Added the Bochner integral.

src/fa/lc/continuous.tex

@@ -35,3 +35,4 @@
 \begin{proof}
     $(1) \Rightarrow (2)$: By continuity of $T$, there exists continuous seminorms $\seqf{[\cdot]_j}$ on $\seqf{E_j}$ such that for any $x \in \prod_{j = 1}^n E_j$, $\max_{1 \le j \le n}[x_j]_{E_j} < 1$ implies that $[Tx]_F < 1$. In which case, the inequality follows from linearity.
 \end{proof}
+

src/fa/lp/definition.tex

@@ -100,19 +100,31 @@
     By \hyperref[Minkowski's Inequality]{theorem:minkowski}.
 \end{proof}
 
+
+\begin{proposition}
+\label{proposition:dct-lp}
+    Let $(X, \cm, \mu)$ be a measure space, $E$ be a Banach space over $K \in \RC$, $p \in [1, \infty)$, $\seq{f_n} \subset L^p(X; E)$, and $f \in L^p(X; E)$. If
+    \begin{enumerate}
+        \item[(a)] $f_n \to f$ strongly pointwise.
+        \item[(b)] There exists $g \in L^p(X) \cap L^+(X)$ such that $\norm{f_n}_E \le g$ for all $n \in \natp$.
+    \end{enumerate}
+    
+    then $f_n \to f$ in $L^p(X; E)$.
+\end{proposition}
+\begin{proof}
+    By assumptions $a$ and $b$, $\norm{f_n - f}_E \le 2g$ for all $n \in \natp$. Since $\seq{f_n} \subset L^p(X; E)$, $f \in L^p(X; E)$, and $g \in L^p(X)$, $\norm{f_n - f}_E^p, g^p \in L^1(X)$ for all $n \in \natp$. By the \hyperref[Dominated Convergence Theorem]{theorem:dct}, 
+    \[
+    \limv{n}\norm{f_n - f}_{L^p(X; E)}^p = \limv{n}\int \norm{f_n - f}_E d\mu  = 0
+    \]
+\end{proof}
+
+
 \begin{proposition}
 \label{proposition:lp-simple-dense}
     Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed space, and $p \in [1, \infty)$, then $\Sigma(X, \cm; E) \cap L^p(X; E)$ is dense in $L^p(X; E)$.
 \end{proposition}
 \begin{proof}
-    Let $f \in L^p(X; E)$. By \autoref{definition:strongly-measurable}, there exists $\seq{f_n} \subset \Sigma(X, \cm; E)$ such that $\norm{f_n}_E \le \norm{f}_E$ and $\norm{f_n - f}_E \to 0$ pointwise as $n \to \infty$.
-
-    For each $n \in \nat$, $\norm{f_n}_E \le \norm{f}_E$, so $\norm{f_n}_{L^p(X; E)} \le \norm{f}_{L^p(X; E)} < \infty$, and $\norm{f_n - f}_E \le 2\norm{f}_E$. By the \hyperref[Dominated Convergence Theorem]{theorem:dct},
-    \[
-    \limv{n}\int \norm{f_n - f}_E^p d\mu = \int \limv{n}\norm{f_n - f}_E^p d\mu = 0
-    \]
-
-    Therefore $\norm{f_n - f}_{L^p(X; E)} \to 0$ as $n \to \infty$.
+    Let $f \in L^p(X; E)$. By \autoref{definition:strongly-measurable}, there exists $\seq{f_n} \subset \Sigma(X, \cm; E)$ such that $\norm{f_n}_E \le \norm{f}_E$ and $\norm{f_n - f}_E \to 0$ strongly pointwise as $n \to \infty$. By the \hyperref[Dominated Convergence Theorem]{proposition:dct-lp}, $f_n \to f$ in $L^p(X; E)$.
 \end{proof}
 
 \begin{theorem}[Markov's Inequality]

