# Why is the Fourier Transform of a Lévy Process a continuous function? What about the inverse? (Bochners Theorem)

I was confronted with this question when reading "Stochastic Integration and Differential Equations" by Protter. Just after the definition of a Lévy process he says the following:

If $X_t$ is a Lévy-process and we consider the function $f_t(u)=\mathbb{E}(e^{iuX_t})$ where $f_0(u)=1$ and $f_{t+s}(u)=f_t(u)f_s(u)$, and $f_t(u) \neq 0$ for every $(t,u)$. Then, using the right continuity in probability we conclude that there exists a continuous function $\psi$ with $\psi(0)=0$ such that $f_t(u)=\text{exp}(-t\psi(u))$.

How can one prove this? (Right) continuity in probability seems a rather weak notion to me for the existence of a fully continuous $\psi(u)$. It would mean that also $f_t(u)$ is continuous right? So what we need is that the Fourier transform of a Lévy process is continuous i think. Any hints on that? (Probably its a well-known fact and I am missing something obvious here)

In the same section the so-called "Bochners Theorem" is also mentioned. Could anyone share a resource for me with the details and the sketch of proof?

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## 1 Answer

From the fact that $X_t$ is continuous in probability and from $|f_t(u)|\leq1$ you can deduce continuity of $f_t(u)$ (except at point $t=0$ where you only have right continuity).

To do it, you only need to use a version of Lebesgue dominated convergence theorem where you have convergence in probability and domination by $1$ which is an integrable function here
NB : This version can be deduced from equivalence between convergence of a sequence of integral and (the convergence in probablity + uniform integrability of the sequence of integrands) which is a standard result in measure theory.

Now if you have a look at the Poisson process section of Protter's book, he uses also the same fact that the only function with the properties :

-right continuity
-semigroup property

is an exponential or equal to 0. As $f_0(u)=1$ it is an exponential and there is a constant $\psi(u)$ that does the job.

I don't have the proof of this "real analysis" theorem. But I believe it's fairly standard, is the sense that this characterizes the exponential function. Edit : wiki gives a closely related proof here : http://en.wikipedia.org/wiki/Characterizations_of_the_exponential_function

Best regards

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Hi! Ok I get the argument with the semigroup generator $\psi$ but could you elaborate a little bit on the convergence in probability please? I still dont get it. Do you need a result like "in the space ... $L^1$-convergence and convergence in measure are the same"? – vanguard2k Dec 6 '12 at 7:35

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@Vanguard : You have the following theorem which states equivalence between ( $E[f_n]\to E[f]$) and (convergence in proba + uniform integrability of the sequence). As $f_n$ is bounded by 1, you have uniform integrability so convergence of $E[f_n]\to E[f]$ by the theorem if I am not mistaken. For the rest of the question, you can go along those steps. 1 - Use Kolmogoroff existence theorem to show that there is a processus with the characterisitc functions $e^{t.\psi(u)}$.2 Show indepedence of increments of this process. 3 Show right continuity in proba the hard part the hard part – TheBridge Dec 6 '12 at 8:17
Sounds good to me :-). I wasnt aware of the first result (anymore). I am new to the theory of Lévy processes so now and then I stumble a bit. Thank you! – vanguard2k Dec 6 '12 at 10:42
@Vanguard2k : I wonder if Bochner's theorem doesn't give the conclusion directly I should try to find an "instance" of it. For the measure theory theorem about Uniform integrability, it is stated on wiki's page on uniform integrability section "Relation to convergence of random variables". Best Regards. – TheBridge Dec 6 '12 at 16:31
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