Progress over the past week.

src/fa/lc/convex.tex

@@ -1,6 +1,7 @@
 \section{Seminorms}
 \label{section:seminorms}
 
+
 \begin{definition}[Convex]
 \label{definition:convex}
     Let $E$ be a vector space over $K \in \RC$, then $A \subset E$ is \textbf{convex} if for any $x, y \in A$, $\bracs{\lambda x + (1 - \lambda) y| \lambda \in [0, 1]} \subset A$.

src/fa/lc/hahn-banach.tex

@@ -0,0 +1,66 @@
+\section{The Hahn-Banach Theorem}
+\label{section:hahn-banach}
+
+\begin{lemma}
+\label{lemma:hahn-banach}
+    Let $E$ be a vector space over $\real$, $\rho: E \to \real$ be a sublinear functional, $F \subset E$ be a subspace, $\phi \in \hom(F; \real)$ with $\phi(x) \le \rho(x)$ for all $x \in F$.
+
+    Let $x_0 \in E \setminus F$, $\lambda \in \real$, and define
+    \[
+    \phi_{x_0, \lambda}: F + \real x_0 \quad (y + tx_0) \mapsto \phi(y) + \lambda t
+    \]
+    then there exists $\lambda \in \real$ such that $\phi_{x_0, \lambda}(x) \le \rho(x)$ for all $x \in F + \real x_0$.
+\end{lemma}
+\begin{proof}
+    Let $x, y \in F$, then
+    \begin{align*}
+        \phi(x) + \phi(y) = \phi(x + y) &\le \rho(x + y) \le \rho(x - x_0) + \rho(y + x_0) \\
+        \phi(x) - \rho(x - x_0) &\le \rho(y + x_0) - \phi(y) \\
+        \sup_{x \in F}[\phi(x) - \rho(x - x_0)] &\le \inf_{x \in F}[\rho(x + x_0) - \phi(x)]
+    \end{align*}
+    Let
+    \[
+    \lambda \in \braks{\sup_{x \in F}[\phi(x) - \rho(x - x_0)], \inf_{x \in F}[\rho(x + x_0) - \phi(x)]}
+    \]
+    then for any $x \in F$ and $t > 0$,
+    \begin{align*}
+        \phi(x + tx_0) &= \phi(x) + t\lambda = t\braks{\phi(t^{-1}x) + \lambda} \\
+        &\le t\braks{\rho(t^{-1}x + x_0) - \phi(t^{-1}x) + \phi(t^{-1}x)} \\
+        &= t\rho(t^{-1}x + x_0) = \rho(x + tx_0) \\
+        \phi(x - tx_0) &= \phi(x) - t\lambda = t\braks{\phi(t^{-1}x) - \lambda} \\
+        &\ge t\braks{\rho(t^{-1}x - x_0) + \phi(t^{-1}x) - \phi(t^{-1}x)} \\
+        &= t\rho(t^{-1}x - x_0) = \rho(x - tx_0)
+    \end{align*}
+\end{proof}
+
+\begin{theorem}[Hahn-Banach, Analytic Form {{\cite[Theorem 5.6, 5.7]{Folland}}}]
+\label{theorem:hahn-banach}
+    Let $E$ be a vector space over $K \in \RC$, $\rho: E \to \real$ be a sublinear functional, $F \subsetneq E$ be a subspace,
+    \begin{enumerate}
+        \item For any $\phi \in \hom(F; \real)$ with $\phi \le \rho|_F$, there exists $\Phi \in \hom(E; \real)$ such that $\Phi \le \rho$ and $\Phi|_F = \phi$.
+        \item If $\rho$ is a seminorm, then for any $\phi \in \hom(F; \complex)$ with $\abs{\phi} \le \rho|_F$, there exists $\Phi \in \hom(E; \complex)$ such that $\abs{\Phi} \le \rho$ and $\Phi|_F = \phi$.
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    (1): Let $x_0 \in E \setminus F$, then by \ref{lemma:hahn-banach}, there exists $\lambda \in \real$ such that if
+    \[
+    \phi_{x_0, \lambda}: F + \real x_0 \to \real \quad (x + tx_0) \mapsto x + \lambda t
+    \]
+    then $\phi_{x_0, \lambda}(x) \le \rho(x)$ for all $x \in F + \real x_0$. Thus for any $F \subsetneq E$, $\phi$ admits an extension to a subspace that strictly contains $F$.
+
+    Let
+    \[
+    \mathbf{F} = \bracs{\Phi \in \hom(F'; \real)|F' \supset F, \Phi|_F = \phi, \Phi \le \rho|_{F'}}
+    \]
+    For any $\Phi, \Phi' \in \mathbf{F}$, define $\Phi \le \Phi'$ if $\Phi'$ is an extension of $\Phi$. Let $\cf \subset \mathbf{F}$ be a chain. For any $\Phi \in \cf$, let $\cd(\Phi)$ be its domain. Since $\bigcup_{\phi \in \cf}\cd(\Phi)$ is a subspace and $\cf$ is a chain, by \ref{lemma:glue-linear}, there exists $\Phi \in \hom\paren{\bigcup_{\Phi' \in \cf}\cd(\Phi'); \real}$ such that $\Phi|_{\cd(\Phi')} = \Phi'$ for all $\Phi' \in \cf$. Thus $\Phi \in \mathbf{F}$ is an upper bound of $\cf$.
+
+    By Zorn's lemma, $\mathbf{F}$ admits a maximal element $\Phi$. If $\cd(\Phi) \subsetneq E$, then $\Phi$ is not maximal by the preceding discussion. Therefore $\cd(\Phi) = E$ and $\Phi$ is a desired extension.
+
+    (1): Given that $\rho$ is a seminorm, for any $u \in \hom(E; \real)$, $u \le \rho$ if and only if $\abs{u} \le \rho$.
+
+    Let $u = \re{\phi}$, then $u \in \hom(E; \real)$ by \ref{proposition:polarisation-linear}. By (1), there exists $U \in \hom(E; \real)$ such that $\abs{U} \le \rho$ and $U|_F = u$. For each $x \in E$, let $\Phi(x) = U(x) - iU(ix)$, then $\Phi \in \hom(E; \complex)$ and $\Phi|_F = \phi$ by \ref{proposition:polarisation-linear}. In addition, for any $x \in E$, if $\alpha = \overline{\sgn(\Phi(x))}$, then
+    \[
+    |\Phi(x)| = \alpha\Phi(x) = \Phi(\alpha x) = U(\alpha x) \le \rho(\alpha x) = \rho(x)
+    \]
+    so $\abs{\Phi} \le \rho$.
+\end{proof}

src/fa/lc/index.tex

@@ -4,3 +4,4 @@
 
 \input{./src/fa/lc/convex.tex}
 \input{./src/fa/lc/continuous.tex}
+\input{./src/fa/lc/hahn-banach.tex}