I know that the above statement is true, but I can't demonstrate it. It's a pretty powerful theorem, here is its mathematical formulation:
Theorem: The only continuous martingales with stationary increments are Brownian motions
(1) being for all $u$, $e$, $X(u+e) - X(u) = X(e)$ (equal in distribution)
I would greatly appreciate if you could point me in the right direction or link me to some material :)
Thanks!
This question keeps getting bumped up, so (at long last) I'll convert my comments above to an answer.
It is true that every continuous martingale $X$ with stationary independent increments is a Brownian motion or, to be precise, $X=X_0+\sigma B_t$ for a standard Brownian motion $B$ and constant $\sigma$. This is because any such process is a Lévy process, and Brownian motions (possibly with drift) are the only continuous Lévy processes. This is standard, and most decent book on continuous-time stochastic processes should show this. I also have a proof of this on my blog here.
However, you do not mention the necessary condition that the increments of $X$ are independent. So, the statement in the question is false. There do in fact exist continuous martingales with stationary increments which are not Brownian motions. You can take $$ X_t=\int_0^t Y\,dB $$ for a standard Brownian motion $B$ and independent stationary process $Y$. Then $X$ has stationary increments, but is not a Brownian motion unless $Y$ is constant. For example, $Y$ could be an Ornstein-Uhlenbeck process started in its stationary distribution.
[Finally, the link to the theorem in the question is not very useful, as it is not complete and references some undefined "(1)" that the process is supposed to follow.]
