Defining the Polar set

Question

For a subset P of R^n (real numbers) the polar set is defined by: $$ P^*:= \{ y\in \Bbb R^n\mid y\cdot x \leq 1 \text{ for all } x\in P \}. $$

Can someone break the definition into plain english as im struggling to understand the notations used and such a sets structure.

The best guess ive got is: $P*$ is defined as the set of all (x,y) which follows that, for all real y ($\in \mathbb {R^n})$ there exists an x in P that satisfies $y·x \le 1$

Even if have this correct how is this useful in general geometry? The original question is to show that if $P$ is a convex set that contains the origin then (P*)*=P

Answer 1

One way to read the definition literally is to say the following:

$P^*$ is the set of all vectors $y$ for which $y \cdot x \leq 1$ holds for all $x \in P$.

For more information on how to make sense of definitions like these, see the wiki page for "set-builder notation". The idea behind a polar set is that it provides a "test" by which we can check whether or not an element fails to be in $P$.

To see how this works, let's consider an example. Take the set $$ P = \{(x_1,x_2) \in \Bbb R^2 \mid 0 \leq x_2 \leq x_1\}. $$ That is, $P$ is the region in the $xy$-plane satisfying $0 \leq y \leq x$. As it turns out, the polar set will be equal to $$ P^* = \{(x_1,x_2) \in \Bbb R^2 \mid x_1 \leq 0 \text{ and } x_2 \leq -x_1\}. $$ Now, consider the point $v = (1,2)$, which we can see fails to be an element of $P$. The polar set gives us another way to show that $v$ fails to be an element of $P$: if there is an element $w \in P^*$ for which $w \cdot x > 1$, then we can be sure (by the definition of a polar cone) that $v$ is not an element of $P$.

In this case, it is useful to consider the element $w = (-2,2) \in P^*$. We find that $$ v \cdot w = 1 \cdot (-2) + 2 \cdot (2) = 2 > 1. $$ If $v$ were an element of $P$, then by the definition of a polar cone, any $w \in P^*$ would have to satisfy $v \cdot w \leq 1$. Because that fails to happen here, we can be sure that $v$ is not an element of $P$.

Answer 2

Say $P$ is a subset of $\mathbb{R}^n$ ( perhaps convex) that contains $0$ (in its interior). Consider a hyperplane is given by the equation in $x$ $$\sum_{i=1}^n a_i x_i = b$$

Assume now $b\ne 0$ (that is, the hyperplane does not containg $0$). Dividing by $b$ we get an equivalent equation $$\sum a'_i x_i = 1$$ where $a'_i = \frac{a_i}{b}$. So we can write the equation of a hyperplane that does not pass through the origin as $$\mathbf{a} \cdot \mathbf{x} = 1$$ There is also an advantage in that now the equation is unique.

Now, we want all these hyperplanes (not containing $0$) that leave $P$ on ones side. Now, a hyperplane of equation $\mathbf{a} \cdot \mathbf{x} = 1$ divides the space into two halfspaces $\mathbf{a} \cdot \mathbf{x} \le (\ge) 1$. Since $P$ contains $0$, and $ \mathbf{a} \cdot \mathbf{0}=0 < 1$, for $P$ to be on one side of the hyperplane is equivalent to $$\mathbf{a} \cdot \mathbf{x} \le 1$$ for all $x \in P$.

Now, assume that $P$ is closed, convex, containing $0$ and $z$ a point of $\mathbb{R}^n$ not contained in $P$. Intuitively it should be clear ( and can be proved) that there exists a hyperplane separating strictly $P$ and $z$. In other words, there exist $y\in \mathbb{R}^n$ such that $$y \cdot x < 1$$ for all $x \in P$ and $$y \cdot z>1$$

A moment's thought shows that $y\in P^{\star}$, and $z \not \in (P^{\star})^{\star}$. The conclusion is that $(P^{\star})^{\star}$ must be included in $P$. However, it is easy to see that $P$ is contained in $(P^{\star})^{\star}$. By double inclusion we get the equality $P=(P^{\star})^{\star}$.