MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

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


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

(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


$^1$ (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

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

share|cite|improve this question
Your formula goes outside the margins, at least in my display, so it's illegible. Also, you have not explained your motivation for the question (whatever it is). You presumably have a reason to know that the illegible formula is true, 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
Use induction based on an inequality like $\text{var}(fg)\le \|f\|_\infty\text{var}(g)+\|g\|_\infty\text{var}(f)$? – Anthony Quas Feb 12 '11 at 16:09

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.