To prove, that Bernoulli shift is sensitive to initial conditions

Question

I have a following function:

$$ B(x) = 2x\bmod 1\\ x \in \left<0; 1\right) $$

How could I prove, that this function is sensitive to initial conditions? I know it has something to do with Lyapunov exponent, but I only know following formula, which is used for discrete systems:

$$ \lambda(x_0) = \lim_{n\rightarrow\infty} \frac{1}{n} \sum^{n-1}_{i=0} \ln \left| f'(x_i) \right| $$

But, as far as I know, it's not useful here, because this problem is defined on a continuous domain.

How can I prove this?

Answer 1

Write $x$ in binary form: $$ x=\sum_{k=1}^\infty\frac{a_k}{2^k},\quad x_k=0\text{ or }1. $$ Identify $x$ with the sequence $(x_1,x_2,x_3,\dots)$. Then $$ B(x)=(x_2,x_3,\dots). $$ Now let $x$ and $y$ be two very close numbers; say that their first $n$ binary digits are the same, and that $x_{n+1}\ne y_{n+1}$. Then the firs binary digit of $B^n(x)$ and $B^n(y)$ are different.