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Consider a Q-matrix on a countable state space. (A Q-matrix is a matrix whose rows sum up to $0$, with nonpositive finite diagonal entries and nonnegative offdiagonal entries).

As explained for example in the book of Norris on Markov chains, every Q-matrix (without any further assumptions) defines a transition function (given by the minimal nonnegative solution of the corresponding backward equation).

Is the the semigroup associated to this minimal transition fucntion always Feller? (definition of Feller: functions vanishing at infinity are maped into functions vanishing at infinty).

I know that in general, not every continuous time Markov chain is Feller, but I guess that for this minimal chains it is always true without further assumptions on $Q$. Though I didn't find a statement like this in the books I consulted.

Has anybody a reference/proof/counterexample?

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Isn't it that any transition function there is a correspondent $Q$-matrix? – Ilya Feb 13 '12 at 10:09
Why should this imply Feller property? – Filippo Feb 13 '12 at 14:08
It shouldn't: I was thinking that you can take a process which is not a Feller and deduce a $Q$-matrix from it and then show that the minimal solution for the backward problem with such $Q$-matrix is not Feller. However, for this purpose one has to show that the original process is the least solution. I don't know though if there are sufficient conditions for the backward equation has the unique solution among non-negative ones. – Ilya Feb 13 '12 at 14:13
I see waht you mean now. There is also another problem: in general the Q-matrix of a (standard) transition function is not as nice as I defined it (the diagonal entries could be infinite, or the row sum positive). This restricts the possible non Feller counterexamples even more. A typical condition which implies uniquness for the backward equation is the rates to be uniformly bounded. But this implies also the Feller property, so it wouldn't help. Let me know if you have further ideas! thanks. – Filippo Feb 13 '12 at 14:58

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