Added nuclear operators.

.gitignore

@@ -14,3 +14,5 @@ codebook.toml
 *.out
 *.pdf
 spec.db
+spec.db-shm
+spec.db-wal

.vscode/project.code-snippets

@@ -176,5 +176,10 @@
   "scope": "latex",
     "prefix": "cproof",
     "body": ["[Proof, {{\\cite[$1]{$2}}}. ]$0"]
+  },
+  "Scaffold": {
+    "scope": "latex",
+    "prefix": "scaf", 
+    "body": ["\\hyperref[scaffolded]{definition:measure-scaffold}$0"]
   }
 }

document.tex

@@ -4,7 +4,11 @@
 
 \begin{document}
 
-Hello this is all my notes. 
+Hi, welcome to my digital garden, where I collect math results that I learn. 
+
+Despite the contents being presented in a linear order by the table of contents, I will frequently reference things between chapters and sections. 
+
+Occasionally, I make up some definitions to play with. These definition blocks will always have a * at the end of its title to indicate that it lives mostly in my head. These terms will always be referenced with a link to their definition block. 
 
 \input{./src/cat/index}
 \input{./src/topology/index}

preamble.sty

@@ -37,6 +37,7 @@
 \newcommand{\parens}[1]{\left(#1\right)}
 \newcommand{\bracs}[1]{\left\{#1\right\}}
 \newcommand{\braks}[1]{\left[#1\right]}
+\newcommand{\braksn}[1]{[#1]}
 \newcommand{\angles}[1]{\left\langle#1\right\rangle}
 \newcommand{\abs}[1]{\left|#1\right|}
 \newcommand{\bracsn}[1]{\{{#1}\}}

refs.bib

@@ -214,4 +214,47 @@
   pages   = {263--269},
   year    = {1971},
   doi     = {10.1007/BF02771592}
+}
+
+@misc {StackRadonDual,
+    title = {How to understand C(X)'' = bounded Borel measurable functions?},
+    author = {Edgar, G.A.},
+    howpublished = {Mathematics Stack Exchange},
+    note = {URL:https://math.stackexchange.com/q/392719 (version: 2013-05-15)},
+    eprint = {https://math.stackexchange.com/q/392719},
+    url = {https://math.stackexchange.com/q/392719}
+}
+
+@book{Clarke,
+  author    = {Clarke, Francis},
+  title     = {Functional Analysis, Calculus of Variations and Optimal Control},
+  series    = {Graduate Texts in Mathematics},
+  volume    = {264},
+  publisher = {Springer},
+  address   = {London},
+  year      = {2013},
+  doi       = {10.1007/978-1-4471-4820-3},
+  isbn      = {978-1-4471-4820-3}
+}
+
+@book{CohnMeasure,
+  author    = {Cohn, Donald L.},
+  title     = {Measure Theory},
+  edition   = {2nd},
+  series    = {Birkhäuser Advanced Texts Basler Lehrbücher},
+  publisher = {Birkhäuser},
+  address   = {New York},
+  year      = {2013},
+  isbn      = {978-1-4614-6955-1},
+  doi       = {10.1007/978-1-4614-6956-8}
+}
+
+@book{Munkres,
+  author    = {Munkres, James R.},
+  title     = {Topology},
+  edition   = {2nd},
+  publisher = {Prentice Hall},
+  address   = {Upper Saddle River, NJ},
+  year      = {2000},
+  isbn      = {0-13-181629-2}
 }

src/cat/gluing/gluing.tex

@@ -0,0 +1,47 @@
+\section{Gluing Lemmas}
+\label{section:gluing}
+
+
+
+\begin{lemma}[Gluing for Functions]
+\label{lemma:glue-function}
+    Let $X, Y$ be sets, $\seq{U_i} \subset 2^X$, and $\seqi{f}$ with $f_i: U_i \to Y$ for all $i \in I$. If:
+    \begin{enumerate}
+        \item[(a)] $\bigcup_{i \in I}U_i = X$.
+        \item[(b)] For each $i, j \in I$, either $U_i \cap U_j = \emptyset$, or $f_i|_{U_i \cap U_j} = f_j|_{U_i \cap U_j}$.
+    \end{enumerate}
+
+    then there exists a unique $f: X \to Y$ such that $f|_{U_i} = f_i$ for all $i \in I$.
+\end{lemma}
+\begin{proof}
+    For each $i \in I$, let $\Gamma_i \subset U_i \times Y$ be the graph of $f_i$. Let $\Gamma = \bigcup_{i \in I}\Gamma_i$, then:
+    \begin{enumerate}
+        \item By assumption (a), $\bracs{x|(x, y) \in \Gamma} = \bigcup_{i \in I}U_i = X$.
+        \item For any $x \in X$, there exists $y \in Y$ with $(x, y) \in \Gamma$, and $i \in I$ such that $(x, y) \in \Gamma_i$. If $(x, y') \in \Gamma_j \subset \Gamma$, then $x \in U_i \cap U_j \ne \emptyset$. By assumption (b), $y = y'$.
+    \end{enumerate}
+
+    Thus $\Gamma$ is the graph of a function $f: X \to Y$ with $f|_{U_i} = f_i$ for all $i \in I$.
+\end{proof}
+
+
+\begin{lemma}[Gluing for Linear Functions]
+\label{lemma:glue-linear}
+    Let $E, F$ be vector spaces over a field $K$, $\fF$ be a family of subspaces of $E$, and $\bracs{T_V}_{V \in \fF}$ with $T_V \in \hom(V; F)$ for all $V \in \fF$. If:
+    \begin{enumerate}
+        \item[(a)] $\bigcup_{V \in \fF}V = E$.
+        \item[(b)] For each $V, W \in \fF$, $T_V|_{V \cap W} = T_W|_{V \cap W}$.
+        \item[(c)] $\fF$ is upward-directed with respect to includion.
+    \end{enumerate}
+
+    then there exists a unique $T \in \hom(E; F)$ such that $T|_{V} = T_V$ for all $V \in \fF$.
+\end{lemma}
+\begin{proof}
+    By (a), (b), and \autoref{lemma:glue-function}, there exists a unique $T: E \to F$ such that $T|_V = T_V$ for all $V \in \fF$.
+
+    Let $x, y \in E$ and $\lambda \in \fF$. By assumption (a), there exists $V_x, V_y \in \fF$ with $x \in V_x$ and $y \in V_y$. By assumption (3), there exists $V \in \fF$ with $V \supset V_x \cup V_y$. Hence
+    \[
+    T(\lambda x + y) = T_V(\lambda x + y) = \lambda T_Vx + T_Vy = \lambda Tx + Ty
+    \]
+
+    and $T \in \hom(E; F)$.
+\end{proof}

src/cat/gluing/index.tex

@@ -1,46 +1,8 @@
-\chapter{Gluing Lemmas}
+\chapter{Functions}
 \label{chap:gluing}
 
-
+The following chapter contains certain tricks in working with functions in an abstract setting. 
-\begin{lemma}[Gluing for Functions]
-\label{lemma:glue-function}
-    Let $X, Y$ be sets, $\seq{U_i} \subset 2^X$, and $\seqi{f}$ with $f_i: U_i \to Y$ for all $i \in I$. If:
-    \begin{enumerate}
-        \item[(a)] $\bigcup_{i \in I}U_i = X$.
-        \item[(b)] For each $i, j \in I$, either $U_i \cap U_j = \emptyset$, or $f_i|_{U_i \cap U_j} = f_j|_{U_i \cap U_j}$.
-    \end{enumerate}
-
-    then there exists a unique $f: X \to Y$ such that $f|_{U_i} = f_i$ for all $i \in I$.
-\end{lemma}
-\begin{proof}
-    For each $i \in I$, let $\Gamma_i \subset U_i \times Y$ be the graph of $f_i$. Let $\Gamma = \bigcup_{i \in I}\Gamma_i$, then:
-    \begin{enumerate}
-        \item By assumption (a), $\bracs{x|(x, y) \in \Gamma} = \bigcup_{i \in I}U_i = X$.
-        \item For any $x \in X$, there exists $y \in Y$ with $(x, y) \in \Gamma$, and $i \in I$ such that $(x, y) \in \Gamma_i$. If $(x, y') \in \Gamma_j \subset \Gamma$, then $x \in U_i \cap U_j \ne \emptyset$. By assumption (b), $y = y'$.
-    \end{enumerate}
-
-    Thus $\Gamma$ is the graph of a function $f: X \to Y$ with $f|_{U_i} = f_i$ for all $i \in I$.
-\end{proof}
 
 
-\begin{lemma}[Gluing for Linear Functions]
+\input{./gluing.tex}
-\label{lemma:glue-linear}
+\input{./level.tex}
-    Let $E, F$ be vector spaces over a field $K$, $\fF$ be a family of subspaces of $E$, and $\bracs{T_V}_{V \in \fF}$ with $T_V \in \hom(V; F)$ for all $V \in \fF$. If:
-    \begin{enumerate}
-        \item[(a)] $\bigcup_{V \in \fF}V = E$.
-        \item[(b)] For each $V, W \in \fF$, $T_V|_{V \cap W} = T_W|_{V \cap W}$.
-        \item[(c)] $\fF$ is upward-directed with respect to includion.
-    \end{enumerate}
-
-    then there exists a unique $T \in \hom(E; F)$ such that $T|_{V} = T_V$ for all $V \in \fF$.
-\end{lemma}
-\begin{proof}
-    By (a), (b), and \autoref{lemma:glue-function}, there exists a unique $T: E \to F$ such that $T|_V = T_V$ for all $V \in \fF$.
-
-    Let $x, y \in E$ and $\lambda \in \fF$. By assumption (a), there exists $V_x, V_y \in \fF$ with $x \in V_x$ and $y \in V_y$. By assumption (3), there exists $V \in \fF$ with $V \supset V_x \cup V_y$. Hence
-    \[
-    T(\lambda x + y) = T_V(\lambda x + y) = \lambda T_Vx + T_Vy = \lambda Tx + Ty
-    \]
-
-    and $T \in \hom(E; F)$.
-\end{proof}

src/cat/gluing/level.tex

@@ -0,0 +1,42 @@
+\section{Preimages}
+\label{section:preimage}
+
+\begin{definition}[Preimage Function*]
+\label{definition:preimage-function}
+    Let $X, Y$ be sets and $P: 2^Y \to 2^X$, then $P$ is a \textbf{preimage function} if
+    \begin{enumerate}[label=(PF\arabic*)]
+        \item $P(\emptyset) = \emptyset$. 
+        \item For each $\mathcal{S} \subset 2^Y$, $\bigcup_{S \in \mathcal{S}}P(S) = P\paren{\bigcup_{S \in \mathcal{S}}S}$. 
+        \item For each $\mathcal{S} \subset 2^Y$, $\bigcap_{S \in \mathcal{S}}P(S) = P\paren{\bigcap_{S \in \mathcal{S}}}$. 
+    \end{enumerate}
+
+    and $P$ is \textbf{total} if
+    \begin{enumerate}
+        \item[(T)] $P(Y) = X$. 
+    \end{enumerate}
+\end{definition}
+
+\begin{proposition}
+\label{proposition:preimage-gymnastics}
+    Let $X$ and $Y$ be sets, then:
+    \begin{enumerate}
+        \item For any $f: X \to Y$, the mapping $S \mapsto f^{-1}(S)$ is a total \hyperref[preimage function]{definition:preimage-function}.
+        \item For any total preimage function $P: 2^Y \to 2^X$, there exists a unique $f: X \to Y$ such that $P(S) = f^{-1}(S)$ for all $S \in 2^Y$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (2): Let $x, y \in Y$ with $x \ne y$, then by (PF1) and (PF3), 
+    \[
+    P(\bracs{x}) \cap P(\bracs{y}) = P(\bracs{x} \cap \bracs{y}) = P(\emptyset) = \emptyset
+    \]
+
+    By (T) and (PF2), $X = P(Y) = P\paren{\bigcup_{y \in Y}\bracs{y}} = \bigcup_{y \in Y}P(\bracs{y})$. Therefore $X = \bigsqcup_{y \in Y}P(\bracs{y})$. 
+
+    Therefore for each $x \in X$, there exists a unique $f(x) \in Y$ such that $x \in P(f(x))$. The association $x \mapsto f(x)$ then is the unique desired function. 
+\end{proof}
+
+
+
+
+
+

src/cat/tricks/product.tex

@@ -26,3 +26,34 @@
     \]
 
 \end{proof}
+
+\begin{lemma}
+\label{lemma:power-difference}
+    Let $p \in [1, \infty)$, then for each $a, b \ge 0$, 
+    \[
+    |a^p - b^p| \le p|a - b|(a \vee b)^{p - 1}
+    \]
+\end{lemma}
+\begin{proof}
+    Assume without loss of generality that $0 < a < b$, then
+    \[
+        b^p - a^p = p\int_{a}^b t^{p - 1}dt \le p|a - b|(a \vee b)^{p - 1}
+    \]
+\end{proof}
+
+\begin{lemma}
+\label{lemma:power-sum}
+    Let $p \in [1, \infty)$, then for each $a, b \ge 0$, 
+    \[
+    (a + b)^p \le 2^{p-1}(a^p + b^p)
+    \]
+\end{lemma}
+\begin{proof}
+    Since $t \mapsto t^p$ is convex, 
+    \begin{align*}
+        \braks{\frac{a + b}{2}}^p &\le \frac{a^p}{2} + \frac{b^p}{2} \\
+        \frac{(a+b)^p}{2^p} &\le \frac{a^p}{2} + \frac{b^p}{2} \\
+        (a + b)^p &\le 2^{-1}(a^p + b^p)
+    \end{align*}
+    
+\end{proof}

src/dg/complex/entire.tex

@@ -74,7 +74,7 @@
 
     then $f = 1$. 
 \end{proposition}
-\begin{proof}[Proof, {{\cite[Lemma I.4.4]{Zhu}}}. ]
+\begin{proof}[Proof, {{\cite[Lemma 4.4]{Zhu}}}. ]
     By (c) and \autoref{proposition:entire-logarithm}, there exists $g \in H(\complex; \complex)$ such that $f = e^g$. Since $f(0) = 1$, $g(0) = 0$. As $f'(0) = g'(0)f(0) = 0$, $g'(0) = 0$ as well. 
 
     From (c), $|g(z)| \le e^{|z|}$ for each $z \in \complex$. Thus for every $r > 0$ and $z \in B_\complex(0, r)$, $|g(z)| \le |g(z) - 2r|$, and

src/dg/complex/runge.tex

@@ -16,7 +16,7 @@
     \end{enumerate}
 \end{proposition}
 \begin{proof}[Proof, {{\cite[Proposition VIII.1.1]{ConwayComplex}}}. ]
-    Via \hyperref[fattening]{proposition:distance-compact}, let $V \in \cn_\complex^o(K)$ precompact with $\ol V \subset U$. Identify $\complex = \real^2$, then since $V$ is precompact, there exists $\delta > 0$ and $\seqf{(x_j, y_j)} \subset U$ such that:
+    Via \hyperref[fattening]{proposition:distance-compact}, let $V \in \cn_\complex^o(K)$ relatively compact with $\ol V \subset U$. Identify $\complex = \real^2$, then since $V$ is relatively compact, there exists $\delta > 0$ and $\seqf{(x_j, y_j)} \subset U$ such that:
     \begin{enumerate}
         \item For each $1 \le j \le n$, $R_j = [x_j, x_j + \delta] \times [y_j, y_j + \delta] \subset U$. 
         \item $\bigcup_{j = 1}^n R_j \supset V$. 

src/dg/complex/space.tex

@@ -24,8 +24,8 @@
 
     and the following are equivalent:
     \begin{enumerate}[label=(C\arabic*)]
-        \item $\cf$ is precompact in $H(U; E)$. 
+        \item $\cf$ is relatively compact in $H(U; E)$. 
-        \item $\cf$ is bounded in $H(U; E)$, and for each $x \in U$, $\cf(x) = \bracs{f(x)|f \in \cf}$ is precompact.  
+        \item $\cf$ is bounded in $H(U; E)$, and for each $x \in U$, $\cf(x) = \bracs{f(x)|f \in \cf}$ is relatively compact.  
     \end{enumerate}
 \end{theorem}
 \begin{proof}

src/dg/complex/zero.tex

@@ -96,6 +96,35 @@
 \end{proof}
 
 
+
+\begin{lemma}
+\label{lemma:maximum-modulus-strip}
+    Let $S = \bracs{z \in \complex| \text{Re}(z) \in (0, 1)}$, $E$ be a Banach space over $\complex$, and $f \in H(S; E) \cap BC(\ol S; E)$, then
+    \[
+    \norm{f}_{u} = \sup_{z \in \partial S}\norm{f(z)}_E
+    \]
+\end{lemma}
+\begin{proof}
+    Let $\eps > 0$ and $\phi_\eps(z) = e^{\eps(z^2 - 1)}$, then $\phi f \in H(S; E) \cap BC(\ol S; E)$. 
+    
+    For each $R > 0$, let $S_R = \bracs{z \in \complex| \text{Re}(z) \in (0, 1), |\text{Im}(z)| < R}$, then by the \hyperref[Maximum Modulus Theorem]{theorem:maximum-modulus-theorem}, 
+    \[
+    \norm{\phi_\eps f}_u = \lim_{R \to \infty} \sup_{z \in \partial S_R}\norm{\phi_\eps f(z)}_E
+    \]
+
+    However, since $\phi_\eps f(z) \to 0$ as $|\text{Im}(z)| \to \infty$, 
+    \[
+    \norm{\phi_\eps f}_u = \lim_{R \to \infty} \sup_{z \in \partial S_R}\norm{\phi_\eps f(z)}_E =  \sup_{z \in \partial S}\norm{\phi_\eps f(z)}_E
+    \]
+
+    Therefore
+    \[
+    \norm{f}_u = \sup_{\eps > 0} \norm{\phi_\eps f}_u = \sup_{\eps > 0} \sup_{z \in \partial S}\norm{\phi_\eps f(z)}_E = \sup_{z \in \partial S}\norm{f(z)}_E
+    
+    \]
+\end{proof}
+
+
 \begin{lemma}[Hadamard's Three Lines Lemma]
 \label{lemma:three-lines}
     Let $S = \bracs{z \in \complex| \text{Re}(z) \in [0, 1]}$, $E$ be a Banach space over $\complex$, and $f \in H(S; E) \cap BC(\ol{S}; E)$. For each $s \in [0, 1]$, let
@@ -121,7 +150,7 @@
     h: \ol S \to E \quad z \mapsto \frac{f(z)}{g(z)}
     \]
 
-    then $h \in H(S; E) \cap BC(\ol S; E)$ with $\norm{h(z)}_E \le 1$ for all $z \in \partial S$. By the \hyperref[Maximum Modulus Theorem]{theorem:maximum-modulus-theorem}, $\norm{h(z)}_E \le 1$ for all $z \in S$. Thus for every $z \in S$,
+    then $h \in H(S; E) \cap BC(\ol S; E)$ with $\norm{h(z)}_E \le 1$ for all $z \in \partial S$. By the \hyperref[Maximum Modulus Theorem]{lemma:maximum-modulus-strip}, $\norm{h(z)}_E \le 1$ for all $z \in S$. Thus for every $z \in S$,
     \[
     f(z) \le M(0)^{\text{Re}(z)} M(1)^{1-\text{Re}(z)}
     \]

src/dg/derivative/sets.tex

@@ -43,7 +43,7 @@
 
 \begin{definition}
 \label{definition:derivative-garden}
-    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma^E_{\text{Fin}}, \sigma^E_{c}, \sigma^E_b \subset 2^E$ be the collection of all finite, precompact, and bounded subsets, respectively, then differentiability with respect to $\sigma^E_{\text{Fin}}, \sigma^E_{c}, \sigma^E_b$ correspond to \textbf{Gateaux}, \textbf{Hadamard}, and \textbf{Fréchet} differentiability.
+    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma^E_{\text{Fin}}, \sigma^E_{c}, \sigma^E_b \subset 2^E$ be the collection of all finite, relatively compact, and bounded subsets, respectively, then differentiability with respect to $\sigma^E_{\text{Fin}}, \sigma^E_{c}, \sigma^E_b$ correspond to \textbf{Gateaux}, \textbf{Hadamard}, and \textbf{Fréchet} differentiability.
 \end{definition}
 
 
@@ -84,7 +84,7 @@
 \label{proposition:chain-rule-sets-conditions}
     Let $E$, $F$, $G$, be TVSs over $K \in \RC$ with $F, G$ being separated. If $\sigma \subset \mathfrak{B}(E)$ and $\tau \subset \mathfrak{B}(F)$ correspond to the following families of sets on $E$ and $F$:
     \begin{enumerate}
-        \item Precompact sets.
+        \item Relatively compact sets.
         \item Bounded sets.
     \end{enumerate}
 

src/fa/convex/def.tex

@@ -0,0 +1,109 @@
+\section{Convexity}
+\label{section:convex-functions}
+
+\begin{definition}[Epigraph]
+\label{definition:epigraph}
+    Let $E$ be a vector space over $\real$ and $f: E \to (-\infty, \infty]$, then the \textbf{epigraph} of $f$ is the set
+    \[
+    \text{epi}(f) = \bracs{(x, y) \in E \times \real| y \ge f(x)}
+    \]
+\end{definition}
+
+
+\begin{definition}[Convex Function]
+\label{definition:convex-function}
+    Let $E$ be a vector space over $\real$ and $f: E \to (-\infty, \infty]$, then the following are equivalent:
+    \begin{enumerate}
+        \item For every $x, y \in E$ and $t \in [0, 1]$, 
+        \[
+        f((1 - t)x + ty) \le (1 - t)f(x) + tf(y)
+        \]
+        \item $\text{epi}(f)$ is convex. 
+    \end{enumerate}
+    
+    If the above holds, then $f$ is \textbf{convex}. 
+\end{definition}
+\begin{proof}
+    (1) $\Rightarrow$ (2): Let $(x, \alpha), (y, \beta) \in \text{epi}(f)$ and $t \in [0, 1]$, then
+    \begin{align*}
+        (1 - t)\alpha + t\beta &\ge (1 - t)f(x) + tf(y) \\
+        &\ge f((1 - t)x + ty)
+    \end{align*}
+
+    so $((1 - t)x + ty, (1 - t)\alpha + t\beta) \in \text{epi}(f)$. 
+
+    (2) $\Rightarrow$ (1): Let $x, y \in E$ and $t \in [0, 1]$, then
+    \[
+    ((1 - t)x + ty, (1 - t)f(x) + tf(y)) \in \text{epi}(f)
+    \]
+
+    so $f((1 - t)x + ty) \le (1 - t)f(x) + tf(y)$. 
+    
+\end{proof}
+
+\begin{lemma}
+\label{lemma:convex-domain}
+    Let $E$ be a vector space over $\real$ and $f: E \to (-\infty, \infty]$, then $f$ is convex if and only if $\bracs{f < \infty}$ is convex and $f|_{\bracs{f < \infty}}$ is a convex function. 
+\end{lemma}
+\begin{proof}
+    If $f$ is convex, then for any $x, y \in \bracs{f < \infty}$ and $t \in [0, 1]$, 
+    \[
+    f((1 - t)x + ty) \le (1 - t)f(x) + tf(y) < \infty
+    \]
+
+    so $\bracs{f < \infty}$ is convex, and the restriction $f|_{\bracs{f < \infty}}$ is a convex function. 
+
+    On the other hand, for any $x, y \in E$ and $t \in [0, 1]$, if $f(x) = \infty$ or $f(y) = \infty$, then
+    \[
+    f((1 - t)x + ty) \le = \infty (1 - t)f(x) + tf(y)
+    \]
+
+    Otherwise, $x, y \in \bracs{f < \infty}$, and
+    \[
+    f((1 - t)x + ty) \le (1 - t)f(x) + tf(y)
+    \]
+
+    by convexity of $f|_{\bracs{f < \infty}}$. 
+\end{proof}
+
+\begin{lemma}
+\label{lemma:convex-reverse}
+    Let $E$ be a vector space over $\real$ and $f: E \to (-\infty, \infty]$ be convex, then for any $x, y \in E$ and $t \in \real \setminus [0, 1]$, 
+    \[
+    f((1 - t)x + ty) \ge (1 - t)f(x) + tf(y)
+    \]
+\end{lemma}
+\begin{proof}
+    Via an affine transformation, assume without loss of generality that $x = 0$ and $f(x) = 0$. By exchanging $x$ and $y$, assume without loss of generality that $t > 1$. In which case, since $f$ is convex and $y = t^{-1}ty$, $f(y) \le t^{-1}f(ty)$ and $tf(y) \le f(ty)$. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:convex-differential}
+    Let $E$ be a vector space over $\real$, $f: E \to (-\infty, \infty]$ be convex, and $x \in \bracs{f < \infty}$, then for any $h \in E$,
+    \[
+    \lim_{t \downto 0} \frac{f(x + th) - f(x)}{t} = \inf_{t > 0} \frac{f(x + th) - f(x)}{t}
+    \]
+
+    exists in $[-\infty, \infty]$. 
+\end{proposition}
+\begin{proof}
+    Let $0 < s \le t$, then since $f$ is convex, 
+    \begin{align*}
+        f(x + sh) &\le \paren{1 - \frac{s}{t}}f(x) + \frac{s}{t}f(x + th) \\
+        f(x + sh) - f(x) &\le \frac{s}{t}[f(x + th) - f(x)] \\
+        \frac{f(x + sh) - f(x)}{s} &\le \frac{f(x + th) - f(x)}{t}
+    \end{align*}
+\end{proof}
+
+
+\begin{proposition}
+\label{proposition:convex-extension}
+    Let $E$ be a vector space over $\real$, then:
+    \begin{enumerate}
+        \item For any $f, g: E \to (-\infty, \infty]$ convex and $\lambda \ge 0$, $\lambda f + g$ is convex. 
+        \item For any convex functions $\cf \subset (-\infty, \infty]^E$, $\sup_{f \in \cf}f$ is convex. 
+    \end{enumerate}
+\end{proposition}
+
+
+

src/fa/convex/del.tex

@@ -0,0 +1,59 @@
+\section{Subgradients}
+\label{section:subgradient}
+
+\begin{definition}[Subdifferential]
+\label{definition:subgradient}
+    Let $E$ be a locally convex space over $\real$, $f: E \to (-\infty, \infty]$, $x \in \bracs{f < \infty}$, and $\phi \in E^*$, then $\phi$ is a \textbf{subgradient} of $f$ if for any $h \in E$, 
+    \[
+    f(x + h) \ge f(x) + \dpn{h, \phi}{E}
+    \]
+
+    The set $\partial f(x)$ of all subgradients of $f$ at $x$ is the \textbf{subdifferential of $f$ at $x$}, and the mapping $\partial f$ is the \textbf{subdifferential} of $f$.  
+\end{definition}
+
+\begin{proposition}
+\label{proposition:subgradient-gymnastics}
+    Let $E$ be a locally convex space over $\real$, $f, g: E \to (-\infty, \infty]$, and $x \in \bracs{f, g < \infty}$, then:
+    \begin{enumerate}
+        \item $\partial f(x)$ is a weak*-closed convex set. 
+        \item $\partial(f + g)(x) \supset \partial f(x) + \partial g(x)$. 
+    \end{enumerate}
+\end{proposition}
+% Proof omitted. 
+
+\begin{proposition}
+\label{proposition:subgradient-existence}
+    Let $E$ be a locally convex space over $\real$, $f: E \to (-\infty, \infty]$ be convex, and $x \in \bracs{f < \infty}$, then:
+    \begin{enumerate}
+        \item For any $\phi \in \partial f(x)$ and $h \in E$, 
+        \[
+        \dpn{h, \phi}{E} \le |f(x + h) - f(x)|
+        \]
+        \item If $f$ is continuous at $x$, then $\partial f(x)$ is non-empty, equicontinuous, and weak*-compact. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition 4.6]{Clarke}}}. ]
+    (1): By the subgradient inequality, 
+    \[
+    \dpn{h, \phi}{E} \le f(x + h) - f(x) \le |f(x + h) - f(x)|
+    \]
+
+    (2): Since $f$ is continuous at $x$, $\text{epi}(f)^o \ne \emptyset$. Since $(x, f(x)) \not\in \text{epi}(f)^o$, by the \hyperref[Hahn-Banach Theorem]{theorem:hahn-banach-geometric-1}, there exists $\phi \in E^*$ and $\lambda \in \real$ such that for any $(y, \alpha) \in \text{epi}(f)^o$, 
+    \[
+    \dpn{y, \phi}{E} + \lambda \alpha < \dpn{x, \phi}{E} + \lambda f(x)
+    \]
+
+    By continuity of $f$ at $x$, there exists $\alpha_0 \in \real$ such that $\bracs{x} \times [\alpha_0, \infty) \subset \text{epi}(f)^o$. Thus $\lambda < 0$. 
+
+    By \autoref{proposition:convex-interior-closure}, $\text{epi}(f) \subset \ol{\text{epi}(f)^o}$. Therefore for any $(y, \alpha) \in \text{epi}(f)$, 
+    \[
+    \lambda^{-1}\dpn{y, \phi}{E} - \alpha \le \dpn{x, \phi}{E} - f(x)
+    \]
+
+    so $\phi \in \partial f(x) \ne \emptyset$.  
+
+    (3): By \autoref{proposition:subgradient-gymnastics} and the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu}. 
+\end{proof}
+
+
+

src/fa/convex/index.tex

@@ -0,0 +1,7 @@
+\chapter{Convex Functions}
+\label{chap:convex-functions}
+
+\input{./def.tex}
+\input{./del.tex}
+\input{./legendre.tex}
+\input{./support.tex}

src/fa/convex/legendre.tex

@@ -0,0 +1,294 @@
+\section{Conjugate Functions}
+\label{section:legendre}
+
+\begin{definition}[Affine Minorant]
+\label{definition:affine-minorant}
+    Let $\dpn{E, F}{\lambda}$ be a duality over $\real$, $f: E \to (-\infty, \infty]$, and $(\phi, \alpha) \in F \times \real$, then the pair $(\phi, \alpha)$ is an \textbf{affine minorant} of $f$, denoted $(\phi, \alpha) \le f$, if
+    \[
+    \dpn{x, \phi}{\lambda} - \alpha \le f(x) \quad \forall x \in E
+    \]
+\end{definition}
+
+
+\begin{definition}[Conjugate Function]
+\label{definition:conjugate-function}
+    Let $\dpn{E, F}{\lambda}$ be a duality over $\real$ and $f: E \to (-\infty, \infty]$ with $f \ne \infty$, then for each $\phi \in F$, 
+    \[
+    \sup_{x \in E}\dpn{x, \phi}{\lambda} - f(x) = \inf\bracsn{\alpha \in \real| (\phi, \alpha) \le f}
+    \]
+
+    The mapping
+    \[
+    f^*: F \to (-\infty, \infty] \quad \phi \mapsto \sup_{x \in E}\dpn{x, \phi}{\lambda} - f(x)
+    \]
+
+    is the \textbf{conjugate function} of $f$ with respect to the duality $\dpn{E, F}{\lambda}$. 
+\end{definition}
+\begin{proof}
+    Fix $\phi \in F$. Let $\alpha \in  \real$ such that $\dpn{x, \phi}{\lambda} - \alpha \le f(x)$ for all $x \in E$, then for any $x \in E$,  
+    \[
+    \dpn{x, \phi}{\lambda} - f(x) \le \dpn{x, \phi}{\lambda} - \dpn{x, \phi}{\lambda} + \alpha = \alpha
+    \]
+
+    so
+    \[
+    \sup_{x \in E}\dpn{x, \phi}{\lambda} - f(x) \le \inf\bracsn{\alpha \in \real| (\phi, \alpha) \le f}
+    \]
+
+    On the other hand, suppose that $\alpha = \sup_{x \in E} \dpn{x, \phi}{\lambda} - f(x) < \infty$, then
+    \[
+    \dpn{x, \phi}{\lambda} - \alpha \le f(x)
+    \]
+
+    for all $x \in E$. Therefore
+    \[
+    \sup_{x \in E}\dpn{x, \phi}{\lambda} - f(x) = \inf\bracsn{\alpha \in \real| (\phi, \alpha) \le f}
+    \]
+\end{proof}
+\begin{lemma}
+\label{lemma:conjugate-minorant}
+    Let $\dpn{E, F}{\lambda}$ be a duality over $\real$ and $f: E \to (-\infty, \infty]$, then the following are equivalent:
+    \begin{enumerate}
+        \item $f^* \ne \infty$. 
+        \item There exists $(\phi, \alpha) \in F \times \real$ with $(\phi, \alpha) \le f$. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    (1) $\Rightarrow$ (2): Let $\phi \in \bracsn{f^* < \infty}$, then by \autoref{definition:conjugate-function}, $(\phi, f^*(\phi)) \le f$. 
+
+    (2) $\Rightarrow$ (1): By \autoref{definition:conjugate-function}, $f^*(\phi) \le \alpha$.   
+\end{proof}
+
+
+\begin{lemma}
+\label{lemma:conjugate-function-gymnatics}
+    Let $\dpn{E, F}{\lambda}$ be a duality over $\real$ and $f, g: E \to (-\infty, \infty]$ with $f, g \ne \infty$, then
+    \begin{enumerate}
+        \item $f^*$ is convex and lower semicontinuous. 
+        \item If $f \le g$, then $f^* \ge g^*$. 
+        \item If $f^* \ne \infty$, then $f^{**} \le f$. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    (1): By \autoref{proposition:convex-extension} and \autoref{proposition:semicontinuous-properties}. 
+
+    (3): For each $x \in E$ and $\phi \in F$, 
+    \begin{align*}
+        \dpn{x, \phi}{\lambda} - f^*(\phi) &= \dpn{x, \phi}{\lambda} - \braks{\sup_{z \in E}\dpn{z, \phi}{\lambda} - f(z)} \\
+        &= \dpn{x, \phi}{\lambda} + \braks{\inf_{z \in E}f(z) - \dpn{z, \phi}{\lambda}} \\
+        &\le \dpn{x, \phi}{\lambda} + f(x) - \dpn{x, \phi}{\lambda} = f(x)
+    \end{align*}
+
+    As the above holds for all $\phi \in F$, $f^{**} \le f$.     
+\end{proof}
+
+\begin{theorem}[Fenchel's Inequality]
+\label{theorem:fenchel-inequality}
+    Let $\dpn{E, F}{\lambda}$ be a duality over $\real$ and $f: E \to (-\infty, \infty]$ with $f \ne \infty$, then for any $x \in E$ and $\phi \in F$, 
+    \[
+    \dpn{x, \phi}{\lambda} \le f(x) + f^*(\phi)
+    \]
+
+    with equality if and only if $\phi \in \partial f(x)$. 
+\end{theorem}
+\begin{proof}
+    Let $x \in E$ and $\phi \in F$. Assume without loss of generality that $f(x) < \infty$, then
+    \[
+        f(x) + f^*(\phi) \ge f(x) + \dpn{x, \phi}{\lambda} - f(x) = \dpn{x, \phi}{\lambda}
+    \]
+
+    Now let $x \in E$ and $\phi \in F$ such that $\dpn{x, \phi}{\lambda} = f(x) + f^*(\phi)$, then $f(x), f^*(\phi) < \infty$ and
+    \begin{align*}
+        f(x) + f^*(\phi) &= \dpn{x, \phi}{\lambda} \\
+        f^*(\phi) &= \dpn{x, \phi}{\lambda} - f(x) 
+    \end{align*}
+
+    By definition, 
+    \[
+    \dpn{x, \phi}{\lambda} - f(x)  = f^*(\phi) = \sup_{h \in E}\dpn{x+h, \phi}{\lambda} - f(x+h)
+    \]
+
+    so for every $h \in E$, 
+    \begin{align*}
+        \dpn{x, \phi}{\lambda} - f(x) &\ge \dpn{x+h, \phi}{\lambda} - f(x+h) \\
+        f(x + h) - f(x) &\ge \dpn{h, \phi}{\lambda} 
+    \end{align*}
+
+    Thus $\phi$ satisfies the subgradient inequality, and $\phi \in \partial f(x)$. 
+
+    On the other hand, if $f(x) < \infty$ and $\phi \in \partial f(x)$, then $f(x + h) - f(x) \ge \dpn{h, \phi}{\lambda}$ for all $h \in E$, and
+    \begin{align*}
+    f^*(\phi) &= \sup_{h \in E}\dpn{x+h, \phi}{\lambda} - f(x+h) = \dpn{x, \phi}{\lambda} - f(x) \\
+    f^*(\phi) + f(x) &=  \dpn{x, \phi}{\lambda}
+    \end{align*}
+    
+\end{proof}
+
+\begin{lemma}
+\label{lemma:closed-convex-epigraph}
+    Let $E$ be a locally convex space over $\real$ and $f: E \to (-\infty, \infty]$ with $f \ne \infty$, then for each $(x, \alpha) \in \ol{\text{Conv}}(\text{epi}(f))$, $\bracs{x} \times [\alpha, \infty) \subset \ol{\text{Conv}}(\text{epi}(f))$. 
+\end{lemma}
+\begin{proof}
+    Let
+    \[
+    A = \bracsn{(\phi, \alpha) \in \ol{\text{Conv}}(\text{epi}(f))| \bracs{x} \times [\alpha, \infty) \subset \ol{\text{Conv}}(\text{epi}(f))}
+    \]
+
+    For each $(x, \alpha), (y, \beta) \in A$, $t \in [0, 1]$, and $\gamma \ge (1 - t)\alpha + t\beta$, there exists $\alpha' \ge \alpha$ and $\beta' \ge \beta$ such that $\gamma = (1 - t)\alpha' + t\beta'$. In which case, 
+    \[
+    ((1 - t)x + ty, \gamma) = ((1 - t)x + ty, (1 - t)\alpha' + t\beta') \in \ol{\text{Conv}}(\text{epi}(f))
+    \]
+
+    so $\bracs{(1 - t)x + ty} \times [\gamma, \infty) \subset \ol{\text{Conv}}(\text{epi}(f))$, $(1 - t)(x + \alpha) + t(y, \beta) \in A$, and $A$ is convex. 
+
+    Let $(x, \alpha) \in \ol A$, then there exists a net $\langle (x_\gamma, \alpha_\gamma) \rangle_{\gamma \in C} \subset A$ with $(x_\gamma, \alpha_\gamma) \to (x, \alpha)$. In which case, for each $r > 0$, $\langle (x_\gamma, \alpha_\gamma + r) \rangle_{\gamma \in C} \subset A$ and $(x_\gamma, \alpha_\gamma + r) \to (x, \alpha + r)$, so $(x, \alpha + r) \in \ol{\text{Conv}}(\text{epi}(f))$ and
+    \[
+    \bracs{x} \times [\alpha, \infty) \subset \ol{\text{Conv}}(\text{epi}(f))
+    \]
+
+    Thus $(x, \alpha) \in A$ and $A$ is closed. 
+
+    Since $A$ is a closed convex set containing $\text{epi}(f)$, $A = \ol{\text{Conv}}(\text{epi}(f))$. 
+\end{proof}
+
+\begin{lemma}[Almost Subgradient]
+\label{lemma:lsc-affine-minorant}
+    Let $\dpn{E, F}{\lambda}$ be a duality over $\real$, $f: E \to (-\infty, \infty]$ be convex and $\sigma(E, F)$-lower semicontinuous, $x \in \bracs{f < \infty}$, and $\alpha < f(x)$, then there exists $(\phi, \gamma) \in E \times \real$ such that:
+    \begin{enumerate}
+        \item $(\phi, \gamma) \le f$. 
+        \item $\dpn{x, \phi}{\lambda} - \gamma = \alpha$. 
+    \end{enumerate}
+
+    In particular, $f^{*} \ne \infty$. 
+\end{lemma}
+\begin{proof}
+    (1): Since $f$ is convex and $\sigma(E, F)$-lower semicontinuous, $\text{epi}(f)$ is $\sigma(E \times \real, F \times \real)$-closed and convex. By the \hyperref[Hahn-Banach Theorem]{theorem:hahn-banach-geometric-2}, there exists $\phi \in F$ and $\mu \in \real$ such that
+    \[
+    \sup_{(y, \beta) \in \text{epi}(f)}\dpn{y, \phi}{E} - \mu \beta < \dpn{x, \phi}{E} - \mu \alpha
+    \]
+
+    For any $(y, \beta) \in \text{epi}(f)$, $\beta$ may be arbitrarily large, $\mu \ge 0$. In particular, since $(x, f(x)) \in \text{epi}(f)$, the strict inequality implies that $\mu > 0$. Thus for each $y \in \bracs{f < \infty}$, 
+    \begin{align*}
+        \dpn{x, \phi}{E} - \mu\alpha &> \dpn{y, \phi}{E} - \mu f(y) \\
+        -\dpn{y, \phi}{E} + \dpn{x, \phi}{E} - \mu\alpha &> -\mu f(y)\\
+        \dpn{y, \mu^{-1}\phi}{E} - \dpn{x, \mu^{-1}\phi}{E} + \alpha &< f(y)
+    \end{align*}
+
+    so $(\mu^{-1}\phi, \dpn{x, \mu^{-1}\phi}{E} - \alpha) \le f$ and
+    \[
+    \dpn{x, \mu^{-1}\phi}{E} - \dpn{x, \mu^{-1}\phi}{E} + \alpha = \alpha
+    \]
+\end{proof}
+
+
+\begin{theorem}[Fenchel-Moreau]
+\label{theorem:fenchel-moreau}
+    Let $\dpn{E, F}{\lambda}$ be a duality over $\real$, and $f: E \to (-\infty, \infty]$ with $f \ne \infty$ and $f^{*} \ne \infty$, then:
+    \begin{enumerate}
+        \item For each $x \in E$,  
+        \[
+        f^{**}(x) = \sup\bracs{\dpn{x, \phi}{\lambda} - \alpha|(\phi, \alpha) \in F \times \real, (\phi, \alpha) \le f}
+        \]
+        \item $\text{epi}(f^{**})$ is the $\sigma(E \times \real, F \times \real)$-closed convex hull of $\text{epi}(f)$. 
+        \item $f^{**}$ is the greatest convex and $\sigma(E, F)$-lower semicontinuous function bounded above by $f$. 
+        \item $f = f^{**}$ if and only if $f$ is convex and $\sigma(E, F)$-lower semicontinuous.  
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    (1): Let $\phi \in F$ such that $f^*(\phi) < \infty$, then for each $x \in E$, 
+    \[
+    \dpn{x, \phi}{\lambda} - f^*(\phi) \le f(x)
+    \]
+
+    so $\dpn{\cdot, \phi}{\lambda} - f^*(\phi) \le f$, and
+    \begin{align*}
+    f^{**}(x) &= \sup_{\phi \in F} \dpn{x, \phi}{\lambda} - f^*(\phi) = \sup_{\substack{\phi \in F \\ f^*(\phi) < \infty}} \dpn{x, y}{\lambda} - f^*(\phi) \\
+    &\le \sup\bracs{\dpn{x, \phi}{\lambda} - \alpha|(\phi, \alpha) \in F \times \real, (\phi, \alpha) \le f}
+    \end{align*}
+    
+    On the other hand, let $\phi \in F$ and $\alpha \in \real$ such that $(\phi, \alpha) \le f$, then $f^*(\phi) \le \alpha$, and
+    \[
+    f^{**}(x) \ge \dpn{x, \phi}{\lambda} - f^*(\phi) \ge \dpn{x, \phi}{\lambda} - \alpha
+    \]
+
+    Therefore
+    \[
+        f^{**}(x) \ge \sup\bracs{\dpn{x, \phi}{\lambda} - \alpha| \phi \in F, \alpha \in \real, \dpn{\cdot, y}{\lambda} - \alpha \le f}
+    \]
+
+    (2): By \autoref{lemma:conjugate-function-gymnatics}, $f^{**}$ is lower semicontinuous and convex with $f^{**} \le f$, so $\text{epi}(f^{**}) \supset \text{epi}(f)$ and $\text{epi}(f^{**}) \supset \ol{\text{Conv}}(\text{epi}(f))$. Thus it is sufficient to show that $\text{epi}(f^{**}) \subset \ol{\text{Conv}}(\text{epi}(f))$, or equivalently, 
+    \[
+    E \times \real \setminus \ol{\text{Conv}}(\text{epi}(f)) \subset E \times \real \setminus \text{epi}(f)
+    \]
+    
+    To this end, let $A = \ol{\text{Conv}}(\text{epi}(f))$ and $(x, \alpha) \in E \times \real \setminus A$. By the \hyperref[Hahn-Banach Theorem]{theorem:hahn-banach-geometric-2}, there exists $\phi \in F$, $\mu \in \real$, and $\alpha_0 \in (\alpha, \infty)$ such that
+    \[
+    \sup_{(y, \beta) \in A}\dpn{y, \phi}{\lambda} - \mu \beta \le \dpn{x, \phi}{\lambda} - \mu \alpha_0 \le \dpn{x, \phi}{\lambda} - \mu \alpha
+    \]
+
+    For any $(y, \beta) \in A$, $\beta$ may be arbitrarily large by \autoref{lemma:closed-convex-epigraph}, so $\mu \ge 0$. 
+
+    In the case that $\mu > 0$, for each $y \in \bracs{f < \infty}$, 
+    \begin{align*}
+        \dpn{x, \phi}{\lambda} - \mu\alpha_0 &\ge \dpn{y, \phi}{\lambda} - \mu f(y) \\
+        -\dpn{y, \phi}{\lambda} + \dpn{x, \phi}{\lambda} - \mu\alpha_0 &\le - \mu f(y) \\
+        \dpn{y, \mu^{-1}\phi}{\lambda} - \dpn{x, \mu^{-1}\phi}{\lambda} + \alpha_0 &\le f(y)
+    \end{align*}
+
+    so $(\mu^{-1}\phi, \dpn{x, \mu^{-1}\phi}{\lambda} - \alpha_0) \le f$ and
+    \[
+    f^{**}(x) \ge \dpn{x, \mu^{-1}\phi}{\lambda} - \dpn{x, \mu^{-1}\phi}{\lambda} + \alpha_0 > \alpha
+    \]
+
+    Now suppose that $\mu = 0$. Given that $f^* \ne \infty$, there exists at least one pair $(\phi_0, \gamma_0) \in F \times \real$ such that $(\phi_0, \gamma_0) \le f$. Let
+    \[
+    \gamma = \sup_{(y, \beta) \in A}\dpn{y, \phi}{\lambda} < \dpn{x, \phi}{\lambda}
+    \]
+
+    For each $t > 0$, let $\Phi_t = \phi_0 + t\phi$ and $\Gamma_t = \gamma_0 + t\gamma$, then for each $y \in \bracs{f < \infty}$, 
+    \[
+    \dpn{y, \Phi_t}{\lambda} - \Gamma_t \le f(y) + t\underbrace{(\dpn{y, \phi}{\lambda} - \gamma)}_{\le 0} \le f(y)
+    \]
+
+    so $(\Phi_t, \Gamma_t) \le f$. By (1), 
+    \[
+        f^{**}(x) \ge \dpn{x, \Phi_t}{\lambda} - \Gamma_t = \dpn{x, \phi_0}{\lambda} - \gamma_0  + t\underbrace{(\dpn{x, \phi}{\lambda} - \gamma)}_{> 0}
+    \]
+
+    As the above holds for all $t > 0$, $f^{**}(x) = \infty > \alpha$. 
+    
+    Thus $f^{**}(x) > \alpha$ and $(x, \alpha) \not\in \text{epi}(f^{**})$ for all $(x, \alpha) \in E \times \real \setminus A$. Therefore
+    \[
+    E \times \real \setminus A \subset E \times \real \setminus \text{epi}(f^{**})
+    \]
+    
+    and $\text{epi}(f^{**}) \subset \ol{\text{Conv}}(\text{epi}(f))$. 
+
+     
+\end{proof}
+
+\begin{corollary}
+\label{corollary:separable-legendre}
+    Let $\dpn{E, F}{\lambda}$ be a duality over $\real$ and $f: E \to (-\infty, \infty]$ with $f \ne \infty$ be convex and lower semicontinuous, then there exists $\seq{(\phi_n, \alpha_n)} \subset F \times \real$ such that for each $x \in E$, 
+    \[
+    f(x) = \sup_{n \in \natp} \dpn{x, \phi_n}{\lambda} - \alpha_n
+    \]
+\end{corollary}
+\begin{proof}
+    For each $(\phi, \alpha) \in F \times \real$, denote $(\phi, \alpha) \le f$ if $\dpn{\cdot, \phi}{\lambda} - \alpha \le f$. By the \hyperref[Fenchel-Moreau Theorem]{theorem:fenchel-moreau},  
+    \[
+    f(x) = f^{**}(x) = \sup\bracs{\dpn{x, \phi}{\lambda} - \alpha|(\phi, \alpha) \in F \times \real, (\phi, \alpha) \le f}
+    \]
+
+    for all $x \in E$. By \autoref{proposition:separable-dual},
+    \[
+    S = \bracs{(\phi, \alpha) \in F \times \real| (\phi, \alpha) \le f}
+    \]
+
+    is separable with respect to $\sigma(F \times \real, E \times \real)$. Therefore there exists $\seq{(\phi_n, \alpha_n)} \subset S$ such that for each $x \in E$, 
+    \[
+    f(x) = \sup_{n \in \natp} \dpn{x, \phi_n}{\lambda} - \alpha_n
+    \]
+\end{proof}
+
+

src/fa/convex/support.tex

@@ -0,0 +1,135 @@
+\section{Support Functions}
+\label{section:support-function}
+
+\begin{definition}[Support Function]
+\label{definition:support-function}
+    Let $\dpn{E, F}{\lambda}$ be a duality over $\real$ and $A \subset E$ be non-empty, then the mapping
+    \[
+    H_A: F \to (-\infty, \infty] \quad y \mapsto \sup_{x \in A}\dpn{x, y}{\lambda}
+    \]
+    
+    the \textbf{support function} of $A$ with respect to $\dpn{E, F}{\lambda}$. 
+\end{definition}
+
+\begin{definition}[Indicator Function]
+\label{definition:infinity-characteristic-function}
+    Let $E$ be a vector space over $\real$ and $A \subset E$, then the mapping
+    \[
+    I_A: E \to (-\infty, \infty] \quad x \mapsto \begin{cases}
+        \infty &x \not\in A \\
+        0 & x \in A
+    \end{cases}
+    \]
+
+    is the \textbf{infinity characteristic function/indicator function} of $A$. 
+\end{definition}
+
+
+
+\begin{lemma}
+\label{lemma:support-function-gymnastics}
+    Let $\dpn{E, F}{\lambda}$ be a duality over $\real$, then:
+    \begin{enumerate}
+        \item For any non-empty $A \subset E$, let $\ol{\conv}(A)$ be the $\sigma(E, F)$-closed convex hull of $A$, then $H_A = H_{\ol{\conv}(A)}$. 
+        \item For any non-empty $A, B \subset E$, $H_A \le H_B$ if and only if $A$ is contained in the $\sigma(E, F)$-closed convex hull of $B$.
+        \item For any non-empty $A \subset E$, $I_A^* = H_A$. 
+        \item For any $A \subset E$ non-empty, $\sigma(E, F)$-closed, and convex, $H_A^* = I_A$. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    (1): Since $\ol{\conv}(A) \supset A$, $H_A \le H_{\ol{\conv}(A)}$. On the other hand, for each $\phi \in F$, $A \subset \bracs{\phi \le H_A(\phi)}$. Since $\bracs{\phi \le H_A(\phi)}$ is a $\sigma(E, F)$-closed convex set, it contains $\ol{\conv}(A)$. Thus $\ol{\conv}(A) \subset \bracs{\phi \le H_A(\phi)}$ and $H_{\ol{\conv}(A)}(\phi) \le H_A(\phi)$. 
+
+    (2): Using (1), assume without loss of generality that $B$ is $\sigma(E, F)$-closed and convex. Suppose that $H_A \le H_B$, then $A \subset \bigcap_{\phi \in F}\bracs{\phi \le H_B(\phi)}$. By the \hyperref[Hahn-Banach Theorem]{theorem:hahn-banach-geometric-2}, $B = \bigcap_{\phi \in F}\bracs{\phi \le H_B(\phi)}$, so $A \subset B$. 
+
+    (3): Let $\phi \in F$, then since $I_A|_{A^c} = \infty$, 
+    \begin{align*}
+    I_A^*(\phi) &= \sup_{x \in E}\dpn{x, \phi}{\lambda} - I_A(x) = \sup_{x \in A}\dpn{x, \phi}{\lambda} - I_A(x) \\
+    &= \sup_{x \in A}\dpn{x, \phi}{\lambda} = H_A(\phi)
+    \end{align*}
+    
+    (4): Given that $A$ is $\sigma(E, F)$-closed and convex, $I_S$ is convex and lower semicontinuous. By the \hyperref[Fenchel-Moreau Theorem]{theorem:fenchel-moreau} and (3), $I_A = I_A^{**} = H_A^*$. 
+\end{proof}
+
+\begin{theorem}
+\label{theorem:support-function-seminorm}
+    Let $\dpn{E, F}{\lambda}$ be a duality over $\real$, $f: X \to (-\infty, \infty]$ be a $\sigma(E, F)$-lower semicontinuous, subadditive, and positively homogeneous function with $f(0) = 0$, and
+    \[
+    \Sigma = \bracs{\phi \in F| (\phi, 0) \le f}
+    \]
+
+    then:
+    \begin{enumerate}
+        \item $f^* = I_\Sigma$. 
+        \item $\Sigma$ is the unique non-empty $\sigma(F, E)$-closed convex subset of $F$ such that $f = H_\Sigma$. 
+        \item $\Sigma$ is equicontinuous if and only if there exists $U \in \cn_E(0)$ such that $\sup_{x \in U}f(x) < \infty$. 
+    \end{enumerate}
+
+    Conversely, 
+    \begin{enumerate}[start=2]
+        \item For any non-empty $\Sigma \subset A$, $H_\Sigma$ is a lower semicontinuous, subadditive, and positively homogeneous function with $H_\Sigma(0) = 0$. 
+    \end{enumerate}
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem 4.25]{Clarke}}}. ]
+    (1): Let $\phi \in \Sigma$, then since $\dpn{x, \phi}{\lambda} \le f(x)$ for all $x \in E$, 
+    \[
+    0 \ge f^*(\phi) = \sup_{x \in E}\dpn{x, \phi}{\lambda} - f(x) = \dpn{0, \phi}{\lambda} - f(0) = 0
+    \] 
+
+    where the supremum is achieved at $0$. On the other hand, if $\phi \in F \setminus \Sigma$, then there exists $x \in E$ such that $\dpn{x, \phi}{\lambda} - f(x) > 0$. In which case, by positive homogeneity of $f$, 
+    \[
+    f^*(\phi) \ge \sup_{\mu > 0}\dpn{\mu x, \phi}{\lambda} - f(\mu x) = \sup_{\mu > 0}\mu\braks{\dpn{ x, \phi} - f(x)} = \infty
+    \]
+
+    Thus $f^{*} = I_\Sigma$. 
+    
+    (2): By \autoref{lemma:lsc-affine-minorant}, $I_\Sigma \ne \infty$, so $\Sigma \ne \emptyset$. Since
+    \[
+    \Sigma = \bigcap_{x \in E}\bracs{\phi \in F| \dpn{x, \phi}{\lambda} \le f(x)}
+    \]
+
+    is an intersection of $\sigma(F, E)$-closed and convex sets, it is also $\sigma(F, E)$-closed. 
+    
+    Now, let $x, y \in E$ and $t \in [0, 1]$, then since $f$ is subadditive and positively homogeneous, 
+    \[
+        f((1 - t)x + ty) \le f((1 - t)x) + f(ty) = (1 - t)f(x) + tf(y)
+    \]
+
+    so $f$ is convex. Given that $f$ is also $\sigma(E, F)$-lower semicontinuous, the \hyperref[Fenchel-Moreau Theorem]{theorem:fenchel-moreau} and (4) of \autoref{lemma:support-function-gymnastics} imply that
+    \[
+    f = f^{**} = I_\Sigma^* = H_\Sigma
+    \]
+
+    By (2) of \autoref{lemma:support-function-gymnastics}, $\Sigma$ is the unique closed $\sigma(F, E)$-convex set such that $f = H_\Sigma$. 
+
+    (3): Let $U \in \cn_E(0)$ be circled such that $M = \sup_{x \in U}f(x) < \infty$, then
+    \[
+    \sup_{\substack{y \in \Sigma \\ x \in U}}\dpn{x, y}{\lambda} \le \sup_{x \in U}f(x) = M < \infty
+    \]
+
+    Since $U$ is circled, $\bigcup_{y \in \Sigma}\dpn{M^{-1}U, y}{\lambda} \subset \ol{B_\real(0, 1)}$, so $\Sigma$ is equicontinuous by \autoref{proposition:equicontinuous-linear}. 
+
+    Conversely, if $\Sigma$ is equicontinuous, then there exists $U \in \cn_E(0)$ such that $M = \sup_{y \in \Sigma, x \in U}\dpn{x, y}{\lambda} < \infty$. In which case, (2) implies that
+    \[
+    \sup_{x \in U}f(x) = \sup_{x \in U}H_\Sigma(x) = \sup_{\substack{y \in \Sigma \\ x \in U}} \dpn{x, y}{\lambda} = M < \infty
+    \]
+
+    (4): By \autoref{proposition:semicontinuous-properties}, $H_\Sigma$ is lower semicontinuous.
+
+    Let $x, y \in E$ and $\mu > 0$, then
+    \begin{align*}
+        H_\Sigma(x + y) &= \sup_{z \in \Sigma}\dpn{x + y, z}{\lambda} \\
+        &\le \sup_{z \in \Sigma}\dpn{x, z}{\lambda} + \sup_{z \in \Sigma}\dpn{y, z}{\lambda} = H_\Sigma(x) + H_\Sigma(y)
+    \end{align*}
+
+    and
+    \[
+    H_\Sigma(\mu x) = \sup_{z \in \Sigma}\dpn{\mu x, z}{\lambda} = \mu\sup_{z \in \Sigma}\dpn{x, z}{\lambda} = \mu H_\Sigma(x)
+    \]
+
+    so $H_\Sigma$ is subadditive and positively homogeneous. 
+    
+    
+\end{proof}
+
+
+

src/fa/duality/definitions.tex

@@ -14,6 +14,15 @@
     In the context of a dual system, $E$ and $F$ are identified as subspaces of each others' algebraic duals. 
 \end{definition}
 
