Definition 13.4.1 (Polar Body).label Let $C\suf R^{n}$ be a compact set containing $0$ then the polar of $C$ is
with several properties
- (1)
$C^{o}$ is convex
- (2)
$L\suf K$ then $K^{o}\suf L^{o}$
- (3)
For any $t>0$, $(tC)^{o}=\frac{1}{t}C^{o}$
- (4)
For $R>0$, $\parens{\cl{B}(0,R)}^{o}=\cl{\parens{B\parens{0,\frac{1}{R}}}}$
- (5)
For $A$ positive definite symmetric $n\ti n$ matrix we have $E\define \curl{x\in\R^n:\inner{x}{Ax}\leq 1},E^{o}=\curl{y\in\R^n:\inner{A\inv y}{y}\leq 1}$
- (6)
$C^{oo}=\cl{Conv(C\cup\curl{0})}$. In particular for $C$ is convex $C^{oo}=C$
- (7)
If $P=Conv\curl{v_1,...,v_m}$ is a polytope with $0\in int(P)$ then $P^{o}=\curl{y\in\R^n:\inner{y}{v_i}\leq 1,\fall i=\cl{1,m}}$
- (8)
For a polytope containing the origin $P=\curl{x\in\R^n:\inner{x}{y_i}\leq d_i,i=\cl{1,m}}$ then $P^{o}=Conv\curl{\frac{y_{i}}{d_{i}},i=\cl{i,m}}$
Proof.
- (1)
For $\lam\in (0,1)$ and $y_{1},y_{2}\in C^{o}$ we have
\begin{align*}\inner{(1-\lam)y_1+\lam y_2}{x}=(1-\lam)\inner{y_1}{x}+\lam\inner{y_2}{x}\leq 1-\lam+\lam=1\end{align*} - (2)
$K^{o}$ has more constraints
- (3)
By linearity of $\inners$ in one variable
- (4)
Observe $\inner{x}{y}\leq\norm{x}{}\norm{y}{}\cos(0)=R\norm{y}{}$
- (5)
$1\geq\inner{A\inv A x}{Ax}$ then $y=Ax$
- (6)
We will prove for $K$ closed and convex with $0\in K$ then $K^{oo}=K$. By definition of $K^{o}$, $K\suf K^{oo}$. Suppose for contradiction $\exists z\in K^{oo}\del K$ then $\exists u\in \mathbb{S}^{n-1},\inner{z}{u}>\SUP{x\in K}{x}{u}\geq 0$ where second inequality comes from $0\in K$. If it is positive then $y\define \frac{u}{\SUP{x\in K}\inner{x}{u}}\in K^{o}$ but $\inner{z}{y}>1$ so $z\nin K^{oo}$. If the supremum is $0$ then pick $t>1/\inner{z}{u}>0$. Then
\begin{align*}\inner{tu}{x}=t\inner{u}{x}\leq t\SUP{x\in K}\inner{u}{x}=0\leq 1&&\fall x\in K\end{align*}so $tu\in K^{o}$ and
\begin{align*}\inner{z}{tu}>\inner{z}{\frac{u}{\inner{z}{u}}}=1\end{align*}showing $z\nin K^{oo}$.
- (7)
If the inequalities hold for each $v_{i}$ then for any $x=\sums{}{i}\lam_{i}v_{i}\in P$ we have $\inner{x}{y}=\sums{}{i}\lam_{i}\inner{v_i}{y}\leq\sums{}{i}\lam_{i}=1$
- (8)
Set $Q\define Conv\curl{\frac{y_{i}}{d_{i}},i=\cl{i,m}}$ then $P$ is bounded so $0$ lies in the interior of $Q$. Indeed, otherwise a separating hyperplane through the origin gives a nonzero vector $w$ with $\inner{w}{y_i/d_i}\leq 0$ for every $i$ (after reversing $w$ is needed) so for every $t\geq 0$ we have $\inner{u_i}{tw}\leq 0\leq d_{i}$ so the ray $\curl{tw:t\geq 0}$ lie in $P$, contradicting boundedness. By (6,7) we are done.
$\square$
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