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). 



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> My question is: to show that $\mu$ is the unique invariant probability distribution, is it sufficient to show that $(T\_t)$ is irreducible?
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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](http://www.springer.com/mathematics/algebra/book/978-3-540-16454-8)), from which, if I understand well, I can just conclude that $\text{dim (ker } L) = 1$, but not $\text{dim (ker } L^\*) = 1$. 



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


