hyperplane separation

Let $X$ be a vector space, and $\mathrm{\Phi }$ be any subspaceof 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 topologyconsists of neigborhoods of the formfor $y\in X$, $p\in \mathrm{\Phi }$ and $ϵ>0$.Since $X\setminus S$ is weakly open,there exist $f_1,\mathrm{\dots },f_n\in \mathrm{\Phi }$ and $ϵ>0$such that

In other words, if $x\in S$ then at least one of$|f_i(x)-f_i(a)|$ is $\ge ϵ$.

Define a map $F:X\to {ℝ}^n$by $F(x)=(f_1(x),\mathrm{\dots },f_n(x))$.The set $\overline{F(S)}$ is evidentlyclosed and convex in ${ℝ}^n$, a Hilbert spaceunder 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 $⟨y-b,b-F(a)⟩\ge 0$ for all $y\in \overline{F(S)}$;for otherwise we can attain a smaller value of the normby moving from the point $b$ along a line towards $y$.(Formally, we have$0\le {\frac{d}{dt}|}_{t=0}{\parallel ty+(1-t)b-F(a)\parallel }^2=2⟨y-b,b-F(a)⟩$.)

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)$$=⟨b-F(a),F(x-a)⟩$

$=⟨b-F(a),b-F(a)⟩+⟨b-F(a),y-b⟩,y=F(x)\in \overline{F(S)}$

$\ge {\parallel b-F(a)\parallel }^2+0\ge {ϵ}^2.\mathit{∎}$

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 toobtain a $\varphi \in \mathrm{\Phi }$ suchthat

The desired conclusion follows at once.∎