# Is irreducibility sufficient for uniqueness of invariant distribution for a Feller semigroup?

Let $$(T_t)$$ be a strongly continuous semigroup of positive operators on $$C(K)$$, where $$K$$ is a compact space. Assume also that $$T_t1 =1$$ for every $$t\geq 0$$. (This is also called a Feller semigroup.)

Since $$K$$ is compact we know that there exists a probability measure $$\mu$$ on $$K$$ satisfying $$\mu T^*_t = \mu$$ for every $$t\geq 0$$ (i.e. $$\mu$$ is invariant).

My question is: to show that $$\mu$$ is the unique invariant probability distribution, is it sufficient to show that $$(T_t)$$ is irreducible?

Recall that a semigroup is by definition irreducible if the resolvent $$R_\lambda=(\lambda-L)^{-1}$$ ($$L$$ is the generator of $$(T_t)$$) maps for sufficiently large $$\lambda$$ nonnegative nonzero functions into strictly positive functions.

I thought this should be true by applying some version of the Krein-Rutman theorem, but did not find a suitable reference.

The closest I found is Proposition 3.5. on p. 185 of this book (link at Springer site), from which, if I understand well, I can just conclude that $$\text{dim (ker } L) = 1$$, but not $$\text{dim (ker } L^*) = 1$$.

Take $K = \{0,1\}^{\mathbf{Z}^2}$ and take for $T_t$ the Glauber dynamic for the Ising model below the critical temperature. Then $T_t$ is Feller and irreducible, but it has two distinct ergodic invariant measures.
If however you know that $T_t$ is strong Feller (or asymptotically strong Feller) then irreducibility does imply uniqueness of the invariant measure.