+\begin{definition}[Norming Duality]
+\label{definition:norming-duality}
+    Let $K \in \RC$ and $\dpn{E, F}{\lambda}$ be a duality of normed vector spaces over $K$, then $\dpn{E, F}{\lambda}$ is \textbf{norming} if:
+    \begin{enumerate}
+        \item For each $x \in E$, $\norm{x}_E = \sup_{y \in F, \norm{y}_F \le 1}\dpn{x, y}{\lambda}$. 
+        \item For each $y \in F$, $\norm{y}_F = \sup_{x \in E, \norm{x}_E \le 1}\dpn{x, y}{\lambda}$. 
+    \end{enumerate}
+\end{definition}
+
 \begin{definition}[Weak Topology]
 \label{definition:duality-weak-topology}
     Let $K \in \RC$ and $\dpn{E, F}{\lambda}$ be a duality over $K$, then the weak topology generated by $F$, denoted $\sigma(E, F)$, is the \textbf{weak topology} of the duality on $E$. 

src/fa/duality/index.tex

@@ -3,5 +3,6 @@
 
 \input{./definitions.tex}
 \input{./polar.tex}
+\input{./mackey.tex}
 
 

src/fa/duality/mackey.tex

@@ -0,0 +1,84 @@
+\section{The Mackey-Arens Theorem}
+\label{section:mackey-arens}
+
+\begin{definition}[Consistent]
+\label{definition:consistent-lct-dual}
+    Let $\dpn{E, F}{\lambda}$ be a duality over $K \in \RC$ and $\mathcal{T} \subset 2^E$ be a locally convex topology, then $\mathcal{T}$ is \textbf{consistent} with $\dpn{E, F}{\lambda}$ if $(E, \mathcal{T})^* = F$. 
+\end{definition}
+
+\begin{lemma}
+\label{lemma:consistent-same-convex}
+    Let $\dpn{E, F}{\lambda}$ be a duality over $K \in \RC$, $\mathcal{T} \subset 2^E$ be a locally convex topology consistent with $\dpn{E, F}{\lambda}$, then for any $A \subset E$ convex, the $\mathcal{T}$-closure of $A$ and the $\sigma(E, F)$-closure of $A$ coincide. 
+\end{lemma}
+\begin{proof}
+    By the \hyperref[Hahn-Banach theorem]{theorem:hahn-banach-geometric-2}. 
+\end{proof}
+
+
+\begin{theorem}[Mackey-Arens]
+\label{theorem:mackey-arens}
+    Let $\dpn{E, F}{\lambda}$ be a duality over $K \in \RC$. 
+    \begin{enumerate}
+        \item[(E)] For any locally convex topology $\mathcal{T} \subset 2^E$ consistent with $\dpn{E, F}{\lambda}$, $\mathcal{T}$ is the topology of uniform convergence on all $\mathcal{T}$-equicontinuous subsets of $F$. 
+        \item[(C)] For any saturated covering ideal $\sigma \subset 2^{F}$ consisting of relatively $\sigma(F, E)$-compact sets, the $\sigma$-uniform topology on $E$ is consistent with $\dpn{E, F}{\lambda}$. 
+    \end{enumerate}
+
+    Moreover, 
+    \begin{enumerate}
+        \item[(M)] The topology of uniform convergence on relatively $\sigma(F, E)$-compact, convex, and circled sets is the finest locally convex topology on $E$ consistent with $\dpn{E, F}{\lambda}$. 
+    \end{enumerate}
+
+    and strongest locally convex topology on $E$ consistent with $\dpn{E, F}{\lambda}$ is the \textbf{Mackey topology} of $\dpn{E, F}{\lambda}$ on $E$, denoted $\tau(E, F)$. 
+\end{theorem}
+\begin{proof}
+    (E): Let $\cf \subset F$ be a $\mathcal{T}$-equicontinuous subset, then $\cf^\circ \in \cn_{\mathcal{T}}(0)$. If $\cf$ is circled, then
+    \[
+    \cf^\circ = \bracsn{x \in E|\ |\dpn{x, y}{\lambda}| \le 1 \forall y \in \cf}
+    \]
+
+    Thus $\mathcal{T}$ is finer than the topology of uniform convergence on all $\mathcal{T}$-equicontinuous subsets of $F$. 
+    
+    Conversely, for any convex and closed neighbourhood $U \in \cn_{\mathcal{T}}(0)$, $U^\circ$ is equicontinuous. By the \hyperref[Bipolar Theorem]{theorem:bipolar}, $U^{\circ \circ} = U$, so the family
+    \[
+    \fB = \bracsn{\cf^\circ| \cf \subset F, \cf \text{ is } \mathcal{T}\text{-equicontinuous}}
+    \]
+
+    is a fundamental system of neighbourhoods at $0$ for $\mathcal{T}$. Hence $\mathcal{T}$ is coarser than the topology of uniform convergence on all $\mathcal{T}$-equicontinuous subsets of $F$. 
+
+    (C): Let $\mathcal{T}$ be the $\sigma$-uniform topology on $E$, and $E^*$ be the dual of $(E, \mathcal{T})$. Since $\sigma$ is covering, it contains all singletons, so $\mathcal{T} \supset \sigma(E, F)$, and $F \subset E^*$. Thus it is sufficient to show that $F = E^*$. 
+
+    Let $\phi \in E^*$ and $\bracs{\phi}^\circ$ be the polar of $\phi$ with respect to $\dpn{E, E^*}{E}$. Since $\phi$ is continuous, $\bracs{\phi}^\circ \in \cn_{\mathcal{T}}(0)$. Thus there exists $A \in \sigma$ and $\mu > 0$ such that
+    \[
+    \bracsn{x \in E|\dpn{x, \psi}{E} \le \mu \forall \psi \in A} \subset \bracsn{x \in E|\dpn{x, \phi}{E} \le 1} = \bracs{\phi}^\circ
+    \]
+
+    Since $\sigma$ is covering and saturated, assume without loss of generality that $\mu = 1$ and that $A$ is convex, circled, and $\sigma(F, E)$-compact with $0 \in A$. In which case, let $A^\circ$ be the polar of $A$ with respect to $\dpn{E, E^*}{E}$, then
+    \[
+    A^\circ = \bracsn{x \in E|\dpn{x, \psi}{E} \le 1 \forall \psi \in A} \subset \bracs{\phi}^\circ
+    \]
+
+    Let $A^{\circ\circ}$ and $\bracs{\phi}^{\circ\circ}$ be the bipolar of $A$ and $\bracs{\phi}$ with respect to $\dpn{E, E^*}{E}$, then by \autoref{proposition:polar-gymnastics}, $A^{\circ\circ} \supset \bracs{\phi}^{\circ\circ}$. 
+
+    Now, $A$ is $\sigma(F, E)$-compact, and hence $\sigma(E^*, E)$-compact in $E^*$\footnote{Closedness is insufficient because $F$ is $\sigma(E^*, E)$-dense in $E^*$ by \autoref{lemma:duality-dense}. } by \autoref{proposition:compact-closed}. Thus by the \hyperref[Bipolar Theorem]{theorem:bipolar}, $\phi \in A^{\circ\circ} = A \subset F$. 
+
+    (M): By (C), the topology of uniform convergence on relatively $\sigma(E, F)$-compact, convex, and circled sets is consistent with $\dpn{E, F}{\lambda}$. 
+    
+    On the other hand, let $\mathcal{T} \subset 2^E$ be a locally convex topology consistent with $\dpn{E, F}{\lambda}$. By the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu}, every $\mathcal{T}$-equicontinuous set is relatively $\sigma(F, E)$-compact. Therefore $\mathcal{T}$ is coarser than the topology of uniform convergence on relatively $\sigma(E, F)$-compact, convex, and circled sets. 
+\end{proof}
+
+\begin{definition}[Mackey Space]
+\label{definition:mackey-space}
+    Let $E$ be a separated locally convex space over $K \in \RC$, then $E$ is a \textbf{Mackey space} if $E$ is equipped with the Mackey topology of $\dpn{E, E^*}{E}$. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:barreled-mackey}
+    Let $E$ be a separated barreled space over $K \in \RC$, then $E$ is a Mackey space. 
+\end{proposition}
+\begin{proof}
+    Let $\cf \subset E^*$ be a $\sigma(E^*, E)$-compact set and $U \in \cn_{K}(0)$ be a barrel, then $V = \bigcap_{\phi \in \cf}\phi^{-1}(U)$ is convex, circled, and closed. For each $x \in E$, $\cf(x) = \bracs{\dpn{x, \phi}{E}|\phi \in \cf}$ is bounded. Thus $V$ is absorbing and hence a barrel. Since $E$ is barreled, $V \in \cn_E(0)$. Therefore the Mackey topology is contained in the topology of $E$, and $E$ is a Mackey space. 
+\end{proof}
+
+
+
+

src/fa/duality/polar.tex

@@ -113,7 +113,7 @@
 \begin{proof}[Proof, {{\cite[IV.1.5]{SchaeferWolff}}}. ]
     By \autoref{proposition:polar-properties}, $A^{\circ \circ}$ is a $\sigma(E, F)$-closed, convex set that contains $0$. Since $A^{\circ \circ} \supset A$, it is sufficient to show that $A^{\circ\circ} \subset \ol{\conv}(A \cup \bracs{0})$. 
 
-    Let $x_0 \in E \setminus \ol{\conv}(A \cup \bracs{0})$, then by the \hypreref[Hahn-Banach Theorem]{theorem:hahn-banach-geometric-2}, there exists $\phi: E \to \real$ such that:
+    Let $x_0 \in E \setminus \ol{\conv}(A \cup \bracs{0})$, then by the \hyperref[Hahn-Banach Theorem]{theorem:hahn-banach-geometric-2}, there exists $\phi: E \to \real$ such that:
     \begin{enumerate}
         \item $\phi$ is $\sigma(E, F)$-continuous.
         \item $\phi(\ol{\conv}(A \cup \bracs{0})) \subset (-\infty, 1)$ and $\phi(x_0) > 1$. 
@@ -127,3 +127,6 @@
     Therefore $y \in A^{\circ}$, but $\text{Re}\dpn{x_0, y}{\lambda} > 1$, so $x_0 \not\in A^{\circ\circ}$.
 \end{proof}
 
+
+
+

src/fa/index.tex

@@ -9,5 +9,6 @@
 \input{./lp/index.tex}
 \input{./order/index.tex}
 \input{./duality/index.tex}
+\input{./convex/index.tex}
 \input{./interpolation/index.tex}
 \input{./notation.tex}

src/fa/interpolation/complex.tex

@@ -6,7 +6,7 @@
     Let $S = \bracs{z \in \complex| \text{Re}(z) \in (0, 1)}$ and $(E_0, E_1)$ be a compatible couple of Banach spaces over $\complex$, then the \textbf{Calderón space} $\cf(E_0, E_1)$ is the Banach space of functions $f: \ol S \to E_0 + E_1$ such that:
     \begin{enumerate}
         \item $f$ is holomorphic on $S$. 
-        \item $f$ is continuous on $\ol S$. 
+        \item $f$ is bounded and continuous on $\ol S$. 
         \item For each $t \in \real$, $f(it) \in E_0$, and $\lim_{|t| \to \infty}\norm{f(it)}_{E_0} = 0$. 
         \item For each $t \in \real$, $f(1 + it) \in E_1$, and $\lim_{|t| \to \infty}\norm{f(1 + it)}_{E_1} = 0$. 
     \end{enumerate}
@@ -17,9 +17,9 @@
     \]
 \end{definition}
 \begin{proof}
-    By the \hyperref[Maximum Modulus Theorem]{theorem:maximum-modulus-theorem} applied to $f$ as a function in $H(S; E_0 + E_1)$, $\norm{\cdot}_{\cf(E_0, E_1)}$ is a norm. 
+    By the \hyperref[Maximum Modulus Theorem]{lemma:maximum-modulus-strip} applied to $f$ as a function in $H(S; E_0 + E_1)$, $\norm{\cdot}_{\cf(E_0, E_1)}$ is a norm. 
 
-    By the \hyperref[Maximum Modulus Theorem]{theorem:maximum-modulus-theorem}, \autoref{proposition:holomorphic-complete}, and \autoref{proposition:uniform-limit-continuous}, $\cf(E_0, E_1)$ is complete. 
+    By the \hyperref[Maximum Modulus Theorem]{lemma:maximum-modulus-strip}, \autoref{proposition:holomorphic-complete}, and \autoref{proposition:uniform-limit-continuous}, $\cf(E_0, E_1)$ is complete. 
 \end{proof}
 
 \begin{definition}[The Complex Interpolation Method]
@@ -40,19 +40,19 @@
         \item The mapping $C_\theta$ defined by $(E_0, E_1) \mapsto [E_0, E_1]_\theta$ is an interpolation functor of exact exponent $\theta$. 
     \end{enumerate}
 
-    and functor $C_\theta$ is the \textbf{method of complex interpolation}. 
+    and the functor $C_\theta$ is the \textbf{method of complex interpolation}. 
 \end{definition}
 \begin{proof}[Proof, {{\cite[Theorem 4.1.2]{BerghInterpolation}}}. ]
     (1): Let $\seq{x_n} \subset [E_0, E_1]_\theta$ with $\sum_{n \in \natp}\norm{x_n}_{[E_0, E_1]_\theta} < \infty$, then there exists $\seq{f_n} \subset \cf(E_0, E_1)$ such that for each $n \in \natp$, $f_n(\theta) = x_n$ and $\norm{f_n}_{\cf(E_0, E_1)} \le 2\norm{x_n}_{[E_0, E_1]_\theta}$. Since $\cf(E_0, E_1)$ is complete, there exists $f \in \cf(E_0, E_1)$ such that $f = \sum_{n = 1}^\infty f_n$. Let $x = f(\theta)$, then since $\sum_{n = 1}^N f_n \to f$ in $\cf(E_0, E_1)$ as $N \to \infty$, $\sum_{n = N}^\infty x_n \to x$ in $[E_0, E_1]_\theta$ as $N \to \infty$. Therefore $[E_0, E_1]_\theta$ is a Banach space by \autoref{lemma:banach-criterion}. 
 
-    For any $x \in E_0 \cap E_1$ and $\delta > 0$, let $f_\delta(z) = x_0 e^{(z - \theta)^2}$, then $f_\delta \in \cf(E_0, E_1)$ with $\norm{f_\delta}_{\cf(E_0, E_1)} \le e^\delta\norm{x}_{E_0 \cap E_1}$. Thus $x \in [E_0, E_1]_\theta$ with
+    For any $x \in E_0 \cap E_1$ and $\delta > 0$, let $f_\delta(z) = x_0 e^{\delta(z - \theta)^2}$, then $f_\delta \in \cf(E_0, E_1)$ with $\norm{f_\delta}_{\cf(E_0, E_1)} \le e^\delta\norm{x}_{E_0 \cap E_1}$. Thus $x \in [E_0, E_1]_\theta$ with
     \[
     \norm{x}_{[E_0, E_1]_\theta} \le \norm{f}_{\cf(E_0, E_1)} \le e^\delta\norm{x}_{E_0 \cap E_1}
     \]
 
     As the above holds for all $\delta > 0$, $E_0 \cap E_1$ is continuously embedded in $[E_0, E_1]_\theta$. 
 
-    Let $x \in [E_0, E_1]_\theta$ and $f \in \cf(E_0, E_1)$ with $f(\theta) = x$, then by the \hyperref[Maximum Modulus Theorem]{theorem:maximum-modulus-theorem}, 
+    Let $x \in [E_0, E_1]_\theta$ and $f \in \cf(E_0, E_1)$ with $f(\theta) = x$, then by the \hyperref[Maximum Modulus Theorem]{lemma:maximum-modulus-strip}, 
     \begin{align*}
     \norm{x}_{E_0 + E_1} &= \norm{f(\theta)}_{E_0 + E_1}  \\
     &\le  \max\braks{\sup_{t \in \real}\norm{f(it)}_{E_0 + E_1}, \sup_{t \in \real}\norm{f(1 + it)}_{E_0 + E_1}} \\ 

src/fa/lc/convex.tex

@@ -1,4 +1,4 @@
-\section{Seminorms}
+\section{Convex Sets and Seminorms}
 \label{section:seminorms}
 
 
@@ -29,7 +29,7 @@
 
 
 
-\begin{lemma}[{{\cite[II.1.1]{SchaeferWolff}}}]
+\begin{lemma}
 \label{lemma:convex-interior}
     Let $E$ be a TVS over $K \in \RC$, $A \subset E$ be convex, $x \in A^o$, and $y \in \ol{A}$, then
     \[
@@ -37,7 +37,7 @@
     \]
 
 \end{lemma}
-\begin{proof}
+\begin{proof}[Proof, {{\cite[II.1.1]{SchaeferWolff}}}. ]
     Fix $t \in (0, 1)$. Using translation, assume without loss of generality that $tx + (1 - t)y = 0$. In which case, $x = \alpha y$ where $\alpha = (1 - t)/t$. By (TVS2), $\alpha A^o \in \cn^o(y)$. Since $y \in \ol{A}$, $\alpha A^o \cap A \ne \emptyset$, so there exists $z \in A^o$ such that $\alpha z \in A$.
 
     Let $\mu = \alpha/(\alpha - 1)$, then since $\alpha < 0$, $\mu \in (0, 1)$ and
@@ -74,11 +74,11 @@
     converges to $tx + (1 - t)y$ by (TVS1) and (TVS2). Hence $tx + (1 - t)y \in \ol{A}$.
 \end{proof}
 
-\begin{proposition}[{{\cite[II.1.3]{SchaeferWolff}}}]
+\begin{proposition}
 \label{proposition:convex-interior-closure}
     Let $E$ be a TVS over $K \in \RC$ and $A \subset E$ be convex. If $A^o \ne \emptyset$, then $\ol{A} = \ol{A^o}$.
 \end{proposition}
-\begin{proof}
+\begin{proof}[Proof, {{\cite[II.1.3]{SchaeferWolff}}}. ]
     Since $A^o \subset A$, $\ol{A^o} \subset \ol{A}$. Let $x \in A^o$, then for any $y \in \ol{A}$,
     \[
     y \in \ol{\bracs{tx + (1 - t)y|t \in (0, 1)}} \subset \ol{A^o}
@@ -159,7 +159,7 @@
         \item For any $x \in E$ and $\lambda \ge 0$, $[\lambda x]_A = \lambda [x]_A$.
         \item If $A$ is convex, then for any $x, y \in E$, $[x + y]_A \le [x]_A + [y]_A$.
         \item If $A$ is circled, then for any $x \in E$ and $\lambda \in K$, $[\lambda x]_A = \abs{\lambda}[x]_A$.
-        \item If $A$ is circled, then $\bracs{\rho < 1} \subseteq A \subseteq \bracs{\rho \le 1} \subseteq \ol A$.
+        \item If $A$ is circled, then $\bracs{\rho < 1} \subseteq A \subseteq \bracs{\rho \le 1} \subseteq \ol A$, with respect to any vector space topology on $E$. 
     \end{enumerate}
 
     In particular,

src/fa/lc/hahn-banach.tex

@@ -26,10 +26,10 @@
 
     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} \\
+        \phi_{x_0, \lambda}(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} \\
+        \phi_{x_0, \lambda}(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*}
@@ -40,7 +40,7 @@
     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}
         \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 $\Phi|_F = \phi$.
+        \item If $\rho$ is a seminorm, then for any $\phi \in \hom(F; K)$ with $\abs{\phi} \le \rho|_F$, there exists $\Phi \in \hom(E; K)$ such that $\abs{\Phi} \le \rho$ and $\Phi|_F = \phi$.
     \end{enumerate}
 \end{theorem}
 \begin{proof}[Proof {{\cite[Theorem 5.6, 5.7]{Folland}}}. ]
@@ -60,7 +60,7 @@
 
     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.
 
-    (2): Given that $\rho$ is a seminorm, for any $u \in \hom(E; \real)$, $u \le \rho$ if and only if $\abs{u} \le \rho$.
+    (2): Given that $\rho$ is a seminorm, for any $u \in \hom(E; \real)$, $u \le \rho$ if and only if $\abs{u} \le \rho$. Assume without loss of generality that $K = \complex$. 
 
     Let $u = \re{\phi}$, then $u \in \hom(E; \real)$ by \autoref{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 \autoref{proposition:polarisation-linear}. In addition, for any $x \in E$, if $\alpha = \overline{\sgn(\Phi(x))}$, then
     \[
@@ -98,12 +98,14 @@
 
 \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 $B \subset \bracs{\phi \ge \alpha}$.
+    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 $A \subset \bracs{\phi < \alpha}$ and $B \subset \bracs{\phi \ge \alpha}$.
 \end{theorem}
 \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 $B \subset \bracs{\phi \ge \alpha}$.
+
+    Since $A$ is open, so is $\phi(A)$. Thus $\phi(A) \subset (-\infty, \alpha]$ implies that $\phi(A) \subset (-\infty, \alpha)$. 
 \end{proof}
 
 \begin{theorem}[Hahn-Banach, Second Geometric Form]

src/fa/lc/index.tex

@@ -13,3 +13,4 @@
 \input{./hahn-banach.tex}
 \input{./spaces-of-linear.tex}
 \input{./tensor.tex}
+\input{./nuclear.tex}

src/fa/lc/nuclear.tex

@@ -0,0 +1,158 @@
+\section{Nuclear Operators}
+\label{section:nuclear-operator}
+
+\begin{definition}[Nuclear Operator Between Banach Spaces]
+\label{definition:nuclear-operator-normed}
+    Let $E, F$ be Banach spaces, $E^*$ be the dual of $E$, equipped with the uniform topology, and $T \in L(E; F)$, then $T$ is \textbf{nuclear} if there exists $\seq{\phi_n} \subset E^*$ and $\seq{y_n} \subset F$ such that:
+    \begin{enumerate}
+        \item For each $x \in E$, $Tx = \sum_{n = 1}^\infty y_n \dpn{x, \phi_n}{E}$. 
+        \item $\sum_{n \in \natp}\norm{y_n}_F\norm{\phi_n}_{E^*} < \infty$. 
+    \end{enumerate}
+
+    The set $N(E; F)$ is the \textbf{space of nuclear operators} from $E$ to $F$. For each $T \in N(E; F)$, let
+    \[
+    \norm{T}_{N(E; F)} = \inf\bracs{\sum_{n \in \natp}\norm{y_n}_F\norm{\phi_n}_{E^*} \bigg | Tx = \sum_{n = 1}^\infty y_n \dpn{x, \phi_n}{E} \forall x \in E}
+    \]
+
+    then $\norm{\cdot}_{N(E; F)}$ is a norm on $N(E; F)$, and $N(E; F)$ is a Banach space. 
+\end{definition}
+
+\begin{lemma}
+\label{lemma:nuclear-operator-normed-tensor}
+    Let $E, F$ be Banach spaces, $E^*$ be the dual of $E$, equipped with the uniform topology, then the mapping
+    \[
+    E^* \otimes F \to N(E; F) \quad \sum_{j = 1}^n \phi_j \otimes y_j \mapsto \sum_{j = 1}^n y_j\dpn{\cdot, \phi_j}{E}
+    \]
+
+    extends continuously into a surjective linear map $E^* \tilde \otimes_\pi F \to N(E; F)$. 
+\end{lemma}
+
+
+
+
+\begin{definition}[Nuclear Operator]
+\label{definition:nuclear-operator}
+    Let $E, F$ be separated locally convex spaces over $K \in \RC$ and $T \in L(E; F)$, then the following are equivalent:
+    \begin{enumerate}
+        \item There exists convex and circled sets $U \in \cn_E(0)$ and $B \in B(F)$ such that:
+        \begin{enumerate}[label=(\alph*)]
+            \item The auxiliary space $F_B$ is a Banach space.
+            \item $T(U) \subset B$. 
+            \item The induced map $\wh E_U \to F_B$ is nuclear. 
+        \end{enumerate}
+        \item There exists an equicontinuous sequence $\seq{\phi_n} \subset E^*$, a convex, circled, and bounded subset $B \subset F$, $\seq{y_n} \subset B$, and $\seq{\lambda_n} \subset K$ such that
+        \begin{enumerate}[label=(\alph*)]
+            \item The auxiliary space $F_B$ is a Banach space. 
+            \item For each $x \in E$, $Tx = \sum_{n = 1}^\infty \lambda_n y_n \dpn{x, \phi_n}{E}$. 
+            \item $\sum_{n \in \natp}|\lambda_n| < \infty$. 
+        \end{enumerate}
+    \end{enumerate}
+    
+    If the above holds, then $T$ is \textbf{nuclear}. The set $N(E; F)$ is the \textbf{space of nuclear operators} from $E$ to $F$. 
+\end{definition}
+\begin{proof}[Proof, {{\cite[Theorem III.7.1]{SchaeferWolff}}}. ]
+    (1) $\Rightarrow$ (2): Let $\pi: E \to E_U$ be the canonical projection map associated with $E_U$ and $\iota: F_B \to F$ be the canonical inclusion map associated with $F_B$. By assumption (1b), there exists an induced map $\hat T: E_U \to F_B$ such that the following diagram commutes:
+    \[
+    \xymatrix{
+    E \ar@{->}[r]^{T} \ar@{->}[d]_{\pi} & F \\
+    E_U \ar@{->}[r]_{\hat T} & F_B \ar@{->}[u]_{\iota}
+    }
+    \]
+
+    By the \hyperref[linear extension theorem]{theorem:linear-extension-theorem-normed}, $E_U^* = (\wh E_U)^*$. Assume without loss of generality that $E_U$ is a Banach space, then (1c) implies that $\hat T \in L(E_U; F_B)$ is a nuclear operator. By \autoref{lemma:nuclear-operator-normed-tensor} and \autoref{theorem:metrisable-tensor-product}, there exists $\seq{\phi_n} \subset E_U^*$, $\seq{y_n} \subset F_B$, and $\seq{\lambda_n} \subset K$ such that:
+    \begin{enumerate}[label=(\roman*)]
+        \item $\sum_{n \in \natp}|\lambda_n| < \infty$. 
+        \item $\limv{n}\phi_n = 0$ and $\limv{n}y_n = 0$. 
+        \item For each $x \in E_U$, $\hat Tx = \sum_{n = 1}^\infty \lambda_n y_n \dpn{x, \phi_n}{E_U}$. 
+    \end{enumerate}
+
+    By (ii), $\sup_{n \in \natp}\norm{\phi_n}_{E_U^*} < \infty$ and $\sup_{n \in \natp}\norm{y_n}_{F_B} < \infty$, so $\seq{\phi_n}$ is equicontinuous, and there exists $R > 0$ such that $\seq{y_n} \subset RB$. After rescaling, assume without loss of generality that $\seq{y_n} \subset B$. By unraveling the factorisation, (iii) shows that for each $x \in E$, 
+    \[
+    Tx = \sum_{n = 1}^\infty \lambda_n y_n \dpn{x, \phi_n \circ \pi}{E}
+    \]
+
+    Therefore the decomposition using $\seq{\phi_n \circ \pi} \subset E^*$, $B \subset F$, $\seq{y_n} \subset B$, and $\seq{\lambda_n} \subset K$ given above satisfies (2).
+
+    (2) $\Rightarrow$ (1): Since $\seq{\phi_n}$ is equicontinuous, $U = \bigcap_{n \in \natp}\phi_n^{-1}(B_K(0, 1))$ is a convex and circled neighbourhood of $0$ in $E$. 
+    
+    (1b): Using assumption (2c) and rescaling, assume without loss of generality that $\sum_{n = 1}^\infty |\lambda_n| < 1$. Let $\rho: F_B \to [0, \infty)$ be the gauge of $B$, then for any $x \in U$, 
+    \begin{align*}
+        \rho(Tx) &= \rho\braks{\sum_{n = 1}^\infty \lambda_n y_n \dpn{x, \phi_n}{E}} \le \sum_{n \in \natp} |\lambda_n| \cdot \underbrace{|\dpn{x, \phi_n}{E}|}_{\le 1} \cdot \underbrace{\rho(y_n)}_{\le 1} \\
+        &\le \sum_{n \in \natp}|\lambda_n| < 1
+    \end{align*}
+
+    so $\rho(Tx) < 1$ and $Tx \in B$. Therefore $T(U) \subset B$. 
+
+    (1c): Let $\pi: E \to E_U$ be the canonical projection map associated with $E_U$ and $n \in \natp$. By construction, $U \subset \phi_n^{-1}(B_K(0, 1))$, so there exists $\hat \phi_n \in E_U^*$ such that the following diagram commutes: 
+    \[
+    \xymatrix{
+    E \ar@{->}[d]_{\pi} \ar@{->}[rd]^{\phi_n} &  \\
+    E_U \ar@{->}[r]_{\hat \phi_n} & K
+    }
+    \]
+
+    Thus for each $x \in E$, 
+    \[
+    Tx = \sum_{n = 1}^\infty \lambda_n y_n \dpn{x, \phi_n}{E} = \sum_{n = 1}^\infty y_n \dpn{\pi(x), \lambda_n \hat \phi_n}{E_U}
+    \]
+    
+    and the induced map $\hat T: \wh E_U \to F_B$ takes the form $\hat Tx = \sum_{n = 1}^\infty y_n \dpn{x, \lambda_n \hat \phi_n}{\wh E_U}$. Finally, for each $n \in \natp$, $U \subset \phi_n^{-1}(B_K(0, 1))$, and $\normn{\hat \phi_n}_{E_U^*} \le 1$. Similarly, since $y_n \in B$, $\norm{y_n}_{F_B} \le 1$ as well. Therefore
+    \[
+        \normn{\hat T}_{N(\wh E_U; F_B)} \le \sum_{n \in \natp}|\lambda_n| \cdot \norm{y_n}_{F_B} \cdot \normn{\hat \phi_n}_{E_U^*} \le \sum_{n \in \natp}|\lambda_n| < \infty
+    \]
+
+    and $\hat T: \wh E_U \to F_B$ is nuclear. 
+    
+    
+\end{proof}
+
+\begin{proposition}
+\label{proposition:nuclear-gymnastics}
+    Let $E, F, G, H$ be separated locally convex spaces and $S \in N(F; G)$, then:
+    \begin{enumerate}
+        \item $S$ is compact.
+        \item For any $T \in L(E; F)$, $S \circ T \in N(E; G)$. 
+        \item For any $R \in L(G; H)$, $R \circ S \in N(F; H)$. 
+        \item There exists a unique $\wh S \in L(\wh F; G)$ such that $\wh S|_{F} = S$. Moreover, $\wh S \in N(\wh F; G)$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Corollary III.7.1.1-III.7.1.3]{SchaeferWolff}}}. ]
+    Let $\seq{\phi_n} \subset F^*$ be an equicontinuous sequence, $B \in B(G)$ be convex and circled, $\seq{y_n} \subset B$, and $\seq{\lambda_n} \subset K$ such that
+    \begin{enumerate}[label=(\alph*)]
+        \item The auxiliary space $G_B$ is a Banach space. 
+        \item For each $x \in F$, $Sx = \sum_{n = 1}^\infty \lambda_n y_n \dpn{x, \phi_n}{F}$. 
+        \item $\sum_{n \in \natp}|\lambda_n| < \infty$. 
+    \end{enumerate}
+
+    (1): Let $U = \bigcap_{n \in \natp}\phi_n^{-1}(B_K(0, 1))$, then since $\seq{\phi_n}$ is equicontinuous, $U$ is a convex and circled neighbourhood of $0$ in $F$. Given that $G_B$ is complete, $S$ is the following composition of continuous maps:
+    \[
+    \begin{CD}
+    U @>{\prod_{n \in \natp} \phi_n}>> \overline{B_K(0,1)}^{\natp} @>{x \mapsto \sum_{n=1}^\infty \lambda_n x_n y_n}>> G_B @>>> G
+    \end{CD}
+    \]
+
+    By \hyperref[Tychonoff's Theorem]{theorem:tychonoff}, $\overline{B_K(0,1)}^{\natp}$ is compact. Since $S(U)$ is contained in its image in the above diagram, $S(U)$ is relatively compact. 
+
+    (2): Since $T \in L(E; F)$, $\seq{\phi_n \circ T} \subset E^*$ is equicontinuous. Thus for any $x \in E$, 
+    \[
+    (S \circ T)x = \sum_{n = 1}^\infty \lambda_n y_n \dpn{x, \phi_n \circ T}{E}
+    \]
+
+    and $S \circ T \in N(E; G)$. 
+
+    (3): Using (1), assume without loss of generality that $B$ is also compact. In which case, $R(B)$ is a convex, circled, and compact set in $H$ containing $0$. Thus $H_{R(B)}$ is a Banach space. For each $x \in F$, 
+    \[
+    (R \circ S)x = \sum_{n = 1}^\infty \lambda_n R(y_n) \dpn{x, \phi_n}{F}
+    \]
+
+    and $R \circ S \in N(F; H)$. 
+
+    (4): By the \hyperref[linear extension theorem]{theorem:linear-extension-theorem-tvs}, such an extension exists and is unique. Moreover, $\seq{\phi_n} \subset F^*$ extend into an equicontinuous family $\bracsn{\wh \phi_n}_1^\infty \subset \wh F^*$. Since $G_B$ is complete, the extension $\wh S \in L(\wh F; G)$ takes the form
+    \[
+    \wh S x = \sum_{n = 1}^\infty \lambda_n y_n \dpn{x, \wh \phi_n}{\wh F}
+    \]
+
+    Therefore $\wh S \in N(\wh F; G)$.
+\end{proof}
+
+

src/fa/lc/tensor.tex

@@ -135,7 +135,7 @@
     (5): By (6) of \autoref{definition:projective-tensor-product}. 
 \end{proof}
 
-\begin{theorem}[{{\cite[III.6.4]{SchaeferWolff}}}]
+\begin{theorem}
 \label{theorem:metrisable-tensor-product}
     Let $E, F$ be metrisable locally convex spaces over $K \in \RC$, then for any $z \in E \td{\otimes}_\pi F$, there exists $\seq{\lambda_n} \subset K$ and $\seq{(x_j, y_j)} \subset E \times F$ such that:
     \begin{enumerate}
@@ -146,7 +146,7 @@
 
     
 \end{theorem}
-\begin{proof}
+\begin{proof}[Proof, {{\cite[III.6.4]{SchaeferWolff}}}.]
     Let $\seq{p_n}$ and $\seq{q_n}$ be increasing sequences of continuous seminorms that induce the topology on $E$ and $F$, respectively. For each $n \in \natp$, let $r_n = p_n \otimes q_n$, and $\td r_n$ be the continuous extension of $r_n$ to $E \td{\otimes}_\pi F$. 
 
     Let $u \in E \td{\otimes}_\pi F$, then there exists $\seq{u_n} \subset E \otimes_\pi F$ such that $\td r_n(u - u_n) < 2^{-n}/n^2$ for all $n \in \natp$. For each $N \in \natp$, let $v_N = u_{N+1} - u_N$, then

src/fa/lp/definition.tex

@@ -23,11 +23,13 @@
 
 \begin{definition}[Hölder conjugates]
 \label{definition:holder-conjugates}
-    Let $p, q \in (1, \infty)$, then $p$ and $q$ are \textbf{Hölder conjugates} if
+    Let $p, q \in [1, \infty]$, then $p$ and $q$ are \textbf{Hölder conjugates} if
     \[
     \frac{1}{p} + \frac{1}{q} = 1
     \]
 
+    under the identification that $1/\infty = 0$. 
+
 \end{definition}
 
 \begin{lemma}
@@ -40,15 +42,15 @@
 \end{lemma}
 
 
-\begin{theorem}[Hölder's Inequality, {{\cite[6.2]{Folland}}}]
+\begin{theorem}[Hölder's Inequality]
 \label{theorem:holder}
-    Let $(X, \cm, \mu)$ be a measure space, $E, F$ be a normed vector spaces, $p, q \in [1, \infty]$. If $p, q$ are Hölder conjugates or if $p = 1$ and $q = \infty$, then for any $f \in \mathcal{L}^p(X; E)$ and $g \in \mathcal{L}^q(X; F)$,
+    Let $(X, \cm, \mu)$ be a measure space, $E, F$ be a normed vector spaces, $p, q \in [1, \infty]$ be Hölder conjugates, then for any $f \in \mathcal{L}^p(X; E)$ and $g \in \mathcal{L}^q(X; F)$,
     \[
     \int \norm{f}_E \norm{g}_F d\mu \le \norm{f}_{L^p(X; E)}\norm{g}_{L^q(X; F)}
     \]
 
 \end{theorem}
-\begin{proof}
+\begin{proof}[Proof, {{\cite[Theorem 6.2]{Folland}}}. ]
     First suppose that $p = 1$ and $q = \infty$. In this case,
     \[
     \int \norm{f}_E \norm{g}_F d\mu \le \norm{g}_{L^\infty(X; F)}\int \norm{f}_Ed\mu = \norm{f}_{L^1(X; E)}\norm{g}_{L^\infty(X; F)}
@@ -167,36 +169,6 @@
 \end{proof}
 
 
-
-\begin{theorem}[Markov's Inequality]
-\label{theorem:markov-inequality}
-    Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed vector space, and $f: X \to E$ be a Borel measurable function, then
-    \begin{enumerate}
-        \item For any $\alpha > 0$, 
-        \[
-        \mu\bracs{|f| \ge \alpha} \le \frac{1}{\alpha}\norm{f}_{L^1(X; E)}
-        \]
-        \item For $\alpha > 0$ and strictly increasing function $\phi: [0, \infty) \to [0, \infty)$,
-        \[
-        \mu\bracs{|f| \ge \alpha} \le \frac{1}{\phi(\alpha)}\norm{\phi \circ |f|}_{L^1(X; E)}
-        \]
-        \item For any $\alpha > 0$,
-        \[
-        \mu\bracs{|f| \ge \alpha} \le \frac{1}{\alpha^p}\norm{f}_{L^p(X; E)}^p
-        \]
-    \end{enumerate}
-\end{theorem}
-\begin{proof}
-    (1): For any $\alpha > 0$, 
-    \begin{align*}
-        \mu\bracs{|f| \ge \alpha} &= \int_{\bracs{f \ge \alpha}} d\mu \le \frac{1}{\alpha}\int |f|d\mu = \frac{\norm{f}_{L^1(X; E)}}{\alpha}
-    \end{align*}
-    
-    (2): By \autoref{lemma:monotone-measurable}, $\phi: [0, \infty) \to [0, \infty)$ is Borel measurable. Since $\phi$ is strictly increasing, $\bracs{|f| \ge \alpha} = \bracs{\phi \circ |f| \ge \phi(\alpha)}$, and the result holds by applying (1) to $\phi \circ |f|$. 
-
-    (3): Since $x \mapsto x^p$ is strictly increasing on $[0, \infty)$, applying (2) to $\phi(x) = x^p$ yields the desired result.
-\end{proof}
-
 \begin{proposition}
 \label{proposition:lp-in-measure}
     Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed vector space, $p \in [1, \infty]$, $\seq{f_n} \subset L^p(X; E)$, and $f \in L^p(X; E)$ such that $f_n \to f$ in $L^p$, then $f_n \to f$ in measure.
@@ -211,5 +183,36 @@
 \end{proof}
 
 
+% Move this to interpolation
+
+\begin{proposition}
+\label{proposition:lp-intersection-interpolation}
+    Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed vector space, and $0 < p < q < r \le \infty$, then $L^p(X; E) \cap L^r(X; E) \subset L^q(X; E)$, where for each $f \in L^p(X; E) \cap L^r(X; E)$, 
+    \[
+    \norm{f}_{L^q(X; E)} \le \norm{f}_{L^p(X; E)}^\lambda\norm{f}_{L^q(X; E)}^{1 - \lambda}
+    \]
+
+    with
+    \[
+    \frac{1}{q} = \frac{\lambda}{p} + \frac{(1 - \lambda)}{r} \quad \lambda = \frac{q^{-1} - r^{-1}}{p^{-1} - r^{-1}}
+    \]
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition 6.10]{Folland}}}. ]
+    If $r = \infty$, then $\norm{f}_{E}^q \le \norm{f}_{L^\infty(X; E)}^{q - p}\norm{f}_E^p$ and $\lambda = p/q$, so
+    \[
+    \norm{f}_{L^q(X; E)} \le \norm{f}_{L^p(X; E)}^{p/q} \cdot \norm{f}_{L^\infty(X; E)}^{1 - p/q} = \norm{f}_{L^p(X; E)}^\lambda\norm{f}_{L^q(X; E)}^{1 - \lambda}
+    \]
+
+    If $r < \infty$, then by \hyperref[Hölder's inequality]{theorem:holder} applied to the pair $p/(\lambda q)$ and $r/[(1 - \lambda)q]$, 
+    \begin{align*}
+        \norm{f}_{L^q(X; E)}^q &= \int \norm{f}_{E}^{\lambda q}\norm{f}_E^{(1 - \lambda)q} d\mu \\
+        &= \norm{\norm{f}_E^{\lambda q}}_{L^{p/(\lambda q)}(X; \real)} \cdot 
+        \norm{\norm{f}_E^{(1 - \lambda) q}}_{L^{r/[(1 - \lambda)q]}(X; \real)} \\
+        &= \norm{f}_{L^p(X; E)}^{\lambda q}\norm{f}_{L^q(X; E)}^{(1 - \lambda)q} \\
+    \norm{f}_{L^q(X; E)} &\le \norm{f}_{L^p(X; E)}^\lambda\norm{f}_{L^q(X; E)}^{1 - \lambda}
+    \end{align*}
+    
+\end{proof}
+
 
 

src/fa/lp/duality.tex

@@ -0,0 +1,230 @@
+\section{Duality of $L^p$ Spaces}
+\label{section:lp-duality}
+
+\begin{lemma}
+\label{lemma:lp-dual-approximation}
+    Let $(X, \cm, \mu)$ be a measure space, $K \in \RC$, $\dpn{E, F}{\lambda}$ be a norming duality of Banach spaces over $K$, and $f: X \to E$ be a strongly measurable function, then there exists $\seq{\phi_n} \subset \Sigma(X, \cm; F)$ such that:
+    \begin{enumerate}
+        \item For each $n \in \natp$, $\norm{\phi_n}_{F} \le 1$.
+        \item For every $n \in \natp$, $|\dpn{f, \phi_n}{\lambda}| \le \norm{f}_{E}$. 
+        \item $|\dpn{f, \phi_n}{\lambda}| \upto \norm{f}_E$ pointwise as $n \to \infty$. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    Since $f(X)$ is separable, assume without loss of generality that $E$ is separable. By \autoref{proposition:separable-dual}, there exists $\seq{z_n} \subset \bracsn{z \in F|\ \norm{z}_F \le 1}$ such that for each $y \in F$, $\norm{y}_F = \sup_{n \in \natp}\dpn{y, z_n}{\lambda}$. 
+
+    For each $N \in \natp$ and $x \in X$, let $F_N(x) = 0 \vee \bigvee_{n = 1}^N |\dpn{f(x), z_n}{E}|$, then $0 \le F_N \le \norm{F}_E$ and $F_N \upto \norm{f}_E$ pointwise as $N \to \infty$. 
+    
+    For every $1 \le n \le N$, inductively define
+    \[
+    A_{N, n} = \bracs{x \in X|F_N(x) = |\dpn{f(x), z_n}{E}|} \setminus \bigcup_{k = 1}^{n - 1}A_{N, k}
+    \]
+    
+    Let $\phi_N = \sum_{n = 1}^N z_n \one_{A_{N, n}}$, then $|\dpn{f, \phi_N}{\lambda}| = F_N$, and
+    \begin{enumerate}
+        \item Since $\norm{z_n}_F \le 1$ for all $n \in \natp$, $\norm{\phi_N}_F \le 1$ for each $N \in \natp$. 
+        \item For every $N \in \natp$, $|\dpn{f, \phi_N}{\lambda}| = F_N$, so $|\dpn{f, \phi_N}{\lambda}| \le \norm{f}_E$.
+        \item As $F_N \upto \norm{f}_E$ pointwise as $N \to \infty$, $|\dpn{f, \phi_N}{\lambda}| \upto \norm{f}_E$ pointwise as $N \to \infty$. 
+    \end{enumerate}
+\end{proof}
+
+
+After the duality of $L^p$ and $L^q$ is established for Hölder conjugate exponents for $p, q \in (1, \infty)$, it may be seen that each continuous functional on $L^p$ is "supported" on a $\sigma$-finite set. However, this fact can be established beforehand without the explicit identification. 
+
+\begin{lemma}
+\label{lemma:lp-functional-support}
+    Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed vector space over $K \in \RC$, $p \in (1, \infty]$, and $\phi \in L^p(X, \cm, \mu; E)^*$, then there exists a $\sigma$-finite set $A \in \cm$ such that
+    \[
+    \dpn{f, \phi}{L^p(X; E)} = \dpn{\one_A \cdot f, \phi}{L^p(X; E)}
+    \]
+
+    for all $f \in L^p(X, \cm, \mu; E)$. 
+\end{lemma}
+\begin{proof}
+    For each $A \in \cm$, define
+    \[
+    \phi_A: L^p(X; E) \to K \quad \dpn{f, \phi_A}{L^p(X; E)} = \dpn{\one_A \cdot f, \phi}{L^p(X; E)}
+    \]
+
+    Let $f, g \in L^p(X; E)$ and $A, B \in \cm$ such that
+    \begin{enumerate}
+        \item $X = A \sqcup B$. 
+        \item $f|_B = 0$, and $g|_A = 0$. 
+        \item $\norm{f}_{L^p(X; E)} = \norm{g}_{L^p(X; E)} = 1$. 
+    \end{enumerate}
+
+    Suppose that $p \in (1, \infty)$, then for each $t \in \real$, 
+    \begin{align*}
+        \norm{(1 - t)f + tg}_{L^p(X; E)}^p &= \norm{(1 - t)f}_{L^p(X; E)}^p + \norm{tg}_{L^p(X; E)}^p \\
+        &= (1 - t)^p + t^p \\
+        &= (1 - t) \cdot (1 - t)^{p - 1} + t \cdot t^{p - 1}
+    \end{align*}
+
+    Since $p > 1$, for each $t \in (0, 1)$, $(1 - t)^{p - 1}, t^{p - 1} < 1$, so $\norm{(1 - t)f + tg}_{L^p(X; E)}^p < 1$. 
+
+    On the other hand, if $p = \infty$, then
+    \[
+    \norm{(1 - t)f + tg}_{L^\infty(X; E)} = (1 - t) \vee t < 1
+    \]
+
+    Therefore for any $A, B \in \cm$ with $A \cap B = \emptyset$, $\norm{\phi_A}_{L^p(X; E)^*} > 0$, and $\norm{\phi_B}_{L^p(X; E)^*} > 0$,
+    \[
+    \norm{\phi_{A \sqcup B}}_{L^p(X; E)^*} > \norm{\phi_A}_{L^p(X; E)^*} \vee \norm{\phi_B}_{L^p(X; E)^*}
+    \]
+    
+    Now, by \hyperref[density of simple functions]{proposition:lp-simple-dense}, 
+    \[
+    \norm{\phi}_{L^p(X; E)^*} = \sup\bracsn{\norm{\phi_A}_{L^p(X; E)^*}| A \in \cm \ \sigma\text{-finite}}
+    \]
+
+    Let $\seq{A_n} \subset \cm$ such that $\mu(A_n) < \infty$ for all $n \in \natp$, and $\norm{\phi_{A_n}}_{L^p(X; E)^*} \upto \norm{\phi}_{L^p(X; E)^*}$ as $n \to \infty$. Let $A = \bigcup_{n \in \natp}A_n$, then $A$ is $\sigma$-finite and $\norm{\phi_A}_{L^p(X; E)^*} = \norm{\phi}_{L^p(X; E)^*}$. By maximality, there exists no $B \in \cm$ with $B \cap A = \emptyset$ and $\norm{\phi_B}_{L^p(X; E)^*} > 0$, so $\phi = \phi_A$.
+\end{proof}
+
+\begin{theorem}
+\label{theorem:lp-dual-function}
+    Let $(X, \cm, \mu)$ be a measure space, $K \in \RC$, $\dpn{E, F}{\lambda}$ be a norming duality of normed vector spaces over $K$, $p, q \in [1, \infty]$ be Hölder conjugates such that one of the following holds:
+    \begin{enumerate}[label=(\alph*)]
+        \item $p \in (1, \infty]$ and $q \in [1, \infty)$.  
+        \item $p = 1$, $q = \infty$, and $\mu$ is semifinite. 
+    \end{enumerate}
+
+    Let $g: X \to F$ be a strongly measurable function such that for every $f \in \Sigma(X, \cm; E) \cap L^p(X; E)$, $\dpn{f, g}{\lambda} \in L^1(X; E)$, and the mapping
+    \[
+    \phi_g: \Sigma(X, \cm; E) \cap L^p(X; E) \to K \quad f \mapsto \int \dpn{f, g}{\lambda} d\mu
+    \]
+
+    is continuous in the $L^p(X; E)$ norm, then:
+    \begin{enumerate}
+        \item If (a) holds, then $\bracs{g \ne 0}$ is $\sigma$-finite. 
+        \item $g \in L^q(X; F)$ with $\norm{g}_{L^q(X; F)} = \norm{\phi_g}_{L^p(X; E)^*}$. 
+        \item The mapping
+        \[
+        L^q(X; F) \to L^p(X; E)^* \quad g \mapsto \phi_g
+        \]
+
+        is isometric. 
+    \end{enumerate}
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Proposition 6.13, Theorem 6.14]{Folland}}}. ]
+    (1): By \autoref{lemma:lp-functional-support}, there exists a $\sigma$-finite set $A \in \cm$ such that for each $f \in L^p(X; E)$, 
+    \[
+    \int \dpn{f, g}{\lambda} d\mu = \int_A \dpn{f, g}{\lambda} d\mu
+    \]
+
+    Therefore $g|_{A^c} = 0$ almost everywhere, and $\bracs{g \ne 0}$ is $\sigma$-finite. 
+
+    (2, truncated): First suppose that $g \in L^q(X; F)$. Assume without loss of generality that $\norm{g}_{L^q(X; F)} = 1$. 
+
+    By \autoref{lemma:lp-dual-approximation}, there exists $\seq{\phi_n} \subset \Sigma(X, \cm; E)$ such that:
+    \begin{enumerate}
+        \item For each $n \in \natp$, $\norm{\phi_n}_{E} \le 1$.
+        \item For every $n \in \natp$, $0 \le |\dpn{g, \phi_n}{\lambda}| \le \norm{g}_{F}$. 
+        \item $|\dpn{g, \phi_n}{\lambda}| \upto \norm{g}_F$ pointwise as $n \to \infty$. 
+    \end{enumerate}
+    
+    (2, a, truncated): Let $\Phi(x) = \norm{g(x)}_F^{q - 1}$\footnote{Under the convention that $0^0 = 1$} for each $x \in X$. If $p < \infty$, then by \autoref{lemma:holder-conjugate-gymnastics}, 
+    \[
+    \norm{\Phi}_{L^p(X; \real)}^p = \int \norm{g}_F^{p(q - 1)}d\mu = \int \norm{g}_F^q d\mu = 1
+    \]
+
+    Otherwise, $\Phi = 1$, and $\norm{\Phi}_{L^\infty(X; \real)} = 1$. 
+
+    Let $\seq{\Phi_n} \subset \Sigma(X, \cm; E) \cap L^p(X; E)$ be non-negative such that $\Phi_n \upto \one_{\bracs{g \ne 0}} \cdot \Phi$. For each $n \in \natp$, $\Phi_n \phi_n \in \Sigma(X, \cm; E) \cap L^p(X; E)$ with $\norm{\Phi_n \phi_n}_{L^p(X; E)} \le 1$. By assumption, 
+    \[
+    |\Phi_n \dpn{\phi_n, g}{H}| = \Phi_n \dpn{\phi_n, g}{H} \cdot \ol{\sgn \dpn{\phi_n, g}{H}} \in L^1(X; \real)
+    \]
+
+    Let $f_n = \Phi_n \phi_n \cdot \ol{\sgn \dpn{\phi_n, g}{H}}$, then by the \hyperref[Monotone Convergence Theorem]{theorem:mct},
+    \begin{align*}
+        \limv{n}\int \dpn{f_n, g}{\lambda}d\mu &= \limv{n}\int \Phi_n \cdot \dpn{\phi_n, g}{\lambda} \cdot \ol{\sgn \dpn{\phi_n, g}{H}}d\mu \\
+        &= \int \Phi \cdot \norm{g}_F d\mu = \int \norm{g}_F^{q - 1}\norm{g}_F d\mu \\
+        &= \norm{g}_{L^q(X; F)}^q = 1
+    \end{align*}
+
+    (2, b, truncated): Let $\alpha \in (0, 1)$, then $\mu\bracs{\norm{g}_F \ge \alpha} > 0$. Since $\mu$ is semifinite, there exists $A \in \cm$ with $A \subset \bracs{\norm{g}_F \ge \alpha}$ and $0 < \mu(A) < \infty$. For each $x \in X$, let $\Phi(x) = \one_A/\mu(A)$, then $\norm{\Phi}_{L^1(X; \real)} = 1$. 
+
+    For every $n \in \natp$, let $f_n = \Phi \phi_n \cdot \ol{\sgn \dpn{\phi_n, g}{H}}$, then $\norm{f_n}_{L^1(X; E)} \le 1$, and by the \hyperref[Monotone Convergence Theorem]{theorem:mct},
+    \begin{align*}
+        \limv{n}\int \dpn{f_n, g}{\lambda}d\mu &= \limv{n}\int \Phi \cdot \dpn{\phi_n, g}{\lambda} \cdot \ol{\sgn \dpn{\phi_n, g}{H}}d\mu \\
+        &= \int \Phi \cdot \norm{g}_F d\mu = \int_A \frac{\norm{g}_F}{\mu(A)} d\mu \ge \alpha
+    \end{align*}
+    
+    As the above holds for all $\alpha \in (0, 1)$, $\norm{\phi_g}_{L^1(X; E)^*} = 1$. 
+
+    (2, general): If (a) holds, then $\bracs{g \ne 0}$ is $\sigma$-finite. In both cases, there exists $\seq{g_n} \subset L^q(X; F)$ such that
+    \begin{enumerate}
+        \item $\norm{g_n}_{F} \upto \norm{g}_F$ as $n \to \infty$.
+        \item For each $n \in \natp$, $\norm{\phi_{g_n}}_{L^p(X; E)^*} \le \norm{\phi_g}_{L^p(X; E)^*}$. 
+    \end{enumerate}
+
+    By the truncated case, for each $n \in \natp$, 
+    \[
+    \norm{g_n}_{L^q(X; F)} = \norm{\phi_{g_n}}_{L^p(X; E)^*} \le \norm{\phi_g}_{L^p(X; E)^*}
+    \]
+
+    If $q < \infty$, then the \hyperref[Monotone Convergence Theorem]{theorem:mct} implies that $\norm{g}_{L^q(X; F)} \le \norm{\phi_g}_{L^p(X; E)^*}$. Otherwise,
+    \[
+    \norm{g}_{L^\infty(X; H)} \le \sup_{n \in \natp}\norm{g_n}_{L^\infty(X; H)} \le \norm{\phi_g}_{L^1(X; H)^*}
+    \]
+    
+    The above argument shows that the truncation argument was technically not required. By applying the truncated case again, $\norm{g}_{L^q(X; F)} = \norm{\phi_g}_{L^p(X; E)^*}$.
+\end{proof}
+
+
+The typical argument for $L^p$ duality requires using the Radon-Nikodym theorem to extract the function. Since I prefer to not present martingales here, I will only include the Hilbert case. 
+
+\begin{theorem}
+\label{theorem:lp-duality}
+    Let $(X, \cm, \mu)$ be a measure space, $K \in \RC$, $H$ be a Hilbert space over $K$, $p, q \in [1, \infty]$ be Hölder conjugates such that one of the following holds:
+    \begin{enumerate}[label=(\alph*)]
+        \item $p \in (1, \infty)$ and $q \in (1, \infty)$.  
+        \item $p = 1$, $q = \infty$, and $\mu$ is $\sigma$-finite. 
+    \end{enumerate}
+
+    For each $g \in L^q(X, \cm, \mu; H)$, let
+    \[
+    \phi_g: L^p(X, \cm, \mu; H) \to K \quad f \mapsto \int \dpn{f, g}{H} d\mu
+    \]
+
+    then the mapping
+    \[
+    L^q(X, \cm, \mu; H) \to L^p(X, \cm, \mu; H)^* \quad g \mapsto \phi_g
+    \]
+
+    is a conjugate linear isometric isomorphism. 
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem 6.15]{Folland}}}. ]
+    By \autoref{theorem:lp-dual-function}, the given map is isometric. Thus it is sufficient to show that it is surjective. Let $\phi \in L^p(X; H)^*$.
+
+    (Finite): First suppose that $\mu$ is finite, then $\Sigma(X, \cm; H) \subset L^p(X; H)$, and $\phi$ induces an $H$-valued measure on $(X, \cm)$, absolutely continuous with respect to $\mu$. By the \hyperref[Radon-Nikodym Theorem]{theorem:lebesgue-radon-nikodym}, there exists $g \in L^1(X; H)$ such that for each $f \in \Sigma(X, \cm; H)$, 
+    \[
+    \int \dpn{f, g}{H} d\mu = \dpn{f, \phi}{L^p(X; H)}
+    \]
+
+    By \autoref{theorem:lp-dual-function}, $g \in L^q(X; H)$. 
+
+    (Arbitrary): In the case of (a), by \autoref{lemma:lp-functional-support}, there exists a $\sigma$-finite set $A \in \cm$ such that for each $f \in L^p(X; H)$, $\dpn{f, \phi}{L^p(X; H)} = \dpn{\one_A \cdot f, \phi}{L^p(X; H)}$. In the case of (b), $A = X$ is a $\sigma$-finite set satisfying the same restriction condition. 
+
+    Let $\seq{A_n} \subset \cm$ such that $\mu(A_n) < \infty$ for all $n \in \natp$, and $A = \bigsqcup_{n \in \natp}A_n$. By the finite case, there exists $\seq{g_n} \subset L^q(X; H)$ such that for each $n \in \natp$ and $f \in L^p(X; H)$, 
+    \[
+    \int \dpn{f, g_n}{H} d\mu = \dpn{\one_{A_n} \cdot f, \phi}{L^p(X; H)}
+    \] 
+
+    Let $g = \sum_{n = 1}^\infty g_n$. If $q < \infty$, then $g \in L^q(X; H)$ by the \hyperref[Monotone Convergence Theorem]{theorem:mct}. Otherwise,
+    \[
+    \norm{g}_{L^\infty(X; H)} \le \sup_{n \in \natp}\norm{g_n}_{L^\infty(X; H)} \le \norm{\phi}_{L^1(X; H)^*}
+    \]
+    
+    For every $f \in L^p(X; H)$, 
+    \begin{align*}
+    \int \dpn{f, g}{H} d\mu &= \sum_{n = 1}^\infty \int \dpn{f, g_n}{H} d\mu = \sum_{n = 1}^\infty \dpn{\one_{A_n} \cdot f, \phi}{L^p(X; H)} \\
+    &= \dpn{f, \phi}{L^p(X; H)}
+    \end{align*}
+    
+    by the \hyperref[Dominated Convergence Theorem]{theorem:dct}.
+
+    Therefore the mapping is surjective, and hence an isomorphism. 
+\end{proof}
+
+
+

