# Limit of a stochastic integral

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

• Why do you think that what you've stated implies the result does not hold? (The variance of your $G$ goes to zero as $\beta$ increases.) – cardinal Jul 16 '13 at 21:59
• This question has been answered on MO: mathoverflow.net/questions/120438/limit-of-a-wiener-integral – Ben Derrett Jul 17 '13 at 8:19
• Very nice solution, thanks so much – Higgs88 Jul 17 '13 at 18:39