Norm which devides $\mathbb R^3$ into cubes

Question

Depending on the norm on a vector space, a surface containing all points with a given distance to a point $P$ takes a specific shape. For example in $\mathbb{R}^2$ the norm $||(x,y)|| = |x| + |y|$ creates the shape of a square. On $\mathbb{R}^3$ this norm creates an octahedron. Is there a norm where this surface takes the shape of a cube?

Answer 1

Yes. It's known as the $\infty$-norm (in the sense of $p$-norms). It's defined as $$\|(x, y, z)\| = \max\{|x|, |y|, |z|\}.$$ Then $\|(x, y, z)\| \le 1$ if and only if $|x| \le 1$ and $|y| \leq 1$ and $|z| \le 1$, i.e. $(x, y, z) \in [-1, 1]^3$, a cube.