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Sharpened the Hahn-Banach theorem.

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Bokuan Li
2026-06-24 14:05:23 -04:00
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src/fa/lc/hahn-banach.tex
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\begin{theorem}[Hahn-Banach, First Geometric Form] \begin{theorem}[Hahn-Banach, First Geometric Form]
\label{theorem:hahn-banach-geometric-1} \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} \end{theorem}
\begin{proof}[Proof {{\cite[Theorem 1.6]{Brezis}}}. ] \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 $A \cap B = \emptyset$, $0 \not\in C$. 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 $B \subset \bracs{\phi \ge \alpha}$. 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} \end{proof}
\begin{theorem}[Hahn-Banach, Second Geometric Form] \begin{theorem}[Hahn-Banach, Second Geometric Form]