estimate of expectation of stochastic integral Can we bound the following expectation by $C\delta^{\frac{1}{2}}$ for some constant $C$:
$$E\sup_{|s-t|\leq \delta,s\in[0,T],t\in[0,T]}|\int_s^t\sigma(r)dB_r|,$$
where $B$ is a standard Brownian motion and $|\sigma|\leq M$.
When $\sigma$ is a constant, I know this can be done by BDG inequality. But when $\sigma$ is a time dependent function or even random, can we have similar estimate?
 A: Suppose we are working on the probability space $(\Omega,\mathcal{F},P)$. It is sufficient to consider $T=1$ and $\delta=1/n$. First of all, let us recall a well known result: Expected maximum absolute value of $n$ iid standard Gaussians?
We claim the  result: $E\sup_{|s-t|\leq \delta,s,t\in[0,1]}|\int_s^t\sigma_rdB_r|\leq 3M\sqrt{2\delta|\log\delta|}$, where $M$ is the essential supremum of $\sigma$ since it is assumed to be bounded.
We proceed by the following two steps.
Step 1: when $\sigma=1$. We have 
$$E\sup_{|s-t|\leq\delta,s,t\in[0,1]}|B_t-B_s|\leq 3E\max_{1\leq i\leq n}\sup_{s,t\in[(i-1)\delta,i\delta]}|B_t-B_s|=3E\max_{1\leq i\leq n}\sup_{s,t\in[(i-1)\delta,i\delta]}|N^i(0,t-s)|,$$  where $\{N^i(0,t-s)\}_{i=1}^n$ are iid normal variables. So it holds that
\begin{equation}\begin{split}&~E\max_{1\leq i\leq n}\sup_{s,t\in[(i-1)\delta,i\delta]}|N^i(0,t-s)|=E\max_i\sup_{s,t\in[(i-1)\delta,i\delta]}\sqrt{t-s}|N^i(0,1)|\\&~\leq\sqrt{\delta}E\max_i|N^i(0,1)|\leq \sqrt{2\delta|\log\delta|},\end{split}\end{equation}
where the last inequality is due to the recalled result at the very beginning. So in this case, we have $E\sup_{|s-t|\leq\delta,s,t\in[0,1]}|B_t-B_s|\leq 3\sqrt{2\delta|\log\delta|}$.
Step 2. Define $\mathcal{H}=\{\sigma\text{ measurable and bounded}:E\sup_{|s-t|\leq\delta,s,t\in[0,1]}|\int_s^t\sigma_rdB_r|\leq 3M\sqrt{2\delta|\log\delta|},\text{M is the essential supremum of }\sigma\}$. 
By step 1, we have $1\in\mathcal{H}$. Similarly, for each $r_1,r_2\in[0,1]$ and each $A\in\mathcal{F}$, we have $1_{[r_1,r_2]\times A}\in\mathcal{H}$. Moreover, for any $f,g\in\mathcal{H}$, we have $af+bg\in\mathcal{H}$ for any scalar $a,b$. Finally, for any sequence $f^n\in\mathcal{H}$ and $0\leq f^n\uparrow f$ with $f$ being bounded, we have $E\sup_{|s-t|\leq\delta,s,t\in[0,1]}|\int_s^tf^n_rdB_r|\leq 3M_n\sqrt{2\delta|\log\delta|}\leq 3M\sqrt{2\delta|\log\delta|}$, where $M_n$ and $M$ are the essential supremum of $f^n$ and $f$ respectively. Note that 
\begin{equation}
\begin{split}
 &~\left|E\sup_{|s-t|\leq\delta,s,t\in[0,1]}|\int_s^tf^n_rdB_r|-E\sup_{|s-t|\leq\delta,s,t\in[0,1]}|\int_s^tf_rdB_r|\right|\\
\leq&~ E\sup_{|s-t|\leq\delta,s,t\in[0,1]}|\int_s^t(f^n_r-f_r)dB_r|\\
\leq &~2E\sup_{0\leq t\leq 1}|\int_0^t(f^n_r-f_r)dB_r|\leq CE\int_0^1(f^n_r-f_r)^2dr\rightarrow 0,
\end{split}
\end{equation}
by dominated convergence. So we have
$$E\sup_{|s-t|\leq\delta,s,t\in[0,1]}|\int_s^tf_rdB_r|\leq 3M\sqrt{2\delta|\log\delta|},$$
which implies $f\in\mathcal{H}$.
So by monotone class theorem, $\mathcal{H}$ contains all $\mathcal{B}([0,T]\times\Omega)$ measurable and bounded $\sigma$.
