PlanetMath: hyperplane separation

Theorem 1 (Hyperplane Separation Theorem I) Given a weakly closed convex subset $S \subset X$ , and $a \in X \setminus S$ . there is $\phi \in \Phi$ such that $$ \phi(a) < \inf_{x \in S} \phi(x)\,. $$

Proof. The weak topology on $X$ can be generated by the semi-norms $x \mapsto \abs{ p(x) }$ for $p \in \Phi$ . A subbasis for the weak topology consists of neigborhoods of the form $\{ x \in X \colon \abs{p(x-y)} < \epsilon \}$ for $y \in X$ , $p \in \Phi$ and $\epsilon > 0$ . Since $X \setminus S$ is weakly open, there exist $f_1, \dotsc, f_n \in \Phi$ and $\epsilon > 0$ such that

$\displaystyle \lvert f_i(x) - f_i(a)\rvert = \lvert f_i(x-a)\rvert < \epsilon \,,$ for all $i=1, &cdots#dotsc;, n$ implies $\displaystyle x \in X \setminus S\,.$

In other words, if $x \in S$ then at least one of $\abs{f_i(x) - f_i(a)}$ is $\geq \epsilon$ .

Define a map $F\colon X \to \real^n$ by $F(x) = ( f_1(x), \dotsc, f_n(x) )$ . The set $\overline{F(S)}$ is evidently closed and convex in $\real^n$ , a Hilbert space under the standard inner product. So there is a point $b \in \overline{F(S)}$ that minimizes the norm $\norm{b - F(a)}$ .

It follows that $\ip{y-b}{b-F(a)} \geq 0$ for all $y \in \overline{F(S)}$ ; for otherwise we can attain a smaller value of the norm by moving from the point $b$ along a line towards $y$ . (Formally, we have $0 \leq \left.\frac{d}{dt}\right|_{t=0} \norm{ty + (1-t)b - F(a)}^2 = 2\ip{y-b}{b-F(a)}$ .)

Take $\phi = \sum_{i=1}^n \lambda_i f_i$ where $\lambda = b-F(a)$ . Then we find, for all $x \in S$ ,

$\displaystyle \phi(x-a)$	$\displaystyle = \langle {b - F(a)}, {F(x-a)} \rangle$
	$\displaystyle = \langle {b-F(a)}, {b-F(a)} \rangle + \langle {b-F(a)}, {y-b} \rangle \,, \quad y = F(x) \in \overline{F(S)}$
	$\displaystyle \geq \lVert b-F(a)\rVert ^2 + 0 \geq \epsilon^2\,. \qedhere$

$\qedsymbol$

Theorem 2 (Hyperplane Separation Theorem II) Let $S \subset X$ be a weakly closed convex subset, and $K \subset X$ a compact convex subset, that do not intersect each other. Then there exists $\phi \in \Phi$ such that $$ \sup_{y \in K} \phi(y) < \inf_{x \in S} \phi(x)\,. $$

Proof. We show that $S - K = \{ x - y \colon x \in S\,, y \in K\}$ is weakly closed in $X$ . Let $\{ z_\alpha = x_\alpha - y_\alpha\} \subseteq A$ be a net convergent to $z$ . Since $K$ is compact, $\{ y_\alpha \}$ has a subnet $\{ y_{\alpha(\beta)} \}$ convergent to $y \in K$ . Then the subnet $x_{\alpha(\beta)} = z_{\alpha(\beta)} + y_{\alpha(\beta)}$ is convergent to $x = z + y$ . The point $x$ is in $S$ since $S$ is closed; therefore $z = x-y$ is in $S - K$ .

Also, $S-K$ is convex since $S$ and $K$ are. Noting that $0 \notin S-K$ (otherwise $S$ and $K$ would have a common point), we apply the previous theorem to obtain a $\phi \in \Phi$ such that $$ 0 = \phi(0) < \inf_{z \in S-K} \phi(z) \leq \phi(x-y) \,, \text{ for all $x \in S$ and $y \in K$. } $$ The desired conclusion follows at once. $\qedsymbol$