Prove that the polar set of a unit disc is itself

Question

There is a similar problem: Find the Polar of a set. However, I still have no idea the best way to do this.

We know the definition of polar of $C$: $$ C^{\circ} = \lbrace y \in \mathbb{R}^{n} |\; \langle x,y \rangle \leq 1 \quad \forall x \in C \rbrace $$

Let $$C = \big\{(x,y)\in \mathbf{R}^2 : \begin{bmatrix} 1+x & y \\ y & 1-x \end{bmatrix} \succeq 0 \big\}$$

How to show $C^{\circ} = C$ in a closed form?

Is $C$ equal to the unit disc? Then the result is immediate from Cauchy-Schwarz. — gerw, Oct 25 '16 at 06:37
@gerw yes, it will be $x^2+y^2\leq 1$, just use the properties of being PSD. How to show it from C-S inequality? — sleeve chen, Oct 25 '16 at 07:11

score 4 · Accepted Answer · answered Oct 25 '16 at 09:32

Let $C = B_1(0)$ be the unit disc in $\mathbb{R}^n$. Let us show "$C = C^\circ$".

$C \subset C^\circ$. You have to show $\langle x,y\rangle \le 1$ for all $x,y \in C$. This is Cauchy Schwarz.
$C \supset C^\circ$. Let $y \in C^\circ$ be given. Then, $x := y/\|y\| \in C$. Using $x \in C$ and $y \in C^\circ$ yields $\|y\| = \langle x,y\rangle \le 1$. Hence, $y \in C$.

1 Answers1