Strong Separation Theorem

Theorem
Let $C \subset \R^\ell$ be closed and convex.

Let $D = \set {\mathbf v} \subset C^c$.

Then $C$ and $D$ can be strongly separated.

Lemma 1
For any $\mathbf y \ne 0$ and any $\mathbf z \in \map {H_{\mathbf y}^<} r$, $r = \mathbf y \mathbf y$, the line segment joining $\mathbf y$ and $\mathbf z$ contains points with lengths strictly less than $\mathbf y$.

Proof of Lemma 1
Since $\mathbf z \in \map {H_\mathbf y} r^<$:
 * $\mathbf z = \mathbf x - s \mathbf y$

for some $\mathbf x \perp \mathbf y$ and $s > 0$.

Let:
 * $\map f \gamma = \norm {\gamma \mathbf y + \paren {1 - \gamma} \mathbf z}^2$

for $\gamma \in \closedint 0 1$.

We show that:
 * $\map f \gamma < \norm {\mathbf y}^2$

as $\gamma \to 1$.

Expand $\map f \gamma$:

Since $\map {f'} 1 = 2 \paren {1 + s} > 0$:
 * $\map f \gamma < \map f 1$

as $\gamma \to 1$.

Since $\map f 1 = \mathbf y \mathbf y = \norm {\mathbf y}^2$:
 * $\map f \gamma < \norm {\mathbf y}^2$

as $\gamma \to 1$.

Proof of Theorem
Translating $C$ and $D = \set {\mathbf v}$ by $-\mathbf v$:

We show that the theorem holds for $\mathbf v = 0$.

Pick $n > 0$ such that:
 * $K = \map \cl {\map {B_n} {\mathbf v} } \cap C \ne \O$

Since $C$ is closed, $K$ is closed.

Since $\map {B_n} {\mathbf v}$ is bounded, $K$ is bounded.

Since $K$ is closed and bounded, $K$ is compact.

Let $\map f {\mathbf x} = \norm {\mathbf x}$.

Since $\norm {\mathbf x}^2$ and $\sqrt x: \R_+ \to \R_+$ are both continuous, $\map f {\mathbf x}$ is continuous.

Since $K$ is compact and $\map f {\mathbf x}$ is continuous, $\map f {\mathbf x}$ achieves its minimum at $\mathbf y \in C$ by the Weierstrass Theorem.

Let $\map L {\mathbf x} = \mathbf y \mathbf x$ and $r = \mathbf y \mathbf y$.

We show that:
 * $C \subset \map {L^{-1} } {\hointr r \to}$

Suppose $C$ is not a subset of $\map {L^{-1} } {\hointr r \to}$.

There exists $\mathbf z \in C \cap \map {L^{-1} } {\closedint {-\infty} r}$

So:
 * $\mathbf z \in \map {H_\mathbf y} r$

As $\gamma \to 1$, there exists a $0 < \gamma < 1$ such that:
 * $\norm {\mathbf y \, \gamma \, \mathbf z} < \norm {\mathbf y}$

by Lemma 1.

Since $C$ is convex, $\mathbf y \in C$ and $\mathbf z \in C$, convex combination $\mathbf y \, \gamma \, \mathbf z \in C$.

Since $\mathbf y $ is a minimmizer of $\norm {\mathbf x}$:
 * $\norm {\mathbf y} \le \norm {\mathbf y \, \gamma \, \mathbf z}$

which is a contradiction.

For strong separation, let $\epsilon < r / 2$.

We have:
 * $C \subset \map {L^{-1} } {\hointr r \to} \subset \map {L^{-1} } {\hointr {r/2 + \epsilon} \infty}$

Since:
 * $\map L {\mathbf 0} = \mathbf y \mathbf 0 = 0$

and:
 * $r/2 - \epsilon > 0$

it follows that:
 * $\mathbf v = \mathbf 0 \subset \map {L^{-1} } {\hointl {-\infty} {r/2 - \epsilon} }$