Let $E = [0,1]$. I want to show that the operator $Af = \frac{1}{2}f''$ defined on $$D(A) = \{f \in C(E) : f'' \in C(E), f'(0) = 0 = f'(1)\} \subset C(E)$$ is a Markov generator. This Markov generator corresponds to Brownian motion with reflecting barriers.
The definition of a Markov generator I am working with is the following.
A Markov generator is a closed Markov pregenerator $A$ for which $R(\operatorname{Id}-\lambda A) = C(E)$ for all $\lambda \geq 0$.
Now, I have shown that $A$ is a closed Markov pregenerator, so I need to show that $R(\operatorname{Id}-\lambda A) = C(E)$ for all $\lambda \geq 0$. So I fix $\lambda \geq 0$ and $g \in C(E)$, and I want to show that there is an $f \in D(A)$ such that $f - \lambda Af = g$.
This is where I am stuck. I can't, for example, use a power series $f = \sum_{n=0}^\infty \lambda^n A^n g$ since $A$ is not everywhere defined and not bounded.
Something that might make this problem easier, which I have not proven myself so I am shying away from using it, is the following fact: For a closed Markov pregenerator $A$, if $R(\operatorname{Id}-\lambda A) = C(E)$ for small enough $\lambda > 0$, then the same is true for all $\lambda \geq 0$.