Proving the Containment Property of Polar Cones for Sets in R^n

[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.

 

[tiny-tim]

welcome to pf!

hi avilaca! welcome to pf!

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.

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

 

[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, which implies S2* \subset S1*.

 

[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*?

 

[tiny-tim]

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, which implies S2* \subset S1*.

looks good!

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*?

show us what you get

 

[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.