# hyperplane separation

Let $X$ be a vector space^{}, and $\mathrm{\Phi}$ be any subspace^{}
of linear functionals^{} on $X$.
Impose on $X$ the weak topology generated by $\mathrm{\Phi}$.

###### Theorem 1 (Hyperplane Separation Theorem I).

Given a weakly closed convex subset $S\mathrm{\subset}X$, and $a\mathrm{\in}X\mathrm{\setminus}S$. there is $\varphi \mathrm{\in}\mathrm{\Phi}$ such that

$$ |

###### Proof.

The weak topology on $X$ can be generated by the semi-norms $x\mapsto |p(x)|$ for $p\in \mathrm{\Phi}$. A subbasis for the weak topology consists of neigborhoods of the form $$ for $y\in X$, $p\in \mathrm{\Phi}$ and $\u03f5>0$. Since $X\setminus S$ is weakly open, there exist ${f}_{1},\mathrm{\dots},{f}_{n}\in \mathrm{\Phi}$ and $\u03f5>0$ such that

$$ |

In other words, if $x\in S$ then at least one of $|{f}_{i}(x)-{f}_{i}(a)|$ is $\ge \u03f5$.

Define a map $F:X\to {\mathbb{R}}^{n}$
by $F(x)=({f}_{1}(x),\mathrm{\dots},{f}_{n}(x))$.
The set $\overline{F(S)}$ is evidently
closed and convex in ${\mathbb{R}}^{n}$, a Hilbert space^{}
under the standard inner product^{}.
So there is a point $b\in \overline{F(S)}$
that minimizes the norm $\parallel b-F(a)\parallel $.

It follows that $\u27e8y-b,b-F(a)\u27e9\ge 0$ 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 $0\le {\frac{d}{dt}|}_{t=0}{\parallel ty+(1-t)b-F(a)\parallel}^{2}=2\u27e8y-b,b-F(a)\u27e9$.)

Take $\varphi ={\sum}_{i=1}^{n}{\lambda}_{i}{f}_{i}$ where $\lambda =b-F(a)$. Then we find, for all $x\in S$,

$\varphi (x-a)$ | $=\u27e8b-F(a),F(x-a)\u27e9$ | ||

$=\u27e8b-F(a),b-F(a)\u27e9+\u27e8b-F(a),y-b\u27e9,y=F(x)\in \overline{F(S)}$ | |||

$\ge {\parallel b-F(a)\parallel}^{2}+0\ge {\u03f5}^{2}.\mathit{\u220e}$ |

###### Theorem 2 (Hyperplane Separation Theorem II).

Let $S\mathrm{\subset}X$ be a weakly closed convex subset, and $K\mathrm{\subset}X$ a compact convex subset, that do not intersect each other. Then there exists $\varphi \mathrm{\in}\mathrm{\Phi}$ such that

$$ |

###### Proof.

We show that $S-K=\{x-y:x\in S,y\in K\}$ 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 $\varphi \in \mathrm{\Phi}$ such that

$$ |

The desired conclusion^{} follows at once.
∎

Title | hyperplane separation |
---|---|

Canonical name | HyperplaneSeparation |

Date of creation | 2013-03-22 17:19:01 |

Last modified on | 2013-03-22 17:19:01 |

Owner | stevecheng (10074) |

Last modified by | stevecheng (10074) |

Numerical id | 4 |

Author | stevecheng (10074) |

Entry type | Theorem |

Classification | msc 46A55 |

Classification | msc 49J27 |

Classification | msc 46A20 |

Synonym | separating hyperplane |

Related topic | HahnBanachgeometricFormTheorem |