Determine a transformation from a octahedron centered at $0$, of height and width $1$, to the unit cube centered at $0$.

Question

I have been working on this problem for a while, and I still do not have a clue as to how to approach this problem. We can determine the points in the octahedron as those points $(x,y,z) \in \mathbb{R}^3$ satisfying: $\max\{|x|,|y|\} + |z| \leq \frac{1}{2}.$

Clearly, $(x,y,z) \in \mathbb{R}^3$ is contained in the unit cube centered at $0$, if $\max\{|x|,|y|,|z|\} \leq 1/2$. However, this doesn't necessarily lend itself to a relatively simple solution involving a radial, bi-Lipschitz map from the octahedron to the cube.

Any thoughts on how to approach this problem?

Answer 1

I'd write the octahedron as $f(x,y,z) = |x|+|y|+|z|\le 1/2$. The cube is $g(x,y,z) = \max(|x|,|y|,|z|) \le 1/2$. You can try mapping $$ (x,y,z) \mapsto \frac{f(x,y,z)}{g(x,y,z)} (x,y,z) $$