Fixed typo.

.vscode/project.code-snippets

@@ -27,6 +27,17 @@
       "$0"
     ]
   },
+  "Summary": {
+  "scope": "latex",
+    "prefix": "summ",
+    "body": [
+      "\\begin{summary}[$1]",
+      "\\label{summary:$2}",
+      "    $3",
+      "\\end{summary}",
+      "$0"
+    ]
+  },
   "Lemma Block": {
   "scope": "latex",
     "prefix": "lem",
@@ -126,6 +137,11 @@
     "prefix": "cal",
     "body": ["\\mathcal{$1}$0"]
   },
+  "Mathscr": {
+  "scope": "latex",
+    "prefix": "scr",
+    "body": ["\\mathscr{$1}$0"]
+  },
   "Mathfrak": {
   "scope": "latex",
     "prefix": "fk",

.vscode/settings.json

@@ -7,5 +7,5 @@
     ],
     "latex.linting.enabled": false,
     "latex-workshop.latex.autoBuild.run": "never",
-    "latex-workshop.latex.texDirs": ["${workspaceFolder}"]
+    "latex-workshop.latex.search.rootFiles.include": ["document.tex"]
 }

document.tex

@@ -1,18 +1,20 @@
-%\documentclass{report}
+\documentclass{}
 \usepackage{amssymb, amsmath, hyperref}
 \usepackage{preamble}
 
 \begin{document}
 
+Hello this is all my notes. 
+
 \input{./src/cat/index}
 \input{./src/topology/index}
 \input{./src/fa/index}
 \input{./src/measure/index}
 \input{./src/dg/index}
+\input{./src/op/index}
 %\input{./src/process/index}
 
 \bibliographystyle{alpha} % We choose the "plain" reference style
 \bibliography{refs.bib} % Entries are in the refs.bib file
 
-
 \end{document}

preamble.sty

@@ -15,6 +15,7 @@
 
 \newtheorem{definition}[theorem]{Definition}
 \newtheorem{example}[theorem]{Example}
+\newtheorem{summary}[theorem]{Summary}
 % \newtheorem{exercise}[subsection]{Exercise}
 % \newtheorem{situation}[subsection]{Situation}
 
@@ -223,3 +224,8 @@
 
 % Real or Complex Numbers
 \newcommand{\RC}{\bracs{\real, \complex}}
+
+% Convex Stuff
+\newcommand{\conv}{\text{Conv}}
+\newcommand{\aconv}{\text{AbsConv}}
+

refs.bib

@@ -103,3 +103,22 @@
   year={1983},
   publisher={Hermann}
 }
+@article{HensgenSinger,
+  title={A simple proof of Singer’s representation theorem},
+  author={Hensgen, Wolfgang},
+  journal={Proceedings of the American Mathematical Society},
+  volume={124},
+  number={10},
+  pages={3211--3212},
+  year={1996}
+}
+@book{ConwayComplex,
+  title={Functions of One Complex Variable I},
+  author={Conway, J.B.},
+  isbn={9780387903286},
+  lccn={lc78018836},
+  series={Functions of one complex variable / John B. Conway},
+  url={https://books.google.ca/books?id=9LtfZr1snG0C},
+  year={1978},
+  publisher={Springer}
+}

spec.toml

@@ -9,8 +9,16 @@ indirectReferences = true
 
 [website]
 font = "roboto"
+fontSize = 16
+lineHeight = 1.3
+textAlign = "left"
+lineWidth = 45
 primaryColour = "violet"
 neutralColour = "grey"
 searchLimit = 16
 maxSearchPages = 48
+recentChanges = 0
+tableOfContentsDepth = 1
+hoverPreview = false
+copyLabelButton = false
 advertiseSpec = true

src/cat/cat/cat-func.tex

@@ -14,6 +14,7 @@
         \item[(CAT2)] For any $A \in \obj{\catc}$, there exists $\text{Id}_A \in \mor{A, A}$ such that $f \circ \text{Id}_A = f$ and $\text{Id}_A \circ g = g$ for all $B, C \in \obj{\catc}$, $f \in \mor{A, B}$, and $g \in \mor{C, A}$.
         \item[(CAT3)] For any $A, B, C, D \in \obj{\catc}$, $f \in \mor{A, B}$, $g \in \mor{B, C}$, and $h \in \mor{C, D}$, $(h \circ g) \circ f = h \circ (g \circ f)$.
     \end{enumerate}
+
     The elements of $\obj{\catc}$ are the \textbf{objects} of $\catc$, and elements of $\mor{A, B}$ are the \textbf{morphisms/arrows} from $A$ to $B$.
 \end{definition}
 

src/cat/cat/index.tex

@@ -5,3 +5,4 @@
 
 \input{./cat-func.tex}
 \input{./universal.tex}
+\input{./tensor.tex}

src/cat/cat/tensor.tex

@@ -0,0 +1,221 @@
+\section{Universal Constructions for Modules}
+\label{section:universal-module}
+
+
+
+\begin{definition}[Product]
+\label{definition:product-module}
+    Let $R$ be a ring and $\seqi{A}$ be $R$-modules, then there exists $(A, \bracsn{\pi_i}_{i \in I})$ such that:
+    \begin{enumerate}
+        \item For each $i \in I$, $\pi_i \in \hom(A; A_i)$. 
+        \item[(U)] For any $(B, \seqi{T})$ satisfying (1), there exists a unique $T \in \hom(B; A)$ such that the following diagram commutes:
+        \[
+        \xymatrix{
+        B \ar@{->}[rd]_{T_i} \ar@{->}[r]^{T} & A \ar@{->}[d]^{\pi_i} \\
+        & A_i
+        }
+        \]
+    \end{enumerate}
+
+    The module $A = \prod_{i \in I}A_i$ is the \textbf{product} of $\seqi{A}$. 
+\end{definition}
+
+\begin{definition}[Direct Sum]
+\label{definition:direct-sum}
+    Let $E$ be a ring and $\seqi{A}$ be $R$-modules, then there exists $(A, \bracsn{\iota_i}_{i \in I})$ such that:
+    \begin{enumerate}
+        \item For each $i \in I$, $\iota_i \in \hom(A_i; A)$. 
+        \item[(U)] For each $(B, \seqi{T})$ satisfying (1), there exists a unique $T \in \hom(A; B)$ such that the following diagram commutes
+        \[
+        \xymatrix{
+        A \ar@{->}[r]^{T} & B \\
+        A_i \ar@{->}[u]^{\iota_i} \ar@{->}[ru]_{T_i} & 
+        }
+        \]
+    \end{enumerate}
+
+    
+    The module $A = \bigoplus_{i \in I}A_i$ is the \textbf{direct sum} of $\seqi{A}$. 
+\end{definition}
+\begin{proof}
+    Let
+    \[
+    A = \bracs{x \in \prod_{i \in I}A_i \bigg | x_i \ne 0 \quad \text{for finitely many}\ i \in I}
+    \]
+
+    For each $i \in I$, let
+    \[
+    \iota_i: A_i \to A \quad (\iota_ix)_j = \begin{cases}
+        x &i = j \\
+        0 &i \ne j
+    \end{cases}
+    \]
+
+    then $\iota_i \in \hom(A_i; A)$. 
+
+    (U): Let
+    \[
+    T: A \to B \quad x \mapsto \sum_{i \in I}T_ix_i
+    \]
+
+    then $T \in \hom(A; B)$ and the diagram commutes. Since $\bigcup_{i \in I}\iota_i(A_i)$ spans $A$, $T$ is the unique linear map making the diagram commute. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:module-direct-limit}
+    Let $R$ be a ring and $(\seqi{A}, \bracsn{T^i_j| i, j \in I, i \lesssim j})$ be an upward-directed system of $R$-modules, then there exists $(A, \bracsn{T^i_A}_{i \in I})$ such that:
+    \begin{enumerate}
+        \item For each $i \in I$, $T^i_A \in \hom({A_i; A})$.
+        \item For any $i, j \in I$ with $i \lesssim j$, the following diagram commutes:
+
+        \[
+        \xymatrix{
+        A_i \ar@{->}[rd]_{T^i_A} \ar@{->}[r]^{T^i_j} & A_j \ar@{->}[d]^{T^j_A} \\
+         & A
+        }
+        \]
+
+
+        \item[(U)] For any pair $(B, \bracsn{S^i_B}_{i \in I})$ satisfying (1) and (2), there exists a unique $S \in \hom({A, B})$ such that the following diagram commutes
+
+        \[
+        \xymatrix{
+        A_i \ar@{->}[d]_{T^i_A} \ar@{->}[rd]^{S^i_B} &  \\
+        A \ar@{->}[r]_{g} & B
+        }
+        \]
+
+
+        for all $i \in I$.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    Let $M = \bigoplus_{i \in I}A_i$. For any $i, j \in I$ with $i \lesssim j$ and $x \in A_i$, let $x_{i, j} \in M$ such that for any $k \in I$,
+    \[
+    \pi_k(x_{i, j}) = \begin{cases}
+        x &k = i \\
+        T^i_j x &k = j \\
+        0 &k \ne i, j
+    \end{cases}
+    \]
+
+    Let $N \subset M$ be the submodule generated by $\bracs{x_{i, j}|i, j \in I, i \lesssim j, x \in A_i}$, $A = M/N$, and $\pi: M \to M/N$ be the canonical map.
+
+    (1): For each $i \in I$, let
+    \[
+    T^i_M: A_i \to M \quad \pi_k T^i_M x = \begin{cases}
+        x &k = i \\
+        0 &k \ne i
+    \end{cases}
+    \]
+
+    and $T^i_A = \pi \circ T^i_M$.
+
+    (2): Let $i, j \in I$ with $i \lesssim j$, then for any $x \in A_i$, $T^i_Mx - T^j_M T^i_j x \in N$. Hence $T^i_Ax = T^j_A T^i_jx$.
+
+    (U): Let
+    \[
+    S_0: M \to B \quad x \mapsto \sum_{i \in I}S^i_B \pi_i x
+    \]
+
+    then $S_0$ is the unique linear map such that $S_0 \circ T^i_M = S^i_B$ for all $i \in I$. For any $i, j \in I$ with $i \lesssim J$, $S^i_B x = S^j_B T^i_j x$, so $\ker S_0 \supset N$. By the first isomorphism theorem, there exists a unique $S \in \hom(A; B)$ such that $S_0 = S \circ \pi$.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:module-inverse-limit}
+    Let $R$ be a ring and $(\seqi{A}, \bracs{T^i_j|i, j \in I, i \lesssim j})$ be a downward-directed system of $R$-modules, then there exists $(A, \bracsn{T^A_i}_{i \in I})$ such that:
+    \begin{enumerate}
+        \item For each $i \in I$, $T^A_i \in \hom(A; A_i)$.
+        \item For any $i, j \in I$ with $i \lesssim j$, the following diagram commutes:
+        \[
+        \xymatrix{
+        A_i \ar@{->}[r]^{T^i_j} & A_j \\
+        A \ar@{->}[u]^{T^A_i} \ar@{->}[ru]_{T^A_j} &
+        }
+        \]
+
+        \item[(U)] For any pair $(B, \bracsn{S^B_i}_{i \in I})$ satisfying (1) and (2), there exists a unique $S \in \hom(B; A)$ such that the following diagram commutes
+
+        \[
+        \xymatrix{
+         & A_i \\
+        B \ar@{->}[r]_{S} \ar@{->}[ru]^{S^B_i} & A \ar@{->}[u]_{T^A_i}
+        }
+        \]
+
+
+        for all $i \in I$.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    Let
+    \[
+    A = \bracs{x \in \prod_{i \in I}A_i \bigg | \pi_j(x) = T^i_j\pi_i(x) \forall i, j \in I, i \lesssim j}
+    \]
+
+    For each $i \in I$, let $T^A_i = \pi_i$, then $(A, \bracsn{T^A_i}_{i \in I})$ satisfies (1) and (2) by definition of $A$.
+
+    (U): Let $(B, \bracsn{S^B_i}_{i \in I})$ satisfying (1) and (2). Let
+    \[
+    S: B \to \prod_{i \in I}A_i \quad \pi_i(Sx) = S^B_i
+    \]
+
+    then for any $x \in B$ and $i, j \in I$ with $i \lesssim j$,
+    \[
+    \pi_j (Sx) = S^B_jx = T^i_j S^B_ix = T^i_j \pi_i(S x)
+    \]
+
+    so $S \in \hom(B; A)$, and the diagram commutes. Since any map $f: B \to A$ is uniquely determined by its composition with the projections, $S$ is unique.
+\end{proof}
+
+\begin{definition}[Tensor Product]
+\label{definition:tensor-product}
+    Let $R$ be a commutative ring and $\seqf{E_j}$ be $R$ modules, then there exists a pair $(\bigotimes_{j = 1}^n E_j, \iota)$ such that:
+    \begin{enumerate}
+        \item $\bigotimes_{j = 1}^n E_j$ is an $R$-module. 
+        \item $\iota: \prod_{j = 1}^n E_j \to \bigotimes_{j = 1}^n E_j$ is a $n$-linear map. 
+        \item[(U)] For any pair $(F, \lambda)$ satisfying (1) and (2), there exists a unique $\Lambda \in \hom(\bigotimes_{j = 1}^n E_j; F)$ such that the following diagram commutes:
+        \[
+        \xymatrix{
+        \prod_{j = 1}^n E_j \ar@{->}[rd]_{\lambda} \ar@{->}[r]^{\iota} & \bigotimes_{j = 1}^n E_j \ar@{->}[d]^{\Lambda} \\
+        & F
+        }
+        \] 
+
+        \item $\bigotimes_{j = 1}^n E_j$ is the linear span of $\iota(\prod_{j = 1}^n E_j)$. 
+    \end{enumerate}
+
+    The module $\bigotimes_{j = 1}^n E_j$ is the \textbf{tensor product} of $\seqf{E_j}$, and $\iota: \prod_{j = 1}^n E_j \to \bigotimes_{j = 1}^n E_j$ is the \textbf{canonical embedding}. For any $(x_1, \cdots, d_n) \in \prod_{j = 1}^n E_j$, the image
+    \[
+    x_1 \otimes \cdots \otimes x_n = \iota(x_1, \cdots, x_n)
+    \]
+
+    is its \textbf{tensor product}.
+     
+\end{definition}
+\begin{proof}
+    Let $M$ be the free module generated by $\prod_{j = 1}^nE_j$, and $N \subset M$ be the submodule generated by elements of the following form:
+    \begin{enumerate}
+        \item For any $1 \le j \le n$, $(x_1, \cdots, x_n) \in \prod_{k = 1}^n E_k$, and $x_j \in E_j$, 
+        \[
+        (x_1, \cdots, x_j + x_j', \cdots, x_n) - (x_1, \cdots, x_j, \cdots, x_n) - (x_1, \cdots, x_j', \cdots, x_n)
+        \]
+
+        \item For any $(x_1, \cdots, x_n) \in \prod_{k = 1}^n E_k$ and $\alpha \in R$, 
+        \[
+        (x_1, \cdots, \alpha x_j, \cdots, x_n) - \alpha(x_1, \cdots, x_n)
+        \]
+    \end{enumerate}
+
+    
+    (1), (2): Let $\bigotimes_{j = 1}^n E_j = M/N$ and
+    \[
+    \iota: \prod_{j = 1}^n E_j \to \bigotimes_{j = 1}^n E_j \quad (x_1, \cdots, x_n) \mapsto (x_1, \cdots, x_n) + N
+    \]
+
+    then by definition of $N$, $\iota$ is $n$-linear. 
+
+    (U): Let $(F, \lambda)$ be a pair satisfying (1) and (2), then $\lambda$ admits a unique extension to a linear map $\ol \lambda: M \to F$. Since $\lambda$ is $n$-linear, $\ker \ol \lambda \supset N$. By the first isomorphism theorem, there exists a unique $\Lambda \in \hom(\bigotimes_{j = 1}^n E_j; F)$ such that the given diagram commutes. 
+
+    (4): Since $M$ is the free module generated by $\prod_{j = 1}^n E_j$, $M/N$ is generated by $\iota(\prod_{j = 1}^n E_j)$.  
+\end{proof}

src/cat/cat/universal.tex

@@ -8,6 +8,7 @@
         \item \textbf{universally attracting} if for every $A \in \obj{\catc}$, there exists a unique $f \in \mor{A, P}$.
         \item \textbf{universally repelling} if for every $A \in \obj{\catc}$, there exists a unique $f \in \mor{P, A}$.
     \end{enumerate}
+
     If $P$ is universally attracting or repelling, then $P$ is a \textbf{universal object}.
 
