Let $\mathbb{R}^n$ and $\mathbb{R}$ be equipped with the Euclidean distance. Consider the interchanging-digit map (for lack of a better name) $f: [0,1]^n \mapsto [0,1]$, defined as follow:
$$f(x_1,x_2,\ldots,x_n) = \sum_{i=1}^{\infty}\sum_{j=1}^{n}10^{-(n.(i-1)+j)}x_{j,i}\, ,$$
where $x_{j,i}$ is the $i^{th}$ digit of the $j^{th}$ variable.
Example: consider $x_1 = (0.437,0.982) \in [0,1]^2$. Then, $f(x_1,x_2) = 0.493872$.
That is, the first digit from $x_1$ is taken, followed by the first digit from $x_2$, followed by the second digit from $x_1$, followed by the second digit from $x_2$, and so on.
It would help me a lot if the map $f$ was $1$-Lipschitz. I have doubts that it is but I haven't been able to prove nor disprove the claim. Moreover, I've tried to show that $f$ wasn't continuous to show that it wasn't Lipschitz but I didn't have any success with that either.
Does someone know how to prove that? Thank you.
PS: If $f$ is 1-Lipschitz for another norm than the Euclidean norm, please feel free to change the norm.
I believe your map is not continuous. $$ f\left(\frac{1}{10},\frac{1}{10}\right) = f(0.1\overline{0}, 0.1\overline{0}) = 0.11\overline{0} = \frac{11}{100} $$ but $$ f\left(\frac{1}{10},\frac{1}{10}\right) =f(0.0\overline{9}, 0.0\overline{9}) = 0.00\overline{9} = \frac{1}{100} $$ If you try to rule out one of the possible expansions of those numbers with two expansions, you will still get a discontinuity by approaching this point from the right compared to from the left.