0
$\begingroup$

Where: $$ K^{\circ}=\{z\in\mathbb{R}^3 \lvert \forall x\in K : z^tx\leq 0\}$$ And: $$\begin{equation*} K:=\left \{ \begin{pmatrix} x_1\\x_2\\x_3 \end{pmatrix} \in \mathbb{R}^3 \Bigg{|} \ x_3^2 \geq x_1^2+x_2^2, x_3 \geq 0 \right \} \end{equation*}$$ Where: $$q= \begin{pmatrix} p_1\\p_2\\p_3 \end{pmatrix} \in \mathbb{R}^3 \setminus -K \text{ and } p_1^2+p_2^2 \neq 0 \text{ and } p_3 < 0$$

And use this to show that $-K=K^{\circ}$.

How should I start the proof? Any pointers?

$\endgroup$
0
$\begingroup$

Edited one. Hint: If $q \in K^{\circ}$ but $q \notin - K$ then you can strongly separate $q$ and $ -K $ via hyperpaln, and then see what will happen...

$\endgroup$
  • $\begingroup$ The last sentence is wrong. There are closed convex cones $K \subset \mathbb R^n$ with $-K \ne K^\circ$. $\endgroup$ – gerw Jun 28 '17 at 6:30
  • $\begingroup$ Right , got confused with double dual... Thanks @gerw $\endgroup$ – Red shoes Jun 28 '17 at 6:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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