src/fa/lp/index.tex

@@ -2,3 +2,8 @@
 \label{chap:lp}
 
 \input{./definition.tex}
+\input{./ineq.tex}
+\input{./duality.tex}
+\input{./ui.tex}
+\input{./seq.tex}
+\input{./local.tex}

src/fa/lp/ineq.tex

@@ -0,0 +1,50 @@
+\section{$L^p$ Inequalities}
+\label{section:lp-inequalities}
+
+
+\begin{theorem}[Markov's Inequality]
+\label{theorem:markov-inequality}
+    Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed vector space, and $f: X \to E$ be a Borel measurable function, then
+    \begin{enumerate}
+        \item For any $\alpha > 0$, 
+        \[
+        \mu\bracs{\norm{f}_E \ge \alpha} \le \frac{1}{\alpha}\norm{f}_{L^1(X; E)}
+        \]
+        \item For $\alpha > 0$ and strictly increasing function $\phi: [0, \infty) \to [0, \infty)$,
+        \[
+        \mu\bracs{\norm{f}_E \ge \alpha} \le \frac{1}{\phi(\alpha)}\norm{\phi \circ |f|}_{L^1(X; E)}
+        \]
+        \item For any $\alpha > 0$,
+        \[
+        \mu\bracs{\norm{f}_E \ge \alpha} \le \frac{1}{\alpha^p}\norm{f}_{L^p(X; E)}^p
+        \]
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    (1): For any $\alpha > 0$, 
+    \begin{align*}
+        \mu\bracs{\norm{f}_E \ge \alpha} &= \int_{\bracs{\norm{f}_E \ge \alpha}} d\mu \le \frac{1}{\alpha}\int |\norm{f}_Ed\mu = \frac{\norm{f}_{L^1(X; E)}}{\alpha}
+    \end{align*}
+    
+    (2): By \autoref{lemma:monotone-measurable}, $\phi: [0, \infty) \to [0, \infty)$ is Borel measurable. Since $\phi$ is strictly increasing, $\bracs{\norm{f}_E \ge \alpha} = \bracs{\phi \circ \norm{f}_E \ge \phi(\alpha)}$, and the result holds by applying (1) to $\phi \circ \norm{f}_E$. 
+
+    (3): Since $x \mapsto x^p$ is strictly increasing on $[0, \infty)$, applying (2) to $\phi(x) = x^p$ yields the desired result.
+\end{proof}
+
+
+\begin{lemma}
+\label{lemma:lp-p-diff}
+    Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed vector space, $p \in [1, \infty)$, and $f, g \in L^p(X; E)$, then
+    \[
+    |\norm{f}_{L^p(X; E)}^p - \norm{g}_{L^p(X; E)}^p| \le p\norm{f - g}_{L^p(X; E)}(\norm{f}_{L^p(X; E)} \vee \norm{g}_{L^p(X; E)})^{p-1}
+    \]
+\end{lemma}
+\begin{proof}
+    By \autoref{lemma:power-difference}, 
+    \begin{align*}
+    |\norm{f}_{L^p(X; E)}^p - \norm{g}_{L^p(X; E)}^p| &\le p|\norm{f}_{L^p(X; E)} - \norm{g}_{L^p(X; E)}|\\
+    &\times (\norm{f}_{L^p(X; E)} \vee \norm{g}_{L^p(X; E)})^{p-1} \\
+    &\le p\norm{f - g}_{L^p(X; E)}(\norm{f}_{L^p(X; E)} \vee \norm{g}_{L^p(X; E)})^{p-1}
+    \end{align*}
+    
+\end{proof}

src/fa/lp/local.tex

@@ -0,0 +1,44 @@
+\section{Locally Integrable Functions}
+\label{section:locally-integrable}
+
+\begin{definition}[Locally Integrable*]
+\label{definition:locally-integrable}
+    Let $(X, \cm, \cf, \mu)$ be a \hyperref[scaffolded]{definition:measure-scaffold} measure space, $E$ be a normed vector space over $K \in \RC$, $f: X \to E$ be strongly measurable, and $p \in [1, \infty]$, then $f$ is \textbf{locally $p$-integrable} if for every $A \in \cf$, $\int_A \norm{f}_A^p d\mu < \infty$. 
+
+    The set $\mathcal{L}^p_\cf(X, \cm, \mu; E) = \mathcal{L}^p_\cf(X; E) = \mathcal{L}^p_\cf(\mu; E)$ is the space of all locally $p$-integrable $E$-valued functions on $X$. 
+\end{definition}
+
+\begin{definition}[Locally Bounded*]
+\label{definition:locally-bounded-measure}
+    Let $(X, \cm, \cf, \mu)$ be a \hyperref[scaffolded]{definition:measure-scaffold} measure space, $E$ be a normed vectorr space over $K \in \RC$, and $f: X \to E$ be strongly measurable, then $f$ is \textbf{essentially bounded} if for every $A \in \cf$, $\norm{\one_A f}_{L^\infty(A; E)} < \infty$. 
+
+    The set $\mathcal{L}^\infty_\cf(X, \cm, \mu; E) = \mathcal{L}^\infty_\cf(X; E) = \mathcal{L}^\infty_\cf(\mu; E)$ is the space of all locally bounded $E$-valued functions on $X$. 
+\end{definition}
+
+\begin{definition}[Local $L^p$ Space*]
+\label{definition:local-lp-space}
+    Let $(X, \cm, \cf, \mu)$ be a \hyperref[scaffolded]{definition:measure-scaffold} measure space, $E$ be a normed vector space over $K \in \RC$, and $p \in [1, \infty]$. For each $A \in \cf$ and $f \in \mathcal{L}^p_\cf(X, \cm, \mu; E)$, let $[f]_{L^p_A(X; \mu)} = \norm{\one_A f}_{L^p(X; \mu)}$, then the $[\cdot]_{L^p_A(X; \mu)}$ is a seminorm on \hyperref[$\mathcal{L}^p_\cf(X; E)$]{definition:locally-integrable}. The set
+    \[
+    L^p_\cf(X, \cm, \mu; E) = \mathcal{L}^p_\cf(X, \cm, \mu; E)/\bracs{f|f = 0\text{ a.e.}}
+    \]
+
+    equipped with the seminorms $\bracsn{[\cdot]_{L^p_A(X;\mu)}|A \in \cf}$ is a separated locally convex space, and the \textbf{local $L^p$ space} of $(X, \cm, \mu)$. 
+\end{definition}
+
+\begin{lemma}
+\label{lemma:gluing-local-lp}
+    Let $(X, \cm, \cf, \mu)$ be a \hyperref[scaffolded]{definition:measure-scaffold} localisable measure space, $E$ be a normed vector space over $K \in \RC$, $p \in [1, \infty]$, and $\bracsn{f_A}_{A \in \cf}$ such that:
+    \begin{enumerate}[label=(\alph*)]
+        \item For each $A \in \cf$, $f_A \in \mathcal{L}^p(A; E)$. 
+        \item For each $A, B \in \cf$, $f_A|_{A \cap B} = f_B|_{A \cap B}$ almost everywhere.
+        \item $\bigcup_{A \in \cf}f_A(A)$ is a separable subset of $E$. 
+    \end{enumerate}
+
+    then there exists a unique $f \in L^p_\cf(X; E)$ such that $f|_A = f_A$ for all $A \in \cf$. 
+\end{lemma}
+\begin{proof}
+    By the \hyperref[gluing lemma for measurable functions]{lemma:gluing-measurable}.
+\end{proof}
+
+
+

src/fa/lp/seq.tex

@@ -0,0 +1,100 @@
+\section{$l^p$ Direct Sums}
+\label{section:lp-direct-sum}
+
+\begin{definition}[$l^p$-Direct Sum]
+\label{definition:lp-direct-sum}
+    Let $\seqi{X}$ be normed vector spaces over $K \in \RC$ and $p \in [1, \infty)$, then the \textbf{$l^p$-direct sum} of $\seqi{X}$ is the space
+    \[
+    [l^p(I); X_i] = \bracs{x \in \prod_{i \in I}X_i \bigg | \sum_{i \in I}\norm{x_i}_{X_i}^p < \infty}
+    \]
+
+    equipped with the norm
+    \[
+    \norm{x}_{[l^p(I); X_i]} = \braks{\sum_{i \in I}\norm{x_i}_{X_i}^{p}}^{1/p}
+    \]
+\end{definition}
+
+\begin{definition}[$l^\infty$-Direct Product]
+\label{definition:l-infty-direct-product}
+    Let $\seqi{X}$ be normed vector spaces over $K \in \RC$ and $p \in [1, \infty)$, then the \textbf{$l^\infty$-direct product} of $\seqi{X}$ is the space
+    \[
+    [l^\infty(I); X_i] = \bracs{x \in \prod_{i \in I}X_i \bigg | \sup_{i \in I}\norm{x_i}_{X_i} < \infty}
+    \]
+
+    equipped with the norm
+    \[
+    \norm{x}_{[l^\infty(I); X_i]} = \sup_{i \in I}\norm{x_i}_{X_i}
+    \]
+\end{definition}
+
+\begin{proposition}
+\label{proposition:lp-direct-sum-gymnastics}
+    Let $\seqi{X}$ be normed vector spaces over $K \in \RC$ and $p \in [1, \infty]$. 
+    \begin{enumerate}
+        \item (\textbf{Hölder's Inequality}) Let $q \in [1, \infty]$ be the Hölder conjugate of $p$, $\seqi{Y}, \seqi{Z}$ be normed spaces, and $\seqi{\lambda}$ such that for each $i \in I$, $\lambda \in L^2(X_i, Y_i; Z)$.
+        
+        For each $x \in [l^p(I); X_i]$ and $y \in [l^q(I); Y_i]$, let $\lambda(x, y)_i = \lambda_i(x_i, y_i)$, then
+        \[
+        \norm{\lambda(x, y)}_{[l^1(I); Z_i]} \le \norm{x}_{[l^p(I); X_i]} \cdot \norm{y}_{[l^q(I); X_i]} \cdot \sup_{i \in I}\norm{\lambda_i}_{L^2(X_i, Y_i; Z_i)}
+        \]
+        \item (\textbf{Minkowski's Inequality}) For each $x, y \in [l^p(I); X_i]$,
+        \[
+        \norm{x + y}_{[l^p(I); X_i]} \le \norm{x}_{[l^p(I); X_i]} + \norm{y}_{[l^p(I); X_i]}
+        \]
+
+        \item (\textbf{Markov's Inequality}) If $p < \infty$, then for each $\alpha > 0$ and $x \in [l^p(I); X_i]$, 
+        \[
+        |\bracsn{i \in I|\ \norm{x}_{X_i} \ge \alpha}| \le \frac{1}{\alpha^p}\norm{f}_{[l^p(I); X_i]}^p
+        \]
+
+        In particular, $\bracs{i \in I|x_i \ne 0}$ is countable. 
+        \item For any $q \in [p, \infty]$, $[l^p(I); X_i] \subset [l^q(I); X_i]$, where for any $x \in [l^p(I); X_i]$, $\norm{x}_{[l^q(I); X_i]} \le \norm{x}_{[l^p(I); X_i]}$. 
+        \item If $X_i$ is a Banach space for all $i \in I$, then so is $[l^p(I); X_i]$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1), (2), (3): By the classical \hyperref[Hölder's inequality]{theorem:holder}, \hyperref[Minkowski's inequality]{theorem:minkowski}, and \hyperref[Markov's inequality]{theorem:markov-inequality}.
+    
+    (4): For each $i \in I$, $\norm{x_i}_{X_i} \le \norm{x}_{[l^p(I); X_i]}$, so the result holds when $q = \infty$. 
+
+    If $q < \infty$, then by \autoref{proposition:lp-intersection-interpolation}, there exists $\lambda \in [p, q]$ such that 
+    \[
+    \norm{x}_{[l^p(I); X_i]} \le \norm{x}_{[l^p(I); X_i]}^{\lambda}\norm{x}_{[l^\infty(I); X_i]}^{1 - \lambda} \le \norm{x}_{[l^p(I); X_i]}
+    \]
+\end{proof}
+
+\begin{theorem}
+\label{theorem:lp-sum-dual}
+    Let $\seqi{X}$ be normed vector spaces over $K \in \RC$ and $p \in [1, \infty)$ and $q \in (1, \infty]$ be Hölder conjugates. For each $y \in [l^q(I); X_i^*]$, let
+    \[
+    \phi_y: [l^p(I); X_i] \to K \quad x \mapsto \sum_{i \in I}\dpn{x_i, y_i}{X_i}
+    \]
+
+    then the mapping
+    \[
+    [l^q(I); X_i^*] \to [l^p(I); X_i]^* \quad y \mapsto \phi_y
+    \]
+
+    is an isometric isomorphism. 
+\end{theorem}
+\begin{proof}
+    Let $\phi \in [l^p(I); X_i]^*$, then there exists $y \in [l^\infty(I); X_i^*]$ such that for each $i \in I$ and $x \in X_i$, $\dpn{x_i \cdot \one_{\bracs{i}}, \phi}{[l^p(I); X_i]} = \dpn{x_i, y_i}{X_i}$. 
+
+    Since the $q = \infty$ case has been ruled out, assume that $q \in (1, \infty)$. For each $\alpha \in (0, 1)$, there exists $x \in [l^\infty(I); X_i]$ with $\norm{x_i}_{X_i} \le 1$ and $\dpn{x_i, y_i}{X_i} \ge \alpha \norm{y_i}_{X_i^*}$. For each $J \subset I$ finite and $i \in I$, let $F_J(i) = \one_{J}(i) \cdot \norm{y_i}_{X_i^*}^{q - 1}$, then by \autoref{lemma:holder-conjugate-gymnastics},
+    \[
+    \norm{F_J x}_{[l^p(I); X_i]}^p \le \sum_{j \in J}\norm{y_j}_{X_j^*}^{p(q - 1)}= \sum_{j \in J}\norm{y_j}_{X_j^*}^{q}
+    \]
+
+    so 
+    \begin{align*}
+    \alpha \sum_{j \in J} \norm{y_j}_{X_j^*}^q & \le 
+    \sum_{i \in I}F_J(i)\dpn{x_i, y_i}{X_i} = 
+    \dpn{F_J x, \phi}{[l^p(I); X_i]} \\
+    &\le \norm{\phi}_{[l^p(I); X_i]^*} \cdot \braks{\sum_{j \in J}\norm{y_j}_{X_j^*}^{q}}^{1/p} \\
+    \alpha \braks{\sum_{j \in J}\norm{y_j}_{X_j^*}^q}^{1/q} &\le \norm{\phi}_{[l^p(I); X_i]^*}
+    \end{align*}
+
+    As the above holds for all $\alpha \in (0, 1)$ and $J \subset I$ finite, $y \in [l^q(I); X_i^*]$ with $\norm{y}_{[l^q(I); X_i^*]} = \norm{\phi}_{[l^p(I); X_i]^*}$. 
+    
+\end{proof}
+

src/fa/lp/ui.tex

@@ -0,0 +1,260 @@
+\section{Uniform Integrability}
+\label{section:uniform-integrable}
+
+
+
+\begin{definition}[Uniform Integrability]
+\label{definition:uniform-integrable}
+    Let $(X, \cm, \mu)$ be a measure space, $p \in [1, \infty)$, $E$ be a normed vector space over $K \in \RC$, and $\cf \subset L^p(X; E)$, then $\cf$ is \textbf{uniformly $p$-integrable} if
+    \[
+    \lim_{M \to \infty}\sup_{f \in \cf} \int_{\bracs{\norm{f}_E \ge M}} \norm{f}^p d\mu = 0
+    \]
+\end{definition}
+
+\begin{proposition}
+\label{proposition:ui-gymnastics}
+    Let $(X, \cm, \mu)$ be a measure space, $p \in [1, \infty)$, $E$ be a normed vector space over $K \in \RC$, $\cf \subset L^p(X; K)$, and let $|\cf|^p = \bracsn{\norm{f}_E^p| f \in \cf}$, then:
+    \begin{enumerate}
+        \item If $\cf$ is uniformly $p$-integrable, then $|\cf|^p$ is uniformly absolutely continuous with respect to $\mu$. 
+        \item If $\sup_{f \in \cf}\norm{f}_{L^p(X; E)} < \infty$ and $|\cf|^p$ is uniformly absolutely continuous with respect to $\mu$, then $\cf$ is uniformly $p$-integrable.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): Let $\eps > 0$, then there exists $M \ge 0$ such that
+    \[
+    \sup_{f \in \cf}\int_{\bracs{\norm{f}_E \ge M}} \norm{f}^p d\mu < \eps/2
+    \]
+
+    Thus for any $A \in \cm$, 
+    \begin{align*}
+        \sup_{f \in \cf}\int_{A}\norm{f}^p d\mu &\le \sup_{f \in \cf}\int_{\bracs{\norm{f}_E \le M} \cap A} \norm{f}^p d\mu+ \sup_{f \in \cf}\int_{\bracs{\norm{f}_E \ge M} \cap A} \norm{f}^p d\mu \\
+        &\le M \mu(A) + \eps/2
+    \end{align*}
+
+    so if $\mu(A) \le M\eps/2$, then $\sup_{f \in \cf}\int_{A}\norm{f}^p d\mu \le \eps$. 
+    
+    (2): Let $\eps > 0$, then there exists $\delta > 0$ such that for any $A \in \cm$ with $\mu(A) < \delta$, $\sup_{f \in \cf}\int_A \norm{f}^p d\mu < \eps$. By \hyperref[Markov's inequality]{theorem:markov-inequality}, 
+    \[
+    \sup_{f \in \cf}\mu(\bracs{\norm{f}_E \ge M}) \le \frac{\sup_{f \in \cf}\norm{f}_{L^p(X; E)}^p}{M^p}
+    \]
+
+    Thus for sufficiently large $M$, $\sup_{f \in \cf}\mu(\bracs{\norm{f}_E \ge M}) < \delta$, and 
+    \[
+    \sup_{f \in \cf}\int_{\bracs{\norm{f}_E \ge M}} \norm{f}^p d\mu < \eps
+    \]
+\end{proof}
+
+
+
+\begin{theorem}[Vitali Convergence Theorem]
+\label{theorem:vitali-convergence}
+    Let $(X, \cm, \mu)$ be a measure space, $p \in [1, \infty)$, $E$ be a normed vector space over $K \in \RC$, and $\fF \subset 2^{L^p(X; E)}$ be a filter, then $\fF$ is Cauchy in $L^p(X; E)$ if and only if:
+    \begin{enumerate}
+        \item[(M)] $\fF$ is locally Cauchy in measure. 
+        \item[(UI)] For each $\eps > 0$, there exists $M \ge 0$ and $F \in \fF$ such that
+        \[
+        \sup_{f \in F}\int_{\bracs{\norm{f}_E \ge M}}\norm{f}_E^p d\mu < \eps
+        \]
+        \item[(T)] For each $\eps > 0$, there exists $A \in \cm$ and $F \in \fF$ with $\mu(A) < \infty$ and $\sup_{f \in F}\int_{A^c}\norm{f}_E^p < \eps$.
+    \end{enumerate}
+
+    % M stands for measure, UI stands for Uniform Integrability, T stands for tightness. 
+\end{theorem}
+\begin{proof}
+    ($L^p$) $\Rightarrow$ (M): By \hyperref[Markov's inequality]{theorem:markov-inequality}. 
+
+    ($L^p$) $\Rightarrow$ (UI): Let $F \in \fF$ such that for each $f, g \in F$, $\norm{f - g}_{L^p(X; E)} < \eps$. Fix $f \in F$, then by the \hyperref[Dominated Convergence Theorem]{theorem:dct}, there exists $M > 0$ such that $\int_{\bracs{\norm{f}_E \ge M}}\norm{f}_E^p d\mu < \eps$. For any $g \in F$, 
+    \[
+    \bracs{\norm{g}_E \ge 2M} \subset \bracs{\norm{f}_E \ge M} \cup \bracs{\norm{g}_E \ge 2M, \norm{f}_E \le M}
+    \]
+
+    so by \autoref{lemma:lp-p-diff},
+    \begin{align*}
+        \int_{\bracs{\norm{g}_E \ge 2M}}\norm{g}_E^pd\mu &\le \int_{\bracs{\norm{f}_E \ge M}}\norm{g}_E^pd\mu + \int_{\bracs{\norm{g}_E \ge 2M, \norm{f}_E \le M}}\norm{g}_E^pd\mu \\
+        &\le \int_{\bracs{\norm{f}_E \ge M}}\norm{f}_E^pd\mu + \int_{\bracs{\norm{g}_E \ge 2M, \norm{f}_E \le M}}\norm{f}_E^pd\mu \\
+        &+ 2p\norm{f - g}_{L^p(X; E)}(\norm{f}_{L^p(X; E)} \vee \norm{g}_{L^p(X; E)})^{p-1}
+    \end{align*}
+
+    Since $\bracs{\norm{g}_E \ge 2M, \norm{f}_E \le M} \subset \bracs{\norm{f - g}_E \ge M}$, by \hyperref[Markov's inequality]{theorem:markov-inequality}, 
+    \[
+    \int_{\bracs{\norm{g}_E \ge 2M, \norm{f}_E \le M}}\norm{f}_E^pd\mu \le M^p\mu\bracs{\norm{f - g}_E \ge M} \le \norm{f - g}_{L^p(X; E)}^p
+    \]
+
+    Therefore
+    \[
+    \sup_{g \in F}\int_{\bracs{\norm{g}_E \ge 2M}}\norm{g}_E^p d\mu \le 2p \eps (\norm{f}_{L^p(X; E)} + \eps)^{p - 1} + \eps + \eps^p
+    \]
+
+    ($L^p$) $\Rightarrow$ (T): Let $F \in \fF$ such that for each $f, g \in F$, $\norm{f - g}_{L^p(X; E)} < \eps$. Fix $f \in F$, then by the \hyperref[Dominated Convergence Theorem]{theorem:dct}, there exists $A \in \cm$ such that $\mu(A) < \infty$ and $\norm{\one_{A^c}f}_{L^p(X; E)} < \eps$. In which case, for any $g \in F$, 
+    \[
+        \norm{\one_{A^c}g}_{L^p(X; E)} \le \norm{\one_{A^c}f}_{L^p(X; E)} + \norm{f - g}_{L^p(X; E)}
+        \le \norm{\one_{A^c}f}_{L^p(X; E)} + \eps
+    \]
+    
+
+
+    (M) + (UI) + (T) $\Rightarrow$ ($L^p$): Let $\eps > 0$. By (T), there exists $A \in \cm$ and $F_1 \in \fF$ with $\mu(A) < \infty$ and $\sup_{f \in F_1}\int_{A^c}\norm{f}_E^p < \eps^p$. Thus for every $f, g \in F_1$, 
+    \begin{align*}
+        \norm{f - g}_{L^p(X; E)} &\le \norm{\one_A(f - g)}_{L^p(X; E)}  + \norm{\one_{A^c}f}_{L^p(X; E)}  + \norm{\one_{A^c}g}_{L^p(X; E)} \\
+        &\le \norm{\one_A(f - g)}_{L^p(X; E)} + 2\eps
+    \end{align*}
+
+    By (UI), there exists $M > 0$ and $F_2 \in \fF$ with $F_2 \subset F_1$ such that
+    \[
+    \sup_{h \in F_2} \int_{\bracs{\norm{h}_E \ge M}}\norm{h}_E^p < \eps^p
+    \]
+    
+    Assume without loss of generality that $\mu(A) > 0$ and let $\delta = \eps\mu(A)^{-1/p}$. By (M), there exists $F_3 \in \fF$ with $F_3 \subset F_2$, such that for any $f, g \in F_3$, 
+    \[
+    \mu(A \cap \bracsn{\norm{f - g}_E \ge \delta}) \le \paren{\frac{\eps}{2M}}^p
+    \]
+    
+    In which case,
+    \begin{align*}
+        \norm{\one_{A}(f - g)}_{L^p(X; E)} &\le \normn{\one_{A \cap \bracs{\norm{f - g}_E \le \delta}}(f - g)}_{L^p(X; E)} \\
+        &+ \normn{\one_{A \cap \bracsn{\norm{f - g}_E \ge \delta}}(f - g)}_{L^p(X; E)} \\
+        &\le \delta\mu(A)^{1/p} + \normn{\one_{A \cap \bracsn{\norm{f - g}_E \ge \delta}}(f - g)}_{L^p(X; E)}  \\
+        &\le \normn{\one_{A \cap \bracsn{\norm{f - g}_E \ge \delta}}(f - g)}_{L^p(X; E)} + \eps
+    \end{align*}
+
+    then for any $f, g \in F_3$, 
+    \begin{align*}
+        \normn{\one_{A \cap \bracsn{\norm{f - g}_E \ge \delta}}(f - g)}_{L^p(X; E)} &\le \normn{\one_{A \cap \bracsn{\norm{f - g}_E \ge \delta}}f}_{L^p(X; E)} \\
+        &+ \normn{\one_{A \cap \bracsn{\norm{f - g}_E \ge \delta}}g}_{L^p(X; E)}
+    \end{align*}
+    
+    Now, 
+    \begin{align*}
+        \normn{\one_{A \cap \bracsn{\norm{f - g}_E \ge \delta}}f}_{L^p(X; E)} &\le \normn{\one_{\bracsn{\norm{f}_E \ge M}}f}_{L^p(X; E)} \\
+        &+ \normn{\one_{A \cap \bracsn{\norm{f - g}_E \ge \delta, \norm{f}_E \le M}}f}_{L^p(X; E)} \\
+        &\le \eps + \eps/2 = 3\eps/2
+    \end{align*}
+
+    Similarly, $\normn{\one_{A \cap \bracsn{\norm{f - g}_E \ge \delta}}g}_{L^p(X; E)} \le 3\eps/2$. Thus
+    \[
+    \normn{\one_{A \cap \bracsn{\norm{f - g}_E \ge \delta}}(f - g)}_{L^p(X; E)} \le 3\eps
+    \]
+    
+    
+    Therefore for any $f, g \in F_3$, $\norm{f - g}_{L^p(X; E)} \le 6 \eps$, so $\fF$ is Cauchy in $L^p(X; E)$.  
+    
+\end{proof}
+
+\begin{corollary}[Dominated Convergence Theorem (In Measure)]
+\label{corollary:dct-filter}
+    Let $(X, \cm, \mu)$ be a measure space, $p \in [1, \infty)$, $E$ be a normed vector space over $K \in \RC$, $\fF \subset 2^{L^p(X; E)}$ be a filter, and $g, h \in L^p(X; \real)$ such that:
+    \begin{enumerate}[label=(\alph*)]
+        \item[(M)] $\fF \to g$ locally in measure.
+        \item[(D)] There exists $F \in \fF$ such that $|f| \le h$ for all $f \in F$. 
+    \end{enumerate}
+    
+    then $\fF \to f$ in $L^p(X; E)$. In particular, if $p = 1$, then
+    \[
+    \lim_{f, \fF}\int f d\mu = \int g d\mu
+    \]
+\end{corollary}
+\begin{proof}
+    Since (D) implies (UI) and (T) of the \hyperref[Vitali Convergence Theorem]{theorem:vitali-convergence}, the result follows from the Vitali Convergence Theorem. 
+\end{proof}
+
+\begin{lemma}[Scheffé]
+\label{lemma:scheffe}
+    Let $(X, \cm, \mu)$ be a measure space, $p \in [1, \infty)$, $E$ be a separable normed vector space over $K \in \RC$, $\fF \subset 2^{L^p(X; E)}$ be a filter, and $g \in L^p(X; E)$, then $\fF \to g$ in $L^p(X; E)$ if and only if:
+    \begin{enumerate}
+        \item[(M)] $\fF \to g$ locally in measure. 
+        \item[(N)] $\lim_{f, \fF} \norm{f}_{L^p(X; E)} = \norm{g}_{L^p(X; E)}$. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    It is sufficient to show conditions (UI) and (T) of the \hyperref[Vitali Convergence Theorem]{theorem:vitali-convergence}. 
+
+    (T): Let $\eps > 0$. By (N), there exists $F_1 \in \fF$ such that $|\norm{f}_{L^p(X; E)}^p - \norm{g}_{L^p(X; E)}^p| < \eps$ for all $f \in F_1$. By the \hyperref[Dominated Convergence Theorem]{theorem:dct}, there exists $A \in \cm$ with $\mu(A) < \infty$ such that:
+    \begin{enumerate}[label=(\roman*)]
+        \item $\int_{A^c} \norm{g}_E^p < \eps$. 
+    \end{enumerate}
+    
+    Since $\norm{g}_E^p d\mu \ll \mu$, by \autoref{proposition:ac-epsilon-delta}, there exists $\delta > 0$ such that:
+    \begin{enumerate}[label=(\roman*), start=1]
+        \item For each $B \in \cm$ with $\mu(B) < \delta$, $\int_{B}\norm{g}_E^p d\mu < \eps$. 
+        \item $\int_{A}[(\norm{g}_E - \delta) \vee 0]^p d\mu > \int_A \norm{g}_E^p d\mu - \eps$. 
+    \end{enumerate}
+
+    By (M), there exists $F_2 \in \fF$ with $F_2 \subset F_1$ such that for each $f \in F_2$, $\mu(A \cap \bracs{\norm{f - g}_E > \delta}) < \delta$. In which case, for any $f \in F_2$, 
+    \begin{align*}
+        \int_A \norm{f}_E^p d\mu &\ge \int_{A \cap \bracs{\norm{f - g}_E \le \delta}}\norm{f}_E^p d\mu \\
+        &\ge \int_{A}[(\norm{g}_E - \delta) \vee 0]^p d\mu - \int_{A \cap \bracs{\norm{f - g}_E > \delta}}\norm{g}_E^p d\mu
+    \end{align*}
+
+    By (iii), 
+    \[
+    \int_A \norm{f}_E^p d\mu \ge \int_A \norm{g}_E^p d\mu - \eps - \int_{A \cap \bracs{\norm{f - g}_E > \delta}}\norm{g}_E^p d\mu
+    \]
+
+    and by (ii), $\int_A \norm{f}_E^p d\mu \ge \int_A\norm{g}_E^p d\mu - 2\eps$. Finally, by (i), 
+    \[
+    \int_A \norm{f}_E^p d\mu \ge \int \norm{g}_E^p d\mu - 3\eps \ge \int \norm{f}_E^p d\mu - 4\eps
+    \]
+
+    and $\int_{A^c}\norm{f}_E^p d\mu \le 4\eps$. 
+    
+    
+    (UI): Let $\eps > 0$. By (N) and (T), there exists $F_1 \in \fF$ and $A \in \cm$ with $\mu(A) < \infty$ such that for every $f \in F_1$, 
+    \begin{enumerate}[label=(\roman*)]
+        \item $|\norm{f}_{L^p(X; E)}^p - \norm{g}_{L^p(X; E)}^p| < \eps$. 
+        \item $\int_{A^c}\norm{f}_E^p d\mu, \int_{A^c}\norm{g}_E^p d\mu < \eps$. 
+    \end{enumerate}
+    
+    By (i) and (ii), 
+    \begin{enumerate}[label=(\roman*), start=2]
+        \item $\abs{\int_{A}\norm{f}_E^p d\mu - \int_{A}\norm{g}_E^p}d\mu \le 3\eps$. 
+    \end{enumerate}
+    
+
+    Since $\norm{g}_E^p d\mu \ll \mu$, by \autoref{proposition:ac-epsilon-delta}, there exists $\delta > 0$ such that:
+    \begin{enumerate}[label=(\roman*), start=3]
+        \item For each $B \in \cm$ with $\mu(B) < \delta$, $\int_{B}\norm{g}_E^p d\mu < \eps$. 
+        \item $\int_{A}[(\norm{g}_E - \delta) \vee 0]^p d\mu > \int_A \norm{g}_E^p d\mu - \eps$. 
+    \end{enumerate}
+
+    By (M), there exists $F_2 \in \fF$ with $F_2 \subset F_1$ such that for every $f \in F_2$,  
+    \begin{enumerate}[label=(\roman*), start=5]
+        \item $\mu(A \cap \bracs{\norm{f - g}_E > \delta}) < \delta$. 
+    \end{enumerate}
+    
+    Let $B \in \cm$ with $B \subset A$ and $\mu(B) < \delta$, then for any $f \in F_2$, 
+    \begin{align*}
+        \int_{A \setminus B}\norm{f}_E^p d\mu &\ge \int_{(A \setminus B) \cap \bracs{\norm{f - g}_E \le \delta}}\norm{f}_E^p d\mu \\
+        &\ge \int_{(A \setminus B) \cap \bracs{\norm{f - g}_E \le \delta}}(\norm{g}_E - \delta \vee 0)^pd\mu \\
+        &\ge \int_{(A \setminus B)}(\norm{g}_E - \delta \vee 0)^pd\mu - \int_{A \cap \bracs{\norm{f - g}_E > \delta}} \norm{g}_E^p d\mu
+    \end{align*}
+
+    By (vi) and (iv), $\int_{A \cap \bracs{\norm{f - g}_E > \delta}} \norm{g}_E^p d\mu < \eps$, so
+    \[
+    \int_{A \setminus B}\norm{f}_E^p d\mu \ge \int_{(A \setminus B)}(\norm{g}_E - \delta \vee 0)^pd\mu - \eps
+    \]
+
+    By (v), 
+    \[
+    \int_{A \setminus B}\norm{f}_E^p d\mu \ge \int_{(A \setminus B)}\norm{g}_E^pd\mu - 2\eps
+    \]
+
+    Since $\mu(B) < \delta$, by (iv),
+    \[
+    \int_{A \setminus B}\norm{f}_E^p d\mu \ge \int_{A}\norm{g}_E^pd\mu - 3\eps
+    \]
+
+    and by (iii), 
+    \[
+    \int_{A \setminus B}\norm{f}_E^p d\mu \ge \int_{A}\norm{f}_E^pd\mu - 6\eps
+    \]
+
+    so $\int_{B}\norm{f}_E^p d\mu \le 6\eps$ for all $B \in \cm$ with $B \subset A$ and $\mu(B) < \delta$.
+    
+    Finally, by (i) and \hyperref[Markov's Inequality]{theorem:markov-inequality}, there exists $M \ge 0$ such that $\mu\bracs{\norm{f}_E \ge M} < \delta$ for all $f \in F_2$. Therefore for any $f \in F_2$, 
+    \[
+        \int_{\bracs{\norm{f}_E \ge M}}\norm{f}_E^p d\mu \le \int_{A \cap \bracs{\norm{f}_E \ge M}}\norm{f}_E^p d\mu + \int_{A^c}\norm{f}_E^p d\mu \le 7\eps
+    \]
+    
+    
+\end{proof}
+

src/fa/norm/hilbert.tex

@@ -83,20 +83,23 @@
 
 \begin{definition}[Inner Product]
 \label{definition:inner-product}
-    Let $E$ be a vector space over $K$ and $\inp_E: E \times E \to K$, then $\inp_E$ is an \textbf{inner product} if:
+    Let $E$ be a vector space over $K$ and $\inp_E: E \times E \to K$, then $\inp_E$ is a \textbf{pseudo inner product} if:
     \begin{enumerate}[label=(H\arabic*)]
         \item For each $x, y, z \in E$, $\angles{x + y, z}_E = \dpn{x, z}{E} + \dpn{y, z}{E}$. 
         \item For any $x, y \in E$ and $\mu \in K$, $\dpn{\mu x, y}{E} = \mu \dpn{x, y}{E}$. 
         \item For every $x, y \in E$, $\dpn{x, y}{E} = \ol{\dpn{y, x}{E}}$. 
-        \item[(I)] For each $x \in E$, $\dpn{x, x}{E} \ge 0$, with equality if and only if $x = 0$. 
+        \item[(I)] For each $x \in E$, $\dpn{x, x}{E} \ge 0$. 
     \end{enumerate}
 
+    and an \textbf{inner product} if for each $x \in E$, $\dpn{x, x}{E} = 0$ if and only if $x = 0$. 
+    
+
 \end{definition}
 
 
 \begin{proposition}[Cauchy-Schwarz Inequality]
 \label{proposition:cauchy-schwarz}
-    Let $H$ be a vector space over $K \in \RC$ and $\inp_H: E \times E \to K$ be an inner product, then for any $x, y \in H$, $\dpn{x, y}{H} \le \norm{x}_H\norm{y}_H$. 
+    Let $H$ be a vector space over $K \in \RC$ and $\inp_H: E \times E \to K$ be a pseudo inner product, then for any $x, y \in H$, $\dpn{x, y}{H} \le \norm{x}_H\norm{y}_H$. 
 \end{proposition}
 \begin{proof}[Proof, {{\cite[Theorem 5.19]{Folland}}}. ]
     Assume without loss of generality that $\dpn{x, y}{H} > 0$, then for each $t \in \real$, 
@@ -352,7 +355,7 @@ A significant property of Hilbert spaces is that every closed subspace is comple
     \phi_x: H \to \complex \quad \phi_x(y) = \dpn{y, x}{E}
     \]
 
-    then the mapping $H \to H^*$ defined by $x \mapsto \phi_x$ is an isometric conjugate linear isomorphism. 
+    then the mapping $H \to H^*$ defined by $x \mapsto \phi_x$ is an conjugate linear isometric isomorphism. 
 \end{theorem}
 \begin{proof}
     By the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz} and definition of the norm, $x \mapsto \phi_x$ is an isometric conjugate linear map. 

src/fa/norm/normed.tex

@@ -68,7 +68,7 @@
 \end{proof}
 
 
-\begin{theorem}[Successive Approximation]
+\begin{theorem}[Successive Approximations]
 \label{theorem:successive-approximation}
     Let $E, F$ be normed vector spaces, $T \in L(E; F)$, $C \ge 0$, and $\gamma \in (0, 1)$. If for all $y \in F$, there exists $x \in E$ such that:
     \begin{enumerate}
@@ -84,7 +84,7 @@
 
     In particular, if $E$ is a Banach space, then for every $y \in F$, there exists $x \in E$ such that $\norm{x}_E \le C\norm{y}_F/(1 - \gamma)$ and $Tx = y$.
 \end{theorem}
-\begin{proof}
+\begin{proof}[Proof, learned from Anson Li (https://ansonli0.com/). ]
     Let $y_1 = y \in F$. Let $n \in \natp$ and suppose inductively that $\bracs{x_k| 0 \le k < n}$ and $\bracs{y_k| 0 \le k \le n}$ has been constructed. By (a) and (b), there exists $x_n \in E$ such that $\norm{x_k}_E \le C\norm{y_k}_F$ and $\norm{y_n - Tx_n}_F \le \gamma \norm{y_{n}}_F$. Let $y_{n+1} = y_n - Tx_n$, then $\norm{y_{n+1}}_F \le \gamma \norm{y_n}_F$.
 
     For each $n \in \nat$,

src/fa/norm/separable.tex

@@ -3,22 +3,32 @@
 
 \begin{proposition}
 \label{proposition:separable-dual}
-    Let $E$ be a separable normed vector space, then $E^*$ is separable with respect to the weak*-topology. 
+    Let $K \in \RC$ and $\dpn{E, F}{\lambda}$ be a duality of normed vector spaces over $K$ with $E$ being separable, and $S = \bracsn{y \in F|\ \norm{y}_F \le 1}$ be the closed unit ball of $F$, then
+    \begin{enumerate}
+        \item $S$ is separable with respect to the $\sigma(F, E)$-topology. 
+        \item $S$ is metrisable with respect to the $\sigma(F, E)$-topology. 
+        \item For any $A \subset F$, $A$ is separable with respect to the $\sigma(F, E)$-topology. 
+        \item If the duality is norming, then there exists $\seq{y_n} \subset F$ such that for each $x \in E$, $\norm{x}_E = \sup_{n \in \natp}|\dpn{x, y_n}{\lambda}|$. 
+    \end{enumerate}
 \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
+    (1): Let $\seq{x_n} \subset E$ be a dense subset. For each $N \in \natp$, let
     \[
-    T_N: S \to \real^N \quad \phi \mapsto (\dpn{x_1, \phi}{E}, \cdots, \dpn{x_N, \phi}{E})
+    T_N: S \to \real^N \quad y \mapsto (\dpn{x_1, y}{\lambda}, \cdots, \dpn{x_N, y}{\lambda})
     \]
 
-    Since $\real^N$ is separable, $T_N(S)$ is separable by \autoref{proposition:separable-metric-space}. Thus 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 $T_N(S)$.
+    Since $\real^N$ is separable, $T_N(S)$ is separable by \autoref{proposition:separable-metric-space}. Thus there exists $\bracs{y_{N, k}}_{k = 1}^\infty \subset S$ such that $\bracs{T_Ny_{N, k}}_{k = 1}^\infty$ 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 $1 \le n \le N$, 
+    Let $y \in S$, then for each $N \in \natp$, there exists $k_N \in \natp$ such that for each $1 \le n \le N$, 
     \[
-    |\dpn{x_n, \phi_{N, k_N}}{E} - \dpn{x_n, \phi}{E}| \le \frac{1}{N}
+    |\dpn{x_n, y_{N, k_N}}{\lambda} - \dpn{x_n, y}{\lambda}| \le \frac{1}{N}
     \] 
 
-    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}. 
+    Thus for each $N \in \natp$, $\dpn{x_n, y_{N, k_N}}{\lambda} \to \dpn{x_n, y}{\lambda}$ as $N \to \infty$. Since $y_{N, k_N} \to y$ pointwise on a dense subset of $E$ and $\bracsn{y_{N, k_N}|N \in \natp} \subset S$ is uniformly equicontinuous, $y_{N, k_N} \to y$ in the $\sigma(F, E)$-topology by \autoref{proposition:strong-operator-dense}. 
+
+    (2): Let $\seq{x_n} \subset E$ be a dense subset, then by \autoref{proposition:strong-operator-dense}, the $\sigma(F, E)$-topology on $S$ is induced by $\seq{x_n}$, and hence metrisable by \autoref{theorem:uniform-metrisable}. 
+
+    (3): For any $A \subset E$, $A = \bigcup_{n \in \natp}A \cap nS$. By \autoref{proposition:separable-metric-space}, $A \cap nS$ is separable for each $n \in \natp$. Therefore $A$ is also separable. 
 \end{proof}
 
 \begin{proposition}

src/fa/notation.tex

@@ -26,6 +26,7 @@
     $E \otimes_\pi F$ & Projective tensor product of $E$ and $F$. & \autoref{definition:projective-tensor-product} \\
     $E \,\widetilde{\otimes}_\pi F$ & Projective completion of $E$ and $F$. & \autoref{definition:projective-tensor-product} \\
     $p \otimes q$ & Cross seminorm of $p$ and $q$. & \autoref{definition:cross-seminorm} \\
+    $N(E; F)$ & Nuclear mappings from $E$ to $F$. & \autoref{definition:nuclear-operator-normed} \\ 
     % ---- Order Structures ----
     $x \vee y$, $x \wedge y$ & $\sup$ and $\inf$ in vector lattice. & \autoref{definition:vector-lattice} \\
     $|x|$ & Absolute value $x \vee (-x)$ in a vector lattice. & \autoref{definition:order-absolute-value} \\
@@ -48,7 +49,13 @@
     $\mathrm{Reg}([a,b], G; E)$ & Regulated functions w.r.t.\ $G$ on $[a,b]$. & \autoref{definition:regulated-function} \\
     $\mu_G$ & Lebesgue-Stieltjes measure associated with $G$. & \autoref{definition:riemann-lebesgue-stieltjes} \\
     $\int_\gamma f$, $\int_\gamma f(z)dz$ & Path integral of $f$ with respect to $\gamma$. & \autoref{definition:path-integral} \\
-    $PI([a, b], \gamma; E)$ & Space of path integrable functions with respect to $\gamma$. & \autoref{definition:path-integral}
+    $PI([a, b], \gamma; E)$ & Space of path integrable functions with respect to $\gamma$. & \autoref{definition:path-integral} \\ 
+    % ---- Convex Functions ---- \\
+    $\partial f(x)$ & Subdifferential of $f$ at $x$. & \autoref{definition:subgradient} \\
+    $(\phi, \alpha) \le f$ & $\phi - \alpha \le f$. $(\phi, \alpha)$ is an affine minorant of $f$. & \autoref{definition:affine-minorant} \\ 
+    $f^*$ & Conjugate function of $f$. & \autoref{definition:conjugate-function} \\ 
+    $I_A$ & Indicator/infinity characteristic function of $A$. & \autoref{definition:infinity-characteristic-function} \\
+    $H_A$ & Support function of $A$ &\autoref{definition:support-function} \\ 
     % ---- Interpolation Spaces ---- \\
     $\catc_1$ & Category of compatible couples in $\catc$. & \autoref{definition:compatible-category} \\
     $\cf(E_0, E_1)$ & Calderón space of $(E_0, E_1)$ & \autoref{definition:calderon-space} \\

src/fa/order/index.tex

@@ -1,5 +1,7 @@
 \chapter{Order Structures}
 \label{chap:order-structure}
 
+\input{./order.tex}
+\input{./positive.tex}
 \input{./lattice.tex}
 \input{./norm.tex}

src/fa/order/lattice.tex

@@ -1,63 +1,6 @@
 \section{Vector Lattices}
 \label{section:vector-lattice}
 
-\begin{definition}[Ordered Vector Space]
-\label{definition:ordered-vector-space}
-    Let $E$ be a vector space over $\real$ and $\le$ be a partial order on $E$, then $(E, \le)$ is a \textbf{ordered vector space} if
-    \begin{enumerate}
-        \item[(LO1)] For any $x, y, z \in E$ with $x \le y$, $x + z \le y + z$.
-        \item[(LO2)] For any $x, y \in E$ and $\lambda > 0$, $x \le y$ implies that $\lambda x \le \lambda y$.
-    \end{enumerate}
-
-    
-\end{definition}
-
-\begin{proposition}
-\label{proposition:ordered-vector-space-properties}
-    Let $(E, \le)$ be an ordered vector space and $A, B \subset E$ such that $\sup(A)$ and $\sup (B)$ exist, then 
-    \begin{enumerate}
-        \item $\sup(A + B) = \sup(A) + \sup(B)$.
-        \item $\sup(A) = -\inf (-A)$
-    \end{enumerate}
-
-    
-\end{proposition}
-\begin{proof}
-    (1): For any $a \in A$ and $b \in B$, $a + b \le \sup(A) + \sup(B)$ by (LO1). Let $c \in E$ such that $c \ge a + b$ for all $a \in A$ and $b \in B$, then $c \ge a + \sup(B)$ for all $a \in A$, so $c \ge \sup(A) + \sup(B)$. Therefore $\sup(A) + \sup(B) = \sup(A + B)$.
-
-    (2): For any $a \in A$, $\sup(A) \ge a = -(-a)$, so $-\sup(A) \le \inf(-A)$, and $\sup(A) \ge -\inf(-A)$. On the other hand, for any $a \in A$, $-\inf(-A) \ge a$. Therefore $\sup(A) \le -\inf(-A)$. 
-\end{proof}
-
-
-\begin{definition}[Interval]
-\label{definition:ordered-vector-space-interval}
-    Let $(E, \le)$ be an ordered vector space and $x, y \in E$, then
-    \[
-    [x, y] = \bracs{z \in E| x \le z \le y}
-    \]
-
-    is the \textbf{order interval} with endpoints $x$ and $y$.
-\end{definition}
-
-\begin{definition}[Order Bounded]
-\label{definition:ordered-vector-space-bounded}
-    Let $(E, \le)$ be an ordered vector space and $A \subset E$, then $A$ is \textbf{order bounded} if there exists $x, y \in E$ such that $A \subset [x, y]$.
-\end{definition}
-
-\begin{definition}[Order Complete]
-\label{definition:order-vector-complete}
-    Let $(E, \le)$ be an ordered vector space, then $E$ is \textbf{order complete} if for any order bounded set $A \subset E$, $\sup (A)$ and $\inf (A)$ exist.
-\end{definition}
-
-\begin{definition}[Order Bounded Dual]
-\label{definition:order-bounded-dual}
-    Let $(E, \le)$ be an ordered vector space, then the space $E^b$ consisting of all linear functionals on $E$ that are bounded on order bounded sets is the \textbf{order bounded dual} of $E$.
-\end{definition}
-
-\begin{definition}[Order Dual]
-\label{definition:order-dual}
-    Let $(E, \le)$ be an ordered vector space and $\Phi^+ \in \hom(E; \real)$, then $\Phi^+$ is \textbf{positive} if for any $x \in E$ with $x \ge 0$, $\Phi^+(x) \ge 0$. The subspace $E^+ \subset \hom(E; \real)$ generated by the positive linear functionals on $E$ is the \textbf{order dual} of $E$. 
-\end{definition}
 
 
 \begin{definition}[Vector Lattice]
@@ -65,7 +8,6 @@
     Let $(E, \le)$ be an ordered vector space, then $E$ is a \textbf{vector lattice} if for any $x, y \in E$, $x \vee y = \sup\bracs{x, y}$ and $x \wedge y = \inf\bracs{x, y}$ exist. 
 \end{definition}
 
-
 \begin{definition}[Absolute Value]
 \label{definition:order-absolute-value}
     Let $(E, \le)$ be a vector lattice and $x \in E$, then $|x| = x \vee -x$ is the \textbf{absolute value} of $x$. 
@@ -88,41 +30,79 @@
     Let $(E, \le)$ be a vector lattice and $x, y \in E$, then $x$ and $y$ are \textbf{disjoint}, denoted $x \perp y$, if $|x| \wedge |y| = 0$.
 \end{definition}
 
+\begin{lemma}
+\label{lemma:lattice-gymnastics}
+    Let $E$ be a vector lattice and $a, b, c, d \in E$, then
+    \begin{enumerate}
+        \item $a + b = a \vee b + a \wedge b$.
+        \item $b \le a + |b - a|$.
+        \item If $c \ge 0$, then $(a + c) \vee b \le a \vee b + c$.
+        \item $|a \vee c - b \vee c| \le |a - b|$.
+        \item $|a \vee c - b \vee d| \le |a - b| \vee |c- d|$
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    (1): By translation invariance, 
+    \begin{align*}
+        x \vee y + x \wedge y - x - y &= 0 \vee (y - x) + (x - y) \wedge 0 \\
+        &= 0 \vee (y - x) - 0 \vee (x - y) = 0
+    \end{align*}
 
-\begin{proposition}[{{\cite[V.1.1]{SchaeferWolff}}}]
+    (2): $b = a + b - a \le a + |b - a|$.
+
+    (3): Since $c \ge 0$, 
+    \[
+    (a + c) \vee b = (a + c) \vee (b - c + c) = a \vee (b - c) + c \le a \vee b + c
+    \]
+
+    (4): By (2) and then (3), 
+    \begin{align*}
+        a \vee c - b \vee c &\le (b + |a - b|) \vee c - b \vee c \\
+        &\le b \vee c + |a - b| - b \vee c \le |a - b|
+    \end{align*}
+    
+    Similarly, $b \vee c - a \vee c \le |b - a| = |a - b|$.
+
+    (5): Finally, 
+    \begin{align*}
+        a \vee c - b \vee d &\le (b + |a - b|) \vee (d + |c- d|) - b \vee d \\
+        &\le (b + |a - b| \vee |c- d|) \vee (d + |a - b| \vee |c- d|) - b \vee d \\
+        &= b \vee d + |a - b| \vee |c- d| - b \vee d \\
+        &\le |a - b| \vee |c- d| 
+    \end{align*}
+
+    Similarly, 
+    \[
+    b \vee d - a \vee c \le  |b - a| \vee |d - c| = |a - b| \vee |c- d|  
+    \]
+    
+\end{proof}
+
+
+
+\begin{proposition}
 \label{proposition:lattice-properties}
     Let $(E, \le)$ be a vector lattice, then:
     \begin{enumerate}
-        \item For any $x, y \in E$,
-        \[
-        x + y = x \vee y + x \wedge y
-        \]
-
         \item Let $x \in E$, $x^+ = x \vee 0 \ge 0$, and $x^- = -(x \wedge 0) \ge 0$, then $x = x^+ - x^-$ and $|x| = x^+ + x^-$. Moreover, $(x^+, x^-)$ are the unique disjoint non-negative elements of $E$ such that $x = x^+ - x^-$.
     \end{enumerate}
 
     For any $x, y \in E$ and $\lambda \in \real$, 
-    \begin{enumerate}
+    \begin{enumerate}[start=1]
-        \item[(3)] $|\lambda x| = |\lambda| \cdot |x|$
+        \item $|\lambda x| = |\lambda| \cdot |x|$
-        \item[(4)] $|x + y| \le |x| + |y|$.
+        \item $|x + y| \le |x| + |y|$.
     \end{enumerate}
 
     
     Finally, for any $x, y \in E$ with $x, y \ge 0$, 
-    \begin{enumerate}
+    \begin{enumerate}[start=3]
-        \item[(5)] $[0, x] + [0, y] = [0, x + y]$.
+        \item $[0, x] + [0, y] = [0, x + y]$.
     \end{enumerate}
 
     
 \end{proposition}
-\begin{proof}
+\begin{proof}[Proof, {{\cite[V.1.1]{SchaeferWolff}}}. ]
-    (1): By \autoref{proposition:ordered-vector-space-properties},
+    (1): By \autoref{lemma:lattice-gymnastics},
-    \begin{align*}
-        x \vee y + x \wedge y - x - y &= 0 \vee (y - x) + (x - y) \wedge 0 \\
-        &= 0 \vee (y - x) - 0 \vee (y - x) = 0
-    \end{align*}
-    
-    (2): By (1),
     \[
     x  = x + 0 = x \vee 0 + x \wedge 0 = x^+ - x^-
     \]
@@ -151,12 +131,12 @@
 
     and $x^- \le z$. If $y \perp z$, then $y - x^+ \perp z - x^-$. Since $y - x^+ = z - x^-$, $y - x^+ = z - x^- = 0$, so $y = x^+$ and $z = x^-$. 
 
