I am looking at the example below from Schilling's Brownian Motion. Here the generator of a Brownian transition semigroup is proven to be the Laplacian. What I don't understand from the proof below is the last part. Namely, why does the expression $|\partial_j \partial_k u(x+\theta B_t)-\partial_j \partial_k u(x)|^2$ converge to $0$ as $t \to 0$ uniformly in x? I don't see how to get this uniform convergence. Also, how do we use this to get, by dominated convergence, uniform convergence of $E[\frac{u(B_t +x)-u(x)}{t} - \frac{1}{2} \sum_{j=1}^d \partial_j^2 u(x)]$?
It seems like we can use dominated convergence theorem on $(\sum_{j,k=1}^d E(|\partial_j \partial_k u(x+\theta B_t) - \partial_j \partial_k u(x)|^2))^{1/2}$. But what we actually need to show is $\sup_{x\in \mathbb{R}^d}(\sum_{j,k=1}^d E(|\partial_j \partial_k u(x+\theta B_t) - \partial_j \partial_k u(x)|^2))^{1/2} \to 0$ as $t \to 0$. I do not know how to account for this supremum outside of this. I would greatly appreciate any help.
