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Apr 9, 2020 at 22:47 comment added Jochen Glueck @MateuszKwaśnicki: I see, thank you very much! I sometimes tend to see things through my "semigroup filter" - so when I hear "Markov process with countable state space", I just think of a positive $C_0$-semigroup on $\ell^1$ which is norm-preserving on the positive cone.
Apr 9, 2020 at 20:55 comment added Mateusz Kwaśnicki @JochenGlueck: That likely depends on your favourite definition of a Markov chain. As I understand, if we assume that the paths are càdlàg in the discrete topology on $\mathbb{Z}$, then the indicator functions are in the domain of the generator.
Apr 7, 2020 at 22:43 comment added Jochen Glueck I just found this paper by Ornstein; in the middle of the first page, he claims that there are examples where not all the indicator functions $1_n$ are in the domain of $G$: "if $i=j$ there are examples where it [it = the derivative of the transition probability $P_{ij}(t)$ at $t=0$] is infinite".
Apr 3, 2020 at 23:18 comment added Jochen Glueck Well yes, but my question is: why are all of these indicator functions in the domain? I'm not really convinced that this follows from your assumptions. (But I might be overlooking something simple or well-known.)
Apr 3, 2020 at 21:39 answer added Mateusz Kwaśnicki timeline score: 1
Apr 3, 2020 at 21:02 comment added Ribhu I think it should be straight forward. Since the functions of the form $1_n$ (indicator functions) are in the domain, so is their span.
Apr 3, 2020 at 19:54 comment added Jochen Glueck Thanks for your response! How do you know that $C_c(\mathbb{Z})$ is contained in the domain of $G$?
Apr 3, 2020 at 19:40 comment added Ribhu Yes, it is the set of functions with finite support. I have edited it now.
Apr 3, 2020 at 19:39 history edited Ribhu CC BY-SA 4.0
added 70 characters in body
Apr 3, 2020 at 19:25 comment added Jochen Glueck Could you specifiy what you mean by $C_c(\mathbb{Z})$? The functions with finite support?
Apr 3, 2020 at 18:33 history asked Ribhu CC BY-SA 4.0