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?
 A: 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
