How to get $ E[\int_0^{t\land T_n}\frac{\partial u}{\partial x_i}(B_s)dB_i(s)]=0 $?

Question

Let $u$ be a solution of the Poisson problem on $U$. Define open sets $U_n=\{x\in U: |x-y|>1/n , y \in\partial U\}$. Let $T_n$ be the first exit time of $U_n$. As $\frac 12 \Delta u(x)=-g(x)$ for all $x\in U$. From multi-dimensional Ito formula, we get $$ u(B_{t\land T_n})=u(B_0)+\sum_{i=1}^d\int_0^{t\land T_n}\frac{\partial u}{\partial x_i}(B_s)dB_i(s)-\int_0^{t\land T_n} g(B_s)ds $$ Why the expectation of the second term is zero? That is $$ E[\int_0^{t\land T_n}\frac{\partial u}{\partial x_i}(B_s)dB_i(s)]=0 $$

Answer 1

It follows because the integral of a predictable process with respect to Brownian motion is a local martingale, and in this case we can show that it is in fact a true martingale. We know that $\frac{\partial u}{\partial x_i}$ is continuous on $\overline U_n$, and since $U_n$ is bounded we know $\frac{\partial u}{\partial x_i}$ is also bounded on $\overline U_n$. By definition of $T_n$, we have that $B_s \in U_n$ for $s \le T_n$, so $\frac{\partial u}{\partial x_i}(B_s)$ is bounded for $s \le T_n$. Since we are integrating up to time $t \wedge T_n$, we are integrating a bounded process over a bounded time interval, and hence the stochastic integral is a true martingale.