Let $H$ be a separable $\mathbb R$-Hilbert space, $Q\in\mathfrak L(U)$ be nonnegative and self-adjoint with $\operatorname{tr}Q<\infty$ and $(W_t)_{t\ge0}$ be a $H$-valued Wiener process on a probability space $(\Omega,\mathcal A,\operatorname P)$ with covariance $Q$.
Moreover, let $U^2_b(H)$ denote the space of twice Fréchet differentiable functions $f:H\to\mathbb R$ such that $f$, ${\rm D}f$ and ${\rm D}^2f$ are bounded and uniformly continuous.
On p. 79 of Stochastic Partial Differential Equations with Lévy Noise, the authors prove (in a rather complicated, brute-force, way) that $U^2_b(H)$ is contained in the domain of the generator $(\mathcal D(A),A)$ of $W$ and$^1$ $$(Af)(x)=\frac12{\rm D}^2f(x)Q\;\;\;\text{for all }x\in H\text{ and }f\in U^2_b(H)\tag1.$$
I wonder whether we can show $(1)$ on a larger class of functions.
I think it's way easier to establish the desired result by constructing a solution to a martingale problem. By the Ito formula, we know that if $f\in C^2(H)$ is such that $f$, ${\rm D}f$ and ${\rm D}^2f$ are uniformly continuous on bounded subsets of $H$, then $$f(W_t)-f(W_0)=\frac12\int_0^t{\rm D}^2f(W_s)Q\:{\rm d}s+\int_0^t{\rm D}f(W_s)\:{\rm d}W_s\tag3$$ almost surely for all $t\ge0$. If $f$ is bounded, then the stochastic integral on the rhs is a martingale. So, we should be able to immediately conclude that $f\in\mathcal D(A)$ and $f$ satisfies $(1)$. Am I missing something?
$^1$ in $(1)$, I'm identifying ${\rm D}^2f(x)$ with an element of $\left(H\hat\otimes_\pi H\right)'$. If you don't like that, you can equivalently treat ${\rm D}^2f(x)$ as an element of $\mathfrak L(H)$ so that $${\rm D}^2f(x)Q=\operatorname{tr}Q{\rm D}^2f(x)\tag2.$$
