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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 relate to this problem even though it was briefly covered in my Functional Analysis course. It remarks that 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 number of vertices?

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 $\| \cdot \|_{\infty}$?.

ViktorStein
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    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. – Calvin Khor Jun 01 '19 at 22:37
  • As a starting point for your answer, what type of explicit answer would you want for the unit circle of $|\cdot|_p$? – Michael Burr Jun 02 '19 at 10:49
  • 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. –  Jun 02 '19 at 11:01
  • @Rahul thanks for the info, I edited. – ViktorStein Jun 03 '19 at 06:34
  • @MichaelBurr I just looking for norms "producing" polygon unit balls, and in my last remark the question I suspect that can only be the case when we "compose" infinity or 1-norms. – ViktorStein Jun 03 '19 at 06:41
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    You must have a bug in your $|\cdot|\infty$ implementation. [The unit ball for $|\cdot|\infty$ is a square.](https://en.wikipedia.org/wiki/Norm_(mathematics)#Maximum_norm_(special_case_of:_infinity_norm,_uniform_norm,_or_supremum_norm)) 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 $|x|=\inf{t>0:x/t\in A}$. –  Jun 03 '19 at 13:03
  • @Rahul, you're right, I forgot some absolute values :/ I edited accordingly. And for the second part of your comment. This is what I have been looking for and this is also the connection to the Minkowski functional, right? Is there an more explicit representation of this norm of, let's say regular polygons with an even number of vertices? – ViktorStein Jun 03 '19 at 14:06
  • I can't because (1) I don't know the answer to the questions in your last comment, and (2) your actual question as stated above is "My question is if (and how) one can find out which shape (polygon?) $\mathscr B$ corresponds for a composite norm..." which I also don't know the answer to. I was just commenting to correct some errors in your question, but I don't intend to answer it because I don't know how. –  Jun 03 '19 at 18:19
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    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 $| (x,y) |{1,4}=| (x,y) |{1}$. – pregunton Jul 12 '19 at 15:22
  • @pregunton Why? (and in what angle?) – Emolga Feb 17 '20 at 03:19
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    @Emolga As the norm is invariant under rotations of $\frac{2\pi}{m}$ radians, we can restrict ourselves to the set of points whose argument is strictly between $0$ and $\frac{2\pi}{m}$. Within that sector none of the summands become zero, so we can replace $|\cdots|$ by either $(\cdots)$ or $-(\cdots)$. It is then obvious that the set of points with norm $1$ is a line segment, and the result follows by symmetry. The constant $2\cot\left(\frac{\pi}{m}\right)$ is just a normalization so that the resulting polygon has a vertex at $(1,0)$, see here. – pregunton Feb 17 '20 at 07:20

2 Answers2

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At OP's request, I expand my comments into an answer.

Even though it doesn't have the form stated in the question, a particularly simple example of a norm whose unit circle is an $m$-sided regular polygon for even $m>2$ 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 we have that $\| (x,y) \|_{1,4}=|x|+|y|=\| (x,y) \|_{1}$ is the usual taxicab norm. As was pointed out in the question, a similar norm for odd-sided regular polygons cannot exist since they don't have the required central symmetry.

Proof: As the norm is invariant under rotations of $\frac{2\pi}{m}$ radians, we can restrict ourselves to the set of points whose argument is strictly between $0$ and $\frac{2\pi}{m}$. Within that sector none of the summands become zero, so we can replace $|\cdots|$ by either $(\cdots)$ or $−(\cdots)$, according to whether the argument inside the bars is positive or negative.

Since then the expression turns into an ordinary linear function of $x$ and $y$, it is obvious that the set of points with norm $1$ will be a straight line segment. The result follows by symmetry and the fact that the unit circle must be connected.

Finally, to make the resulting regular polygon have a vertex at $(1,0)$, we divide the sum by the normalization constant $\sum_{n=1}^m\left|1\cdot \sin\left(\frac{2\pi n}{m}\right)+0\cdot \cos\left(\frac{2\pi n}{m}\right)\right| = 2\cot\left(\frac{\pi}{m}\right)$.

This link shows the unit circle corresponding to these norms for even $m$ between $4$ and $12$, together with a circumscribed (ordinary) circle for comparison purposes. The slider can be used to change the value of $m$.

pregunton
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In Matrix Analysis by Horn and Johnson one can find on page 337 the following characterisation:

Theorem 5.5.8 A set $B$ in a finite-dimensional real or complex vector space $V$ with positive dimension is the unit ball of a norm if and only if $B$

  1. is compact,
  2. is convex,
  3. is equilibrated, that is, $x \in B \implies \alpha x \in B$ if $| \alpha | = 1$ (also see balanced set),
  4. has 0 as an interior point.
ViktorStein
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