Baker Map
The baker map $f: \Omega \rightarrow \Omega$ is given as follows,
\begin{equation} f(x_1, x_2) = \left( 2x_1 \mod{1}, \frac{1}{2}(x_2 + [2x_1]) \right) \equiv (x_1',x_2') \end{equation} where $[\xi]$ is the integer part of $\xi \in \mathbb{R}$ and $\Omega = [0,1[$.
We introduce the dyadic representation, \begin{equation} \label{eq:dyadic} \vec{x} = (x_1, x_2) = \underbrace{\dots b_{-3}b_{-2}b_{-1}}_{x_1}.\underbrace{b_0b_1b_2 \dots}_{x_2} \in \Omega \end{equation}
and the "Manhattan" norm, $$ ||\vec{x}|| = |x_1| + |x_2| $$
Question
Let $\vec{x}$, $\vec{x}' \in \Omega$ et $N > 0$. I want to show that if $b_n(\vec{x}) = b_n(\vec{x}')$ for $-N \leq n \leq (N - 1)$ then \begin{equation} ||\vec{x} - \vec{x}'|| \leq 2^{-(N-1)} \end{equation}
using the "Manhattan" norm and dyadic representation.
Answers
note: I edited the post since I eventually found the answer (I think)
note : $x$ and $x'$ are two numbers $\in \Omega$ i.e. $x' \neq f(x)$ in that case.
We can write, \begin{align*} x_1 &= .b_0b_1b_2\dots = \sum_{n=1}^{+\infty} \frac{b_{n-1}}{2^n} \\ x_2 &= .b_{-1}b_{-2}b_{-3}\dots = \sum_{n=1}^{+\infty} \frac{b_{-n}}{2^n} \end{align*}
Let
\begin{align*}
b_n(\vec{x}) &= (\dots,b_{-2},b_{-1},b_0,b_1,b_2,b_3,\dots), \quad b_i \in \{0,1\} \\
b_n(\vec{x}') &= (\dots,b'_{-2},b'_{-1},b'_0,b'_1,b'_2,b'_3,\dots), \quad b'_i \in \{0,1\} \\
\end{align*}
Let $b_n = b_n(\vec{x})$ and $b'_n = b_n(\vec{x}')$.
The distance between $\vec{x}$ and $\vec{x}'$ is given by,
\begin{align*} ||\vec{x} - \vec{x}'|| &= |x'_1 - x_1| + |x'_2 - x_2| \\ &= \Big| \sum_{n=1}^{+\infty} \frac{b'_{n-1}}{2^n} - \sum_{n=1}^{+\infty} \frac{b_{n-1}}{2^n} \Big| + \Big| \sum_{n=1}^{+\infty} \frac{b'_{-n}}{2^n} - \sum_{n=1}^{+\infty} \frac{b_{-n}}{2^n} \Big| \\ &= \Big| \sum_{n=1}^{+\infty} \frac{b'_{n-1} - b_{n-1}}{2^n} \Big| + \Big| \sum_{n=1}^{+\infty} \frac{b'_{-n} - b_{n}}{2^n} \Big| \\ &= \sum_{n=1}^{+\infty} \frac{|b'_{n-1} - b_{n-1}|}{|2^n|} + \sum_{n=1}^{+\infty} \frac{|b'_{-n} - b_{n}|}{|2^n|} \end{align*}
Because $b_n$ and $b'_n \in [0,1[$, the distance $|b'_n - b_n| \leq 1$ and because $b_n = b'_n$ for all $-N \leq n \leq N - 1$, $|b'_n - b_n| = 0$ for all $-N \leq n \leq N - 1$, \begin{align*} ||\vec{x} - \vec{x}'|| &\leq \sum_{n=N+1}^{+\infty} \frac{1}{|2^n|} + \sum_{n=N+1}^{+\infty} \frac{1}{|2^n|} \\ &= 2 \sum_{n=N+1}^{+\infty} \frac{1}{2^n} \\ &= \sum_{n=N+1}^{+\infty} \frac{1}{2^{n-1}} = \frac{1}{2^N} + \frac{1}{2^{N+1}} + \dots \\ &\leq \frac{1}{2^{N-1}} = 2^{-(N-1)} \end{align*}