Progress over the past week.

src/fa/index.tex

@@ -1,5 +1,5 @@
 \part{Functional Analysis}
-\label{part:part-fa}
+\label{part:fa}
 
 
 \input{./src/fa/tvs/index.tex}

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}

src/fa/rs/bv.tex

@@ -47,13 +47,13 @@
     \]
     By assumption (b), $[0, M_\rho]$ is in the filter generated by $V_{\rho, P}(\fF)$. Thus $V_{\rho, P}(f) \le M_\rho$. As this holds for all $P \in \scp([a, b])$, $V_{\rho, P}(f) \le M_\rho$, and $f \in BV([a, b]; E)$.
 
-    (5): For each $n \in \nat$, let
+    (5): For each $n \in \nat^+$, let
     \[
     D_n = \bracs{x \in [a, b]|\forall \eps > 0, \exists y \in (x - \eps, x + \eps): \norm{f(x) - f(y)}_E \ge 1/n}
     \]
-    then $D = \bigcup_{n \in \nat}D_n$ is the set of discontinuity points of $f$. If $D$ is uncountable, then there exists $N \in \nat$ such that $D_n$ is infinite.
+    then $D = \bigcup_{n \in \nat^+}D_n$ is the set of discontinuity points of $f$. If $D$ is uncountable, then there exists $N \in \nat^+$ such that $D_n$ is infinite.
 
-    Fix $N \in \nat$. Let $E_1 = D_n \cap (a, b)$ and $I_1 = (a, b)$, then
+    Fix $N \in \nat^+$. Let $E_1 = D_n \cap (a, b)$ and $I_1 = (a, b)$, then
     \begin{enumerate}
         \item[(a)] $|E_k| \ge N - k$.
         \item[(b)] $E_k \subset I_k^o$.
@@ -62,5 +62,5 @@
 
     Let $k \le N$ and suppose inductively that $E_k, I_k$ have been constructed. Let $x_k \in E_k$, then by (b), there exists $\eps > 0$ such that $[x_k - \eps, x_k + \eps] \subset I_k$ and $|E_k \setminus [x_k - \eps, x_k + \eps]| \ge N - k$. Let $y_k \in [x_k - \eps, x_k + \eps]$ such that $\norm{f(x_k) - f(y_k)} \ge 1/n$, $I_{k + 1} = I_k \setminus [x_k - \eps, x_k + \eps]$, and $E_{k+1} = E_k \setminus [x_k - \eps, x_k + \eps]$, then $I_k$ and $E_k$ satisfies (a) and (b).
 
-    Therefore there exists pairs $\bracs{(x_k, y_k)|1 \le k \le N}$ such that $\norm{f(x_k) - f(y_k)}_E \ge 1/n$ for all $n$, and the smallest interval containing each $(x_k, y_k)$ are pairwise disjoint. Thus $[f]_{\var} \ge N/n$ for all $N \in \nat$, so $[f]_{\var} = \infty$.
+    Therefore there exists pairs $\bracs{(x_k, y_k)|1 \le k \le N}$ such that $\norm{f(x_k) - f(y_k)}_E \ge 1/n$ for all $n$, and the smallest interval containing each $(x_k, y_k)$ are pairwise disjoint. Thus $[f]_{\var} \ge N/n$ for all $N \in \nat^+$, so $[f]_{\var} = \infty$.
 \end{proof}

src/fa/rs/rs-bv.tex

@@ -1,4 +1,4 @@
-\section{Riemann-Stieltjes Integrals and Functions of Bounded Variationo}
+\section{Riemann-Stieltjes Integrals and Functions of Bounded Variation}
 \label{section:rs-bv}
 
 \begin{proposition}
@@ -36,7 +36,7 @@
     then $f \in RS([a, b], G)$ and $\int_a^b f dG = \lim_{\alpha \in A}\int_a^b f_\alpha dG$. In particular,
     \begin{enumerate}
         \item If $H$ is complete, then condition (a) may be omitted.
-        \item If $H$ is sequentially complete and $A = \nat$, then condition (b) may be omitted.
+        \item If $H$ is sequentially complete and $A = \nat^+$, then condition (b) may be omitted.
     \end{enumerate}
 \end{proposition}
 \begin{proof}

