0
$\begingroup$

Possible Duplicate:
Dual of a dual cone

I try to prove the following statement:

Let $V$ be a finite-dimensional ordered topological vector space ($V^{**} \cong V$) with a closed positive cone $C$. Then, then dual cone of the dual cone of C is equal to the original cone, $C^{**} = C$.

I tried to prove it as follows: Obviously, $C \subset C^{**}$. Therefore, it suffices to show that $w \notin C$ implies $w \notin C^{**}$. So let $w \in V, w \notin C$. Then, by the Hahn-Banach-Theorem, there is a linear functional $f \in V^*$ with $f(w) < \inf \{f(v) \mid v \in C\}$. What I fail to show is that $f$ can be chosen to be non-negative on $C$ (therefore lying in $C^*$ ) but negative at $w$ (therefore $w$ not lying in $C^{**}$).

I've seen that someone asked (almost) the same question, but the only answer to his post did not answer this particular question: Dual of a dual cone

Can anyone help?

  • 1
    The question you linked to says that $C^{**}$ is the closure of $C$. You assume that $C$ is closed. Am I missing something?2011-09-27
  • 0
    The question there has not been fully answered (there is still an open question concerning the proof there, and if I understand things correctly, it's the same question that I need to be answered to complete the proof).2011-09-27
  • 0
    I see. Then the following meta question arises: should we close questions as exact duplicates if the duplicated question has not been answered properly? I assume that the answer is yes, because otherwise, a hard question could occur multiple times, remaining unanswered.2011-09-27
  • 1
    I wrote an answer to the duplicate question Rasmus linked to. The point is that $0$ is in $C$, so your $f$ has the property that its infimum $M = \inf \{f(v) \mid v \in C\}$ is $\leq 0$. We can't have $M \lt 0$ since otherwise there were $v \in C$ such that $f(v) \lt 0$. Multiplying $v$ by a large enough scalar $\alpha \gt 0$ you'd have $f(\alpha v) \lt f(w)$ while $\alpha v \in C$, contradicting the choice of $f$, hence $f(v) \geq 0$ for all $v \in C$.2011-09-30
  • 0
    Good, that's the kind of argument I was looking for. Thank you!2011-09-30

0 Answers 0