Let $E$ be a locally compact separable metric space, $$C_0(E):=\left\{f\in C(E):\left\{|f|\ge\varepsilon\right\}\text{ is compact for all }\varepsilon>0\right\},$$ $\kappa$ be a Markov kernel on $(E,\mathcal B(E))$, $$\kappa f:=\int\kappa(\;\cdot\;,{\rm d}y)f(y)\;\;\;\text{for bounded Borel measurable}f:E\to\mathbb R,$$ $\lambda\ge0$ and $$Af:=\lambda(\kappa f-f)\;\;\;\text{for }f\in C_0(E).$$
Are we able to show that $A$ is the generator of a Feller semigroup (as defined in my other question) on $C_0(E)$?
Note that $(\mathcal D(A),A)$ is the generator of the uniformly continuous semigroup $$S(t)f:=e^{tA}f\;\;\;\text{for }f\in C_0(E)\text{ and }t\ge0.$$
EDIT: For whatever it may be useful, we may note that since $\kappa$ is contractive, $$\lambda\left\|\kappa f\right\|_\infty\le\lambda\left\|f\right\|_\infty\le(\lambda+\mu)\left\|f\right\|_\infty\tag2$$ and hence $$\left\|A_\mu f\right\|_\infty\ge\left|(\lambda+\mu)\left\|f\right\|_\infty-\lambda\left\|\kappa f\right\|_\infty\right|\ge(\lambda+\mu)\left\|f\right\|_\infty-\lambda\left\|f\right\|_\infty=\mu\left\|f\right\|_\infty\tag3$$ by the reverse triangle inequality for all bounded Borel measurable $f:E\to\mathbb R$ (in particular, for $f\in C_0(E)$) and $\mu\ge0$. So, $(\mathcal D(A),A)$ is dissipative.
EDIT2: It's trivial to see that $A$ satisfies the nonnegative maximum principle: If $f\in C_0(E)$ and $x_0\in E$ with $$f(x_0)=\sup_{x\in E}f(x)$$ (it's not necessary to assume $f(x_0)\ge0$ for the following conclusion), then $$(\kappa f)(x_0)-f(x_0)=\int\kappa(x_0,{\rm d}y)\underbrace{\left(f(y)-f(x_0)\right)}_{\le\:0}\le0\tag4.$$