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}
+
+

src/fa/tvs/continuous.tex

@@ -51,3 +51,30 @@
     \end{enumerate}
     The uniformity $\fU$ and its induced topology are the \textbf{product uniformity/topology}, and $E$ equipped with $\fU$ is the \textbf{product TVS} of $\seqi{E}$.
 \end{definition}
+
+\begin{theorem}[Linear Extension Theorem (TVS)]
+\label{theorem:linear-extension-theorem-tvs}
+    Let $E$ be a TVS over $K \in \RC$, $F$ be a complete Hausdorff TVS 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[(U)] For any $S \in C(E; F)$ satisfying (1), $S = \ol T$.
+    \end{enumerate}
+
+\end{theorem}
+\begin{proof}
+    By (3) of \autoref{definition:continuous-linear}, $T \in UC(D; F)$. By \autoref{theorem:uniform-continuous-extension}, there exists a unique $\ol T \in C(E; F)$ such that $\ol T|_D = T$. 
+
+    It remains to show that $\ol T$ is linear. Since $\ol T$ is continuous, the maps
+    \[
+    A: E \times E \to E \quad (x, y) \mapsto \ol Tx + \ol Ty - \ol T(x + y)
+    \]
+
+    and
+    \[
+    M: K \times E \to E \quad (\lambda x) \mapsto \lambda \ol Tx - Tx
+    \]
+
+    are continuous. Thus $\bracs{A = 0} \supset D \times D$ and $\bracs{M = 0} \supset K \times D$ are both closed. By density of $D$, $\bracs{A = 0} = E \times E$ and $\bracs{M = 0} = K \times E$. Therefore $T$ is linear. 
+\end{proof}
+
+

src/fa/tvs/spaces-of-linear.tex

@@ -153,14 +153,34 @@
 
 \begin{definition}[Strong Operator Topology]
 \label{definition:strong-operator-topology}
-    Let $E, F$ be TVSs over $K \in \RC$, $\fF \subset 2^E$ be the collection of finite subsets of $E$, then the $\fF$-uniform topology on $L(E; F)$ is the \textbf{strong operator topology}.
+    Let $E, F$ be TVSs over $K \in \RC$, $\fF \subset 2^E$ be the collection of finite subsets of $E$, then the $\fF$-uniform topology on $F^E$ is the \textbf{strong operator topology}.
 
     The space $L_s(E; F)$ denotes $L(E; F)$ equipped with the strong operator topology.
 \end{definition}
 
+
+\begin{proposition}
+\label{proposition:strong-operator-dense}
+    Let $E, F$ be TVSs over $K \in \RC$ and $\net{T} \subset L(E; F)$ and $T \in L_s(E; F)$. If
+    \begin{enumerate}
+        \item[(a)] There exists a dense subset $S \subset E$ such that $T_\alpha x \to Tx$ strongly for all $x \in S$.
+        \item[(b)] $\bracs{T_\alpha|\alpha \in A}$ is uniformly equicontinuous. 
+    \end{enumerate}
+    
+    then $T_\alpha \to T$ in $L_s(E; F)$.
+\end{proposition}
+\begin{proof}
+    Let $x \in E$, $U \in \cn_F(Tx)$, and $V \in \cn_F(Tx)$ be balanced such that $V + V + V \subset U$. By (b), there exists a balanced neighbourhood $W \in \cn_E(0)$ such that $T(W) \cup \bigcup_{\alpha \in A}T_\alpha(W) \subset V$. By (a), there exists $y \in S \cap (x + W)$ and $\alpha_0 \in A$ such that for all $\alpha \ge \alpha_0$, $T_\alpha y - Ty \in V$. In which case, for any $\alpha \ge \alpha_0$,
+    \[
+    T_\alpha x - Tx = \underbrace{T_\alpha x - T_\alpha y}_{\in V} + \underbrace{T_\alpha y - Ty}_{\in V} + \underbrace{Ty - Tx}_{\in V} \in U
+    \]
+\end{proof}
+
+
+
 \begin{definition}[Weak Operator Topology]
 \label{definition:weak-operator-topology}
-    Let $E, F$ be TVSs over $K \in \RC$, $\fF \subset 2^E$ be the collection of finite subsets of $E$, then the $\fF$-uniform topology on $L(E; F_w)$ is the \textbf{weak operator topology}.
+    Let $E, F$ be TVSs over $K \in \RC$, $\fF \subset 2^E$ be the collection of finite subsets of $E$, then the $\fF$-uniform topology on $F_w^E$ is the \textbf{weak operator topology}.
 
     The space $L_w(E; F) = L_s(E; F_w)$ denotes $L(E; F)$ equipped with the weak operator topology.
 \end{definition}