Fourier transform of Lévy process

Question

I have the following definition of a Lévy process:

An adapted process $X=(X_t)$ with $X_0=0$ a.s. is a Lévy process if

1. it has independent increments of the past, i.e. $X_t-X_s$ is independent of $\mathcal{F}_s$
2. it has stationary increments, i.e. $X_t-X-s$ has the same distribution as $X_{t-s}$
3. $X_t$ is continuous in probability, i.e. $\lim_{t\to s}X_t=X_s $ in probability.

Let $f_t(u):=E[e^{iuX_t}]$, then it is easily seen $f_{t+s}(u)=f_t(u)f_s(u)$. Now I want to show the continuity of $f_t(u)$. So here is, what I've done:

Let $t_n\to t$ and we need to prove $f_{t_n}(u)\to f_t(u)$. By $3$. we can choose a subsequence (also denoted by $t_n$) such that $X_{t_n}\to X_t$ a.s. Applying dominated convergence and using continuity of $\exp(x)$ we get

$$\lim_n f_{t_n}(u)=E[\lim_ne^{iuX_{t_n}}]=f_t(u)$$

is this correct? Now I know the theorem that any continuous function $h$ which solves $h(x+y)=h(x)h(y)$ can be written of the form $h(x)=e^{a x}$ for some unique $a$. How can I use this result to conclude that $f_t(u)=e^{-t\phi(u)}$ for some function $\phi(u)$?

Answer 1

For any $n$ which is a positive integer, by stationary independent increment. Let $f_n = (f_1)^n$. Moreover for $m,n$ integers, we get $(f_1)^m = f_m = (f_{m/n})^n$

This means $f_{m/n} = (f_1)^\frac{m}{n}$ or $f_t = (f_1)^t$ for $t\in\mathbb{Q}$.

Then for an irrational $t$, take $t_n\downarrow t$, where $t_n\in\mathbb{Q}$, by almost sure right continuity of $X_t$ and the dominated convergence theroem.

$(f_1)^t = \lim\limits_{n\rightarrow\infty} (f_1)^{t_n} = \lim\limits_{n\rightarrow\infty} Ee^{iuX_{t_n}} = E \lim\limits_{n\rightarrow\infty} e^{iuX_{t_n}}= Ee^{iuX_t} = f_t$.

From this, $f_{s+t}=f_sf_t$ follows.

PS it is might actually better to define $\psi_t = - \log f_t$ instead and do it directly: i.e. for $m,n$ integers,

$m\psi_1 = \phi_m = n\psi_{m/n}$ ect.