Let $\exp( i u X_{t}+\frac{1}{2}u^{2}t)$ be a local martingale for any $u$, show that $X_{t}$ is a Brownian motion Let $X$ be an $\mathcal{F}_{t},\; t\geq 0$, adapted a.s. continuous process with $X_{0}=0$ and    $M^{u,X}_{t}:=\exp( i u X_{t}+\frac{1}{2}u^{2}t)$ be a local martingale for any $u$, show that $X_{t}$ is a $\mathcal{F}_{t}$-Brownian motion.
My idea:
We need to show that $\operatorname{cov}(X_{t},X_{s})=t \land s$ and that $X$ is a Gaussian process.
Let $0 \leq t_{0} < ... < t_{n} <\infty$. I know that for $T > 0, \; u \in \mathbb R$ the process $M_{t\land T}^{u,X}=\exp( i u X_{t\land T}+\frac{1}{2}u^{2}(t\land T))$ is a local martingale too. Furthermore, $\lvert M^{u,X}_{t\land T}\rvert\leq \exp(\frac{1}{2}u^{2}T)< \infty$.
Therefore, $(M_{t\land T}^{u,X})_{t\geq 0}$ is a martingale. By Doob's optional sampling Theorem, we obtain for $t < T$,  $$E[M_{t}^{u,X}]=E[M_{t\land T}^{u,X}]=E[M_{0\land T}^{u,X}]=E[M_{0}^{u,X}]=1$$
This would then imply that $$ E[\exp(iuX_{t})]=\exp(-\frac{u^{2}t}{2})$$ which is the characteristic function of a normal random variable $\mathcal{N}(0,t)$ evaluated at $u$. But I am struggling to do the same for the random vector $(X_{t_{0}},...,X_{t_{n}})$ and $0 \leq t_{0} < ... < t_{n}< \infty$. I think I need to use induction and the tower property to reduce it all to the random variable $X_{t_{0}}$, but that seems less efficient.
On the issue of showing $ \operatorname{cov}(X_{t},X_{s})= t\land s$, I am completely stuck. Maybe I can set $u:=\log(E[X_{t}X_{s}])$ and by assumption: $ M^{u,X}_{r}:=\exp( i \log(E[X_{t}X_{s}]) X_{r}+\frac{1}{2}\log(2E[X_{t}X_{s}])r)$ is a local martingale
We then have:
$$ E[\exp( i \log(E[X_{t}X_{s}]) X_{r}]=\exp(-\frac{\log(E[X_{t}X_{s}])r}{2})$$
Then
$$ E[\exp( i \log(E[X_{t}X_{s}]) X_{r}]=E[X_{t}X_{s}]^{-\frac{r}{2}}$$
I do not see how I get to $t\land s $ from here. Any ideas?
Or is something simply missing in the question's assumptions?
 A: You've made a good start. An other approach to the covariance and the fact that $X$ is a Gaussian process would be to use the martingale property of $M^{u,X}$ and $M^{v,X}$ to
compute $E[\exp(iuX_s+ivX_t)]$ for $0<s<t$. This will show that $X_u$ and $X_v$ are jointly Gaussian with the anticipated covariance. Similar reasoning handles the joint law of $X_{t_1}, X_{t_2},\ldots , X_{t_n}$.
A: (Out of laziness I'm just going to write $M^u$,  rather than $M^{u,X}$, etc.)
What I had in mind: for $0<s<t$,
$$
E[M^u_sM^v_t] =E[M^u_sM^v_s]
$$
by the martingale property of $M^v$ and the adaptedness of $M^u$. Writing this out in detail:
$$
E[\exp(iuX_s+ivX_t]\cdot\exp(su^2/2+tv^2/2) = E[\exp(i(u+v)X_s)]\cdot\exp(su^2/2+sv^2/2).
$$
or, simplifying,
$$
\eqalign{
E[\exp(iuX_s+ivX_t] 
&= E[\exp(i(u+v)X_s)]\cdot\exp(-(t-s)v^2/2)\cr
&=\exp(-s(u+v)^2/2)\cdot\exp((t-s)v^2/2)\cr
&=\exp( -[su^2+2uvs+tv^2]/2).
}
$$
This joint characteristic function of $(X_s, X_t)$ is that of a bivariate normal distribution with zero means and covariance matrix
$$
\left[\matrix{s&s\cr s&t\cr}\right],
$$
as hoped for. The calculation for the joint law of $X_{t_1}$, $X_{t_2},\ldots, X_{t_n}$ will be similar but more tedious.
