Note: This is very similar to this question of mine on Math.Se, which hasn't received any answers yet.

It is known, that for any centrally symmetric convex polygon $A$ there exists a norm $\| \cdot \|$, whose unit ball is $A$, namely $\| x \| = \inf \{ t>0: \frac{x}{t} \in A\}$. But how about the other way around, given a "composite" norm $$ \| \cdot \| := \sum_{k \in \mathbb{N}} \alpha_k \| \cdot \|_k, $$ on a vector space $V$, where $\sum_{k \in \mathbb{N}} \alpha_k = 1$, such that the norm converges and the $\{i_k\}_{k \in \mathbb{N}}$ just being some index set for the norms used, what can we say about the unit ball centred at zero with respect to $\| \cdot \|$, $\mathscr{B} := \{ \omega \in V: \| \| \omega \| < 1 \}$.

It is clear that that a a polygon with a odd number of vertices can not occur because of the symmetry of the norm.

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    $\begingroup$ Are we talking about $\mathbb{R}^n$? Can the $\|\cdot\|_k$ be any norms? Is the notation $i_k$ used anywhere? Do you allow $\alpha_0 =1$ and all other $\alpha_k = 0$? $\endgroup$ – Nik Weaver Jul 12 at 15:12
  • $\begingroup$ We can assume $V = \mathbb{R}^n$. For the definition of the "composite norm" it shall be just a sum of arbitrary norms with coefficients chosen such that the series converges, no other restrictions apply. $\endgroup$ – Viktor Glombik Jul 13 at 12:18
  • $\begingroup$ Take $\alpha_0 =1$ and all other $\alpha_k = 0$. Then $\|\cdot\| = \|\cdot\|_0$. If $\|\cdot\|_0$ is arbitrary then so is $\|\cdot\|$. $\endgroup$ – Nik Weaver Jul 13 at 12:43

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