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 $\text{dim (ker } L^*) = 1$.


1 Answer 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.


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