Malliavin derivative of a gaussian

calculus-of-variationsmalliavin-calculusprobability theorystochastic-analysisstochastic-calculus

Let $W$ be an $H$-isonormal Gaussian process and $H$ is a real separable Hilbert Space.
Set $$X=f\big(W(h_1),\ldots, W(h_n)\big) $$ for $f$ an infinite differentiable with their partial derivatives have polynomial growth.

We define the Malliavin derivative as
$$DX=\sum_{i=1}^n \partial_i f\big(W(h_1),\ldots, W(h_n)\big)h_i.$$

I would like to compute the malliavin derivation of $X=W(h)$ where $h\in L^2(0,\infty) $. Its $DX=h$ but I don't see how can we prove this.

Best Answer

Simply apply the definition with $n=1$, $h_1=h$, $f(x)=x$. Note that $f$ is smooth and $\partial_1 f = f' = 1$ has polynomial growth so the hypotheses are satisfied. Now your formula says exactly that $DX = h$.