Let $C \subset \mathbb{R}^n$ be a convex body (i.e., full-dimensional, compact) that includes the origin in its interior. Its $polar$ convex body is defined as

$C^\circ = \{ y \in \mathbb{R}^n : \langle y , x \rangle \leq 1, \forall x \in C\}$.

(a) Show that $C^\circ$ is a convex body.

(b) Let $C$ be the triangle with vertices $(-1,1)$, $(-1,-1)$ and $(a,0$, where $a>0$. Draw $C$ and $C^{\circ}$ as a function of the parameter $a$.

(c) Let $C$ be an axis-aligned ellipse with semiaxes $a$ and $b$. What is $C^\circ$?

(d) Let $C=\left\{x \in \mathbb{R}^n \, : \, ||x||_p:=\left(\sum_{i=1}^n |x_i|^p\right)^{\frac{1}{p}} \leq 1 \right\}$. Find a nice description of $C^{\circ}$. Hint: Use Hölder's inequality.

### Solution:

(a) It is clear from the defining inequality that $C^\circ$ is both closed and convex. To see that it's also bounded, take $\epsilon >0$ such that $x \in C$ whenever $\|x\| \leq \epsilon$ (there exists such an $\epsilon$ because $0 \in int(C)$). If $0\neq y \in C^\circ$, then $\frac{\epsilon}{\|y\|}y \in C$ and we get $\|y\| \leq \frac{1}{\epsilon}$. We have shown that $C^\circ$ is a compact, convex set.

To see that it is full-dimensional, let $M>0$ be a bound on $C$. Then whenever $\|y\| \leq \frac{1}{M}$ we have $y \in C^\circ$ since $\langle y, x \rangle \leq \|y\| \|x\| \leq \frac{1}{M}M = 1$ for all $x \in C$.-MH

Remark on this argument: The polar of a disk of radius $r$ is a disk of radius $\frac{1}{r}$. This is true for every $n\in\mathbb{N}$.

(c) First, let's prove a general statement: If $T: \mathbb{R}^n \to \mathbb{R}^n$ is an invertible and symmetric linear transformation, then $(T(K))^\circ = T^{-1}(K^\circ)$. This is true because $y \in T(K)^\circ \iff \langle y, Tx\rangle \leq 1, \forall x \in K \iff \langle Ty, x\rangle \leq 1, \forall x \in K \iff Ty \in K^\circ \iff y \in T^{-1}(K^\circ)$.

Now notice that the given ellipse is $T(S^1)$ with $T$ defined by $T(x,y) = (ax,by)$. Since the unit ball is the polar of the unit circle, the polar of this ellipse is the convex hull of another axis-aligned ellipse of semiaxes $\frac{1}{a}, \frac{1}{b}$ -MH

(b) The vertices of the triangle give the inequalities $ax \leq 1$, $-x-y \leq 1$, $-x+y \leq 1$, so the polar is a triangle with vertices $(-1,0), (1/a,1+1/a), (1/a, -1-1/a)$.

(d) Let $q = p/(p-1)$ and let $D = \{x \in \mathbb{R}^n : ||x||_q \leq 1 \}$. For all $y \in D$, and $x \in C$ we have $\langle y, x \rangle \leq ||y||_q ||x||_p \leq 1$ by Hölder's inequality so $D \subset C^\circ$. For $y \notin D$, let $y' = y/||y||_q \in D$. Choose $x = (y'_1^{q/p},\ldots, y'_n^{q/p})$ which is in $C$. Then $\langle y', x \rangle = 1$ and so $\langle y, x \rangle > 1$ showing that $y \notin C^\circ$. Therefore $C^\circ = D$.