-    (3): For any $\lambda > 0$, by (LO2),
+    (2): For any $\lambda > 0$, by (LO2),
     \[
     |\lambda x| = (\lambda x) \vee (-\lambda x) = \lambda (x \vee -x) = \lambda |x|
     \]
 
-    (4): For any $x, y \in E$, by \autoref{proposition:ordered-vector-space-properties},
+    (3): For any $x, y \in E$, by \autoref{proposition:ordered-vector-space-properties},
     \begin{align*}
     x^+ + y^+ &= (x \vee 0) + (y \vee 0) = x \vee y \vee (x + y) \vee 0 \\
     &\ge (x + y) \vee 0 = (x + y)^+
@@ -167,7 +147,7 @@
     |x+y| = (x+y)^+ + (x + y)^- \le x^+ + y^+ + x^- + y^- = |x| + |y|
     \]
 
-    (5): Let $x, y \in E$ with $x, y \ge 0$, then $[0, x] + [0, y] \subset [0, x + y]$. For any $z \in [0, x + y]$, let $u = z \wedge x$ and $v = z - u$, then
+    (4): Let $x, y \in E$ with $x, y \ge 0$, then $[0, x] + [0, y] \subset [0, x + y]$. For any $z \in [0, x + y]$, let $u = z \wedge x$ and $v = z - u$, then
     \[
         v = z - z \wedge x = z + (-z \vee -x) = (0 \vee z - x) \le z - x \le y
     \]
@@ -212,7 +192,7 @@
 \end{proof}
 
 
-\begin{proposition}[{{\cite[V.1.4]{SchaeferWolff}}}]
+\begin{proposition}
 \label{proposition:order-vector-dual}
     Let $(E, \le)$ be an ordered vector space. If for any $x, y \in E$ with $x, y \ge 0$, $[0, x] + [0, y] = [0, x + y]$, then:
     \begin{enumerate}
@@ -227,7 +207,7 @@
 
     
 \end{proposition}
-\begin{proof}
+\begin{proof}[Proof, {{\cite[V.1.4]{SchaeferWolff}}}. ]
     (1): Let $C = \bracs{x \in E|x \ge 0}$, $\Phi^+ \in E^b$, and
     \[
     \Phi^+: C \to [0, \infty) \quad x \mapsto \sup(f([0, x]))

src/fa/order/order.tex

@@ -0,0 +1,56 @@
+\section{Ordered Vector Spaces}
+\label{section:ovs}
+
+\begin{definition}[Ordered Vector Space]
+\label{definition:ordered-vector-space}
+    Let $E$ be a vector space over $K \in \RC$ and $\le$ be a partial order on $E$, then $(E, \le)$ is a \textbf{ordered vector space} if
+    \begin{enumerate}
+        \item[(LO1)] For any $x, y, z \in E$ with $x \le y$, $x + z \le y + z$.
+        \item[(LO2)] For any $x, y \in E$ and $\lambda > 0$, $x \le y$ implies that $\lambda x \le \lambda y$.
+    \end{enumerate}
+
+    The set $C = \bracs{x \in E|x \ge 0}$ is the \textbf{positive cone} of $E$. 
+\end{definition}
+
+\begin{definition}[Ordered Topological Vector Space]
+\label{definition:ordered-tvs}
+    Let $(E, \le)$ be an ordered vector space over $K \in \RC$, and $\topo$ be a vector space topology on $E$, then the triple $(E, \topo, \le)$ is an \textbf{ordered topological vector space} if the positive cone $C = \bracs{x \in E|x \ge 0}$ is closed. 
+\end{definition}
+
+
+\begin{proposition}
+\label{proposition:ordered-vector-space-properties}
+    Let $(E, \le)$ be an ordered vector space and $A, B \subset E$ such that $\sup(A)$ and $\sup (B)$ exist, then 
+    \begin{enumerate}
+        \item $\sup(A + B) = \sup(A) + \sup(B)$.
+        \item $\sup(A) = -\inf (-A)$
+    \end{enumerate}
+
+    
+\end{proposition}
+\begin{proof}
+    (1): For any $a \in A$ and $b \in B$, $a + b \le \sup(A) + \sup(B)$ by (LO1). Let $c \in E$ such that $c \ge a + b$ for all $a \in A$ and $b \in B$, then $c \ge a + \sup(B)$ for all $a \in A$, so $c \ge \sup(A) + \sup(B)$. Therefore $\sup(A) + \sup(B) = \sup(A + B)$.
+
+    (2): For any $a \in A$, $\sup(A) \ge a = -(-a)$, so $-\sup(A) \le \inf(-A)$, and $\sup(A) \ge -\inf(-A)$. On the other hand, for any $a \in A$, $-\inf(-A) \ge a$. Therefore $\sup(A) \le -\inf(-A)$. 
+\end{proof}
+
+
+\begin{definition}[Interval]
+\label{definition:ordered-vector-space-interval}
+    Let $(E, \le)$ be an ordered vector space and $x, y \in E$, then
+    \[
+    [x, y] = \bracs{z \in E| x \le z \le y}
+    \]
+
+    is the \textbf{order interval} with endpoints $x$ and $y$.
+\end{definition}
+
+\begin{definition}[Order Bounded]
+\label{definition:ordered-vector-space-bounded}
+    Let $(E, \le)$ be an ordered vector space and $A \subset E$, then $A$ is \textbf{order bounded} if there exists $x, y \in E$ such that $A \subset [x, y]$.
+\end{definition}
+
+\begin{definition}[Order Complete]
+\label{definition:order-vector-complete}
+    Let $(E, \le)$ be an ordered vector space, then $E$ is \textbf{order complete} if for any order bounded set $A \subset E$, $\sup (A)$ and $\inf (A)$ exist.
+\end{definition}

src/fa/order/positive.tex

@@ -0,0 +1,34 @@
+\section{The Order Dual}
+\label{section:order-dual}
+
+
+\begin{definition}[Order Bounded Dual]
+\label{definition:order-bounded-dual}
+    Let $(E, \le)$ be an ordered vector space, then the space $E^b$ consisting of all linear functionals on $E$ that are bounded on order bounded sets is the \textbf{order bounded dual} of $E$.
+\end{definition}
+
+
+\begin{definition}[Order Dual]
+\label{definition:order-dual}
+    Let $(E, \le)$ be an ordered vector space over $K \in \RC$ and $\Phi^+ \in \hom(E; K)$, then $\Phi^+$ is \textbf{positive} if for any $x \in E$ with $x \ge 0$, $\text{Re}\dpn{x, \Phi^+}{E} \ge 0$. The subspace $E^+ \subset \hom(E; K)$ generated by the positive linear functionals on $E$ is the \textbf{order dual} of $E$. 
+\end{definition}
+
+
+\begin{theorem}[Bauer-Namioka]
+\label{theorem:bauer-namioka}
+    Let $E$ be an ordered vector space over $K \in \RC$ with positive cone $C$, $\topo$ be a vector space topology on $E$\footnote{The order and the topology need not to be compatible. }, $F \subset E$ be a subspace, and $\phi \in (F, \topo)^*$, then the following are equivalent:
+    \begin{enumerate}
+        \item There exists a continuous positive linear functional $\Phi \in (E, \topo)^*$ such that $\Phi|_F = \phi$. 
+        \item There exists $U \in \cn_\topo(0)$ convex such that
+        \[
+        \sup\bracs{\text{Re}\dpn{x, \phi}{F}|x \in F \cap (U - C)} < \infty
+        \]
+    \end{enumerate}
+\end{theorem}
+\begin{proof}[Proof, {{\cite[V.5.4]{SchaeferWolff}}}. ]
+    (1) $\Rightarrow$ (2): Let $U = \bracs{\text{Re}(\Phi) < 1}$, then $U \in \cn_\topo(0)$ is convex. Since $\Phi$ is positive, for any $x \in U$ and $y \in C$, $\text{Re}\dpn{x - y, \Phi}{E} \le \text{Re}\dpn{x, \Phi}{E} < 1$. 
+
+    (2) $\Rightarrow$ (1): Assume without loss of generality that $K = \real$. Let $\alpha > 0$ such that $F \cap (U - C) \subset \bracs{\phi < \alpha}$. Since $U$ is convex and open, and $C$ is convex, $U - C \subset E$ is an open convex set, disjoint from the convex set $\bracs{\phi = \alpha}$. By the \hyperref[Hahn-Banach Theorem]{theorem:hahn-banach-geometric-1}, there exists $\Phi \in E^*$ such that $U - C \subset \bracs{\Phi < \alpha}$, and $\bracs{\phi = \alpha} \subset \bracs{\Phi \ge \alpha}$. 
+
+    After rescaling, assume without loss of generality that $\bracs{\phi = \alpha} \subset \bracs{\Phi = \alpha}$, then $\Phi \in E^*$ is an extension of $\phi$. For each $x \in C$ and $\lambda > 0$, $-\lambda x \in U - C$, and $-\lambda\dpn{x, \Phi}{E} < \alpha$. As this holds for all $\lambda > 0$, $\dpn{x, \Phi}{E} \ge 0$, so $\Phi$ is the desired extension.
+\end{proof}

src/fa/tvs/complexify.tex

@@ -84,6 +84,18 @@
     
 \end{proof}
 
+\begin{definition}[Hermitian]
+\label{definition:hermitian-functional}
+    Let $E$ be a vector space over $\real$, $*: \complex(E) \to \complex(E)$ be the canonical complex conjugation map, and $\phi \in \hom(\complex(E); \complex)$, then the following are equivalent: 
+    \begin{enumerate}
+        \item $\phi|_E \in \hom(E; \real)$.
+        \item For each $x \in E$, $\dpn{x, \phi}{\complex(E)} = \ol{\dpn{x^*, \phi}{\complex(E)}}$. 
+    \end{enumerate}
+
+    If the above holds, then $\phi$ is \textbf{Hermitian}.     
+\end{definition}
+
+
 
 
 
@@ -107,8 +119,7 @@
     The pair $(\complex(E), \iota)$ is the \textbf{complexification} of $E$ as a topological vector space, and
     \begin{enumerate}[start=4]
         \item If $E$ is locally convex, then so is $\complex(E)$. 
-        \item If $E$ is normed, then $\complex(E)$ is normable, and there exists a norm $\norm{\cdot}_{\complex(E)}: \complex(E) \to [0, \infty)$ such that $\iota: E \to \complex(E)$ is isometric. 
+        \item[(F)] For any topological vector space $F$ over $\real$ and continuous $\real$-linear map $T: E \to F$, there exists a unique continuous $\complex$-linear map $\complex(T): \complex(E) \to \complex(F)$ such that the following diagram commutes:
-        \item[(F)] For any vector space $F$ over $\real$ and continuous $\real$-linear map $T: E \to F$, there exists a unique continuous $\complex$-linear map $\complex(T): \complex(E) \to \complex(F)$ such that the following diagram commutes:
         \[
         \xymatrix{
         \mathbb{C}(E) \ar@{->}[r]^{\mathbb{C}(T)} & \mathbb{C}(F) \\
@@ -120,6 +131,8 @@
         \[
         \complex(T)(x + iy) = Tx + iTy
         \]
+
+        Moreover, if $E$ and $F$ are normed, then $\norm{\complex(T)}_{L(\complex(E); \complex(F))} = \norm{T}_{L(E; F)}$. 
     \end{enumerate}
     
 \end{definition}
@@ -129,25 +142,81 @@
     (U): By (U) of the \hyperref[complexification]{definition:complexification}, there exists a $\complex$-linear map $\complex(T): \complex(E) \to F$ such that the given diagram commutes. Since $T \circ \iota$ and $iT \circ \iota$ are continuous, $T$ is continuous by (U) of the \hyperref[direct sum]{definition:tvs-direct-sum}.
 
     (4): By \autoref{proposition:finite-lc-product}, the direct sum and product of finitely many locally convex spaces coincide. By \autoref{proposition:lc-projective-topology}, this topology is locally convex. 
+    
 
-    (5): Let $\norm{\cdot}_E: E \to [0, \infty)$ be the norm of $E$, and define
+    (F): Existence of $\complex(T)$ is given by (U) applied to $\iota \circ T$. 
+
+    For the isometry, 
+    \begin{align*}
+    \norm{\complex(T)}_{L(\complex(E); \complex(F))} &= \norm{\complex(T)}
+    \end{align*}
+    
+\end{proof}
+
+\begin{definition}[Complexification of Normed Spaces]
+\label{definition:complexification-of-normed-spaces}
+    Let $E$ be a normed vector space over $\real$, then
     \[
     \norm{\cdot}_{\complex(E)}: \complex(E) \to [0, \infty) \quad (x, y) \mapsto \sup_{\theta \in [0, 2\pi]}\norm{\cos(\theta)x + \sin(\theta)y}_E
     \]
 
-    then for any $\phi \in [0, 2\pi]$ and $x, y \in E$, 
+    is a norm on $\complex(E)$ such that the inclusion map $\iota: E \to \complex(E)$ is isometric. 
+    
+    Moreover, for any normed vector space $F$ over $\real$ and $T \in L(E; F)$, $\norm{\complex(T)}_{L(\complex(E); \complex(F))} = \norm{T}_{L(E; F)}$. 
+\end{definition}
+\begin{proof}
+    For any $\phi \in [0, 2\pi]$ and $x, y \in E$, 
     \begin{align*}
         \normn{e^{i \phi}(x, y)}_{\complex(E)} &= \normn{(\cos(\phi)x - \sin(\phi)y, \sin(\phi)x + \cos(\phi)y)}_{\complex(E)} \\
         &= \sup_{\theta \in [0, 2\pi]}\norm{\cos(\theta - \phi)x + \sin(\theta - \phi)y}_E \\
         &= \norm{(x, y)}_{\complex(E)}
     \end{align*}
 
-    so $\norm{(x, y)}_{\complex(E)}$ is a norm. For any $x \in E$, 
+    so $\norm{\cdot}_{\complex(E)}$ is a norm on $\complex(E)$. For any $x \in E$, 
     \[
     \norm{\iota x}_{\complex(E)} = \sup_{\theta \in [0, 2\pi]}\norm{\cos(\theta)x}_E = \norm{x}_E \\
     \]
 
     Therefore $\iota: E \to \complex(E)$ is isometric. 
 
-    (F): By (U) applied to $\iota \circ T$.
+    Now, let $F$ be a normed vector space over $\real$ and $T \in L(E; F)$, then
+    \begin{align*}
+        \norm{\complex(T)}_{L(\complex(E); \complex(F))} &= \sup\bracsn{\norm{\complex(T)(x,y)}_{\complex(F)}|(x, y) \in \complex(E), \norm{(x, y)}_{\complex(E)} = 1} \\
+        &\ge \sup\bracsn{\norm{Tx}|x \in E, \norm{x}_E = 1} = \norm{T}_{L(E; F)}
+    \end{align*}
+
+    On the other hand, let $(x, y) \in \complex(E)$, then there exists $\theta \in [0, 2\pi]$ such that
+    \begin{align*}
+    \normn{\complex(T)(x, y)}_{\complex(F)} &= \normn{(Tx, Ty)}_{\complex(F)} = \norm{\cos(\theta)Tx + \sin(\theta)Ty}_F \\
+    &\le \norm{T}_{L(E; F)} \cdot \norm{\cos(\theta)x + \sin(\theta)y}_E \\
+    &\le \norm{T}_{L(E; F)} \cdot \sup_{\phi \in [0, 2\pi]}\norm{\cos(\phi)x + \sin(\phi)y}_E \\
+    &= \norm{T}_{L(E; F)} \cdot \norm{(x, y)}_E
+    \end{align*}
+    
+    
 \end{proof}
+
+
+\begin{proposition}
+\label{proposition:hermitian-functional-norm}
+    Let $E$ be a normed vector space over $\complex$, $*: E \to E$ be a complex conjugation map such that $\norm{x}_E = \normn{x^*}_E$ for all $x \in E$, and $\phi \in E^*$ be a Hermitian functional, then
+    \[
+    \norm{\phi}_{E^*} = \sup\bracsn{\dpn{x, \phi}{E}|x \in E, x = x^*, \norm{x}_E = 1}
+    \]
+\end{proposition}
+\begin{proof}
+    Since $\bracsn{x \in E|x = x^*} \subset E$, $\norm{\phi}_{E^*} \ge \sup\bracsn{\dpn{x, \phi}{E}|x \in E, x = x^*, \norm{x}_E = 1}$. 
+
+    On the other hand, let $x \in E$ with $\norm{x}_E = 1$. Assume without loss of generality that $\dpn{x, \phi}{E} \in \real$, then
+    \begin{align*}
+        \dpn{x, \phi}{E} &= \dpn{\text{Re}(x), \phi}{E} + \underbrace{i\dpn{\text{Im}(x), \phi}{E}}_{\in \real} =
+\dpn{\text{Re}(x), \phi}{E} \\
+&\le  \norm{\text{Re}(x)}_E \cdot \sup\bracsn{\dpn{y, \phi}{E}|y \in E, y = y^*, \norm{y}_E = 1}
+    \end{align*}
+
+    where $\norm{\text{Re}(x)}_E = \norm{{(x + x^*)}/{2}}_E \le \norm{x}_E$. As the above holds for all $x \in E$, 
+    \[
+    \norm{\phi}_{E^*} \le \sup\bracsn{\dpn{x, \phi}{E}|x \in E, x = x^*, \norm{x}_E = 1}
+    \]
+\end{proof}
+

src/fa/tvs/definition.tex

@@ -6,18 +6,13 @@
 \label{definition:tvs}
     Let $E$ be a vector space over $K \in \bracs{\real, \complex}$ and $\topo \subset 2^E$ be a topology. If
     \begin{enumerate}
-        \item[(TVS1)] $E \times E \to E$ with $(x, y) \mapsto x + y$ is continuous.
+        \item[(TVS1)] The addition map $E \times E \to E$ with $(x, y) \mapsto x + y$ is continuous.
-        \item[(TVS2)] $K \times E \to E$ with $(\lambda, x) \mapsto \lambda x$ is continuous.
+        \item[(TVS2)] The scalar multiplication map $K \times E \to E$ with $(\lambda, x) \mapsto \lambda x$ is continuous.
     \end{enumerate}
 
     then the pair $(E, \topo)$ is a \textbf{topological vector space}.
 \end{definition}
 
-\begin{definition}[Translation-Invariant Topology]
-\label{definition:translation-invariant-topology}
-    Let $E$ be a vector space and $\topo$ be a topology on $E$, then $\topo$ is \textbf{translation-invariant} if for any $U \in \topo$ and $y \in E$, $U + y \in \topo$.
-\end{definition}
-
 \begin{lemma}
 \label{lemma:tvs-translation-invariant}
     Let $E$ be a TVS over $K \in \RC$, then the topology of $E$ is translation-invariant.
@@ -26,26 +21,8 @@
     Let $U \subset E$ open and $y \in E$, then $U + y$ is the preimage of $U$ by the map $x \mapsto x - y$. By (TVS1), $U + y$ is open.
 \end{proof}
 
-\begin{definition}[Translation-Invariant Uniformity]
-\label{definition:translation-invariant-uniformity}
-    Let $E$ be a vector space, $\fU$ be a uniformity on $E$, and $U \in \fU$, then $U$ is \textbf{translation-invariant} if for every $z \in E$,
-    \[
-    U = \bracs{(x + z, y + z)|(x, y) \in U}
-    \]
 
-    and $\fU$ is \textbf{translation-invariant} if there exists a fundamental system of translation-invariant entourages.
+\begin{proposition}
-\end{definition}
-
-\begin{lemma}
-\label{lemma:translation-invariant-symmetric}
-    Let $E$ be a vector space and $\fU$ be a translation-invariant uniformity, then $\fU$ admits a fundamental system of symmetric, translation-invariant entourages.
-\end{lemma}
-\begin{proof}
-    Let $z \in E$, then the map $(x, y) \mapsto (x + z, y + z)$ is a bijection. Thus for any translation-invariant entourages $U, V \in \fU$, $(U \cap V) + z = (U + z) \cap (V + z)$, and $U \cap V$ is translation-invariant. By \autoref{lemma:symmetricfundamentalentourage}, $\fU$ admits a fundamental system of symmetric, translation-invariant entourages.
-\end{proof}
-
-
-\begin{proposition}[{{\cite[I.1.4]{SchaeferWolff}}}]
 \label{proposition:tvs-uniform}
     Let $E$ be a TVS over $K \in \bracs{\real, \complex}$, then:
     \begin{enumerate}
@@ -56,73 +33,7 @@
     The space $E$ will always be assumed to be equipped with its translation-invariant uniformity.
 \end{proposition}
 \begin{proof}
-    (2): Firstly, for any $V \in \fB_0$, $U_V$ is translation-invariant.
+    By \autoref{definition:group-translation-invariant-uniformity}. 
-    \begin{enumerate}
-        \item[(FB1)] For any $V, V' \in \fB_0$, there exists $W \in \fB_0$ such that $W \subset V \cap V'$. In which case, $U_V \cap U_{V'} \supset U_W \in \fB$.
-        \item[(UB1)] For any $V \in \fB_0$, $0 \in V$, so $\Delta \subset U_V$.
-        \item[(UB2)] Let $V \in \fB_0$, then by (TVS1), there exists $W \in \fB_0$ such that $W + W \subset V$. For any $(x, y), (y, z) \in U_W$, $(x - y), (y - z) \in W$, so $(x - z) \in V$. Thus $U_W \circ U_W \subset U_V$.
-    \end{enumerate}
-
-    By \autoref{proposition:fundamental-entourage-criterion}, $\fB$ forms a fundamental system of entourages for a translation-invariant uniformity $\fU$ on $E$.
-
-    (1): Let $\mathfrak{V}$ be a translation-invariant uniformity on $E$ inducing the topology. For any symmetric, translation-invariant entourage $V \in \mathfrak{V}$, $V(x) = V(0) + x$ for all $x \in E$, and $(x, y) \in V$ if and only if $y - x \in V$, if and only if $x - y \in V$. Thus $V = U_{V(0)}$.
-
-    Let $W \in \cn(0)$, then by \autoref{lemma:translation-invariant-symmetric}, there exists a symmetric, translation-invariant entourage $V \in \mathfrak{V}$ such that $V(0) \subset W$, and $V \subset U_W$. Thus $\mathfrak{V} \supset \fU$.
-
-    Let $V \in \mathfrak{V}$. Using \autoref{lemma:translation-invariant-symmetric}, assume without loss of generality that $V$ is symmetric and translation-invariant, then there exists $W \in \fB_0$ with $W \subset V(0)$. In which case, $U_W \subset V$, and $\fU \supset \mathfrak{V}$.
-\end{proof}
-
-\begin{proposition}
-\label{proposition:tvs-closure}
-    Let $E$ be a TVS over $K \in \RC$, $A \subset E$, and $\fB \subset \cn(0)$ be a fundamental system of neighbourhoods, then
-    \[
-    \ol{A} = \bigcap_{U \in \fB}\bracs{A + U| U \in \fB}
-    \]
-
-\end{proposition}
-\begin{proof}
-    Let $V \in \cn(0)$ be balanced and $U_V = \bracs{(x, y) \in E \times E| x - y \in V}$, then $y \in U_V(A)$ if and only if there exists $x \in A$ such that $(x, y) \in U_V$. This is equivalent to $x - y \in V$ and $y - x \in V$, so $U_V(A) = A + V$.
-
-    Assume without loss of generality that $\fB$ consists of symmetric entourages. By \autoref{proposition:tvs-uniform}, $\bracs{U_V|V \in \fB}$ forms a fundamental system of entourages for $E$, and \autoref{proposition:uniformclosure} implies that
-    \[
-    \ol{A} = \bigcap_{V \in \fB}\bracs{U_V(A)| U \in \fB} = \bigcap_{V \in \fB}U + A
-    \]
-
-\end{proof}
-
-\begin{proposition}[{{\cite[I.1.1]{SchaeferWolff}}}]
-\label{proposition:tvs-set-operations}
-    Let $E$ be a TVS over $K \in \RC$ and $A, B \subset E$, then:
-    \begin{enumerate}
-        \item If $A$ is open, then $A + B$ is open.
-        \item If $A$ is closed and $B$ is compact, then $A + B$ is closed.
-    \end{enumerate}
-\end{proposition}
-\begin{proof}
-    $(1)$: For every $x \in B$, $A + x$ is open by \autoref{definition:translation-invariant-topology}, so
-    \[
-    A + B = \bigcup_{x \in B}(A + x)
-    \]
-
-    is open.
-
-    $(2)$: Let $x \in \overline{A + B}$, then there exists a filter $\fF \subset 2^{A \cup B}$ converging to $x$. For any $U \in \fF$, $U \cap (A + B) \ne \emptyset$, so $(U - B) \cap A \ne \emptyset$, and $\fB = \bracs{U - B| U \in \fF}$ is a filter base in $A$. By compactness of $A$, there exists $y \in A$ such that
-    \[
-    y \in \bigcap_{U \in \fF}\overline{U - B} = \bigcap_{U \in \fF}\overline{(-B) + U}
-    \]
-
-
-    By \autoref{proposition:tvs-closure}, $\overline{(-B) + U} \subset (-B) + U + U$, so
-    \[
-    y \in \bigcap_{U \in \fF}\overline{(-B) + U} \subset \bigcap_{U \in \fF}[(-B) + U + U]
-    \]
-
-    Since $\fF$ converges to $x$, (TVS1) implies that $\bracs{U + U| U \in \fF}$ contains a neighbourhood base of $x$. Thus
-    \[
-    y \in \bigcap_{U \in \fF}[(-B) + U + U] \subset \bigcap_{V \in \cn(0)}[(x-B) + V] = \overline{x - B} = x - B
-    \]
-
-    so $x \in y + B \subset A + B$.
 \end{proof}
 
 \begin{definition}[Balanced/Circled]
@@ -158,7 +69,7 @@
     Let $U \in \cn(0)$ be closed, then there exists a balanced neighbourhood $V \in \cn^o(0)$ such that $V \subset U$. In which case, for any $\lambda \in K$ with $0 < \abs{\lambda} \le 1$, $\lambda \overline{V} = \overline{\lambda V} \subset \overline{V}$ by (TVS2). Therefore $\overline{V} \subset U$ is balanced as well.
 \end{proof}
 
-\begin{proposition}[{{\cite[I.1.2]{SchaeferWolff}}}]
+\begin{proposition}
 \label{proposition:tvs-0-neighbourhood-base}
     Let $E$ be a vector space over $K \in \RC$, and $\topo$ be a vector space topology on $E$, then there exists a fundamental system of neighbourhoods $\fB \subset \cn_E(0)$ such that:
     \begin{enumerate}
@@ -177,7 +88,7 @@
         \item[(3)] $(E, \topo)$ is a TVS.
     \end{enumerate}
 \end{proposition}
-\begin{proof}
+\begin{proof}[Proof, {{\cite[I.1.2]{SchaeferWolff}}}. ]
     \textbf{Forward:} By \autoref{proposition:tvs-good-neighbourhood-base}, there exists a fundamental system of neighbourhoods $\fB \subset \cn_E(0)$ consisting of circled and radial sets. By (TVS1), $\fB$ satisfies (TVB1).
 
     \textbf{Converse:} For each $V \in \fB$, let $U_V = \bracs{(x, y) \in E|x - y \in V}$, then $U_V$ is symmetric and translation-invariant by (TVB1). Let
@@ -189,7 +100,7 @@
     \begin{enumerate}
         \item[(FB1)] For any $V, V' \in \fB$, there exists $W \in \fB$ with $W \subset V \cap V'$. In which case, $U_{V} \cap U_{V'} \supset U_W \in \mathfrak{V}$.
         \item[(UB1)] For any $x \in E$ and $V \in \fB$, $x - x = 0 \in V$, so $\Delta \subset U_V$.
-        \item[(UB2)] For any $V \in \fB$, by (TVB1), there exists $W \in \fB$ such that $W + W \subset V$. In which case, for any $x, y, z \in E$ with $x - y, y - z \in W$, $x - z \in V$. Therefore $U_W \circ U_W \subset U_V$.
+        \item[(UB3)] For any $V \in \fB$, by (TVB1), there exists $W \in \fB$ such that $W + W \subset V$. In which case, for any $x, y, z \in E$ with $x - y, y - z \in W$, $x - z \in V$. Therefore $U_W \circ U_W \subset U_V$.
     \end{enumerate}
 
     By \autoref{proposition:fundamental-entourage-criterion}, there exists a unique uniformity $\fU$ on $E$ for which $\mathfrak{V}$ is a fundamental system of entourages.

src/fa/tvs/equicontinuous.tex

@@ -1,7 +1,7 @@
 \section{Equicontinuous Families of Linear Maps}
 \label{section:equicontinuous-linear}
 
-\begin{proposition}[{{\cite[IV.4.2]{SchaeferWolff}}}]
+\begin{proposition}
 \label{proposition:equicontinuous-linear}
     Let $E, F$ be TVSs over $K \in RC$ and $\alg \subset \hom(E; F)$, then the following are equivalent:
     \begin{enumerate}
@@ -12,7 +12,7 @@
         \item For each $V \in \cn_F(0)$, $\bigcap_{T \in \alg}T^{-1}(V) \in \cn_E(0)$. 
     \end{enumerate}
 \end{proposition}
-\begin{proof}
+\begin{proof}[Proof, {{\cite[IV.4.2]{SchaeferWolff}}}. ]
     (5) $\Rightarrow$ (1): Let $V \in \cn_F(0)$, then $U = \bigcap_{T \in \alg}T^{-1}(V) \in \cn_E(0)$. Thus for any $x, y \in E$ with $x - y \in U$, $Tx - Ty \in V$ for all $T \in \alg$. 
 \end{proof}
 
@@ -43,7 +43,7 @@
 
     and that
     \begin{enumerate}
-        \item[(B2')] For each $x \in E$, $\alg(x) = \bracs{Tx|T \in \alg}$ is bounded in $F$. 
+        \item[(B2)] For each $x \in E$, $\alg(x) = \bracs{Tx|T \in \alg}$ is bounded in $F$. 
     \end{enumerate}
 
     then
@@ -113,10 +113,10 @@
 % TODO: Replace this with a more general version involving polars in the future. 
 \begin{theorem}[Banach-Alaoglu]
 \label{theorem:alaoglu}
-    Let $E$ be a locally convex space over $K \in \RC$ and $\alg \subset E^*$ be equicontinuous, then $\alg$ is precompact with respect to $\sigma(E^*, E)$.
+    Let $E$ be a locally convex space over $K \in \RC$ and $\alg \subset E^*$ be equicontinuous, then $\alg$ is relatively compact with respect to $\sigma(E^*, E)$.
 \end{theorem}
 \begin{proof}
-    For each $x \in E$, $\alg(x) = \bracsn{\dpn{x, \phi}{E}|\phi \in \alg}$ is precompact by \autoref{proposition:equicontinuous-bounded}. By the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}, 
+    For each $x \in E$, $\alg(x) = \bracsn{\dpn{x, \phi}{E}|\phi \in \alg}$ is relatively compact by \autoref{proposition:equicontinuous-bounded}. By the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}, 
     \begin{enumerate}
         \item[(C2)] The $\sigma(E^*, E)$-closure of $\alg$ in $\prod_{x \in E}K$ is equicontinuous. 
         \item[(C3)] The $\sigma(E^*, E)$-closure of $\alg$ in $\prod_{x \in E}K$ is compact. 

src/measure/bochner-integral/strongly.tex

@@ -24,7 +24,7 @@
 
     By the separable case, $f$ is $(\cm, \cb_{F})$-measurable. Let $B \in \cb_E$, then $B \cap F \in \cb_F$ by \autoref{lemma:borel-induced}. Therefore $\bracs{f \in B} = \bracs{f \in B \cap F} \in \cm$, and $f$ is $(\cm, \cb_E)$-measurable. 
 
-    (2) $\Rightarrow$ (3): By \autoref{proposition:measurable-simple-separable-norm}.
+    (2) $\Rightarrow$ (3): By \autoref{corollary:measurable-simple-separable-norm}.
 
     (3) $\Rightarrow$ (1): For each $\phi \in E^*$, $\phi \circ f = \limv{n}\phi \circ f_n$ is measurable by \autoref{proposition:limit-measurable}. Since
     \[

src/measure/index.tex

@@ -8,5 +8,5 @@
 \input{./measurable-maps/index.tex}
 \input{./lebesgue-integral/index.tex}
 \input{./bochner-integral/index.tex}
-\input{./differentiation/index.tex}
+\input{./lcg/index.tex}
 \input{./notation.tex}

src/measure/lcg/haar.tex

@@ -0,0 +1,254 @@
+\section{Haar Measures}
+\label{section:haar}
+
+\begin{definition}[Haar Measure]
+\label{definition:haar-measure}
+    Let $G$ be a locally compact group and $\mu: \cb_G \to [0, \infty]$ be a non-zero Radon measure, then $\mu$ is a \textbf{left Haar measure} if
+    \begin{enumerate}
+        \item[(LH)] For each $g \in G$ and $A \in \cb_G$, $\mu(gA) = \mu(A)$. 
+    \end{enumerate}
+
+    and a \textbf{right Haar measure} if
+    \begin{enumerate}
+        \item[(RH)] For each $g \in G$ and $A \in \cb_G$, $\mu(Ag) = \mu(A)$. 
+    \end{enumerate}
+\end{definition}
+
+\begin{lemma}
+\label{lemma:lc-sigma-compact}
+    Let $G$ be a locally compact group, then there exists an open and closed subgroup $H$ that is $\sigma$-compact. 
+\end{lemma}
+\begin{proof}
+    Let $K \in \cn_G(1)$ and $K^{(1)} = K$. For each $n \in \natp$, let $K^{(n+1)} = KK^{(n)}$, then $K^{(n+1)}$ is compact by \autoref{proposition:compact-extensions} with $K^{(n+1)} \in \cn_G(K^{(n)})$. Let $H = \bigcup_{n \in \natp}K^{(n)}$, then $H$ is a subgroup of $G$, which is open by \autoref{lemma:openneighbourhood}. Since $H$ admits an exhaustion by compact sets, it is $\sigma$-compact. 
+
+    Finally, since
+    \[
+    G \setminus H = G \setminus \bigcup_{n \in \natp} K^{(n)} = \bigcap_{n \in \natp}G \setminus K^{(n)}
+    \]
+
+    and $K^{(n)}$ is closed for each $n \in \natp$ by \autoref{proposition:compact-closed}, $G \setminus H$ is closed, and hence $H$ is open. 
+\end{proof}
+
+
+\begin{definition}[Covering Ratio]
+\label{definition:lcg-covering-ratio}
+    Let $G$ be a locally compact group and $f, g \in C_c^+(G)$, then
+    \[
+    (f: g) = \inf\bracs{\sum_{j = 1}^n c_j \bigg | \seqf{c_j} \subset [0, \infty), \seqf{x_j} \subset G, f \le \sum_{j = 1}^n c_j L_{x_j}g}
+    \]
+
+    is the \textbf{covering ratio} of $f$ by $g$. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:covering-ratio-gymnastics}
+    Let $G$ be a locally compact group and $f, h, g \in C_c^+(G)$, then:
+    \begin{enumerate}
+        \item If $g \ne 0$, then $(f: g) < \infty$. 
+        \item $(f + h: g) \le (f: g) + (h: g)$. 
+        \item For each $\lambda \ge 0$, $(\lambda f: g) = \lambda(f: g)$. 
+        \item If $f \le h$, then $(f: g) \le (h: g)$. 
+        \item $(f: g) \le (f: h)(h: g)$. 
+        \item $(f: g) \ge \norm{f}_u/\norm{g}_h$. 
+        \item For each $x \in G$, $(L_xf: g) = (f: g)$. 
+    \end{enumerate}
+\end{proposition}
+% Proof omitted due to obviousness. 
+
+\begin{lemma}
+\label{lemma:haar-approx}
+    Let $G$ be a locally compact group, $f, f'\in C_c^+(G)$, and $\eps > 0$, then there exists $V \in \cn_G(1)$ such that for any $g \in C_c^+(V)$ with $g \ne 0$, 
+    \[
+    (f: g) + (f': g) \le (f + f': g) + \eps
+    \]
+\end{lemma}
+\begin{proof}[Proof, {{\cite[Lemma 2.18]{FollandHarmonic}}}. ]
+    By \hyperref[Urysohn's Lemma]{lemma:lch-urysohn}, there exists $\eta \in C_c^+(G; [0, 1])$ such that $\eta|_{\supp{f} \cup \supp{f'}} = 1$. 
+
+    Let $\delta > 0$, and define
+    \[
+    H = f + f' + \delta \eta \quad h = \frac{f}{H} \quad h' = \frac{f'}{H}
+    \]
+    
+    By \autoref{proposition:lcg-cc-uc}, there exists $V \in \cn_G(1)$ such that for any $x, y \in G$ with $x^{-1}y \in V$, 
+    \[
+    |h(x) - h(y)|, |h'(x) - h'(y)| < \delta
+    \] 
+
+    Let $g \in C_c^+(V)$, $\seqf{c_j} \subset [0, \infty)$, and $\seqf{x_j} \subset G$ such that $H \le \sum_{j = 1}^n c_j L_{x_j}\phi$, then for each $x \in G$, 
+    \begin{align*}
+        f(x) &= H(x)h(x) \le \sum_{j = 1}^n c_j L_{x_j}g(x)h(x) = \sum_{j = 1}^n c_jg(x_j^{-1}x)h(x) \\
+        &\le \sum_{j = 1}^n c_j[h(x_j) + \delta] \cdot  L_{x_j}g(x)
+    \end{align*}
+
+    Likewise, 
+    \[
+    f'(x) \le \sum_{j = 1}^n c_j[h'(x_j) + \delta] \cdot L_{x_j}g(x)
+    \]
+
+    As $h + h' \le 1$, 
+    \[
+        (f: g) + (f': g) \le \sum_{j = 1}^n c_j[h(x_j) + h'(x_j) + 2\delta]
+    \]
+    
+    Since the above holds for all such $\seqf{c_j} \subset [0, \infty)$ and $\seqf{x_j} \subset G$, 
+    \begin{align*}
+    (f: g) + (f': g) &\le (1 + 2\delta)(H: g) \\
+    &\le (1 + 2\delta)[(f + f': g) + \delta(\eta: g)]
+    \end{align*}
+\end{proof}
+
+\begin{theorem}[Haar]
+\label{theorem:haar}
+    Let $G$ be a locally compact group, then:
+    \begin{enumerate}
+        \item There exists a left/right Haar measure on $G$. 
+        \item For any two left/right Haar measures $\mu$ and $\nu$ on $G$, there exists $\lambda > 0$ such that $\mu = \lambda \nu$. 
+    \end{enumerate}
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem 2.10, 2.20]{FollandHarmonic}}}. ]
+    (1): Fix $h \in C_c^+(G)$ with $h \ne 0$. For each $g \in C_c^+(G)$ with $g \ne 0$, let
+    \[
+    I_g: C_c^+(G) \to [0, \infty) \quad f \mapsto \frac{(f: g)}{(h: g)}
+    \]
+
+    then by (5) of \autoref{proposition:covering-ratio-gymnastics}, for each $f \in C_c^+(G)$ with $f \ne 0$, 
+    \[
+    \frac{1}{(h: f)} = \frac{(f: g)}{(h: f)(f: g)} \le I_g(f) \le \frac{(f: h)(h: g)}{(h: g)} = (f: h)
+    \]
+
+    Thus $\mathcal{I}(f) = \bracs{I_g(f)|g \in C_c^+(G) \setminus \bracs{0}}$ is relatively compact for each $f \in C_c^+(G)$. 
+
+    For each $V \in \cn_G(1)$, let $E_V = \bracs{I_g|g \in C_c^+(V) \setminus \bracs{0}}$, then $\fF = \bracs{E_V|V \in \cn_G(1)}$ is a filter on the product space $\prod_{f \in C_c^+(G)}\ol{\mathcal{I}(f)}$. By \hyperref[Tychonoff's Theorem]{theorem:tychonoff}, $\prod_{f \in C_c^+(G)}\ol{\mathcal{I}(f)}$ is compact, and $\bigcap_{V \in \cn_G(1)}\ol{E_V} \ne \emptyset$. 
+
+    Let $I \in \bigcap_{V \in \cn_G(1)}\ol{E_V}$, then by continuity, 
+    \begin{enumerate}[label=(\roman*)]
+        \item For each $f \in C_c^+(G) \setminus \bracs{0}$, $I(f) \in [(h: f)^{-1}, (f: h)]$. 
+        \item For every $\lambda \ge 0$ and $f \in C_c^+(G)$, $I(\lambda f) = \lambda I(f)$.
+        \item For any $x \in G$, $I(L_xf) = I(f)$. 
+        \item For each $f, f' \in C_c^+(G)$, $I(f + f') \le I(f) + I(f')$. 
+    \end{enumerate}
+    
+    Let $f, f' \in C_c^+(G)$ and $\eps > 0$. By \autoref{lemma:haar-approx}, there exists $V \in \cn_G(1)$ such that for each $g \in E_V$, 
+    \begin{enumerate}[label=(\alph*)]
+        \item $|I_g(f) - I(f)|, |I_g(f') - I(f')| < \eps$. 
+        \item $I_g(f) + I_g(f') \le I_g(f + f') + \eps$. 
+    \end{enumerate}
+
+    In which case, $I(f) + I(f') \le I(f + f') + 3\eps$. Since this holds for all $\eps > 0$, 
+    \begin{enumerate}[start=4, label=(\roman*)]
+        \item For each $f, f' \in C_c^+(G)$, $I(f + f') \ge I(f) + I(f')$. 
+    \end{enumerate}
+
+    Using \autoref{lemma:positive-functional-extension}, $I$ extends to a positive linear functional on $C_c(G; \real)$, with $I(f) > 0$ for all $f \in C_c^+(G) \setminus \bracs{0}$, and
+    \begin{enumerate}
+        \item[(LH)] For each $f \in C_c(G)$ and $x \in G$, $I(L_xf) = I(f)$. 
+    \end{enumerate}
+    
+    By (i) and the \hyperref[Riesz Representation Theorem]{theorem:riesz-radon}, there exists a unique non-zero Radon measure $\mu: \cb_G \to [0, \infty]$ such that for each $f \in C_c^+(G)$, $I(f) = \int_G f d\mu$. Finally, by \hyperref[density of $C_c(G; \real)$ in $L^1(\mu; \real)$]{proposition:radon-cc-dense} and (LH), $\mu$ is a left Haar measure. 
+
+    (2): Let $f, g \in C_c^+(G) \setminus \bracs{0}$ and $\eps > 0$, then by \autoref{proposition:lcg-cc-uc}, there exists a symmetric neighbourhood $V \in \cn_G(1)$ such that for any $x \in G$ and $y \in V$, 
+    \[
+    |f(xy) - f(yx)|, |g(xy) - g(yx)| < \eps
+    \]
+
+    By \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, there exists $h \in C_c^+(V) \setminus \bracs{0}$ such that $h(x) = h(x^{-1})$ for all $x \in V$. Since $f$ and $h$ are both compactly supported and $\mu, \nu$ are locally finite, by \hyperref[Tonelli's Theorem]{theorem:fubini-tonelli},
+    \begin{align*}
+        \paren{\int f d\mu}\paren{\int h d\nu} &= \iint f(x)h(y) \mu(dx)\nu(dy) \\
+        &= \iint f(yx)h(y) \nu(dx)\mu(dy) \\
+    \end{align*}
+
+    Similarly, by symmetry of $h$, 
+    \begin{align*}
+        \paren{\int h d\mu}\paren{\int f d\nu} &= \iint h(x)f(y) \mu(dx)\nu(dy) \\
+        &= \iint h(y^{-1}x)f(y) \mu(dx)\nu(dy) \\
+        &= \iint h(x^{-1}y)f(y) \nu(dy)\mu(dx) \\
+        &= \iint h(y)f(xy) \nu(dy)\mu(dx) \\
+        &= \iint h(y)f(xy) \mu(dx)\nu(dy)
+    \end{align*}
+
+    Thus there exists $C > 0$ such that
+    \begin{align*}
+    &\abs{\paren{\int f d\mu}\paren{\int h d\nu} - \paren{\int h d\mu}\paren{\int f d\nu}} \\ 
+    &\le \abs{\iint h(y)[f(xy) - f(yx)]\mu(dx)\nu(dy)} \le C\eps
+    \end{align*}
+
+    and
+    \[
+    \abs{\paren{\int g d\mu}\paren{\int h d\nu} - \paren{\int h d\mu}\paren{\int g d\nu}}
+    \le C\eps
+    \]
+
+    so there exists $C' > 0$ such that
+    \[
+    \abs{\frac{\int h d\nu}{\int h d\mu} - \frac{\int g d\nu}{\int g d\mu}}
+    \le C'\eps
+    \]
+
+    and
+    \[
+    \abs{\frac{\int f d\nu}{\int f d\mu} + \frac{\int h d\nu}{\int h d\mu}}
+    \le C'\eps
+    \]
+
+    Therefore
+    \[
+    \abs{\frac{\int f d\nu}{\int f d\mu} - \frac{\int g d\nu}{\int g d\mu}} \le 2C' \eps
+    \]
+
+    As the above holds for all $\eps > 0$, $\int f d\nu/\int f d\mu = \int g d\nu/\int g d\mu$. By uniqueness from the \hyperref[Riesz Representation Theorem]{section:riesz-radon}, there exists $\lambda > 0$ such that $\mu = \lambda \nu$. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:haar-measure-charge}
+    Let $G$ be a locally compact group and $\mu: \cb_G \to [0, \infty]$ be a left/right Haar measure, then:
+    \begin{enumerate}
+        \item For each $U \subset G$ open with $U \ne \emptyset$, $\mu(U) > 0$. 
+        \item For each $f \in C_c^+(G) \setminus \bracs{0}$, $\int \phi d\mu > 0$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}[Proof, for left Haar measures. ]
+    Since $\mu \ne 0$, there exists $g \in C_c^+(G) \setminus \bracs{0}$ such that $\int g d\mu > 0$. By compactness of $\supp{g}$, there exists $\seqf{x_j} \subset G$ such that:
+    \[
+    \supp{g} \subset \bigcup_{j = 1}^n x_j^{-1}U \quad g \le \sum_{j = 1}^n L_{x_j}f
+    \]
+
+    Thus $0 < \int g d\mu \le n\mu(U)$ and $0 < \int g d\mu \le n \int f d\mu$. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:haar-translation}
+    Let $G$ be a locally compact group, $\mu: \cb_G \to [0, \infty]$ be a left Haar measure, $p \in [1, \infty)$, and $E$ be a normed vector space over $K \in \RC$, then the mapping
+    \[
+    G \times L^p(\mu; E) \quad (x, f) \mapsto L_xf
+    \]
+
+    is jointly continuous. Similarly, if $\nu: \cb_G \to [0, \infty]$ is a right Haar measure, then
+    \[
+    G \times L^p(\mu; E) \quad (x, f) \mapsto R_xf
+    \]
+
+    is also jointly continuous.
+\end{proposition}
+\begin{proof}[Proof of the left case. ]
+    Let $\eps > 0$, $x, y \in G$, and $f, g \in L^p(\mu; E)$, then
+    \begin{align*}
+        \norm{L_xf - L_yg}_{L^p(\mu; E)} &\le \norm{L_xf - L_yf}_{L^p(\mu; E)} + \norm{L_yf - L_y g}_{L^p(\mu; E)} \\
+        &= \norm{L_xf - L_yf}_{L^p(\mu; E)} + \norm{f - g}_{L^p(\mu; E)}
+    \end{align*}
+
+    By \autoref{proposition:radon-cc-dense}, there exists $\phi \in C_c(G)$ such that $\norm{\phi - f}_{L^p(\mu; E)} < \eps$. In which case, 
+    \begin{align*}
+        \norm{L_xf - L_yf}_{L^p(\mu; E)} &\le \norm{L_xf - L_x \phi}_{L^p(\mu; E)} + \norm{L_x\phi - L_y\phi}_{L^p(\mu; E)} \\
+        &+ \norm{L_yf - L_y \phi}_{L^p(\mu; E)} \\
+        &= 2\norm{f - \phi}_{L^p(\mu; E)} + \norm{L_x\phi - L_y\phi}_{L^p(\mu; E)} \\
+        &\le 2\eps + \normn{L_{x^{-1}y}\phi - \phi}_u\mu\bracs{\phi \ne 0}
+    \end{align*}
+
+    By \autoref{proposition:lcg-cc-uc}, there exists $V \in \cn_G(1)$ such that if $x^{-1}y \in V$, then $\norm{L_{x^{-1}y}\phi - \phi}_u < \eps/\mu\bracs{\phi \ne 0}$. Thus if $x^{-1}y \in V$, then
+    \[
+    \norm{L_xf - L_yf}_{L^p(\mu; E)} \le 3\eps + \norm{f - g}_{L^p(\mu; E)}
+    \]
+\end{proof}
+
+

src/measure/lcg/index.tex

@@ -0,0 +1,6 @@
+\chapter{Integration on Locally Compact Groups}
+\label{chap:locally-compact-integration}
+
+\input{./lcg.tex}
+\input{./haar.tex}
+\input{./modular.tex}

src/measure/lcg/lcg.tex

@@ -0,0 +1,17 @@
+\section{Locally Compact Groups}
+\label{section:lcg}
+
+\begin{definition}[Locally Compact Group]
+\label{definition:lcg}
+    Let $G$ be a topological group, then $G$ is \textbf{locally compact} if $G$ is an LCH space. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:lcg-cc-uc}
+    Let $G$ be a locally compact group, $E$ be a TVS over $K \in \RC$, and $\phi \in C_c(G; E)$, then $\phi$ is left and right uniformly continuous. 
+\end{proposition}
+\begin{proof}
+    By \autoref{lemma:lch-compact-neighbour}, there exists a compact neighbourhood $U \in \cn_G(\text{supp}(\phi))$. By \autoref{proposition:uniform-continuous-compact}, $\phi|_{U}$ and $\phi|_{\supp(\phi)^c}$ are both left and right uniformly continuous. Therefore $\phi$ is left and right uniformly continuous. 
+\end{proof}
+
+

src/measure/lcg/modular.tex

@@ -0,0 +1,49 @@
+\section{The Modular Function}
+\label{section:modular-function}
+
+\begin{definition}[Modular Function]
+\label{definition:modular-function}
+    Let $G$ be a locally compact group and $\mu$ be a left Haar measure on $G$, then
+    \begin{enumerate}
+        \item For any $f, g \in C_c^+(G; \real) \setminus \bracs{0}$, $A, B \in \cb_G$ with $\mu(A), \mu(B) > 0$, and $y \in G$, 
+        \[
+        \Delta_G(y) = \frac{\int R_{y^{-1}} f d\mu}{\int f d\mu} = \frac{\int R_{y^{-1}} g d\mu}{\int g d\mu} = \frac{\mu(Ay)}{\mu(A)} = \frac{\mu(By)}{\mu(B)} > 0
+        \]
+        \item For each $y \in G$ and $A \in \cb_G$, denote $\mu_y(A) = \mu(Ay)$, then $\mu_y(dx) = \Delta_G(y)\mu(dx)$. 
+        \item For any choice of $f \in C_c^+(G; \real) \setminus \bracs{0}$, the mapping $\Delta_G: G \to (0, \infty)$ defined by $y \mapsto \int R_{y^{-1}}f d\mu$ is a continuous group homomorphism. 
+        \item For each $A \in \cb_G$, let $\nu(A) = \mu(A^{-1})$, then $\nu(dx) = \Delta(x^{-1})\mu(dx)$. 
+    \end{enumerate}
+    
+    The homomorphism $\Delta_G: G \to (0, \infty)$ is the \textbf{modular function} of $G$. 
+\end{definition}
+\begin{proof}[Proof, {{\cite[Proposition 2.24, Proposition 2.31]{FollandHarmonic}}}. ]
+    (1), (2): For each $y \in G$, $\mu_y$ is also a left Haar measure. By \hyperef[Haar's Theorem]{theorem:haar}, there exists $\lambda > 0$ such that $\mu_y = \lambda \mu$. In which case, 
+    \[
+        \Delta_G(y^{-1}) = \frac{\int R_y f d\mu}{\int f d\mu} = \frac{\int R_y g d\mu}{\int g d\mu} = \frac{\mu(Ay)}{\mu(A)} = \frac{\mu(By)}{\mu(B)}
+    \]
+
+    (3): By \autoref{proposition:haar-translation}, the mapping $y \mapsto \int R_y f d\mu$ is continuous. For any $x, y \in G$ and $A \in \cb_G$ with $\mu(A) > 0$, 
+    \[
+    \Delta_G(xy)\mu(A) = \mu(Axy) = \Delta_G(y)\mu(Ax) = \Delta_G(y)\Delta_G(x)\mu(A)
+    \]
+
+    (4): Let $f \in C_c^+(G)$ and $y \in G$, then
+    \begin{align*}
+        \int R_{y}f(x) \Delta_G(x^{-1})\mu(dx) &= \Delta_G(y)\int f(xy)\Delta_G[(xy)^{-1}]\mu(dx) \\
+        &= \int f(x)\Delta(x^{-1})\mu(dx)
+    \end{align*}
+
+    so $\Delta_G(x^{-1})\mu(dx)$ is a right Haar measure. By Haar's Theorem, there exists $\lambda > 0$ such that $\nu(dx) = \lambda \Delta_G(x^{-1})\mu(dx)$. 
+
+    If $\lambda \ne 1$, then there exists $U \in \cn_G(1)$ symmetric and compact such that $|\Delta_G(x^{-1}) - 1| \le 2^{-1}|\lambda - 1|$ on $U$. In which case, by symmetry, $\mu(U) = \nu(U)$, and
+    \begin{align*}
+        |\lambda - 1|\mu(U) &= |\lambda \nu(U) - \mu(U)| = \abs{\int_U \Delta_G(x^{-1})-1 \mu(dx)} \\
+        &\le \frac{1}{2}|\lambda - 1|\mu(U)
+    \end{align*}
+
+    which contradicts the fact that $\lambda \ne 1$. Therefore $\lambda = 1$, and $\nu(dx) = \Delta_G(x^{-1})\mu(dx)$. 
+    
+    
+\end{proof}
+
+

src/measure/lebesgue-integral/complex.tex

@@ -122,9 +122,11 @@
     and $\int f d\mu = \limv{n}\int f_n d\mu$.
 \end{proof}
 
-\begin{remark}[There is no dominated convergence theorem for nets]
+\begin{remark}[Dominated Convergence Theorem for Nets?]
 \label{remark:dct-no-net}
     In analysis, one frequently encounters places where only sequential continuity is provided or required. It is my opinion that a good portion of this comes from the lack of an extension of the \hyperref[dominated convergence theorem]{theorem:dct} to nets. This limitation arises from the \hyperref[monotone convergence theorem]{theorem:mct}, where continuity from below is used.
 
