I have to show that for the generator $A:L^1 \rightarrow L^1$ of a sub-Markov semigroup and a non-negative $f_* \in L^1$ (with $L^1$ Set of Lebesgue-integrable functions) with $\int_{-\infty}^\infty f_* = 1 $ the following operator $A_*$ is the generator of a Markov semigroup with $A_* f = Af - (\int_{-\infty}^\infty Af)f_*$.
By Hille-Yosida we know that $A$ fulfills the following conditions: i) $A$ is densely defined ii) $\forall g \in L^1, \lambda > 0 \exists! f \in D(A): \lambda f - A f = g$ and iii) $\int_{-\infty}^\infty Af \leq 0$ for all non-negative $f \in D(A)$ [for a Markov semigroup (without the sub-) iii) is an equality for all $f \in D(A)$ we call this condition iii*)].
So now I have to show i),ii) and iii*) for $A_*$ i) follows directly from the definition I assume since the domain of $A_*$ doesn't change and iii*) is simple arithmetic.
But I struggle with ii), I tried plugging in the definition of $A_*f$ in the equation $g = \lambda f - A_* f = \lambda f - Af + (\int_{-\infty}^\infty Af)f_*$ and thought that maybe since the last part is just a constant we still have the property but I'm not sure.
I also thought about using Trotters Product Formula and Phillips Perturbation Theorem but I have difficulties seeing how to piece it all together.
Could anyone help, a little intuition wouldn't hurt either.
