4
$\begingroup$

Denote by $P_0(\mathbb{R}^d)$ the sets of continuous paths over $[0,1]$ started at $x=0$ with values in $\mathbb{R}^d$, we equip this space with the sup-norm and make it into a probability space by endowing it with the $\sigma$-algebra generated by open sets and Wiener-measure.

Let $F:P_0(\mathbb{R}^d)\rightarrow \mathbb{R}$ be a cylinder function, i.e., $F(\omega)=f(\omega_{s_1},...,\omega_{s_n})$, where $0<s_1<...<s_n<1$ and $f$ and all its derivatives have at most polynomial growth.

How can I see that the limit for a Cameron Martin path $h\in\mathcal{H}$

$\lim_{t\rightarrow 0} \frac{F(\omega+th)-F(\omega)}{t}$ exists in at least $L^1$?

Or put differently, why can I interchange the limit with the integral sign in

$\lim_{t\rightarrow 0}\int_{P_0(\mathbb{R}^d)}\frac{F(\omega+th)-F(\omega)}{t}d\mathbb{P}(\omega)$?

APPROACH

In order to interchange the limit with the integral it would suffice to find an integrable function $G$ such that $F'(\omega+th)\leq G(\omega)$ for $t\in (-\epsilon, \epsilon)$. But I only see that

$F'(\omega+th)=\sum_{j=1}^n \nabla^jf(\omega_{s_1}+th_{s_1},...,\omega_{s_l}+th_{s_l})\cdot h_{s_j} $ $\leq\sup_{t\in (-\epsilon,\epsilon)} \sum_{j=1}^n\nabla^jf(\omega_{s_1}+th_{s_1},...,\omega_{s_l}+th_{s_l})\cdot h_{s_j}$

But how do I see that the last line in integrable? I understand that without the sup this line in integrable as $f$ is polynomial and $B_{s_i}$ has moments of all orders but this does not seem to suffice here.

$\endgroup$

1 Answer 1

2
$\begingroup$

We can assume that all the first derivatives of $f$ are bounded by some polynomial function, let's say $|\nabla^j f(x)|\le C(1+|x|^p)$. Then we have $|F'(\omega + th)| \le n C(1 + (\|\omega\|_\infty + t \|h\|_\infty)^p)$ which can be bounded, independent of $t$, by something like $C_1(1 + \|\omega\|_\infty^p)$. But by Fernique's or Skorohod's theorem, $\|B\|_\infty$ has finite moments of all orders, i.e. $\int \|\omega\|_\infty^p \,d\mathbb{P}(\omega) < \infty$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.