From c1b7a1a78ec5c08843f82c9852109df0b244607b Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Wed, 24 Jun 2026 14:05:23 -0400 Subject: [PATCH] Sharpened the Hahn-Banach theorem. --- src/fa/lc/hahn-banach.tex | 4 +++- 1 file changed, 3 insertions(+), 1 deletion(-) diff --git a/src/fa/lc/hahn-banach.tex b/src/fa/lc/hahn-banach.tex index 53fbd84..afa9941 100644 --- a/src/fa/lc/hahn-banach.tex +++ b/src/fa/lc/hahn-banach.tex @@ -98,12 +98,14 @@ \begin{theorem}[Hahn-Banach, First Geometric Form] \label{theorem:hahn-banach-geometric-1} - Let $E$ be a TVS over $\real$, $A, B \subset E$ be non-empty convex sets, such that $A \cap B = \emptyset$ and $A$ is open, then there exists $\phi \in E^* \setminus \bracs{0}$ and $\alpha \in \real$ with $A \subset \bracs{\phi \le \alpha}$ and $B \subset \bracs{\phi \ge \alpha}$. + Let $E$ be a TVS over $\real$, $A, B \subset E$ be non-empty convex sets, such that $A \cap B = \emptyset$ and $A$ is open, then there exists $\phi \in E^* \setminus \bracs{0}$ and $\alpha \in \real$ with $A \subset \bracs{\phi < \alpha}$ and $B \subset \bracs{\phi \ge \alpha}$. \end{theorem} \begin{proof}[Proof {{\cite[Theorem 1.6]{Brezis}}}. ] Let $C = A - B$, then $C \subset E$ is an open convex set by \autoref{lemma:convex-gymnastics}. Since $A \cap B = \emptyset$, $0 \not\in C$. By \autoref{lemma:hahn-banach-separation}, there exists $\phi \in E^*$ such that $C \subset \bracs{\phi < 0}$. In which case, for any $a \in A$ and $b \in B$, $\dpb{a - b, \phi}{E} < 0$ implies that $\dpb{a, \phi}{E} < \dpb{b, \phi}{E}$. Let $\alpha \in [\sup_{a \in A}\dpb{a, \phi}{E}, \inf_{b \in B}\dpb{b, \phi}{E}]$, then $A \subset \bracs{\phi \le \alpha}$ and $B \subset \bracs{\phi \ge \alpha}$. + + Since $A$ is open, so is $\phi(A)$. Thus $\phi(A) \subset (-\infty, \alpha]$ implies that $\phi(A) \subset (-\infty, \alpha)$. \end{proof} \begin{theorem}[Hahn-Banach, Second Geometric Form]