most recent 30 from http://mathoverflow.net 2013-05-20T17:05:38Z http://mathoverflow.net/feeds/question/120438 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/120438/limit-of-a-wiener-integral Paul 2013-01-31T18:19:57Z 2013-02-01T19:42:32Z

<p>How to show that </p> <p>$$ \lim_{\alpha \rightarrow \infty} \sup_{t \in \left [0,T \right]} \left | e^{-\alpha t} \int _ 0 ^t e^{\alpha s} ~ dB_s \right | =0, \ \ \text{a.e.}$$ </p> <p>where $\left (B_s \right)_{s\geq 0}$ is a real standard brownian motion starting from zero ? </p> <p>I'd like to have some ideas to deal with this problem. After all, I'll show some solutions that I propose and somme colegues also but that i believe be all wrong. (I just don't show know to don't interffer in your ideas. </p> <p>Thank you all.</p>

http://mathoverflow.net/questions/120438/limit-of-a-wiener-integral/120537#120537 Yuri Bakhtin 2013-02-01T19:42:32Z 2013-02-01T19:42:32Z

<p>Here's one way of dealing with it. Integrate by parts to see that the expression under the sup is $$ \Bigl|B(t)-\alpha\int_0^t e^{\alpha(s-t)}B(s)ds\Bigr|\le\alpha\int_0^t e^{\alpha(s-t)}|B(t)-B(s)|ds +e^{-\alpha t}|B(t)|. $$</p> <p>Now the result follows since $B$ is a.s.-bounded and a.s.-Holder on [0,T].</p>