Let $W$ be a standard one-dimensional Brownian motion and $0<T<+\infty$. Then $$\lim_{\beta\to+\infty}\sup_{0\le t\le T}|e^{-\beta t}\int_0^te^{\beta s}\mathrm{d}W_s|=0\quad \text{a.s.}$$
Could anyone give some hints or proofs?
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Sign up to join this communityLet $W$ be a standard one-dimensional Brownian motion and $0<T<+\infty$. Then $$\lim_{\beta\to+\infty}\sup_{0\le t\le T}|e^{-\beta t}\int_0^te^{\beta s}\mathrm{d}W_s|=0\quad \text{a.s.}$$
Could anyone give some hints or proofs?
As Nate Eldredge says, integration by parts implies $$ e^{ - \beta t } \int_0^t e^{ \beta s } dW_s = W_t - \beta \int_0^t e^{ - \beta ( t - s ) } W_s ds \\ = \beta \int_0^t e^{ - \beta ( t - s ) } ( W_t - W_s ) ds + e^{ -\beta t }W_t. $$ The supremum of the two terms above tend to zero. The second one, for example, follows because the maximum of the function $ t \mapsto e^{-\beta t } t$ is attained at $\beta^{-1}$, and $\sup_{t \in (0,T )} t^{-1}|W_t|$ is in $L^1$. The first term converges uniformly to zero from the fact that brownian motion is uniformly Holder of order $\alpha\in(0,1/2)$ and because $$ \lim_{ \beta \to \infty }\sup_{ t \in (0,T) } \beta \int_0^te^ss^\alpha ds = 0. $$ The last fact can be proved from the standard integration by parts theorem.