Ruelle Perron Frobenius Operator Hello everybody! Recently in my research, I came across the Perron-Frobenius operator .. I would like to intuitive interpretation of this operator, ie, physical interpretations are possible, articles (can be physical).
I wonder how this operator originated.
 A: I'll try to complement Vaughn's answer. In general the (Ruelle-) Perron-Frobenius or transfer operator $\mathcal{F}$ is given by
$\int_X \mathcal{F}f = \int_{T^{-1}(X)} f$ 
for $f$ and $X$ generic and nice, and $T$ some invertible transformation of the ambient space. (The "transfer" nomenclature is related to the "transfer matrix" method in statistical physics.)
So let's think about what this means on a finite space $X \equiv \{1,\dots,n\}$ with (probability) measure $p$ corresponding to a row vector, and with functions corresponding to column vectors. Let $\mathcal{P}_X$ be the projection onto $X$: then $\int_X g \equiv p \mathcal{P}_X g$ and $\mathcal{F}$ is defined via $p \mathcal{P}_X \mathcal{F}f = p \mathcal{P}_{T^{-1}(X)} f$. 
W/l/o/g, let $f = e_j$ for some $j$. Then $p \mathcal{P}_X \mathcal{F}e_j = p \mathcal{P}_{T^{-1}(X)} e_j$ and the RHS is either an entry of $p$ or zero according to whether or not $j \in T^{-1}(X)$, respectively. 
In the special case where $T$ is a permutation, $\mathcal{F}$ is the corresponding permutation matrix. The general idea is similar: $\mathcal{F}$ is an operator that represents the action of $T$ on a suitable function space.
A: The place to start is with the Perron-Frobenius theorem, which (in its most basic form) says that a $d\times d$ matrix $A$ with only positive entries has exactly one positive eigenvector $\vec{v}$, which corresponds to the largest eigenvalue (which is real and positive).  Furthermore, all the other eigenvalues are strictly smaller in absolute value, and the iterates of any non-negative vector under the matrices $A^n$ converge exponentially to the eigendirection spanned by $\vec{v}$.  This can be interpreted very intuitively by looking at the action on the unit simplex $\Delta = \{\vec{w}\in \mathbb{R}^d \mid w_i \geq 0, \|\vec{w}\|_1 = 1\}$ given by $f(\vec{w}) = A\vec{w} / \|\vec{w}\|_1$.  This is discussed at more length in the books by Brin and Stuck and by Katok and Hasselblatt.
Once you understand that case, you have at least a general picture for what you're trying to prove in the more general setting, when $\vec{w} \in \mathbb{R}^d$ is replaced by a function $\phi$ in some Banach space $X$.  The idea in a wide variety of settings is that you have a dynamical system which defines a natural action on spaces of functions, and you'd like to find a Banach space of functions on which the properties of that action (which is the Ruelle Perron-Frobenius operator) mimic those of the positive matrix $A$ above.  The two cases are linked by the fact that if $A$ is a $0$-$1$ matrix that determines the admissible transitions for a mixing topological Markov chain on $d$ symbols, then the matrix $A$ actually is the Perron-Frobenius operator.  (Note that the mixing property says that some power of $A$ is positive, so then you can apply the standard Perron-Frobenius theorem.)
A: I like to explain the Ruelle Perron Frobenius operator this way:
Suppose that $T$ is a map from some space $X$ to itself. Suppose also that $X$ has some distinguished ambient measure $\lambda$ on it (think of Lebesgue measure). One more assumption: the ambient measure is non-singular with respect to $T$ (i.e. if $A$ is a set of measure 0, then $T^{-1}A$ has measure 0 also).
Given a measure $\mu$ on $X$, the "push-forward" of $\mu$ is defined to be $\mu\circ T^{-1}$. Seems counter-intuitive, but in the case where $\mu$ is concentrated at a single point $a$ you can check that the push-forward is concentrated at $T(a)$.
The non-singularity condition implies that if $\mu$ is a measure absolutely continuous with respect to $\lambda$, then its push-forward is also absolutely continuous with respect to $\lambda$. By the Radon-Nikodym theorem, measures absolutely continuous with respect to $\lambda$ have essentially unique densities.
The Ruelle-Perron-Frobenius operator applied to $f$ gives the density of the push-forward of the measure whose density is $f$.
In more picturesque language: if $X$ is a random variable with density $f(x)$, then $T(X)$ is a random variable with density $L[f](x)$.
