I would like to know if $B_t=W_t-\int_0^t \frac{W_u}{u}du$ is a brownian motion. I know that $W_t$ is a brownian motion. For that i would like to use Levy's characterization, so I have to show that $[B]_t=t$ and that $B_t$ is a continuous local martingale. How can I show that? Is $B_t$ an Ito process or can I rewrite $B_t$ in order to see that it is an Ito process?
I tried it with partial integration: $ \int_0^t \frac{W_u}{u}du=W_t-\int_0^t u d(\frac{W_u}{u})$. Because then I could use that $[B]_t=\int_0^t 1^2 du=t$. Could you please help me?
It is not a Brownian motion. You can write $dB_t = dW_t - \frac {W_t}{t} dt$. You are right that $[B_t] = 1$, but the following argument shows that $B_t$ is not a martinglae.
Note that the coefficient $\frac {W_t}{t}$ of $dt$ is nonzero and $\int_0^t \mathbb E\big[|\frac{W_u}u|\big] du <\infty$. So, the uniqueness in the martingale representation theorem implies that $B_t$ is not a martingale w.r.t. the filtration of $W_t$. However, to show that $B_t$ is not a martingale w.r.t. its own filtration, I have the following solution. You directly check that $\mathbb E[B_t|\mathcal F_s] = B_s - W_s (\ln t - \ln s) $. Using the fact that the pair $(B_t,W_t)$ is Gaussian, it is not hard to obtain $\mathbb E[W_s|B_s]\neq 0$. Therefore, $\mathbb E[B_t|B_s]\neq B_s$ which implies that $B_t$ is not a martinglae w.r.t its own filtration.