Added the Bochner integral.

src/fa/norm/index.tex

@@ -4,4 +4,5 @@
 \input{./normed.tex}
 \input{./absolute.tex}
 \input{./linear.tex}
+\input{./separable.tex}
 \input{./multilinear.tex}

src/fa/norm/linear.tex

@@ -12,3 +12,25 @@
 \begin{proof}
     By \autoref{proposition:lc-spaces-linear-map}.
 \end{proof}
+
+
+\begin{theorem}[Linear Extension Theorem (Normed)]
+\label{theorem:linear-extension-theorem-normed}
+    Let $E$ be a normed vector space over $K \in \RC$, $F$ be a Banach space over $K$, $D \subset E$ be a dense subspace, and $T \in L(D; F)$, then:
+    \begin{enumerate}
+        \item There exists an extension $\ol T \in L(E; F)$ such that $\ol T|_D = T$.
+        \item $\normn{\ol T}_{L(E; F)} = \normn{T}_{L(D; F)}$.
+        \item[(U)] For any $S \in C(E; F)$ satisfying (1), $S = \ol T$.
+    \end{enumerate}
+
+\end{theorem}
+\begin{proof}
+    (1), (U): By \autoref{theorem:linear-extension-theorem-tvs}.
+
+    (2): Since $\ol{T}$ is continuous, the function
+    \[
+    N: E \to \real \quad x \mapsto \normn{\ol T}_{F} - \norm{T}_{L(D; F)}\cdot \norm{x}_E
+    \]
+
+    is continuous, so $\bracs{N \le 0} \supset D$ is closed. By density of $D$, $\bracs{N \le 0} = E$. Therefore $\normn{\ol T}_{L(E; F)} = \normn{T}_{L(D; F)}$.
+\end{proof}

src/fa/norm/normed.tex

@@ -86,3 +86,17 @@
 
     so $\sup_{T \in \mathcal{T}}\norm{T}_{L(E; F)} \le 2n/r$.
 \end{proof}
+
+
+\begin{proposition}
+\label{proposition:dual-norm}
+    Let $E$ be a normed space, then for any $x \in E$, 
+    \[
+    \norm{x}_E = \sup_{\substack{\phi \in E^* \\ \norm{\phi}_{E^*} = 1}}\dpn{x, \phi}{E}
+    \]
+\end{proposition}
+\begin{proof}
+    For any $\phi \in E^*$ with $\norm{\phi}_{E^*} = 1$, $\dpn{x, \phi}{E} \le \norm{x}_E \cdot \norm{\phi}_{E^*} = \norm{x}_E$. On the other hand, by the \hyperref[Hahn-Banach Theorem]{proposition:hahn-banach-utility}, there exists $x \in E^*$ such that $\dpn{x, \phi}{E} = \norm{x}_E$ and $\norm{\phi}_{E^*} \le 1$. 
+\end{proof}
+
+

src/fa/norm/separable.tex

@@ -0,0 +1,48 @@
+\section{Separable Normed Spaces}
+\label{section:separable-banach-space}
+
+\begin{proposition}
+\label{proposition:separable-dual}
+    Let $E$ be a separable normed space, then $E^*$ is separable with respect to the weak*-topology. 
+\end{proposition}
+\begin{proof}
+    Let $\seq{x_n} \subset E$ be a dense subset and $S = \bracsn{\phi \in E^*| \norm{\phi}_{E^*} \le 1}$. For each $N \in \natp$, let
+    \[
+    T_N: S \to \real^N \quad \phi \mapsto (\dpn{x_1, \phi}{E}, \cdots, \dpn{x_N, \phi}{E})
+    \]
+
+    then since $\real^N$ is separable, there exists $\bracs{\phi_{N, k}}_{k = 1}^\infty \subset S$ such that $\bracs{T_N\phi_{N, k}}_{k = 1}^\infty$ is dense in $T_N(S)$.
+
+    Let $\phi \in S$, then for each $N \in \natp$, there exists $k_N \in \natp$ such that for each $1 \le n \le N$, 
+    \[
+    |\dpn{x_n, \phi_{N, k_N}}{E} - \dpn{x_n, \phi}{E}| \le \frac{1}{n}
+    \] 
+
+    Thus for each $N \in \natp$, $\dpn{x_n, \phi_{N, k_N}}{E} \to \dpn{x_n, \phi}{E}$ as $N \to \infty$. Since $\phi_{N, k_N} \to \phi$ pointwise on a dense subset of $E$, and $\bracsn{\phi_{N, k_N}|N \in \natp} \subset S$ is uniformly equicontinuous, $\phi_{N, k_N} \to \phi$ in the weak*-topology by \autoref{proposition:strong-operator-dense}. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:separable-banach-borel-sigma-algebra}
+    Let $E$ be a separable normed space, then the Borel $\sigma$-algebra on $E$ is generated by the following families of sets:
+    \begin{enumerate}
+        \item Open sets in $E$ with respect to the strong topology. 
+        \item $\bracs{B(x, r)|x \in E, r > 0}$.
+        \item $\bracsn{\ol{B(x, r)}|x \in E, r > 0}$. 
+        \item Open sets in $E$ with respect to the weak topology. 
+    \end{enumerate}
+    
+\end{proposition}
+\begin{proof}
+    (1) $\Leftrightarrow$ (2) $\Leftrightarrow$ (3): By \autoref{proposition:separable-metric-borel-sigma-algebra}.
+
+    (4) $\subset$ (1): Every weakly open set is strongly open. 
+
+    (2) $\subset$ (4): By \autoref{proposition:seminorm-lsc}, $\norm{\cdot}_E: E \to [0, \infty)$ is Borel measurable with respect to the weak topology. For any $x \in E$, let
+    \[
+    \phi_x: E \to [0, \infty) \quad y \mapsto \norm{x - y}_E
+    \]
+
+    then $\phi_x$ is Borel measurable with respect to the weak topology, so $B(x, r) = \bracs{\phi_x < r}$ is a Borel set with respect to the weak topology. 
+\end{proof}
+
+