Let $A$ be a Markov pregenerator, $E$ a compact and complete space. That is, $A$ is a linear operator satisfying:
$D(A)$ is dense in $C(E)$
$A1 = 0$
For all $f \in D(A)$, $\underset{x \in E}{min}(f(x) - \lambda Af(x)) \leq \underset{x \in E}{min} f(x)$
I want to show that if the range of $I - \lambda A$ is $C(E)$ for sufficiently small positive $\lambda$, then this holds for all positive $\lambda$. I know that this is true and have tried to prove it for the last half hour with no luck. Thanks in advance.
Equivalently, we can assume that the range of $\lambda - A$ is $C(E)$ for some $\lambda > 0$. This will be a notational convenience for applying functional analytic results.
Recall that a markov pregenerator satisfies for $f \in D(A)$, $\|f\| \leq \|(I - \mu^{-1}A)f \|$ for all $\mu > 0$. In particular, if $(\lambda - A)$ is surjective then it is invertible and its inverse $R(\lambda, A)$ satisfies $\|R(\lambda, A) g\| \leq \lambda^{-1} \|g\|$ for every $g \in C(E)$. If $\lambda - A$ is invertible we say $\lambda$ is in the resolvent set $\rho(A)$ of $A$ and call $R(\lambda,A)$ the resolvent of $A$ at $\lambda$. In particular, we know that $\|R(\lambda,A)\| \leq \lambda^{-1}$.
By a Neumann series argument, if $\lambda \in \rho(A)$ and $|\mu - \lambda| < \|R(\lambda,A)\|^{-1}$ then $\mu \in \rho(A)$ also. So fix $\mu \in (0, 2 \lambda)$. Then $|\mu - \lambda| < \lambda \leq \|R(\lambda,A)\|^{-1}$ so $\mu \in \rho(A)$. Since invertible operators are surjective, we then have that the range of $\mu - A$ is $C(E)$.
To finish, we iterate this argument. Replace your initial $\lambda$ with $\frac32 \lambda \in (0, 2 \lambda)$. Then the above argument yields the same conclusion for $\mu \in (0, 3\lambda)$. Now start with $\frac52 \lambda$ to conclude for $\mu \in (0,5 \lambda)$ and so on.