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Suppose I have proved this version of the separation theorem :

Let $K \subseteq R^n$ be a convex, closed set. If $x^* \notin K$, $x^* \in \Bbb R^n$, then $\exists a \in \Bbb R^n$, $\beta \in \Bbb R$ such that $a^Ty \leq \beta \ \forall y \in K$ and $a^Tx^* \gt \beta$.

From this theorem, is it possible to deduce the stronger version separating two closed, convex sets $K_1$, $K_2$, one of which is compact ?

Bernard
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Desura
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1 Answers1

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Yes you can.

Hint:

Define $K := K_2 - K_1$ then $x^* := 0 \notin K$ because $K_1 \cap K_2 = \emptyset$.

$K$ is closed since it is written as an addition of a compact set and closed set in $\Bbb{R^n},$ and also it is clearly convex.

Now apply your separation theorem to these newly defined $K$ and $x^*$!

Red shoes
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  • Yeah, I get it. Say $K_2$ is the compact one. We get $a \in \Bbb R^n$ and $\beta \lt 0$ such that $a^Tk_2 - a^Tk_1 \leq \beta$ for any $k_2 \in K_2$ and for any $k_1 \in K_1$. Take $y^* \in K_2$ such that $a^Tk_2$ is maximal by compacity. Then $a^Tk_2 \leq a^Ty^* \lt a^Ty^* - \beta \leq a^Tk_1$. Thank you ! – Desura Jul 04 '17 at 08:23