     If $P, Q \in \obj{\catc}$ are both universally attracting/repelling, then they are isomorphic.
@@ -63,11 +64,13 @@
         \item For any $i \in I$, $i \lesssim i$.
         \item For any $i, j, k \in I$ such that $i \lesssim j$ and $j \lesssim k$, $i \lesssim k$.
     \end{enumerate}
+
     and one of the following holds:
     \begin{enumerate}
         \item[(3U)] For any $i, j \in I$, there exists $k \in I$ with $i, j \lesssim k$.
         \item[(3D)] For any $i, j \in I$, there exists $k \in I$ with $k \lesssim i, j$.
     \end{enumerate}
+
     The directed set is \textbf{upward-directed} if it satisfies (3U), and \textbf{downward-directed} if it satisfies (3D).
 \end{definition}
 
@@ -86,6 +89,7 @@
         \item For each $i, j \in I$ with $i \lesssim j$, $f^i_j \in \mor{A_i, A_j}$.
         \item For each $i, j, k \in I$ with $i \lesssim j \lesssim k$, $f^j_k \circ f^i_j = f^i_k$.
     \end{enumerate}
+
     If $I$ is upward/downward-directed, then $(\seqi{A}, \bracsn{f^i_j| i, j \in I, i \lesssim j})$ is upward/downward-directed.
 \end{definition}
 
@@ -147,109 +151,5 @@ Let $\catc$ be a category and $(\seqi{A}, \bracsn{f^i_j| i, j \in I, i \lesssim
 \end{enumerate}
 \end{definition}
 
-\begin{proposition}
-\label{proposition:module-direct-limit}
-    Let $R$ be a ring and $(\seqi{A}, \bracsn{T^i_j| i, j \in I, i \lesssim j})$ be an upward-directed system of $R$-modules, then there exists $(A, \bracsn{T^i_A}_{i \in I})$ such that:
-    \begin{enumerate}
-        \item For each $i \in I$, $T^i_A \in \hom({A_i, A})$.
-        \item For any $i, j \in I$ with $i \lesssim j$, the following diagram commutes:
-
-        \[
-        \xymatrix{
-        A_i \ar@{->}[rd]_{T^i_A} \ar@{->}[r]^{T^i_j} & A_j \ar@{->}[d]^{T^j_A} \\
-         & A
-        }
-        \]
 
 
-        \item[(U)] For any pair $(B, \bracsn{S^i_B}_{i \in I})$ satisfying (1) and (2), there exists a unique $S \in \hom({A, B})$ such that the following diagram commutes
-
-        \[
-        \xymatrix{
-        A_i \ar@{->}[d]_{T^i_A} \ar@{->}[rd]^{S^i_B} &  \\
-        A \ar@{->}[r]_{g} & B
-        }
-        \]
-
-
-        for all $i \in I$.
-    \end{enumerate}
-\end{proposition}
-\begin{proof}
-    Let $M = \bigoplus_{i \in I}A_i$. For any $i, j \in I$ with $i \lesssim j$ and $x \in A_i$, let $x_{i, j} \in M$ such that for any $k \in I$,
-    \[
-    \pi_k(x_{i, j}) = \begin{cases}
-        x &k = i \\
-        T^i_j x &k = j \\
-        0 &k \ne i, j
-    \end{cases}
-    \]
-
-    Let $N \subset M$ be the submodule generated by $\bracs{x_{i, j}|i, j \in I, i \lesssim j, x \in A_i}$, $A = M/N$, and $\pi: M \to M/N$ be the canonical map.
-
-    (1): For each $i \in I$, let
-    \[
-    T^i_M: A_i \to M \quad \pi_k T^i_M x = \begin{cases}
-        x &k = i \\
-        0 &k \ne i
-    \end{cases}
-    \]
-
-    and $T^i_A = \pi \circ T^i_M$.
-
-    (2): Let $i, j \in I$ with $i \lesssim j$, then for any $x \in A_i$, $T^i_Mx - T^j_M T^i_j x \in N$. Hence $T^i_Ax = T^j_A T^i_jx$.
-
-    (U): Let
-    \[
-    S_0: M \to B \quad x \mapsto \sum_{i \in I}S^i_B \pi_i x
-    \]
-
-    then $S_0$ is the unique linear map such that $S_0 \circ T^i_M = S^i_B$ for all $i \in I$. For any $i, j \in I$ with $i \lesssim J$, $S^i_B x = S^j_B T^i_j x$, so $\ker S_0 \supset N$. By the first isomorphism theorem, there exists a unique $S \in \hom(A; B)$ such that $S_0 = S \circ \pi$.
-\end{proof}
-
-\begin{proposition}
-\label{proposition:module-inverse-limit}
-    Let $R$ be a ring and $(\seqi{A}, \bracs{T^i_j|i, j \in I, i \lesssim j)}$ be a downward-directed system of $R$-modules, then there exists $(A, \bracsn{T^A_i}_{i \in I})$ such that:
-    \begin{enumerate}
-        \item For each $i \in I$, $T^A_i \in \hom(A; A_i)$.
-        \item For any $i, j \in I$ with $i \lesssim j$, the following diagram commutes:
-        \[
-        \xymatrix{
-        A_i \ar@{->}[r]^{T^i_j} & A_j \\
-        A \ar@{->}[u]^{T^A_i} \ar@{->}[ru]_{T^A_j} &
-        }
-        \]
-
-        \item[(U)] For any pair $(B, \bracsn{S^B_i}_{i \in I})$ satisfying (1) and (2), there exists a unique $S \in \hom(B; A)$ such that the following diagram commutes
-
-        \[
-        \xymatrix{
-         & A_i \\
-        B \ar@{->}[r]_{S} \ar@{->}[ru]^{S^B_i} & A \ar@{->}[u]_{T^A_i}
-        }
-        \]
-
-
-        for all $i \in I$.
-    \end{enumerate}
-\end{proposition}
-\begin{proof}
-    Let
-    \[
-    A = \bracs{x \in \prod_{i \in I}A_i \bigg | \pi_j(x) = T^i_j\pi_i(x) \forall i, j \in I, i \lesssim j}
-    \]
-
-    For each $i \in I$, let $T^A_i = \pi_i$, then $(A, \bracsn{T^A_i}_{i \in I})$ satisfies (1) and (2) by definition of $A$.
-
-    (U): Let $(B, \bracsn{S^B_i}_{i \in I})$ satisfying (1) and (2). Let
-    \[
-    S: B \to \prod_{i \in I}A_i \quad \pi_i(Sx) = S^B_i
-    \]
-
-    then for any $x \in B$ and $i, j \in I$ with $i \lesssim j$,
-    \[
-    \pi_j (Sx) = S^B_jx = T^i_j S^B_ix = T^i_j \pi_i(S x)
-    \]
-
-    so $S \in \hom(B; A)$, and the diagram commutes. Since any map $f: B \to A$ is uniquely determined by its composition with the projections, $S$ is unique.
-\end{proof}

src/cat/gluing/index.tex

@@ -9,6 +9,7 @@
         \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}
@@ -17,9 +18,11 @@
         \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:
@@ -28,6 +31,7 @@
         \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}

src/cat/index.tex

@@ -4,3 +4,4 @@
 \input{./cat/index.tex}
 \input{./gluing/index.tex}
 \input{./tricks/index.tex}
+\input{./notation/index.tex}

src/cat/notation/index.tex

@@ -0,0 +1,19 @@
+\chapter{Notations}
+\label{chap:notations}
+
+\begin{tabular}{lll}
+    \textbf{Notation} & \textbf{Description} & \textbf{Source} \\
+    \hline
+    % ---- Category Theory ----
+    $\obj{\catc}$ & Objects of category $\catc$. & \autoref{definition:category} \\
+    $\mor{A, B}$ & Morphisms from $A$ to $B$ in a category. & \autoref{definition:category} \\
+    $\text{Id}_A$ & Identity morphism on $A$. & \autoref{definition:category} \\
+    $E \otimes F$, $x_1 \otimes \cdots \otimes x_n$ & Tensor product of modules; image of $(x_1,\ldots,x_n)$ under canonical embedding. & \autoref{definition:tensor-product} \\
+    $\lim_{\longrightarrow} A_i$ & Direct limit of an upward-directed system. & \autoref{definition:direct-limit} \\
+    $\lim_{\longleftarrow} A_i$ & Inverse limit of a downward-directed system. & \autoref{definition:inverse-limit} \\
+    $\mathbb{D}_n$, $\mathbb{D}$ & Dyadic rationals of level $n$; all dyadic rationals. & \autoref{definition:dyadic} \\
+    $\mathrm{rk}(q)$ & Dyadic rank of $q \in \mathbb{D}$. & \autoref{definition:dyadic-rank} \\
+    $M(x)$ & Unique $M(x) \subset \mathbb{N}^+ \cap [1, \mathrm{rk}(x)]$ such that $x = \sum_{n \in M(x)} 2^{-n}$. & \autoref{proposition:dyadic-subset} \\
+    $[n]$ & $\bracs{1, \cdots, n}$ & N/A
+\end{tabular}
+

src/cat/tricks/dyadic.tex

@@ -38,9 +38,16 @@
 
 \begin{proposition}
 \label{proposition:dyadic-semigroup-order}
-    Let $G$ be a commutative ordered semigroup, and $\seq{g_n} \subset G$ such that for each $n \in \natp$, $g_{n+1} + g_{n+1} \le g_n$. For each $x \in \mathbb{D} \cap [0, 1)$, let $\phi(x) = \sum_{n \in M(x)}g_n$, then
+    Let $G$ be a commutative ordered semigroup, and $\seq{g_n} \subset G$ such that 
     \begin{enumerate}
-        \item For any $x, y \in \mathbb{D} \cap [0, 1)$ such that $x + y \in [0, 1)$, $\phi(x) + \phi(y) \le \phi(x + y)$.
+        \item[(a)] for each $n \in \natp$, $g_{n+1} + g_{n+1} \le g_n$.
+        \item[(b)] For each $x, y \in G$, $x + y \ge x, y$. 
+    \end{enumerate}
+
+    
+    For each $x \in \mathbb{D} \cap [0, 1)$, let $\phi(x) = \sum_{n \in M(x)}g_n$, then
+    \begin{enumerate}
+        \item For any $x, y \in \mathbb{D} \cap [0, 1)$ such that $x + y \in (0, 1)$, $\phi(x) + \phi(y) \le \phi(x + y)$.
         \item For any $x, y \in \mathbb{D} \cap [0, 1)$ with $x \le y$, $\phi(x) \le \phi(y)$.
     \end{enumerate}
 \end{proposition}

