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A mapping $f: \mathbb{R}^n\to \mathbb{R}^n$ is said to be $K$-bi-Lipschitz, $K>1$, if \begin{equation*} \dfrac{1}{K}\leqslant \dfrac{|f(x)-f(y)|}{|x-y|}\leqslant K, \end{equation*} for any $x,y\in \mathbb{R}^n$.

I aim to construct a specific mapping $f$, which maps a cube $Q$ to the ball $B(x_Q,l(Q)/2)$ with $f(x_Q)=x_Q$, where $x_Q$ is the center of the cube, and $l(Q)$ is the side length of the cube. The notation $B(x_Q,l(Q)/2)$ denotes the ball centered at $x_Q$ with radius $l(Q)/2$. The cube is like $[-1,1]^n$, which means the edges of the cube are parallel to the coordinate axes.

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Sorry, maybe the above is not very clear. My question is, could you show me an exact bi-Lipschitz mapping $f$ for some $K>1$, which maps a cube $Q$ to the ball $B(x_Q,l(Q)/2)$?

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    $\begingroup$ What is the question? Maybe you want the minimum $K$? You may take wlog $Q$ and $B$ to be the unit balls in the max norm resp. the Euclidean norm, with center the origin. $\endgroup$
    – Pietro Majer
    Commented Sep 15 at 18:41
  • $\begingroup$ This is not a research-level question. Do you know how to find a homeomorphism between two convex solids? Did you try to check that it is bilipschitz? $\endgroup$
    – Moishe Kohan
    Commented Sep 16 at 2:24
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    $\begingroup$ @Moishe Kohan Sorry, I don't know. I read a paper, which claims that such a bi-Lipschitz mapping exists, but it doesn't say more. I am not familar with those concepts, so could you provide me with some references about what you mentioned? $\endgroup$
    – Javier
    Commented Sep 16 at 3:15
  • $\begingroup$ I believe this question would be more appropriate in Math Stackexchange than here. However, you can define your mapping as follows (assuming wlog that $x_Q = 0$ for convenience of notation): $$f(x) = \frac{\max\{|x_1|,\ldots,|x_n|\}}{\sqrt{x_1^2+\ldots+x_n^2}}x.$$ $\endgroup$
    – Clara Torres-Latorre
    Commented Sep 16 at 7:31
  • $\begingroup$ It is a totally general fact about two norms $||x||, ||x||'$ on $\mathbb R^n$. You verify that the map $f(x)=\frac{||x||}{||x||'}x$ sends homeomorphically the unit ball in the first norm to the unit ball in the second norm. Verifying the Lipschitz property of $f$ is a calculus exercise. (You can use partial derivatives because each norm function is Lipschitz, you just have to check uniform boundedness of partial derivatives of $f$ wherever they are defined.) $\endgroup$
    – Moishe Kohan
    Commented Sep 16 at 14:25

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