src/fa/tvs/continuous.tex

@@ -84,3 +84,41 @@
     \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{definition}[Uniform/Bounded Operator Topology]
+\label{definition:uniform-op-topo}
+    Let $E, F$ be TVSs over $K \in \RC$. For each bounded set $B \subset E$ and entourage $V \subset F \times F$, let
+    \[
+    U(B, V) = \bracs{(S, T) \in L(E; F) \times L(E; F)|(Sx, Tx) \in V \forall x \in B}
+    \]
+    then
+    \[
+    \fB = \bracs{U(B, V)| B \subset E \text{ bounded}, U \subset F \times F \text{ entourage}}
+    \]
+    is a fundamental system of entourages for a vector space uniformity on $L(E; F)$, which induces the \textbf{uniform/bounded operator topology} on $L(E; F)$.
+\end{definition}
+\begin{proof}
+    Firstly,
+    \begin{enumerate}
+        \item[(FB1)] Let $U(B, V), U(B', V') \in \fB$, then $U(B, V) \cap U(B', V') \supset U(B \cup B', V \cap V') \in \fB$.
+        \item[(UB1)] For any $U(B, V) \in \fB$, the diagonal is contained in $V$, so the diagonal is also contained in $U(B, V)$.
+        \item[(UB2)] For any $U(B, V) \in \fB$, there exists an entourage $W \subset F \times F$ such that $W \circ W \subset V$. So $U(B, W) \in \fB$ with
+        \[
+        U(B, W) \circ U(B, W) \subset U(B, W \circ W) \subset U(B, V)
+        \]
+    \end{enumerate}
+    By \ref{proposition:fundamental-entourage-criterion}, $\fB$ forms a fundamental system of entourages for a uniformity on $L(E; F)$.
+
+    TODO SCHAEFER WOLFF PAGE 79
+\end{proof}
+
+
+\begin{definition}[Strong Operator Topology]
+\label{definition:strong-op-topo}
+    Let $E, F$ be TVSs over $K \in \RC$, then the \textbf{strong operator topology} on $L(E; F)$ is the initial topology generated by the maps $\bracs{T \mapsto Tx| x \in E}$, where $F$ is equipped with the strong topology.
+\end{definition}
+
+\begin{definition}[Weak Operator Topology]
+\label{definition:weak-op-topo}
+    Let $E, F$ be TVSs over $K \in \RC$, then the \textbf{weak operator topology} on $L(E; F)$ is the initial topology generated by the maps $\bracs{T \mapsto Tx| x \in E}$, where $F$ is equipped with the weak topology.
+\end{definition}

src/fa/tvs/dual.tex

@@ -0,0 +1,34 @@
+\section{The Dual Space}
+\label{section:tvs-dual}
+
+\begin{proposition}[Polarisation, {{\cite[Proposition 5.5]{Folland}}}]
+\label{proposition:polarisation-linear}
+    Let $E$ be a vector space over $\complex$, $\phi \in \hom(E; \complex)$, and $u = \re{\phi}$, then
+    \begin{enumerate}
+        \item $u \in \hom(E; \real)$ when $E$ is viewed as a vector space over $\real$.
+        \item For any $x \in E$, $\dpb{x, \phi}{E} = \dpb{x, u}{E} - i \dpb{ix, u}{E}$.
+    \end{enumerate}
+    Conversely, if $u \in \hom(E; \real)$ and $\phi \in \hom(E; \complex)$ is defined by $\dpb{x, \phi}{E} = \dpb{x, u}{E} - i \dpb{ix, u}{E}$ for all $x \in E$, then $f \in \hom(E; \complex)$.
+\end{proposition}
+\begin{proof}
+    (1): Given that $\phi \in \hom(E; \real)$, $u \in \hom(E; \real)$. For any $x \in E$,
+    \[
+    \im{\dpb{x, \phi}{E}} = \re{-i \dpb{x, \phi}{E}} = -\dpb{ix, u}{E}
+    \]
+    so (2) holds.
+
+    (2): Since $u \in \hom(E; \real)$, so is $\phi$. On the other hand, for any $x \in E$,
+    \[
+    \dpb{ix, \phi}{E} = \dpb{ix, u}{E} - i\dpb{-x, u}{E} = i(\dpb{x, u}{E} - i \dpb{ix, u}{E}) = i \dpb{x, \phi}{E}
+    \]
+\end{proof}
+
+\begin{definition}[Topological Dual]
+\label{definition:topological-dual}
+    Let $E$ be a TVS over $K \in \RC$, then the \textbf{topological dual} $E^*$ of $E$ is the set of all \textit{continous} linear functionals on $E$.
+\end{definition}
+
+\begin{definition}[Weak Topology]
+\label{definition:weak-topology}
+    Let $E$ be a TVS over $K \in \RC$, then the \textbf{weak topology} on $E$ is the initial topology generated by $E^*$. The space $E$ equipped with its weak topology is denoted $E_w$.
+\end{definition}

src/fa/tvs/index.tex

@@ -3,4 +3,6 @@
 
 \input{./src/fa/tvs/definition.tex}
 \input{./src/fa/tvs/bounded.tex}
+\input{./src/fa/tvs/dual.tex}
 \input{./src/fa/tvs/continuous.tex}
+\input{./src/fa/tvs/completion.tex}