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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

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    $\begingroup$ 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.) $\endgroup$ – cardinal Jul 16 '13 at 21:59
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    $\begingroup$ This question has been answered on MO: mathoverflow.net/questions/120438/limit-of-a-wiener-integral $\endgroup$ – Ben Derrett Jul 17 '13 at 8:19
  • $\begingroup$ Very nice solution, thanks so much $\endgroup$ – Higgs88 Jul 17 '13 at 18:39
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It has been answered on Math Overflow. The key was an integration by parts argument.

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