Now also asked on MathOverflow.SE.

When answering this question about finding the open unit ball $\mathscr{B} := \{ x \in \mathbb{R}^2: \| x \| < 1\}$ of the "composite" norm $$ \| \cdot \|: \mathbb{R}^2 \to \mathbb{R}, \ (x,y) \mapsto a \| (x,y) \|_1 + \frac{b}{2} \| (x,y) \|_{\infty}. $$ I thought of the following question. In the above question one has $\Omega := \mathbb{R}^2$, $a := \frac{1}{3}$ and $b := \frac{4}{3}$ but those aren't important for my question. All that matters is $a,b > 0$, as verified in this question.

It turns out that $\mathscr{B}$ is a octagon (as intersection of two rotated squares, as they are the geometric interpretations of $\| \cdot \|_1$ and $\| \cdot \|_{\infty}$ (is that really true?), which can be seen in the diagram appended to my answer to the first mentioned question).

My question is if (and how) one can find out which shape (polygon?) $\mathscr{B}$ corresponds for a composite norm of the form $$ \| \cdot \| := \sum_{k = 1}^{\infty} \alpha_k \| \cdot \|_{x_k}, \qquad \text{where } \alpha_k \ge 0, x_k \in [1, \infty]. $$ As @CalvinKohr points out in the comments, we can normalize this representation: $\sum_{k} \alpha_k = 1$ such that the sum is well defined i.e. converges.

This question seems to be related but I don't know how the Minkowski functional would related to this problem even though it was briefly covered in my functional analysis course. It remarks that a a polygon with a odd number of vertices can not occur because of the symmetry of the norm. As you can see in the last example below, other shapes than octagons are possible. Can $\mathscr{B}$ be another polygon with an even vertices count?

Maybe this is related to the concept of polyhedral norms?

One special case Cosider the norm $\mathfrak{p}_n(x,y) := \sum_{k = 1}^{n} \| (x,y) \|_{k}$. If we graph it and intersect it with a plane $z = \ell$ for $\ell > 0$ we obtain the the shape of $\mathscr{B}$. I graphed $\mathfrak{p}_n$ for $n \in \{1, \ldots, 5\}$ and one observes that shapes of $\mathscr{B}$ are 4-gons that "get more convex" and converge to some circle.

This suggests it might by only interesting to at norms whose $\mathscr{B}$ is a polygon i.e. $\mathscr{B}$s with straight lines. Are those just produced by $\| \cdot \|_1$ and $\| \cdot \|_{\infty}$?.

  • 1
    $\begingroup$ You can normalise $\sum \alpha_k = 1$ so that (its convergent, and) the standard basis vectors are always in the closed unit ball. As noted in one of the linked question's answers for $\ell^1,\ell^\infty$, actually all the $\ell^p$ unit balls have the same 8-fold symmetries as a square around 0 parallel to the axes, so whatever it is, its completely determined by the part in the octant $y\ge x\ge 0$. I don't know how to make this any more explicit, however. $\endgroup$ – Calvin Khor Jun 1 at 22:37
  • $\begingroup$ As a starting point for your answer, what type of explicit answer would you want for the unit circle of $\|\cdot\|_p$? $\endgroup$ – Michael Burr Jun 2 at 10:49
  • $\begingroup$ Yes, $\mathscr B$ is an octagon, but it is not a regular octagon (except for specific choices of $a$ and $b$). For example, both extreme cases $a=1,b=0$ and $a=0,b=1$ yield octagons that degenerate to squares. $\endgroup$ – Rahul Jun 2 at 11:01
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    $\begingroup$ You must have a bug in your $\|\cdot\|_\infty$ implementation. The unit ball for $\|\cdot\|_\infty$ is a square. By the way, regarding your last comment: for any centrally symmetric convex polygon $A$ there exists a norm $\|\cdot\|$ whose unit ball is $A$, namely $\|x\|=\inf\{t>0:x/t\in A\}$. $\endgroup$ – Rahul Jun 3 at 13:03
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    $\begingroup$ It doesn't have the form you state in your question, but a particularly simple example of a norm whose unit circle is an $m$-sided regular polygon for $m$ even is $\| (x,y) \|_{1,m} = \frac{1}{2\cot\left(\frac{\pi}{m}\right)}\sum_{n=1}^m\left|x\sin\left(\frac{2\pi n}{m}\right)+y\cos\left(\frac{2\pi n}{m}\right)\right|$. In particular $\| (x,y) \|_{1,4}=\| (x,y) \|_{1}$. $\endgroup$ – pregunton Jul 12 at 15:22

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