src/dg/complex/derivative.tex

@@ -0,0 +1,283 @@
+\section{Complex Differentiability}
+\label{section:complex-derivative}
+
+\begin{lemma}
+\label{lemma:complex-analytic}
+    Let $E$ be a separated locally convex space over $\complex$, $U \subset \complex$, and $f: U \to E$, then the following are equivalent:
+    \begin{enumerate}
+        \item $f \in C^1(U; E)$. 
+        \item Under the identification of $C = \real^2$, $\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \in C(U; E)$ and
+        \[
+        \frac{\partial f}{\partial x} = i\frac{\partial f}{\partial y}
+        \]
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    (1) $\Rightarrow$ (2): Let $x_0 \in U$, then
+    \[
+    \frac{\partial f}{\partial x} = \lim_{\substack{h \to 0 \\ h \in \real}}\frac{f(x_0 + h) - f(x_0)}{h} 
+    = \lim_{h \to 0}\lim_{\substack{h \to 0 \\ h \in \real}}\frac{f(x_0 + ih) - f(x_0)}{ih} = \frac{1}{i} \frac{\partial f}{\partial y}
+    \]
+
+    (2) $\Rightarrow$ (1): Let $x_0 \in U$ and
+    \[
+    L: \complex \to E \quad a + bi \mapsto a \frac{\partial f}{\partial x}(x_0) + b \frac{\partial f}{\partial y}(x_0)
+    \]
+
+    by assumption and \autoref{proposition:polarisation-linear}, $L \in L(\complex; E)$. By \autoref{proposition:partial-total-derivative}, $f \in C^1(U \subset \real^2; E)$, where for any $(a, b) \in \real^2$, 
+    \[
+    Df(x_0)(a, b) = a \frac{\partial f}{\partial x}(x_0) + b \frac{\partial f}{\partial y}(x_0)
+    \]
+
+    so by definition of differentiability, $f$ is complex-differentiable at $x_0$ with derivative $L$. 
+\end{proof}
+
+\begin{theorem}[Cauchy]
+\label{theorem:cauchy-homotopy}
+    Let $E$ be a complete separated locally convex space over $\complex$, $U \subset \complex$ be open, $f \in C^1(U; E)$, and $\gamma, \mu \in C([a, b]; U)$ be closed rectifiable paths. If $\gamma$ and $\mu$ are homotopic, then
+    \[
+    \int_\gamma f = \int_\mu f
+    \]
+\end{theorem}
+\begin{proof}[Proof of smooth case. ]
+    Let $\Gamma \in C^\infty([0, 1] \times [a, b]; U)$ be a smooth homotopy of loops from $\gamma$ to $\mu$, and
+    \[
+    F: [0, 1] \to E \quad t \mapsto \int_{\Gamma (t, \cdot)}f = \int_a^b (f \circ \Gamma)(t, s) \Gamma(t, ds)
+    \]
+
+    then for any $t \in [0, 1]$, by the \hyperref[change of variables formula]{theorem:rs-change-of-variables}, 
+    \begin{align*}
+        F(t) &= \int_a^b (f \circ \Gamma)(t, s) \Gamma(t, ds) \\
+        &= \int_a^b (f \circ \Gamma)(t, s) \frac{\partial \Gamma}{\partial s}(t, s) ds
+    \end{align*}
+
+    Now, by \autoref{proposition:difference-quotient-compact}, 
+    \[
+        \frac{dF}{dt}(t) = \int_a^b \frac{\partial}{\partial t}\braks{(f \circ \Gamma)(t, s) \frac{\partial \Gamma}{\partial s}(t, s)}ds
+    \]
+
+    Under the identification that $\complex = \real^2$, by the \hyperref[power rule]{theorem:power-rule} and the \hyperref[chain rule]{proposition:chain-rule-sets-conditions},
+    \[
+    \frac{\partial }{\partial t}\braks{(f \circ \Gamma) \frac{\partial \Gamma}{\partial s}} = (Df \circ \Gamma)\paren{\frac{\partial\Gamma}{\partial t}} \frac{\partial \Gamma}{\partial s} + (f \circ \Gamma) \frac{\partial^2\Gamma}{\partial t \partial s}
+    \]
+    
+    Now, since $f \in C^1(U; E)$ satisfies the \hyperref[Cauchy-Riemann equations]{lemma:complex-analytic},
+    \begin{align*}
+    (Df \circ \Gamma)\paren{\frac{\partial\Gamma}{\partial t}} \frac{\partial \Gamma}{\partial s} &= 
+    (Df \circ \Gamma)\frac{\partial\Gamma}{\partial t} \frac{\partial \Gamma}{\partial s} = (Df \circ \Gamma)\paren{\frac{\partial \Gamma}{\partial s}}\frac{\partial\Gamma}{\partial t} 
+    \end{align*}
+
+    so
+    \begin{align*}
+        \frac{\partial }{\partial t}\braks{(f \circ \Gamma) \frac{\partial \Gamma}{\partial s}} &= (Df \circ \Gamma)\paren{\frac{\partial\Gamma}{\partial s}} \frac{\partial \Gamma}{\partial t} + (f \circ \Gamma) \frac{\partial^2\Gamma}{\partial s \partial t} \\
+        &= \frac{\partial }{\partial s}\braks{(f \circ \Gamma) \frac{\partial \Gamma}{\partial t}}
+    \end{align*}
+
+    Hence by the \hyperref[Fundamental Theorem of Calculus]{theorem:ftc-riemann}, 
+    \begin{align*}
+        \frac{dF}{dt}(t) &= \int_a^b \frac{\partial}{\partial s}\braks{(f \circ \Gamma)(t, s) \frac{\partial \Gamma}{\partial t}(t, s)}ds \\
+        &= (f \circ \Gamma)(t, b)\frac{\partial \Gamma}{\partial t}(t, b) - (f \circ \Gamma)(t, a)\frac{\partial \Gamma}{\partial t}(t, a)
+    \end{align*}
+
+    Since $\Gamma(t, a) = \Gamma(t, b)$ for all $t \in [0, 1]$, the above expression evaluates to $0$, so
+    \[
+    \int_\gamma f = F(0) = F(1) = \int_\mu f
+    \]
+
+    by \autoref{proposition:zero-derivative-constant}. 
+\end{proof}
+\begin{proof}[Proof of general case. ]
+    Let $\Gamma \in C([0, 1] \times [a, b]; \complex)$ be a homotopy of loops from $\gamma$ to $\mu$. By augmenting $\Gamma$ and using \autoref{lemma:rectifiable-piecewise-linear}, assume without loss of generality that:
+    \begin{enumerate}[label=(\alph*)]
+        \item $\mu$, $\gamma$ are piecewise linear. 
+    \end{enumerate}
+    
+    Furthermore, by passing through a reparametrisation, assume without loss of generality that:
+    \begin{enumerate}[label=(\alph*),start=1]
+        \item For each $t \in [0, \eps)$, $\Gamma(t, \cdot) = \gamma$.
+        \item For each $t \in (1 - \eps, 1]$, $\Gamma(t, \cdot) = \mu$.
+        \item For each $t \in [0, 1]$, $\Gamma$ is constant on $\bracs{t} \times ([a, a + \eps] \cup [b - \eps, b])$.
+    \end{enumerate}
+
+    Extend $\Gamma$ to $[0, 1] \times \real$ by
+    \[
+    \Gamma_0: \real^2 \to \complex \quad (t, s) \mapsto \begin{cases}
+        \Gamma(t, s) &t \in k(b-a) + [a, b], k \in \integer \\
+    \end{cases}
+    \]
+
+    then extend $\Gamma_0$ to $\real^2$ by
+    \[
+    \ol \Gamma: \real^2 \to \complex \quad (t, s) \mapsto \begin{cases}
+        \Gamma(t, s) &t \in [0, 1] \\
+        \Gamma(1, s) &t \ge 1 \\
+        \Gamma(0, s) &t \le 0
+    \end{cases}
+    \]
+
+    Let $\varphi \in C_c^\infty(\real^2; \real)$ with $\int_{\real^2} \varphi = 1$. For each $\delta \ge 0$, let
+    \[
+    \Gamma_\delta: [0, 1] \times [a, b] \to \complex \quad (t, s) \mapsto \frac{1}{\delta^2}\int_{\real^2} \Gamma(y) \varphi\paren{\frac{(t, s) - y}{\delta}}dy
+    \]
+
+    Since for each $k \in \integer$ and $(t, s) \in \real^2$, $\Gamma(t, s + k(b - a)) = \Gamma(t, s)$, $\Gamma_\delta(t, a) = \Gamma_\delta(t, b)$ for all $t \in [0, 1]$. Therefore $\Gamma_\delta$ is a homotopy of loops. Since $\Gamma$ is continuous, $\Gamma([0, 1] \times [a, b])$ is compact, so $\Gamma_\delta$ lies in $U$ for sufficiently small 
+
+    By assumptions (b) and (c), for sufficiently small $\delta$, there exists $\psi \in C_c^\infty(\real; \real)$ with $\int_{\real} \psi = 1$ such that
+    \[
+    \Gamma_\delta(0, s) = \frac{1}{\delta}\int_{\real^2} \Gamma(0, y) \psi\paren{\frac{s - y}{\delta}}dy
+    \]
+
+    and
+    \[
+    \Gamma_\delta(1, s) = \frac{1}{\delta}\int_{\real^2} \Gamma(1, y) \psi\paren{\frac{s - y}{\delta}}dy
+    \]
+
+    By assumption (a), (d), and \autoref{lemma:rectifiable-smooth}, 
+    \[
+    \int_\gamma f = \lim_{\delta \downto 0} \int_{\Gamma_\delta(0, \cdot)}f = \lim_{\delta \downto 0} \int_{\Gamma_\delta(1, \cdot)}f = \int_\mu f 
+    \]
+\end{proof}
+
+\begin{definition}
+\label{definition:winding-number-1}
+    Let $U \subset \complex$, $z_0 \in U$, and $r > 0$ such that $\ol{B(z_0, r)} \subset U$, then the path
+    \[
+    \omega_{z_0, r}: [0, 2\pi] \to U \quad \theta \mapsto a + re^{i\theta}
+    \]
+
+    is the \textit{standard path of winding number $1$} at $a$ with radius $r$. 
+\end{definition}
+
+
+\begin{theorem}[Cauchy's Integral Formula]
+\label{theorem:cauchy-formula}
+    Let $E$ be a complete separated locally convex space over $\complex$, $U \subset \complex$ be open, $z_0 \in U$, $r > 0$ such that $\ol{B(z_0, r)} \subset U$, $\gamma \in C([a, b]; \complex)$ be a closed, rectifiable path homotopic to $\omega_{z_0, r}$ on $U \setminus \bracs{z_0}$, and $f \in C^1(U; E)$, then
+    \begin{enumerate}
+        \item $\int_\gamma f = 0$.
+        \item $f(z) = \frac{1}{2\pi i}\int_\gamma \frac{f(z)}{z - z_0}dz$. 
+    \end{enumerate}
+
+    More over, for any $g \in C(U; E)$ that satisfies (2) for all $z_0 \in U$, $r > 0$ with $\ol{B(z_0, r)} \subset U$, closed rectifiable curve $\gamma \in C([a, b]; \complex)$ homotopic to $\omega_{z_0, r}$ on $U \setminus \bracs{z_0}$, 
+    \begin{enumerate}[start=2]
+        \item $g \in C^\infty(U; E)$, where for each $k \in \natz$, 
+        \[
+        D^kg(z_0) = \frac{k!}{2\pi i}\int_{\gamma} \frac{g(z)}{(z - z_0)^{k+1}}dz
+        \]
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    By \autoref{theorem:cauchy-homotopy} and the \hyperref[change of variables formula]{theorem:rs-change-of-variables}, for any $g \in C^1(U \setminus \bracs{z_0}; E)$, 
+    \[
+    \int_\gamma g = \lim_{s \downto 0} \int_{\omega_{z_0, s}} g = \int_0^{2\pi} 
+    = \lim_{s \downto 0}\frac{s}{2\pi} \int_{0}^{2\pi} g \circ \omega_{z_0, s}(\theta) e^{i\theta} d\theta
+    \]
+
+    (1): Since $f \in C(U; E)$, $f$ is bounded on $\ol{B(z_0, r)}$, so for any $s \in (0, r)$,
+    \[
+    \frac{s}{2\pi} \int_{0}^{2\pi} f \circ \omega_{z_0, s}(\theta) e^{i\theta} d\theta \in s\ol{\text{Conv}}(f(\ol{B(z_0, r)}))
+    \]
+
+    As $E$ is locally convex, 
+    \[
+    \int_\gamma g = \lim_{s \downto 0} \int_{\omega_{z_0, s}} g = 0
+    \]
+
+    (2): Since $f \in C(U; E)$, 
+    \begin{align*}
+        \frac{1}{2\pi i}\int_{\gamma} \frac{f(z)}{z - z_0}dz &= \lim_{s \downto 0}\frac{s}{2\pi} \int_{0}^{2\pi} \frac{f \circ \omega_{z_0, s}(\theta)}{\omega_{z_0, s}(\theta) - z_0} e^{i\theta} d\theta \\
+        &= \lim_{s \downto 0}\frac{1}{2\pi}\int_0^{2\pi} f \circ \omega_{z_0, s}(\theta) d\theta = f(z_0)
+    \end{align*}
+
+    (3): Suppose inductively that (3) holds for $k \in \natz$. For sufficiently small $h \in \complex$, 
+    \[
+    \frac{D^kg(z_0 + h) -D^kg(z_0)}{h} = \frac{k!}{2\pi ih} \int_\gamma \frac{g(z)}{(z - z_0-h)^{k+1}} - \frac{g(z)}{(z- z_0)^{k+1}}dz
+    \]
+
+    By \autoref{proposition:difference-quotient-compact}, 
+    \[
+    \lim_{h \to 0}\frac{D^kg(z_0 + h) -D^kg(z_0)}{h} = \frac{(k+1)!}{2\pi i} \int_\gamma \frac{g(z)}{(z - z_0)^{k+2}} dz
+    \]
+
+    Therefore $g \in C^{k+1}(U; E)$ with
+    \[
+    D^{k+1}g(z_0) = \frac{(k+1)!}{2\pi i} \int_\gamma \frac{g(z)}{(z - z_0)^{k+2}} dz
+    \]    
+\end{proof}
+
+\begin{corollary}[Cauchy's Estimate]
+\label{corollary:cauchy-estimate}
+    Let $E$ be a complete separated locally convex space over $\complex$, $U \subset \complex$ be open, $z_0 \in U$, $r > 0$ such that $\ol{B(z_0, r)} \subset U$, then for any $k \in \natz$ and continuous seminorm $[\cdot]_E: E \to [0, \infty)$, 
+    \[
+    [D^kf(z_0)]_E \le \frac{k!}{r^k} \sup_{z \in \ol{B(z_0, r)}}[f(z)]_E
+    \]
+\end{corollary}
+\begin{proof}
+    By \autoref[Cauchy's Integral Formula]{theorem:cauchy-formula} and \autoref{proposition:rs-bound}, 
+    \begin{align*}
+        D^kf(z_0) &= \frac{k!}{2\pi i}\int_{\omega_{z_0, r}} \frac{f(z)}{(z - z_0)^{k+1}}dz \\
+        [D^kf(z_0)]_E &\le \frac{k!}{2\pi i}\int_0^{2\pi}\frac{[f(z)]_E}{|z - z_0|^{k+1}}dz \\
+        &= \frac{k!}{2\pi i}\int_0^{2\pi}\frac{[f(z)]_E}{r^{k+1}}dz \le \frac{k!}{r^k} \sup_{z \in \ol{B(z_0, r)}}[f(z)]_E
+    \end{align*}
+\end{proof}
+
+
+\begin{definition}[Complex Analytic]
+\label{definition:complex-analytic}
+    Let $E$ be a complete separated locally convex space over $\complex$, $U \subset \complex$ be open, and $f \in C(U; E)$, then the following are equivalent:
+    \begin{enumerate}
+        \item (\textbf{Complex Differentiability}) $f \in C^1(U; E)$. 
+        \item (\textbf{Cauchy-Riemann Equations}) Under the identification of $C = \real^2$, $\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \in C(U; E)$ and
+        \[
+        \frac{\partial f}{\partial x} = i\frac{\partial f}{\partial y}
+        \]
+        \item (\textbf{Cauchy's Integral Formula}) For each $z_0 \in U$, $r > 0$ such that $\ol{B(z_0, r)} \subset U$, and closed rectifiable path $\gamma \in C([a, b]; U)$ homotopic to $\omega_{z_0, r}$ on $U \setminus \bracs{z_0}$, 
+        \[
+        f(z_0) = \frac{1}{2\pi i} \int_\gamma \frac{f(z)}{z - z_0}dz
+        \]
+        \item (\textbf{Analyticity}) For each $z_0 \in U$ and $r > 0$ such that $\ol{B(z_0, r)} \subset U$, there exists $\seq{a_n} \subset E$ such that $f$ may be expressed as a power series
+        \[
+        f(z) = \sum_{n = 0}^\infty a_n(z - z_0)^n
+        \]
+
+        with radius of convergence at least $r$.
+        \item (\textbf{Weak Holomorphy}) For each $\phi \in E^*$, $\phi \circ f$ satisfies the above. 
+    \end{enumerate}
+
+    If the above holds, then $f$ is \textbf{complex analytic}. 
+\end{definition}
+\begin{proof}
+    (1) $\Leftrightarrow$ (2): \autoref{lemma:complex-analytic}. 
+
+    (1) + (2) $\Rightarrow$ (3): See \hyperref[Cauchy's Integral Formula]{theorem:cauchy-formula}. 
+
+    (3) $\Rightarrow$ (4): By \hyperref[Cauchy's Integral Formula]{theorem:cauchy-formula}, $f \in C^\infty(U; E)$ where for each $k \in \natz$, 
+    \[
+        D^kf(z_0) = \frac{k!}{2\pi i}\int_{\gamma} \frac{f(z)}{(z - z_0)^{k+1}}dz
+    \]
+    
+    Let
+    \[
+    g(z) = \sum_{k = 0}^\infty \frac{1}{k!} D^kf(z_0)(z - z_0)^n
+    \]
+
+    then by \hyperref[Cauchy's Estimate]{corollary:cauchy-estimate}, for any $k \in \natz$ and continuous seminorm $[\cdot]_E: E \to [0, \infty)$, 
+    \[
+        [D^kf(z_0)]_E \le \frac{k!}{r^k} \sup_{z \in \ol{B(z_0, r)}}[f(z)]_E = \frac{Ck!}{r^k}
+    \]
+
+    Thus $[D^kf(z_0)/k!]_E \le C/r^k$ for all $k \in \natz$, and the radius of convergence of $g$ is at least $r$. 
+    
+    Let $z \in B(z_0, r/2)$, $s = |z - z_0|$, and $n \in \natp$, then by \hyperref[Taylor's Formula]{theorem:taylor-lagrange} and \hyperref[Cauchy's Estimate]{corollary:cauchy-estimate}, 
+    \begin{align*}
+        \braks{f(z) - \sum_{k = 0}^n \frac{1}{k!} D^kf(z_0)(z - z_0)^n}_E &\le s^{n+1} \cdot \sup_{z' \in \ol{B(z_0, s)}} [D^{n+1}f(z')]_E \\
+        &\le \frac{Cs^{n+1}}{(r-s)^{n+1}}
+    \end{align*}
+
+    which tends to $0$ as $n \to \infty$. 
+
+    (4) $\Rightarrow$ (1): By \autoref{theorem:termwise-differentiation}. 
+
+    (5) $\Rightarrow$ (3): By the equivalence of the prior points, for any $\phi \in E^*$, $\phi \circ f$ satisfies (3). By the \hyperref[Hahn-Banach Theorem]{proposition:hahn-banach-utility}, $f$ also satisfies (3). 
+\end{proof}
+
+

src/dg/complex/index.tex

@@ -0,0 +1,8 @@
+\chapter{Complex Analysis}
+\label{chap:complex-analysis}
+
+
+\input{./derivative.tex}
+\input{./log.tex}
+
+

src/dg/complex/log.tex

@@ -0,0 +1,35 @@
+\section{The Complex Logarithm}
+\label{section:complex-log}
+
+\begin{definition}[Branch of Logarithm]
+\label{definition:branch-of-log}
+    Let $U \subset \complex$ be a connected open set with $0 \not\in U$ and $f \in C(U; \complex)$, then $f$ is a \textbf{branch of the logarithm} if for every $z \in U$, $z = \exp(f(z))$. 
+\end{definition}
+
+\begin{lemma}
+\label{lemma:branch-of-log-shift}
+    Let $U \subset \complex$ be a connected open set with $0 \not\in U$, and $f, g \in C(U; \complex)$ be two branches of the logarithm, then there exists $k \in \integer$ such that $f - g = 2\pi k i$. 
+\end{lemma}
+\begin{proof}[Proof, {{\cite[Proposition 2.19]{ConwayComplex}}}. ]
+    For each $x \in U$, there exists $k \in \integer$ such that $f(x) - g(x) = 2\pi k i$. Thus $f - g \in C(U; 2\pi i\integer)$. Since $U$ is connected, $(f - g)(U)$ must be a singleton. Therefore there exists $k \in \integer$ such that $f - g = 2\pi k i$. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:branch-of-log-analytic}
+    Let $U \subset \complex$ be a connected open set with $0 \not\in U$, and $f \in C(U; \complex)$ be a branch of the logartihm, then $f$ is analytic. 
+\end{proposition}
+\begin{proof}
+    By the \autoref{theorem:inverse-function-theorem}. 
+\end{proof}
+
+
+\begin{definition}[Principal Logarithm]
+\label{definition:principal-logarithm}
+    Let $U = \complex \setminus \bracs{z \in \real|z \le 0}$, then there exists a unique mapping $\ell: U \to \complex$ such that:
+    \begin{enumerate}
+        \item $\ell$ is a branch of the complex logarithm. 
+        \item For each $re^{i\theta} \in U$, $\ell(r^{i\theta}) = \ln r + i\theta$. 
+    \end{enumerate}
+
+    The function $\ell$ is the \textbf{principal logarithm} on $U$. 
+\end{definition}

src/dg/derivative/euclid.tex

@@ -0,0 +1,57 @@
+\section{Derivatives on $\mathbb R^n$}
+\label{section:derivatives-euclidean}
+
+\begin{proposition}
+\label{proposition:derivative-sets-real}
+    Let $E$ be a separated topological vector space and $\sigma \subset \mathfrak{B}(\real)$ be a covering ideal, then
+    \begin{enumerate}
+        \item $\mathcal{R}_{\sigma}(\real; E) = \mathcal{R}_{\mathfrak{B}(\real)}(\real; E)$. Hence, all forms of $\sigma$-differentiability on $\real$ are equivalent.
+        \item For any $U \subset \real$ open, $f: U \to E$, and $x_0 \in U$, $f$ is differentiable at $x_0$ if and only if
+        \[
+        \lim_{t \to 0}\frac{f(x + t) - f(x)}{t}
+        \]
+
+        exists. In which case, the above limit is identified with the derivative of $f$ at $0$.
+        \item For any $U \subset \real$ open, $f: U \to E$, and $x_0 \in U$, if $f$ is differentiable at $x_0$, then $f$ is continuous at $x_0$.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): Let $r \in \mathcal{R}_\sigma(\real; E)$. For any $R > 0$ and $U \in \cn_E(0)$, there exists $\delta > 0$ such that $t^{-1}r(tR), t^{-1}r(-tR) \in U$ for all $t \in (0, \delta)$. Thus $t^{-1}r(tB(0, R)) \subset U$, and $r \in \mathcal{R}_{\mathfrak{B}(\real)}(\real; E)$.
+
+    (2): Suppose that $f$ is differentiable at $x_0$, then there exists $r \in \mathcal{R}_\sigma$ such that for any $t \in \real$ with $x_0 + t \in U$,
+    \begin{align*}
+        f(x_0 + t) - f(x_0) &= Df(x_0)(t) + r(t) \\
+        \frac{f(x_0 + t) - f(x_0)}{t} &= Df(x_0)(1) + t^{-1}r(t) \\
+        \lim_{t \to 0}\frac{f(x_0 + t) - f(x_0)}{t} &= Df(x_0)(1)
+    \end{align*}
+    Now suppose that $v = \lim_{t \to 0}\frac{f(x + t) - f(x)}{t}$ exists. Let $T: \real \to E$ be defined by $t \mapsto tv$, then
+    \[
+    \lim_{t \to 0}\frac{f(x_0 + t) - f(x_0) - Tt}{t} = \lim_{t \to 0}\frac{f(x_0 + t) - f(x_0)}{t} - v = 0
+    \]
+
+    and $Df(x_0) = T$.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:difference-quotient-compact}
+    Let $E$ be a separated locally convex space over $K \in \RC$, $U \subset K$ be open, $Y$ be a Hausdorff space, and $f: U \times Y \to E$. If $f$ is differentiable in the first variable and $\frac{df}{dx} \in C(U \times Y; E)$, then
+    \[
+    \frac{f(x + h, y) - f(x, y)}{h} \to \frac{df}{dx}(x, y)
+    \]
+
+    as $h \to 0$, uniformly on compact sets. 
+\end{proposition}
+\begin{proof}
+    Let $A \subset U$ and $B \subset Y$ be compact, then by the \hyperref[Mean Value Theorem]{theorem:mean-value-theorem-line}, for any $(x, y) \in A \times B$ and $h \in \real$ with $x + h$, 
+    \begin{align*}
+        &\frac{f(x + h, y) - f(x, y)}{h} - \frac{df}{dx}(x, y) \\
+        &\in \overline{\text{Conv}}\bracs{\frac{df}{dx}(x + k, y) - \frac{df}{dx}(x, y) \bigg | k \in B_K(0, |h|)}
+    \end{align*}
+
+    Let $\eps > 0$ such that $A + B_K(0, |\eps|) \subset U$, then since  $\frac{df}{dx} \in C(U \times Y; E)$, $\frac{df}{dx}|_{(A + B_K(0, |\eps|)) \times B}$ is uniformly continuous\footnote{$K$ is a compact Hausdorff space, which comes with a \hyperref[unique uniform structure]{proposition:compact-uniform-structure}. }. Since $E$ is locally convex,
+    \[
+    \frac{f(x + h, y) - f(x, y)}{h} - \frac{df}{dx}(x, y) \to 0
+    \]
+
+    uniformly on $A \times B$.
+\end{proof}

src/dg/derivative/higher.tex

@@ -1,36 +1,72 @@
 \section{Higher Derivatives}
 \label{section:higher-derivatives}
 
-\begin{definition}[$n$-Fold Differentiability]
-\label{definition:n-differentiable-sets}
-    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset B(E)$ be an upward-directed family that contains all finite sets, $\mathcal{H} \subset B_\sigma(E; F)$ be a subspace, and $\mathcal{R}_\sigma = \mathcal{R}_\sigma(E; F)$.
 
