Bernoulli Map: $f'(x) = 2$ Almost Everywhere and "Local Separation" Increases as $2^n$

Question

The Bernoulli map is

$$x_{n + 1} = f(x_n)= \begin{cases} 2x_n, & 0 \leq x_n < 0.5\\ 2x_n - 1, & 0.5 \le x_n \le 1 \end{cases}$$

I am told that (1) $f'(x) = 2$ almost everywhere and so (2) local separation increases as $2^n$.

1. With regards to (1), why is $f'(x) = 2$ almost everywhere instead of just everywhere?

2. And what is meant with by (2)? I can't find any information about what "local separation" means in this context.

I would greatly appreciate it if people could please take the time to clarify this.

Answer 1

Let's calculate $f'(0.5)$: $$\begin{align} f'(0.5)&=\lim_\limits{h \to 0} \frac{f(0.5+h)-f(0.5)}{h}\\ &=\lim_\limits{h \to 0} \frac{f(0.5+h)-0}{h}\\ &=\lim_{h \to 0} \frac{f(0.5+h)}{h} \end{align}$$ But this limit does not exists, because $$\begin{align}\lim_{h \to 0+0} \frac{f(0.5+h)}{h}&=\lim_{h \to 0+0} \frac{2(0.5+h)-1}{h}\\ &=\lim_{h \to 0+0} \frac{1+2h-1}{h}\\ &=\lim_{h \to 0+0} \frac{2h}{h}\\ &=2 \end{align}$$ But $$\begin{align}\lim_{h \to 0-0} \frac{f(0.5+h)}{h}&=\lim_{h \to 0-0} \frac{2(0.5+h)}{h}\\ &=\lim_{h \to 0-0} \frac{1+2h}{h}\\ &=\lim_{h \to 0-0} \frac{2h}{h}+\frac{1}{h} \end{align}$$ Which is not finite, but $f'(x)=2 \forall x \in (0, 1) \setminus\{0.5\}$

Answer 2

What 2. means is that $|f(x)-f(y)|=2\,|x-y|$ if both $x$ and $y$ are on the same side of $1/2$. Iterating, we get $|f^n(x)-f^n(y)|=2^n\,|x-y|$ as long as all the iterates are on the same side of $1/2$; $x$ and $y$ are separated by a factor of $2^n$ when $f$ is applied $n$ times.