Convergence of stochastic integral to Brownian motion

Question

Let $a \in \mathbb R$, $W(t)$ a standard Brownian motion, and

$$ V(t) = a \int_0^{t} e^{-a(t-s)} d W_s. $$

Is it true that $$ \int_0^t V(u) \, du = W(t) - W(0) \quad \text{as} \quad a \to \infty $$ in a certain sense?

Answer 1

First, \begin{align} \int_0^t V_u du & = \int_0^t\int_0^t 1_{s\le u}ae^{-a(u-s)}dW_sdu\\ & = \int_0^t\int_0^t 1_{s\le u}ae^{-a(u-s)}dudW_s,\quad\text{stochastic Fubini}. \end{align} Thus (assuming $W_0=0$ for simplicity), \begin{align} E\left[\left(\int_0^t V_u du-W_t\right)^2\right] & = \int_0^t\left( \int_0^t 1_{s\le u}ae^{-a(u-s)}du-1 \right)^2 ds,\quad\text{Ito isometry},\\ & = \frac{1}{2a}(1-e^{-2at})\rightarrow 0\text{ as }a\rightarrow\infty. \end{align} That is, $$\int_0^tV_udu\rightarrow W_t\text{ in }L^2\text{ for each }t>0.$$ In fact, $\{X^n:n=1,2,\ldots\}$, $X^n_t:=\int_0^t n\int_0^u e^{-n(u-s)}dW_sdu$, converges uniformly to $W$ on compact intervals.