It seems strange to me that all we can find about Schilder's theorem in the literature is on a finite interval of Brownian path. 


If we equip the space of continuous function starting from $0$, defined on $\mathbb{R}\_+$ with the topology of uniform convergence on compacts. Can we have the similar large deviation principle? With of course the rate function


$$I(f) = \frac{1}{2}\int\_0^{\infty} \dot{f}(t)^2 dt.$$


I don't see any objection so far but I don't have any reference to confirm my guess. If it is wrong can you tell me why?


The same question for the Cameron-Martin formula. If we write $\mu$ the Wiener measure on $C\_0([0,\infty))$, and $\mu^h$ the measure of $B+h$ where $B \sim \mu$. Can we have, for $h$ of finite functional value $I(h) < \infty$, that $\mu^h$ is absolutely continuous with respect to $\mu$, of density


$$\frac{d\mu^h}{d\mu} = \exp(\int\_0^{\infty} \dot{h}\_t dx\_t - \frac{1}{2}\int\_0^{\infty} \dot{h}\_t^2 dt),$$


where $\int\_0^{\infty} \dot{h}\_t dx\_t$ is defined in $L^2(\mu)$ by Wiener Integral. 


Does that make any sense?


Thank you in advance.



Large deviation for Brownian path on $[0,\infty)$
It is not in the uniform topology but with a topology tapered off at infinity it is correct. It is done that way in the book of Deuschel and Stroock on large deviations.


