# Given a composite norm what polygon describes its unit ball?

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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• 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 $\alpha_k = 0$? – Nik Weaver Jul 12 at 15:12
• 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. – Viktor Glombik Jul 13 at 12:18
• Take $\alpha_0 =1$ and all other $\alpha_k = 0$. Then $\|\cdot\| = \|\cdot\|_0$. If $\|\cdot\|_0$ is arbitrary then so is $\|\cdot\|$. – Nik Weaver Jul 13 at 12:43