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
$I(f) = \frac{1}{2}\int_0^{\infty} \dot{f}(t)^2 dt$.$$I(f) = \frac{1}{2}\int_0^{\infty} \dot{f}(t)^2 dt.$$
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
$\frac{d\mu^h}{d\mu} = \exp(\int_0^{\infty} \dot{h}_t dx_t - \frac{1}{2}\int_0^{\infty} \dot{h}_t^2 dt)$,$$\frac{d\mu^h}{d\mu} = \exp(\int_0^{\infty} \dot{h}_t dx_t - \frac{1}{2}\int_0^{\infty} \dot{h}_t^2 dt),$$
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.