-    Let $U \subset E$ be open, $f: U \to F$, $x_0 \in U$, and $n > 1$, then $f$ is \textbf{$n$-fold $\sigma$-differentiable at $x_0$} if
+\begin{definition}[Codomain of Derivatives]
-    \begin{enumerate}
+\label{definition:higher-derivatives-codomain}
-        \item There exists $V \in \cn_E(x_0)$ such that $f$ is $(n-1)$-fold differentiable on $V$.
+    Let $E, F$ be TVSs over $K \in \RC$, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, and $L^{(0)}_\sigma(E; F) = F$. For each $n \in \natp$, inductively define
-        \item The derivative $D_\sigma^{n-1}f: U \to B^{n-1}_\sigma(E; F)$ is derivative at $x_0$.
-    \end{enumerate}
-    In which case, $D_\sigma(D_\sigma^{n-1}f)(x_0) \in L(E; B^{n-1}_\sigma(E; F))$ is the \textbf{$n$-fold $\sigma$-derivative of $f$ at $x_0$}.
-
-    The mapping $f: U \to F$ is \textbf{$n$-fold $\sigma$-differentiable on $U$} if it is $n$-fold $\sigma$-differentiable at every point in $U$. Under the identification $B_\sigma(E; B^{n-1}_\sigma(E; F)) = B_\sigma^{n}(E; F)$ given by \autoref{proposition:multilinear-identify},
     \[
-    D_\sigma^{n}f: U \to B^{n-1}_\sigma(E; F)
+    L^{(n)}_\sigma(E; F) = L(E; L^{(n-1)}_\sigma(E; F)) \subset B_\sigma^n(E; F)
     \]
 
-    is the \textbf{$n$-fold $\sigma$-derivative of $f$ at $x_0$}.
+    and equip it with the $\sigma$-uniform topology, then under the identification
+    \[
+    I: L^{(n)}_\sigma(E; F) \to B_\sigma^n(E; F) \quad I\lambda(x_1, \cdots, x_n) = \lambda(x_1)\cdots(x_n)
+    \]
+
+    the space $L^{(n)}_\sigma(E; F)$ is a subspace of $B_\sigma^n(E; F)$. 
+\end{definition}
+\begin{proof}
+    By \autoref{proposition:multilinear-identify}. 
+\end{proof}
+
+
+
+\begin{definition}[$n$-Fold Differentiability]
+\label{definition:n-differentiable-sets}
+    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, and $\mathcal{R}_\sigma = \mathcal{R}_\sigma(E; F)$.
+
+    Let $U \subset E$ be open, $f: U \to F$, $x_0 \in U$, and $n > 1$, then $f$ is \textbf{$n$-fold $\tilde \sigma$-differentiable at $x_0$} if
+    \begin{enumerate}
+        \item There exists $V \in \cn_E(x_0)$ such that $f$ is $(n-1)$-fold $\tilde \sigma$-differentiable on $V$.
+        \item The derivative $D_\sigma^{n-1}f: U \to B^{(n-1)}_\sigma(E; F)$ is $\tilde \sigma$-differentiable at $x_0$.
+    \end{enumerate}
+
+    In which case, $D_\sigma(D_\sigma^{n-1}f)(x_0) = D_\sigma^{n}f(x_0)$ is the \textbf{$n$-fold $\tilde \sigma$-derivative of $f$ at $x_0$}.
+
+    If $f: U \to F$ is $n$-fold $\tilde \sigma$-differentiable at every point in $U$, then $f$ is \textbf{$n$-fold $\tilde \sigma$-differentiable on $U$}. Under the \hyperref[identification]{proposition:multilinear-identify} $B_\sigma(E; B_\sigma^{n}(E; F)) = B_\sigma^{(n)}(E; F)$, the mapping
+    \[
+    D_\sigma^{n}f: U \to B^{(n-1)}_\sigma(E; F)
+    \]
+
+    is the \textbf{$n$-fold $\tilde \sigma$-derivative of $f$}.
+
+    If for each $1 \le k \le n$, $D_\sigma^{k}f$ takes value in $L^{(k)}_\sigma(E; F)$, then $f$ is \textbf{$n$-fold $\sigma$-differentiable}, and $D_\sigma^{n}f$ is the \textbf{$n$-fold $\sigma$-derivative of $f$}.
 \end{definition}
 
-\begin{theorem}[{{\cite[Theorem 5.1.1]{Cartan}}}]
+
+\begin{definition}[Space of Differentiable Functions]
+\label{definition:differentiable-space}
+    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, $U \subset E$ be open, and $n \in \natp$, then $D_\sigma^k(U; F)$/$\tilde D_\sigma^k(U; F)$ is the \textbf{space of $n$-fold $\sigma$/$\tilde \sigma$-differentiable functions} from $U$ to $F$. 
+\end{definition}
+
+\begin{definition}[Space of Continuously Differentiable Functions]
+\label{definition:continuously-differentiable-space}
+    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, $U \subset E$ be open, and $n \in \natp$, then $C_\sigma^k(U; F)$/$\tilde C_\sigma^k(U; F)$ is the \textbf{space of $n$-fold continuously $\sigma$/$\tilde \sigma$-differentiable functions} from $U$ to $F$. 
+\end{definition}
+
+
+\begin{theorem}[Symmetry of Higher Derivatives]
 \label{theorem:derivative-symmetric-frechet}
     Let $E, F$ be Banach spaces, $U \subset E$ be open, $n \in \natp$, and $f: U \to F$ be a function $n$-times Fréchet-differentiable at $x \in U$, then $D^nf(x) \in L^n(E; F)$ is symmetric.
 \end{theorem}
-\begin{proof}
+\begin{proof}[Proof {{\cite[Theorem 5.1.1]{Cartan}}}. ]
     First suppose that $n = 2$. Let $r > 0$ such that $B(x, 2r) \subset U$, and define $A: B_E(0, r) \times B_E(0, r) \to F$ by
     \[
     A(h, k) = f(x + h + k) - f(x + h) - f(x + k) + f(x)
     \]
 
-    then there exists $r_1 \in \mathcal{R}_{B(E)}$ such that
+    then there exists $r_1 \in \mathcal{R}_{\mathfrak{B}(E)}$ such that
     \begin{align*}
         A(h, k) &= Df(x + h)(k) + Df(x)(k) \\
         &+ [f(x + h + k) - f(x + h) - Df(x + h)(k)] \\
@@ -54,7 +90,7 @@
     \end{align*}
     
 
-    Now, there exists $r_2, r_3 \in \mathcal{R}_{B(E)}$ such that for any $k \in B(0, r)$,
+    Now, there exists $r_2, r_3 \in \mathcal{R}_{\mathfrak{B}(E)}$ such that for any $k \in B(0, r)$,
     \begin{align*}
     DB_h(k) &= Df(x + h + k) - Df(x + k) - Df(x + h) + Df(x) \\
     &= D^2f(x)(h + k) + Df(x) - D^2f(x)(h) \\
@@ -85,57 +121,81 @@
     Since any element $\sigma \in S_n$ that does not fix $x_1$ is the composition of the transposition $(12)$ and an element that fixes $x_1$, $Df(x)$ is symmetric.
 \end{proof}
 
-\begin{theorem}[{{\cite[Proposition 4.5.14]{Bogachev}}}]
+\begin{theorem}[Symmetry of Higher Derivatives]
 \label{theorem:derivative-symmetric}
-    Let $E$ be a topological vector space over $K \in \RC$, $\sigma \subset B(E)$ be an upward-directed system that includes all bounded sets contained in finite-dimensional spaces, $F$ be a separated locally convex space over $K$, $U \subset E$ be open, and $f: E \to F$ be $n$-fold $\sigma$-differentiable at $x_0 \in U$, then $D_\sigma^nf(x_0) \in B_\sigma^n(E; F)$ is symmetric.
+    Let $E$ be a topological vector space over $K \in \RC$, $\sigma \subset \mathfrak{B}(E)$ be an ideal that includes all bounded sets contained in finite-dimensional spaces, $F$ be a separated locally convex space over $K$, $U \subset E$ be open, and $f: E \to F$ be $n$-fold $\tilde \sigma$-differentiable at $x_0 \in U$, then $D_\sigma^nf(x_0) \in L_\sigma^{(n)}(E; F)$ is symmetric.
 \end{theorem}
-\begin{proof}
+\begin{proof}[Proof {{\cite[Proposition 4.5.14]{Bogachev}}}. ]
     Let $\seqf{h_j} \subset E$, $E_0$ be the subspace generated by $\seqf{h_j}$, and $g = f|_{E_0 \cap U}: E_0 \cap U \to F$. Since $\sigma$ includes all bounded sets contained in finite-dimensional spaces, for any $\phi \in F^*$, the mapping $\phi \circ g: E_0 \cap U \to K$ is $n$-times Fréchet-differentiable, with
     \[
-    D_{B(E_0)}^n(\phi \circ g)(x_0) = \phi \circ D_\sigma^n g(x_0)
+    D_{\mathfrak{B}(E_0)}^n(\phi \circ g)(x_0) = \phi \circ D_\sigma^n g(x_0)
     \]
 
-    by the \hyperref[chain rule]{proposition:chain-rule-sets-conditions}. By \autoref{theorem:derivative-symmetric-frechet}, $\phi \circ D_\sigma^n g(x_0) \in L^n(E_0; K)$ is symmetric. As this holds for any $\seqf{h_j} \subset E$ and $\phi \in F^*$, $D_{\sigma}^n g(x_0) \in B_\sigma^n(E; F)$ is symmetric by the \hyperref[Hahn-Banach theorem]{proposition:hahn-banach-utility}.
+    by the \hyperref[chain rule]{proposition:chain-rule-sets-conditions}. By \autoref{theorem:derivative-symmetric-frechet}, $\phi \circ D_\sigma^n g(x_0) \in L^n(E_0; K)$ is symmetric. As this holds for any $\seqf{h_j} \subset E$ and $\phi \in F^*$, $D_{\sigma}^n g(x_0) \in B_\sigma^{(n)}(E; F)$ is symmetric by the \hyperref[Hahn-Banach theorem]{proposition:hahn-banach-utility}.
 \end{proof}
 
-\begin{proposition}[Power Rule]
+\begin{theorem}[Power Rule]
-\label{proposition:multilinear-derivative}
+\label{theorem:power-rule}
-    Let $E$ be a topological vector space, $\sigma \subset B(E)$ be an upward-directed family that includes bounded sets contained in finite-dimensional subspaces, $F$ be a Hausdorff locally convex space, and
+    Let $E$ be a topological vector space, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, $F$ be a separated locally convex space, $T \in B^n_\sigma(E; F)$.
+
+    For each $1 \le m \le n$, let $\text{Inj}([m]; [n])$ be the set of injective mappings from $[m]$ to $[n]$. For any $\phi \in \text{Inj}([m]; [n])$, denote $\phi^c \in \text{Inj}([n-m]; [n] \setminus \phi([m]))$ as the unique increasing injective map. For each $h \in E^{n-m}$, $k \in E^m$, and $1 \le j \le n$, write
     \[
-    T \in \underbrace{L(E; L(E; \cdots L(E; F) \cdots ))}_{n \text{ times}} \subset B^n(E; F)
+    (h, k)_\phi = \begin{cases}
+        k_{\phi^{-1}(j)} &j \in \phi([m]) \\
+        h_{(\phi^c)^{-1}(j)} &j \not\in \phi([m])
+    \end{cases}
     \]
 
-    be symmetric. For any $x \in E$ and $1 \le k \le n$, let $x^{(k)}$ denote the tuple of $x$ repeated $k$ times, then:
+    For any $x \in E$, denote $x^{(m)}$ as the tuple of $x$ repeated $m$ times, then the mapping $f: E \to F$ defined by $x \mapsto T(x^{(n)})$ is infinitely $\tilde\sigma$-differentiable on $E$, where
     \begin{enumerate}
-        \item The mapping $f: E \to F \quad x \mapsto T(x^{(n)})$ is infinitely $\sigma$-differentiable on $E$.
+        \item For each $1 \le m \le n$, $x \in E$, and $h \in E^m$, 
-        \item For each $1 \le k \le n$ and $x, h \in E$,
         \[
-        Df(x)(h_1, \cdots, h_k) = \frac{n!}{(n-k)!} T(x^{(n-k)}, h_1, \cdots, h_k)
+        D^{m}_\sigma f(x)(h) = \sum_{\phi \in \text{Inj}([m]; [n])}T((x^{(n-m)}, h)_\phi]
         \]
 
-        In particular, $D^kf = n! \cdot T$.
+        In particular, 
-        \item For each $k > n$ and $x \in E$, $Df(x) = 0$.
+        \[
+        D^{n}_\sigma f(x)(h) = \sum_{\phi \in S_n}T(h_{\phi(1)}, \cdots, h_{\phi(n)})
+        \]
+
+        \item For each $m > n$ and $x \in E$, $D^m_\sigma f(x) = 0$.
     \end{enumerate}
-\end{proposition}
+
+    Notably, if $T \in L^{(n)}(E; F)$ is symmetric or $T \in L^n(E; F)$, then $T$ is infinitely $\sigma$-differentiable on $E$. 
+\end{theorem}
 \begin{proof}
-    Suppose inductively that (2) holds for $0 \le k \le n$. Let $G =  B^{k}_\sigma(E; F)$, then $D^k_\sigma f \in B^{n-k}_\sigma(E; G)$ under the identification $B^n_\sigma(E; F) = B^{n-k}_\sigma(E; B^k_\sigma(E; F))$ in \autoref{proposition:multilinear-identify}. By \autoref{theorem:derivative-symmetric}, $D^k_\sigma f$ is also symmetric, so using the Binomial formula,
+    (1): Let $0 \le m \le n - 1$ and suppose inductively that (1) holds for $m$. For each $x, h \in E$, $S \subset [n-m]$, and $1 \le j \le n - m$,
+    \[
+    [(x, h)_S]_j = \begin{cases}
+        h &j \in S \\
+        x &j \not\in S
+    \end{cases}
+    \]
+    
+    By the Binomial formula, for each $h \in E$ and $k \in E^{m}$,
     \begin{align*}
-        D^k_\sigma f(x + h) &= \sum_{r = 0}^{n-k}{n - k \choose r}D^k_\sigma f(x^{(n-k-r)}, h^{(r)}) \\
+        D^{m}_\sigma f(x + h)(k) &= \sum_{\phi \in \text{Inj}([m]; [n])}T[((x+h)^{(n-m)}, k)_\phi] \\
-        &= f(x) + (n-k)D^k_\sigma f(x^{(n-k-1)}, h) \\
+        &= \sum_{\phi \in \text{Inj}([m]; [n])}\sum_{S \subset [n-m]}T[((x, h)_S, k)_\phi]
-        &+ \underbrace{\sum_{r = 2}^{n-k}{n - k \choose r}D^k_\sigma f(x^{(n-k-r)}, h^{(r)})}_{r(h)}
     \end{align*}
 
-    For each $k \ge 2$, let $A \in \sigma$ and $U \in \cn_F(0)$, then since $D^k_\sigma f \in B^{n-k}_\sigma(E; F)$, there exists $t > 0$ such that
+    For $\ell \ge 2$, maps in $B_\sigma^{\ell}(E; F)$ are $\sigma$-small, so
     \[
-    \frac{D^k_\sigma f(x^{(n-k)}, (sA)^{(k)})}{t} = s^{k-1}D^k_\sigma f(x^{(n-k)}, A^{(k)}) \subset U
+    r(h) = \sum_{\phi \in \text{Inj}([m]; [n])}\sum_{\substack{S \subset [n-m] \\ |S| \ge 2}}T[((x, h)_S, k)_\phi]
     \]
 
-    for all $s \in (0, t)$. Hence $r \in \mathcal{R}_\sigma(E; G)$, and
+    is $\sigma$-small. Hence
+    \begin{align*}
+        &D^{(m)}_\sigma f(x + h)(k) - D_\sigma^{(m)}f(x)(k) \\
+        &= \sum_{\phi \in \text{Inj}([m]; [n])}\sum_{j = 1}^{n-m}T[((x, h)_{\bracs{j}}, k)_\phi] + r(h) \\
+        &= \sum_{\phi \in \text{Inj}([m+1]; [n])}T[(x^{(n-m-1)}, (h, k))_\phi] + r(h)
+    \end{align*}
+
+    and for any $h \in E^{m+1}$,
     \[
-    D^{k+1}_\sigma f(x + h) = f(x) + \frac{n!}{(n-k-1)!}T(x^{(n-k-1)}, h_1, \cdots, h_{k+1}) + r(h)
+    D_\sigma^{(m+1)}f(x)(h) = \sum_{\phi \in \text{Inj}([m+1]; [n])}T[(x^{(n-m-1)}, h)_\phi]
     \]
 
