There are probably some of you guys who already know some of the terms that I am going to use so in order to be not so boring I will put the definition to the end.

Let $f$ be a piecewise expanding map$^1$ on the interval $I=[0,1]$ and let $g:I \to C$ be a complex valued function which is of bounded variation. Let $L=L_g$ be the transfer operator$^2$ associated to $f$ defined on functions of bounded variation. Let $0=a_0< a_1<\cdots < a_n=1$ be the singular points$^1$ of $f$ and define $J_i=[a_i,a_{i-1}]$ for $i=1,\dots ,n$. Let $\psi_i$ be the inverse of $f$ restricted to $J_i$. Set

$$\phi_{i_1, \dots ,i_k}(x)=g(\psi_{i_1} x)\cdot g(\psi_{i_2} \psi_{i_1} x) \cdots g(\psi_{i_k} \cdots \psi_{i_1} x)$$

Then,

$$L^m \Phi(x) = \sum_{i_1,\dots i_{m}} \phi_{i_1,\dots , i_{m}} (x) \Phi(\psi_{i_m} \cdots \psi_{i_1} x)$$

(you can check with $m=1,2$ and convince yourself)

Here is my question: If $Var(\cdot )$ is defined by

$$Var(\Phi)=\sup \left( |\Phi(c_0)| + \sum_{i=1}^n |\Phi(c_i)-\Phi(c_{i-1})|+ |\Phi(c_n)| \right)$$

where the supremum is taken over all finite subsets of points of $I$ such that $c_0 < c_1 < \cdots < c_n$.

Why is the following equality true:

$$Var(L^m \Phi)= \sum_{k=1}^m \sup_x |g(f^{k-2} x) \cdots g(fx)g(x)| \cdot \sum_{i_k} Var (g \circ \psi_{i_k})\cdot \sup_y \left| \sum_{i_{k+1},\dots , i_m } \phi_{i_{k+1},\dots , i_m} (\psi_{i_k} y) \Phi (\psi_{i_m} \dots \psi_{i_k} y ) \right| + \sup_x |g(f^{m-1}x) \cdots g(fx)g(x)| \cdot Var \Phi $$

?

Thank you in advance

Definitions:

$^1$ (Piecewise Expanding maps) A function $f: I \to I$ is called a piecewise expanding map if there exists points $0=a_0 < a_1 < \dots < a_n=1$, called singular points, such that

(i) for each $i=1, \dots , n, f$ is $C^1$ on $[a_i, a_{i-1}]$ and the map on $[a_i, a_{i-1}]$ given by $x \to |f'(x)|^{-1}$ is of bounded variation

(ii) $|f'|>1$ on $[a_i,a_{i-1}]$ for all $i=1,\dots , n$

$^2$ The transfer operator $L_g$ associated to $f$ is the linear operator on $BV$ defined by

$$[L_g(\Phi)](x)= \sum_{f(y)=x} g(y) \Phi (y) $$

istrue, perhaps from reading it in a an article or book where it may or may not have been explained, but you haven't given any clues about where it comes from, or what it might be good for. – Bill Thurston Feb 12 '11 at 14:13