Malliavin Calculus: directional derivatives of cylinder functions exist in what sense?

Question

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.

Answer 1

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. $\int \|\omega\|_\infty^p \,d\mathbb{P}(\omega) < \infty$.