-    by the inductive hypothesis.
+    (2): By (1), $D^n_\sigma f$ is constant. 
-
-    (3): Since $D^n_\sigma f$ is constant, $D^k_\sigma f = 0$ for all $k > n$.
 \end{proof}
+
+

src/dg/derivative/index.tex

@@ -6,3 +6,7 @@
 \input{./mvt.tex}
 \input{./higher.tex}
 \input{./taylor.tex}
+\input{./partial.tex}
+\input{./power.tex}
+\input{./inverse.tex}
+\input{./euclid.tex}

src/dg/derivative/inverse.tex

@@ -0,0 +1,52 @@
+\section{Inverse Mappings}
+\label{section:inverse-function-theorem}
+
+\begin{theorem}[Inverse Function Theorem]
+\label{theorem:inverse-function-theorem}
+    Let $E$ be a Banach space, $U \subset E$ be open, $p \ge 1$, $f \in C^p(U; E)$ be $p$-times continuously Fréchet-differentiable, and $x_0 \in U$. If $Df(x_0)$ is an isomorphism, then:
+    \begin{enumerate}
+        \item There exists $V \in \cn_E(x_0)$ such that $f|_V$ is a $C^p$-isomorphism. 
+        \item Let $f^{-1}: f(V) \to V$ be the local inverse of $f$ on $V$, then $Df^{-1}(x_0) = [Df(x_0)]^{-1}$. 
+    \end{enumerate}
+    
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem XIV.1.2]{Lang}}}. ]
+    By translation, assume without loss of generality that $x_0 = f(x_0) = 0$ and $Df(x_0) = Df(0) = I$. 
+
+    \textit{Existence and Uniqueness of Inverse}: Since $f \in C^1$, there exists $r > 0$ such that $\norm{Df(x) - I}_{L(E; E)} < 1/2$ for all $x \in \ol{B_E(0, r)}$. In which case, by \autoref{lemma:neumann-series}, $Df(x)$ is an isomorphism for all $x \in B(0, r)$. Let
+    \[
+    g: \overline{B_E(0, r)} \to E \quad x \mapsto x - f(x)
+    \]
+
+    For any $x, y \in \overline{B_E(0, r)}$, by the \hyperref[Mean Value Theorem]{theorem:mean-value-theorem}, 
+    \[
+    \norm{g(x) - g(y)}_E \le \norm{x}_E \cdot \sup_{y \in \overline{B_E(0, r)}}\norm{Dg(y)}_E \le \frac{\norm{x - y}_E}{2}
+    \]
+    
+    In particular, for any $x \in \overline{B_E(0, r)}$, $\norm{g(x)}_E = \norm{g(x) - g(0)}_E \le \norm{x}_E/2$, so $g: \ol{B_E(0, r)} \to \ol{B_E(0, r/2)}$ is a contraction. 
+
+    For each $y \in B(0, r/2)$, the mapping
+    \[
+    g_y: \overline{B(0, r)} \to \overline{B(0, r)} \quad x \mapsto x - f(x) + y
+    \]
+
+    is also a contraction. By \hyperref[Banach's Fixed Point Theorem]{theorem:banach-fixed-point}, there exists a unique $x \in B(0, r)$ such that $g_y(x) = x$. In which case, $f(x) = y$. Therefore $f$ restricted to $V = f^{-1}(B(0, r))$ is invertible. 
+
+    \textit{Differentiability of Inverse}: Let $f^{-1}: f(V) \to V$ be the local inverse of $f$ on $V$. By assumption, it is sufficient to show that $Df^{-1}(0) = I$ as well. For each $y \in \overline{B(0, r/2)}$, 
+    \begin{align*}
+        \norm{f^{-1}(y) - y}_E &= \norm{f^{-1}(y) - f(f^{-1}(y))}_E \\
+        &= \norm{f^{-1}(y) - f^{-1}(y) - r(f^{-1}(y))}_E = \norm{r(f^{-1}(y))}_E
+    \end{align*}
+
+    where $r(x)/\norm{x}_E \to 0$ as $x \to 0$. In addition, 
+    \begin{align*}
+        \norm{f^{-1}(y)}_E &= \norm{f^{-1}(y) - y + y}_E \\
+        &\le \norm{g(f^{-1}(y))}_E + \norm{y}_E \le 2\norm{y}_E
+    \end{align*}
+
+    so $[f^{-1}(y) - y]/\norm{y}_E \to 0$ as $y \to 0$. Therefore $f^{-1}$ is differentiable at $0$ with $Df^{-1} = I$. 
+
+    \textit{Smoothness of Inverse}: By the above argument, the inverse is differentiable on every point in $B(0, r/2)$, and $Df^{-1}(f(x)) = [Df(x)]^{-1}$ for all $x \in V$. By \autoref{proposition:banach-algebra-inverse}, the inversion map $T \mapsto T^{-1}$ is smooth. Therefore if $Df \in C^{p - 1}$, then $f \in C^{p - 1}$ as well. 
+\end{proof}
+
+

src/dg/derivative/mvt.tex

@@ -18,6 +18,7 @@
         \item[(a)] $f, g$ are right-differentiable on $[a, b] \setminus N$.
         \item[(b)] For every $x \in [a, b] \setminus N$, $D^+f(x) \le D^+g(x)$.
     \end{enumerate}
+
     then for any $x \in [a, b]$, $f(x) - f(a) \le g(x) - g(a)$.
 \end{lemma}
 \begin{proof}
@@ -52,6 +53,7 @@
         \item For each $x \in [a, b] \setminus N$, $D^+f(x) \in D^+g(x)B$.
         \item $g$ is non-decreasing.
     \end{enumerate}
+
     then
     \[
     f(b) - f(a) \in [g(b) - g(a)]B
@@ -103,7 +105,7 @@
 
 \begin{theorem}[Mean Value Theorem]
 \label{theorem:mean-value-theorem}
-    Let $E$ be a topological vector space, $F$ be a separated locally convex space, $V \subset E$ be open and star shaped at $x \in V$, $f: V \to F$ be Gateau-differentiable on $V$, then for any $y \in V$,
+    Let $E$ be a topological vector space over $K \in \RC$, $F$ be a separated locally convex space over $K$, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, $V \subset E$ be open and star shaped at $x \in V$, and $f: V \to F$ be $\tilde \sigma$-differentiable on $V$, then for any $y \in V$,
     \[
     f(y) - f(x) \in \overline{\text{Conv}\bracs{Df(z)(y - x)|z \in (x, y)}}
     \]
@@ -111,18 +113,18 @@
     where $[x, y] = \bracs{(1 - t)x + ty|y \in [0, 1]}$.
 \end{theorem}
 \begin{proof}
-    Let $g: [0, 1] \to F$ be defined by $g(t) = f((1 - t)x + ty)$. Since $f$ is Gateaux-differentiable, $g$ is differentiable by the chain rule \autoref{proposition:chain-rule-sets-conditions} with $Dg(t) = Df((1 - t)x + ty)(y - x)$, and continuous by \autoref{proposition:derivative-sets-real}.
+    Let $g: [0, 1] \to F$ be defined by $g(t) = f((1 - t)x + ty)$, then $g$ is differentiable with $Dg(t) = Df((1 - t)x + ty)(y - x)$, and continuous by \autoref{proposition:derivative-sets-real}.
 
-    By the \hyperref[Mean Value Theorem]{theorem:mean-value-theorem-line},
+    By the \hyperref[Mean Value Theorem]{theorem:mean-value-theorem-line}, $f(y) - f(x) = g(1) - g(0)$ is contained in
     \[
-    f(y) - f(x) = g(1) - g(0) \in \overline{\text{Conv}\bracs{Dg(t)|t \in [0, 1]}} = \overline{\text{Conv}\bracs{Df(z)(y - x)|z \in (x, y)}}
+    \overline{\text{Conv}\bracs{Dg(t)|t \in [0, 1]}} = \overline{\text{Conv}\bracs{Df(z)(y - x)|z \in (x, y)}}
     \]
 
 \end{proof}
 
 \begin{proposition}
 \label{proposition:zero-derivative-constant}
-    Let $E$ be a topological vector space, $F$ be a separated locally convex space, $V \subset E$ be open and connected, $f: V \to F$ be Gateaux-differentiable on $V$ such that $Df(x) = 0$ for all $x \in V$, then $f$ is constant.
+    Let $E$ be a topological vector space over $K \in \RC$, $F$ be a separated locally convex space over $K$, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, $V \subset E$ be open and connected, and $f: V \to F$ be $\tilde \sigma$-differentiable on $V$ with $Df = 0$, then $f$ is constant.
 \end{proposition}
 \begin{proof}
     Let $x \in V$, then for any $U \in \cn(0)$ circled with $U + x \subset V$ and $y \in U + x$,

src/dg/derivative/partial.tex

@@ -0,0 +1,61 @@
+\section{Partial Derivatives}
+\label{section:partial-derivatives}
+
+\begin{definition}[Partial Derivative]
+\label{definition:partial-derivative}
+    Let $E_1, E_2$ be TVSs over $K \in \RC$, $\sigma_1 \subset \mathfrak{B}(E_1)$ and $\sigma_2 \subset \mathfrak{B}(E_2)$ be covering ideals, $F$ be a separated TVS over $K$, $U \subset E_1 \times E_2$ be open, and $f: U \to F$. For each $(x_0, y_0) \in E$, let $f_{x_0}(y) = f(x_0, y)$ and $f_{y_0}(x) = f(x, y_0)$ be the partial maps of $f$. If $f_{x_0}$ is $\tilde \sigma_1$-differentiable for each $x_0$, and $f_{y_0}$ is $\tilde \sigma_2$-differentiable for each $y_0$, then
+    \[
+    D_1f: U \to B_{\sigma_1}(E_1; F) \quad (x, y) \mapsto D_{\sigma_1}f_{x}(y)
+    \]
+
+    and
+    \[
+    D_2f: U \to B_{\sigma_2}(E_2; F) \quad (x, y) \mapsto D_{\sigma_2}f_{y}(x)
+    \]
+
+    are the \textbf{partial derivatives} of $f$.
+\end{definition}
+
+\begin{proposition}
+\label{proposition:partial-total-derivative}
+    Let $E_1, E_2$ be TVSs over $K \in \RC$, $\sigma_1 \subset \mathfrak{B}(E_1)$ and $\sigma_2 \subset \mathfrak{B}(E_2)$ be covering ideals, $F$ be a separated locally convex space over $K$, $U \subset E_1 \times E_2$ be open, $f: U \to F$, and $p \ge 1$, then the following are equivalent:
+    \begin{enumerate}
+        \item $f \in \tilde C_{\sigma_1 \otimes \sigma_2}^p(U; F)$. 
+        \item $D_1 f \in \tilde C_{\sigma_1 \otimes \sigma_2}^{p-1}(U; B_{\sigma_1}(E; F))$ and $D_2 f \in \tilde C_{\sigma_1 \otimes \sigma_2}^{p-1}(U; B_{\sigma_2}(E; F))$
+    \end{enumerate}
+
+    If the above holds, then for any $x \in U$ and $(h_1, h_2) \in E_1 \times E_2$, 
+    \[
+    D_{\sigma_1 \otimes \sigma_2}f(x)(h_1, h_2) = D_1f(x)(h_1) + D_2f(x)(h_2)
+    \]
+    
+\end{proposition}
+\begin{proof}
+    (2) $\Rightarrow$ (1): For each $(x, y) \in U$ and $(h_1, h_2) \in E_1 \times E_2$, 
+    \begin{align*}
+        f(x + h_1, y + h_2) - f(x, y) &= f(x + h_1, y + h_2) - f(x + h_1, y) \\
+        &+ f(x + h_1, y) - f(x, y) \\
+        &= f(x + h_1, y + h_2) - f(x + h_1, y) \\
+        &+ D_1f(x, y)(h_1) + r_1(h_1)
+    \end{align*}
+
+    where $r_1 \in \mathcal{R}_{\sigma_1}(E_1; F)$. On the other hand, by the \hyperref[Mean Value Theorem]{theorem:mean-value-theorem},
+    \begin{align*}
+        &f(x + h_1, y + h_2) - f(x + h_1, y) - Df_2(x, y)(h_2) \\
+        &\in h_2\ol{\text{Conv}}\bracs{D_2f(x + h_1, y + th_2) - Df_2(x, y)|t \in [0, 1]}
+    \end{align*}
+
+    Since $D_2f$ is continuous and $F$ is locally convex, 
+    \[
+    f(x + h_1, y + h_2) - f(x + h_1, y) - Df_2(x, y)(h_2) = r_2(h_1, h_2)
+    \]
+
+    where $r_2 \in \mathcal{R}_{\sigma_1 \otimes \sigma_2}(E_1 \times E_2; F)$. Therefore
+    \begin{align*}
+    f(x + h_1, y + h_2) - f(x, y) &= D_1f(x, y)(h_1) + D_2f(x, y)(h_2) \\
+    &+ r_1(h_1) + r_2(h_1, h_2)
+    \end{align*}
+    
+    
+\end{proof}
+

src/dg/derivative/power.tex

@@ -0,0 +1,102 @@
+\section{Power Series}
+\label{section:power-series}
+
+\begin{definition}[Power Series]
+\label{definition:power-series}
+    Let $E, F$ be locally convex spaces $K \in \RC$ with $F$ being complete, $\bracsn{T_n}_0^\infty$ with $T_n \in L^n(E; F)$ for each $n \in \natz$, and $a \in E$, then the \textbf{power series} of $\bracsn{T_n}_0^\infty$ about $a$ is the function
+    \[
+    f(x) = \sum_{n = 0}^\infty T_n(x - a)^{(n)}
+    \]
+
+    defined on points on which the series converges. 
+\end{definition}
+
+\begin{definition}[Radius of Convergence]
+\label{definition:radius-of-convergence}
+    Let $E$ be a normed space over $K \in \RC$, $F$ be a complete locally convex space over $K$, and $\sum_{n = 0}^\infty T_n(x - a)^{(n)}$ be a power series about $a \in E$, $\rho: F \to [0, \infty)$ be a continuous seminorm on $F$. For each $T \in L^n(E; F)$, let
+    \[
+    [T]_{L^n(E; F), \rho} = \sup_{x \in B_E(0, 1)^n}\rho(Tx)
+    \]
+    
+    then $R_\rho \in [0, \infty]$ be defined by\footnote{Under the abuse that $1/\infty = 0$ and $1/0 =\infty$.}
+    \[
+    \frac{1}{R_\rho} = \limsup_{n \to \infty}\norm{T_n}_{L^n(E; F)}^{1/n}
+    \]
+
+    is the \textbf{radius of convergence of $\sum_{n = 0}^\infty T_n(x - a)^{(n)}$} with respect to $\rho$, and
+    \begin{enumerate}
+        \item For each $0 < r < R$, the series converges uniformly and absolutely on $B_E(a, r)$ with respect to $\rho$. 
+        \item Let
+        \[
+        R = \inf\bracs{R_\rho| \rho: F \to [0, \infty) \text{ is a continuous seminorm}}
+        \]
+        
+        the series converges uniformly and absolutely on $B_E(a, R)$, and $R$ is the \textbf{radius of convergence of $\sum_{n = 0}^\infty T_n(x - a)^{(n)}$}.
+    \end{enumerate}
+\end{definition}
+\begin{proof}
+    For all $x \in B_E(a, r)$, 
+    \[
+        \sum_{n = 0}^\infty \rho(T_n(x - a)^{(n)}) \le \sum_{n \in \natz} [T_n]_{L^n(E; F), \rho} \norm{x - a}_E^n \le \sum_{n \in \natz}  r^n[T_n]_{L^n(E; F), \rho}
+    \]
+
+    For any $s \in (r, R)$, there exists $N \in \natp$ such that $\norm{T_n}_{L^n(E; F)}^{1/n} \le 1/s$ for all $n \ge N$. In which case, 
+    \[
+    \sum_{n = 0}^\infty r^n[T_n]_{L^n(E; F), \rho} \le \sum_{n = 0}^N r^n[T_n]_{L^n(E; F), \rho} + \sum_{n \ge N}\frac{r^n}{s^n} < \infty
+    \]
+
+    As this estimate holds uniformly on $B_E(a, r)$, the series converges uniformly and absolutely on $B_E(a, r)$ with respect to $\rho$. 
+\end{proof}
+
+\begin{remark}
+\label{remark:radius-of-convergence}
+    In \autoref{definition:radius-of-convergence}, the radius of convergence appears to be an arbitrary lower bound on the domain of convergence. However, in the more specialised case of power series from $\complex$ to $\complex$ or in a Banach algebra, $R$ is the largest constant such that the series converges uniformly and absolutely on all $B(0, r)$ where $0 < r < R$. The lack of this "maximum" claim is why the above statement is a definition. 
+\end{remark}
+
+
+\begin{theorem}[Termwise Differentiation]
+\label{theorem:termwise-differentiation}
+    Let $E$ be a normed space over $K \in \RC$, $F$ be a complete locally convex space over $K$, $f(x) = \sum_{n = 0}^\infty T_n(x - a)^{(n)}$ a power series about $a \in E$, and $R$ be its radius of convergence, then 
+    \begin{enumerate}
+        \item $f \in C^\infty(B(a, R); F)$ is infinitely Fréchet differentiable. 
+        \item For each $x \in B(a, R)$ and $h \in E$, 
+        \[
+        Df(x)(h) = \sum_{n = 0}^\infty \sum_{k = 1}^{n+1}T_{n+1}(((x-a)^{(n)}, h)^{\bracs{k}})
+        \]
+
+        \item The radius of convergence of the above series is at least $R$. 
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    (3): Let $\rho: F \to [0, \infty)$ be a continuous seminorm. For each $n \in \natz$ and $T \in L^n(E; F)$, let
+    \begin{align*}
+    [T]_{L^n(E; F), \rho} &= \sup_{x \in B_E(0, 1)^n}\rho(Tx) \\
+    [T]_{L^n(E; L(E; F)), \rho} &= \sup_{x \in B_E(0, 1)^n}[Tx]_{L(E; F), \rho}
+    \end{align*}
+
+    and
+    \[
+    S_n(x_1, \cdots, x_{n})(h) = \sum_{k = 1}^{n+1}T_{n+1}(((x_1, \cdots, x_n), h)^{\bracs{k}})
+    \]
+
+    then $[S_n]_{L^n(E; L(E; F)), \rho} \le (n+1)[T_{n+1}]_{L^{n+1}(E; F), \rho}$. Since $(n+1)^{1/n}$ is convergent and $\{[T_{n+1}]_{L^{n+1}(E; F), \rho}\}_1^\infty$ is bounded, 
+    \[
+    \limsup_{n \to \infty} [S_n]_{L^n(E; L(E; F)), \rho}^{1/n} \le \limsup_{n \to \infty}(n+1)^{1/n}[T_{n+1}]_{L^{n+1}(E; F), \rho}^{1/n} \le \frac{1}{R}
+    \]
+
+    so the radius of convergence of the proposed series is at least $R$. 
+    
+    (2): By the \autoref{theorem:power-rule}, for each $N \in \natp$, 
+    \[
+    D\braks{\sum_{n = 0}^N T_n(x - a)^{(n)}}(h) = \sum_{n = 0}^N \sum_{k = 1}^{n+1}T_{n+1}(((x-a)^{(n)}, h)^{\bracs{k}})
+    \]
+
+    By \autoref{definition:radius-of-convergence}, the proposed series converges uniformly on $B(a, r)$ for each $0 < r < R$. Thus by \autoref{theorem:differentiable-uniform-limit}, $f$ is differentiable on $B(a, R)$ with 
+    \[
+    Df(x)(h) = \sum_{n = 0}^\infty S_n(x - a)(h) = \sum_{n = 0}^\infty \sum_{k = 1}^{n+1}T_{n+1}(((x-a)^{(n)}, h)^{\bracs{k}})
+    \]
+
+    (1): By (2), (3) applied inductively to $D^nf$. 
+\end{proof}
+
+

src/dg/derivative/sets.tex

@@ -3,12 +3,13 @@
 
 \begin{definition}[Small]
 \label{definition:differentiation-small}
-    Let $E, F$ be TVSs over $K \in \RC$, $\sigma \subset B(E)$ be an upward-directed family that contains all finite sets, $r: E \to F$, and $n \in \natz$, then the following are equivalent:
+    Let $E, F$ be TVSs over $K \in \RC$, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, $r: E \to F$, and $n \in \natz$, then the following are equivalent:
     \begin{enumerate}
         \item For each $A \in \sigma$, $r(th)/t^n \to 0$ uniformly on $A$.
         \item If $r_t(x) = r(tx)/t^n$, then $r_t \to 0$ as $t \to 0$ with respect to the $\sigma$-uniform topology on $F^E$.
         \item For each $A \in \sigma$, $\seq{a_k} \subset A$, and $\seq{t_k} \subset K \setminus \bracs{0}$ with $t_k \to 0$ as $n \to \infty$, $r(t_ka_k)/t_k^n \to 0$ as $n \to \infty$.
     \end{enumerate}
+
     If the above holds, then $r$ is \textbf{$\sigma$-small of order $n$}.
 
