Let $f: I \to I$ be a one-dimensional differentiable function of bounded distortion with distortion constant $M$, where $I$ is a compact interval in $\mathbb{R}$. That is, $I$ can be partitioned such that for each $J$ in the partition there exists an interger $n_J$ such \begin{align} \max_{x, y \in J} \log \frac{Df^k(x)}{Df^k(y)} \leq M, \quad 1 \leq k \leq n_J. \end{align} I've been told that this implies for any subintervals $A, B \subset I$ \begin{align} \frac{|A|}{|B|} \leq e^M \frac{|f(A)|}{|f(B)|} \end{align} where $|A|$ denotes the size of the interval $A$. The problem is that I don't know how to go about showing this true, does anyone know how to show this?
5
-
1$\begingroup$ Although I think I can guess what all the symbols mean, could you be more explicit about the domain and range of $f$, what $k$ is in the first equation, what $A$ and $B$ are in the second equation, and what is $|A|$? $\endgroup$– Deane YangCommented Nov 19, 2020 at 20:25
-
$\begingroup$ It looks as if you can simplify the left hand-side of the first inequality by cancelling $D$ in the numerator and the denominator =). $\endgroup$– Luc GuyotCommented Nov 19, 2020 at 21:49
-
$\begingroup$ Luc Guyot: apologies, the $Df^k$ is meant to mean the derivative of the $k$-th iterate of $f$. I've changed the constant to $M$ to make that clearer. $\endgroup$– Andrew LarkinCommented Nov 19, 2020 at 22:05
-
$\begingroup$ I think it should be $e^M$ in the second inequality? $\endgroup$– Anthony QuasCommented Nov 20, 2020 at 2:49
-
$\begingroup$ Anthony Quas: Yes you're right, I'll change that now $\endgroup$– Andrew LarkinCommented Nov 20, 2020 at 11:00
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
Ah, I found the solution, it's just the mean value theorem.
By the mean value theorem we know that there exist points $\eta_1, \eta_2$ such that \begin{align} Df(\eta_1) = \frac{|f(A)|}{|A|}, \quad Df(\eta_2) = \frac{|f(B)|}{|B|} \end{align} Thus we have \begin{align} \frac{|f(B)|}{|B|} \Big/ \frac{|f(A)|}{|A|} = \frac{|Df(\eta_2)|}{|Df(\eta_1)|} \leq e^M, \end{align} which rearranges to \begin{align} \frac{|A|}{|B|} \leq e^M \frac{|f(A)|}{|f(B)|}.\end{align}