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How to show that

$$ \lim_{\alpha \rightarrow \infty} \sup_{t \in \left [0,T \right]} \left | e^{-\alpha t} \int _ 0 ^t e^{\alpha s} ~ dB_s \right | =0, \ \ \text{a.e.}$$

where $\left (B_s \right)_{s\geq 0}$ is a real standard brownian motion starting from zero ?

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.

Thank you all.

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  • $\begingroup$ LaTeX fixed${}{}{}$ $\endgroup$ – Gerald Edgar Jan 31 '13 at 18:24
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Here's one way of dealing with it. Integrate by parts to see that the expression under the sup is $$ \Bigl|B(t)-\alpha\int_0^t e^{\alpha(s-t)}B(s)ds\Bigr|\le\alpha\int_0^t e^{\alpha(s-t)}|B(t)-B(s)|ds +e^{-\alpha t}|B(t)|. $$

Now the result follows since $B$ is a.s.-bounded and a.s.-Holder on [0,T].

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  • $\begingroup$ @Yuri Bakhtin: Thank you very much for your answer! Nice solution. $\endgroup$ – Paul Feb 18 '13 at 0:50
  • $\begingroup$ As the fact that B is Holder concludes? $\endgroup$ – Luana Amaral Gurgel Nov 2 '16 at 13:46

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