal of this discussion is to show that (1) and (2) are equivalent whenever ${\nu}$ is symmetric, but not in general.

Proposition 1 ${C^{2}_{b}}$ is a subset of the domain of ${\mathcal I}$, i.e. ${C^{2}_{b} \subset \mathcal D( \mathcal I)}$.

Proof: This is the consequence of Taylor expansion. $\Box$

Exercise 1 Find ${\phi \in C^{1}_{b}\setminus \mathcal D(\mathcal I)}$.

Now we can rewrite ${\mathcal I = \mathcal I_{1,1} + \mathcal I_{2,1}}$, where

$\displaystyle \mathcal I_{1,r} \phi(x) = \int_{B_{r}} (\phi(x+y) - \phi(x) - \phi'(x) y) \nu(dy)$

and

$\displaystyle \mathcal I_{2,r} \phi(x) = \int_{\mathbb R \setminus B_{r}} (\phi(x+y) - \phi(x)) \nu(dy)$

Lemma 2 ${\lim_{\epsilon \rightarrow 0^{+}} \mathcal I_{1,\epsilon} \phi(x) = 0}$ for all ${\phi \in C^{2}_{b}}$.

Proof: By Taylor expansion, one can write

$\displaystyle | \mathcal I_{1,\epsilon} \phi(x) | \le \int_{-\epsilon}^{\epsilon} |\frac 1 2 \phi''(x) y^{2} + o(y^{2})| \nu (dy) \le K \int_{B_{\epsilon}} y^{2} \nu(dy) = K f(\epsilon),$

where the function ${f(\epsilon) := \int_{B_{\epsilon}} y^{2} \nu(dy)}$ is a non-negative monotone function with ${f(0) = 0}$. Furthermore, since ${f(x)/f(1)}$ is a CDF of a probability measure on ${[0,1]}$, ${f}$ is right continuous. Therefore, ${\lim_{\epsilon \rightarrow 0^{+}} f(\epsilon) = f(0) = 0}$. $\Box$

Proposition 3 If the Levy measure ${\nu}$ is symmetric, i.e. ${\nu(A) = \nu(-A)}$ for all Borel sets ${A}$, then ${\mathcal I\phi = \mathcal L \phi}$ for all ${\phi\in C^{2}_{b}}$.

Proof: It’s equivalent to show that ${\mathcal I\phi(x) = \lim_{\epsilon \rightarrow 0^{+}} \mathcal I_{2,\epsilon} \phi(x)}$. Note for any ${\epsilon \in (0,1)}$, we can write

$\displaystyle \mathcal I\phi(x) = \mathcal I_{1, \epsilon} \phi(x) + \mathcal I_{2, \epsilon} \phi(x) + \int_{B_{1}\setminus B_{\epsilon}} (- \phi'(x) y) \nu(dy).$

The last term of the above equation is zero due to the symmetry of ${\nu}$, and it yields

$\displaystyle \mathcal I\phi(x) = \mathcal I_{1, \epsilon} \phi(x) + \mathcal I_{2, \epsilon} \phi(x).$

Finally, one can take the limit as ${\epsilon \rightarrow 0}$ and apply Lemma 2 to conclude the conclusion. $\Box$