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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.

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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

1 Answer 1

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 $n>\frac\kappa{\sqrt 2}$.

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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

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