This question was previously posted on MathOverflow.
Consider the iterated function system (IFS) $X_n$ on $I = [0,1] $generated by the functions $\Phi = \{f_1,f_2,f_3\}$ and the probability vector $P = (p/2,p/2,1-p),$ where:
- $f_1,f_2: I\to I$, where $f_1(x) = x/2$ and $f_2(x) = (1+x)/2$;
- $f_3:I\to I$, where:
- $f_3(0)=0$, $f_3(1/2) = 1,$ $|f_3'|(x) > 1;$ and
- $f_3(x) = f_3(1-x)$ for every $x\in [0,1/2].$
- $0< 1- p<\varepsilon$.
Under the above conditions, it is relatively easy to show that if $\varepsilon>0$ is small enough $X_n$ admits a unique stationary measure $\mu$ on [0,1] and $$W(\mathcal P^n \nu, \mu) \xrightarrow{n\to 0} 0,\ \text{exponentially fast}, $$ where $W(\cdot,\cdot)$ is the Wasserstein metric and $$\mathcal P \nu = \sum_{i=1}^{3}p_i \nu(f_i^{-1}(\cdot)). $$
My question:
Given $x\in \mathrm{supp}\mu$, define $$d_-(\mu,x) = \liminf_{r\to 0}\frac{\log \mu(B_r(x))}{\log r}\ \text{and}\ d_+(\mu,x) = \limsup_{r\to 0}\frac{\log \mu(B_r(x))}{\log r}. $$
In case $d_{-}(x) = d_{+}(x)$ we say that $d(x) = d_{+}(x).$
Quesiton: Is it possible to show that there exists $d>0$ such that $d(x) = d$ for $\mu$-almost every $x\in [0,1]$. Or at least, to show that $d_-(x) \geq d$ for $\mu$-almost every $x\in [0,1]$.
In the literature, I can only find results that either assume that all $f_i$ functions are contracting or that the $f_i$ functions are contracting on average and have disjoint images, which is not my case. Also, I found this this preprint, which solves the second part of my question, assuming that each $f_i$ is a bi-Lipschitz homeomorphism (with its image) and some mild non-degenerated conditions (sadly, I was not able to replicate the method proposed in this paper to my example).
Any references or comments are highly appreciated.