Let $W_t$ be a one-dimensional Brownian motion and I would like to prove
$$\lim_{\beta\rightarrow+\infty}\sup_{0\leq t\leq T}\left|e^{-\beta t} \int_0^te^{\beta s}\mathrm dW_s\right|=0$$
This is an exercise in Chapter 3 of Karatzas&Shreve, but I don't think this proposition is true since
$$\sup_{0\leq t\leq T}\left|e^{-\beta t} \int_0^te^{\beta s}\mathrm dW_s\right|\geq \left|e^{-\beta T} \int_0^Te^{\beta s}\mathrm dW_s\right|:=|G|$$
where $G$ is a centered Gaussian random variable of variance $(1-e^{-2\beta T})/2\beta$, which implies the precedent proposition can not be true. Could someone tell me where I am wrong or this proposition is not true? Thanks a lot
It has been answered on Math Overflow. The key was an integration by parts argument.