     For an example, consider the Lebesgue measure on $[0, 1]$. Let $A$ be the net of all finite subsets of $[0, 1]$, directed by inclusion, then $\lim_{\alpha \in A}\one_\alpha = 1$ pointwise. However, $\int \one_\alpha = 0$ for all $\alpha \in A$.
+
+    At least, that is how I thought back in 2025. While arbitrary pointwise convergence has no reason to cooperate with the structure of a measure space, additionally supplying convergence in measure bridges the above gap. The corresponding Monotone Convergence Theorem is given at \autoref{theorem:mct-measure}, and the Dominated Convergence Theorem is given at \autoref{corollary:dct-filter} following the Vitali Convergence Theorem. 
 \end{remark}

src/measure/lebesgue-integral/non-negative.tex

@@ -5,10 +5,10 @@
 \label{definition:measurable-non-negative}
     Let $(X, \cm)$ be a measure space, then
     \[
-    \mathcal{L}^+(X, \cm) = \bracs{f: X \to \real| f \ge 0, f \text{ is } (\cm, \cb_\real) \text{-measurable}}
+    \mathcal{L}^+(X, \cm) = \bracs{f: X \to [0, \infty]| f \ge 0, f \text{ is } (\cm, \cb_\real) \text{-measurable}}
     \]
 
-    is the space of non-negative $\real$-valued measurable functions on $(X, \cm)$.
+    is the space of non-negative $\ol \real$-valued measurable functions on $(X, \cm)$.
 \end{definition}
 
 \begin{definition}[Integral of Non-Negative Function]
@@ -23,19 +23,36 @@
 
 \begin{lemma}
 \label{lemma:lebesgue-non-negative-strict}
-    Let $(X, \cm, \mu)$ be a measure space and $f \in \mathcal{L}^+(X, \cm)$, then
+    Let $(X, \cm, \mu)$ be a measure space and $f \in \mathcal{L}^+(X, \cm)$. 
-    \[
+    \begin{enumerate}
-    \int f d\mu = \sup\bracs{\int \phi d\mu \bigg | \phi \in \Sigma^+(X, \cm), \phi|_{\bracs{f > 0}} < f|_{\bracs{f > 0}}, \phi \le f}
+        \item For each $\phi \in \Sigma^+(X, \cm)$, denote $\phi \le_u f$ if there exists $\delta > 0$ such that $\phi + \delta \le f$ on $\bracs{\phi > 0}$, then
-    \]
+        \[
+        \int f d\mu = \sup\bracs{\int \phi d\mu \bigg | \phi \in \Sigma^+(X, \cm), \phi \le_u f}
+        \]
 
+        \item If $\mu$ is semifinite, then
+        \[
+        \int f d\mu = \sup\bracs{\int \phi d\mu \bigg | \phi \in \Sigma^+(X, \cm) \cap L^1(X; \real), \phi \le_u f}
+        \]
+    \end{enumerate}
 \end{lemma}
 \begin{proof}
-    Let $\phi \in \Sigma^+(X, \cm)$ with $\phi \le f$, then for any $\alpha \in (0, 1)$, $\alpha \phi|_{\bracs{f > 0}} < f|_{\bracs{f > 0}}$. Since
+    (1): Let $\phi \in \Sigma^+(X, \cm)$ with $\phi \le f$ and $\alpha \in (0, 1)$. Since $\phi(X) \setminus \bracs{0} \subset (0, \infty)$ is finite, $\delta = \min_{y \in \phi(X) \setminus \bracs{0}}(1 - \alpha)y > 0$. Thus $\alpha \phi + \delta \le \phi \le f$ on $\bracs{\phi > 0} = \bracs{\alpha \phi > 0}$, and $\alpha \phi \le_u f$. 
+    
+    By \hyperref[linearity on simple functions]{proposition:lebesgue-simple-properties}, 
     \[
     \int \phi d\mu = \sup_{\alpha \in (0, 1)}\alpha \int \phi d\mu = \sup_{\alpha \in (0, 1)}\int \alpha \phi d\mu
     \]
 
-    the two sides are equal.
+    Thus
+    \[
+    \int \phi d\mu \le \sup\bracs{\int \psi d\mu \bigg | \psi \in \Sigma^+(X, \cm), \psi \le_u f}
+    \]
+
+    As the above holds for all $\phi \in \Sigma^+(X, \cm)$, 
+    \[
+    \int f d\mu = \sup\bracs{\int \psi d\mu \bigg | \psi \in \Sigma^+(X, \cm), \psi \le_u f}
+    \]
 \end{proof}
 
 \begin{theorem}[Monotone Convergence Theorem]
@@ -79,7 +96,7 @@
     Let $(X, \cm)$ be a measurable space and $f \in \mathcal{L}^+(X, \cm)$, then there exists $\seq{f_n} \subset \Sigma^+(X, \cm)$ such that $f_n \upto f$ pointwise.
 \end{lemma}
 \begin{proof}
-    By \autoref{proposition:measurable-simple-separable-norm}, there exists $\seq{g_n} \subset \Sigma^+(X, \cm)$ such that $0 \le g_n \le f$ for each $n \in \natp$, and $g_n \to f$ pointwise. For each $n \in \natp$, let $f_n = \max_{1 \le k \le n}g_k$, then $f_n \in \Sigma^+(X, \cm)$, $0 \le f_n \le f$, and $f_n \upto f$ pointwise.
+    By \autoref{corollary:measurable-simple-separable-norm}, there exists $\seq{g_n} \subset \Sigma^+(X, \cm)$ such that $0 \le g_n \le f$ for each $n \in \natp$, and $g_n \to f$ pointwise. For each $n \in \natp$, let $f_n = \max_{1 \le k \le n}g_k$, then $f_n \in \Sigma^+(X, \cm)$, $0 \le f_n \le f$, and $f_n \upto f$ pointwise.
 \end{proof}
 
 
@@ -98,7 +115,7 @@
 \begin{proof}
     (1): By \autoref{lemma:lebesgue-simple-monotone}, there exists $\seq{f_n}, \seq{g_n} \subset \Sigma^+(X, \cm)$ with $0 \le f_n \le f$ and $0 \le g_n \le g$ for each $n \in \natp$, $f_n \upto f$, and $g_n \upto g$. By \autoref{proposition:lebesgue-simple-properties} and the \hyperref[Monotone Convergence Theorem]{theorem:mct},
     \begin{align*}
-    \int \alpha f + g d\mu = \limv{n}\int \alpha f_n + g_n d\mu = \alpha\limv{n}\int f_n d\mu + \limv{n}\int g_n d\mu \\
+    \int \alpha f + g d\mu &= \limv{n}\int \alpha f_n + g_n d\mu = \alpha\limv{n}\int f_n d\mu + \limv{n}\int g_n d\mu \\
     &= \alpha \int f d\mu + \int g d\mu
     \end{align*}
 

src/measure/lebesgue-integral/simple.tex

@@ -16,9 +16,9 @@
     is the \textbf{Lebesgue integral} of $f$.
 \end{definition}
 
-\begin{proposition}[{{\cite[Proposition 2.13]{Folland}}}]
+\begin{proposition}
 \label{proposition:lebesgue-simple-properties}
-    Let $(X, \cm, \mu)$ be a measure space and $f, g \in \Sigma^*(X, \cm)$, then:
+    Let $(X, \cm, \mu)$ be a measure space and $f, g \in \Sigma^+(X, \cm)$, then:
     \begin{enumerate}
         \item For any $\alpha \ge 0$, $\int \alpha f d\mu = \alpha \int f d\mu$.
         \item $\int f + g d\mu = \int f d\mu + \int g d\mu$.
@@ -26,7 +26,7 @@
         \item The mapping $A \mapsto \int \one_A \cdot f d\mu$ is a measure on $(X, \cm)$.
     \end{enumerate}
 \end{proposition}
-\begin{proof}
+\begin{proof}[Proof, {{\cite[Proposition 2.13]{Folland}}}. ]
     (1): If $\alpha = 0$, then $\int \alpha f d\mu = \int 0 d\mu = 0 = 0 \cdot \int f d\mu$. Otherwise, the mapping $y \mapsto \alpha y$ is a bijection. Hence
     \[
     \alpha f = \sum_{y \in f(X)} (\alpha y) \cdot \one_{\bracs{f = y}}

src/measure/measurable-maps/approx.tex

@@ -0,0 +1,184 @@
+\section{Approximations with Simple Functions}
+\label{section:simple-approx}
+
+\begin{definition}[Admissible Approximant Function*]
+\label{definition:admissible-approximant-function}
+    Let $X$ be a topological space and $\mathcal{A}: X \to 2^X$, then $\mathcal{A}$ is an \textbf{admissible approximant function} on $X$ if:
+    \begin{enumerate}[label=(AA\arabic*)]
+        \item For each $x \in X$, $x \in \overline{\mathcal{A}(x)^o}$. 
+        \item $\bigcap_{x \in X}\mathcal{A}(x) \ne \emptyset$. 
+    \end{enumerate}
+
+    and $\mathcal{A}$ is \textbf{Borel measurable} if:
+    \begin{enumerate}[label=(AA\arabic*), start=2]
+        \item[(B)] For any $x_0 \in X$, $\bracs{x \in X|x_0 \in \mathcal{A}(x)} \in \cb_X$.
+    \end{enumerate}
+\end{definition}
+
+\begin{lemma}
+\label{lemma:admissible-approximant-existence}
+    Let $X$ be a topological space, and $\mathcal{A}: X \to 2^X$ be defined by $x \mapsto X$, then $\mathcal{A}$ is an \hyperref[admissible approximant function]{definition:admissible-approximant-function}. 
+\end{lemma}
+
+
+\begin{definition}[Approximation of the Identity*]
+\label{definition:approximation-id-measure}
+    Let $X$ be a topological space and $\net{I} \subset X^X$ be a net, then $\net{I}$ is an \textbf{approximation of the identity} if:
+    \begin{enumerate}[label=(AI\arabic*)]
+        \item For each $x \in X$, $I_\alpha(x) \to x$. 
+    \end{enumerate}
+
+    For any \hyperref[admissible approximant function]{definition:admissible-approximant-function} $\mathcal{A}: X \to 2^X$, $\net{I}$ is \textbf{$\mathcal{A}$-admissible} if:
+    \begin{enumerate}[label=(AI\arabic*), start=1]
+        \item For each $x \in X$ and $\alpha \in A$, $I_\alpha(x) \in \mathcal{A}(x)$. 
+    \end{enumerate}
+
+    The approximation $\net{I}$ is \textbf{simple} if $I_\alpha$ is finitely-valued for all $\alpha \in A$, and \textbf{Borel measurable} if $I_\alpha$ is Borel measurable for all $\alpha \in A$. 
+\end{definition}
+
+\begin{lemma}[Existence of Simple Approximations of the Identity]
+\label{lemma:separable-metric-space-approx-identity}
+    Let $X$ be a separable metric space, $\mathcal{A}: X \to 2^X$ be a Borel measurable \hyperref[admissible approximant function]{definition:admissible-approximant-function}, and $\seq{x_n} \subset X$ be a dense subset with $x_1 \in \bigcap_{x \in X}\mathcal{A}(x)$, then there exists $\seq{I_n} \subset X^X$ such that:
+    \begin{enumerate}
+        \item $\seq{I_n}$ is a Borel measurable, $\mathcal{A}$-admissible \hyperref[approximation of the identity]{definition:approximation-id-measure}. 
+        \item For each $N \in \natp$, $I_N(X) \subset \bracsn{x_n|1 \le n \le N}$. 
+        \item For each $N \in \natp$ and $x \in X$, 
+        \[
+        d(x, I_N(x)) = \min\bracs{d(x, x_n)| 1 \le n \le N, x_n \in \mathcal{A}(x)}
+        \]
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    By removing duplicate elements from the sequence, assume without loss of generality that for each $m, n \in \natp$ with $m \ne n$, $x_m \ne x_n$. 
+
+    Let $N \in \natp$. For each $x \in X$, let
+    \[
+    C_N(x) = \bracs{1 \le n \le N| x_n \in \mathcal{A}(x)}
+    \]
+
+    Since $x_1 \in \bigcap_{y \in X}\mathcal{A}(y)$, $1 \in C_N(x)$ and $C_N(x) \ne \emptyset$. Now, let
+    \[
+    k_N(x) = \min\bracs{n \in C_N(x) \bigg | d(x, x_n) = \min_{m \in C_N(x)}d(x, x_m)}
+    \]
+
+    be the minimum $n \in C_N(x)$ on which the minimal distance from $x$ to $\bracs{x_m|m \in C_N(x)}$ is achieved. Define
+    \[
+    I_N: X \to X \quad x \mapsto x_{k_N(x)}
+    \]
+
+    (2): For each $x \in X$, $k_N(x) \in [N]$, so $I_N(x) \in \bracsn{x_n|1 \le n \le N}$. 
+
+    (3): Let $x \in X$, then by definition of $k_N$ and $C_N$, 
+    \begin{align*}
+    d(x, I_N(x))& = d(x, x_{k_N(x)}) =  \min_{n \in C_N(x)}d(x, x_n) \\
+    &= \min\bracsn{d(x, x_n)|1 \le n \le N, x_n \in \mathcal{A}(x)}
+    \end{align*}
+    
+    
+    (1, Borel Measurable): Fix $N \in \natp$, then for each $n \in \natp$, 
+    \begin{align*}
+        \bracs{k_N \le n} &= \bigcup_{j = 1}^n \bracs{x \in X \bigg | j \in C_N(x), d(x, x_j) = \min_{m \in C_N(x)}d(x, x_m)} \\
+        &= \bigcup_{j = 1}^n \bracs{j \in C_N} \cap \bracs{x \in X \bigg | d(x, x_j) = \min_{m \in C_N(x)}d(x, x_m)} \\
+        &= \bigcup_{j = 1}^n\bigcup_{J \subset [N]} \bracs{j \in C_N, J = C_N} \cap \bracs{x \in X \bigg | d(x, x_j) = \min_{m \in J}d(x, x_m)} 
+    \end{align*}
+
+    Given that $\mathcal{A}$ is Borel measurable, $\bracs{n \in C_N} = \bracs{x_n \in \mathcal{A}(x)}$ is a Borel set for each $1 \le n \le N$. As a result, $\bracs{J = C_N}$ is Borel for each $J \subset [N]$. Thus $\bracs{j \in C_N, J = C_N}$ is Borel for each $1 \le j \le n$ and $J \subset [N]$. 
+
+    On the other hand, for each $1 \le n \le N$, the function $x \mapsto d(x, x_n)$ is continuous and hence Borel measurable. Similarly, for each $J \subset [N]$, the mapping $\real^J \to \real$ with $\alpha \mapsto \min_{j \in J}\alpha_j$ is also Borel measurable. 
+    
+    
+    The above facts combined show that $\bracs{k_N \le n}$ is a Borel set, and $k_N: X \to [N]$ is a Borel measurable function. Now, let
+    \[
+    I_N: X \to \bracsn{x_n|1 \le n \le N} \quad x \mapsto x_{k_N(x)}
+    \]
+
+    By assumption that $\seq{x_n}$ are distinct, $\bracs{I_N = x_n} = \bracs{k_N = n}$ is a Borel set for each $1 \le n \le N$. Therefore $I_N$ is Borel measurable. 
+
+    (1, $\mathcal{A}$-Admissible): Let $N \in \natp$ and $x \in X$, then
+    \[
+    I_N(x) = x_{k_N(x)} \in \bracs{x_n|n \in C_N(x)} \subset \mathcal{A}(x)
+    \]
+
+    (1, Approximation): Let $x \in X$ and $\eps > 0$. Since $x \in \ol{\mathcal{A}(x)^o}$ and $\seq{x_n}$ is dense in $X$, there exists $N_0 \in \natp$ such that $x_{N_0} \in \mathcal{A}(x)$ and $d(x, x_{N_0}) < \eps$. By (3), 
+    \[
+    \limv{N}d(x, I_N(x)) = \limv{N}\min\bracs{d(x, x_n)| 1 \le n \le N, x_n \in \mathcal{A}(x)} = 0
+    \]
+\end{proof}
+
+\begin{remark}
+\label{remark:separable-metric-space-approx-identity}
+    In \autoref{lemma:separable-metric-space-approx-identity}, if $X$ is compact and $\mathcal{A} \equiv X$, then $I_N \to I$ \textit{uniformly}. 
+\end{remark}
+
+\begin{corollary}
+\label{corollary:measurable-simple-separable}
+    Let $(X, \cm)$ be a measurable space, $Y$ be a separable metric space, and $\mathcal{A}: Y \to 2^Y$ be a Borel measurable \hyperref[admissible approximant function]{definition:admissible-approximant-function}, then for any $f: X \to Y$, the following are equivalent:
+    \begin{enumerate}
+        \item $f$ is $(\cm, \cb_Y)$-measurable.
+        \item For any dense subset $\seq{y_n} \subset Y$ with $y_1 \in \bigcap_{y \in Y}\mathcal{A}(y)$, there exists a sequence $\seq{f_n}$ of $(\cm, \cb_Y)$-measurable simple functions such that
+        \begin{enumerate}
+            \item[(i)] For each $x \in X$ and $N \in \natp$,
+            \[
+            f_N(x) \in \mathcal{A}(f(x)) \cap \bracsn{y_n|1 \le n \le N}
+            \]
+            \item[(ii)] $f_n \to f$ pointwise as $n \to \infty$.
+        \end{enumerate}
+
+        \item There exists a sequence $\seq{f_n}$ of $(\cm, \cb_Y)$-measurable simple functions such that $f_n \to f$ pointwise.
+    \end{enumerate}
+\end{corollary}
+\begin{proof}
+    (1) $\Rightarrow$ (2): Let $\seq{y_n} \subset Y$ be a dense subset with $y_1 \in \bigcap_{y \in Y}\mathcal{A}(y)$. By \autoref{lemma:separable-metric-space-approx-identity}, there exists $\seq{I_n} \subset Y^Y$ such that:
+    \begin{enumerate}
+        \item $\seq{I_n}$ is an $\mathcal{A}$-admissible \hyperref[approximation of the identity]{definition:approximation-id-measure}. 
+        \item For each $N \in \natp$, $I_N$ is Borel measurable with $I_N(Y) \subset \bracsn{y_n|1 \le n \le N}$. 
+    \end{enumerate}
+
+    For each $n \in \natp$, let $f_n = I_N \circ f_n$, then:
+    \begin{enumerate}
+        \item[(i)] For each $x \in X$ and $N \in \natp$,
+        \[
+        f_N(x) = I_N(f(x)) \in \mathcal{A}(f(x)) \cap \bracsn{y_n|1 \le n \le N}
+        \]
+        \item[(ii)] Since $I_n \to \text{Id}$ pointwise as $n \to \infty$, $f_n \to f$ pointwise as $n \to \infty$. 
+    \end{enumerate}
+
+    (3) $\Rightarrow$ (1): By \autoref{proposition:metric-measurable-limit}.
+\end{proof}
+
+\begin{corollary}
+\label{corollary:measurable-simple-separable-norm}
+    Let $(X, \cm)$ be a measurable space, $(E, \norm{\cdot}_E)$ be a separable normed vector space, and $f: X \to E$, then the following are equivalent:
+    \begin{enumerate}
+        \item $f$ is $(\cm, \cb_E)$-measurable.
+        \item There exists simple functions $\seq{f_n}$ such that $\abs{f_n} \le \abs{f}$ for all $n \in \natp$, and $f_n \to f$ pointwise.
+    \end{enumerate}
+\end{corollary}
+\begin{proof}
+    (1) $\Rightarrow$ (2): Let
+    \[
+    \mathcal{A}: E \to 2^E \quad y \mapsto \begin{cases}
+        B_E(0, \norm{y}_E) & y \ne 0 \\ 
+        E & y = 0
+    \end{cases}
+    \]
+
+    then
+    \begin{enumerate}
+        \item[(AA1)] For each $y \in E$, $y \in \ol{\mathcal{A}(y)^o}$.
+        \item[(AA2)] $0 \in \bigcap_{y \in E}\mathcal{A}(y)$.
+        \item[(B)] For any fixed $y_0 \in E \setminus \bracs{0}$,
+        \[
+        \bracs{y \in E|y_0 \in \mathcal{A}(y)} = \bracs{y \in E|\norm{y_0}_E < \norm{y}_E} \cup \bracs{0} \in \cb_E
+        \]
+
+        and $\bracs{y \in E|0 \in \mathcal{A}(y)} = E$. 
+    \end{enumerate}
+
+    so $\mathcal{A}$ is a Borel measurable \hyperref[admissible approximant function]{definition:admissible-approximant-function}. 
+
+    By (2) of \autoref{corollary:measurable-simple-separable}, there exists simple functions $\seq{f_n}$ such that $|f_n| \le |f|$ on $\bracs{f \ne 0}$ for all $n \in \natp$ and $f_n \to f$ pointwise. In which case, $|\one_{\bracs{f \ne 0}}f_n| \le |f|$ globally for all $n \in \natp$ and $\one_{\bracs{f \ne 0}}f_n \to f$ pointwise as $n \to \infty$. 
+
+    (2) $\Rightarrow$ (1): By \autoref{proposition:metric-measurable-limit}.
+\end{proof}
+

src/measure/measurable-maps/in-measure.tex

@@ -1,25 +1,82 @@
 \section{Convergence in Measure}
 \label{section:convergence-in-measure}
 
-\begin{definition}[Convergence in Measure]
+\begin{definition}[In Measure]
-\label{definition:convergence-in-measure}
+\label{definition:in-measure}
-    Let $(X, \cm, \mu)$ be a measure space, $(Y, d)$ be a metric space, and $\seq{f_n}$ and $f$ be Borel measurable functions from $X$ to $Y$, then $f_n \to f$ \textbf{in measure} if for every $\eps > 0$,
+    Let $(X, \cm, \mu)$ be a measure space and $(Y, d)$ be a separable metric space. For each $\eps, \delta > 0$, let
     \[
-    \lim_{n \to \infty}\mu(\bracs{d(f_n, f) > \eps}) = 0
+    U(\delta, \eps) = \bracs{(f, g) \in \mathcal{L}^0(X; Y)| \mu\bracs{d(f, g) > \delta} < \eps}
     \]
-\end{definition}
 
-\begin{definition}[Cauchy in Measure]
+    then
-\label{definition:cauchy-in-measure}
-    Let $(X, \cm, \mu)$ be a measure space, $(Y, d)$ be a metric space, and $\seq{f_n}$ be Borel measurable functions from $X$ to $Y$, then $\seq{f_n}$ is \textbf{Cauchy in measure} if for every $\eps > 0$,
     \[
-    \mu(\bracs{d(f_m, f_n) > \eps}) \to 0 \quad \text{as}\ m, n \to \infty
+    \fB = \bracs{U(\delta, \eps)|\eps, \delta > 0}
     \]
+
+    forms a fundamental system of entourages for a uniformity. The uniformity induced by $\fB$ is the \textbf{uniform structure of convergence in measure} on $\mathcal{L}^0(X; Y)$.
 \end{definition}
+\begin{proof}
+    It is sufficient to check the conditions of \autoref{proposition:fundamental-entourage-criterion}:
+    \begin{enumerate}
+        \item[(FB1)] For each $\eps, \eps', \delta, \delta' > 0$,  
+        \[
+        U(\delta \wedge \delta', \eps \wedge \eps') \subset U(\delta, \eps) \cap U(\delta', \eps')
+        \]
+        \item[(UB3)] For each $\eps, \delta > 0$ and $f, g, h \in \mathcal{L}^0(X; Y)$, 
+        \[
+        \bracs{d(f, h) > \delta} \subset \bracs{d(f, g) > \delta} \cup \bracs{d(g, h) > \delta}
+        \]
+
+        so $U(\delta/2, \eps/2) \circ U(\delta/2, \eps/2) \subset U(\delta, \eps)$. 
+    \end{enumerate}
+\end{proof}
+
+\begin{definition}[Ky Fan Metric]
+\label{definition:ky-fan}
+    Let $(X, \cm, \mu)$ be a measure space, $(Y, d)$ be a separable metric space, and
+    \[
+    \alpha: L^0(X; Y)^2 \to [0, \infty) \quad (f, g) \mapsto \inf\bracs{\eps > 0| \mu\bracs{d(f, g) > \eps} \le \eps} \wedge 1
+    \]
+
+    then:
+    \begin{enumerate}
+        \item $\alpha$ is a metric on $L^0(X; Y)$.
+        \item $\alpha$ induces the uniform structure of convergence in measure on $L^0(X; Y)$. 
+    \end{enumerate}
+
+    The mapping $\alpha$ is the \textbf{Ky Fan metric} on $L^0(X; Y)$. 
+\end{definition}
+\begin{proof}
+    (1): Let $f, g, h \in \mathcal{L}^0(X; Y)$, then
+    \begin{enumerate}
+        \item[(M)] If $\alpha(f, g) = 0$, then by \hyperref[continuity from above]{proposition:measure-properties}, 
+        \[
+        \mu\bracs{d(f, g) > 0} = \limv{n}\mu\bracs{d(f, g) > 1/n} \le \limv{n}\frac{1}{n} = 0
+        \]
+
+        so $f = g$ almost everywhere. 
+        \item[(PM3)] For each $\eps > 0$, 
+        \[
+        \bracs{d(f, h) > \eps} \subset \bracs{d(f, g) > \eps/2} \cup \bracs{d(g, h) > \eps/2}
+        \]
+
+        so $\alpha(f, h) \le \alpha(f, g) + \alpha(g, h)$. 
+    \end{enumerate}
+
+    so $\alpha$ is a metric on $\mathcal{L}^0(X; Y)$, modulo almost everywhere equality. 
+
+    (2): Let $f, g \in \mathcal{L}^0(X; Y)$. For any $\eps, \delta > 0$, if $\alpha(f, g) < \eps \wedge \delta$, then there exists $r \in (0, \eps \wedge \delta]$ such that $\mu\bracs{d(f, g) > r} \le r$. Thus
+    \[
+    \bracs{(f, g) \in \mathcal{L}^0(X; Y)|\alpha(f, g) < \eps \wedge \delta} \subset 
+    \bracs{(f, g) \in \mathcal{L}^0(X; Y)|\mu\bracs{d(f, g) >\delta} < \eps}
+    \]
+
+    On the other hand, if $\mu\bracs{d(f, g) > \eps} \le \eps$, then $d(f, g) \le \eps$. Therefore $\alpha$ induces the uniform structure of convergence in measure. 
+\end{proof}
 
 \begin{lemma}
 \label{lemma:ae-in-measure}
-    Let $(X, \cm, \mu)$ be a finite measure space, $(Y, d)$ be a metric space, and $\seq{f_n}$ and $f$ be Borel measurable functions from $X$ to $Y$ such that $f_n \to f$ almost everywhere, then $f_n \to f$ in measure.
+    Let $(X, \cm, \mu)$ be a finite measure space, $(Y, d)$ be a separable metric space, and $\seq{f_n}$ and $f$ be Borel measurable functions from $X$ to $Y$ such that $f_n \to f$ almost everywhere, then $f_n \to f$ in measure.
 \end{lemma}
 \begin{proof}
     Let $\eps > 0$, then for almost every $x \in X$, there exists $N \in \natp$ such that $d(f_n(x), f(x)) < \eps$ for all $n \ge N$, so
@@ -33,18 +90,16 @@
     \]
 \end{proof}
 
-
+\begin{theorem}
-\begin{proposition}[{{\cite[Theorem 2.30]{Folland}}}]
+\label{theorem:cauchy-in-measure-limit}
-\label{proposition:cauchy-in-measure-limit}
+    Let $(X, \cm, \mu)$ be a measure space and $(Y, d)$ be a Polish space, then:
-    Let $(X, \cm, \mu)$ be a measure space, $(Y, d)$ be a complete metric space, and $\seq{f_n} \subset Y^X$ be a sequence Borel measurable functions from $X \to Y$ that is Cauchy in measure, then:
     \begin{enumerate}
-        \item There exists a Borel measurable function $f: X \to Y$ such that $f_n \to f$ in measure.
+        \item For any $\seq{f_n} \subset L^0(X; Y)$ that is Cauchy in measure, there exists $f \in L^0(X; Y)$ and a subsequence $\seq{n_k}$ such that $f_{n_k} \to f$ almost everywhere and in measure.
-        \item There exists a subsequence $\seq{n_k}$ such that $f_{n_k} \to f$ almost everywhere.
+        \item $L^0(X; Y)$ equipped with the uniform structure of convergence in measure is complete. 
-        \item For any Borel measurable function $g: X \to Y$ such that $f_n \to g$ in measure, $f = g$ almost everywhere.
     \end{enumerate}
-\end{proposition}
+\end{theorem}
-\begin{proof}
+\begin{proof}[Proof, {{\cite[Theorem 2.30]{Folland}}}. ]
-    (2): Since $\seq{f_n}$ is Cauchy in measure, there exists a subsequence $\seq{n_k}$ such that for each $k \in \natp$, $\mu(\bracsn{d(f_{n_k}, f_{n_{k+1}}) > 2^{-k}}) \le 2^{-k}$.
+    (1): Since $\seq{f_n}$ is Cauchy in measure, there exists a subsequence $\seq{n_k}$ such that for each $k \in \natp$, $\mu(\bracsn{d(f_{n_k}, f_{n_{k+1}}) > 2^{-k}}) \le 2^{-k}$.
     
     In this case, for any $K \in \natp$ and $j \ge k \ge K$, 
     \[
@@ -52,7 +107,7 @@
     \subset \bigcup_{\ell = K}^{\infty}\bracsn{d(f_{n_\ell}, f_{n_{\ell + 1}}) > 2^{-\ell}}
     \]
 
-    By monotonicity and subadditivity (\autoref{proposition:measure-properties}), 
+    By \hyperref[monotonicity and subadditivity]{proposition:measure-properties}, 
     \[
     \mu\paren{\bigcup_{j, k \ge K}\bracsn{d(f_{n_j}, f_{n_k}) > 2^{-K+1}}} 
     \le \sum_{\ell \ge K}\mu\bracsn{d(f_{n_\ell}, f_{n_{\ell + 1}}) > 2^{-\ell}}
@@ -69,15 +124,16 @@
     \mu\braks{\limsup_{K \to \infty}\paren{\bigcap_{j, k \ge K}\bracsn{d(f_{n_j}, f_{n_k}) \le 2^{-K+1}}}^c} = 0
     \]
 
-    Thus, for almost every $x \in X$, there exists $K \in \natp$ such that $d(f_{n_j}(x), f_{n_k}(x)) < 2^{-K+1}$ for all $j, k \ge K$. Therefore $\seq{f_n(x)}$ is Cauchy for almost every $x$, and converges to a Borel measurable function $f: X \to Y$.
+    Thus, for almost every $x \in X$, there exists $K \in \natp$ such that $d(f_{n_j}(x), f_{n_k}(x)) < 2^{-K+1}$ for all $j, k \ge K$. Therefore $\seq{f_n(x)}$ is Cauchy for almost every $x$, and converges almost everywhere to a Borel measurable function $f \in L^0(X; Y)$.
 
-    (1): Let $\eps, \delta > 0$, then there exists $N \in \natp$ such that $\mu(\bracs{d(f_m, f_n) > \delta/2}) < \eps/2$ for all $m, n \ge N$. Let $k \in \natp$ such that $n_k \ge N$ and $\mu(\bracs{d(f, f_{n_k}) > \delta/2}) < \eps/2$, then for any $m \ge N$,
+    Finally, for each $K \in \natp$,  
     \[
-    \mu\bracs{d(f_m, f) > \delta} \le \mu\bracs{d(f_m, f_{n_k}) > \delta/2} + \mu\bracs{d(f, f_{n_k}) > \delta/2} < \eps
+        \mu\bracs{d(f_{n_K}, f) > 2^{-K+1}} \le \sum_{k \ge K}\mu\bracs{d(f_{n_k}, f_{n_K}) > 2^{-k}} \le \sum_{k \ge K}2^{-k}
     \]
 
-    (3): By (2), there exists a subsequence $\seq{n_k}$ such that $f_{n_k} \to f$ and $f_{n_k} \to g$ almost everywhere, so $f = g$ almost everywhere. 
+    so $f_{n_k} \to f$ in measure as well. 
+    
+
+    (2): Since the uniform structure of convergence in measure on $L^0(X; Y)$ is defined by the \hyperref[Ky Fan metric]{definition:ky-fan}, completeness follows from (1) and \autoref{proposition:complete-metric-space}. 
 \end{proof}
 
-
-

src/measure/measurable-maps/index.tex

@@ -6,4 +6,6 @@
 \input{./real-valued.tex}
 \input{./simple.tex}
 \input{./metric.tex}
+\input{./approx.tex}
 \input{./in-measure.tex}
+\input{./locally-in-measure.tex}

src/measure/measurable-maps/local-in-measure.tex

@@ -0,0 +1,116 @@
+\section{Local Convergence in Measure}
+\label{section:locally-in-measure}
+
+\begin{definition}[Locally In Measure*]
+\label{definition:locally-in-measure}
+    Let $(X, \cm, \cf, \mu)$ be a \hyperref[scaffolded]{definition:measure-scaffold} measure space and $(Y, d)$ be a separable metric space. For each $\eps, \delta > 0$ and $A \in \cf$, let
+    \[
+    U(A, \delta, \eps) = \bracs{(f, g) \in \mathcal{L}^0(X; Y)| \mu(A \cap \bracs{d(f, g) > \delta}) < \eps}
+    \]
+
+    then
+    \[
+    \fB = \bracs{U(A, \delta, \eps)|\eps, \delta > 0, A \in \cm, \mu(A) < \infty}
+    \]
+
+    forms a fundamental system of entourages for a uniformity. 
+    
+    The uniformity defined by $\fB$ is the \textbf{uniform structure of local convergence in measure}, and $\mathcal{L}_\cf^0(X; Y)$ denotes $\mathcal{L}^0(X; Y)$ equipped with this uniformity.
+\end{definition}
+\begin{proof}
+    It is sufficient to check the conditions of \autoref{proposition:fundamental-entourage-criterion}:
+    \begin{enumerate}
+        \item[(FB1)] For each $\eps, \eps', \delta, \delta' > 0$ and $A, A' \in \cm$ with $\mu(A), \mu(A') < \infty$, 
+        \[
+        U(A \cup A', \delta \wedge \delta', \eps \wedge \eps') \subset U(A, \delta, \eps) \cap U(A', \delta', \eps')
+        \]
+        \item[(UB3)] For each $\eps, \delta > 0$, $A \in \cf$, and $f, g, h \in \mathcal{L}^0(X; Y)$, 
+        \[
+        \bracs{d(f, h) > \delta} \subset \bracs{d(f, g) > \delta} \cup \bracs{d(g, h) > \delta}
+        \]
+
+        so $U(A, \delta/2, \eps/2) \circ U(A, \delta/2, \eps/2) \subset U(A, \delta, \eps)$. 
+    \end{enumerate}
+\end{proof}
+
+\begin{proposition}
+\label{proposition:convergence-in-measure}
+    Let $(X, \cm, \cf, \mu)$ be a \hyperref[scaffolded]{definition:measure-scaffold} measure space, $(Y, d)$ be a separable metric space, and $\fF$ be a filter of $(\cm, \cb_Y)$-measurable functions, then $\fF$ is Cauchy in measure if and only if:
+    \begin{enumerate}
+        \item[(L)] $\fF$ is \hyperref[definition:locally-in-measure]{definition:locally-in-measure}. 
+        \item[(T)] For each $\eps, \delta > 0$, there exists $F \in \fF$ and $A \in \cf$ such that
+        \[
+        \sup_{f, g \in F}\mu(A^c \cap \bracs{d(f, g) > \delta}) < \eps
+        \]
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (L) + (T) $\Rightarrow$ (In Measure): Let $\eps, \delta > 0$. By (T) then there exists $F_1 \in \fF$ and $A \in \cf$ such that 
+    \[
+    \sup_{f, g \in F_1}\mu(A^c \cap \bracs{d(f, g) > \delta}) < \eps
+    \]
+
+    By (L), there exists $F_2 \in \fF$ with $F_2 \subset F_1$ such that
+    \[
+    \sup_{f, g \in F_2}\mu(A \cap \bracs{d(f, g) > \delta}) < \eps
+    \]
+
+    Therefore
+    \[
+    \sup_{f, g \in F_2}\mu\bracs{d(f, g) > \delta} < 2\eps
+    \]
+\end{proof}
+
+\begin{theorem}
+\label{theorem:locally-in-measure-complete}
+    Let $(X, \cm, \cf, \mu)$ be a \hyperref[scaffolded]{definition:measure-scaffold} localisable measure space and $(Y, d)$ be a Polish space, then \hyperref[$\mathcal{L}^0_\cf(X; Y)$]{definition:locally-in-measure} is complete. 
+\end{theorem}
+\begin{proof}
+    Let $\fF \subset \mathcal{L}^0_\cf(X; Y)$ be a Cauchy filter. By \autoref{theorem:cauchy-in-measure-limit}, for each $A \in \cf$, there exists an almost everywhere unique $f_A \in \mathcal{L}^0(A; Y)$ such that $\fF$ converges to $f_A$ when restricted to $A$. Thus for any $A, B \in \cf$, $f_{A \cup B}|_{A \cap B} = f_A|_{A \cap B} = f_B|_{B \cap A}$ almost everywhere. By the \hyperref[gluing lemma for measurable functions]{lemma:gluing-measurable}, there exists $f \in \mathcal{L}^0(X; Y)$ such that $f|_A = f_A$ for all $A \in \cf$. Thus $\fF \to f$ locally in measure, and $\mathcal{L}^0(X; Y)$ is complete. 
+\end{proof}
+
+\begin{theorem}[Monotone Convergence Theorem (in Measure)]
+\label{theorem:mct-measure}
+    Let $(X, \cm, \mu)$ be a semifinite measure space, $\net{f} \subset \mathcal{L}^+(X, \cm)$, and $f \in \mathcal{L}^+(X, \cm)$ such that
+    \begin{enumerate}[label=(\alph*)]
+        \item For each $x \in X$, $f_\alpha(x) \upto f(x)$. 
+        \item $f_\alpha \to f$ locally in measure. 
+    \end{enumerate}
+
+    then
+    \[
+    \lim_{\alpha \in A}\int f_\alpha d\mu = \int f d\mu
+    \]
+\end{theorem}
+\begin{proof}
+    By \autoref{definition:lebesgue-non-negative}, $\int f_\alpha d\mu \le \int f d\mu$ for each $\alpha \in A$. 
+
+    On the other hand, using \autoref{lemma:lebesgue-non-negative-strict}, it is sufficient to show that
+    \[
+    \lim_{\alpha \in A}\int f_\alpha d\mu = \sup_{\alpha \in A}\int f_\alpha d\mu \ge \int \phi d\mu
+    \]
+    
+    for any $\phi \in \Sigma^+(X, \cm)$ satisfying:
+    \begin{enumerate}[label=(\roman*)]
+        \item There exists $\delta > 0$ such that $\phi + \delta \le f$ on $\bracs{\phi > 0}$. 
+        \item $\phi \in L^1(X, \cm)$. 
+    \end{enumerate}
+
+    To this end, let $\eps > 0$. Since $\mu\bracs{\phi > 0} < \infty$, by (b), there exists $\alpha \in A$ such that
+    \[
+    \mu\bracs{\phi > 0, f_\alpha + \delta < \phi} \le \mu\bracs{\phi > 0, f_\alpha + \delta < f} < \frac{\eps}{\norm{\phi}_u}
+    \]
+
+    In which case, by \hyperref[linearity]{proposition:lebesgue-simple-properties},
+    \begin{align*}
+        \int \phi d\mu &= \int_{\bracs{\phi > 0}}\phi d\mu = \int_{\bracs{\phi > 0, f_\alpha + \delta \ge \phi}} \phi d\mu + \int_{\bracs{\phi > 0, f_\alpha + \delta < \phi}} \phi d\mu \\
+        &\le \int_{\bracs{\phi > 0, f_\alpha + \delta \ge \phi}} f_\alpha d\mu +\norm{\phi}_u\mu \bracs{\phi > 0, f_\alpha + \delta < \phi} \\
+        &\le \int f_\alpha d\mu + \norm{\phi}_u \frac{\eps}{\norm{\phi}_u} = \int f_\alpha d\mu + \eps
+    \end{align*}
+
+    As the above holds for all $\eps > 0$, $\int \phi d\mu \le \sup_{\alpha \in A}\int f_\alpha d\mu$. Therefore
+    \[
+    \int f d\mu = \sup_{\alpha \in A}\int f_\alpha d\mu = \lim_{\alpha \in A}\int f_\alpha d\mu
+    \]
+\end{proof}
+

src/measure/measurable-maps/measurable-maps.tex

@@ -6,6 +6,14 @@
     Let $(X, \cm)$ and $(Y, \cn)$ be measurable spaces and $f: X \to Y$ be a mapping, then $f$ is \textbf{$(\cm, \cn)$-measurable} if $f^{-1}(E) \in \cm$ for all $E \in \cn$.
 \end{definition}
 
+\begin{definition}[Space of Measurable Functions]
+\label{definition:measurable-function-space}
+    Let $(X, \cm)$ and $(Y, \cn)$ be measurable spaces, then the set $\mathscr{L}^0(X, \cm; Y) = \mathcal{L}^0(X; Y)$ is the \textbf{space of measurable functions} from $X$ to $Y$. 
+
+    For any measure $\mu$ on $(X, \cm)$, the space $L^0(X, \cm, \mu; Y) = L^0(X; Y)$ is the space of measurable functions from $X$ to $Y$, modulo almost everywhere equality. 
+\end{definition}
+
+
 
 \begin{definition}[Borel Measurable]
 \label{definition:borel-measurable-function}

src/measure/measurable-maps/metric.tex

@@ -21,7 +21,7 @@
 
 \begin{proposition}
 \label{proposition:metric-measurables}
-    Let $(X, \cm)$ be a measurable space, $Y$ be a metrisable topological space, $f, g: X \to Y$ be $(\cm, \cb_Y)$-measurable functions, then the following functions are measurable:
+    Let $(X, \cm)$ be a measurable space, $Y$ be a separable and metrisable topological space, $f, g: X \to Y$ be $(\cm, \cb_Y)$-measurable functions, then the following functions are measurable:
     \begin{enumerate}
         \item For any metric $d$ on $Y$, $x \mapsto d(f(x), f(y))$.
         \item If $Y$ is a TVS over $K \in \RC$ and $\lambda \in K$, $\lambda f + g$. In particular, $\bracs{f = g} \in \cm$.
@@ -40,7 +40,7 @@
         \item For each $\phi \in C(X; [0, 1])$, $\phi \circ f$ is $(\cm, \cb_\real)$-measurable.
     \end{enumerate}
 \end{proposition}
-\begin{proof}
+\begin{proof}[Proof, {{\cite[Lemma 8.1.9]{CohnMeasure}}}. ]
     (2) $\Rightarrow$ (1): For each $U \subset X$ open, the function
     \[
     d_{U^c}: X \to [0, 1] \quad x \mapsto d(x, U^c) \wedge 1
@@ -53,11 +53,11 @@
 \label{proposition:metric-measurable-limit}
     Let $(X, \cm)$ be a measurable space, $Y$ be a metrisable topological space, and $\seq{f_n}$ be $(\cm, \cb_Y)$-measurable functions, then:
     \begin{enumerate}
-        \item If $Y$ is completely metrisable, then $\bracsn{\limv{n}f_n \text{ exists}} \in \cm$.
+        \item If $Y$ is Polish, then $\bracsn{\limv{n}f_n \text{ exists}} \in \cm$.
         \item If $f = \limv{n}f_n$ exists, then it is $(\cm, \cb_Y)$-measurable.
     \end{enumerate}
 \end{proposition}
-\begin{proof}
+\begin{proof}[Proof, {{\cite[Proposition 8.1.10-8.1.11]{CohnMeasure}}}. ]
     (1): Let $d$ be a complete metric on $Y$, then for any $x \in X$, $\limv{n}f_n(x)$ exists if and only if $\seq{f_n(x)}$ is Cauchy. In which case,
     \[
     \bracs{\limv{n}f_n \text{ exists}} = \bigcap_{k \in \natp}\bigcup_{N \in \natp}\bigcap_{m \in \natp}\bigcap_{n \in \natp}\bracsn{d(f_m, f_n) < 1/k}
@@ -69,86 +69,3 @@
 \end{proof}
 