     The set $\mathcal{R}_\sigma^n(E; F)$ is the $K$-vector space of all $\sigma$-small functions of order $n$ from $E$ to $F$. For simplicity, $\mathcal{R}_\sigma(E; F)$ denotes $\mathcal{R}_\sigma^1(E; F)$.
@@ -16,50 +17,51 @@
 
 \begin{proposition}
 \label{proposition:differentiation-sets}
-    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset B(E)$ be an upward-directed family contains all finite sets, $B_\sigma(E; F)$ be the space of linear operators bounded on sets in $\sigma$, and $\mathcal{R}_\sigma(E; F)$ be the space of $\sigma$-small functions, then $(B_\sigma(E; F), \mathcal{R}_\sigma(E; F))$ is a system of derivatives and remainders.
+    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, and $\mathcal{R}_\sigma(E; F)$ be the space of $\sigma$-small functions, and $\mathcal{H} \subset B_\sigma(E; F)$ be a subspace, $(\mathcal{H}, \mathcal{R}_\sigma(E; F))$ is a system of derivatives and remainders.
 \end{proposition}
 \begin{proof}
     Let $T \in B_\sigma(E; F)$ and suppose that there exists $V \in \cn_E(0)$ circled and $r \in \mathcal{R}_\sigma(E; F)$ such that $T|_V = r|_V$. For any $x \in V$, $\bracs{x} \in \sigma$, so $T(x) = \lim_{t \downto 0}T(tx)/t = 0$ as $F$ is separated.
 \end{proof}
 
 
-\begin{definition}[Derivative]
+\begin{definition}[$\sigma$-Derivative]
 \label{definition:derivative-sets}
-    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset B(E)$ be an upward-directed family that contains all finite sets, $U \subset E$ be open, $f: U \to F$, and $x_0 \in U$, then $f$ is \textbf{$\sigma$-differentiable at $x_0$} if there exists $V \in \cn_E(0)$, $T \in L(E; F)$, and $r \in \mathcal{R}_\sigma(E; F)$ such that
+    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, $U \subset E$ be open, $f: U \to F$, and $x_0 \in U$, then $f$ is \textbf{$\tilde \sigma$-differentiable at $x_0$} if there exists $V \in \cn_E(0)$, $T \in B_\sigma(E; F)$, and $r \in \mathcal{R}_\sigma(E; F)$ such that
     \[
     f(x_0 + h) = f(x_0) + Th + r(h)
     \]
 
     for all $h \in V$.
 
-    The linear map $T \in L(E; F)$ is the \textbf{$\sigma$-derivative of $f$ at $x_0$}, denoted $D_{\sigma}f(x_0)$.
+    The linear map $T \in B_\sigma(E; F)$ is the \textbf{$\tilde \sigma$-derivative of $f$ at $x_0$}, denoted $D_{\tilde \sigma}f(x_0)$. If $T \in L(E; F)$, then $f$ is \textbf{$\sigma$-differentiable at $x_0$}, and $T$ is the \textbf{$\sigma$-derivative of $f$ at $x_0$}.
 \end{definition}
 
 \begin{definition}[Differentiable]
 \label{definition:differentiable-sets}
-    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset B(E)$ be an upward-directed family that contains all finite sets, $U \subset E$ be open, and $f: U \to F$, then $f$ is \textbf{$\sigma$-differentiable on $U$} if it is $\sigma$-differentiable at every point in $U$. In which case, the map $D_\sigma f: U \to L(E; F)$ is the \textbf{$\sigma$-derivative} of $f$.
+    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, $U \subset E$ be open, and $f: U \to F$, then $f$ is \textbf{$\sigma$/$\tilde \sigma$-differentiable on $U$} if it is $\sigma$/$\tilde \sigma$-differentiable at every point in $U$. In which case, the map $D_\sigma f: U \to B_\sigma(E; F)$ is the \textbf{$\sigma$/$\tilde \sigma$-derivative} of $f$.
 \end{definition}
 
 \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, 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.
+    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.
 \end{definition}
 
 
-\begin{proposition}[{{\cite[Proposition 4.5.2]{Bogachev}}}]
+\begin{proposition}[Chain Rule]
 \label{proposition:chain-rule-sets}
-    Let $E$, $F$, $G$, be TVSs over $K \in \RC$ with $F, G$ being separated, $\sigma \subset B(E)$ and $\tau \subset B(F)$ be upward-directed families that contain all finite sets. If:
+    Let $E$, $F$, $G$, be TVSs over $K \in \RC$ with $F, G$ being separated, $\sigma \subset \mathfrak{B}(E)$ and $\tau \subset \mathfrak{B}(F)$ be covering ideals. If:
     \begin{enumerate}
         \item[(a)] For any $r \in \mathcal{R}_\sigma(E; F)$ and $T \in L(F; G)$, $T \circ r \in \mathcal{R}_\sigma(E; G)$.
         \item[(b)] For any $r \in \mathcal{R}_\sigma(E; F)$, $T \in L(E; F)$, and $s \in \mathcal{R}_\tau(F; G)$, $s \circ (T + r) \in \mathcal{R}_\sigma(E; G)$.
     \end{enumerate}
+
     then for any $U \subset E$ and $V \subset F$ open, $f: U \to V$ $\sigma$-differentiable at $x_0 \in U$, $g: V \to F$ $\tau$-differentiable at $f(x_0) \in V$, $g \circ f: U \to F$ is $\sigma$-differentiable at $x_0$ with
     \[
     D_\sigma(g \circ f)(x_0) = D_\tau g(f(x_0)) \circ D_\sigma f(x_0)
     \]
 
 \end{proposition}
-\begin{proof}
+\begin{proof}[Proof {{\cite[Proposition 4.5.2]{Bogachev}}}. ]
     Since $g$ is $\tau$-differentiable at $f(x_0)$, there exists $s \in \mathcal{R}_\tau(F; G)$ such that
     \[
     g(f(x_0) + h) = g \circ f (x_0) + D_\tau g(f(x_0))h + s(h)
@@ -80,16 +82,18 @@
 
 \begin{proposition}[{{\cite[Corollary 4.5.4]{Bogachev}}}]
 \label{proposition:chain-rule-sets-conditions}
-    Let $E$, $F$, $G$, be TVSs over $K \in \RC$ with $F, G$ being separated. If $\sigma \subset B(E)$ and $\tau \subset B(F)$ correspond to the following families of sets on $E$ and $F$:
+    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 Compact sets.
+        \item Precompact sets.
         \item Bounded sets.
     \end{enumerate}
+
     then
     \begin{enumerate}
         \item For any $r \in \mathcal{R}_\sigma(E; F)$ and $T \in L(F; G)$, $T \circ r \in \mathcal{R}_\sigma(E; G)$.
         \item For any $r \in \mathcal{R}_\sigma(E; F)$, $T \in L(E; F)$, and $s \in \mathcal{R}_\tau(F; G)$, $s \circ (T + r) \in \mathcal{R}_\sigma(E; G)$.
     \end{enumerate}
+
     and by \autoref{proposition:chain-rule-sets}, $\sigma$-derivatives and $\tau$-derivatives satisfy the Chain rule.
 \end{proposition}
 \begin{proof}
@@ -124,35 +128,84 @@
     A method of extending this sense of differentiability is to require that \textit{every} extension of the function to some open set, or to the entire space is differentiable. Given that this paves way to confusion for related definitions of differentiability, this definition is not formally included here.
 \end{remark}
 
-
+\begin{theorem}[Interchange of Limits and Derivatives]
-
+\label{theorem:differentiable-uniform-limit}
-\begin{proposition}
+    Let $E$ be a TVS over $K \in \RC$, $F$ be a separated locally convex space over $K$, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, $U \subset E$ be open, and $n \in \natp$. Let $\fF \subset 2^{\tilde D_\sigma^n(U; F)}$ be a filter such that:
-\label{proposition:derivative-sets-real}
+    \begin{enumerate}[label=(\alph*)]
-    Let $E$ be a separated topological vector space and $\sigma \subset B(\real)$ be an upward-directed system that contains finite sets, then
+        \item There exists $f: U \to F$ such that $\fF \to f$ pointwise. 
-    \begin{enumerate}
+        \item For each $1 \le k \le n$, there exists $f^{(k)}: U \to B^{k}_\sigma(E; F)$ such that for all $x \in U$, and $A \in \sigma$ with $x + [0, 1]A \subset U$, $D_\sigma^k(\fF) \to f^{(k)}$ uniformly on $x + [0, 1]A$. 
-        \item $\mathcal{R}_{\sigma}(\real; E) = \mathcal{R}_{B(\real)}(\real; E)$. Hence, all forms of $\sigma$-differentiability on $\real$ are equivalent.
-        \item For any $U \subset \real$ open, $f: U \to E$, and $x_0 \in U$, $f$ is differentiable at $x_0$ if and only if
-        \[
-        \lim_{t \to 0}\frac{f(x + t) - f(x)}{t}
-        \]
-
-        exists. In which case, the above limit is identified with the derivative of $f$ at $0$.
-        \item For any $U \subset \real$ open, $f: U \to E$, and $x_0 \in U$, if $f$ is differentiable at $x_0$, then $f$ is continuous at $x_0$.
     \end{enumerate}
-\end{proposition}
-\begin{proof}
-    (1): Let $r \in \mathcal{R}_\sigma(\real; E)$. For any $R > 0$ and $U \in \cn_E(0)$, there exists $\delta > 0$ such that $t^{-1}r(tR), t^{-1}r(-tR) \in U$ for all $t \in (0, \delta)$. Thus $t^{-1}r(tB(0, R)) \subset U$, and $r \in \mathcal{R}_{B(\real)}(\real; E)$.
 
-    (2): Suppose that $f$ is differentiable at $x_0$, then there exists $r \in \mathcal{R}_\sigma$ such that for any $t \in \real$ with $x_0 + t \in U$,
+    then $f \in \tilde D_\sigma^n(U; F)$ and $D^k_\sigma f = f^{(k)}$ for all $1 \le k \le n$. In particular, if $\sigma$ is saturated, then $(b)$ may be replaced by
+    \begin{enumerate}
+        \item[(b)] For each $1 \le k \le n$, there exists $f^{(k)}: U \to B^{k}_\sigma(E; F)$ such that $D_\sigma^k(\fF) \to f^{(k)}$ uniformly on every $A \in \sigma$. 
+    \end{enumerate}
+    
+\end{theorem}
+\begin{proof}
+    Assume without loss of generality that $n = 1$. For any $\varphi \in \tilde D^1_\sigma(U; F)$, $x \in U$, and $h \in E$ such that $x + h \in U$, 
     \begin{align*}
-        f(x_0 + t) - f(x_0) &= Df(x_0)(t) + r(t) \\
+        f(x + h) - f(x) - f^{(1)}(x)h &= \underbrace{\varphi(x + h) - \varphi(x) - D_\sigma\varphi(x)h}_{\in \mathcal{R}_\sigma(E; F)} \\
-        \frac{f(x_0 + t) - f(x_0)}{t} &= Df(x_0)(1) + t^{-1}r(t) \\
+        &+ (f - \varphi)(x + h) - (f - \varphi)(x) \\
-        \lim_{t \to 0}\frac{f(x_0 + t) - f(x_0)}{t} &= Df(x_0)(1)
+        &+ (D_\sigma\varphi - f^{(1)})(x)h 
     \end{align*}
-    Now suppose that $v = \lim_{t \to 0}\frac{f(x + t) - f(x)}{t}$ exists. Let $T: \real \to E$ be defined by $t \mapsto tv$, then
+
+    Since $\fF \to f$ pointwise, for any $S \in \fF$, 
     \[
-    \lim_{t \to 0}\frac{f(x_0 + t) - f(x_0) - Tt}{t} = \lim_{t \to 0}\frac{f(x_0 + t) - f(x_0)}{t} - v = 0
+    (f - \varphi)(x + h) - (f - \varphi)(x) \in \overline{\bracs{(g - \varphi)(x + h) - (g - \varphi)(x)|g \in S}}
     \]
     
-    and $Df(x_0) = T$.
+    By the \hyperref[Mean Value Theorem]{theorem:mean-value-theorem}, for any $g \in \tilde D^1_\sigma(U; F)$, 
+    \[
+    (g - \varphi)(x + h) - (g - \varphi)(x) \in \ol{\conv}\bracs{D_\sigma(g - \varphi)(x + th)h|t \in [0, 1]}
+    \]
+
+    Hence
+    \[
+    (f - \varphi)(x + h) - (f - \varphi)(x) \in  \ol{\conv}\bracs{D_\sigma(g - \varphi)(x + th)h|g \in S, t \in [0, 1]}
+    \]
+
+    so for any $t \in (0, 1)$ and $A \in \sigma$, 
+    \begin{align*}
+    &(f - \varphi)(x + tA) - (f - \varphi)(x) \\
+    &\subset \ol{\conv}\bracs{D_\sigma(g - \varphi)(x + sth)th|g \in S, s \in [0, 1], h \in A} \\
+    &= t\ol{\conv}\bracs{D_\sigma(g - \varphi)(x + sth)h|g \in S, s \in [0, 1], h \in A}
+    \end{align*}
+
+    and
+    \begin{align*}
+    &t^{-1}[(f - \varphi)(x + tA) - (f - \varphi)(x)] \\
+    &\subset \ol{\conv}\bracs{D_\sigma(g - \varphi)(x + sth)h|g \in S, s \in [0, 1], h \in A} \\
+    &\subset \ol{\conv}\bracs{D_\sigma(g - \varphi)(x + sh)h|g \in S, s \in [0, 1], h \in A}
+    \end{align*}
+
+    In addition, since $D_\sigma(\fF) \to f^{(1)}$ pointwise, 
+    \[
+    t^{-1}(f^{(1)} - D_\sigma\varphi)(x)(tA) \subset \ol{\conv}\bracs{D_\sigma(g - \varphi)(x + sh)h|g \in S, s \in [0, 1], h \in A}
+    \]
+
+    as well. 
+
+    Now, let $V \in \cn_F(0)$ be convex and circled. Using assumption (b), let $S \in \fF$ such that for any $\varphi \in S$, 
+    \[
+    \ol{\conv}\bracs{D_\sigma(g - \varphi)(x + sh)h|g \in S, s \in [0, 1], h \in A} \subset V
+    \]
+    
+    Fix $\varphi \in S$, then as $\varphi$ is differentiable at $x$, there exists $\delta \in (0, 1)$ such that
+    \[
+    t^{-1}[\varphi(x + tA) - \varphi(x) - D_\sigma\varphi(x)(tA)] \subset V
+    \]
+
+    for all $t \in (0, \delta)$. 
+
+    So
+    \[
+    t^{-1}[f(x + tA) - f(x) - f^{(1)}(x)(tA)] \subset 3V
+    \]
+
+    for all $t \in (0, \delta)$. Therefore $f$ is $\tilde \sigma$-differentiable at $x$ with $D_\sigma f(x) = f^{(1)}(x)$. 
+    
 \end{proof}
+
+
+

src/dg/derivative/taylor.tex

@@ -1,20 +1,20 @@
 \section{Taylor's Formula}
 \label{section:taylor}
 
-\begin{theorem}[Taylor's Formula, Lagrange Remainder {{\cite[Theorem 4.7.1]{Bogachev}}}]
+\begin{theorem}[Taylor's Formula, Lagrange Remainder]
 \label{theorem:taylor-lagrange}
     Let $-\infty < a < b < \infty$, $E$ be a separated locally convex space, $S \subset [a, b]$ be at most countable, $n \in \natp$, and $f \in C^{n}([a, b]; E)$ be $(n+1)$-fold differentiable on $[a, b] \setminus N$, then
     \[
     f(b) - f(a) - \sum_{k = 1}^n \frac{1}{k!}D^kf(a)(b - a)^k
     \]
 
-    is contained in the closed convex hull\footnote{It may be possible to sharpen the below claim to include the $1/(n+1)!$ factor. However, I was not able to follow the proof for this.} of
+    is contained in the closed convex hull of
     \[
     \bracs{D^{n+1}f(s)(t - a)^{n+1}  | s \in (a, b) \setminus N, t \in [a, b]}
     \]
 
 \end{theorem}
-\begin{proof}
+\begin{proof}[Proof {{\cite[Theorem 4.7.1]{Bogachev}}}. ]
     If $n = 0$, then the theorem is the \hyperref[Mean Value Theorem]{theorem:mean-value-theorem-line}.
 
