Let $d\mu$ be a Gaussian measure on the space of tempered distributions $\mathcal{S}'(\mathbb{R}^n)$. The characteristic functional of this measure is computed as $$S[f] = \int e^{i\phi(f)}d\mu(\phi), \quad f \in \mathcal{S}(\mathbb{R}^n)$$ where $\mathcal{S}(\mathbb{R}^n)$ denotes the Schwartz space of functions.
To compute the moments of $\mu$ we may differentiate $S$. For example, for the first moment we have $$\int \phi(f) d\mu(\phi) = \Big(-i \frac{d}{d\lambda}\Big) S[\lambda f] \Big|_{\lambda = 0}. \tag{1} $$
I would like to verify the above. By differentiating under the integral sign, ordinary calculus and the chain rule gives $$\Big(-i \frac{d}{d\lambda}\Big) S[\lambda f] = \int (-i)i\phi(\lambda f) \frac{d\phi}{d\lambda} f e^{i\phi(\lambda f)}d\mu(\phi) = \int \phi(\lambda f) \frac{d\phi}{d\lambda} f e^{i\phi(\lambda f)}d\mu(\phi).$$ After evaluating at $\lambda = 0$ we have $$\Big(-i \frac{d}{d\lambda}\Big) S[\lambda f]\Big|_{\lambda = 0} = \int \phi(0) \frac{d\phi}{d\lambda}\Big|_{\lambda = 0} f d\mu(\phi)$$ but there seems to be some problems with this. Namely,
- How can we compute $\frac{d\phi}{d\lambda}$?
- We have $\phi(0)$ on the RHS above but $\phi(f)$ in the LHS of (1).
- How do we get $f$ to disappear so that the above can equal the LHS of (1)?
I think I am missing something very simple here but I cannot spot it.