Adjusted citation formats. Moved citation off of named theorems if possible.

src/fa/lc/hahn-banach.tex

@@ -35,7 +35,7 @@
     \end{align*}
 \end{proof}
 
-\begin{theorem}[Hahn-Banach, Analytic Form {{\cite[Theorem 5.6, 5.7]{Folland}}}]
+\begin{theorem}[Hahn-Banach, Analytic Form]
 \label{theorem:hahn-banach}
     Let $E$ be a vector space over $K \in \RC$, $\rho: E \to \real$ be a sublinear functional, and $F \subsetneq E$ be a subspace, then
     \begin{enumerate}
@@ -43,7 +43,7 @@
         \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}
+\begin{proof}[Proof {{\cite[Theorem 5.6, 5.7]{Folland}}}. ]
     (1): Let $x_0 \in E \setminus F$, then by \autoref{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
@@ -96,21 +96,21 @@
     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}}}]
+\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 $B \subset \bracs{\phi \ge \alpha}$.
 \end{theorem}
-\begin{proof}
+\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 $B \subset \bracs{\phi \ge \alpha}$.
 \end{proof}
 
-\begin{theorem}[Hahn-Banach, Second Geometric Form {{\cite[Theorem 1.7]{Brezis}}}]
+\begin{theorem}[Hahn-Banach, Second Geometric Form]
 \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}
+\begin{proof}[Proof {{\cite[Theorem 1.7]{Brezis}}}. ]
     Let $C = A - B$, then $C$ is closed by \autoref{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 \autoref{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 $b \in B$,