Hyperplane separation theorem, continuous linear functional

Question

One version of the Hyperplane separation theorem states that two non-intersecting convex sets in a normed linear space can be separated by a linear functional. In the proof I know they use the gauge (or Minkowsky functional) to show this. There's another version of this theorem that states that two such convex sets can be separated by a continuous linear functional. I wonder if somebody can explain to me how. Is the gauge bounded by the norm ( $ p_K(x)\leq C\lVert x \rVert$)?Is this true for all $x$ in a normed linear space?

Answer 1

This cannot be true. Since If you cosider any infinite dimensional normed space $(X, ||\cdot ||_{*} )$ and any other norm $||\cdot ||^{*} :X\to \mathbb{R} $ stronger than $||\cdot ||_{*} ,$ then taking $K=\{ x\in X: ||x||^{*} \leqslant 1\}$ you obtain $p_K (x) =||x||^{*} \nleqslant C||x||_{*}$ for any $C\in\mathbb{R}.$

Answer 2

It is not true that two non-intersecting convex sets can be separated by a continuous functional without further assumptions (Such as one is closed and the other compact, or one of the convex subsets has nonempty interior).

Here is a counterexample (c.f exercise 1.14 of Brezis). In $\ell^1(\mathbb{R})$ define the two closed subspaces \begin{align*} A &= \{x\in \ell^1 | x_{2n}=0 \text{ for all } n\in\mathbb{N}\} \\ B &= \{x\in \ell^1 | x_{2n}=\frac{1}{2^n}x_{2n-1} \text{ for all } n\in\mathbb{N}\}. \end{align*}

Observe that $\overline{A+B}=\ell^1$. Let $p$ be the point defined by $$ p_{2n}=\frac{1}{2^n}, \quad p_{2n-1}=0.$$ We have that $p \notin A + B$ (we can see this by working backwards from the definitions and seeing the only possible way this could be is if there exists a point such that $x_{2n-1}=1$ for all $n$; but no such point exists.)

Now, set $C = A+p$, we have that $B\cap C = \emptyset$ where $B$ and $C$ are closed convex subsets, with empty interior. Suppose the linear functional $f$ separates $B$ and $C$. So for all $b\in B$, $f(b) < \alpha$ and for all $a\in A$, $f(a+p) > \alpha$. Now, as $B$ is a subspace, we must have $f(b)=0$ for all $b\in B$ and similarly, $f(a)=0$ for all $a\in A$. However, such a linear functional is not continuous, since $A+B$ is dense in $\ell^1$ and so $f$ must be zero if continuous. Therefore $B$ and $C$ is an example of two convex subsets which cannot be separated by a closed hyperplane.