 
-\begin{proposition}
-\label{proposition:measurable-simple-separable}
-    Let $(X, \cm)$ be a measurable space, $Y$ be a separable metric space, and $N: Y \to 2^Y$ such that
-    \begin{enumerate}
-        \item[(a)] For each $y \in Y$, $y \in \ol{N(y)^o}$.
-        \item[(b)] $\bigcap_{y \in Y}N(y) \ne \emptyset$.
-        \item[(c)] For any $y_0 \in Y$, $\bracs{y \in Y|y_0 \in N(y)} \in \cb_Y$.
-    \end{enumerate}
-
-    Then, for any $f: X \to Y$, the following are equivalent:
-    \begin{enumerate}
-        \item $f$ is $(\cm, \cb_Y)$-measurable.
-        \item There exists a sequence $\seq{f_n}$ of $(\cm, \cb_Y)$-measurable simple functions such that
-        \begin{enumerate}
-            \item[(i)] For each $x \in X$ and $n \in \natp$, $f_n(x) \in N(f(x))$.
-            \item[(ii)] $f_n \to f$ pointwise.
-        \end{enumerate}
-
-        \item There exists a sequence $\seq{f_n}$ of $(\cm, \cb_Y)$-measurable simple functions such that $f_n \to f$ pointwise.
-    \end{enumerate}
-\end{proposition}
-\begin{proof}
-    (1) $\Rightarrow$ (2): Let $\seq{y_n} \subset Y$ be a dense subset of $Y$. Assume without loss of generality that $y_1 \in \bigcap_{y \in Y}N(y)$. For each $N \in \nat$ and $x \in X$, let
-    \[
-    C(N, x) = \bracs{1 \le n \le N|y_n \in N(f(x))}
-    \]
-
-    By assumption (b), $1 \in C(N, x) \ne \emptyset$. Let
-    \[
-    k(N, x) = \min\bracs{n \in C(N, x) \bigg | d(f(x), y_n) = \min_{m \in C(N, x)}d(f(x), y_m)}
-    \]
-
-    then for any $k \in \natp$, $\bracs{x \in X|k(n, x) \le k}$ is equal to
-    \[
-     \bigcup_{j = 1}^k\bracs{x \in X \bigg |y_j \in C(n, x), d(f(x), y_j) = \min_{m \in C(N, x)}d(f(x), y_m)}
-    \]
-
-    For each $1 \le m \le N$, $y \mapsto d(y, y_m)$ is continuous. Thus $x \mapsto d(f(x), y_m)$ and $x \mapsto \min_{m \in C(N, x)}d(f(x), y_m)$ are $(\cm, \cb_\real)$-measurable by \autoref{proposition:limit-measurable} and assumption (c). By \autoref{proposition:metric-measurables},
-    \[
-    \bracs{x \in X \bigg | d(f(x), y_j) = \min_{m \in C(N, x)}d(f(x), y_m)} \in \cm
-    \]
-
-    for each $1 \le j \le k$. Combining this with assumption (c) shows that $\bracs{x \in X|k(n, x) \le k} \in \cm$.
-
-    Let $f_N(x) = y_{k(N, x)}$, then for each $N \in \natp$, $f_N(x) \subset \bracs{y_n|1 \le n \le N}$, so $f_N(x)$ is finitely valued. Since $x \mapsto k{(N, x)}$ is $(\cm, 2^\nat)$-measurable, $f_N$ is $(\cm, \cb_Y)$-measurable.
-
-    Fix $x \in X$, then
-    \[
-    f(x) \in \ol{N^o(f(x))} = \ol{\bracs{y_n| n \in \natp, y_n \in N^o(f(x))}}
-    \]
-
-    by assumption (a) and \autoref{definition:dense}. Thus for any $\eps > 0$, there exists $N \in \nat$ such that $y_N \in N^o(f(x))$ and $d(f(x), y_N) < \eps$. In which case, $d(f(x), f_n(x)) < \eps$ for all $n \ge N$. Therefore $f_n \to f$ pointwise.
-
-    (3) $\Rightarrow$ (1): By \autoref{proposition:metric-measurable-limit}.
-\end{proof}
-
-\begin{proposition}
-\label{proposition:measurable-simple-separable-norm}
-    Let $(X, \cm)$ be a measurable space, $(E, \norm{\cdot}_E)$ be a separable normed vector space, and $f: X \to E$, then the following are equivalent:
-    \begin{enumerate}
-        \item $f$ is $(\cm, \cb_E)$-measurable.
-        \item There exists simple functions $\seq{f_n}$ such that $\abs{f_n} \le \abs{f}$ for all $n \in \natp$, and $f_n \to f$ pointwise.
-    \end{enumerate}
-\end{proposition}
-\begin{proof}
-    Let
-    \[
-    N: E \to 2^E \quad y \mapsto B_E(0, \norm{y}_E)
-    \]
-
-    then
-    \begin{enumerate}
-        \item[(a)] $y \in \ol{B_E(0, \norm{y}_E)}$.
-        \item[(b)] $0 \in \bigcap_{y \in E}N(y)$.
-        \item[(c)] For any fixed $y_0 \in E$,
-        \[
-        \bracs{y \in E|y_0 \in N(y)} = \bracs{y \in E|\norm{y_0}_E \le \norm{y}_E} \in \cb_E
-        \]
-
-    \end{enumerate}
-
-    By \autoref{proposition:measurable-simple-separable}, (1) and (2) are equivalent.
-\end{proof}

src/measure/measure/index.tex

@@ -4,7 +4,9 @@
 \input{./measure.tex}
 \input{./complete.tex}
 \input{./semifinite.tex}
+\input{./scaffold.tex}
 \input{./sigma-finite.tex}
+\input{./localisable.tex}
 \input{./regular.tex}
 \input{./outer.tex}
 \input{./lebesgue-stieltjes.tex}

src/measure/measure/localisable.tex

@@ -0,0 +1,240 @@
+\section{Localisable Measures}
+\label{section:localisable-measure}
+
+\begin{definition}[Essential Supremum]
+\label{definition:esssup-measure-space}
+    Let $(X, \cm, \mu)$ be a measure space, $\ce \subset \cm$, and $S \in \cm$, then $S$ is an \textbf{essential upper bound} of $\ce$ if for any $E \in \ce$, $\mu(E \setminus S) = 0$. If in addition, for any essential upper bound $T$ of $\ce$, $\mu(S \setminus T) = 0$, then $S$ is an \textbf{essential supremum} of $\ce$.  
+\end{definition}
+
+
+\begin{definition}[Localisable]
+\label{definition:localisable-measure}
+    Let $(X, \cm, \mu)$ be a measure space, then $X$ is \textbf{localisable} if:
+    \begin{enumerate}
+        \item $\mu$ is semifinite. 
+        \item For every $\ce \subset \cm$, there exists an essential supremum $A$ of $\ce$. 
+    \end{enumerate}
+\end{definition}
+
+\begin{definition}[Decomposable]
+\label{definition:decomposable-measure}
+    Let $(X, \cm, \mu)$ be a measure space and $\seqi{A} \subset \cm$, then $\seqi{A}$ is a \textbf{decomposition} of $(X, \cm, \mu)$ if:
+    \begin{enumerate}
+        \item For each $i \in I$, $\mu(A_i) < \infty$. 
+        \item $X = \bigsqcup_{i \in I}X_i$. 
+        \item $\cm = \bracs{E \subset X|E \cap A_i \in \cm \forall i \in I}$. 
+        \item For each $E \in \cm$, $\mu(E) = \sum_{i \in I}\mu(E \cap A_i)$. 
+    \end{enumerate}
+
+    If $(X, \cm, \mu)$ admits a decomposition, then it is \textbf{decomposable}. 
+\end{definition}
+
+\begin{lemma}
+\label{lemma:essential-upperbound-finite}
+    Let $(X, \cm, \mu)$ be a finite measure space, $\ce \subset \cm$, and $\mathcal{S}$ be the set of all essential upper bounds of $\ce$, then:
+    \begin{enumerate}
+        \item $\mathcal{S} \ne \emptyset$. 
+        \item For any $S \in \cm$, $S \in \mathcal{S}$ if and only if $\mu(S \cap E) = \mu(E)$ for all $E \in \ce$. 
+        \item For any $\seq{S_n} \subset \mathcal{S}$, $\bigcap_{n \in \natp}S_n \in \mathcal{S}$. 
+        \item There exists $S \in \mathcal{S}$ such that $\mu(S) = \inf\bracs{\mu(T)|T \in \mathcal{S}}$. 
+        \item For any $S \in \mathcal{S}$ with $\mu(S) = \inf\bracs{\mu(T)|T \in \mathcal{S}}$, $S$ is an essential supremum of $\ce$. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    (1): $X \in \mathcal{S}$. 
+
+    (2): Since $\mu$ is finite, for any $E \in \ce$, $\mu(S \cap E) = \mu(E)$ if and only if $\mu(E \setminus S) = 0$. 
+
+    (3): Firstly, for any $S, T \in \mathcal{S}$ and $E \in \ce$, 
+    \[
+    \mu(E \setminus (S \cap T)) \le \mu(E \setminus S) + \mu(E \setminus T) = 0
+    \]
+    
+    so $S \cap T \in \mathcal{S}$. Now let $\seq{S_n} \subset X$ and $E \in \ce$, then since $\mu$ is finite, 
+    \[
+    \mu(E) = \limv{N}\mu\paren{E \cap \bigcap_{n = 1}^N S_n} = \mu\paren{E \cap \bigcap_{n \in \natp}S_n}
+    \]
+    
+    by \hyperref[continuity from above]{proposition:measure-properties}. By (2), $\bigcap_{n \in \natp}S_n \in \mathcal{S}$. 
+
+    (4): Let $M = \inf\bracs{\mu(T)|T \in \mathcal{S}}$ and $\seq{S_n} \subset \mathcal{S}$ such that $\limv{n}\mu(S_n) = M$, then by continuity from above,
+    \[
+    M \le \mu\paren{\bigcap_{n \in \natp}S_n} \le \limv{n}\mu(S_n) = M
+    \]
+
+    By (3), $\bigcap_{n \in \natp}S_n \in \mathcal{S}$, therefore the minimum is achieved. 
+
+    (5): Let $R \in \mathcal{S}$. By (3), $S \cap R \in \mathcal{S}$ with $\mu(S \cap R) = \inf\bracs{\mu(T)|T \in \mathcal{S}} = \mu(S)$, so
+    \[
+    \mu(S \setminus R) = \mu(S \setminus (S \cap R)) = \mu(S) - \mu(S \cap R) = 0
+    \]
+
+    and $S$ is the essential supremum of $\ce$. 
+\end{proof}
+
+
+\begin{proposition}
+\label{proposition:decomposable-localisable}
+    Let $(X, \cm, \mu)$ be a decomposable measure space, then $X$ is localisable. 
+\end{proposition}
+\begin{proof}
+    Let $\ce \subset \cm$ and $\seqi{A} \subset \cm$ be a decomposition of $X$. 
+    
+    For each $i \in I$, let $\ce_i = \bracs{E \cap A_i|E \in \ce}$. By \autoref{lemma:essential-upperbound-finite}, there exists an essential upper bound $S_i \in \cm$ of $\ce_i$ contained in $A_i$ with respect to the restricted measure $\mu|_{A_i}$. In other words,  
+    \begin{enumerate}[label=(\roman*)]
+        \item $S_i$ is an essential upper bound of $\ce_i$. 
+        \item For any essential upper bound $T \in \cm$ of $\ce_i$ with $T \subset A_i$, $\mu(S_i \setminus T) = 0$. 
+    \end{enumerate}
+
+    Now, let $S = \bigsqcup_{i \in I}S_i$, then since $X = \bigsqcup_{i \in I}A_i$ and $S_i \subset A_i$ for all $i \in I$, $S \cap A_i = S_i \in \cm$ for all $i \in I$, so $S \in \cm$. For any $E \in \ce$, $E \cap A_i \in \ce_i$. Passing through the decomposition, (i) implies that, 
+    \begin{align*}
+        \mu(E \setminus S) &= \sum_{i \in I}\mu((E \setminus S) \cap A_i) = \sum_{i \in I}\mu((E \cap A_i) \setminus (S \cap A_i)) \\
+        &= \sum_{i \in I}\mu((E \cap A_i) \setminus S_i) = \sum_{i \in I}0 = 0
+    \end{align*}
+
+    and $S$ is an essential upper bound of $\ce$. 
+    
+    Finally, let $T \in \cm$ be an essential upper bound of $\ce$, then $T \cap A_i$ is an essential upper bound of $\ce_i$ for all $i \in I$. The decomposition and (ii) then shows that
+    \[
+    \mu(S \setminus T) = \sum_{i \in I}\mu((S \cap A_i) \setminus (T \cap A_i)) = \sum_{i \in I}\mu(S_i \setminus (T \cap A_i)) = 0
+    \]
+
+    therefore $S$ is an essential supremum of $\ce$. 
+\end{proof}
+
+
+\begin{lemma}
+\label{lemma:gluing-measurable-sets}
+    Let $(X, \cm, \cf, \mu)$ be a \hyperref[scaffolded]{definition:measure-scaffold} localisable measure space and $\bracs{(E_A, F_A)}_{A \in \cf}$ be pairs of measurable sets such that:
+    \begin{enumerate}[label=(\alph*)]
+        \item For each $A \in \cf$, $E_A, F_A \in \cm$, $E_A, F_A \subset A$, and $E_A \cap F_A = \emptyset$. 
+        \item For each $A, B \in \cf$, $\mu((E_A \cap B) \Delta (E_B \cap A)) = 0$ and $\mu((F_A \cap B) \Delta (F_B \cap A)) = 0$.
+    \end{enumerate}
+
+    Let $E$ and $F$ be essential suprema of $\bracsn{E_A}_{A \in \cf}$ and $\bracsn{F_A}_{A \in \cf}$, respectively, then
+    \begin{enumerate}
+        \item For each $B \in \cf$, $\mu(E \cap F_B) = 0$. 
+        \item $\mu(E \cap F) = 0$.
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    (1): Let $A, B \in \cf$, then
+    \begin{align*}
+        \mu(E_A \setminus F_B^c) &= \mu(E_A \cap F_B) = \mu(E_A \cap F_B \cap A \cap B) \\
+        &\le \mu(E_A \cap F_A) + \mu((F_A \cap B) \Delta (F_B \cap A)) = 0
+    \end{align*}
+
+    so $F_B^c$ is an essential upper bound of $\bracs{E_A}_{A \in \cf}$. Since $E$ is an essential supremum of $\bracs{E_A}_{A \in \cf}$, $\mu(E \setminus F_B^c) = \mu(E \cap F_B) = 0$. 
+    
+    (2): For any $B \in \cf$, $\mu(F_B \setminus E^c) = \mu(F_B \cap E) = \mu(E \cap F_B) =0$. Thus $E^c$ is an essential upper bound of $\bracs{F_B}_{B \in \cf}$. Given that $F$ is an essential supremum of $\bracsn{F_B}_{B \in \cf}$, $\mu(F \cap E) = \mu(F \setminus E^c) = 0$. 
+\end{proof}
+
+
+\begin{lemma}[Gluing Lemma for Measurable Functions]
+\label{lemma:gluing-measurable}
+    Let $(X, \cm, \cf, \mu)$ be a \hyperref[scaffolded]{definition:measure-scaffold} localisable measure space, $Y$ be a Polish space, and $\bracsn{f_A: A \to Y|A \in \cf}$ such that:
+    \begin{enumerate}[label=(\alph*)]
+        \item For each $A \in \cf$, $f_A \in \mathcal{L}^0(A; Y)$. 
+        \item For each $A, B \in \cf$, $f_A|_{A \cap B} = f_B|_{A \cap B}$ almost everywhere.  
+    \end{enumerate}
+    
+    then there exists $f: X \to Y$ such that:
+    \begin{enumerate}
+        \item $f \in \mathcal{L}^0(X; Y)$. 
+        \item For each $A \in \cf$, $f|_A = f_A$ almost everywhere. 
+        \item[(U)] For any $g: X \to Y$ satisfying (1) and (2), $f = g$ almost everywhere. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    First suppose that $Y$ is finite. For each $y \in Y$, let $P(y)$ be an essential supremum of $\bracs{f_A^{-1}(y)|A \in \cf}$. By \autoref{lemma:gluing-measurable-sets}, for any $x, y \in Y$ with $x \ne y$, $\mu(P(x) \cap P(y)) = 0$. After modification by null sets, assume without loss of generality that $X = \bigsqcup_{y \in Y}P(y)$. 
+
+    For each $x \in X$, let $f(x) \in Y$ be the unique element of $Y$ such that $x \in P(f(x))$, then:
+    \begin{enumerate}
+        \item For each $y \in Y$, $f^{-1}(y) = P(y)$, so $f \in \mathcal{L}^0(X; Y)$ is measurable. 
+        \item Let $A \in \cf$, then for each $y \in Y$, $\mu(f_A^{-1}(y) \setminus P(y)) = 0$. On the other hand, 
+        \begin{align*}
+            \mu\braks{(P(y) \cap A) \setminus f_A^{-1}(y)} &\le \sum_{z \in Y \setminus \bracs{y}}\mu(P(y) \cap f_A^{-1}(z) ) \\
+            &\le \sum_{z \in Y \setminus \bracs{y}}\mu(P(y) \cap P(z)) \\
+            &+ \sum_{z \in Y \setminus \bracs{y}}\mu(f_A^{-1}(z) \setminus P(z)) \\
+            &= 0
+        \end{align*}
+        
+        so $f|_A = f_A$ almost everywhere on $A$. 
+        \item[(U)] For all $A \in \cf$, $f|_A = g|_A$ almost everywhere. Since $\cf$ is a scaffold for $\mu$, $f = g$ almost everywhere. 
+    \end{enumerate}
+    
+    Therefore $f$ is the desired function. 
+
+    Now suppose that $Y$ is an arbitrary separable metrisable space. By \autoref{lemma:separable-metric-space-approx-identity}, there exists $\seq{I_n} \subset Y^Y$ such that:
+    \begin{enumerate}[label=(\roman*)]
+        \item $I_n \to \text{Id}$ pointwise as $n \to \infty$. 
+        \item For each $n \in \natp$, $I_n(Y)$ is finite and Borel measurable. 
+    \end{enumerate}
+
+    For each $n \in \natp$, let $f_{A, n} = I_n \circ f_A$, then 
+    \begin{enumerate}[label=(\alph*)]
+        \item For each $A \in \cf$, $f_{A, n} \in \mathcal{L}^0(A; Y)$. 
+        \item For each $A, B \in \cf$, $f_{A, n}|_{A \cap B} = f_{B, n}|_{A \cap B}$ almost everywhere. 
+        \item For each $A \in \cf$, $f_{A, n}(A) \subset I_n(Y)$. 
+    \end{enumerate}
+
+    By the finite case, there exists $f_n: X \to Y$ such that:
+    \begin{enumerate}
+        \item $f_n \in \mathcal{L}^0(X; Y)$. 
+        \item For each $A \in \cf$, $f_n|_A = f_{A, n}$ almost everywhere. 
+    \end{enumerate}
+    
+    Since $Y$ is Polish, \autoref{proposition:metric-measurable-limit} implies that $\bracsn{\limv{n}f_n \text{ exists}} \in \cm$. For each $A \in \cf$, $f_{A, n} \to f_A$ pointwise by (i). For each$n \in \natp$, $f_{A, n} = f_n|_A$ almost everywhere on $A$ by (2). Thus
+    \begin{align*}
+        \bracs{\limv{n}f_n \text{ exists}} \cap A &\supset \bigcap_{n \in \natp}\bracs{f_{A, n} = f_n|_A} \\ 
+        \mu\paren{\bracs{\limv{n}f_n \text{ exists}} \cap A} &= \mu\paren{\bigcap_{n \in \natp}\bracs{f_{A, n} = f_n|_{A}}} = \mu(A) \\
+        \mu\paren{\bracs{\limv{n}f_n \text{ does not exist}} \cap A} &= 0
+    \end{align*}
+
+    As $\cf$ is a scaffold for $\mu$, $\mu\bracsn{\limv{n}f_n \text{ does not exist}} = 0$. By (1) and \autoref{proposition:metric-measurable-limit}, there exists $f \in \mathcal{L}^0(X; Y)$ such that $f = \limv{n}f_n$ almost everywhere. In which case, 
+    \begin{enumerate}
+        \item $f \in \mathcal{L}^0(X; Y)$. 
+        \item For each $A \in \cf$, $f|_A = \limv{n}f_n|_A = \limv{n}f_{A, n} = f_A$ almost everywhere. 
+        \item[(U)] For all $A \in \cf$, $f|_A = g|_A$ almost everywhere. Since $\cf$ is a scaffold for $\mu$, $f = g$ almost everywhere. 
+    \end{enumerate}
+\end{proof}
+
+\begin{corollary}
+\label{corollary:l-infty-dedekind-complete}
+    Let $(X, \cm, \mu)$ be a localisable measure space, then $L^\infty(X; \real)$ is order complete. 
+\end{corollary}
+\begin{proof}
+    Let $\seqi{f} \subset L^\infty(X; \real)$ and $M \in \real$ such that $f_i \le M$ almost everywhere for all $i \in I$. 
+
+    Fix $A \in \cm$ with $\mu(A) < \infty$, and let
+    \[
+    \mathcal{S}_A = \bracs{g \in L^\infty(A; \real)| f_i|_A \le g \text{ almost everywhere }\forall i \in I}
+    \]
+
+    then since $f_i \le M$ almost everywhere for all $i \in I$, $\mathcal{S}_A \ne \emptyset$, and $m_A = \inf_{g \in \mathcal{S}_A}\int g d\mu \in \real$.
+    
+    Let $\seq{g_{A, n}} \subset \mathcal{S}_A$ such that $\seq{g_{A, n}}$ is decreasing pointwise and $\limv{n}\int_A g_{A, n} d\mu \downto m_A$. Take $g_A = \limv{n}g_{A, n}$, then by the \hyperref[Dominated Convergence Theorem]{theorem:dct}, $\int g_A d\mu = m_A$. 
+
+    For each $i \in I$, since $g_{A, n} \ge f_i|_A$ almost everywhere for all $n \in \natp$, $g_A \ge f_i|_A$ almost everywhere as well. Thus $g_A \in \mathcal{S}_A$. For any $h \in \mathcal{S}_A$, $g_A \wedge h \in \mathcal{S}_A$ with
+    \[
+    m_A \le \int_A g_A \wedge h d\mu \le \int_A g_A d\mu = m_A 
+    \]
+
+    Thus $g_A \wedge h = g_A$ almost everywhere, so $g_A \le h$ almost everywhere, and $g_A$ is an essential supremum of $\bracsn{f_i|_A}_{i \in I}$. 
+
+    Now, let $A, B \in \cm$ with $\mu(A), \mu(B) < \infty$, then $\one_{A \cap B} g_B + \one_{A \setminus B}M \in \mathcal{S}_A$, and
+    \[
+    m_A \le \int_A g_A \wedge (\one_{A \cap B} g_B + \one_{A \setminus B}M)d\mu \le \int_A g_A d\mu = m_A
+    \]
+
+    Thus $g_A \wedge (\one_{A \cap B} g_B + \one_{A \setminus B}M) = g_A$ almost everywhere, so $g_A|_{A \cap B} \le g_B|_{A \cap B}$ almost everywhere. As the argument is symmetric, $g_A|_{A \cap B} = g_B|_{A \cap B}$ almost everywhere. 
+
+    By the \hyperref[gluing lemma for measurable functions]{lemma:gluing-measurable}, there exists a measurable function $g:X \to \real$ such that $g|_A = g_A$ for all $A \in \cm$ with $\mu(A) < \infty$. 
+
+    Let $h \in L^\infty(X; \real)$ with $h \ge f_i$ almost everywhere for all $i \in I$, then for any $A \in \cm$ with $\mu(A) < \infty$, 
+    \[
+    \mu(\bracs{h < g} \cap A) \le \mu(\bracs{h|_A < g_A} \cup \bracs{g|_A \ne g_A}) = 0
+    \]
+    
+    As $\mu$ is semifinite, $\mu(\bracs{h < g}) = 0$. Finally, since $g_A \le M$ almost everywhere for all $A \in \cm$ with $\mu(A) < \infty$, $g \le M$ almost everywhere. Therefore $g \in L^\infty(X; \real)$ is indeed the essential supremum of $\seqi{f}$. 
+\end{proof}

src/measure/measure/product.tex

@@ -127,7 +127,7 @@
 
     Therefore $\alg = \cm \otimes \cn$. 
 
-    Now let $F \subset L^+(X \times Y)$ be the set of functions satisfying (1), then $F \supset \Sigma^+(X, \cm)$ by linearity. Let $f \in L^+(X \times Y)$, then by \autoref{proposition:measurable-simple-separable}, there exists $\seq{\phi_n} \subset \Sigma^+(X \times Y)$ such that $\phi_n \upto f$. In which case, by the Monotone Convergence Theorem, (1) also holds for $f$. 
+    Now let $F \subset L^+(X \times Y)$ be the set of functions satisfying (1), then $F \supset \Sigma^+(X, \cm)$ by linearity. Let $f \in L^+(X \times Y)$, then by \autoref{corollary:measurable-simple-separable}, there exists $\seq{\phi_n} \subset \Sigma^+(X \times Y)$ such that $\phi_n \upto f$. In which case, by the Monotone Convergence Theorem, (1) also holds for $f$. 
 \end{proof}
 
 

src/measure/measure/regular.tex

@@ -28,7 +28,7 @@
 \label{theorem:sigma-compact-regular-measure}
     Let $X$ be a topological space and $\mu: \cb_X \to [0, \infty]$ be a Borel measure. If:
     \begin{enumerate}
-        \item[(a)] $X$ is a LCH space.
+        \item[(a)] $X$ is an LCH space.
         \item[(b)] Every open set of $X$ is $\sigma$-compact.
         \item[(c)] For any $K \subset X$ compact, $\mu(K) < \infty$.
     \end{enumerate}

src/measure/measure/scaffold.tex

@@ -0,0 +1,122 @@
+\section{Scaffolds}
+\label{section:scaffold}
+
+\begin{definition}[Scaffold*]
+\label{definition:measure-scaffold}
+    Let $(X, \cm, \mu)$ be a measure space and $\cf \subset \cm$, then $\cf$ is a \textbf{scaffold} for $\mu$ if:
+    \begin{enumerate}[label=(S\arabic*)]
+        \item For each $A \in \cf$, $\mu(A) < \infty$. 
+        \item For all $E \in \cm$, $\mu(E) = \sup\bracs{\mu(E \cap A)|A \in \cf}$. 
+        \item For any $A, B \in \cf$, $A \cup B \in \cf$. 
+    \end{enumerate}
+
+    and the quadruple $(X, \cm, \cf, \mu)$ is a \textbf{scaffolded measure space}. 
+
+    For any semifinite measure space $(X, \cm, \mu)$, $\cf = \bracs{A \in \cm|\mu(A) < \infty}$ is the \textbf{canonical scaffold} for $\mu$, and $(X, \cm, \mu)$ will be equipped with this scaffold unless specified otherwise. 
+\end{definition}
+
+\begin{example}
+    Let $X$ be an LCH space, $\mu$ be a semifinite Radon measure, and $\mathcal{K}$ be the collection of compact subsets of $X$, then $\mathcal{K}$ is a scaffold for $\mu$. 
+\end{example}
+% Omitted
+
+
+\begin{lemma}[Gluing Lemma for Measures]
+\label{lemma:gluing-measure}
+    Let $(X, \cm)$ be a measurable space, $\cf \subset \cm$, and $\bracsn{\mu_A}_{A \in \cf}$ such that:
+    \begin{enumerate}[label=(\alph*)]
+        \item For each $A \in \cf$, $\mu_A$ is a finite measure on $A$.
+        \item For each $A, B \in \cf$ and $E \in \cm$, $\mu_A(E \cap A \cap B) = \mu_B(E \cap A \cap B)$. 
+        \item For each $A, B \in \cf$, $A \cup B \in \cf$. 
+    \end{enumerate}
+
+    Let
+    \[
+    \mu: \cm \to [0, \infty] \quad E \mapsto \sup\bracsn{\mu_A(E \cap A)|A \in \cf}
+    \]
+
+    then:
+    \begin{enumerate}
+        \item $\mu$ is a measure.
+        \item $\cf$ is a \hyperref[scaffold]{definition:measure-scaffold} for $(X, \cm, \mu)$.
+        \item For each $E \in \cm$, $\mu(A \cap E) = \mu_A(A \cap E)$. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    (3): Let $E \in \cm$, then $\mu_A(A \cap E) \le \mu(A \cap E)$ by definition. On the other hand, for any $B \in \cf$, 
+    \[
+    \mu_B(A \cap B \cap E) = \mu_A(A \cap B \cap E) \le \mu_A(A \cap E)
+    \]
+
+    Since the above holds for all $B \in \cf$, $\mu(A \cap E) \le \mu_A(A \cap E)$. 
+
+    (1): Let $\seq{E_n} \subset \cm$ be pairwise disjoint, then for each $A \in \cm$, 
+    \[
+    \mu\paren{A \cap \bigsqcup_{n \in \natp}E_n} = \sum_{n \in \natp}\mu_A(A \cap E_n) \le \sum_{n \in \natp}\mu(E_n)
+    \]
+
+    so $\mu\paren{\bigsqcup_{n \in \natp}E_n} \le \sum_{n \in \natp}\mu(E_n)$. 
+
+    On the other hand, let $n \in \natp$, $\seqf{A_k} \subset \cf$, and $A = \bigcup_{k = 1}^n A_k \in \cf$, then
+    \begin{align*}
+        \sum_{k = 1}^n \mu(A_k \cap E_k) &= \sum_{k = 1}^n \mu_{A_k}(A_k \cap E_k)  \\
+        &\le \mu_A\paren{A \cap \bigsqcup_{k = 1}^n E_k} \\
+        &\le \mu\paren{\bigsqcup_{k \in \natp}E_k}
+    \end{align*}
+
+    As this holds for all choice of $\seq{A_k} \subset \cf$, 
+    \[
+    \sum_{k = 1}^n \mu(E_k) \le \mu\paren{\bigsqcup_{k \in \natp}E_k}
+    \]
+
+    and as the above holds for all $n \in \natp$, $\sum_{n \in \natp}\mu(E_n) \le \mu\paren{\bigsqcup_{n \in \natp}E_n}$. 
+    
+    (2): By definition of $\mu$. 
+\end{proof}
+
+\begin{corollary}
+\label{corollary:scaffolded-part}
+    Let $(X, \cm, \mu)$ be a measure space, $\cf \subset \bracs{A \in \cm|\mu(A) < \infty}$ be an ideal, and
+    \[
+    \mu_\cf: \cm \to [0, \infty] \quad E \mapsto \sup\bracs{\mu(A \cap E)|A \in \cf}
+    \]
+
+    then
+    \begin{enumerate}
+        \item $\mu_\cf$ is a measure on $(X, \cm)$. 
+        \item $\cf$ is a \hyperref[scaffold]{definition:measure-scaffold} for $\mu_\cf$/
+    \end{enumerate}
+    
+    and $\mu_\cf$ is the \textbf{$\cf$-scaffolded part} of $\mu$. 
+\end{corollary}
+\begin{proof}
+    For each $A \in \cf$ and $E \in \cm$, let $\mu_A(E) = \mu(E \cap A)$, then $\bracsn{\mu_A}_{A \in \cf}$ is a family of measures satisfying \autoref{lemma:gluing-measure}. Therefore $\mu_\cf$ as defined is a measure, and $\cf$ is a scaffold for $\mu_\cf$. 
+\end{proof}
+
+\begin{lemma}
+\label{lemma:scaffolded-ac}
+    Let $(X, \cm, \cf, \mu)$ be a \hyperref[scaffolded]{definition:measure-scaffold} measure space and $f \in L^+(X)$, then for any $E \in \cm$, 
+    \[
+    \int_E f d\mu = \sup_{A \in \cf}\int_{E \cap A} f d\mu
+    \]
+\end{lemma}
+\begin{proof}
+    For any $F \in \cm$, 
+    \[
+        \int_{E} \one_F d\mu = \mu(E \cap F) = \sup_{A \in \cf}\mu(A \cap E \cap F) = \sup_{A \in \cf}\int_{E \cap A} \one_F d\mu
+    \]
+
+    so by linearity, the above holds for all simple functions in $L^+(X)$. 
+
+    Now, for each simple function $\phi \in \Sigma^+(X)$ with $\phi \le f$, 
+    \[
+    \int_E \phi d\mu = \sup_{A \in \cf}\int_{E \cap A}\phi d\mu \le \sup_{A \in \cf} \int_{E \cap A} f d\mu
+    \]
+
+    As the above holds for all $\phi \in \Sigma^+(X)$ with $\phi \le f$, 
+    \[
+    \int_E f d\mu = \sup_{A \in \cf}\int_{E \cap A} f d\mu
+    \]
+    
+\end{proof}
+

src/measure/measure/semifinite.tex

@@ -59,3 +59,4 @@
     \]
 
 \end{proof}
+

src/measure/notation.tex

@@ -8,10 +8,13 @@
     $\sigma(\mathcal{E})$ & $\sigma$-algebra generated by $\mathcal{E}$. & \autoref{definition:generated-sigma-algebra} \\
     $\lambda(\mathcal{E})$ & $\lambda$-system generated by $\mathcal{E}$. & \autoref{definition:generated-lambda-system} \\
     $\sigma \otimes \tau$ & Product of ideals. & \autoref{definition:product-ideal} \\
+    $\text{epi}(f)$ & Epigraph of $f$. & \autoref{definition:epigraph} \\ 
     % ---- Measure Theory ----
     $\mathcal{B}_X$ & Borel $\sigma$-algebra on $X$. & \autoref{definition:borel-sigma-algebra} \\
     $\sigma(\{f_i \mid i \in I\})$ & $\sigma$-algebra generated by the maps $\{f_i\}$. & \autoref{definition:generated-sigma-algebra-function} \\
     $\bigotimes_{i \in I} \mathcal{M}_i$ & Product $\sigma$-algebra. & \autoref{definition:product-sigma-algebra} \\
+    $\mathcal{L}^0(X; Y)$ & Space of measurable functions from $X$ to $Y$. & \autoref{definition:measurable-function-space} \\
+    $L^0(X; Y)$ & Space of measurable functions from $X$ to $Y$, modulo almost everywhere equality. & \autoref{definition:measurable-function-space} \\
     $\chi_E = \mathbf{1}_E$ & Indicator function of $E$. & \autoref{definition:indicator-function} \\
     $\Sigma(X, \mathcal{M}; E)$ & Space of $E$-valued simple functions on $(X, \mathcal{M})$. & \autoref{definition:simple-function-standard-form} \\
     $\Sigma^+(X, \mathcal{M})$ & Space of non-negative simple functions. & \autoref{definition:simple-function-scalar} \\

src/measure/radon/c0.tex

@@ -20,14 +20,14 @@
 
 \begin{definition}[Radon Measure]
 \label{definition:radon-measure-extended}
-    Let $X$ be a LCH space and $\mu$ be a Borel signed/vector measure on $X$, then $\mu$ is \textbf{Radon} if $|\mu|$ is Radon. 
+    Let $X$ be an LCH space and $\mu$ be a Borel signed/vector measure on $X$, then $\mu$ is \textbf{Radon} if $|\mu|$ is Radon. 
 \end{definition}
 
 
 
 \begin{definition}[Space of Finite Radon Measures]
 \label{definition:space-radon-measures}
-    Let $X$ be a LCH space and $E$ be a normed vector space over $K \in \RC$, then $M_R(X; E)$ is the \textbf{space of finite Radon measures} on $X$, which forms a vector space over $K$. 
+    Let $X$ be an LCH space and $E$ be a normed vector space over $K \in \RC$, then $M_R(X; E)$ is the \textbf{space of finite Radon measures} on $X$, which forms a vector space over $K$. 
 \end{definition}
 \begin{proof}
     Let $\mu, \nu \in M_R(X; E)$, then for any $A \in \cb_X$, $|\mu + \nu|(A) \le |\mu|(A) + |\nu|(A)$. Let $\eps > 0$, then by outer regularity and \autoref{proposition:radon-measurable-description}, there exists $K \subset A$ compact and $U \in \cn^o(A)$ such that $(|\mu| + |\nu|)(A \setminus K), (|\mu| + |\nu|)(U \setminus A) < \eps$. Therefore $|\mu + \nu|$ is regular on all Borel sets, and hence Radon. 
@@ -132,7 +132,7 @@
 
     then $T$ is maps $C_0(X; E)$ continuously into a subspace of $C_0(X \times B; K)$. 
 
-    Let $I \in C_0(X; E)^*$, then by the \hyperref[Hahn-Banach theorem]{theorem:hahn-banach}, there exists $\ol{I} \in C_0(X \times B; K)^*$ such that $\ol I \circ T = I$. By \hyperref[Alaoglu's Theorem]{theorem:alaoglu}, $B$ is a compact Hausdorff space. Therefore $X \times B$ is a LCH space by \autoref{proposition:lch-product}. By the \hyperref[Riesz Representation Theorem]{theorem:riesz-radon-c0}, there exists $\mu \in M_R(X \times B; K)$ such that for any $f \in C_0(X \times B; K)$, 
+    Let $I \in C_0(X; E)^*$, then by the \hyperref[Hahn-Banach theorem]{theorem:hahn-banach}, there exists $\ol{I} \in C_0(X \times B; K)^*$ such that $\ol I \circ T = I$. By \hyperref[Alaoglu's Theorem]{theorem:alaoglu}, $B$ is a compact Hausdorff space. Therefore $X \times B$ is an LCH space by \autoref{proposition:lch-product}. By the \hyperref[Riesz Representation Theorem]{theorem:riesz-radon-c0}, there exists $\mu \in M_R(X \times B; K)$ such that for any $f \in C_0(X \times B; K)$, 
     \[
     \dpn{f, \ol I}{C_0(X \times B; K)} = \int_{X \times B} f d\mu
     \]

src/measure/radon/radon.tex

@@ -3,7 +3,7 @@
 
 \begin{definition}[Radon Measure]
 \label{definition:radon-measure}
-    Let $X$ be a LCH space and $\mu: \cb_X \to [0, \infty]$ be a Borel measure, then $\mu$ is a \textbf{Radon measure} if:
+    Let $X$ be an LCH space and $\mu: \cb_X \to [0, \infty]$ be a Borel measure, then $\mu$ is a \textbf{Radon measure} if:
     \begin{enumerate}
         \item[(R1)] For any $K \subset X$ compact, $\mu(K) < \infty$.
         \item[(R2)] $\mu$ is outer regular on all Borel sets.
@@ -13,7 +13,7 @@
 
 \begin{proposition}
 \label{proposition:radon-measure-cc}
-    Let $X$ be a LCH space and $\mu: \cb_X \to [0, \infty]$ be a Radon measure, then
+    Let $X$ be an LCH space and $\mu: \cb_X \to [0, \infty]$ be a Radon measure, then
     \begin{enumerate}
         \item For any $U \subset X$ open,
         \[
@@ -51,15 +51,15 @@
 
 \end{proof}
 
-\begin{proposition}[{{\cite[Proposition 7.5]{Folland}}}]
+\begin{proposition}
 \label{proposition:radon-regular-sigma-finite}
-    Let $X$ be a LCH space and $\mu: \cb_X \to [0, \infty]$ be a Radon measure, then
+    Let $X$ be an LCH space and $\mu: \cb_X \to [0, \infty]$ be a Radon measure, then
     \begin{enumerate}
         \item $\mu$ is inner regular on all its $\sigma$-finite sets.
         \item If $X$ is $\sigma$-compact or $\mu$ is $\sigma$-finite, then $\mu$ is regular.
     \end{enumerate}
 \end{proposition}
-\begin{proof}
+\begin{proof}[Proof, {{\cite[Proposition 7.5]{Folland}}}. ]
     (1): Let $E \in \cb_X$ with $\mu(E) < \infty$ and $\eps > 0$, then there exists
     \begin{itemize}
     \item $U \in \cn^o(E)$ with $\mu(U \setminus E) < \eps/2$.
@@ -85,15 +85,15 @@
 
 \end{proof}
 
-\begin{proposition}[{{\cite[Proposition 7.7]{Folland}}}]
+\begin{proposition}
 \label{proposition:radon-measurable-description}
-    Let $X$ be a LCH space, $\mu: \cb_X \to [0, \infty]$ be a $\sigma$-finite Radon measure, and $E \in \cb_X$, then
+    Let $X$ be an LCH space, $\mu: \cb_X \to [0, \infty]$ be a $\sigma$-finite Radon measure, and $E \in \cb_X$, then
     \begin{enumerate}
         \item For every $\eps > 0$, there exists $U \in \cn^o(E)$ and $F \subset E$ closed such that $\mu(U \setminus F) < \eps$.
         \item There exists a $F_\sigma$ set $A$ and a $G_\delta$ set $B$ such that $A \subset E \subset B$ and $\mu(B \setminus A) = 0$.
     \end{enumerate}
 \end{proposition}
-\begin{proof}
+\begin{proof}[Proof, {{\cite[Proposition 7.7]{Folland}}}. ]
     (1): Let $\seq{E_n} \subset \cb_X$ such that $\mu(E_n) < \infty$. By outer regularity, there exists $\seq{U_n}$ open such that $U_n \in \cn^o(E_n)$ and $\mu(U_n) < \mu(E_n) + \eps/2^n$ for all $n \in \natp$. In which case,
     \[
     \mu\paren{\bigcup_{n \in \natp}U_n \setminus E} \le \sum_{n \in \natp}\mu(U_n \setminus E_n) < \eps
@@ -111,7 +111,7 @@
 
 \begin{proposition}
 \label{proposition:finite-compact-regular}
-    Let $X$ be a LCH space and $\mu: \cb_X \to [0, \infty]$ be a Borel measure such that:
+    Let $X$ be an LCH space and $\mu: \cb_X \to [0, \infty]$ be a Borel measure such that:
     \begin{enumerate}
         \item[(a)] For any $U \subset X$ open, $U$ is $\sigma$-compact.
         \item[(b)] For any $K \subset X$ compact, $\mu(K) < \infty$.
@@ -151,7 +151,7 @@
 
 \begin{lemma}
 \label{lemma:radon-compact-project}
-    Let $X$ be a LCH space, $Y$ be a compact Hausdorff space, $\mu$ be a finite Radon measure on $X \times Y$, and $\nu$ be a measure on $X$. If for each $A \in \cb_X$, $\nu(A) \le \mu(A \times Y)$, then $\nu$ is also a Radon measure. 
+    Let $X$ be an LCH space, $Y$ be a compact Hausdorff space, $\mu$ be a finite Radon measure on $X \times Y$, and $\nu$ be a measure on $X$. If for each $A \in \cb_X$, $\nu(A) \le \mu(A \times Y)$, then $\nu$ is also a Radon measure. 
 \end{lemma}
 \begin{proof}
     Let $A \in \cb_X$ and $\eps > 0$. By outer regularity of $\mu$, there exists $U \in \cn_{X \times Y}(A \times Y)$ such that $\mu(U \setminus (A \times Y)) < \eps$. By the \hyperref[Tube Lemma]{lemma:tube-lemma}, there exists $V \in \cn_X(A)$ such that $V \times Y \subset U$. In which case, 
@@ -172,30 +172,32 @@
 
 \begin{proposition}
 \label{proposition:radon-cc-dense}
-    Let $X$ be a LCH space, $\mu: \cb_X \to [0, \infty]$ be a Radon measure, $E$ be a normed vector space, and $p \in [1, \infty)$, then $C_c(X; E)$ is dense in $L^p(X; E)$.
+    Let $X$ be an LCH space, $\mu: \cb_X \to [0, \infty]$ be a Radon measure, $E$ be a normed vector space, and $p \in [1, \infty)$, then $C_c(X; E)$ is dense in $L^p(X; E)$.
 \end{proposition}
 \begin{proof}
     By \autoref{proposition:lp-simple-dense}, $\Sigma(X, \cm; E) \cap L^p(X; E)$ is dense in $L^p(X; E)$. Using linearity, it is sufficient to approximate indicator functions of Borel sets with finite measure.
 
     Let $A \in \cb_X$ and $\eps > 0$. By \autoref{proposition:radon-regular-sigma-finite}, there exists $U \in \cn^o(A)$ and $K \subset A$ compact such that $\mu(U \setminus A), \mu(A \setminus K) < \eps/2$. By \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, there exists $f \in C_c(X; [0, 1])$ such that $f|_K = 1$ and $\supp{f} \subset U$. In which case, for any $y \in E$,
-    \[
+    \begin{align*}
-    \norm{x \cdot \one_A - x \cdot f}_{L^p(X; E)} \le \norm{x}_E \mu(\bracs{f \ne \one_A})^{1/p} \le \norm{x}_E\mu(U \setminus K)^{1/p} < \eps^{1/p}\norm{x}_E
+    \norm{x \cdot \one_A - x \cdot f}_{L^p(X; E)} &\le \norm{x}_E \mu(\bracs{f \ne \one_A})^{1/p} \\
-    \]
+    &\le \norm{x}_E\mu(U \setminus K)^{1/p} < \eps^{1/p}\norm{x}_E
+    \end{align*}
+    
 \end{proof}
 
 \begin{theorem}[Lusin]
 \label{theorem:lusin}
-    Let $X$ be a LCH space, $\mu$ be a Radon measure on $X$, $E$ be a normed vector space, and $f: X \to E$ be a measurable function with $\mu\bracs{f \ne 0} < \infty$, then for any $\eps > 0$,
+    Let $X$ be an LCH space, $\mu$ be a Radon measure on $X$, $E$ be a normed vector space, and $f: X \to E$ be a measurable function with $\mu\bracs{f \ne 0} < \infty$, then for any $\eps > 0$,
     \begin{enumerate}
         \item There exists $A \subset \bracs{f \ne 0}$ such that $f|_A$ is continuous and $\mu(\bracs{f \ne 0} \setminus A) < \eps$
         \item If $E = \complex$, then there exists $\phi \in C_c(X; E)$ such that $\mu\bracs{f \ne \phi} < \eps$.
         \item If $E = \complex$ and $f$ is bounded, then $\phi$ can be taken such that $\norm{\phi}_u \le \norm{f}_u$.
     \end{enumerate}
 \end{theorem}
-\begin{proof}[Proof {{\cite[Theorem 7.10]{Folland}}}. ]
+\begin{proof}[Proof, {{\cite[Theorem 7.10]{Folland}}}. ]
     First assume that $f$ is bounded.
 
-    (1, bounded): If $f$ is bounded, then $f \in L^1(X; E)$. By \autoref{proposition:radon-cc-dense}, there exists $\seq{\phi_n} \subset C_c(X)$ such that $\phi_n \to f$ in $L^1(\mu)$. Since $\phi_n \to f$ in $L^1(\mu)$, $\phi_n \to f$ in measure by \autoref{proposition:lp-in-measure}. By taking a subsequence using \ref{proposition:cauchy-in-measure-limit}, assume without loss of generality that $\phi_n \to f$ almost everywhere. 
+    (1, bounded): If $f$ is bounded, then $f \in L^1(X; E)$. By \autoref{proposition:radon-cc-dense}, there exists $\seq{\phi_n} \subset C_c(X)$ such that $\phi_n \to f$ in $L^1(\mu)$. Since $\phi_n \to f$ in $L^1(\mu)$, $\phi_n \to f$ in measure by \autoref{proposition:lp-in-measure}. By taking a subsequence using \autoref{theorem:cauchy-in-measure-limit}, assume without loss of generality that $\phi_n \to f$ almost everywhere. 
 
     By \hyperref[Egoroff's Theorem]{theorem:egoroff}, there exists $A \subset \bracs{f \ne 0}$ such that $\phi_n \to f$ uniformly and $\mu(\bracs{f \ne 0} \setminus A) < \eps/3$. By \autoref{proposition:radon-regular-sigma-finite}, there exists $K \subset A$ compact such that $\mu(A \setminus K) < \eps/3$. 
 
@@ -224,14 +226,14 @@
     (2, unbounded): By (1) applied to $g$, there exists $\phi \in C_c(X; \complex)$ such that $\mu\bracs{\phi \ne g} < \eps/2$, and $\mu(\bracs{\phi \ne f}) < \eps$.
 \end{proof}
 
-\begin{proposition}[Monotone Convergence Theorem for Lower Semicontinuous Functions]
+\begin{proposition}[Monotone Convergence Theorem (LSC)]
 \label{proposition:mct-radon}
-    Let $X$ be a LCH space, $\net{f}$ and $f: X \to [0, \infty]$ be non-negative lower semicontinuous functions such that $f_\alpha \upto f$, then for any Radon measure $\mu$ on $X$, 
+    Let $X$ be an LCH space, $\net{f}$ and $f: X \to [0, \infty]$ be non-negative lower semicontinuous functions such that $f_\alpha \upto f$, then for any Radon measure $\mu$ on $X$, 
     \[
     \int f d\mu = \sup_{\alpha \in A}\int f_\alpha d\mu
     \]
 \end{proposition}
-\begin{proof}[Proof {{\cite[Proposition 7.12]{Folland}}}. ]
+\begin{proof}[Proof, {{\cite[Proposition 7.12]{Folland}}}. ]
     Assume without loss of generality that $\int f d\mu < \infty$. By \autoref{proposition:semicontinuous-properties}, $f$ is Borel measurable, so $f \ge f_\alpha$ for all $\alpha \in A$ implies that
     \[
     \int f d\mu \ge \sup_{\alpha \in A}\int f_\alpha d\mu

src/measure/radon/riesz.tex

@@ -8,7 +8,7 @@
 
 \begin{proposition}
 \label{proposition:positive-linear-functional-cc-property}
-    Let $X$ be a LCH space and $I \in \hom(C_c(X; \real); \real)$ be a positive linear functional, then
+    Let $X$ be an LCH space and $I \in \hom(C_c(X; \real); \real)$ be a positive linear functional, then
     \begin{enumerate}
         \item For any $f, g \in C_c(X; \real)$ with $f \le g$, $\dpb{f, I}{C_c(X; \real)} \le \dpb{g, I}{C_c(X; \real)}$.
         \item For any $K \subset X$ compact, there exists $C_K \ge 0$ such that for all $f \in C_c(X; \real)$ with $\supp{f} \subset K$, $|{\dpb{f, I}{C_c(X; \real)}}| \le \norm{f}_u$.
@@ -28,7 +28,7 @@
 
 \begin{theorem}[Riesz Representation Theorem]
 \label{theorem:riesz-radon}
-    Let $(X, \topo)$ be a LCH space and $I \in \hom(C_c(X; \real); \real)$ be a positive linear functional, then there exists a Borel measure $\mu: \cb_X \to [0, \infty]$ such that:
+    Let $(X, \topo)$ be an LCH space and $I \in \hom(C_c(X; \real); \real)$ be a positive linear functional, then there exists a Borel measure $\mu: \cb_X \to [0, \infty]$ such that:
     \begin{enumerate}
         \item For any $U \subset X$ open, $\mu(U) = \sup_{f \prec U}\dpb{f, I}{C_c(X; \real)}$.
         \item For any $K \subset X$ compact, $\mu(K) = \inf_{f \in C_c(X; [0, 1]), f \ge \one_K}\dpb{f, I}{C_c(X; \real)}$.

src/measure/vector/ac.tex

@@ -38,6 +38,8 @@
 \begin{definition}[Uniformly Absolutely Continuous]
 \label{definition:u-ac}
     Let $(X, \cm)$ be a measurable space, $\mu$ be a positive measure on $(X, \cm)$, $E$ be a normed space over $K \in \RC$, and $\mathcal{U} \subset M(X, \cm; E)$ be a family of $E$-valued vector measures on $(X, \cm)$, then $\mathcal{U}$ is \textbf{uniformly absolutely continuous} with respect to $\mu$ if for every $\eps > 0$, there exists $\delta > 0$ such that for all $A \in \cm$ with $\mu(A) < \delta$, $\norm{\nu(A)}_E < \eps$ for all $\nu \in \mathcal{U}$. 
+
+    For any family $\cf$ of measurable functions, $\cf$ is \textbf{uniformly absolutely continuous} with respect to $\mu$ if the measures $\mathcal{U} = \bracs{fd\mu|f \in \cf}$ are uniformly absolutely continuous with respect to $\mu$. 
 \end{definition}
 
 \begin{theorem}[Vitali-Hahn-Saks]
@@ -59,7 +61,7 @@
 
     (1): Let $\eps > 0$. For each $N \in \natp$, let
     \[
-    A_N = \bracs{A \in \cm_0 \bigg | \sup_{n \ge N}\norm{\nu_n(A) - \nu_N(A)} \le \eps}
+    A_N = \bracs{A \in \cm_0 \bigg | \sup_{n \ge N}\norm{\nu_n(A) - \nu_N(A)}_E \le \eps}
     \]
 
     then since $\seq{\nu_n} \subset UC(\cm_0; E)$, $A_N \subset \cm_0$ is closed. By assumption (b), $\cm_0 = \bigcup_{N \in \natp}A_N$. By the \hyperref[Baire Category Theorem]{theorem:baire}, there exists $N \in \natp$, $A \in \cm_0$, and $\delta > 0$ such that for every $m, n \ge N$ and $B \in \cm_0$ with $\mu(A \Delta B) \le \delta$, $\norm{\nu_m(B) - \nu_n(B)}_E \le \eps$. 
@@ -73,13 +75,22 @@
     and $\seq{\mu_n}$ is uniformly absolutely continuous with respect to $\mu$. 
 
     (2): Let $\seq{A_n} \subset \cm$ and $A \in \cm$ with $A = \bigsqcup_{n \in \natp}A_n$, then for each $N \in \natp$, 
+    \begin{align*}
+        \norm{\nu(A) - \sum_{n = 1}^N \nu(A_n)}_E &= 
+        \norm{\nu\paren{A \setminus \bigsqcup_{n = 1}^N A_n}}_E \\
+        &=
+        \limv{k}\norm{\nu_k\paren{A \setminus \bigsqcup_{n = 1}^N A_n}}_E
+    \end{align*}
+    
+    Since $\seq{\nu_k}$ is uniformly absolutely continuous, 
     \[
-        \norm{\nu(A) - \sum_{n = 1}^N \nu(A_n)}_E = 
+    \lim_{N \to \infty}\sup_{k \in \natp}\norm{\nu_k\paren{A \setminus \bigsqcup_{n = 1}^N A_n}}_E = 0
-        \nu\paren{A \setminus \bigsqcup_{n = 1}^N A_n} =
-        \limv{k}\nu_k\paren{A \setminus \bigsqcup_{n = 1}^N A_n}
     \]
     
-    so $\nu$ is a vector measure. Absolute continuity is equivalent to uniform continuity as a mapping $\cm_n \to E$, and uniform absolute continuity is equivalent to uniform equicontinuity as mappings $\cm \to E$. By the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}, $\nu \in UC(\cm_0; E)$, so $\nu \ll \mu$. 
+    so $\nu$ is a vector measure. 
+    
+    
+    Absolute continuity is equivalent to uniform continuity as a mapping $\cm_0 \to E$, and uniform absolute continuity is equivalent to uniform equicontinuity as mappings $\cm_0 \to E$. By the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}, $\nu \in UC(\cm_0; E)$, so $\nu \ll \mu$. 
     
