Showing the existence of separating hyperplanes using the Hahn-Banach theorem

Question

I am trying to show the existence of separating hyperplanes for points $v$ outside of any compact, convex, equilibrated set $X \subseteq V$ that contains $0$ in its interior. I know that since $X$ has these properties, a norm $\nu_X: X → \mathbb{R}$ with unit ball $X$ can be defined by $\nu_X(x) = \inf\{\alpha > 0 | \frac1\alpha x ∈ X\}$. So $\max\{\nu_X(x)|x \in X\} = 1$, then by the Hahn-Banach theorem, $\nu_X$ can be extended to a linear functional on $V$ whose maximum magnitude on the unit $\nu$-sphere in $V$ is $1$. Since $v \notin X$, then $\nu_X(v) > 1$, so $\nu_X(v) > \nu_X(x)$ for all $x \in X$, which implies the existence of a separating hyperplane. Is my work correct? I feel like I never really used the Hahn-Banach theorem. Can someone explain why exactly I need Hahn-Banach? Thanks a lot in advance!

Answer 1

then by the Hahn-Banach theorem, $\nu_X$ can be extended to a linear functional on $V$

The Hahn-Banach theorem definitely doesn't say that, because no such linear functional exists. You can see this from the fact $\nu_X \geq 0$ on a neighborhood of $0$, and the only linear functional $f$ satisfying $f \geq 0$ in a neighborhood of zero is $f=0$.

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Instead, the proof usually follows these steps:

1. Since $X$ is a compact, convex, balanced neighborhood of $0$ [plus necessary assumptions on the real topological vector space $V$], there is a norm $\nu_X$ on $V$ such that $X = \{x \in V : \nu_X(x) \leq 1\}$
2. For $v \in V \setminus X$, define a linear functional $f_v : \mathbb{R}v \to \mathbb{R}$ by $f_v(tv) = t\nu_X(v).$ Note that $$|f_v(tv)| = |t\nu_X(v)| = |t|\nu_X(v) = \nu_X(tv)$$ so in particular $|f_v| \leq \nu_X$ and $f_v(v) = \nu_X(v)$.
3. By Hahn-Banach, there is a linear functional $g_v : V \to \mathbb{R}$ extending $f_v$ such that $|g_v| \leq \nu_X$.

With that setup, we can conclude, for all $x\in X$, $$g_v(x) \leq \nu_X(x) \leq 1 < \nu_X(v) = g_v(v)$$ from which we can find a separating hyperplane.