Let $(X_t)_{t\ge0}$ be a real-valued Lévy process. Note that $$\mu_t:=\mathcal L(X_t)\;\;\;\text{for }t\ge0$$ is a continuous convolution semigroup$^1$. Let $$\tau_x:\mathbb R\to\mathbb R\;,\;\;\;y\mapsto y+x.$$ $(X_t)_{t\ge0}$ is a time-homogeneous Markov process with transiton semigroup $$\kappa_t(x,B)=\tau_x(\mu_t)(B)=\mu_t(B-x)\;\;\;\text{for }(x,B)\in\mathbb R\times\mathcal B(\mathbb R)\text{ and }t\ge0.$$ If $f:\mathbb R\to\mathbb R$ is bounded and uniformly continuous, it's easy to see that $$\left\|\kappa_tf-f\right\|_\infty\xrightarrow{t\to0+}0\tag1.$$ So, $(\kappa_t)_{t\ge0}$ is a strongly continuous contraction semigroup on the space $U$ of those $f$ equipped with the supremum norm.

Now assume the characteristic function $\varphi_\mu$ of $\mu:=\mu_1$ has the form $\varphi_\mu=e^\psi$, where $$\psi(\xi)=-\frac{\sigma^2}2\xi^2+{\rm i}b\xi+\int e^{{\rm i}\xi }x-1-1_{(-1,\:1)}(x){\rm i}\xi x\:\nu({\rm d}x)\;\;\;\text{for all }\xi\in\mathbb R$$ for some $b,\sigma\in\mathbb R$ and a $\sigma$-finite measure $\nu$ on $\mathbb R$ with $\nu(\{0\})=0$.

Let $$(Lf)(x):=\frac{\sigma^2}2f''(x)+bf'(x)+\int f(x+y)-f(x)-1_{(-1,\:1)}(x)yf'(x)\;\nu({\rm d}y)$$ for $f\in C^2(\mathbb R)\cap\mathcal L^1(\nu)$.

Let $A$ denote the generator of $(\kappa_t)_{t\ge0}$ and $f\in C^2(\mathbb R)$ such that $f,f',f''\in U$. I know several references showing that $f\in\mathcal D(A)$ and $Af=Lf$ either using an appropriate decomposition of $(X_t)_{t\ge0}$ or by considering Fourier transforms.

I would really like to know if we are able to prove the claim by showing that $\left(f(X_t)-\int_0^t(Lf)(X_s)\:{\rm d}s\right)_{t\ge0}$ is a martingale$^2$. Or maybe by a more semigroup-theoretic approach.

$^1$ i.e. $\mu_{s+t}=\mu_s\ast\mu_t$ for all $s,t\ge0$ and $$\int f\:{\rm d}\mu_s\xrightarrow{s\to t}\int f\:{\rm d}\mu_t\;\;\;\text{for all }f\in C_b(\mathbb R)\text{ and }t\ge0.$$

$^2$ Maybe one can use that a process $M:=X-\int_0^{\;\cdot}Y_s\:{\rm d}s$ is a martingale iff $N_t:=e^{-\lambda t}X_t+\int_0^te^{-\lambda s}(\lambda X_s-Y_s)\:{\rm d}s$ is a martingale.


2 Answers 2


I think you can do it by computing the expectation of the Laplace transform of both the function f(X_t) and the integral (the one with the martingale vanishes) and then identity the Laplace transform of your semi-group with the résolvent of both L and A. This should be sufficient to identify A and L.

This kind of approach is used by T. Kurtz on his works on the martingale problem and his body of works is probably the best place to find an answer to your question.

  • $\begingroup$ I'm sorry for my late response; I wasn't able to think about this question in the past. Which work(s) of Kurtz do you've got in mind? Do you have a reference? $\endgroup$
    – 0xbadf00d
    Apr 15, 2021 at 8:11
  • $\begingroup$ I'm still interested in this question. Could you elaborate on the approach you've suggested or even provide a more detailed outline? $\endgroup$
    – 0xbadf00d
    Sep 3, 2021 at 15:55

Sorry for the delay.

If I understand well your question, you are willing to show that $A$ extends $L$. Here is an attemps, with still one point left open.

Let $f$ be in $\mathcal{L}$, the set of functions for which $(f(X_t)-\int_0^t Lf(X_s)ds)_t$ is a martingale for any starting point $X_0=x$. For any $\alpha>0$, $$ \int_0^{+\infty} e^{-\alpha t}E_x[f(X_t)]dt =\int_0^{+\infty} e^{-\alpha t}\int_0^t E_x[Lf(X_s)]ds dt =\frac{1}{\alpha}\int_0^{+\infty} e^{-\alpha t}E_x[Lf(X_t)]dt +\frac{1}{\alpha}f(x). $$ Define $$G_\alpha f(x)=\int_0^{+\infty} e^{-\alpha t}E[f(X_t)]dt.$$ The above equality (of course, there are some Fubini hidden here that should be justified) shows that $$(\alpha -L)G_\alpha f(x)=f(x)\text{ for }f\in\mathcal{L}.$$ In particular, $\alpha-L$ is invertible on $\mathcal{L}$ (with the uniform norm, since $G_\alpha$ is bounded on $U$) and $G_\alpha=(\alpha-L)^{-1}$.

On the other hand, from the very definition of $\kappa_t$, when $f\in\mathcal{L}$, $$G_\alpha f(x)=R_\alpha f(x):=\int_0^{+\infty} e^{-\alpha t}\kappa_t f(x)dt,$$ where $(R_\alpha)_\alpha$ is the resolvent associated to $\kappa$. The only difference between $R_\alpha$ and $G_\alpha$ lies potentialy in their domains.

Therefore, since $(\alpha-A)R_\alpha f=f$ for any $f\in U$, we obtain that for $f\in\mathcal{L}$, $AG_\alpha f$ is well defined and thus belongs to the domain of $A$ (which is nothing more than the image of $U$ by $R_\alpha$). Besides, $$(\alpha-A)G_\alpha f=f=(\alpha-L)G_\alpha f$$ and $A=L$ on $G_\alpha(\mathcal{L})$. This proves that $A$ is actually an extension of $L$ restricted to $G_\alpha(\mathcal{L})$.

There is a last point to satisfy, that one may replace $G_\alpha(\mathcal{L})$ by $\mathcal{L}$, but it seems to me that one may use again the properties of the resolvent by letting $\alpha\to\infty$. Here, we have to be very precise about the spaces and so one but the global reasonning is pretty general.


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