Let $(W_t)_{t\geq0}$ denote a standard Brownian motion and $I=\left[a,b\right]$ a compact interval. Show that $P\left[\frac {W_{t+h}-W_t} {h} \in I\right] \rightarrow 0$ as $h\rightarrow 0$. What does this precisely mean for the differentiability of the Brownian paths?
My idea of proof is rather intuitive. The expression $P\left[\frac {W_{t+h}-W_t} {h} \in I\right] \rightarrow 0$ as $h\rightarrow 0$ reminds me of differentiability od real function of one variable. I see that for $h=0$ the fraction is $\frac {0} {0}$, which means the limit is undefined and therefore cannot belong to $I$, as it is compact. However I think it is not sufficient. Could you please help me?
The fraction is distributed like $\mathcal{N}(0,1/h)$, after invoking the definition of Brownian motion. As $h$ gets small, this distribution gets wider-and-wider. Meaning that any finite interval will have less-and-less probability (formally justify this by writing down the normal density and showing it goes to zero everywhere).