 \end{proof}
 

src/measure/vector/fin.tex

@@ -0,0 +1,162 @@
+\section{Spaces of Finite Measures}
+\label{section:space-finite-measure}
+
+
+\begin{definition}[Space of Finite Measures]
+\label{definition:vector-measure-finite-space}
+    Let $(X, \cm)$ be a measurable space, $E$ be a normed vector space over $K \in \RC$, and $M(X, \cm; E)$ be the set of all finite $E$-valued vector measures on $(X, \cm)$. For each $\mu \in M(X, \cm; E)$, let $\norm{\mu}_{\text{var}} = |\mu|(X)$, then:
+    \begin{enumerate}
+        \item $M(X, \cm; E)$ equipped with $\norm{\cdot}_{\text{var}}$ is a normed vector space over $K$.
+        \item If $E$ is a Banach space, then so is $M(X, \cm; E)$.
+    \end{enumerate}
+
+\end{definition}
+\begin{proof}
+    (1): For any $\mu, \nu \in M(X, \cm; E)$ and $\seqf{A_j} \subset \cm$ such that $X = \bigsqcup_{j = 1}^n A_j$, 
+    \[
+    |\mu|(X) + |\nu|(X) \ge \sum_{j = 1}^n \norm{\mu(A_j)}_E + \norm{\nu(A_j)}_E \ge \sum_{j = 1}^n \norm{(\mu + \nu)(A_j)}_E
+    \]
+
+    As this holds for all choices of $\seqf{A_j}$, $\norm{\mu + \nu}_{\text{var}} \le \norm{\mu}_{\text{var}} + \norm{\nu}_{\text{var}}$. 
+
+    (2): Let $\seq{\mu_n} \subset M(X, \cm; E)$ such that $\sum_{n \in \natp}\norm{\mu_n}_{\text{var}} < \infty$. For each $A \in \cm$, since $E$ is complete, let
+    \[
+    \mu(A) = \sum_{n = 1}^\infty \mu_n(A)
+    \]
+
+    then for any $\seq{A_k} \subset \cm$ and $A \in \cm$ with $A = \bigsqcup_{k \in \natp}A_k$, 
+    \[
+        \mu(A) = \sum_{n = 1}^\infty \mu_n(A) = \sum_{n = 1}^\infty \sum_{k = 1}^\infty \mu_n(A_k) = \sum_{k = 1}^\infty \mu(A_k)
+    \]
+
+    by \hyperref[Fubini's theorem]{theorem:fubini-tonelli}.
+\end{proof}
+
+
+While the space of bounded Borel functions on $(X, \cm)$ forms a subspace of the dual of $M(X, \cm; \complex)$, it may not be immediately clear that they are insufficient. Before moving on to an explicit description of this dual, it is beneficial to consider the following "example". 
+
+Let $X = [0, 1]$, equipped with its Borel $\sigma$-algebra, and $\mu$ be the Lebesgue measure on $X$, then for any $x \in [0, 1]$, $\mu$ is mutually singular with the delta mass at $x$. Therefore the closure of $\text{span}\bracs{\delta_x|x \in X}$ in $M_R(X, \cm; \complex)$ is a proper closed subspace of $M(X, \cm; \complex)$. By the \hyperref[Hahn-Banach Theorem]{proposition:hahn-banach-utility}, there exists $\phi \in M(X, \cm; \complex)^*$ such that $\dpn{\delta_x, \phi}{M(X, \cm; \complex)} = 0$ for all $x \in X$, but $\dpn{\mu, \phi}{M(X, \cm; \complex)} = 1$. If there exists a bounded Borel function $f: X \to \complex$ such that $\dpn{\nu, \phi}{M(X, \cm; \complex)} = \int f d\nu$ for all $\nu \in M(X, \cm; \complex)$, then $f(x) = \dpn{\delta_x, \phi}{M(X, \cm; \complex)} = 0$ for all $x \in [0, 1]$, which is impossible. Therefore $\phi$ cannot be represented as a bounded Borel function. 
+
+Aside from abusing the Hahn-Banach theorem, a more explicit construction of such a functional is via the atomic decomposition. However, it is not included here due to time constraints. 
+
+In any case, the above example shows that a linear functional on $M(X, \cm; \complex)$ may act as \textit{different} Borel functions on different families of measures that are mutually singular to each other. The following theorem will make this relation precise. 
+
+\begin{theorem}
+\label{theorem:hilbert-measures-dual}
+    Let $(X, \cm)$ be a measurable space, $H$ be a Hilbert space over $K \in \RC$, and $\mathscr{M} \subset M(X, \cm; H)$ be a closed subspace such that:
+    \begin{enumerate}
+        \item[(A)] For each $\mu \in \mathscr{M}$ and $\nu \in M(X, \cm; H)$ with $\nu \ll \mu$, $\nu \in \mathscr{M}$.
+    \end{enumerate}
+
+    Let $|\mathscr{M}| = \bracsn{|\mu|: \mu \in \mathscr{M}}$, then for any maximal family $\seqi{\mu} \subset |\mathscr{M}|$ of pairwise mutually singular measures, 
+    \begin{enumerate}
+        \item For each $f \in [l^1(I); L^1(\mu_i; H)]$, let
+        \[
+        \mu_f: \cm \to H \quad A \mapsto \sum_{i \in I} \int_A f_i d\mu_i
+        \]
+
+        then the mapping 
+        \[
+        [l^1(I); L^1(\mu_i; H)] \to \mathscr{M} \quad f \mapsto \mu_f
+        \]
+
+        is an isometric isomorphism. 
+        \item $\mathscr{M}^*$ is isometrically isomorphic to $[l^\infty(I); L^\infty(\mu_i; H)]$. 
+    \end{enumerate}
+    
+    In other words,
+    \begin{enumerate}[start=2]
+        \item There exists a decomposable measure space $(\Omega, \cn, \mu)$ such that $\mathscr{M} \iso L^1(\Omega; H)$ and $\mathscr{M}^* \iso L^\infty(\Omega; H)$. 
+    \end{enumerate}
+\end{theorem}
+\begin{proof}[Proof, {{\cite{StackRadonDual}}}. ]
+    (1): By (A), $\mu_f \in \mathscr{M}$ for all $f \in [l^1(I); L^1(\mu_i; H)]$. 
+    
+    On the other hand, let $\nu \in \mathscr{M}$. For each $i \in I$, let $\nu = \nu_a^{(i)} + \nu_s^{(i)}$ be the Lebesgue decomposition of $\nu$ with respect to $\mu_i$, then by the \hyperref[Radon-Nikodym theorem]{theorem:lebesgue-radon-nikodym}, there exists $f_i \in L^1(\mu_i; H)$ such that $\nu_a^{(i)}(dx) = f_i \mu_i(dx)$. 
+    
+    For each countable $J \subset I$, define
+    \[
+    \nu_J: \cm \to H \quad A \mapsto \sum_{j \in J}\int_A f_j d\mu_j
+    \]
+
+    Let $M = \sup\bracs{\norm{\nu_J}_{\text{var}}| J \subset I \text{ countable}}$, then there exists $J \subset I$ countable such that $\norm{\nu_J}_{\text{var}} = M$. For each $i \in I \setminus J$, 
+    \[
+    \norm{\nu_J}_{\text{var}} + \normn{\nu_a^{(i)}}_{\text{var}}
+    = \normn{\nu_{J} + \nu_a^{(i)}}_{\text{var}} = \normn{\nu_{J \cup \bracs{i}}}_{\text{var}} \le M
+    \]
+
+    By maximality of $M$, $\normn{\nu_a^{(i)}}_{\text{var}} = 0$, and by maximality of $\seqi{\mu}$, $\nu - \nu_J = 0$. Let $g \in [l^1(I); L^1(\mu_i; H)]$ be defined by $g_i = f_i$ for each $i \in I$, then $\nu = \nu_J = \mu_g$, and the mapping is surjective. 
+
+    (2): By \autoref{theorem:lp-sum-dual} and \autoref{theorem:lp-duality}, 
+    \[
+    [l^1(I); L^1(\mu_i; H)]^* \iso [l^\infty(I); L^1(\mu_i; H)^*] \iso [l^\infty(I); L^\infty(\mu_i; H)] 
+    \]
+\end{proof}
+
+Despite not covering the full dual space, the bounded Borel functions still form a sequentially weak-* closed subspace with a convenient description for sequential convergence. 
+
+\begin{proposition}
+\label{proposition:measure-l-infinity-dominated-convergence}
+    Let $(X, \cm)$ be a measurable space, $E$ be a normed vector space over $K \in \RC$, and $\mathscr{M} \subset M(X, \cm; E)$ be a closed subspace such that:
+    \begin{enumerate}
+        \item[(P)] For each $x \in X$, $\bracs{x} \in \cm$, and the delta mass $\delta_x$ is in $\mathscr{M}$.
+    \end{enumerate}
+
+    then for any sequence $\seq{f_n: X \to E^*}$ of bounded strongly measurable functions, the following are equivalent:
+    \begin{enumerate}
+        \item For each $\mu \in \mathscr{M}$, $\limv{n}\int f_n d\mu$ exists. 
+        \item There exists a bounded strongly measurable function $f: X \to E^*$ such that $f_n \to f$ pointwise and $\sup_{n \in \natp}\norm{f_n}_u < \infty$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1) $\Rightarrow$ (2): By (P), for each $x \in X$, $\limv{n}f_n(x)$ exists. By the \hyperref[Uniform Boundedness Principle]{theorem:uniform-boundedness},
+    \[
+    \sup_{n \in \natp}\norm{f_n}_u \le \sup_{n \in \natp}\norm{f_n}_{\mathscr{M}^*} < \infty
+    \]
+
+    (2) $\Rightarrow$ (1): By the \hyperref[Dominated Convergence Theorem]{theorem:dct-bochner-vector}. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:space-of-measures-extreme-points}
+    Let $X$ be an LCH space and $\cm \subset \overline{B_{M_R(X; \complex)}(0, 1)}$ be a compact convex set such that:
+    \begin{enumerate}[label=(\alph*)]
+        \item For each $\mu \in \cm$ and $A, B \in \cb_X$, let $\mu_A(B) = \mu(A \cap B)$, then $\mu_A \in \cm$.
+        \item For each $\mu \in \cm \setminus \bracs{0}$ and $t \in [0, 1/\norm{\mu}_{\text{var}}]$, $t\mu \in \cm$. 
+    \end{enumerate}
+
+    then for any $\mu \in \cm \setminus \bracs{0}$, the following are equivalent: 
+    \begin{enumerate}
+        \item $\norm{\mu}_{\text{var}} = 1$ and $\mu$ takes on exactly two distinct values. 
+        \item There exists $x \in X$ and $\lambda \in \partial B_\complex(0, 1)$ such that $\mu = \lambda \delta_x$. 
+        \item $\mu$ is an extreme point of $\cm$. 
+    \end{enumerate}
+
+    Moreover, $\cm$ is the closed convex hull of $(\bracsn{\lambda \delta_x|x \in X, \lambda \in \partial B_\complex(0, 1)} \cup\bracs{0}) \cap \cm$. 
+\end{proposition}
+\begin{proof}
+    (1) $\Rightarrow$ (2): Assume without loss of generality that $\mu$ is positive and $\mu(\cb_X) = \bracs{0, 1}$. By inner regularity on open sets, there exists at least one compact set $K \subset X$ such that $\mu(K) = 1$. 
+
+    Let $\mathcal{F} = \bracs{K \subset X|K \text{ compact}, \mu(K) = 1}$, then $\mathcal{F}$ is a $\pi$-system that does not contain $\emptyset$, and as such satisfies the finite intersection property. Thus $A = \bigcap_{K \in \mathcal{F}}K \ne \emptyset$. 
+
+    Let $U \in \cn_X(A)$ and $K \in \cf$, then $K \setminus U$ is compact. Since $K \setminus U \cap A = \emptyset$, $K \setminus U \not\in \cf$, and $\mu(K \setminus U) = 0$. Thus $\mu(U) = \mu(K \cap U) = 1$. As this holds for all $U \in \cn_X(A)$, $\mu(A) = 1$ by outer regularity. 
+
+    Finally, let $x \in A$ and $U \in \cn_X(x)$, then $A \setminus U \subsetneq A$, so $A \setminus U \not\in \cf$. As such, $A \subset A \cap \ol U$ for all $U \in \cn_X(x)$. Since $\bigcap_{U \in \cn_X(x)}\ol{U} = \bracs{x}$, $A = \bracs{x}$, and $\mu = \delta_x$. 
+    
+    (2) $\Rightarrow$ (3): Assume without loss of generality that $\mu = \delta_x$. 
+    
+    Let $\nu, \rho \in \cm$ and $t \in (0, 1)$ such that $\mu = (1 - t)\nu + t\rho$, then $1 = \mu(\bracs{x}) = (1 - t)\nu(\bracs{x}) + t\rho(\bracs{x})$. Since $\mu(\bracs{x}) = 1$ and $|\nu(\bracs{x})|, |\rho(\bracs{x})| \le 1$, $\nu(\bracs{x}) = \rho(\bracs{x}) = 1$. As $\norm{\nu}_{\text{var}}, \norm{\rho}_{\text{var}} \le 1$, $\nu = \rho = \delta_x = \mu$. Therefore $\mu$ is an extreme point of $\cm$. 
+
+    (3) $\Rightarrow$ (1): If $\norm{\mu}_{\text{var}} \in (0, 1)$, then $\mu$ is a convex combination of $0$ and $\mu/\norm{\mu}_{\text{var}}$, so $\norm{\mu}_{\text{var}}$ must be $1$. 
+    
+    Suppose that $\mu$ takes on at least three distinct values, then there exists $A \in \cb_X$ such that $|\mu|(A), |\mu|(X \setminus A) > 0$. For each $B \in \cb_X$, let $\nu(B) = \mu(B \cap A)$ and $\rho(B) = \mu(B \setminus A)$, then $\mu = \nu + \rho$, $\nu, \rho \ne 0$, $\nu \perp \rho$, and $\norm{\nu}_{\text{var}} + \norm{\rho}_{\text{var}} = \norm{\mu}_{\text{var}}$. In which case, 
+    \[
+    \mu = \frac{\norm{\nu}_{\text{var}}}{\norm{\mu}_{\text{var}}} \cdot \underbrace{\frac{\norm{\mu}_{\text{var}} \cdot \nu}{\norm{\nu}_{\text{var}}}}_{\in \cm} + \frac{\norm{\rho}_{\text{var}}}{\norm{\mu}_{\text{var}}} \cdot \underbrace{\frac{\norm{\mu}_{\text{var}} \cdot \rho}{\norm{\rho}_{\text{var}}}}_{ \in \cm}
+    \]
+
+    is a convex combination of $\mu$ in terms of two other elements of $\cm$. 
+
+    Finally, by the \hyperref[Krein-Milman Theorem]{theorem:krein-milman}, $\cm$ is the closed convex hull of $(\bracsn{\lambda \delta_x|x \in X, \lambda \in \partial B_\complex(0, 1)} \cup\bracs{0}) \cap \cm$. 
+\end{proof}
+
+

src/measure/vector/index.tex

@@ -7,4 +7,5 @@
 \input{./ac.tex}
 \input{./ms.tex}
 \input{./rn.tex}
+\input{./fin.tex}
 

src/measure/vector/rn-draft.tex

@@ -0,0 +1,99 @@
+\begin{theorem}[Lebesgue-Radon-Nikodym (Localisable)]
+\label{theorem:lebesgue-radon-nikodym-localisable}
+    Let $(X, \cm)$ be a measurable space, $\mu, \nu: \cm \to [0, \infty]$ be positive measures, and $\cf \subset \cm$ be an ideal such that:
+    \begin{enumerate}[label=(\alph*)]
+        \item $(X, \cm, \mu + \nu)$ and $(X, \cm, \mu)$ are localisable. 
+        \item $\cf$ is a \hyperref[scaffold]{definition:measure-scaffold} for both $\mu$ and $\nu$. 
+    \end{enumerate}
+    
+    then there exists a unique pair of positive measures $\nu_a, \nu_s: \cm \to [0, \infty]$ such that:
+    \begin{enumerate}
+        \item $\nu = \nu_a + \nu_s$. 
+        \item $\nu_a$ is absolutely continuous with respect to $\mu$. 
+        \item $\nu_s$ is mutually singular with $\mu$. 
+        \item $\cf$ is a scaffold for $\nu_a$ and $\nu_s$. 
+    \end{enumerate}
+
+    The pair $(\nu_a, \nu_s)$ is the \textbf{Lebesgue decomposition} of $\nu$ with respect to $\mu$. Moreover, there exists a unique $\frac{d\nu_a}{d\mu} \in L^+(X, \cm, \mu)$ such that for every $A \in \cm$, 
+    \[
+    \nu_a(A) = \int_{A} \frac{d\nu_a}{d\mu} d\mu
+    \]
+
+    If $\nu \ll \mu$, then $\nu_s = 0$ and $\nu(dx) = \frac{d\nu_a}{d\mu}\mu(dx) = \frac{d\nu_a}{d\mu}\mu(dx)$. In which case, the function $\frac{d\nu}{d\mu}$ is the \textbf{Radon-Nikodym derivative} of $\nu$ with respect to $\mu$.
+\end{theorem}
+\begin{proof}
+    For each $A \in \cf$ and $E \in \cm$, let
+    \[
+    \mu^A(E) = \mu(A \cap E) \quad \nu^A(E) = \nu(A \cap E)
+    \]
+
+    then $\mu^A$ and $\nu^A$ are finite measures on $A$. By the \hyperref[finite case]{theorem:lebesgue-radon-nikodym}, there exists an a.e. unique $f_A \in L^+(A, \mu)$ and $\nu_s^A: \cm \to [0, \infty]$ such that:
+    \begin{enumerate}
+        \item $d\nu^A = f_A\mu^A + \nu_s^A = f_A\mu + \nu_s^A$. 
+        \item $\nu_s^A$ is mutually singular with $\mu$. 
+    \end{enumerate}
+
+    The uniqueness given by the finite case implies that for any $A, B \in \cf$, $f_A|_{A \cap B} = f_B|_{A \cap B}$ $\mu$-almost everywhere. By the \hyperref[gluing lemma for measurable functions]{lemma:gluing-measurable}, there exists $f \in L^+(X, \mu)$ such that $f|_A = f_A$ $\mu$-almost everywhere for all $A \in \cf$. 
+
+    By the \hyperref[gluing lemma for measures]{lemma:gluing-measure}, 
+    \[
+    \nu_s: \cm \to [0, \infty] \quad E \mapsto \sup_{A \in \cf}\nu_s^A(E \cap A)
+    \]
+
+    is a measure. 
+
+    (4): By definition, $\cf$ is a scaffold for $\nu_s$. By \autoref{lemma:scaffolded-ac}, it is also a scaffold for $f d\mu$. 
+
+    (1): Let $E \in \cm$, then since $\cf$ is a scaffold for $\nu$, 
+    \begin{align*}
+    \nu(E) &= \sup_{A \in \cf}\nu(A \cap E) = \sup_{A \in \cf}\nu^A(A \cap E) \\
+    &= \sup_{A \in \cf}\braks{\int_{A \cap E}f_A d\mu + \nu_s^A(A \cap E)}
+    \end{align*}
+
+    By (S3), for any $A, B \in \cf$, $A \cup B \in \cf$, and
+    \begin{align*}
+        \int_{A \cap E}f_A d\mu \vee \int_{B \cap E}f_B d\mu &\le \int_{(A \cup B) \cap E}f_{A \cup B} d\mu \\
+        \nu_s^A(A \cap E) \vee \nu_s^B(B \cap E) &\le \nu_s^{A \cup B}((A \cup B) \cap E)
+    \end{align*}
+
+    Thus the sum and the supremum may be interchanged, so
+    \[
+    \nu(E) = \sup_{A \in \cf}\int_{A \cap E}f_A d\mu + \sup_{A \in \cf}\nu_s^A(A \cap E)
+    \]
+
+    Now, since $\cf$ is a scaffold for $f\mu$, 
+    \[
+    \sup_{A \in \cf}\int_{A \cap E}f_A d\mu = \int_A f d\mu
+    \]
+
+    By definition of $\nu_s$, 
+    \[
+    \sup_{A \in \cf}\nu_s^A(A \cap E) = \nu_s(E)
+    \]
+
+    Therefore $\nu(E) = \int_E f d\mu + \nu_s(E)$, so $\nu(dx) = f(x)\mu(dx) + \nu_s(dx)$. 
+
+    (2): $\nu_a(dx) = f(x)\mu(dx)$. 
+
+    (3): For each $A \in \cf$, $f_A d\mu \perp \nu_s^A$, so there exists $E_A, F_A \in \cm$ such that:
+    \begin{enumerate}[label=(\roman*)]
+        \item $A = E_A \sqcup F_A$. 
+        \item $\mu(F_A) = 0$ and $\nu_s^A(E_A) = 0$. 
+    \end{enumerate}
+
+    Let $F$ be an essential supremum of $\bracs{F_A}_{A \in \cf}$ with respect to $(\mu + \nu)$. By \autoref{lemma:gluing-measurable-sets}, for each $A \in \cf$,
+    \[
+    \mu(F \cap A) = \mu(F \cap E_A) \le (\mu + \nu)(F \cap E_A) = 0
+    \]
+
+    Since $\cf$ is a scaffold for $\mu$, $\mu(F) = 0$. On the other hand, for each $A \in \cf$, 
+    \begin{align*}
+    \nu_s(A \setminus F) &= \nu_s(E_A \setminus F) + \nu_s(F_A \setminus F) \\
+    &\le \nu_s(E_A) + (\mu + \nu)(F_A \setminus F) = 0
+    \end{align*}
+
+    By definition of $\nu_s$, $\nu(F^c) = 0$. Therefore $\mu \perp \nu_s$. 
+
+    (Uniqueness): By uniqueness of the finite case, the restriction of $(\nu_a, \nu_s)$ to each set in $\cf$ is unique. Since $\cf$ is a scaffold for both measures, they are uniquely determined by their restrictions to each set in $\cf$.  
+    
+\end{proof}

src/measure/vector/rn.tex

@@ -1,7 +1,7 @@
 \section{The Lebesgue-Radon-Nikodym Theorem}
 \label{section:lebesgue-radon-nikodym}
 
-\begin{theorem}[Lebesgue-Radon-Nikodym]
+\begin{theorem}[Lebesgue-Radon-Nikodym (Finite)]
 \label{theorem:lebesgue-radon-nikodym}
     Let $(X, \cm)$ be a measurable space, $\mu$ be a $\sigma$-finite positive measure on $(X, \cm)$, $H$ be a Hilbert space over $K \in \RC$, and $\nu: \cm \to H$ be a finite vector measure, then there exists a unique pair of finite vector measures $\nu_a, \nu_s: \cm \to H$ such that:
     \begin{enumerate}
@@ -15,7 +15,7 @@
     \nu_a(A) = \int_{A} \frac{d\nu_a}{d\mu} d\mu
     \]
 
-    If $\nu \ll \mu$, then $\nu_s = 0$ and $\nu(dx) = \frac{d\nu}{d\mu}\mu(dx)$. In which case, the function $\frac{d\nu}{d\mu}$ is the \textbf{Radon-Nikodym derivative} of $\nu$ with respect to $\mu$.
+    If $\nu \ll \mu$, then $\nu_s = 0$ and $\nu(dx) = \frac{d\nu_a}{d\mu}\mu(dx) = \frac{d\nu_a}{d\mu}\mu(dx)$. In which case, the function $\frac{d\nu}{d\mu}$ is the \textbf{Radon-Nikodym derivative} of $\nu$ with respect to $\mu$.
 \end{theorem}
 \begin{proof}[Proof, {{\cite[Exercise 6.18]{Folland}}}\footnote{I decided to abuse Hilbert spaces for this theorem because it is more fun, and because I will use the Riesz representation theorem twice.}. ]
     (Finite + Positive): First suppose that $\mu$ is finite and $\nu$ is positive. Let $\lambda = \mu + \nu$, then the mapping
@@ -96,3 +96,6 @@
     (Uniqueness): For any decomposition $\nu = \rho_a + \rho_s$ satisfying the above, then $\rho_a - \nu_a = \nu_s - \rho_s$ with $\rho_a - \nu_a \perp \nu_s - \rho_s$. Therefore $\rho_a = \nu_a$, $\rho_s = \nu_s$, and the decomposition is unique. 
 \end{proof} 
 
+
+
+

src/measure/vector/variation.tex

@@ -90,34 +90,24 @@
     Let $(X, \cm)$ be a measurable space and $\mu$ be a signed/vector measure, then $\mu$ is \textbf{finite} if $|\mu|$ is finite. 
 \end{definition}
 
-\begin{definition}[Space of Finite Measures]
+\begin{lemma}
-\label{definition:vector-measure-finite-space}
+\label{lemma:variation-bound}
-    Let $(X, \cm)$ be a measurable space, $E$ be a normed vector space over $K \in \RC$, and $M(X, \cm; E)$ be the set of all finite $E$-valued vector measures on $(X, \cm)$. For each $\mu \in M(X, \cm; E)$, let $\norm{\mu}_{\text{var}} = |\mu|(X)$, then:
+    Let $(X, \cm)$ be a measurable space, $K \in \RC$, $\mu: \cm \to K$ be a signed/complex measure, then:
     \begin{enumerate}
-        \item $M(X, \cm; E)$ equipped with $\norm{\cdot}_{\text{var}}$ is a normed vector space over $K$.
+        \item If $K = \real$, then $|\mu|(X) \le 2\sup_{A \in \cm}\norm{\mu(A)}_E$.
-        \item If $E$ is a Banach space, then so is $M(X, \cm; E)$.
+        \item If $K = \complex$, then $|\mu|(X) \le 4\sup_{A \in \cm}\norm{\mu(A)}_E$. 
     \end{enumerate}
-
+\end{lemma}
-    
-\end{definition}
 \begin{proof}
-    (1): For any $\mu, \nu \in M(X, \cm; E)$ and $\seqf{A_j} \subset \cm$ such that $X = \bigsqcup_{j = 1}^n A_j$, 
+    (1, scalar): Let $X = P \sqcup N$ with $P, N \in \cm$ be a \hyperref[Hahn decomposition]{theorem:hahn-decomposition} of $\mu$, then
     \[
-    |\mu|(X) + |\nu|(X) \ge \sum_{j = 1}^n \norm{\mu(A_j)}_E + \norm{\nu(A_j)}_E \ge \sum_{j = 1}^n \norm{(\mu + \nu)(A_j)}_E
+    |\mu|(X) = |\mu(X \cap P)| + |\mu(X \cap N)| \le 2\sup_{A \in \cm}\norm{\mu(A)}_E
     \]
 
-    As this holds for all choices of $\seqf{A_j}$, $\norm{\mu + \nu}_{\text{var}} \le \norm{\mu}_{\text{var}} + \norm{\nu}_{\text{var}}$. 
+    (2, scalar): By (1), 
-
-    (2): Let $\seq{\mu_n} \subset M(X, \cm; E)$ such that $\sum_{n \in \natp}\norm{\mu_n}_{\text{var}} < \infty$. For each $A \in \cm$, since $E$ is complete, let
     \[
-    \mu(A) = \sum_{n = 1}^\infty \mu_n(A)
+    |\mu|(X) \le |\text{Re}(\mu)|(X) + |\text{Im}(\mu)|(X) \le 4\sup_{A \in \cm}\norm{\mu(A)}_E
     \]
-
-    then for any $\seq{A_k} \subset \cm$ and $A \in \cm$ with $A = \bigsqcup_{k \in \natp}A_k$, 
-    \[
-        \mu(A) = \sum_{n = 1}^\infty \mu_n(A) = \sum_{n = 1}^\infty \sum_{k = 1}^\infty \mu_n(A_k) = \sum_{k = 1}^\infty \mu(A_k)
-    \]
-
-    by \hyperref[Fubini's theorem]{theorem:fubini-tonelli}.
 \end{proof}
 
+

src/op/banach/definitions.tex

@@ -29,9 +29,16 @@
 
 \begin{definition}[Unital Homomorphism]
 \label{definition:banach-algebra-unital-homomorphism}
-    Let $A, B$ be unital Banach algebras and $\phi: A \to B$ be a homomorphism, then $\phi$ is a \textbf{unital homomorphism} if $\phi(1) = 1$. 
+    Let $A, B$ be unital Banach algebras and $\phi: A \to B$ be a homomorphism, then $\phi$ is a \textbf{unital homomorphism} if $\phi(1_A) = 1_B$. 
 \end{definition}
 
+\begin{definition}[Representation]
+\label{definition:banach-algebra-representation}
+    Let $A$ be a Banach algebra, then a \textbf{representation} of $A$ is a pair $(E, \pi)$ where $E$ is a Banach space, and $\pi: A \to L(E; E)$ is a continuous homomorphism. 
+\end{definition}
+
+
+
 \begin{definition}[Unitisation]
 \label{definition:unitisation}
     Let $A$ be a Banach algebra over $\complex$, and $\tilde A = \complex \oplus A$ with

src/op/banach/invertible.tex

@@ -52,11 +52,20 @@
     (3): Since the inversion map is locally a power series, it is $C^\infty$ by \autoref{theorem:termwise-differentiation}. 
 \end{proof}
 
+\begin{corollary}
+\label{corollary:invertible-boundary-explode}
+    Let $A$ be a unital Banach algebra, $x \in A \setminus G(A)$, and $r > 0$, then $\normn{y^{-1}}_A > 1/r$ for all $y \in B(x, r) \cap G(A)$. 
+\end{corollary}
+\begin{proof}
+    Let $y \in B(x, r)$, then $B(y, \normn{y^{-1}}_A^{-1}) \subset G(A)$ by \autoref{proposition:banach-algebra-inverse}. Since $x \not\in G(A)$, $r > \norm{x - y}_A \ge \normn{y^{-1}}_A^{-1}$, and $1/r < \normn{y^{-1}}_A$. 
+\end{proof}
+
+
 \begin{proposition}
 \label{proposition:swap-invertible}
     Let $A$ be a unital Banach algebra and $x, y \in A$, then $1 - xy \in G(A)$ if and only if $1 - yx \in G(A)$. 
 \end{proposition}
-\begin{proof}[Proof, {{\cite[Proposition I.3.4]{Zhu}}}. ]
+\begin{proof}[Proof, {{\cite[Proposition 3.4]{Zhu}}}. ]
     If $1 - xy \in G(A)$, then
     \begin{align*}
         (1 - yx)[y(1 - xy)^{-1}x + 1] &= y(1 - xy)^{-1}x + 1 \\
@@ -74,5 +83,3 @@
     \end{align*}
     
 \end{proof}
-
-

src/op/banach/multiplicative.tex

@@ -37,7 +37,7 @@
     Let $A$ be a Banach algebra, then $\Omega(A)$ is the \textbf{space of multiplicative linear functionals}, and with respect to the weak-* topology, 
     \begin{enumerate}
         \item If $A$ is unital, then $\Omega(A)$ is a compact Hausdorff space. 
-        \item $\Omega(A) \cup \bracs{0}$ is a compact Hausdorff space, and $\Omega(A)$ is a LCH space.
+        \item $\Omega(A) \cup \bracs{0}$ is a compact Hausdorff space, and $\Omega(A)$ is an LCH space.
     \end{enumerate}
 \end{definition}
 \begin{proof}
@@ -69,7 +69,7 @@
         \item $\phi(1) = 1$ and $\phi(G(A)) \subset \complex \setminus \bracs{0}$. 
     \end{enumerate}
 \end{theorem}
-\begin{proof}[Proof, {{\cite[Theorem I.4.5]{Zhu}}}. ]
+\begin{proof}[Proof, {{\cite[Theorem 4.5]{Zhu}}}. ]
     (1) $\Rightarrow$ (2): \autoref{proposition:multiplicative-unit}. 
 
     (2) $\Rightarrow$ (1): Let $x \in A$ and $\lambda \in \complex$ with $|\lambda| > [x]_{sp}$, then $\lambda - x \in G(A)$ and $\phi(\lambda - x) \ne 0$. Therefore $\phi(x) \subset \ol{B(0, [x]_{sp})}$, and $\norm{\phi}_{A^*} = 1$. 

src/op/banach/spectrum.tex

@@ -77,7 +77,7 @@
     Let $x \in A$. By \autoref{proposition:spectrum-non-empty}, there exists $\lambda \in \sigma_A(x)$. Since $\lambda \cdot 1 - x$ is not invertible and every non-zero element of $A$ is invertible, $x = \lambda \cdot 1$. Therefore the mapping $\complex \to A$ defined by $\lambda \mapsto \lambda \cdot 1$ is an isometric isomorphism. 
 \end{proof}
 
-\begin{proposition}[Spectral Radius Formula]
+\begin{proposition}[Beurling's Spectral Radius Formula]
 \label{proposition:spectral-radius-hadamard}
     Let $A$ be a unital Banach algebra and $x \in A$, then $[\cdot]_{sp} = \limsup_{n \to \infty}\normn{x^n}_A^{1/n}$. 
 \end{proposition}
@@ -128,7 +128,7 @@
 
 \begin{proposition}
 \label{proposition:commutative-spectrum-gymnastics}
-    Let $A$ be a unital Banach algebra and $x, y \in A$ with $x = y$, then
+    Let $A$ be a commutative unital Banach algebra and $x, y \in A$ with $xy = yx$, then
     \begin{enumerate}
         \item $\sigma_A(x + y) \subset \sigma_A(x) + \sigma_A(y)$.
         \item $\sigma_A(xy) \subset \sigma_A(x)\sigma_A(y)$.
@@ -146,4 +146,29 @@
     The above holds for $x$ and $y$ with respect to $\sigma_A$. 
 \end{proof}
 
+\begin{proposition}
+\label{proposition:spectrum-subalgebra-gymnastics}
+    Let $A$ be a unital Banach algebra, $B \subset A$ be a closed subalgebra containing $1$, and $x \in B$, then:
+    \begin{enumerate}
+        \item $\sigma_A(x) \subset \sigma_B(x)$. 
+        \item $\partial \sigma_B(x) \subset \sigma_A(x)$. 
+        \item $\sigma_B(x)$ is the union of $\sigma_A(x)$ and some bounded components of $\complex \setminus \sigma_A(x)$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): $G(B) \subset G(A)$. 
+
+    (2): Let $\lambda \in \partial \sigma_B(x)$, then there exists $\seq{\lambda_n} \subset \complex \setminus \sigma_B(x)$ such that $\lambda_n - x \in G(B)$ for all $n \in \natp$, and $\lambda_n \to \lambda$ as $n \to \infty$. By \autoref{corollary:invertible-boundary-explode}, $\norm{(\lambda_n - x)^{-1}}_A \to \infty$ as $n \to \infty$. If $\lambda - x \in G(A)$, then $(\lambda_n - x)^{-1} \to (\lambda - x)^{-1}$ as $n \to \infty$. Thus $\norm{(\lambda - x)^{-1}}_A = \infty$, which is impossible. Therefore $\lambda - x \not\in G(A)$, and $\lambda \in \sigma_A(x)$. 
+\end{proof}
+
+\begin{theorem}[Runge]
+\label{theorem:spectrum-subalgebra-sufficiency}
+    Let $A$ be a unital Banach algebra, $x \in A$, $P \subset \complex \setminus \sigma_A(x)$ such that $P$ intersects every bounded component of $\complex \setminus \sigma_A(x)$, and $B \subset A$ be a closed algebra containing $1$, $x$, and $\bracsn{(\lambda - x)^{-1}|\lambda \in P}$, then $\sigma_A(x) = \sigma_B(x)$. 
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem 4.9]{MarcouxNotes}}}. ]
+    By construction, $P \subset \complex \setminus \sigma_B(x)$. In addition, for any polynomial $p \in \complex[z]$, $p(x) \in B$. Thus for every rational function $f \in \complex(z) \cap H(\complex_\infty \setminus (P \cup \bracs{\infty}); \complex)$, $f(x) \in B$. 
+
+    By \hyperref[Runge's Theorem]{theorem:runge}, $H(\complex_\infty \setminus (P \cup \bracs{\infty}); \complex)$ is dense in $H(\sigma_A(x); \complex)$. The continuity of the \hyperref[holomorphic functional calculus]{definition:holomorphic-functional-calculus} then implies that $f(x) \in B$ for all $f \in H(\sigma_A(x); \complex)$. In particular, $(\lambda - x)^{-1} \in B$ for all $\lambda \in \complex \setminus \sigma_A(x)$. Therefore $\sigma_B(x) \subset \sigma_A(x)$, and $\sigma_B(x) = \sigma_A(x)$ by \autoref{proposition:spectrum-subalgebra-gymnastics}.
+\end{proof}
+
 

src/op/c-star/cont.tex

@@ -0,0 +1,145 @@
+\section{The Continuous Functional Calculus}
+\label{section:continuous-functional-calculus}
+
+\begin{theorem}[Spectral Theorem for $C^*$-Algebras]
+\label{theorem:spectral-c-star}
+    Let $A$ be a unital $C^*$-algebra and $x \in A$ be normal, then the mapping
+    \[
+    \Phi: \Omega(A[x]) \to \sigma_A(x) \quad \Phi(\psi) = \Gamma_{A[x]}(x)(\psi) = \psi(x)
+    \]
+
+    is a homeomorphism. 
+\end{theorem}
+\begin{proof}
+    Firstly, $A[x]$ is commutative by \autoref{proposition:generated-subalgebra-dense}. Thus \autoref{corollary:c-star-algebra-preserve-spectrum} and (3) of \autoref{proposition:gelfand-transform-gymnastics} imply that
+    \[
+    \Phi(\Omega(A[x])) = \Gamma_{A[x]}(\Omega(A[x])) = \sigma_{A[x]}(x) = \sigma_A(x)
+    \]
+
+    and $\Phi$ is a surjection onto $\sigma_A(x)$. 
+
+    On the other hand, the \hyperref[Gelfand-Naimark Theorem]{theorem:gelfand-naimark} implies that $\psi(x^*) = \ol{\psi(x)}$ for any $\psi \in \Omega(A[x])$. Since $A[x]$ is the smallest $C^*$-algebra containing $x$, any element $\psi \in \Omega(A[x])$ is uniquely determined by $\psi(x)$. Therefore $\Phi$ is injective. 
+
+    Finally, since $\Omega(A[x])$ is equipped with the weak* topology and $\Phi$ is the evaluation map at $x$, it is continuous. 
+
+    By \autoref{proposition:compact-hausdorff-homeomorphism}, $\Phi$ is a homeomorphism. 
+\end{proof}
+
+\begin{definition}[Continuous Functional Calculus]
+\label{definition:continuous-functional-calculus}
+    Let $A$ be a unital $C^*$-algebra and $x \in A$ be normal, then there exists a unique continuous unital *-homomorphism
+    \[
+    C(\sigma_A(x); \complex) \to A[x] \quad f \mapsto f(x)
+    \]
+
+    such that:
+    \begin{enumerate}
+        \item $\one(x) = 1_A$. 
+        \item $\text{Id}(x) = x$.
+        \item $\overline{\text{Id}}(x) = x^*$.
+    \end{enumerate}
+
+    Moreover,
+    \begin{enumerate}[start=3]
+        \item Under the identification $\Omega(A[x]) = \sigma_A(x)$, for each $f \in C(\sigma_A(x); \complex)$, $f(x) = \Gamma^{-1}_{A[x]}(f)$.
+        \item The mapping $f \mapsto f(x)$ is a unital *-isomorphism.  
+        \item For each $U \subset \complex$ open and $f \in C(U; \complex)$, the mapping
+        \[
+        \bracs{x \in A|\sigma_A(x) \subset U} \to A \quad x \mapsto f(x)
+        \]
+
+        is continuous. 
+        \item Let $B$ be a unital $C^*$-algebra and $\Phi: A \to B$ be a unital *-homomorphism, then for any $x \in A$ and $f \in C(\sigma_A(x); \complex)$, $\Phi(f(x)) = f(\Phi(x))$.
+    \end{enumerate}
+\end{definition}
+\begin{proof}
+    (1), (4), (5): By the \hyperref[Spectral Theorem]{theorem:spectral-c-star}, $\Omega(A[x])$ and $\sigma_A(x)$ may be identified. For each $f \in C(\sigma_A(x); \complex)$, define $f(x) = \Gamma^{-1}_{A[x]}(f)$, then the mapping $f \mapsto f(x)$ is a unital *-isomorphism by the \hyperref[Gelfand-Naimark Theorem]{theorem:gelfand-naimark}. 
+
+    (2): The identification 
+    \[
+    \Phi: \Omega(A[x]) \to \sigma_A(x) \quad \Phi(\psi) = \Gamma_{A[x]}(x)(\psi) = \psi(x)
+    \]
+
+    given by the Spectral Theorem implies that $\Gamma_{A[x]}(x) = \text{Id}$.
+
+    (3): The mapping $f \mapsto f(x)$ is a *-homomorphism. 
+
+    (6): Fix $x \in A$ and let $K \in \cn_U(\sigma_A(x); \complex)$ be compact. By \autoref{proposition:spectrum-continuous}, there exists $r > 0$ such that $\sigma_A(y) \subset K$ for all $y \in B_A(x, r)$. 
+
+    Let $\eps > 0$. By the \hyperref[Stone-Weierstrass Theorem]{theorem:complex-stone-weierstrass}, there exists $p \in \complex[z, \ol z]$ such that $\sup_{\lambda \in K}|p(\lambda) - f(\lambda)| < \eps$. In which case, 
+    \begin{align*}
+    \norm{f(x) - f(y)}_A &\le \norm{f(x) - p(x)}_A + \norm{f(y) - p(y)}_A + \norm{p(x) - p(y)}_A \\
+    &\le \norm{p(x) - p(y)}_A + 2\eps
+    \end{align*}
+
+    for all $y \in B_A(x, r)$. Since $p$ is a continuous map on $A$, there exists $\delta \in (0, r)$ such that $\norm{p(x) - p(y)}_A < \eps$ for all $y \in B_A(x, \delta)$. Therefore $\norm{f(x) - f(y)}_A < 3\eps$ for all $y \in B_A(x, \delta)$. 
+
+    (Uniqueness): By the \hyperref[Stone-Weierstrass Theorem]{theorem:complex-stone-weierstrass}.
+\end{proof}
+
+\begin{theorem}[Spectral Mapping Theorem (Continuous)]
+\label{theorem:spectral-mapping-continuous}
+    Let $A$ be a unital $C^*$-algebra and $x \in A$ be normal, then:
+    \begin{enumerate}
+        \item For every $f \in C(\sigma_A(x); \complex)$, $\sigma_A(f(x)) = f(\sigma_A(x))$. 
+        \item For every $f \in C(\sigma_A(x); \complex)$ and $g \in C(\sigma_A(f(x)); \complex)$, $(f \circ g)(x) = f(g(x))$.
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    (1): Since $f(x) \in A[x]$, by \autoref{proposition:gelfand-transform-gymnastics} and definition of the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, 
+    \[
+    \sigma_A(f(x)) = \sigma_{A[x]}(f(x)) = \Gamma_{A[x]}(f(x))(\sigma_A(x)) = f(\sigma_A(x))
+    \]
+
+    (2): Firstly, the theorem holds directly if $f, g \in \complex[z, \ol z]$ are polynomials in $z$ and $\ol z$. 
+
+    Suppose that $g \in \complex[z, \ol z]$ but $f$ is arbitrary. By the \hyperref[Stone-Weierstrass Theorem]{theorem:complex-stone-weierstrass}, there exists $\seq{p_n} \subset \complex[z, \ol z]$ such that $p_n \to f$ uniformly on $\sigma_A(x)$. By property (6) of the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, 
+    \[
+    (g \circ f)(x) = \limv{n}(g \circ f_n)(x) = \limv{n}g(f_n(x)) = \limv{n}g(f(x))
+    \]
+
+    Finally, suppose that both $f$ and $g$ are arbitary. By the \hyperref[Stone-Weierstrass Theorem]{theorem:complex-stone-weierstrass}, there exists $\seq{p_n} \subset \complex[z, \ol z]$ such that $p_n \to g$ uniformly on $\sigma_A(f(x))$. By continuity of the continuous functional calculus, 
+    \[
+    (g \circ f)(x) = \limv{n}(g_n \circ f)(x) = \limv{n}g_n(f(x)) = \limv{n}g(f(x))
+    \]
+\end{proof}
+
+
+
+\begin{corollary}
+\label{corollary:normal-spectrum-identity}
+    Let $A$ be a unital $C^*$-algebra and $x \in A$ be normal, then:
+    \begin{enumerate}
+        \item $x$ is self-adjoint if and only if $\sigma(x) \subset \real$. 
+        \item $x$ is unitary if and only if $\sigma(x) \subset \partial B_\complex(0, 1)$. 
+        \item $x$ is a projection if and only if $\sigma(x) \subset \bracs{0, 1}$. 
+    \end{enumerate}
+\end{corollary}
+
+\begin{corollary}
+\label{corollary:sum-of-unitaries}
+    Let $A$ be a unital $C^*$-algebra, then
+    \begin{enumerate}
+        \item For any $x \in A_{sa}$, there exists a unitary element $u \in A$ such that $x = (u + u^*)/(2\norm{x}_A)$. 
+        \item For any $x \in A$, there exists unitary elements $u, v \in A$ such that $x = (u + u^*)/(\norm{x}_A) + i(v + v^*)/(2\norm{x}_A)$. 
+    \end{enumerate}
+\end{corollary}
+\begin{proof}
+    (1): Assume without loss of generality that $\norm{x}_A \le 1$. In which case, $\sigma(x) \subset [-1, 1]$, and $f(\lambda) = \lambda + i\sqrt{1 - \lambda^2}$ is defined and continuous on $[-1, 1]$. Furthermore, $|f(\lambda)| = 1$ for all $\lambda \in [-1, 1]$. Thus \autoref{corollary:normal-spectrum-identity} implies that $f(x)$ is unitary. Finally, since $f + \ol f = 2\text{Id}$, $x = (f(x) + \ol{f(x)})/(2\norm{x}_A)$.
+\end{proof}
+
+\begin{corollary}
+\label{corollary:star-homomorphism-continuous}
+    Let $A, B$ be unital $C^*$-algebras and $\Phi: A \to B$ be an injective unital *-homomorphism, then for each $x \in A$,
+    \begin{enumerate}
+        \item $\sigma_B(\Phi(x)) = \sigma_A(x)$. 
+        \item $\norm{\Phi(x)}_B = \norm{x}_A$.
+    \end{enumerate}
+\end{corollary}
+\begin{proof}[Proof, {{\cite[10.7]{Zhu}}}. ]
+    (1): Since $\Phi(G(A)) \subset G(B)$, $\sigma_B(\Phi(x)) \subset \sigma_A(x)$. If $\sigma_B(\Phi(x)) \subsetneq \sigma_A(x)$, then \hyperref[Urysohn's Lemma]{lemma:urysohn} implies that there exists $C(\sigma_A(x); \complex)$ such that $f|_{\sigma_B(\Phi(x))} = 0$ but $f \ne 0$. In which case, by (7) of the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, $\Phi(f(x)) = f(\Phi(x)) = 0$, which contradicts the fact that $\Phi$ is injective. 
+
+    (2): By \autoref{corollary:c-star-unique-norm}, $\Phi$ is isometric. 
+\end{proof}
+
+

src/op/c-star/gelfand.tex

@@ -0,0 +1,77 @@
+\section{The Gelfand-Naimark Theorem}
+\label{section:gelfand-naimark}
+
+\begin{theorem}[Gelfand-Naimark]
+\label{theorem:gelfand-naimark}
+    Let $A$ be a commutative unital $C^*$-algebra, then the Gelfand transform
+    \[
+    \Gamma_A: A \to C(\Omega(A); \complex) \quad \Gamma_A(x)(\phi) = \phi(x)
+    \]
+
+    is a unital $C^*$-isomorphism. 
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem 9.4]{Zhu}}}. ]
+    By construction $\Gamma_A$ is a unital algebra homomorphism. 
+
+    To see that $\Gamma_A$ preserves involutions, let $y \in A$ be self-adjoint. By \autoref{proposition:gelfand-transform-gymnastics} and \autoref{proposition:self-adjoint-spectrum}, $\Gamma_A(y)(\Omega(A)) = \sigma_A(y) \subset \real$, so $\Gamma_A(y) \in C(\Omega(A); \real)$. For any $x \in A$, write $x = \text{Re}(x) + i\text{Im}(x)$, where $\text{Re}(x)$ and $\text{Im}(x)$ are both self-adjoint, then
+    \begin{align*}
+        \Gamma_A(x^*) &= \Gamma_A(\text{Re}(x) - i\text{Im}(x)) \\
+        &= \Gamma_A(\text{Re}(x)) - i\Gamma_A(\text{Im}(x)) \\
+        &= \overline{\Gamma_A(\text{Re}(x)) + i\Gamma_A(\text{Im}(x))} = \overline{\Gamma_A(x)}
+    \end{align*}
+
+    so $\Gamma_A(x^*) = \Gamma_A(x)^*$. 
+
+    Now, for each $x \in A$, \autoref{corollary:c-star-unique-norm} and \autoref{proposition:gelfand-transform-gymnastics} imply that
+    \begin{align*}
+        \norm{x}_A^2 &= \sup\bracs{|\lambda|\ | \lambda \in \sigma_A(x^*x)} \\
+        &= \sup\bracs{|\Gamma_A(x^*x)(\phi)|\ | \phi \in \Omega(A)} \\
+        &= \sup\bracs{|\Gamma_A(x)(\phi)|^2\ | \phi \in \Omega(A)} \\
+        \norm{x}_A &= \norm{\Gamma_A(x)}_u
+    \end{align*}
+
+    Thus $\Gamma_A$ is an isometry, and $\Gamma_A(A)$ is a closed subalgebra of $C(\Omega(A))$. 
+
+    Since $\Gamma_A(1_A) = 1$, $\Gamma_A(A)$ contains constants. As $\Gamma_A(A)$ separates points and is closed under complex conjugation, $\Gamma_A(A) = C(\Omega(A))$ by the \hyperref[Stone-Weierstrass Theorem]{theorem:complex-stone-weierstrass}.
+\end{proof}
+
+\begin{corollary}
+\label{corollary:gelfand-naimark-converse}
+    Let $A$ be a unital $C^*$-algebra, then the following are equivalent:
+    \begin{enumerate}
+        \item $A$ is commutative. 
+        \item $\Gamma_A$ is a *-isomorphism. 
+        \item $\Gamma_A$ is injective.
+    \end{enumerate}
+\end{corollary}
+
+\begin{corollary}
+\label{corollary:spectrum-characterisation-iff}
+    Let $A$ be a commutative unital $C^*$-algebra and $x \in A$ be normal, then:
+    \begin{enumerate}
+        \item $x$ is self-adjoint if and only if $\sigma_A(x) \subset \real$. 
+        \item $x$ is unitary if and only if $\sigma_A(x) \subset \partial B_\complex(0, 1)$.
+        \item $x$ is positive if and only if $\sigma_A(x) \subset [0, \infty)$. 
+        \item $x$ is a projection if and only if $\sigma_A(x) \subset \bracs{0,1}$. 
+    \end{enumerate}
+\end{corollary}
+% NEEDS WORK
+
+\begin{corollary}
+\label{corollary:stonean-commutative-algebra}
+    Let $A$ be a unital $C^*$-algebra, then $A_{sa}$ is order complete if and only if $\Omega(A)$ is extremely disconnected. 
+\end{corollary}
+\begin{proof}
+    By \autoref{theorem:gelfand-naimark}, $A$ and $C(\Omega(A); \complex)$ are isomorphic as $C^*$-algebras. In particular, $A_{sa}$ and $C(\Omega(A); \real)$ are isomorphic as ordered vector spaces, so $A_{sa}$ is order complete if and only if $C(\Omega(A); \real)$ is order complete. Thus the \hyperref[Stone-Nakano Theorem]{theorem:stone-nakano-extremely-disconnected} implies that $A_{sa}$ is order complete if and only if $\Omega(A)$ is extremely disconnected.
+\end{proof}
+
+\begin{corollary}
+\label{corollary:linfinity-extremely-disconnected}
+    Let $(X, \cm, \mu)$ be a localisable measure space, then $\Omega(L^\infty(X))$ is extremely disconnected. 
+\end{corollary}
+\begin{proof}
+    By \autoref{corollary:l-infty-dedekind-complete}, $L^\infty(X; \real)$ is order complete. By \autoref{corollary:stonean-commutative-algebra}, $\Omega(L^\infty(X))$ is extremely disconnected. 
+\end{proof}
+
+
+

src/op/c-star/gns.tex

@@ -0,0 +1,176 @@
+\section{The GNS Construction}
+\label{section:gns}
+
+\begin{definition}[Cyclic Representation]
+\label{definition:cyclic-representation}
+    Let $A$ be a $C^*$-algebra, $(H, \pi)$ be a representation of $A$, and $\xi \in H$, then $\xi$ is a \textbf{cyclic vector} for $(H, \pi)$ if $\bracsn{\pi(x)(\xi)|x \in A}$ is dense in $H$. The representation $(H, \pi)$ is \textbf{cyclic} if it admits a cyclic vector. 
+\end{definition}
+
+
+\begin{lemma}
+\label{lemma:cstar-state-kernel}
+    Let $A$ be a $C^*$-algebra, $\phi \in S(A)$, and
+    \[
+    N_\phi = \bracsn{x \in A| \dpn{x, x}{\phi} = \dpn{x^*x, \phi}{A} = 0}
+    \]
+
+    then:
+    \begin{enumerate}
+        \item For any $x, y \in A$ with $x \in N_\phi$ or $y \in N_\phi$, $\dpn{x, y}{A} = 0$. 
+        \item $N_\phi$ is a closed left ideal of $A$. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    (1): By the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz}, for any $x, y \in A$, 
+    \[
+    |\dpn{x, y}{\phi}|^2 \le \dpn{x, x}{\phi} \cdot \dpn{y, y}{\phi}
+    \]
+
+    If $x \in N_\phi$ or $y \in N_\phi$, then the above inequality shows that $\dpn{x, y}{\phi} = 0$. 
+
+    (2): As the zero set of a continuous function on $A$, $N_\phi$ is closed.
+
+    For any $x, y \in N_\phi$, 
+    \begin{align*}
+        \dpn{x + y, x + y}{\phi} &= \dpn{x, x}{\phi} + \dpn{x, y}{\phi} + \dpn{y, x}{\phi} + \dpn{y, y}{\phi} \\
+        &= \dpn{x, y}{\phi} + \dpn{y, x}{\phi} 
+    \end{align*}
+
+    By (1), $\dpn{x, y}{\phi} = \dpn{y, x}{\phi} = 0$. Therefore $x + y \in N_\phi$. 
+
+    Finally, for each $x \in N_\phi$ and $y \in A$, 
+    \[
+        \dpn{yx, yx}{\phi} = \dpn{x^*y^*yx, \phi}{A} = \dpn{x^*(y^*yx), \phi}{A} = \dpn{y^*yx, x}{\phi} =0 
+    \]
+
+    by (1). 
+\end{proof}
+
+
+\begin{definition}[GNS Triple]
+\label{definition:gns-triple}
+    Let $A$ be a unital $C^*$-algebra, $\phi \in S(A)$, and
+    \[
+    N_\phi = \bracsn{x \in A| \dpn{x, x}{\phi} = \dpn{x^*x, \phi}{A} = 0}
+    \]
+
+    Let $H_\phi^0 = A/N_\phi$, $H_\phi$ be its completion with respect to $\dpn{\cdot, \cdot}{\phi}$, and
+    \[
+    \pi_\phi^0: A \to B(H_\phi^0) \quad \pi_\phi^0(x)(y + N_\phi) = xy + N_\phi
+    \]
+
+    For each $x \in A$, let $\pi_\phi(x)$ be the continuous extension of $\pi_\phi^0(x)$ to an element of $B(H_\phi)$, then:
+    \begin{enumerate}
+        \item $(H_\phi, \dpn{\cdot, \cdot}{\phi})$ is a Hibert space.
+        \item $(H_\phi, \pi_\phi)$ is a well-defined representation of $A$. 
+        \item $\xi_\phi = 1_A + N_\phi$ is a unit vector in $H_\phi$, and $\bracsn{\pi_\phi(x)\xi_\phi| x \in A}$ is dense in $H_\phi$. Moreover, for each $x, y \in A$,
+        \[
+        \dpn{x, y}{\phi} = \dpn{\pi_\phi(x)\xi_\phi, \pi_\phi(y)\xi_\phi}{H_\phi}
+        \]
+    \end{enumerate}
+    
+    The representation $(H_\phi, \pi_\phi)$ is the \textbf{cyclic representation of $A$ induced by $\phi$}, and the triple $(H_\phi, \pi_\phi, \xi_\phi)$ is the \textbf{Gelfand-Naimark-Segal (GNS) triple associated with $\phi$}. 
+\end{definition}
+\begin{proof}[Proof, {{\cite[Proposition 14.2]{Zhu}}}. ]
+    (2): Fix $x \in A$, then for each $y_1, y_2 \in A$ with $y_1 - y_2 \in N_\phi$, $x(y_1 - y_2) \in N_\phi$ by \autoref{lemma:cstar-state-kernel}, so $\pi_\phi^0(x)$ is well-defined on $A/N_\phi$. 
+
+    By rescaling, assume without loss of generality that $\norm{x}_A \le 1$. In which case, for each $y \in A$, 
+    \[
+        \dpn{y, y}{\phi} - \dpn{xy, xy}{\phi} = \dpn{y^*y, \phi}{A} - \dpn{y^*x^*xy, \phi}{A} = \dpn{y^*(1 - x^*x)y, \phi}{A}
+    \]
+
+    Since $\sigma_A(x^*x) \subset [0, 1]$, $\sigma_A(1 - x^*x) \subset [0, 1]$ and is positive by \autoref{corollary:spectrum-characterisation-iff}. Thus there exists $z \in A$ positive such that $(1 - x^*x) = z^*z$, so
+    \[
+        \dpn{y, y}{\phi} - \dpn{xy, xy}{\phi} = \dpn{y^*z^*zy, \phi}{A} = \dpn{zy, zy}{\phi} \ge 0
+    \]
+
+    and $\dpn{y, y}{\phi} \ge \dpn{xy, xy}{\phi}$. Therefore $\pi_\phi^0(x)$ extends continuously into an element of $B(H_\phi)$ by the \hyperref[linear extension theorem]{theorem:linear-extension-theorem-normed}. 
+    
+
+    Now, let $x, y, z \in A$, then
+    \[
+    \pi_\phi^0(x)[\pi_\phi^0(y)(z + N_\phi)] = \pi_\phi^0(x)(yz + N_\phi) = xyz + N_\phi = \pi_\phi^0(xy)(z + N_\phi)
+    \]
+
+    and by uniqueness of continuous extensions, $\pi_\phi(x)\pi_\phi(y) = \pi_\phi(xy)$, so $\pi_\phi$ is a homomorphism. 
+
+    Finally, 
+    \[
+        \dpn{\pi_\phi^0(x^*)y, z}{\phi} = \dpn{z^*x^*y, \phi}{A} = \dpn{y, xz}{\phi} = \dpn{y, \pi_\phi^0(x)z}{\phi}
+    \]
+
+    By uniqueness of continuous extensions, $\pi_\phi(x^*) = \pi_\phi(x)^*$. Therefore $\pi_\phi$ is a *-homomorphism, and $(H_\phi, \pi_\phi)$ is a representation of $A$. 
+
+    (3): Since $\phi$ is a state, $\dpn{1_A, 1_A}{\phi} = 1$, and $1_A$ is a unit vector. As $H_\phi$ is the completion of $A/N_\phi$ and $A/N_\phi = \bracsn{\pi_\phi(x)(1_A + N_\phi)| x \in A}$, $\bracsn{\pi_\phi(x)(1_A + N_\phi)| x \in A}$ is dense in $H_\phi$. 
+
+    For each $x, y \in A$, $\dpn{x, y}{\phi} = \dpn{\pi_\phi(x)(1_A + N_\phi), \pi_\phi(y)(1_A + N_\phi)}{H_\phi}$ by well-definedness of the inner product on $H_\phi$.
+\end{proof}
+
+\begin{theorem}[Gelfand-Naimark-Segal]
+\label{theorem:gns}
+    Let $A$ be a unital $C^*$-algebra, then:
+    \begin{enumerate}
+        \item For each $\phi \in S(A)$, there exists a triple $(H_\phi, \pi_\phi, \xi_\phi)$ where $(H_\phi, \pi_\phi)$ is a representation of $A$, $\xi_\phi$ is a cyclic unit vector of $(H_\phi, \pi_\phi)$, and
+        \[
+        \dpn{x, y}{\phi} = \dpn{\pi_\phi(x)\xi_\phi, \pi_\phi(y)\xi_\phi}{H_\phi}
+        \]
+        \item For each representation $(H, \pi)$ of $A$ with cyclic unit vector $\xi$, the mapping
+        \[
+        \phi: A \to \complex \quad x \mapsto \dpn{\pi(x)\xi, \xi}{H}
+        \]
+
+        is a state on $A$. Moreover, if $(H_\phi, \pi_\phi, \xi_\phi)$ is the GNS triple associated with $\phi$, then there exists a unitary equivalence $U: H \to H_\phi$ such that $U\xi = \xi_\phi$. 
+        \item For each $\mathcal{S} \subset S(A)$, the mapping
+        \[
+        \pi_{\mathcal{S}}: A \to B([l^2(\mathcal{S}); H_\phi]) \quad \pi_{\mathcal{S}}(x)(\eta)_\phi = \pi_{\phi}(x)(\eta_\phi)
+        \]
+
+        is a representation of $A$, which is injective if for every $x \in A$, there exists $\phi \in \mathcal{S}$ with $\dpn{x^*x, \phi}{A} \ne 0$. 
+
+        In particular, $A$ is isomorphic to a closed subalgebra of $B([l^2(P(A)); H_\phi])$. 
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    (1): By the \hyperref[GNS construction]{definition:gns-triple}. 
+
+    (2): For each $x \in A$, if $x$ is positive, then so is $\pi(x)$, so $\dpn{\pi(x)\xi, \xi}{H} \ge 0$. Since $\xi$ is a unit vector, $\dpn{\pi(1_A)\xi, \xi}{H} = \dpn{\xi, \xi}{H} = 1$, and $\phi$ is a state. 
+
+    Let $H^0 = \bracsn{\pi(x)\xi|x \in A}$ and $H_\phi^0 = \bracsn{\pi_\phi(x)\xi_\phi|x \in A}$. Define
+    \[
+    U: H^0 \to H_\phi^0 \quad \pi(x)\xi \mapsto \pi_\phi(x)\xi_\phi
+    \]
+
+    then for each $x, y \in A$ with $\pi(x - y)\xi = 0$,
+    \begin{align*}
+    0 &= \dpn{\pi(x - y)\xi, \pi(x - y)\xi}{H} = \dpn{(x - y)^*(x - y), \phi}{A} \\
+    &= \dpn{x - y, x- y}{\phi} = \dpn{\pi_\phi(x - y)\xi_\phi, \pi_\phi(x - y)\xi_\phi}{H_\phi}
+    \end{align*}
+    
+
+    and $\pi_\phi(x - y)\xi_\phi = 0$ as well. Thus $U$ is well-defined. Moreover, for each $x \in A$,
+    \[
+        \dpn{\pi(x)\xi, \pi(x)\xi}{H} = \dpn{x^*x, \phi}{A} = \dpn{x^*x, 1_A}{\phi} = \dpn{\pi_\phi(x)\xi_\phi, \pi_\phi(x) \xi_\phi}{H_\phi}
+    \]
+
+    so $U$ is an isometry. For each $x, y \in A$,
+    \begin{align*}
+    U(\pi(x)[\pi(y)\xi]) &= U(\pi(xy)\xi) = \pi_\phi(xy)\xi_\phi \\
+    &= \pi_\phi(x)[\pi_\phi(y)\xi_\phi] = \pi_\phi(x)[U(\pi(y)\xi)]
+    \end{align*}
+    
+
+    so $U$ \hyperref[extends continuously]{theorem:linear-extension-theorem-normed} to a unitary equivalence between $(H, \pi)$ and $(H_\phi, \pi_\phi)$, with $U(\xi) = \xi_\phi$. 
+
+    (3): Suppose that for each $x \in A$, there exists $\phi \in \mathcal{S}$ such that $\dpn{x^*x, \phi}{A} \ne 0$. In which case,
+    \[
+    0 \ne \dpn{x, x}{\phi} = \dpn{x^*x, \phi}{A} = \dpn{\pi_\phi(x)\xi_\phi, \pi_\phi(x)\xi_\phi}{H_\phi}
+    \]
+
+    so $\pi_\phi(x) \ne 0$, and $\pi_{\mathcal{S}}(x) \ne 0$ as well. 
+
+    By \autoref{corollary:cstar-positive-weakstar-dense}, for each $x \in A$, there exists $\phi \in P(A)$ with $\dpn{x^*x, \phi}{A} \ne 0$, so $\pi_{P(A)}$ is injective. By \autoref{theorem:continuity-of-homomorphism-c-star}, $\pi_{P(A)}(A)$ is closed in $B([l^2(P(A)); H_\phi])$. 
+    
+\end{proof}
+
+
+

