Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have been trying to prove this for so long that I pose this to Math.SE: Let D be a bounded (in regards to, for example, the maximum norm), absorbing subset of a finite-dimensional vector space V over a ordered field F and $0\in D$. That means $$\forall x\in V\exists r\in F\;\forall \alpha\in F:|\alpha|\ge r\Rightarrow x\in \alpha D$$ with $\alpha D:=\{\alpha d\mid d\in D\}$. (In other words, this means that D can be "inflated" to contain the whole vector space)

I am trying to prove that D can be used to define a quasinorm by $\|x\|:=\inf\{\alpha\mid x\in\alpha D\}$

However, I am stuck in the proof, there is something obvious i am overlooking. The positive homogeniety follows directly out of the definition, however I dont know how to prove the (weak) version of the triangle equation: $$\exists \kappa\forall a,b\in V:\;\|a+b\|\le\kappa(|a\|+\|b\|)$$ It can be easily be broken down to $$\exists \kappa\forall a,b\in D:\;\|a+b\|\le\kappa(|a\|+\|b\|)$$ and even to $$\exists \kappa\forall a,b\in \partial D:\;\|a+b\|\le\kappa(|a\|+\|b\|)$$(as D is star-shaped).

So basically, I need to show that $\sup_{c=a+b}\{\|c\|\}<\infty$. In $\mathbb{R}^2$ this is easy, but Id like to prove it in general vector spaces.

One Idea was to show that $\partial D$ is always at least some $\varepsilon>0$ away from the origin (as D is absorbing); However, I was unable to make this argument rigorous. The problem is that the border curve could converge toward the origin. I need to show that this is the only thing that would violate the triangle equation and that it violates the absorbency of D.

EDIT: This is unprovable. Today, I found out about the Minkowski Functional for $D\subseteq X$ where X is a topological Vectorspace. $$p(x)=\inf\{\lambda\in F\mid x\in \lambda D\}$$ which is well-defined if D is absorbing. If D is bounded in X (which is well-defined because X is a topological Vectorspace), and if there exists a open neighbourhood $U\subset D$ of 0, it defines a quasinorm. This can be proven rather easily. Thank you anyways.

share|cite|improve this question
In what sense is $D$ bounded? Is $V$ supposed to be a normed vector space? – Hagen von Eitzen Jan 2 '13 at 23:01
@HagenvonEitzen Indeed, that needs to be said. For example, the maximum norm of all elements of D is bounded – CBenni Jan 2 '13 at 23:14
In general vector spaces $V$, there is no maximum norm. – Hagen von Eitzen Jan 2 '13 at 23:25
@HagenvonEitzen if we consider a finite-dimensional vector space over the ordered Field F, i guess there is however? I will have to modify my assumptions... – CBenni Jan 2 '13 at 23:31
up vote 2 down vote accepted

You cannot succeed. Let $V=\ell^2$ be the space of square summable sequences and $$D=\left\{x\colon \lVert x\rVert_2<1\land\forall n\colon \left(|x_n|<\frac1n\lor \exists m\ne n\colon x_m\ne0\right)\right\}.$$ Then $D$ is absorbing: If $x$ has at most one nonzero component, at index $n$, say, then $x\in \alpha D$ if $\alpha \ge n|x_n|$. And in all other cases $x\in \alpha D$ if $\alpha>\lVert x\rVert_2$. But no $\kappa$ as you desire exists: Consider some $n\in\mathbb N$. Let $a= e_n+ e_1$, $b= e_n- e_1$ suitable combination of the first and the $n$th standard basis vector. Then $\lVert a\rVert = \lVert b\rVert =\sqrt 2$, but $\lVert a+b\rVert = \lVert 2 e_n\rVert =2n$. Thus we need $\kappa(\sqrt 2+\sqrt 2)\ge 2n$. But this gives a contradiction if we choose $n>\frac\kappa{\sqrt 2}$.

share|cite|improve this answer
Mmh... Is it possible to do for finite dimensional vector spaces? I was thinking about the polar plot of $$r(\theta):=\begin{cases}\theta & \frac{1}{\theta}\in\mathbb{N}\\1&otherwise\end{cases}$$ I fear this is absorbing, but not norm-inducing aswell. I will probably have to constraint myself to uniformly absorbin sets, where I assume that the supremum always exists. – CBenni Jan 2 '13 at 23:28
D'oh, if you hadn't claimed that "in $\mathbb R^2$ this is easy", I might have seen this counterexample. – Hagen von Eitzen Jan 3 '13 at 13:15
I did a stupid assumption. I am so sorry ;) – CBenni Jan 3 '13 at 14:51

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.