     Suppose inductively that the theorem holds for $n$. Let
@@ -60,21 +60,21 @@
 
 \end{proof}
 
-\begin{theorem}[Taylor's Formula, Peano Remainder {{\cite[Theorem 4.7.3]{Bogachev}}}]
+\begin{theorem}[Taylor's Formula, Peano Remainder]
 \label{theorem:taylor-peano}
-    Let $E$ be a topological vector space, $\sigma \subset B(E)$ be an upward-directed family that includes bounded sets contained in finite-dimensional subspaces, $F$ be a separated locally convex space, $U \subset E$ be open, and $f: U \to F$ be $n$-fold $\sigma$-differentiable at $x_0 \in U$, then there exists $r \in \mathcal{R}_\sigma^n(E; F)$ such that
+    Let $E$ be a topological vector space, $\sigma \subset \mathfrak{B}(E)$ be an ideal that includes bounded sets contained in finite-dimensional subspaces, $F$ be a separated locally convex space, $U \subset E$ be open, and $f: U \to F$ be $n$-fold $\tilde \sigma$-differentiable at $x_0 \in U$, then there exists $r \in \mathcal{R}_\sigma^n(E; F)$ such that
     \[
     g(x_0 + h) = g(x_0) + \sum_{k = 1}^n \frac{1}{k!}D^k_\sigma f(x_0)(h^{(k)}) + r(h)
     \]
 
 \end{theorem}
-\begin{proof}
+\begin{proof}[Proof {{\cite[Theorem 4.7.3]{Bogachev}}}. ]
     Let
     \[
     r(h) = g(x_0 + h) - g(x) - \sum_{k = 1}^n \frac{1}{k!}D^k_\sigma f(x_0)(h^{(k)})
     \]
 
-    For any $1 \le k \le n$, $D^k_\sigma(x_0) \in B^k_\sigma(E; F)$ is symmetric by \autoref{theorem:derivative-symmetric}. Let $T_k(h) = \frac{1}{k!}D^k_\sigma f(x_0)(h^{(k)})$, then by \autoref{proposition:multilinear-derivative}, for any $\bracs{t_j}_1^\ell \in E$,
+    For any $1 \le k \le n$, $D^k_\sigma(x_0) \in B^k_\sigma(E; F)$ is symmetric by \autoref{theorem:derivative-symmetric}. Let $T_k(h) = \frac{1}{k!}D^k_\sigma f(x_0)(h^{(k)})$, then by \autoref{theorem:power-rule}, for any $\bracs{t_j}_1^\ell \in E$,
     \[
     D^\ell_\sigma T_k(h)(t_1, \cdots, t_\ell) = \begin{cases}
         0 &\ell > k \\

src/dg/index.tex

@@ -1,4 +1,6 @@
-\part{Differential Geometry}
+\part{Calculus}
 \label{part:diffgeo}
 
 \input{./derivative/index.tex}
+\input{./complex/index.tex}
+\input{./notation.tex}

src/dg/notation.tex

@@ -0,0 +1,21 @@
+\chapter{Notations}
+\label{chap:dg-notations}
+
+Differential geometry is the study of things invariant under change of notation. 
+
+\begin{tabular}{lll}
+    \textbf{Notation} & \textbf{Description} & \textbf{Source} \\
+    \hline
+    $\mathcal{H}(E;F)$, $\mathcal{R}(E;F)$ & Space of derivatives; space of remainders in an $\mathcal{HR}$-system. & \autoref{definition:derivative-system} \\
+    $D_{\mathcal{HR}} f(x_0)$ & $\mathcal{HR}$-derivative of $f$ at $x_0$. & \autoref{definition:space-differentiability} \\
+    $\mathcal{R}_\sigma^n(E; F)$, $\mathcal{R}_\sigma(E;F)$ & $\sigma$-small functions of order $n$; order 1. & \autoref{definition:differentiation-small} \\
+    $D_\sigma f(x_0)$ & $\sigma$-derivative of $f$ at $x_0$. & \autoref{definition:derivative-sets} \\
+    $D_\sigma^n f$ & $n$-fold $\sigma$-derivative. & \autoref{definition:n-differentiable-sets} \\
+    $D_\sigma^n(U; F)$ & $n$-fold $\sigma$-differentiable functions. & \autoref{definition:differentiable-space} \\
+    $\tilde D_\sigma^n(U; F)$ & $n$-fold $\tilde \sigma$-differentiable functions. & \autoref{definition:differentiable-space} \\
+    $C_\sigma^n(U; F)$ & $n$-fold continuously $\sigma$-differentiable functions. & \autoref{definition:continuously-differentiable-space} \\
+    $\tilde C_\sigma^n(U; F)$ & $n$-fold continuously $\tilde \sigma$-differentiable functions. & \autoref{definition:continuously-differentiable-space} \\
+    $L^{(n)}_\sigma(E; F)$ & Codomain of derivatives. $L^{(0)}_\sigma(E; F) = F$, $L^{(n)}_\sigma(E; F) = L(E; L_\sigma^{(n-1)}(E; F))$, equipped with the $\sigma$-uniform topology. & \autoref{definition:higher-derivatives-codomain} \\
+    $x^{(k)}$ & Tuple of $x$ repeated $k$ times. & \autoref{theorem:taylor-peano} \\
+    $D^+f(x)$ & Right derivative of $f$ at $x$. & \autoref{definition:right-differentiable-mvt}
+\end{tabular}

src/fa/duality/definitions.tex

@@ -0,0 +1,72 @@
+\section{Dual Systems}
+\label{section:dual-systems}
+
+\begin{definition}[Duality]
+\label{definition:duality}
+    Let $K$ be a field, $E, F$ be vector spaces over $K$, and $\lambda: E \times F \to K$ be a bilinear map, then the triple $(E, F, \lambda)$ is a \textbf{dual system/duality} over $K$ if
+    \begin{enumerate}[label=($S_{\arabic*}$)]
+        \item For any $x_0 \in E$, if $\lambda(x_0, y) = 0$ for all $y \in F$, then $x_0 = 0$. 
+        \item For any $y_0 \in E$, if $\lambda(x, y_0) = 0$ for all $x \in E$, then $y_0 = 0$.
+    \end{enumerate}
+
+    The mapping $\lambda: E \times F \to K$ is the \textbf{canonical bilinear form} of the duality, denoted $(x, y) \mapsto \dpn{x, y}{\lambda}$, and the duality $(E, F, \lambda)$ is denoted $\dpn{E, F}{\lambda}$. 
+
+    In the context of a dual system, $E$ and $F$ are identified as subspaces of each others' algebraic duals. 
+\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 $E$. 
+\end{definition}
+
+\begin{lemma}
+\label{lemma:duality-dual}
+    Let $K \in \RC$ and $\dpn{E, F}{\lambda}$ be a duality over $K$, then the dual of $(E, \sigma(E, F))$ is $F$. In other words, for any $\phi \in L(E, \sigma(E, F); K)$, there exists a unique $y \in F$ such that $\dpn{x, \phi}{E} = \dpn{x, y}{\lambda}$ for all $x \in E$.  
+\end{lemma}
+\begin{proof}[Proof, {{\cite[IV.1.2]{SchaeferWolff}}}. ]
+    Since $\phi$ is continuous, there exists $\seqf{y_k} \subset F$ such that for all $x \in E$, 
+    \[
+    |\dpn{x, \phi}{\lambda}| \le \sum_{k = 1}^n |\dpn{x, y_k}{\lambda}|
+    \]
+
+    Assume without loss of generality that $\seqf{y_k}$ is linearly independent, then by the First Isomorphism Theorem, there exists $\Phi \in L(K^n; K)$ such that the following diagram commutes
+    \[
+    \xymatrix{
+    E \ar@{->}[rd]_{\phi} \ar@{->}[r]^{{(y_1, \cdots, y_n)}} & K^n \ar@{->}[d]^{\Phi} \\
+    & K
+    }
+    \]
+    
+    For each $1 \le k \le n$, let $e_k$ be the $k$-th standard basis vector in $K^n$, then for any $x \in E$,
+    \[
+    \dpn{x, \phi}{E} = \sum_{k = 1}^n \Phi(e_k) \dpn{x, y_k}{\lambda}
+    \]
+    
+\end{proof}
+
+\begin{lemma}
+\label{lemma:duality-dense}
+    Let $K \in \RC$ and $\dpn{E, F}{\lambda}$ be a duality over $K$, then for any subspace $F_0 \subset F$, the following are equivalent:
+    \begin{enumerate}
+        \item $\dpn{E, F_0}{\lambda}$ is a duality. 
+        \item For any $y_0 \in F$, $\seqf{x_j} \subset E$, and $\eps > 0$, there exists $y \in F_0$ such that for each $1 \le j \le n$, $\dpn{x_j, y}{\lambda} = \dpn{x_j, y_0}{\lambda}$.
+        \item $F_0$ is $\sigma(F, E)$-dense in $F$.
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    (1) $\Rightarrow$ (2): Let $E_0 = \text{span}\bracs{x_j|1 \le j \le n}$, and
+    \[
+    \phi: E_0 \to K \quad x \mapsto \dpn{x, y_0}{\lambda}
+    \]
+    
+    Since $\dpn{E, F_0}{\lambda}$ is a duality, there exists $\seqf{y_j} \subset F_0$ such that for all $x \in E_0$, 
+    \[
+    |\dpn{x, \phi}{E_0}| \le \sum_{j = 1}^n |\dpn{x, y_j}{\lambda}|
+    \]
+
+    Hence $\phi \in L(E_0, \sigma(E_0, F_0); K)$. By the \hyperref[Hahn-Banach Theorem]{theorem:hahn-banach}, there exists $\Phi \in L(E, \sigma(E, F_0); K)$ such that $\Phi|_{E_0} = \phi$. By \autoref{lemma:duality-dual}, there exists $y \in F_0$ such that $\dpn{x_j, y}{\lambda} = \dpn{x_j, y_0}{\lambda}$ for all $1 \le j \le n$. 
+
+    (3) $\Rightarrow$ (1): Let $x \in E$ such that $\dpn{x, y}{\lambda} = 0$ for all $y \in F_0$, then since $F_0$ is $\sigma(F, E)$-dense in $F$, $\dpn{x, y}{\lambda} = 0$ for all $y \in F$. Hence $x = 0$. 
+\end{proof}
+
+

src/fa/duality/index.tex

@@ -0,0 +1,7 @@
+\chapter{Duality}
+\label{chap:duality}
+
+\input{./definitions.tex}
+\input{./polar.tex}
+
+

src/fa/duality/polar.tex

@@ -0,0 +1,129 @@
+\section{Polars}
+\label{section:polar}
+
+\begin{definition}[Real Polar]
+\label{definition:real-polar}
+    Let $\dpn{E, F}{\lambda}$ be a duality over $K \in \RC$ and $A \subset E$, then
+    \[
+    A^\circ = \bracsn{y \in F| \text{Re}\dpn{x, y}{\lambda} \le 1 \forall x \in A}
+    \]
+
+    is the \textbf{real polar} of $A$. 
+\end{definition}
+
+\begin{definition}[Absolute Polar]
+\label{definition:absolute-polar}
+    Let $\dpn{E, F}{\lambda}$ be a duality over $K \in \RC$ and $A \subset E$, then
+    \[
+    A^\square = \bracsn{y \in F|\ |\dpn{x, y}{\lambda}| \le 1 \forall x \in A}
+    \]
+
+    is the \textbf{absolute polar} of $A$. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:polar-gymnastics}
+    Let $\dpn{E, F}{\lambda}$ be a duality over $K \in \RC$, then:
+    \begin{enumerate}
+        \item $\emptyset^\circ = F$ and $E^\circ = \bracs{0}$. 
+        \item For any $\alpha \in K \setminus \bracs{0}$ and $A \subset E$, $(\alpha A)^\circ = \alpha^{-1} \cdot A^\circ$.
+        \item For any $\seqi{A} \subset E$, $\paren{\bigcup_{i \in I}A_i}^\circ = \bigcap_{i \in I}A_i^\circ$. 
+        \item For any $A \subset B \subset E$, $A^\circ \supset B^\circ$. 
+        \item For any saturated ideal $\sigma \subset \mathfrak{B}(E, \sigma(E, F))$, $\bracs{S^\circ|S \in \sigma}$ is a fundamental system of neighbourhoods at $0$ for the $\sigma$-uniform topology on $F$.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (2): For any $\lambda \in K \setminus \bracs{0}$ and $A \subset E$, 
+    \begin{align*}
+        (\lambda A)^\circ &= \bracs{y \in F|\text{Re}\dpn{x, y}{\lambda} \le 1 \forall x \in \alpha A} \\
+        &= \bracs{y \in F|\text{Re}\dpn{x, y}{\lambda} \le 1 \forall x \in \alpha A} \\
+        &= \bracs{y \in F|\text{Re}\dpn{x, \alpha y}{\lambda} \le 1 \forall x \in A} \\
+        &= \bracs{y \in F|\text{Re}\dpn{\alpha x,  y}{\lambda} \le 1 \forall x \in A} \\
+        &= \bracs{\alpha^{-1} y \in F|\text{Re}\dpn{x,  y}{\lambda} \le 1 \forall x \in A} \\
+        &= \alpha^{-1} \cdot A^\circ
+    \end{align*}
+
+    (5): Let $S \in \sigma$, then
+    \[
+    (\aconv(S))^\circ = \bracs{y \in F|\ |\dpn{x, y}{\lambda}| \le 1 \forall x \in A} \subset S^\circ
+    \]
+
+    Since $\sigma$ is saturated, $\bracs{S^\circ|S \in \sigma}$ is a fundamental system of neighbourhoods at $0$ for the $\sigma$-uniform topology.     
+\end{proof}
+
+\begin{proposition}[{{\cite[IV.1.4]{SchaeferWolff}}}]
+\label{proposition:polar-properties}
+    Let $\dpn{E, F}{\lambda}$ be a duality over $K \in \RC$ and $A \subset E$, then
+    \begin{enumerate}
+        \item $A^\circ$ is a $\sigma(F, E)$-closed convex subset of $F$ containing $0$. 
+        \item If $A$ is circled, then so is $A^\circ$. 
+        \item If $A$ is a subspace of $E$, then
+        \[
+        A^\circ = A^\perp = \bracs{y \in F| \dpn{x, y}{E} = 0 \forall x \in A}
+        \]
+
+        and $A^\circ$ is a subspace of $F$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): For each $x \in E$, 
+    \[
+    \bracs{x}^\circ = \bracs{y \in F|\text{Re}\dpn{x, y}{\lambda} \le 1}
+    \]
+
+    is the sublevel set of a continuous $\real$-linear functional, so it is $\sigma(F, E)$-closed and convex. Since $A^\circ = \bigcap_{x \in A}\bracs{x}^\circ$, $A$ is also $\sigma(F, E)$-closed and convex. 
+
+    (2): If $A$ is circled, then by \autoref{proposition:polar-gymnastics},
+    \[
+    A^\circ = \bigcap_{\substack{\alpha \in K \\ |\alpha| \ge 1}}\alpha A^\circ
+    \]
+
+    For any $y \in A^\circ$ and $\alpha \in K \setminus \bracs{0}$ with $|\alpha| \le 1$. Since $y \in \alpha^{-1}A^\circ$, $\alpha y \in A^\circ$, so $A^\circ$ is circled. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:equicontinuous-polar}
+    Let $E$ be a TVS over $K \in \RC$, $\dpn{E, E^*}{E}$ be the canonical duality, and $A \subset E^*$, then the following are equivalent
+    \begin{enumerate}
+        \item $A$ is equicontinuous.
+        \item $A^\circ \in \cn_E(0)$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    By \autoref{proposition:equicontinuous-linear}, $A$ is equicontinuous if and only if
+    \[
+    \bigcap_{\phi \in A}\phi^{-1}(B_K(0, 1)) \in \cn_E(0)
+    \]
+
+    (1) $\Rightarrow$ (2): $A^\circ \supset \bigcap_{\phi \in A}\phi^{-1}(B_K(0, 1))$.
+
+    (2) $\Rightarrow$ (1): Since $A^\circ \in \cn_E(0)$, there exists $V \in \cn_E(0)$ circled with $V \subset A^\circ$, so $V \subset \bigcap_{\phi \in A}\phi^{-1}(B_K(0, 1))$, and $\bigcap_{\phi \in A}\phi^{-1}(B_K(0, 1)) \in \cn_E(0)$. 
+\end{proof}
+
+
+\begin{theorem}[Bipolar Theorem]
+\label{theorem:bipolar}
+    Let $\dpn{E, F}{\lambda}$ be a duality over $K \in \RC$. For each $A \subset F$,
+    \[
+    A^{\circ\circ} = \ol{\conv}(A \cup \bracs{0})
+    \]
+
+    with respect to the $\sigma(E, F)$-topology. 
+\end{theorem}
+\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:
+    \begin{enumerate}
+        \item $\phi$ is $\sigma(E, F)$-continuous.
+        \item $\phi(\ol{\conv}(A \cup \bracs{0})) \subset (-\infty, 1)$ and $\phi(x_0) > 1$. 
+    \end{enumerate}
+
+    If $K = \real$, let $\Phi = \phi$. If $K = \complex$, then by \autoref{proposition:polarisation-linear}, the mapping $\Phi(x) = \phi(x) - i\phi(ix)$ is a $\sigma(E; F)$-continuous $K$-linear map on $E$ such that $\text{Re}(\Phi) = \phi$. By \autoref{lemma:duality-dual}, there exists $y \in F$ such that $\dpn{x, y}{\lambda} = \Phi(x)$ for all $x \in E$. In which case, for any $x \in E$,
+    \[
+    \text{Re}\dpn{x, y}{\lambda} = \text{Re}\Phi(x) = \phi(x)
+    \]
+
+    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

