Integrating brownian motion times exponential function

Question

I am trying to calculate $$\int_0^tB_se^{\lambda s}ds$$ but I am unsure of how to start the computation. The motivation behind this is that I read (and am now trying to prove) that $$\lim_{\lambda\to\infty}\sup_{t\in[0,T]}| e^{-\lambda t}\int_0^te^{\lambda s}dB_s|=0 $$ So far I have used Ito's formula to simplify the expression inside the absolute values to $$\lambda e^{-\lambda t}\int_0^t e^{\lambda s}B_sds - B_t $$ and am now trying to evaluate the expression so that I can take the supremum and then the limit. Thank you for any help!

Answer 1

I think you are on the right way. Consider $ \ \varphi _{\lambda} (s) = \lambda e ^{-\lambda(t-s)}$ , $s \in [0;t]$. Then $||\varphi _{\lambda}||_{L_1} = 1 - e^{-\lambda t}$, so $||\varphi _{\lambda}||_{L_1} \to 1$ as $\lambda \to \infty$, and at the same time for any $\varepsilon \in (0;t)$

$$ \int _0 ^{t-\varepsilon} \varphi _{\lambda} (s) ds \to 0,\ \lambda \to \infty, $$

which follows from the Lebesgue's dominated convergence theorem. Therefore, one may check that for any continuous on $[0;t]$ function $f$

$$ \int _0 ^{t} \varphi _{\lambda} (s) f(s) ds \to f(t), \ \lambda \to \infty. $$