Define Stopping time considering a Brownian Motion I have not a idea what is asking in this question, can any one give a tip to solve it?
Thanks in advance.
Let $B={B_t,t\geq0}$ be a Brownian motion. Consider a process $u ={u_t,t\in[0,T]}$ that belongs to the class $L^2_T$. Define a sequence of stopping times
$$\tau_n=inf\{t\geq0:\int_0^tu_s^2ds\geq n\}$$
Show that if $p>2$ thenthe inequality
$$E\left[\underset{0\leq t\leq \tau_n\wedge T}{\overset{}{\mathrm sup}}\left|\int_0^tu_sdB_s\right|^p\right]\leq C_pE\left[\underset{0\leq t\leq \tau_n\wedge T}{\overset{}{\mathrm sup}}\left|\int_0^tu_s^2ds\right|^\frac{p}{2}\right] $$
where $C_p$ is a constant that depends on the parameter $p$.
 A: I'm ignoring all technical assumptions on $u$, leaving those details to you. There are three ingredients. 
Step 1. Use Doob's $L^p$-inequality.
Let $M_t = \int_0^t u_s d B_s$, and write $M^*_t =\sup_{s\le t}  |M_s|$. Then by Doob's $L^p$-inequality for the submartingale $(|M_s|:s\ge0)$, we have 
$$ E [ |M^*_t|^p  ] \le c_1(p) \sup_{s\le t}  (E |M_s|^p),$$
where $c_1(p)=  (\frac{p}{p-1})^p$. 
Step 2. Use Ito to compute the RHS (this is the only step where we actually use the fact that $M_t$ is an integral of BM). 
To get from this to the inequality, one needs to work a little more and use Ito's formula: 
$$ |M_t|^p = |M_0|^p + \int_0^t  \mbox{sgn} (M_t) p  |M_s|^{p-1} d M_s + \frac{p(p-1)}{2}\int_0^t |M_s|^{p-2} d <M>_s.$$ 
In our case, $d <M>_s=u_s^2 ds$. Taking expectation, both first summands on RHS are zero, and we're left with 
$$ E [|M_t|^p] =\frac{p(p-1)}{2} \int_0^t |M_s|^{p-2} u_s^2 ds\le 
\frac{p(p-1)}{2} E [ |M^*_t|^{p-2}  \int_0^t u_s^2 ds].$$ 
Step 3. Cauchy-Scwharz.
Now use Cauchy-Schwarz with $p_1 = p/(p-2)$ and $1/q_1= 1-1-p_1 = 1 - \frac{p-2}{p} = \frac{2}{p}$ or $q_1=p/2$, to obtain  
$$  E [|M_t|^p] \le \frac{p(p-1)}{2} (E [ M^*_t|^p])^{\frac{p-2}{p}} \times E [( \int_0^t u_s^2 ds)^{p/2}]^{2/p}.$$ 
Therefore 
Step 4. Wrap-up. 
$$ E [ |M^*_t|^p  ]\le  c_1(p) \sup_{s\le t}  (E |M_s|^p) \le c_2(p) E [ |M_t^*|^p ]^{\frac{p-2}{p}} E [( \int_0^t u_s^2 ds)^{p/2}]^{2/p},$$ 
where $c_2(p)= c_1(p)* p(p-1)/2 $. 
Dividing by $E[|M^*_t|^p]^{\frac{p-2}{p}}$, then raising to the power $p/2$ we obtain the desired inequality with $C_p = c_2(p)^{p/2}$.  
