TITLE: Ruelle-Perron-Frobenius theorem for shift of finite type
QUESTION [6 upvotes]: I know a version of Ruelle's theorem for expansive transformations in a compact metric space that says there is a single equilibrium state for a potential holder. In this Ruelle-Perron-Frobenius theorem from the Bowen book: Equilibrium States and the Ergodic Theory of Anosov Diffeomorphisms, the same thing is said is basically Ruelle's theorem for shift of finite type, but my question is: what does this limit that I marked green intuitively mean?

REPLY [7 votes]: The most intuitive explanation I know is the following: suppose that you have a certain amount of mass (I usually picture a pile of sand) that is distributed over $\Sigma_A^+$ according to the density $g\,d\nu$, where $g$ is an arbitrary continuous function and $\nu$ is the eigenmeasure for the transfer operator $\mathcal{L}$.  Then you move this mass around according to the shift map: a grain of sand at $\underline{x}$ gets moved to $\sigma(\underline{x})$, for each $\underline{x} \in \Sigma_A^+$.  After doing this $n$ times, the sand is distributed according to some new density $g_n \,d\nu$.  Then the new density function $g_n$ is given by $g_n = \lambda^{-n} \mathcal{L}^n g$; in other words, the transfer operator tells you how mass moves around (w.r.t. the underlying reference measure $\nu$).  This much can be proved just by manipulating the definition of the transfer operator, and is a good exercise if you didn't see it yet.  The intuitive content of the limit you marked in green is that as $n\to\infty$, the density functions $g_n$ converge to the eigenfunction $h$, perhaps multiplied by some constant that corresponds to the initial amount of sand; in other words, mass that is spread out according to any initial distribution absolutely continuous w.r.t. $\nu$ will eventually converge towards a distribution with density function given by $h$.