Background:
I know, and can almost prove, that the infinitessimal generator of BM is the Laplacian/2. For me (and I have never heard it done differently) we have Feller Processes, of which BM is one example, semigroups, which are contractive, strongly continuous, positive, and generators which obey the following:
Let $L$ be a linear operator defined on a dense subspace $D$ of $C:=C_0(\mathbb{R}^d)$ equipped with sup norm.
(a) If $f \in D$ and $\lambda \ge 0$ and $g:=f-\lambda Lf$ then $min(g) \le min (f)$.
(b) For all $\lambda>0, (I-\lambda L):D\rightarrow C$ is onto.
(c) There exists $\epsilon>0$ such that $\forall \lambda \in (0, \epsilon) \exists f_n \in D$ such that $g_n:=f_n- \lambda L f_n$ satisfies $sup_n||g_n|| < \infty$ and both $f_n$ and $g_n$ converge to $1$ pointwise.
We know from the general theory of Feller processes that these Markov processes on Euclidean space are in correspondence with the semigroups, and the generators. I can see easily that BM is a Feller process, and that its semigroup, which is a semigroup by definition, is also integration against the gaussian kernel. I know that the spirit of why the Laplacian/2 is the generator is that you can take the time derivative under this integral, and use the fact that the heat equation is satisfied by the kernel. The trouble is the detail.
Question:
(b) is proving especially elusive, as well as making the statement that the Laplacian/2 is the generator in the first place. The missing piece seems to be correctly picking the domain so that (b) works. In analogy with two exercises from Liggett's book, I feel that the proper choice of space may be $D:=\{f \in C|\text{All first and second order partial derivatives exist and are in $C$}\}$ The trouble is (b) then requires me to essentially solve a PDE in the domain described here, and I have no idea how to do it.
I realized at some point in my efforts that it would go a long way towards the solution if I knew that all derivative operators had closed graph, but I failed to find appropriate domains for the action of differentiation to make this statement rigorous and true. Can someone help me with this fact as well?