src/op/c-star/homomorphism.tex

@@ -0,0 +1,77 @@
+\section{*-Homomorphisms}
+\label{section:c-star-homomorphism}
+
+\begin{definition}[*-Homomorphism]
+\label{definition:c-star-homomorphism}
+    Let $A, B$ be involutive algebras over $\complex$ and $\phi: A \to B$, then $\phi$ is a \textbf{*-homomorphism} if:
+    \begin{enumerate}[label=(SH\arabic*)]
+        \item For each $x, y \in A$ and $\lambda \in \complex$, $\phi(\lambda x + y) = \lambda \phi(x) + \phi(y)$. 
+        \item For each $x, y \in A$, $\phi(xy) = \phi(x)\phi(y)$. 
+        \item For each $x \in A$, $\phi(x^*) = \phi(x)^*$.
+    \end{enumerate}
+
+    If $A$ and $B$ are unital, then $\phi$ is \textbf{unital} if:
+    \begin{enumerate}
+        \item[(U)] $\phi(1_A) = 1_B$. 
+    \end{enumerate}
+\end{definition}
+
+
+\begin{proposition}
+\label{proposition:star-homomorphism-contractive}
+    Let $A, B$ be unital $C^*$-algebras and $\phi: A \to B$ be a unital *-homomorphism, then for each $x \in A$, 
+    \begin{enumerate}
+        \item $\sigma_B(\phi(x)) \subset \sigma_A(x)$. 
+        \item $\norm{\phi(x)}_B \le \norm{x}_A$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): Since $\phi$ is unital, $\phi(G(A)) \subset G(B)$, so $\sigma_B(\phi(x)) \subset \sigma_A(x)$. 
+    
+    (2): By (1) and \autoref{corollary:c-star-unique-norm}, 
+    \begin{align*}
+    \norm{\phi(x)}_B^2 &= \sup\bracsn{|\lambda|\ | \lambda \in \sigma_B(\phi(x^*x))} \\
+    &\ge \sup\bracsn{|\lambda|\ | \lambda \in \sigma_A(x^*x)} = \norm{x}_A^2
+    \end{align*}
+    
+\end{proof}
+
+\begin{theorem}
+\label{theorem:continuity-of-homomorphism-c-star}
+    Let $A, B$ be unital $C^*$-algebras and $\Phi: A \to B$ be a unital *-homomorphism, then $\Phi(A)$ is closed. 
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem 11.1]{Zhu}}}. ]
+    Let $y \in \ol{\Phi(A)} \cap B_{sa}$, then there exists $x \in A_{sa}$ such that $\norm{y - \Phi(x)}_B \le \norm{y}_B/2$. Let
+    \[
+    f: \complex \to \complex \quad z \mapsto \begin{cases}
+        z & |z| \le 2\norm{y}_F \\
+        2\norm{y}_F \cdot \sgn z = 2\norm{y}_F \cdot \frac{z}{|z|} & |z| \ge 2\norm{y}_F
+    \end{cases}
+    \]
+
+    then $f \in C(\complex; \complex)$. Since $\norm{\Phi(x)}_B \le \norm{y}_B + \norm{y - \Phi(x)}_B \le 2\norm{y}_B$, $\sigma_B(\Phi(x)) \subset \ol{B_\complex(0, 2\norm{y}_B)}$, and $f|_{\sigma_B(\Phi(x))}$ is the identity. Thus by the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, $\Phi(x) = f(\Phi(x)) = \Phi(f(x))$. By the \hyperref[Spectral Mapping Theorem]{theorem:spectral-mapping-continuous}, $\sigma_A(f(x)) = f(\sigma_A(x))$. By \autoref{theorem:c-star-normal-spectral-radius}, $\norm{f(x)}_A = [f(x)]_{sp} \le \norm{f}_u = 2\norm{y}_F$. 
+
+    The above setup implies that for every $y \in \ol{\Phi(A)} \cap B_{sa}$, there exists $z \in A_{sa}$ such that $\norm{y - \Phi(z)}_{B} \le \norm{y}_B/2$, and $\norm{z}_A \le 2\norm{y}_B$. By the \hyperref[method of successive approximations]{theorem:successive-approximation}, $\phi(A_{sa}) = \ol{\Phi(A)} \cap B_{sa}$. Therefore $\Phi(A) = \ol{\Phi(A)}$. 
+\end{proof}
+
+\begin{definition}[Representation of $C^*$-Algebra]
+\label{definition:representation-cstar-algebra}
+    Let $A$ be a $C^*$-algebra, then a \textbf{representation} of $A$ is a pair $(H, \pi)$, where $H$ is a Hilbert space, and $\pi: A \to B(H)$ is a *-homomorphism. 
+\end{definition}
+
+\begin{definition}[Unitary Equivalence]
+\label{definition:representation-unitary-equivalent}
+    Let $A$ be a $C^*$-algebra and $(H_1, \pi_1), (H_2, \pi_2)$ be representations of $A$, then $(H_1, \pi_1)$ and $(H_2, \pi_2)$ are \textbf{unitarily equivalent} if there exists an isometry $U \in L(H_1; H_2)$ such that the following diagram commutes
+    \[
+    \xymatrix{
+    H_1 \ar@{->}[r]^{U} \ar@{->}[d]_{\pi_1(x)} & H_2 \ar@{->}[d]^{\pi_2(x)} \\
+    H_1 & H_2 \ar@{->}[l]^{U^*}
+    }
+    \]
+
+    for all $x \in A$. 
+\end{definition}
+
+
+
+

src/op/c-star/index.tex

@@ -2,5 +2,13 @@
 \label{chap:c-star-algebras}
 
 \input{./involution.tex}
+\input{./sub.tex}
+\input{./unitary.tex}
 \input{./sa.tex}
+\input{./homomorphism.tex}
+\input{./gelfand.tex}
+\input{./cont.tex}
 \input{./order.tex}
+\input{./positive.tex}
+\input{./state.tex}
+\input{./gns.tex}

src/op/c-star/involution.tex

@@ -41,17 +41,4 @@
     (3): For any $x \in A$, $(x^{-1})^*x^* = (x^{-1}x)^* = 1 = (xx^{-1})^* = x^*(x^{-1})^*$.
 \end{proof}
 
-\begin{definition}[*-Homomorphism]
-\label{definition:star-homomorphism}
-    Let $A, B$ be $C^*$-algebras and $\phi: A \to B$, then $\phi$ is a \textbf{*-homomorphism} if:
-    \begin{enumerate}
-        \item $\phi$ is a homomorphism of Banach algebras.
-        \item For every $x \in A$, $\phi(x^*) = \phi(x)^*$. 
-    \end{enumerate}
-
-    If in addition, $\phi(1) = 1$, then $\phi$ is a \textbf{unital *-homomorphism}. 
-\end{definition}
-
-
-
 

src/op/c-star/order.tex

@@ -6,5 +6,82 @@
     Let $A$ be a $C^*$-algebra and $x \in A$, then $x$ is \textbf{positive} if there exists $y \in A$ such that $x = y^*y$. 
 \end{definition}
 
+\begin{proposition}
+\label{proposition:positive-spectrum}
+    Let $A$ be a unital $C^*$-algebra and $x \in A$ be normal, then $x$ is positive if and only if $\sigma_A(x) \subset [0, \infty)$. 
+\end{proposition}
+\begin{proof}
+    Using the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, $x$ is positive if and only if $\Gamma_{A[x]}(x) = \text{Id}$ is positive in $C(\sigma_A(x); \complex)$, if and only if $\sigma_A(x) = \Gamma_{A[x]}(x)(\Omega(A[x])) \subset [0, \infty)$ by \autoref{proposition:gelfand-transform-gymnastics}. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:positive-norm-inequality}
+    Let $A$ be a unital $C^*$-algebra and $x \in A_{sa}$, then the following are equivalent:
+    \begin{enumerate}
+        \item $x$ is positive.
+        \item $\sigma_A(x) \subset [0, \infty)$. 
+        \item There exists $\lambda \ge \norm{x}_A$ such that $\norm{\lambda - x}_A \le \lambda$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Lemma 11.3]{Zhu}}}. ]
+    (1) $\Leftrightarrow$ (2): \autoref{proposition:positive-spectrum}. 
+
+    (2) $\Leftrightarrow$ (3): By assumption, $\sigma_A(x) \subset \real$, so \autoref{theorem:c-star-normal-spectral-radius} implies that
+    \[
+    \norm{\lambda - x}_A = [\lambda - x]_{sp} = \sup\bracsn{\lambda - \mu|\mu\in \sigma_A(x)}
+    \]
+
+    which is bounded above by $\lambda$ if and only if $\sigma_A(x) \subset [0, \infty)$. 
+\end{proof}
+
+\begin{corollary}
+\label{corollary:positive-ordering}
+    Let $A$ be a unital $C^*$-algebra. For each $x, y \in A$, denote $x \ge y$ if $x - y$ is positive, then $(A, \le)$ is an ordered vector space. 
+\end{corollary}
+\begin{proof}
+    By definition, the ordering is reflexive, antisymmetric, translation-invariant, and invariant under scaling by positive constants. It remains to show that $\le$ is transitive, or equivalently, the sum of two positive elements is positive. 
+
+    Let $x, y \in A$ be positive, then $x + y$ is self-adjoint. Thus there exists $\lambda \ge \norm{x}_A$ and $\mu \ge \norm{y}_A$ such that $\norm{\lambda - x}_A \le \lambda$ and $\norm{\mu - y}_A \le \mu$, so $\norm{(\lambda + \mu) - (x + y)}_A \le \lambda + \mu$, and $x + y$ is positive by \autoref{proposition:positive-norm-inequality}. 
+\end{proof}
+
+
+\begin{definition}[Positive Square Root]
+\label{definition:positive-square-root}
+    Let $A$ be a $C^*$-algebra and $x \in A$ be positive, then there exists a unique positive element $y \in A$ such that $y^2 = x$. The element $y$ is the \textbf{positive square root} of $x$, denoted $\sqrt{x}$. 
+\end{definition}
+\begin{proof}
+    Since $x$ is positive, $\sigma_A(x) \subset [0, \infty)$ by \autoref{proposition:positive-norm-inequality}. Therefore the square root function $f(t) = \sqrt{t}$ is defined and continuous on $\sigma_A(x)$. Using the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, $f(x)$ is a positive element of $A$ such that $f(x)^2 = x$. 
+
+    Let $y \in A$ such that $y^2 = x$, then by the \hyperref[Spectral Mapping Theorem]{theorem:spectral-mapping-continuous}, $y = f(y^2) = f(x)$, so the square root is unique. 
+\end{proof}
+
+
+\begin{definition}[Absolute Value]
+\label{definition:absolute-value-c-star}
+    Let $A$ be a unital $C^*$-algebra and $x \in A$, then $|x| = \sqrt{x^*x}$ is the \textbf{absolute value} of $x$. 
+\end{definition}
+
+\begin{definition}[Positive and Negative Parts]
+\label{definition:positive-negative-cstar-algebra}
+    Let $A$ be a unital $C^*$-algebra and $x \in A$ be self-adjoint, then there exists unique positive elements $x^+, x^- \in A$ such that
+    \begin{enumerate}
+        \item $x = x^+ - x^-$.
+        \item $x^+x^- = x^-x^+ = 0$. 
+    \end{enumerate}
+    
+    The pair $(x^+, x^-)$ are the \textbf{positive and negative parts} of $x$.
+\end{definition}
+\begin{proof}
+    Since $x$ is self-adjoint, $\sigma_A(x) \subset \real$ by \autoref{proposition:self-adjoint-spectrum}. Using the continuous functional calculus, existence is given by the functions $f^+(\lambda) = \lambda \vee 0$ and $f^-(\lambda) = \lambda \wedge 0$ and \autoref{proposition:positive-norm-inequality}. 
+
+    On the other hand, for each $p \in \real[z]$ with $p(0) = 0$, (2) implies that $p(x) = p(x^+) + p(-x^-)$. By the \hyperref[Stone-Weierstrass Theorem]{theorem:stone-weierstrass}, $f(x) = f(x^+) + f(-x^-)$ for all $f \in C(\real; \real)$ with $f(0) = 0$. In particular, (1) then implies that $f^+(x) = f+(x^+) + f^+(-x^-) = f^+(x^+) = x^+$, and likewise $f^-(x) = x^-$. Therefore the decomposition is given uniquely by the continuous functional calculus. 
+\end{proof}
+
+\begin{remark}
+\label{remark:positive-negative-cstar-algebra}
+    The condition in the sign decomposition that $x^+x^- = x^-x^+ = 0$ is essential. Otherwise I may use silly decompositions like $0 = 1 - 1$. 
+\end{remark}
+
+
 
 

src/op/c-star/positive.tex

@@ -0,0 +1,41 @@
+\section{Positive Linear Functionals}
+\label{section:cstar-positive}
+
+\begin{definition}[Positive Linear Functional]
+\label{definition:cstar-positive-functional}
+    Let $A$ be a unital $C^*$-algebra and $\phi \in A^*$, then $\phi$ is \textbf{positive} if $\dpn{x, \phi}{A} \ge 0$ for all positive elements $x \in A$. 
+\end{definition}
+
+\begin{theorem}
+\label{theorem:cstar-positive-algebraic}
+    Let $A$ be a unital $C^*$-algebra and $\phi \in \hom(A; \complex)$, then the following are equivalent:
+    \begin{enumerate}
+        \item $\phi$ is a positive linear functional.
+        \item $\phi \in A^*$ with $\normn{\phi}_{A^*} = \dpn{1, \phi}{A}$. 
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    (1) $\Rightarrow$ (2): For any $x \in A_{sa}$ with $\norm{x}_A \le 1$, $\sigma_A(1 - x) \subset 1 - [-1, 1] = [0, 2]$ by \autoref{proposition:commutative-spectrum-gymnastics}. Thus $1 - x \ge 0$ by \autoref{proposition:positive-spectrum}, and $\dpn{x, \phi}{A} \le \dpn{1, \phi}{A}$. By \autoref{proposition:hermitian-functional-norm}, $\norm{\phi}_{A^*} \le \dpn{1, \phi}{A}$, so $\norm{\phi}_{A^*} = \dpn{1, \phi}{A}$. 
+
+    (2) $\Rightarrow$ (1): For each $x \in A_{sa}$, $\sigma_A(x) = \sigma_{A[x]}(x)$ by \autoref{corollary:c-star-algebra-preserve-spectrum}, so $x \ge 0$ in $A$ if and only if $x \ge 0$ in $A[x]$ by \autoref{proposition:positive-spectrum}. In addition, the \hyperref[Gelfand-Naimark Theorem]{theorem:gelfand-naimark} implies that $x \ge 0$ if and only if $\Gamma_{A[x]}(x) \ge 0$. Moreover, since $1 \in A[x]$, the norm of $\phi$ alongside (2) is preserved by restricting to $A[x]$. By considering each commutative subalgebra, assume without loss of generality that $A$ is commutative. 
+    
+    By the \hyperref[Riesz Representation Theorem]{theorem:riesz-radon-c0}, $\phi$ takes the form of a complex Radon measure $\mu$ on $\Omega(A)$, and $\norm{\phi}_{A^*} = \norm{\mu}_{\text{var}}$. For each Borel set $E \in \cb_{\Omega(A)}$, 
+    \begin{align*}
+        \norm{\mu}_{\text{var}} &= \int_{\Omega(A)} 1 d\mu = \mu(E) + \mu(\Omega(A) \setminus E) \\
+        &\le |\mu(E)| + |\mu(\Omega(A) \setminus E)| \le \norm{\mu}_{\text{var}}
+    \end{align*}
+
+    which is only possible if $\mu(E), \mu(\Omega(A) \setminus E) \ge 0$. As this holds for all $E \in \cb_{\Omega(A)}$, $\mu$ is positive, and $\phi$ then is a positive linear functional. 
+\end{proof}
+
+
+\begin{corollary}
+\label{corollary:positive-linear-functional-extension}
+    Let $A$ be a unital $C^*$-algebra, $B \subset A$ be a closed subspace with $1_A \in B$, and $\phi \in B^*$ with $\norm{\phi}_{B^*} = \dpn{1_A, \phi}{B}$, then there exists a positive linear functional $\Phi \in A^*$ such that $\Phi|_B = A$. 
+\end{corollary}
+\begin{proof}
+    By the \hyperref[Hahn-Banach Theorem]{theorem:hahn-banach}, there exists $\Phi \in A^*$ such that $\Phi|_B = A$ and $\norm{\Phi}_{A^*} = \norm{\phi}_{B^*}$. In which case, $\norm{\Phi}_{A^*} = \dpn{1_A, \Phi}{A}$, and $\Phi$ is also positive by \autoref{theorem:cstar-positive-algebraic}. 
+\end{proof}
+
+
+

src/op/c-star/sa.tex

@@ -6,7 +6,7 @@
     Let $A$ be an involutive algebra over $\complex$ and $x \in A$, then $x$ is \textbf{self-adjoint} if $x = x^*$. The space $A_{sa} = \bracs{x \in A| x = x^*}$ is the \textbf{self-adjoint part} of $A$, and:
     \begin{enumerate}
         \item $A_{sa}$ is a $\real$ subspace of $A$. 
-        \item $A = \complex(A_{sa})$ as a vector space.
+        \item $A = \complex(A_{sa})$, with equivalent norms. 
         \item For each $x \in A$, let
         \[
         \text{Re}(x) = \frac{x + x^*}{2} \quad \text{Im}(x) = \frac{x - x^*}{2i}
@@ -31,6 +31,91 @@
     If the above holds, then $x$ is \textbf{normal}. 
 \end{definition}
 
+\begin{theorem}
+\label{theorem:c-star-normal-spectral-radius}
+    Let $A$ be a unital $C^*$-algebra and $x \in A$ be normal, then $\norm{x}_A = [x]_{sp}$. 
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem 8.1]{Zhu}}}. ]
+    First suppose that $x$ is self-adjoint. In this case, 
+    \begin{align*}
+        \normn{x^2}_A &= \normn{xx^*}_A = \norm{x}_A^2 \\
+        \normn{x^{2^n}}_A &= \norm{x}_A^{2^n}
+    \end{align*}
+
+    for all $n \in \natp$. Thus by the \hyperref[spectral radius formula]{proposition:spectral-radius-hadamard},
+    \[
+    [x]_{sp} = \limsup_{n \to \infty}\norm{x^{n}}_A^{1/n} \ge \limsup_{n \to \infty}\normn{x^{2^n}}_A^{1/2^n} = \norm{x}_A
+    \]
+
+    Now suppose that $x$ is only normal. Since $x$ and $x^*$ commute, $[xx^*]_{sp} \le [x]_{sp}[x^*]_{sp}$ by \autoref{proposition:commutative-spectrum-gymnastics}. Thus
+    \[
+    \norm{x}^2_A = \normn{xx^*}_A = [xx^*]_{sp} \le [x]_{sp}[x^*]_{sp} = [x]_{sp}^2
+    \]
+
+    by (5) of \autoref{proposition:c-star-algebra-gymnastics}. 
+\end{proof}
+
+\begin{corollary}
+\label{corollary:c-star-normal-spectral-radius-corollary}
+    Let $A$ be a unital $C^*$-algebra and $x \in A$ be normal, then:
+    \begin{enumerate}
+        \item There exists $\lambda \in \sigma_A(x)$ such that $|\lambda| = \norm{x}_A$. 
+        \item If there exists $n \in \natp$ such that $x^n = 0$, then $x = 0$ as well. 
+    \end{enumerate}
+\end{corollary}
+\begin{proof}
+    (1): Since $\sigma_A(x)$ is compact, there exists $\lambda \in \sigma_A(x)$ such that $|\lambda| = [x]_{sp}$. By \autoref{theorem:c-star-normal-spectral-radius}, $|\lambda| = [x]_{sp} = \norm{x}_A$. 
+
+    (2): By the \hyperref[Spectral Mapping Theorem]{theorem:spectral-mapping-holomorphic}, 
+    \[
+    \bracs{\lambda^n| \lambda \in \sigma_A(x)} = \bracs{0}
+    \]
+
+    Thus $\sigma_A(x) = \bracs{0}$. By \autoref{theorem:c-star-normal-spectral-radius}, $\norm{x}_A = [x]_{sp} = 0$. 
+\end{proof}
+
+
+\begin{corollary}
+\label{corollary:c-star-unique-norm}
+    Let $A$ be a unital $C^*$-algebra over $\complex$, then for each $x \in A$, 
+    \[
+    \norm{x}_A^2 = \sup\bracs{|\lambda|\ | \lambda \in \sigma_A(x^*x)}
+    \]
+
+    In particular, there exists at most one norm on $A$ making it a $C^*$-algebra. 
+\end{corollary}
+\begin{proof}
+    Since $x^*x$ is self-adjoint, \autoref{theorem:c-star-normal-spectral-radius} implies that
+    \[
+    \norm{x}_A^2 = \norm{x^*x}_A = \sup\bracs{|\lambda|\ | \lambda \in \sigma_A(x^*x)}
+    \]
+
+    which depends only on the algebraic structure of $A$. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:self-adjoint-spectrum}
+    Let $A$ be a unital $C^*$-algebra and $x \in A$ be self-adjoint, then $\sigma_A(x) \subset \real$. 
+\end{proposition}
+\begin{proof}
+    Let
+    \[
+    y = \exp(ix) = \sum_{n = 0}^\infty \frac{i^nx^n}{n!}  
+    \]
+
+    then
+    \[
+    y^*= \sum_{n = 0}^\infty \frac{(-i)^n (x^*)^n}{n!} = \exp(-ix^*) 
+    \]
+
+    Since $x$ is normal, $y$ is also normal. By \autoref{proposition:functional-calculus-exp},,
+    \[
+    y^*y = \exp(-ix^* + ix) = \exp(-ix + ix) = 1
+    \]
+    
+    so $y$ is unitary. By \autoref{proposition:unitary-spectrum} and the \hyperref[Spectral Mapping Theorem]{theorem:spectral-mapping-holomorphic}, $\exp(i\sigma_A(x)) = \sigma_A(y) \subset \partial B_\complex(0, 1)$. Thus $i\sigma_A(x) \subset \bracs{\text{Re} = 0}$, and $\sigma_A(x) \subset \real$. 
+\end{proof}
+
 
 
 

src/op/c-star/state.tex

@@ -0,0 +1,173 @@
+\section{States}
+\label{section:cstar-states}
+
+
+\begin{definition}[State]
+\label{definition:cstar-state}
+    Let $A$ be a unital $C^*$-algebra and $\phi \in A^*$, then $\phi$ is a \textbf{state} if $\phi$ is positive and $\dpn{1, \phi}{A} = 1$. 
+
+    The set of states $S(A) \subset A^*$ of $A$ equipped with the weak* topology is the \textbf{state space} of $A$.
+\end{definition}
+
+\begin{definition}
+\label{definition:cstar-state-pseudo-inner-product}
+    Let $A$ be a unital $C^*$-algebra and $\phi \in A^*$ be a positive linear functional, then the mapping
+    \[
+    A \times A \to \complex \quad (x, y) \mapsto \dpn{x, y}{\phi} := \dpn{y^*x, \phi}{A}
+    \] 
+
+    is a pseudo inner product. In particular, for any $x, y \in A$, 
+    \[
+    |\dpn{y^*x, \phi}{A}|^2 = |\dpn{x, y}{\phi}|^2 \le \dpn{x, x}{\phi} \cdot \dpn{y, y}{\phi}
+    \]
+
+    The pairing $\dpn{\cdot, \cdot}{\phi}$ is the \textbf{pseudo inner product associated with $\phi$}. 
+\end{definition}
+\begin{proof}
+    By the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz}. 
+\end{proof}
+
+
+
+\begin{definition}[Pure State]
+\label{definition:pure-state}
+    Let $A$ be a unital $C^*$-algebra and $\phi \in S(A)$, then $\phi$ is a \textbf{pure state} if $\phi$ is an extreme point of $S(A)$. The set $P(A)$ is the collection of all pure states of $A$. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:state-space-compact-convex}
+    Let $A$ be a unital $C^*$-algebra, then $S(A)$ is a compact convex set, and $S(A)$ is the weak*-closed convex hull of $P(A)$. 
+\end{proposition}
+\begin{proof}
+    Since the evaluation map is weak* continuous and
+    \[
+    S(A) = \bracs{\phi \in A^*|\dpn{1, \phi}{A} = 1} \cap \bigcap_{\substack{x \in A \\ x \ge 0}}\bracs{\phi \in A^*|\dpn{x, \phi}{A} \ge 0}
+    \]
+
+    the state space is an intersection of convex and weak*-closed sets, so it is closed and convex. 
+
+    By \autoref{theorem:cstar-positive-algebraic}, $S(A) \subset \ol{B_{A^*}(0, 1)}$, which is weak* compact by the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu}. Therefore $S(A)$ is compact by \autoref{proposition:compact-extensions}. 
+
+    By the \hyperref[Krein-Milman Theorem]{theorem:krein-milman}, $S(A)$ is the weak*-closed convex hull of $P(A)$. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:multiplicative-pure-state}
+    Let $A$ be a unital $C^*$-algebra, then:
+    \begin{enumerate}
+        \item $\Omega(A) \subset P(A)$. 
+        \item If $A$ is commutative, then $\Omega(A) = P(A)$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): Let $\phi \in \Omega(A)$. By \autoref{proposition:multiplicative-unit}, $\norm{\phi}_{A^*} = \dpn{1, \phi}{A} = 1$. Thus $\phi$ is a state by \autoref{theorem:cstar-positive-algebraic}, and $\Omega(A) \subset S(A)$. 
+
+    Let $\psi, \rho \in S(A)$ and $t \in (0, 1)$ such that $\phi = (1 - t)\psi + t\rho$, then for each $x \in \ker(\phi)$, $x^*x \in \ker(\phi)$ as well. As $t \ne 0$, $x^*x \in \ker(\psi)$ and $x^*x \in \ker(\rho)$. By the \hyperref[Cauchy-Schwarz inequality]{definition:cstar-state-pseudo-inner-product}, 
+    \[
+    |\dpn{x, \psi}{A}|^2 = |\dpn{1^*x, \psi}{A}|^2 \le \dpn{1, \psi}{A} \cdot \dpn{x^*x, \psi}{A} = 0
+    \]
+
+    Likewise, $\dpn{x, \rho}{A} = 0$ as well. Hence $\ker(\psi), \ker(\rho) \supset \ker(\phi)$. Thus there exist scalars $\alpha, \beta \in \complex$ such that $\phi = \alpha \psi = \beta \rho$. However, since $\phi, \psi, \rho \in S(A)$, $\alpha = \beta = 1$, and $\phi = \psi = \rho$. Therefore $\phi$ is a pure state. 
+
+    (2): Using the \hyperref[Gelfand-Naimark Theorem]{theorem:gelfand-naimark}, identify $A$ with $C(\Omega(A); \complex)$ and $S(A)$ as Radon probability measures on $\Omega(A)$. 
+    
+    Let $\cm = \bracs{t\mu|\mu \in S(A), t \in [0, 1]}$. By \autoref{proposition:space-of-measures-extreme-points}, the extreme points of $\cm$ are the delta masses $\bracs{\delta_x|x \in \Omega(A)}$, and possibly $0$. For any $\mu \in S(A)$, $\nu, \rho \in \cm$, and $t \in (0, 1)$, $\mu = (1 - t)\nu + t\rho$ implies that $\nu(\Omega(A)) = \rho(\Omega(A)) = 1$, and $\nu, \rho \in S(A)$ as well. Thus the extreme points of $S(A)$ are exactly the delta masses $\bracs{\delta_x|x \in \Omega(A)}$, which correspond to $\Omega(A)$ itself. 
+\end{proof}
+
+
+\begin{theorem}[Extension of States]
+\label{theorem:cstar-pure-state-extension}
+    Let $A$ be a unital $C^*$-algebra, $B \subset A$ be a $C^*$-subalgebra with $1_A \in B$, and $\phi \in S(B)$, then
+    \begin{enumerate}
+        \item There exists $\Phi \in S(A)$ such that $\Phi|_B = \phi$.
+        \item If $\phi \in P(B)$, then there exists $\Phi \in P(A)$ such that $\Phi|_B = \phi$. 
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    (1): By \autoref{theorem:cstar-positive-algebraic}, $\norm{\phi}_{B^*} = \dpn{1_A, \phi}{B}$. By the \hyperref[Hahn-Banach Theorem]{theorem:hahn-banach}, there exists $\Phi \in A^*$ such that $\Phi|_B = \phi$ and $\norm{\Phi}_{A^*} = \norm{\phi}_{B^*} = \dpn{1_A, \Phi}{A}$. Thus \autoref{theorem:cstar-positive-algebraic} implies that $\Phi \in S(A)$.
+
+    (2): Let $E(\phi) = \bracs{\Phi \in S(A)|\Phi|_B = \phi}$ be the collection of all extensions of $\phi$, then $E(\phi)$ is a weak*-closed convex subset of $S(A)$. By (1), $E(\phi)$ is non-empty, and as such admits an extreme point $\Phi$ by the \hyperref[Krein-Milman Theorem]{theorem:krein-milman}. 
+
+    Let $\psi, \rho \in S(A)$ and $t \in (0, 1)$ such that $\Phi = (1 - t)\psi + t\rho$. In which case, $\phi = (1 - t)\psi|_B + t\rho|_B$. Since $\phi \in P(B)$, $\phi = \psi|_B = \rho|_B$, so $\psi, \rho \in E(\phi)$. As $\Phi$ is an extreme point of $E(\phi)$, $\Phi = \psi = \rho$. Therefore $\Phi \in P(A)$. 
+\end{proof}
+
+
+\begin{corollary}
+\label{corollary:cstar-positive-property-probe}
+    Let $A$ be a unital $C^*$-algebra and $x \in A$ be normal, then\footnote{The crude bound seems kind of tragic, but it wouldn't be true otherwise. }
+    \begin{align*}
+    \sigma_A(x) &\subset \bracs{\dpn{x, \phi}{A}|\phi \in P(A)} \\
+    &\subset \bracs{\dpn{x, \phi}{A}|\phi \in S(A)} = \ol{\text{Conv}}(\sigma_A(x))
+    \end{align*}
+    
+
+    In particular, there exists $\phi \in P(A)$ such that $\norm{x}_A = |\dpn{x, \phi}{A}|$. 
+\end{corollary}
+\begin{proof}
+    Let $\lambda \in \sigma_A(x)$. By \autoref{proposition:gelfand-transform-gymnastics}, there exists $\phi \in \Omega(A[x])$ such that $\dpn{x, \phi}{A[x]} = \lambda$. By \autoref{proposition:multiplicative-pure-state}, $\phi \in P(A[x])$. The \hyperref[pure state extension theorem]{theorem:cstar-pure-state-extension} implies that there exists $\Phi \in P(A)$ such that $\Phi|_{A[x]} = \phi$. Thus $\Phi$ is a pure state with $\dpn{x, \Phi}{A} = \lambda$, and $ \sigma_A(x) \subset \bracs{\dpn{x, \Phi}{A}|\Phi \in P(A)}$. 
+
+    Let $\Phi \in S(A)$ and $\phi = \Phi|_{A[x]}$, then $\phi \in S(A[x])$ as well. By the \hyperref[Gelfand-Naimark Theorem]{theorem:gelfand-naimark}, the \hyperref[Spectral Theorem]{theorem:spectral-c-star}, and the \hyperref[Riesz Representation Theorem]{theorem:riesz-radon}, $\phi$ takes the form of a Radon probability measure $\mu$ on $\sigma_A(x)$. In which case,
+    \[
+    \dpn{x, \Phi}{A} = \dpn{x, \phi}{A[x]} = \int_{\sigma_A(x)}\lambda \mu(d\lambda) \in \ol{\text{Conv}}(\sigma_A(x))
+    \]
+
+    Finally, since $S(A)$ is compact and convex by \autoref{proposition:state-space-compact-convex}, 
+    \begin{align*}
+        \bracs{\dpn{x, \phi}{A}|\phi \in S(A)} &= \ol{\text{Conv}}(\bracs{\dpn{x, \phi}{A}|\phi \in P(A)}) \\
+        &\subset \ol{\text{Conv}}(\sigma_A(x))
+    \end{align*}
+
+    by \autoref{proposition:compact-extensions} and \autoref{proposition:closure-of-image}. 
+\end{proof}
+
+\begin{theorem}
+\label{theorem:cstar-state-existence}
+    Let $A$ be a unital $C^*$-algebra, $x \in A$, and $\lambda \in \sigma_A(x)$, then there exists $\phi \in S(A)$ such that $\dpn{x, \phi}{A} = \lambda$. 
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem 13.7]{Zhu}}}. ]
+    Let $B = \text{span}\bracs{x, 1}$. For each $\alpha x + \beta \in B$, let $\dpn{\alpha x + \beta, \phi_0}{B} = \alpha \lambda + \beta$. Since $\sigma_A(1) = \bracs{1}$, $\phi_0 \in B^*$ is a well-defined linear functional with $\dpn{x, \phi_0}{B} = \lambda$ and $\dpn{1, \phi_0}{B} = 1$. 
+
+    In addition, for each $\alpha x + \beta \in B$, $\alpha \lambda + \beta \in \sigma_A(\alpha x + \beta)$ by \autoref{proposition:commutative-spectrum-gymnastics}, and
+    \[
+    |\alpha \lambda + \beta| \le [\alpha x + \beta]_{sp} \le \norm{\alpha x + \beta}_A
+    \]
+
+    Thus $\norm{\phi_0}_{B^*} = \dpn{1, \phi_0}{B} = 1$. By the \hyperref[Hahn-Banach Theorem]{theorem:hahn-banach}, there exists $\phi \in A^*$ such that $\phi|_B = \phi_0$ and $\norm{\phi}_{A^*} = \norm{\phi_0}_{B^*}$. In which case, $\dpn{x, \phi}{A} = \lambda$ and $\dpn{1, \phi}{A} = \norm{\phi}_{A^*} = 1$. By \autoref{theorem:cstar-positive-algebraic}, $\phi$ is positive and hence a state. 
+\end{proof}
+
+\begin{corollary}
+\label{corollary:cstar-positive-weakstar-dense}
+    Let $A$ be a unital $C^*$-algebra, then:
+    \begin{enumerate}
+        \item For each $x \in A$, $x = 0$ if and only if $\dpn{x, \phi}{A} = 0$ for all $\phi \in P(A)$. 
+        \item The linear span of $P(A)$ is weak*-dense in $A^*$. 
+    \end{enumerate}
+
+    Moreover, for any $x \in A$, 
+    \begin{enumerate}[start=2]
+        \item $x$ is self-adjoint if and only if $\dpn{x, \phi}{A} \in \real$ for all $\phi \in P(A)$. 
+        \item $x$ is positive if and only if $\dpn{x, \phi}{A} \ge 0$ for all $\phi \in P(A)$. 
+    \end{enumerate}
+    
+\end{corollary}
+\begin{proof}[Proof, {{\cite[Theorem 13.9]{Zhu}}}. ]
+    (1): Let $x \in A$ such that $\dpn{x, \phi}{A} = 0$ for all $\phi \in P(A)$. First suppose that $x$ is self-adjoint. By \autoref{theorem:cstar-state-existence}, $\sigma_A(x) = \bracs{0}$, and $\norm{x}_A = [x]_{sp} = 0$ by \autoref{theorem:c-star-normal-spectral-radius}. 
+
+    Now suppose that $x$ is arbitrary. In this case, for each $\phi \in P(A)$, 
+    \[
+    0 = \text{Re}(\dpn{x, \phi}{A}) = \dpn{\text{Re}(x), \phi}{A}
+    \]
+
+    because $\phi$ is Hermitian. Similarly, $\dpn{\text{Im}(x), \phi}{A} = 0$ as well. Thus $\text{Re}(x) = \text{Im}(x) = 0$, and $x = 0$ as well. 
+
+    (2): Since the linear span of $P(A)$ separates points in $A$, it is weak*-dense in $A^*$ by \autoref{lemma:duality-dense}. 
+
+    (3): Let $\phi \in P(A)$, then $\phi$ is Hermitian. If $x$ is self-adjoint, then $\dpn{x, \phi}{A} \in \real$. 
+
+    On the other hand, if $\dpn{x, \phi}{A} \in \real$, then $\dpn{x, \phi}{A} = \dpn{x^*, \phi}{A}$, and $\dpn{x - x^*, \phi}{A} =0 $. If this holds for all $\phi \in P(A)$, then $x - x^* = 0$ by (1), and $x$ is self-adjoint. 
+
+    (4): Let $\phi \in P(A)$, then $\phi$ is positive. Thus if $x$ is positive, $\dpn{x, \phi}{A} \ge 0$. 
+
+    On the other hand, if $\dpn{x, \phi}{A} \ge 0$ for all $\phi \in P(A)$, then $x$ is self-adjoint by (3). By \autoref{corollary:cstar-positive-property-probe}, $\sigma_A(x) \subset [0, \infty)$. As such, $x$ is positive by \autoref{corollary:spectrum-characterisation-iff}. 
+\end{proof}
+

src/op/c-star/sub.tex

@@ -0,0 +1,41 @@
+\section{Subalgebras}
+\label{section:sub-c-star-algebras}
+
+\begin{definition}[Generated Subalgebra]
+\label{definition:generated-subalgebra}
+    Let $A$ be a unital $C^*$-algebra and $S \subset A$, then $A[S]$ is the smallest $C^*$-subalgebra of $A$ containing $1$ and $S$. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:generated-subalgebra-dense}
+    Let $A$ be a unital $C^*$-algebra, $S \subset A$, and $\mathcal{S} = S \cup \bracs{x^*|x \in S}$, then
+    \begin{enumerate}
+        \item The linear span
+        \[
+        \text{span}\bracs{\prod_{j = 1}^n x_j \bigg | \seqf{x_j} \subset \mathcal{S}}
+        \]
+
+        is dense in $A[S]$. 
+        \item If for any $x, y \in \mathcal{S}$, $xy = yx$, then $A[S]$ is commutative.
+        \item For any normal element $x \in A$, $A[x]$ is commutative. 
+    \end{enumerate}
+\end{proposition}
+% Obvious so omitted. 
+
+\begin{proposition}
+\label{proposition:c-star-algebra-preserve-gl}
+    Let $A$ be a unital $C^*$-algebra, $B \subset A$ be a $C^*$-subalgebra containing $1$, then $G(B) = G(A) \cap B$. 
+\end{proposition}
+\begin{proof}
+    Let $x \in G(A) \cap B$, then $x^*x \in G(A) \cap B$ as well. In particular, $0 \not\in \sigma_A(x^*x)$. Since $x^*x \in A_{sa}$, $\sigma_A(x^*x) \subset \real$ by \autoref{proposition:self-adjoint-spectrum}. By \autoref{proposition:spectrum-subalgebra-gymnastics}, $\partial \sigma_B(x^*x) \subset \sigma_A(x^*x) \subset \real$. Thus $\sigma_B(x^*x) \subset \real$ as well, which means that $\partial \sigma_B(x^*x) = \sigma_B(x^*x) = \sigma_A(x^*x)$. Therefore $0 \not\in \sigma_A(x^*x) = \sigma_B(x^*x)$, $x^*x \in G(B)$, and $x \in G(B)$. 
+\end{proof}
+
+\begin{corollary}
+\label{corollary:c-star-algebra-preserve-spectrum}
+    Let $A$ be a unital $C^*$-algebra, $B \subset A$ be a $C^*$-subalgebra containing $1$, and $x \in B$, then $\sigma_A(x) = \sigma_B(x)$. 
+\end{corollary}
+\begin{proof}
+    By \autoref{proposition:c-star-algebra-preserve-gl}. 
+\end{proof}
+
+

src/op/c-star/unitary.tex

@@ -0,0 +1,55 @@
+\section{Unitary Elements}
+\label{section:unitary-c-star}
+
+\begin{definition}[Unitary]
+\label{definition:unitary-element}
+    Let $A$ be a unital $C^*$-algebra and $x \in A$, then $x$ is \textbf{unitary} if $x \in G(A)$ and $x^* = x^{-1}$. 
+\end{definition}
+
+\begin{lemma}
+\label{lemma:unitary-unit}
+    Let $A$ be a unital $C^*$-algebra and $x \in A$ be unitary, then $\norm{x}_A = 1$. 
+\end{lemma}
+\begin{proof}
+    $\normn{x^2}_A = \norm{xx^*}_A = \norm{1}_A = 1$. 
+\end{proof}
+
+\begin{definition}[Unitarily Equivalent]
+\label{definition:unitary-equivalent}
+    Let $A$ be a unital $C^*$-algebra and $x, y \in A$, then $x$ and $y$ are \textbf{unitarily equivalent} if there exists a unitary element $u \in A$ such that $x = uyu^*$. 
+\end{definition}
+
+\begin{lemma}
+\label{lemma:unitary-equivalent-same-stuff}
+    Let $A$ be a unital $C^*$-algebra and $x, y \in A$ be unitarily equivalent, then:
+    \begin{enumerate}
+        \item $\norm{x}_A = \norm{y}_A$. 
+        \item $\sigma_A(x) = \sigma_A(y)$. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    (1): Let $u \in A$ be unitary such that $x = uyu^*$, then $\norm{x}_A \le \norm{u}_A\norm{x}_A\normn{u^*}_A$. By \autoref{lemma:unitary-unit} and (1) of \autoref{proposition:c-star-algebra-gymnastics}, $\norm{u}_A = \normn{u^*}_A = 1$, so $\norm{x}_A \le \norm{y}_A$. As the argument is symmetric, $\norm{x}_A = \norm{y}_A$. 
+
+    (2): Let $\lambda \in \complex$, then
+    \[
+    u(y - \lambda)u^* = uyu^* - \lambda uu^* = uyu^* - \lambda = x - \lambda
+    \]
+
+    Since $u, u^* \in G(A)$, $x - \lambda \in G(A)$ if and only if $y - \lambda \in G(A)$. Therefore $\sigma_A(x) = \sigma_A(y)$. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:unitary-spectrum}
+    Let $A$ be a unital $C^*$-algebra and $x \in A$ be unitary, then $\sigma_A(x) \subset \partial B_\complex(0, 1)$. 
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition 8.2]{Zhu}}}. ]
+    By \autoref{lemma:unitary-unit}, $\norm{x}_A = 1$, so $\sigma_A(x) \subset \ol{B_\complex(0, 1)}$. Thus
+    \[
+    \bracsn{\ol{\lambda}|\lambda \in \sigma_A(x)} = \sigma_A(x^*) = \sigma_A(x^{-1}) \subset \ol{\complex \setminus B_\complex(0, 1)}
+    \] 
+
+    by (4) of \autoref{proposition:c-star-algebra-gymnastics}. For any $\lambda \in \complex$, $\lambda \in \ol{B_\complex(0, 1)}$ and $\ol \lambda \in \ol{\complex \setminus B_\complex(0, 1)}$ if and only if $|\lambda| = 1$. Therefore $\sigma_A(x) \subset \partial B_\complex(0, 1)$. 
+\end{proof}
+
+
+

src/op/example/bc.tex

@@ -17,7 +17,7 @@
 
     is a homeomorphism. Under the identification $\beta X = \Omega(BC(X; \complex))$, $\Gamma_{BC(X; \complex)} = \beta$. 
 \end{theorem}
-\begin{proof}[Proof, {{\cite[Theorem I.6.4]{Zhu}}}. ]
+\begin{proof}[Proof, {{\cite[Theorem 6.4]{Zhu}}}. ]
     Let $\phi \in BC(X; \complex)^* \setminus \ol{E(X)}$, then there exists $\seqf{f_k} \subset BC(X; \complex)$ and $\eps > 0$ such that for every $x \in X$, 
     \[
     f(x) = \sum_{k = 1}^n |f_k(x) - \dpn{f_k, \phi}{BC(X; \complex)}|^2 \ge \eps^2

src/op/example/c0.tex

@@ -3,12 +3,12 @@
 
 \begin{definition}[$C_0(X)$]
 \label{definition:vanishing-infinity-algebra}
-    Let $X$ be a LCH space, then $C_0(X; \complex)$ equipped with pointwise operations and the uniform norm is a $C^*$-algebra. 
+    Let $X$ be an LCH space, then $C_0(X; \complex)$ equipped with pointwise operations and the uniform norm is a $C^*$-algebra. 
 \end{definition}
 
 \begin{theorem}
 \label{theorem:vanishing-infinity-multiplicative-functional}
-    Let $X$ be a LCH space, then the mapping
+    Let $X$ be an LCH space, then the mapping
     \[
     E: X \to \Omega(C_0(X)) \quad E(x)(f) = f(x)
     \]

src/op/example/matrix.tex

@@ -26,3 +26,61 @@
 \begin{proof}
     (1), (2): By \autoref{proposition:matrix-algebra-spectrum}, $\sigma(x)$ is finite. By \autoref{proposition:log-identity-component}, there exists $y \in M_n(\complex)$ such that $x = \exp(y)$. In which case, $x \in G_0(M_n(\complex))$. 
 \end{proof}
+
+
+\begin{proposition}
+\label{proposition:matrix-algebra-state-space}
+    Let $n \in \natp$. For each $y \in M_n(\complex)$, let
+    \[
+    \phi_y: M_n(\complex) \to \complex \quad x \mapsto \dpn{x, y}{F} = \text{tr}(y^*x)
+    \]
+
+    then the following are equivalent:
+    \begin{enumerate}
+        \item $\phi_y \in S(M_n(\complex))$. 
+        \item $y \ge 0$ and $\text{tr}(y) = 1$.
+    \end{enumerate}
+\end{proposition}
+% "Obvious" so won't prove. 
+
+\begin{proposition}
+\label{proposition:matrix-algebra-pure-state}
+    Let $n \in \natp$. For each $y \in M_n(\complex)$, let
+    \[
+    \phi_y: M_n(\complex) \to \complex \quad x \mapsto \dpn{x, y}{F} = \text{tr}(y^*x)
+    \]
+
+    then the following are equivalent:
+    \begin{enumerate}
+        \item $\phi_y$ is a pure state of $M_n(\complex)$. 
+        \item $y$ is a projection operator with rank $1$. 
+        \item There exists $v \in \complex^n$ with $\norm{v}_{\complex^n} = 1$ such that $\dpn{x, \phi_y}{M_n(\complex)} = \dpn{xv, v}{\complex^n}$ for all $x \in M_n(\complex)$.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1) $\Leftrightarrow$ (2): By \autoref{proposition:matrix-algebra-state-space}, $y \ge 0$ with $\text{tr}(y) = 1$. Via an orthogonal change of coordinates, assume without loss of generality that $y$ is diagonal. In which case, $y$ corresponds to an extreme point of $S(M_n(\complex))$ if and only if it is of rank $1$. As $\text{tr}(y) = 1$, $y$ is a projection.
+
+    (2) $\Rightarrow$ (3): Let $v \in \complex^n$ be a unit eigenvector of $y$, then for each $x \in M_n(\complex)$,
+    \[
+        \dpn{x, \phi_y}{M_n(\complex)} = \text{tr}(y^*x) = \text{tr}(yx) = \dpn{xv, v}{\complex^n}
+    \]
+\end{proof}
+
+\begin{example}
+\label{proposition:spectrum-pure-state-counterexample}
+    Let
+    \[
+    A = \begin{bmatrix}
+        1 & 0 \\
+        0 & -1
+    \end{bmatrix}
+    \]
+
+    then $A$ is a self-adjoint element of $M_2(\complex)$. By \autoref{proposition:matrix-algebra-pure-state}, the mapping $T \mapsto \dpn{Tv, v}{\complex^2}$ is a pure state on $M_2(\complex)$ for every unit vector $v \in \complex^2$. In particular, if $v = (\sqrt{2}, \sqrt{2})/2$, then $\dpn{Tv, v}{\complex^2} = 0 \not\in \sigma_{M_2(\complex)}(A)$. Therefore
+    \[
+    \sigma_{M_2(\complex)}(A) \subsetneq \bracsn{\dpn{T, \phi}{M_2(\complex)}|\phi \in P(M_2(\complex))}
+    \]
+\end{example}
+
+
+

src/op/notation.tex

@@ -14,6 +14,11 @@
     $\Omega(A)$ & Space of multiplicative functionals on $A$. & \autoref{definition:multiplicative-functional} \\
     $\cm(A)$ & Maximal ideal space of $A$. & \autoref{definition:maximal-ideal} \\
     $\Gamma = \Gamma_A$ & The Gelfand transform on $A$. & \autoref{definition:gelfand-transform} \\ 
+    $A[S]$ & $C^*$-subalgebra of $A$ generated by $S \subset A$. & \autoref{definition:generated-subalgebra} \\ 
+    $S(A)$ & State space of a $C^*$-algebra $A$. & \autoref{definition:cstar-state} \\ 
+    $P(A)$ & Pure state space of a $C^*$-algebra $A$. & \autoref{definition:pure-state} \\ 
+    $\dpn{x, y}{\phi}$ & Defined as $\dpn{y^*x, \phi}{A}$, the pseudo inner product associated to a positive linear functional. & \autoref{definition:cstar-state-pseudo-inner-product} \\ 
+    $(H_\phi, \pi_\phi, \xi_\phi)$ & GNS triple associated with $\phi \in S(A)$. & \autoref{definition:gns-triple} \\ 
 
     $M_n(\complex)$ & Algebra of $n \times n$ matrices over $\complex$.  & \autoref{definition:matrix-algebra} \\
     $B(H)$ & Algebra of bounded operators on a Hilbert space. & \autoref{definition:hilbert-endomorphism} \\

src/process/markov/definition.tex

@@ -41,7 +41,7 @@
     (\bp_{s+t}\one_A)(x) = \int \one_A(y)P_{s+t}(x, dy) = \int P_t(y, A)P_s(x, dy) = \int \int \one_A(z)P_t(y, dz)P_s(x, dy)
     \]
 
-    Let $f: X \to \complex$ bounded and $(\cm, \cb_\complex)$-measurable. By \autoref{proposition:measurable-simple-separable-norm}, there exists $\seq{f_n} \subset \Sigma(X; \complex)$ such that $\abs{f_n} \le \abs{f}$ for all $n \in\nat$, and $f_n \to f$ pointwise. By the \hyperref[Dominated Convergence Theorem]{theorem:dct},
+    Let $f: X \to \complex$ bounded and $(\cm, \cb_\complex)$-measurable. By \autoref{corollary:measurable-simple-separable-norm}, there exists $\seq{f_n} \subset \Sigma(X; \complex)$ such that $\abs{f_n} \le \abs{f}$ for all $n \in\nat$, and $f_n \to f$ pointwise. By the \hyperref[Dominated Convergence Theorem]{theorem:dct},
     \begin{align*}
     (\bp_{s+t}f)(x) &= \limv{n}\int f_n(y)P_{s+t}(x, dy) = \limv{n}\iint f_n(z)P_t(y, dz)P_s(x, dy) \\
     &= \int \int f_n(z)P_t(y, dz)P_s(x, dy) = (\bp_s\bp_t) f(x)