-2
$\begingroup$

Suppose we have the following relationship, note that $A,B,C$ are closed convex matrix cones,

$A^\ast=C,$

$B^\ast=C,$

can we state that $A=B$? Is the dual cone of a cone is unique?

the definition of dual cone here is:

The dual cone C* of a subset C in a linear space X, e.g. Euclidean space $R^n$, with topological dual space X* is the set

$C^* = \left \{y\in X^*: \langle y , x \rangle \geq 0 \quad \forall x\in C \right \}$,

Improve this question
$\endgroup$
4
  • 1
    $\begingroup$ You should write what you mean by $A^{\ast}$. But under all definitions that I know, it is unique. $\endgroup$
    – Cristóbal Guzmán
    Commented Apr 28, 2014 at 23:02
  • 2
    $\begingroup$ Not only what $A^*$ is, but where does it live. Presumably there are horrible Banach space examples where nothing works... $\endgroup$
    – Igor Rivin
    Commented Apr 28, 2014 at 23:22
  • $\begingroup$ Also closedness might be an issue. At least one possible definition of "dual cone" that I know cannot distinguish between $A$ and $\overline{A}$. $\endgroup$
    – Johannes Hahn
    Commented Apr 28, 2014 at 23:26
  • $\begingroup$ @user3029108: your question is really badly posed. First of all, what is a "matrix cone"? You also haven't addressed Igor Rivin's comment: are you only interested solely in finite-dimensional Euclidean spaces, or are you allowing infinite-dimensionality? I'm downvoting for now. $\endgroup$
    – Alicia Garcia-Raboso
    Commented Apr 28, 2014 at 23:57

1 Answer 1

Reset to default
5
$\begingroup$

The answer to the edited question is yes. For any non-empty, closed, convex cone $C\subseteq V$ for any locally convex and hausdorff topological vector space $V$ the equation $C^{\ast\ast}=C$ holds. One inclusion follows immediately from the definition, the other follows easily by aiming for a contradiction and applying the Hahn-Banach theorem about separation of closed convex sets from points.

Improve this answer
$\endgroup$

Not the answer you're looking for? Browse other questions tagged .