# Polar Cones basic property

1. Nov 1, 2012

### avilaca

Let S1*(S2*) be the polar cone of the set S1(S2) (http://en.wikipedia.org/wiki/Dual_cone_and_polar_cone).

How can I show that if S1 is contained in S2 then S2* is contained in S1*.

It looks obvious (especially if we think in R^2), but I do not find a way to prove it.

2. Nov 1, 2012

### tiny-tim

welcome to pf!

hi avilaca! welcome to pf!
isn't the proof obvious from the definition based on inner product? (see http://en.wikipedia.org/wiki/Dual_cone_and_polar_cone)

3. Nov 3, 2012

### avilaca

Ok, it's easy from the inner product <a,s> = ||a||.||s||cos$\theta$.
<a,s> $\leq$ 0 <=> pi/2 $\leq$ $\theta$ $\leq$ 3pi/2.
This means that if S1 $\subset$ S2, by the above result, the region where the condition {<a,s> $\leq$ 0 , s $\in$ S1 or S2, a $\in$ ℝ$^{n}$} is true for S1 is the same or it's larger than the one for S2, wich implies S2* $\subset$ S1*.

4. Nov 3, 2012

### avilaca

Now another challenge:

Let S = {x $\in$ ℝ$^{n}$: x = Ap, p $\geq$ 0}, where A $\in$ M$_{n*m}$, p $\in$ ℝ$^{m}$.
What is its polar cone S*?

5. Nov 3, 2012

### tiny-tim

looks good!
show us what you get

6. Nov 4, 2012

### avilaca

I didn't achieve a great conclusion.

S* can be defined by {a $\in$ R$^{n}$: x$^{T}$a $\leq$ 0, for all x $\in$ S}.

Now:
x$^{T}$a $\leq$ 0 <=> (Ap)$^{T}$a $\leq$ 0 <=> p$^{T}$A$^{T}$a $\leq$ 0.

But I can not conclude nothing about "a" from here.