Added relevant Hahn-Banach results.

src/fa/lc/hahn-banach.tex

@@ -64,3 +64,77 @@
     \]
     so $\abs{\Phi} \le \rho$.
 \end{proof}
+
+\begin{lemma}[Separation of Point and Convex Set]
+\label{lemma:hahn-banach-separation}
+    Let $E$ be a TVS over $\real$, $A \subset E$ be a non-empty open convex set, and $x \in E \setminus A$, then there exists $\phi \in E^*$ and $\alpha \in \real$ such that $\dpb{x, \phi}{E} = \alpha$ and $A \subset \bracs{\phi < \alpha}$
+\end{lemma}
+\begin{proof}
+    By translation, assume without loss of generality that $0 \in A$. In which case, $A \in \cn^o(0)$ is convex.
+
+    Let $[\cdot]_A: E \to [0, \infty)$ be the gauge (\ref{definition:gauge}) of $A$, then $[\cdot]_A$ is a sublinear functional on $E$. For any $y, z \in E$ and $t > 0$ with $y, z \in tA$,
+    \[
+    \abs{[y]_A - [z]_A} \le [y - z]_A \le t
+    \]
+    Hence $[\cdot]_A$ is continuous on $E$.
+
+    Let $\phi_0: \real x \to \real$ be defined by $\lambda x \mapsto \lambda$. Since $x \not\in A$, $[x]_A > 1$. Hence $\phi_0|_{Kx} \le [\cdot]_A|_{Kx}$. By the Hahn-Banach Theorem (\ref{theorem:hahn-banach}), there exists $\phi \in \hom(E; \real)$ such that $\phi|_{\real x} = \phi_0$ and $\phi(y) \le [y]_A$ for all $y \in E$.
+
+    For any $y \in A \cap (-A)$,
+    \[
+    \dpb{y, \phi}{E} \le [y]_A \le 1 \quad -\dpb{y, \phi}{E} \le [-y]_A < 1
+    \]
+    so $\phi \in E^*$.
+
+    Finally, let $y \in A$, then $\dpb{y, \phi}{E} \le [y]_A < 1 = \dpb{x, \phi}{E}$.
+\end{proof}
+
+\begin{theorem}[Hahn-Banach, First Geometric Form {{\cite[Theorem 1.6]{Brezis}}}]
+\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 $B \subset \bracs{\phi \ge \alpha}$.
+\end{theorem}
+\begin{proof}
+    Let $C = A - B$, then $C \subset E$ is an open convex set by \ref{lemma:convex-gymnastics}. Since $A \cap B = \emptyset$, $0 \not\in C$.
+
+    By \ref{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 $B \subset \bracs{\phi \ge \alpha}$.
+\end{proof}
+
+\begin{theorem}[Hahn-Banach, Second Geometric Form {{\cite[Theorem 1.7]{Brezis}}}]
+\label{theorem:hahn-banach-geometric-2}
+    Let $E$ be a locally convex space over $\real$, $A, B \subset E$ be non-empty convex sets such that $A \cap B = \emptyset$, $A$ is closed, and $B$ is compact, then there exists $\phi \in E^* \setminus \bracs{0}$, $\alpha \in \real$, and $r > 0$ with $A \subset \bracs{\phi \le \alpha - r}$ and $B \subset \bracs{\phi \ge \alpha + r}$.
+\end{theorem}
+\begin{proof}
+    Let $C = A - B$, then $C$ is closed by \ref{proposition:tvs-set-operations} with $0 \not\in C$. Since $E$ is locally convex, there exists $U \in \cn^o(0)$ convex such that $U \cap C \ne \emptyset$.
+
+    By \ref{theorem:hahn-banach-geometric-1}, there exists $\phi \in E^* \setminus \bracs{0}$ and $\alpha \in \real$ such that $U \subset \bracs{f \le \alpha}$ and $C \subset \bracs{f \ge \alpha}$. In which case, for any $u \in U$, $a \in A$, and $b \in B$,
+    \begin{align*}
+    \dpb{u, \phi}{E} &\le \dpb{a, \phi}{E} - \dpb{b, \phi}{E} \\
+    \dpb{b, \phi}{E} + \dpb{u, \phi}{E} &\le \dpb{a, \phi}{E}
+    \end{align*}
+    As $\phi \ne 0$ and $U \in \cn^o(0)$, $r = \sup_{u \in U}\dpb{u, \phi}{E} > 0$. Thus
+    \[
+    \sup_{b \in B}\dpb{b, \phi}{E} + r \le \inf_{a \in A}\dpb{a, \phi}{E}
+    \]
+\end{proof}
+
+
+\begin{proposition}[{{\cite[Theorem 5.8]{Folland}}}]
+\label{proposition:hahn-banach-utility}
+    Let $E$ be a locally convex space over $K \in \RC$, then
+    \begin{enumerate}
+        \item For any subspace $M \subset E$, $x \in M \setminus E$, and continuous seminorm $\rho: E \to [0, \infty)$, there exists $\phi \in E^*$ such that
+        \begin{enumerate}
+            \item $\phi \le \rho$.
+            \item $\dpb{x, \phi}{E} = \inf_{y \in M}\rho(x + y)$.
+        \end{enumerate}
+        \item For any $x \in E$ and continuous seminorm $\rho: E \to [0, \infty)$, there exists $\phi \in E^*$ with $\phi \le \rho$ and $\dpb{x, \phi}{E} = \rho(x)$.
+        \item If $E$ is Hausdorff, then for any $x, y \in E$, there exists $\phi \in E^*$ with $\dpb{x, \phi}{E} \ne \dpb{y, \phi}{E}$.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): Let $\rho_M: E \to [0, \infty)$ be the quotient of $\rho$ by $M$, then $\rho_M \le \rho$ is a continuous seminorm on $E$ by \ref{definition:quotient-norm}. Let $\phi_0: Kx \to K$ be defined by $\lambda x \mapsto \lambda \rho_M(x)$. By the Hahn-Banach theorem (\ref{theorem:hahn-banach}), there exists $\phi \in \hom{E; K}$ such that $\dpb{x, \phi}{E} = \rho_M(x)$ and $\phi \le \rho_M \le \rho$.
+
+    (2): By (1) applied to $M = \bracs{0}$.
+
+    (3): By (2) applied to $x - y$.
+\end{proof}