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$S$ is a closed and convex subset of $\mathbb{R}^n$ with $0 \in S$;

$$S^\circ := \left\{ x \in \mathbb{R}^n : \sum_{i=1}^{n}{x_i s_i} \leq 1, \quad \forall s \in S \right\}$$

Prove that $S = (S^\circ)^\circ$.

I am stuck on this problem. My initial thought was to prove that $S \subseteq (S^\circ)^\circ$ and $(S^\circ)^\circ \subseteq S$.

I got a hint that the hyperplane separation theorem might be useful for one part of this, but as I am not really experienced with this kind of proof. I would be very thankful for some hints.

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sednes
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1 Answers1

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(1) If $y \in S$, then $\langle x, y \rangle \leq 1$ will be held for any $x \in S^{\circ}$. Obviously, we have $y \in S^{\circ\circ}$. Thus $S \subseteq S^{\circ\circ}$.

(2) If $y \in S^{\circ\circ}$, then $\langle x, y \rangle \leq 1$ still will be held for any $x \in S^{\circ}$. we can prove by contradiction. Suppose $y \notin S$, since $S$ is closed and convex set, we can find a hyperplane $\{x \mid \langle a, x \rangle = b\}$ that separate the point $y$ and $S$ strongly. In other words, we can find $a \in \mathbb{R}^n, b \in R$ such that \begin{equation} \begin{aligned} & \langle a, s \rangle \leq b, \quad \forall s \in S. \\ & \langle a, y \rangle > b \end{aligned} \end{equation} Thus, we can find a vector $\bar{a} = \frac{a}{b}$ such that $\bar{a} \in S^{\circ}$, but $\langle \bar{a},y \rangle > 1$. So contradiction. Thus $S^{\circ\circ} \subseteq S$.

By the way, the set $S^{\circ}$ is called the polar of $S$. And one can refer an alternative proof based on gauge function and support function in Rockafaller's book ``Convex Analysis'' (Chapter 14).

Zenan Li
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