$f(x) = 1 - |1 - 2x|$, $a_{n+1} = f(a_n)$, prove convergence of subsequences Let $f(x) = 1 - |1 - 2x|$, $a_1 = a$, $a_{n+1} = f(a_n)$. Prove there exists $a \in [0, 1]$ such that for every $x \in [0, 1]$ there exists a subsequence of $a_n$ convergent to $x$. I've tried to analyze the graph of this function, but couldn't spot anything useful.
 A: Note that $f(x)=2\min\{x,1-x\}$. If the binary expansion of $a$ is
$$a=0.b_1\ldots b_n0x_1x_2\ldots $$
(where not all $x_i$ are $=1$),
we conclude that the binary expansion of $a_{n+2}$ is $$a_{n+2}=0.x_1x_2\ldots$$
So in order to find a subsequence converging to $x$, we only need to make sure that longer and longer prefixes of the binary expansion of $x$ occur after a $0$ in the expansion of $a$ (where we use the expansion $0.1111\ldots$ for $x=1$). In order to achieve this for all possible $x\in[0,1]$, we just need to make sure that all possible bit patterns occur after a $0$. So let 
$$ a=0.0\color{red}00\color{red}10\color{red}{00}0\color{red}{01}0\color{red}{10}0\color{red}{11}0\color{red}{000}0\color{red}{001}0\color{red}{010}0\color{red}{011}0\color{red}{100}0\color{red}{101}0\color{red}{110}0\color{red}{111}0\color{red}{0000}0\color{red}{0001}\ldots$$
A: This isn't (now) an answer, but an idea for a simpler solution.
Let $f^n=f\circ f\circ\dotsb\circ f$ $n$ times and $f^{-n}=f^{-1}\circ f^{-1}\circ\dotsb\circ f^{-1}$ $n$ times, the assertion is equivalent to find an $a\in ]0, 1[$ such that the set
$$
\mathcal{A}=\{f^n(a):n\in\mathbb N\}
$$
is dense in $]0, 1[$. Suppose that for every $a\in ]0, 1[$ exists an interval $I=]x, y[\subseteq \left]0, \frac{1}{2}\right[$ (or $\left]\frac{1}{2}, 1\right[$) with length $\frac{1}{2^k}$ such that
$$
I\cap\mathcal{A}=\emptyset
$$
Let $L=f^{-1}(I)$ observe that
$$
L=\left]\frac x2, \frac y2\right[\sqcup\left]1-\frac y2, 1-\frac x2\right[
$$
then $L$ is symmetric and $L\cap\mathcal A=\emptyset$ because $f^{-1}(\mathcal A)\supseteq\mathcal A$ and
$$
L_1=f^{-1}(L)=\frac{L}{2}\sqcup\left(1-\frac L2\right)
$$
In general

For every $K\subseteq ]0, 1[$ let $K'=f^{-1}(K)$ then
  $$
K'=1-K'=\{1-k:k\in K'\}
$$
  Proof: For every $x\in ]0, 1[$ we have $f(x)=f(1-x)$ then
  $$
x\in K'\Leftrightarrow f(x)=f(1-x)\in K\Leftrightarrow 1-x\in K'
$$

and define $L_n$ by recursion. Then we have
$$
L_n=f^{-1}(L_{n-1})=\frac{L_{n-1}}{2}\sqcup\left(1-\frac {L_{n-1}}2\right)\\
L_n=1-L_n\\
\lvert L_n\rvert = \lvert L\rvert =\frac{1}{2^k}\text{ the Lebesgue measure)}\\
L_n\cap\mathcal A=\emptyset
$$
Now we define this new function:
$$
g:x\in [0, 1]\rightarrow\begin{cases}
2x &\text{ if }2x\leq 1\\
2x -1 &\text{ if }2x > 1
\end{cases}
$$
Then it holds:

If $K$ satisfy $K=1-K$ then
  $$
f^{-1}(K)=g^{-1}(K)
$$
  Proof: let $f(x)\in K$, if $2x\leq 1$ then $f(x)=g(x)$ otherwise $f(x)=2(1-x)$. Because $K$ is symmetric we have 
  $$
1-f(x)=1-2(1-x)=1-2+2x=2x-1=g(x)\in K
$$
  and vice versa.

In particular
$$
f^{-1}(L_n)=g^{-1}(L_n)
$$
so we can work with $g$ instead of $f$, observe that $g$ works simply as a "left shift" of $x$ seen in binary representation then if the assertion isn't true for $f$ then it doesn't value neither for $g$, this function is more simpler than $f$ and we can work more simply on it, for example we can use a simplified version of the preceding solution because the "changing effect" doesn't appear when using $g$.
