What polytopes can be induced by a norm? Let $\|\cdot\|:\mathbb{R}^n\to\mathbb{R}_{\ge0}$ be a norm and define $B_{\|\cdot\|}(0,1):=\{x\in\mathbb{R}^n\mid\|x\|\le1\}$ be the unit circle. For which regular n-dimensional polytopes (relative to the Euclidean norm resp. the Euclidean product) there exists a norm such that $B_{\|\cdot\|}(0,1)$ is equal to the polytope given? For $n=3$ we can easily obtain a cube (wit the max norm) and an oktahedra (with $\|\cdot\|_1$) but can we find a norm withe the unit circle of a tetrahedra? What other forms are possible?
I came up with this question purely out of curiosity, but I am quite clueless how to approach it. So any thoughts are appreciated.
 A: Let $X \subseteq \mathbb{R}^n$ be any compact convex set that is symmetric about the origin and contains an open neighbourhood of $0$. Then we can define the Minkowski functional $p_X : \mathbb{R}^n \to \mathbb{R}_{\geq 0}$ by
$$ p_X(y) = \inf\big\{\lambda \in \mathbb{R}_{> 0} \: : \: \lambda^{-1}y\in X\big\}. $$
One easily shows that $p_X$ is a well-defined norm on $\mathbb{R}^n$ and that $X$ is precisely the closed unit ball for this norm. (Here you use that any open neighbourhood of $0$ is absorbing, so that there always exists some $\lambda > 0$ such that $\lambda^{-1}y \in X$ holds.)
Conversely, let $\lVert\:\cdot\:\rVert$ be any norm on $\mathbb{R}^n$, then one can prove that the closed unit ball with respect to this norm is compact, convex and symmetric about the origin and contains an open neighbourhood of $0$. (Here you need that all norms on $\mathbb{R}^n$ are equivalent, i.e. induce the same topology.) This gives us a bijective correspondence between centrally symmetric bodies in $\mathbb{R}^n$ and norms on $\mathbb{R}^n$.

To answer your question: basically, any form is possible, as long as there is no obvious reason why it shouldn't be. Specifically, any polytope can be realised as (a translation of) the closed unit ball of some norm if and only if it meets the following criteria:


*

*It is symmetric about its center;

*It is convex;

*It is full-dimensional (not contained in some affine hyperplane);

*It is bounded.


Here I implicitly use the following properties that are intuitively clear but nevertheless require some proof.

Proposition 1. A convex set $X \subseteq \mathbb{R}^n$ has empty interior if and only if it is contained in some affine hyperplane.

A proof of this proposition can be found on Mathematics Stack Exchange and elsewhere on the internet.

Proposition 2. Let $X \subseteq \mathbb{R}^n$ be a set that meets all four of the above criteria. Then $X$ contains an open neighbourhood of its center.
Sketch of proof. For this we use that the interior of a convex set is again convex. Assume without loss of generality that the center of $X$ is the origin (translate if necessary). Since $X$ contains an open neighbourhood of some $x\in X$, by symmetry it also contains an open neighbourhood of $-x$ (we can reflect the entire open neighbourhood in the origin). Thus, $x$ and $-x$ are interior points. Since $\text{Int}(X)$ is convex, it follows that $0\in\text{Int}(X)$ holds as well.$\hspace{2mm}\blacksquare$

Finally, I have used the following assumption:

Assumption. Polytopes are already closed to begin with (by most definitions they are).

A: Update: I had forgotten an essential point; see the comments to my answer.
Let $B\subset{\mathbb R}^n$ be any compact convex set which is symmetric to the origin: $-B=B$, and has nonempty interior. Then
$$\|x\|:=\inf\{\lambda\geq0\>|\>x\in\lambda B\}$$
is a norm on ${\mathbb R}^n$ having $B$ as closed unit ball.
A: \begin{equation}
g(x,y):=\max\left(\frac{|x+y|}{\sqrt{2}},\frac{|x-y|}{\sqrt{2}}\right)+\max(|x|,|y|)=1
\end{equation}
gives an octagon in $\mathbb{R}^2$, but I'm not sure if it's a norm...
