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.


1 Answer 1


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.

  • $\begingroup$ Thank you very much for that reference, it is really helpful. Sorry for my ignorance about their book. $\endgroup$
    – yilin wang
    Commented Oct 3, 2015 at 9:33

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