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Definition1: A subset $ A$ of a normed space $E$ is absolutely convex if $ \forall x,y \in A, \quad \forall \alpha ,\beta $ s.t $|\alpha |+ |\beta| \leq 1, \quad $ $ \alpha x +\beta y \in A.$

Definition2: Let $A$ be an absolutely convex subset of a normed space $E$. For $x \in E $ the Minkowski functional of A is defined by $ q_A(x) := \inf \{\lambda \geq 0 : x\in \lambda A \}. $

Theorem: Let $A \subset E$ be absolutely convex.Then, the minkowski functional of A is equivalent to the original norm of $E$.

What I did so far is the following: If $x=0$ , then obviously we have $q_A(x)= ||x||$.

For $0\ne x \in A$, with $||x ||=k$ , since $A$ is absolutely convex, $q_A(x) \leq 1 = || \frac{x}{k} ||$. Then, playing algebraically, we get $ kq_A(x) \leq ||x || $. But , I cannot guarantee the other side, such as $ ||x|| \leq \frac{1}{k} q_A(x) ?$

I have similar arguments for $x \notin A $ but again, I have a problem with one side. How can I proceed from here or was my starting point wrong? (because I could not use the absolute convexity of A)

Esat Koç
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1 Answers1

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This is not true; perhaps something is missing from the assumptions (see the end of this post). Let $q$ be any seminorm on $E$; then the set $A=\{x\in E: q(x)\le 1\}$ is absolutely convex since $$ q(\alpha x+\beta y)\le q(\alpha x)+q(\beta y)=|\alpha | q(x)+|\beta|q(y) \le |\alpha |+|\beta| \le 1 $$ And the Minkowski functional of this $A$ is just $q$ itself: $$\inf \{\lambda \geq 0 : x\in \lambda A \} = \inf \{\lambda \geq 0 : q(x)\le \lambda \} =q(x)$$

But of course, $q$ need not be equivalent to any specific norm. It does not even have to be a norm itself; consider $q(x)=|x_1|$ on $\ell^2$. Or it can be another, inequivalent, norm, like $q(x)=\sup|x_k|$ on $\ell^2$.

Remark

Absolute convexity, as defined above, is equivalent to the requirements that $A$:

  1. is convex
  2. contains $0$
  3. is balanced, meaning $\alpha A=A$ whenever $|\alpha|=1$

Indeed, it's clear that absolute convexity implies 1-2-3. Conversely, if 1-2-3 hold and $|\alpha|+|\beta|\le 1$, $x,y \in A$, then let $$x'= \frac{\alpha}{|\alpha|}x,\quad y'= \frac{\beta}{|\beta|}y$$ Since $A$ is balanced, it contains $x',y'$. Then $$\alpha x +\beta y = |\alpha| x'+|\beta|y'+(1-|\alpha|-|\beta|)0$$ is a convex combination of three points in $A$, and therefore lies in $A$.

With additional assumptions: $A$ is open and bounded

Then the claim is true. Let $r=\inf\{\|x\|:x\in A\}$ and $R=\sup\{\|x\|:x\in A\}$. Note that $r>0$ and $R<\infty$. Using the monotonicity of Minkowski functional with respect to set containment (or just its definition), we get $\|x\|/R\le q_A(x)\le \|x\|/r$.

  • Thank you @Gerry for your nice answer. It is possible that I may miss some assumptions such as the abs. convex set being open (or equivalently absorbing). – Esat Koç Apr 09 '17 at 06:07
  • Still wouldn't be enough; it also needs to be bounded. For example, ${x:|x_1|<1}$ is absolutely convex and open in $\ell^2$. Once you have bounded and open, it's pretty straightforward. –  Apr 09 '17 at 14:31
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    Regarding your last paragraph, surely we get $r=0$? Because as we found above, $A$ contains $0$, so $\inf{||x||:x\in A}$ must be $0$. Then the set of inequalities at the end wouldn't make sense. Is there something wrong here? – user477203 Nov 05 '19 at 14:40
  • @user477203 You're right. Because A is open and contains the origin, you can found an open ball around origin contained in A. – Manuel Bonanno Apr 16 '24 at 17:15