1

Any bounded open symmetric convex $A$ over $\mathbb{R}^n$ induces a norm (see this problem). This gives some geometrical intuition about how many different norms there are in $\mathbb{R}^n$, for example my open ball in the specific norm could be a pentagon in $\mathbb{R}^2$ or an isocaeder in $\mathbb{R}^3$. From my lectures in university I'm only used to the "classical" (or better canonical) norms of $||x||_p$ with $p\in[1,\infty]$. This makes me wonder if these are somewhat special. What is their measure, are they sparse, are they dense? I'll formulate the last one more precisely for you to answer.

Let $A$ be an open subset of $\mathbb{R}^n$ set containing $0$. Let $N$ be the set of all norms of $\mathbb{R}^n$ which are symmetric in each orthant and $P=\{||.||_p:\mathbb{R}^n\rightarrow\mathbb{R} \;|\; p\in[1,\infty]\}$. Furthermore $N|_A=\{f|_A : f\in N\}$ and $P|_A=\{f|_A : f\in P\}$. Let $(\ell^\infty(A), |||.|||_\infty)$ be the normed vector space of the bounded functions on $A$ to $\mathbb{R}$ with the supremum's norm. Show that $P|_A$ is dense in $N|_A$, both taken as subsets of $\ell^\infty(A)$, or find a counter example.

I merely touched functional analysis of $C^r([a,b])$ and $L^r([a,b])$, so this is a bit much for me, but it seems interesting. Feel free to show other properties of $P|_A$ as well or to reference related papers.

José Carlos Santos
  • 440,053
Ayutac
  • 223
  • I think you want to restrict yourself at least to absolute norms. Because as suggested by the answer, the $p$-norms are absolute (or monotonic if you prefer) which is a quite particular property for a norm and so you will have trouble to approximate norms which are not "symmetric on each orthant". – Surb Jan 03 '17 at 21:46
  • @Surb Not sure what "absolute" means, but I see how "symmetric on each orthant" is important. I restricted $N$ to such ones. – Ayutac Jan 03 '17 at 21:53
  • Have a look at that http://www.sciencedirect.com/science/article/pii/002437959190085B Absolute means $| \ |x| \ |=|x|$ for any $x$ (a mathematical way to say: "symmetric on each orthant") – Surb Jan 03 '17 at 22:01
  • I've been thinking a bit further to this problem. Even with the assumption that the norms has to be absolute it is still far to be true. Indeed, if we consider any weighted $p$-norms, then they can not be approximated with non-weighted ones. So you'd probably want to restrict yourself to norms so that $|e_i|=1$ for every $i$, where $e_1,\ldots,e_n$ is the canonical basis. It has a name but I don't remember it. Anyway, it is still not true. There is a large class of non-differentiable norms which can't be approximated. Consider an hexagonal, resp. octagonal, unit ball in $\Bbb R^2$. – Surb Jan 04 '17 at 08:26
  • So probably, you'd need also to add the assumption that the norms are Frechet differentiable. But still in this setting, I don't believe that the $p$-norms are dense. Indeed the $p$-norms in $\Bbb R^2$ are symmetric wrt the axis $(t,t),t \in \Bbb R$ – Surb Jan 04 '17 at 09:22
  • Finally, you could put a matrix norm on vectors. And these are even more crazy (e.g. a matrix $p,q$-norm). – Surb Jan 04 '17 at 09:23

1 Answers1

2

Hint. Consider the set obtained from $[-1,1]\times[-1,1]$ by cutting the edges by lines $y=1-x$, $y=-1-x$. Can it be approximated by $l_2^p$ balls?

Przemysław Scherwentke
  • 13,836
  • Actually, your example is nicely pathologic and could be even worse by doing the same construction but on $[-2,2]\times [-1,1]$ instead. – Surb Jan 04 '17 at 08:27