=== 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 $\mathbf y$. {{explain|Because of the loose use of commas to separate clauses, it is unclear what the above is supposed to mean. Is $r {{=}} \mathbf y \mathbf y$ part of the premise or conclusion? That is, should it be taken to mean "the line segment joining $\mathbf y$ and $\mathbf z$? (If so, then perhaps it ought to be $r {{=}} \mathbf y \mathbf z$.) Or is it one of the conditions on $\mathbf y$ and $\mathbf z$ which needs to be fulfilled in order for the conclusion to follow? (In that case, its meaning needs to be explained.}} {{explain|What is the domain of $\mathbf y$? Presumably it is a vector of some kind, in which case the nature of $0$ also needs to be defined.}} {{explain|$\map {H_{\mathbf y}^<} r$}} === 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 $s > 0$. {{explain|$\mathbf x \perp \mathbf y$}} 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 $\map f \gamma$: {{begin-eqn}} {{eqn | l = \map f \gamma       | r = \norm {\gamma \, \mathbf y + \paren {1 - \gamma} \mathbf z}^2       | c = Definition of $\map f \gamma$  }} {{eqn | r = \paren {\gamma \, \mathbf y + \paren {1 - \gamma} \mathbf z} \cdot \paren {\gamma \mathbf y + \paren {1 - \gamma} \mathbf z}       | c = Definition of norm on vector space $\R^\ell$ }} {{eqn | r = \paren {\gamma \, \mathbf y + \paren {1 - \gamma} \paren {\mathbf x - s \, \mathbf y} } \cdot \paren {\gamma \, \mathbf y + \paren {1 - \gamma} \paren {\mathbf x - s \, \mathbf y} }       | c = as $\mathbf z = \mathbf x - s \, \mathbf y$ }} {{eqn | r = \gamma^2 \mathbf y \mathbf y + 2 \gamma \paren {1 - \gamma} \mathbf y \paren {\mathbf x - s \mathbf y} + \paren {1 - \gamma}^2 \paren {\mathbf x - s \, \mathbf y} \paren {\mathbf x - s \, \mathbf y}       | c = Distributive property of dot product }} {{eqn | r = \gamma^2 \mathbf y \mathbf y - 2 \gamma \paren {1 - \gamma} s \, \mathbf y \mathbf y + \paren {1 - \gamma}^2 s^2 \mathbf y \mathbf y       | c = Distributive property of dot product }} {{eqn | o =        | ro= +       | r = 2 \gamma \paren {1 - \gamma} \mathbf x \mathbf y - 2 \paren {1 - \gamma}^2 s \, \mathbf x \mathbf y + \paren {1 - \gamma}^2 \mathbf x \mathbf x        | c =  }} {{eqn | r = \mathbf y \mathbf y \paren {\gamma^2 - 2 \gamma \paren {1 - \gamma} s + \paren {1 - \gamma}^2 s^2} + \paren {1 - \gamma}^2 \mathbf x \mathbf x        | c = Definition of orthogonal vectors  }} {{eqn | r = \mathbf y \mathbf y \paren {\gamma - s \paren {1 - \gamma} }^2 + \paren {1 - \gamma}^2 \mathbf x \mathbf x        | c = Binomial Theorem }} {{end-eqn}} 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 $\gamma \to 1$. {{qed|lemma}} === 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 $\mathbf y \, \gamma \, \mathbf z \in C$. {{explain|It is not clear which parts of the above compound "sentence" are in the premise and which in the conclusion.}} Since $\mathbf y $ is a minimmizer of $\norm {\mathbf x}$: :$\norm {\mathbf y} \le \norm {\mathbf y \, \gamma \, \mathbf z}$ which is a contradiction. {{explain|Make it plain what it contradicts, and what the conclusion is of that Proof by 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: :$\mathbf v = \mathbf 0 \subset \map {L^{-1} } {\hointl {-\infty} {r/2 - \epsilon} }$ {{explain|Explain how the conclusion of the proof follows from that.}} {{qed}}