How to find the polar dual of $\{(x,y):y\geqslant x^2\}$ in $\Bbb R^2$?

Question

How to find the polar dual of $\{(x,y):y\geqslant x^2\}$ in $\Bbb R^2$?

If the polar dual of a set $A$ is $A^*=\{x\text{ in }\Bbb R^2:ax\leqslant 1\text{ for all }a\text{ in }A\}$.

I study from convexity by Roger Webster, Duality p.99. I tried: if $b>0$ then $ax+bx^2 \le ax+by\le 1$. I don't know if I can take $ax+bx^2=1$ then $x=(-a\pm \sqrt{a^2+4b})/(2b)$ then we require $a^2+4b$ to be $\ge 0$, i.e., $a^2\ge -4b$, hence $-a^2/(4b)\le 1$.

Answer 1

The polar is usually defined as $A^\circ = \{ x | \langle x, a \rangle \le 1, \forall a \in A \}$.

In this case, we have $A^\circ = \{ (a,b) | ax+by \le 1, \forall y \ge x^2 \}$.

Suppose $(a,b) \in A^\circ$. We see that $b \le 0$, since if $b>0$, we would need to have $ax+bx^2 \le 1$ for all $x$ which is impossible.

If $b = 0$, then we see that we must have $a = 0$, otherwise the constraint could be violated for some $x$.

If $b <0$, then we need to find $a$ satisfying $ax+by \le 1$ for all $y \ge x^2$. Since $b<0$, this simplifies to $ax+bx^2 \le 1$ for all $x$. A quick maximization shows that this is true iff $-{a^2 \over 4b} \le 1$.

Combining the above we see $A^\circ = \{(a,b) | a^2 \le -4b \}$.

Answer 2

I would begin with the concept of support function, which is defined as follows: for a unit vector $\xi$, $h(\xi) = \sup_{a\in A}( a\cdot \xi)$. Geometrically, this is what you get by projecting $A$ onto the line with direction $\xi$, and taking the point that's furthest away in direction of $\xi$.

The polar dual is formed by $0$ and the points $x\ne 0$ such that $|x|\le \frac{1}{h(x/|x|)}$. You should convince yourself of the following:

1. $h(\xi)=\infty$ when $\xi_2\ge 0 $
2. When $\xi_2<0$, we have $h(\xi) = \max_{t\in \mathbb R} (t,t^2)\cdot (\xi_1,\xi_2)$. Expand the dot product and find the maximum of the function with the calculus tools: find the derivative, set it to $0$...