@@ -7,3 +7,6 @@
 \input{./norm/index.tex}
 \input{./rs/index.tex}
 \input{./lp/index.tex}
+\input{./order/index.tex}
+\input{./duality/index.tex}
+\input{./notation.tex}

src/fa/lc/barrel.tex

@@ -0,0 +1,67 @@
+\section{Barreled Spaces}
+\label{section:barrel}
+
+\begin{definition}[Barrel]
+\label{definition:barrel}
+    Let $E$ be a TVS over $K \in \RC$ and $D \subset E$, then $D$ is a \textbf{barrel} if it is convex, circled, radial, and closed. 
+\end{definition}
+
+\begin{definition}[Barreled Space]
+\label{definition:barreled-space}
+    Let $E$ be a locally convex space over $K \in \RC$, then the following are equivalent:
+    \begin{enumerate}
+        \item The barrels of $E$ forms a fundamental system of neighbourhoods at $0$. 
+        \item Every barrel in $E$ is a neighbourhood of $0$. 
+        \item Every lower semicontinuous seminorm on $E$ is continuous. 
+    \end{enumerate}
+\end{definition}
+\begin{proof}
+    $(2) \Rightarrow (1)$: Let $\fB \subset \cn_E(0)$ be a fundamental system of neighbourhoods at $0$ consisting of convex, circled, and radial sets, then $\ol{\fB} = \bracsn{\ol U|U \in \fB}$ is a fundamental system of neighbourhoods at $0$ consisting of barrels. 
+
+    $(2) \Rightarrow (3)$: Let $\rho: E \to [0, \infty)$ be a lower semicontinuous seminorm, then $\bracs{\rho > 1}$ is open and $\bracs{\rho \le 1}$ is a Barrel. In which case, $\rho$ is continuous by (4) of \autoref{lemma:continuous-seminorm}. 
+
+    $(3) \Rightarrow (2)$: Let $D \subset E$ be a barrel and $\rho: E \to [0, \infty)$ be its gauge. By (4) of \autoref{definition:gauge}, $D = \bracs{\rho \le 1}$, so $\bracs{\rho > 1}$ is open, and $\rho$ is semicontinuous. By assumption, $\rho$ is continuous, so $D \in \cn_E(0)$ by (5) of \autoref{lemma:continuous-seminorm}. 
+\end{proof}
+
+\begin{summary}
+\label{summary:barreled-space}
+    The following types of locally convex spaces are barrelled:
+    \begin{enumerate}
+        \item Every locally convex space with the Baire property.
+        \item Every Banach space and every Fréchet space. 
+        \item Inductive limits of barrelled spaces. 
+        \item Spaces of type (LB) and (LF). 
+        \item The locally convex direct sum of barrelled spaces. 
+        \item Products of barrelled spaces.
+    \end{enumerate}
+\end{summary}
+\begin{proof}
+    (1), (2): \autoref{proposition:baire-barrel}. 
+
+    (3), (4), (5): \autoref{proposition:barrel-limit}. 
+
+    (6): TODO. 
+\end{proof}
+
+
+\begin{proposition}
+\label{proposition:baire-barrel}
+    Let $E$ be a locally convex space over $K \in \RC$. If $E$ is a Baire space, then $E$ is barrelled. 
+\end{proposition}
+\begin{proof}[Proof, {{\cite[II.7.1]{SchaeferWolff}}}. ]
+    Let $D \subset E$ be a Barrel, then $E = \bigcup_{n \in \natp}nD$ is a countable union of closed sets. Since $E$ is Baire, there exists $n \in \natp$, $U \in \cn_E(0)$ circled, and $x \in E$ such that $x + U \in nB$. In which case, 
+    \[
+    U \subset (x + U) - (x + U) \subset nB - nB = 2nB
+    \]
+
+    so $2nB$ and thus $B$ is a neighbourhood of $0$. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:barrel-limit}
+    Let $\seqi{E}$ be locally convex spaces over $K \in \RC$, $E$ be a vector space over $K$, and $\seqi{T}$ such that $T_i \in \hom(E_i; E)$ for all $i \in I$, then the inductive locally convex topology on $E$ induced by $\seqi{T}$ is barrelled.  
+\end{proposition}
+\begin{proof}[Proof, {{\cite[II.7.2]{SchaeferWolff}}}. ]
+    Let $D \subset E$ be a barrel, then for each $i \in I$, $T_i^{-1}(D) \subset E_i$ is also a barrel, and thus a neighbourhood of $0$ in $E_i$. By (5) of \autoref{definition:lc-inductive}, $D$ is a neighbourhood of $0$ in $E$, so $E$ is barrelled. 
+\end{proof}
+

src/fa/lc/bornologic.tex

@@ -1,15 +1,16 @@
-\section{Bornologic Spaces}
+\section{Bornological Spaces}
-\label{section:bornologic}
+\label{section:bornological}
 
-\begin{definition}[Bornologic Space]
+\begin{definition}[Bornological Space]
-\label{definition:bornologic-space}
+\label{definition:bornological-space}
     Let $E$ be a locally convex space, then the following are equivalent:
     \begin{enumerate}
         \item For any $U \subset E$ convex and balanced, if $U$ absorbs every bounded set of $E$, then $U \in \cn_E(0)$.
         \item For any seminorm $\rho: E \to [0, \infty)$ that is bounded on all bounded sets of $E$, $\rho$ is continuous. 
     \end{enumerate}
 
-    If the above holds, then $E$ is a \textbf{bornologic space}.
+    
+    If the above holds, then $E$ is a \textbf{bornological space}.
 \end{definition}
 \begin{proof}
     (1) $\Rightarrow$ (2): Let $B \subset E$ be bounded, then there exists $R > 0$ such that $\rho(B) \subset [0, R)$. In which case, 
@@ -23,34 +24,68 @@
 \end{proof}
 
 \begin{proposition}
-\label{proposition:metrisable-bornologic}
+\label{proposition:bornological-bounded}
-    Let $E$ be a metrisable locally convex space, then $E$ is bornologic.
+    Let $E$ be a bornological space, $F$ be a locally convex space, and $T \in \hom(E; F)$, then the following are equivalent:
-\end{proposition}
-\begin{proof}
-    Let $U \subset E$ be convex and balanced such that $U$ absorbs every bounded set of $E$. Let $\seq{U_n} \subset \cn^o(0)$ be a decreasing countable fundamental system of neighbourhoods at $0$. If $U_n \setminus nA \ne \emptyset$ for all $n \in \natp$, then there exists $\seq{x_n}$ such that $x_n \in U_n \setminus nA$ for all $n \in \natp$. In which case, $x_n \to 0$ as $n \to \infty$, so $\seq{x_n}$ is bounded. By assumption, there exists $n \in\natp$ such that $nA \supset \seq{x_n}$, which contradicts the fact that $\seq{x_n} \cap A = \emptyset$. 
-\end{proof}
-
-\begin{proposition}
-\label{proposition:bornologic-bounded}
-    Let $E$ be a bornologic space, $F$ be a locally convex space, and $T \in \hom(E; F)$, then the following are equivalent:
     \begin{enumerate}
         \item $T$ is continuous.
+        \item $T$ is sequentially continuous. 
         \item $T$ is bounded.
     \end{enumerate}
-    
 \end{proposition}
 \begin{proof}
-    (1) $\Rightarrow$ (2): By \autoref{proposition:continuous-bounded}. 
+    (2) $\Rightarrow$ (3): Let $B \subset E$ be a bounded set, $\seq{x_n} \subset E$, and $\seq{\lambda_n} \subset K$ such that $\lambda_n \to 0$ as $n \to \infty$, then $\lambda_n x_n \to 0$ as $n \to \infty$. By sequential continuity of $T$, $T(\lambda_n x_n) \to 0$ as $n \to \infty$ as well. Thus $T(B)$ is also bounded. 
 
-    (2) $\Rightarrow$ (1): Let $\rho: F \to [0, \infty)$ be a continuous seminorm, then $\rho \circ T$ is a seminorm on $E$ that is bounded on bounded sets. Since $E$ is bornologic, $\rho \circ T$ is continuous. Therefore $T$ is continuous by \autoref{proposition:tvs-convex-morphism}.
+    (3) $\Rightarrow$ (1): Let $\rho: F \to [0, \infty)$ be a continuous seminorm, then $\rho \circ T$ is a seminorm on $E$ that is bounded on bounded sets. Since $E$ is bornological, $\rho \circ T$ is continuous. Therefore $T$ is continuous by \autoref{proposition:tvs-convex-morphism}.
 \end{proof}
 
 \begin{proposition}
-\label{proposition:bornologic-continuous-complete}
+\label{proposition:metrisable-bornological}
-    Let $E$ be a bornologic space and $F$ be a complete Hausdorff locally convex space, then $L_b(E; F)$ is complete. In particular, $E^*$ equipped with the topology of bounded convergence is complete.
+    Let $E$ be a metrisable locally convex space, then $E$ is bornological.
 \end{proposition}
 \begin{proof}
-    By \autoref{proposition:bornologic-bounded}, $L_b(E; F) = B(E; F)$. By \autoref{proposition:operator-space-completeness}, $B(E; F)$ is complete, so $L_b(E; F)$ is complete as well.
+    Let $A \subset E$ be a convex and circled set that absorbs every bounded set of $E$, and $\seq{U_n} \subset \cn_E(0)$ be a decreasing fundamental system of neighbourhoods at $0$. If $A \not\in \cn_E(0)$, then $U_n \setminus nA \ne \emptyset$ for all $n \in \natp$. For each $n \in \natp$, let $x_n \in U_n \setminus nA$, then $x_n \to 0$ as $n \to \infty$, and $\seq{x_n}$ is bounded. However, since $A$ absorbs every bounded set of $E$, there exists $n \in \natp$ such that $nA \supset \seq{x_n}$, which contradicts the assumption that $A \not\in \cn_E(0)$. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:bornological-limit}
+    Let $\seqi{E}$ be bornological spaces over $K \in \RC$, $E$ be a vector space over $K$, and $\seqi{T}$ such that $T_i \in \hom(E_i; E)$ for all $i \in I$, then $E$ equipped with the inductive topology is bornolgic. 
+\end{proposition}
+\begin{proof}
+    Let $\rho: E \to [0, \infty)$ be a seminorm on $E$ that is bounded on all bounded sets. For each $i \in I$ and $B \subset E_i$ bounded, $T_i(B)$ is bounded by \autoref{proposition:bornological-bounded}, and $\rho \circ T_i(B)$ is bounded by assumption. Thus for every $i \in I$, $\rho \circ T_i$ is continuous, so $\rho$ is continuous by (4) of \autoref{definition:lc-inductive}. 
 \end{proof}
 
 
+\begin{proposition}
+\label{proposition:bornological-continuous-complete}
+    Let $E$ be a bornological space and $F$ be a complete separated locally convex space, then $L_b(E; F)$ is complete. In particular, $E^*$ equipped with the topology of bounded convergence is complete.
+\end{proposition}
+\begin{proof}
+    By \autoref{proposition:bornological-bounded}, $L_b(E; F) = B(E; F)$. By \autoref{proposition:operator-space-completeness}, $B(E; F)$ is complete, so $L_b(E; F)$ is complete as well.
+\end{proof}
+
+\begin{definition}[Associated Bornological Space]
+\label{definition:associated-bornological}
+    Let $(E, \mathcal{T})$ be a separated locally convex space over $K \in \RC$, then there exists a locally convex topology $\mathcal{T}_B \subset 2^E$ such that:
+    \begin{enumerate}
+        \item $\mathfrak{B}(E, \mathcal{T}_B) \supset \mathfrak{B}(E, \mathcal{T})$. 
+        \item[(U)] For every $\mathcal{S}$ satisfying (1), $\mathcal{S} \subset \mathcal{T}_B$. In particular, $\mathcal{T} \subset \mathcal{T}_B$ and $\mathfrak{B}(E, \mathcal{T}_B) = \mathfrak{B}(E, \mathcal{T})$. 
+        \item $(E, \mathcal{T}_B)$ is a bornological space. 
+        \item Let $\mathcal{B} \subset \mathfrak{B}(E, \topo)$ be the collection of all convex, circled, bounded, and closed subsets of $E$, ordered by inclusion. For each $B \in \mathcal{B}$, let $(E_B, \rho_B)$ be the normed space associated with $B$ and $\iota_B: E_B \to E$ be the canonical inclusion, then $\mathcal{T}_B$ is the inductive topology induced by $\bracsn{\iota_B}_{B \in \mathcal{B}}$. 
+    \end{enumerate}
+
+    The space $(E, \mathcal{T}_B)$ is the \textbf{bornological space associated with} $E$. 
+\end{definition}
+\begin{proof}
+    Let $\mathscr{T}$ be the collection of all topologies satisfying (1), then $\mathcal{T} \in \mathscr{T}$, so $\mathscr{T} \ne \emptyset$. Let $\mathcal{T}_B$ be the projective topology induced by $\mathscr{T}$. 
+
+    (1): Since $\mathcal{T}_B \supset \mathcal{T}$, $\mathfrak{B}(E, \mathcal{T}) \supset \mathfrak{B}(E, \mathcal{T}_B)$. Let $B \in (E, \mathcal{T})$ and $U \in \cn_{\mathcal{T}_B}(0)$, then there exists $\mathscr{S} \subset \mathscr{T}$ finite and $\bracsn{U_{\mathcal{S}}}_{\mathcal{S} \in \mathscr{S}}$ such that $\bigcap_{\mathcal{S} \in \mathscr{S}}U_{\mathcal{S}} \subset U$. In which case, since $\mathfrak{B}(E, \mathcal{T}) = \mathfrak{B}(E, \mathcal{S})$ for all $\mathcal{S} \in \mathscr{S}$, $\bigcap_{\mathcal{S} \in \mathscr{S}}U_{\mathcal{S}}$ absorbs $B$, so $U$ absorbs $B$ as well. 
+    
+    (U): By (U) of the \hyperref[projective topology]{definition:tvs-initial}.
+
+    (3): Let $\rho: E \to [0, \infty)$ be a seminorm that is bounded on bounded sets, then the topology induced by $\rho$ satisfies (1). By (U), $\rho$ is continuous with respect to $\mathcal{T}_B$, so $(E, \mathcal{T}_B)$ is a bornological space. 
+
+    (4): Let $\mathcal{S}$ be the inductive topology on $E$ induced by $\bracsn{\iota_B}_{B \in \mathcal{B}}$. Let $B \in \mathfrak{B}(E, \mathcal{T})$, $U \in \cn_{\mathcal{S}}(0)$, and $\rho: E \to [0, \infty)$ be the gauge of $U$. Since $\iota_{\ol B} \in L(E_{\ol B}; E, \mathcal{S})$, there exists $\lambda > 0$ such that $\rho \le \lambda \iota_{\ol B} \circ \rho_{\ol B}$. Thus $2\lambda U \supset B$ and $(E, \mathcal{S})$ satisfies (1). By (U), $\mathcal{T}_B \supset \mathcal{S}$. 
+
+    On the other hand, since $\mathfrak{B}(E, \mathcal{T}_B) = \mathfrak{B}(E, \mathcal{T})$, for each $B \in \mathcal{B}$, $\iota_B: E_B \to (E, \mathcal{T}_B)$ is continuous. Therefore by (U) of the \hyperref[inductive topology]{definition:lc-inductive}, $\mathcal{S} \supset \mathcal{T}_B$. 
+\end{proof}
+

src/fa/lc/continuous.tex

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

src/fa/lc/convex.tex

@@ -7,6 +7,28 @@
     Let $E$ be a vector space over $K \in \RC$, then $A \subset E$ is \textbf{convex} if for any $x, y \in A$, $\bracs{\lambda x + (1 - \lambda) y| \lambda \in [0, 1]} \subset A$.
 \end{definition}
 
+\begin{definition}[Convex Hull]
+\label{definition:convex-hull}
+    Let $E$ be a vector space over $K \in \RC$ and $A \subset E$, then the set
+    \[
+    \text{Conv}(A) = \bracs{\sum_{j = 1}^n t_j x_j \bigg | \seqf{t_j} \subset [0, 1], \seqf{x_j} \subset E, \sum_{j = 1}^n t_j = 1 }
+    \]
+
+    is the \textbf{convex hull} of $A$. 
+\end{definition}
+
+\begin{definition}[Convex Circled Hull]
+\label{definition:convex-circled-hull}
+    Let $E$ be a vector space over $K \in \RC$ and $A \subset E$, then the set
+    \[
+    \aconv(A) = \bracs{\sum_{j = 1}^n t_j x_j \bigg | \seqf{t_j} \subset K, \seqf{x_j} \subset E, \sum_{j = 1}^n |t_j| \le 1 }
+    \]
+
+    is the \textbf{convex circled hull} of $A$. 
+\end{definition}
+
+
+
 \begin{lemma}[{{\cite[II.1.1]{SchaeferWolff}}}]
 \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
@@ -67,10 +89,6 @@
 
 
 
-
-
-
-
 \begin{definition}[Sublinear Functional]
 \label{definition:sublinear-functional}
     Let $E$ be a vector space over $K \in \RC$, then a \textbf{sublinear functional} is a mapping $\rho: E \to \real$ such that:
@@ -94,21 +112,22 @@
 
 \begin{lemma}
 \label{lemma:continuous-seminorm}
-    Let $E$ be a TVS over $K \in \RC$ and $[\cdot]: E \times E \to [0, \infty)$ be a seminorm on $E$, then the following are equivalent:
+    Let $E$ be a TVS over $K \in \RC$ and $[\cdot]: E \to [0, \infty)$ be a seminorm on $E$, then the following are equivalent:
     \begin{enumerate}
         \item $[\cdot]$ is uniformly continuous.
         \item $[\cdot]$ is continuous.
         \item $[\cdot]$ is continuous at $0$.
         \item $\bracs{x \in E| [x] < 1} \in \cn_E(0)$.
+        \item $\bracs{x \in E| [x] \le 1} \in \cn_E(0)$.
     \end{enumerate}
 \end{lemma}
 \begin{proof}
-    $(4) \Rightarrow (1)$: Let $x, y \in E$ and $r > 0$. If
+    $(5) \Rightarrow (1)$: Let $x, y \in E$ and $r > 0$. If
     \[
-    x - y \in \bracs{x \in E|[x] < r} = r\bracs{x \in E|[x] < 1} \in \cn_E(0)
+    x - y \in \bracs{x \in E|[x] \le r} = r\bracs{x \in E|[x] \le 1} \in \cn_E(0)
     \]
     
-    then $[x - y] < r$.
+    then $[x - y] \le r$.
 \end{proof}
 
 
@@ -120,16 +139,17 @@
         \item The topology induced by $\seqi{d}$ makes $E$ a topological vector space.
         \item For each $i \in I$, $[\cdot]_i: E \to [0, \infty)$ is continuous.
     \end{enumerate}
+
     The topology induced by $\seqi{d}$ is the \textbf{vector space topology induced by} $\seqi{[\cdot]}$. In addition,
     \begin{enumerate}
-        \item[(U)] For any family $\seqj{[\cdot]}$ of seminorms continuous on $E$, the vector space topology induced by $\seqj{[\cdot]}$ is contained in the vector space topology induced by $\seqi{[\cdot]}$.
+        \item[(U)] For any family $\bracsn{[\cdot]_j}_{j \in J}$ of continuous seminorms on $E$, the vector space topology induced by $\bracsn{[\cdot]_j}_{j \in J}$ is contained in the vector space topology induced by $\seqi{[\cdot]}$.
     \end{enumerate}
 \end{definition}
 
 
 \begin{definition}[Gauge/Minkowski Functional]
 \label{definition:gauge}
-    Let $E$ be a vector space over $K \in \RC$ and $A \subset E$ be a radial set, then the mapping
+    Let $E$ be a vector space over $K \in \RC$ and $A \subset E$ be radial, then the mapping
     \[
     [\cdot]_A: E \to [0, \infty) \quad x \mapsto \inf\bracsn{\lambda > 0| \lambda^{-1}x \in A}
     \]
@@ -139,11 +159,13 @@
         \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 $\bracs{\rho < 1} \subseteq A \subseteq \bracs{\rho \le 1} \subseteq \ol A$.
     \end{enumerate}
+
     In particular,
-    \begin{enumerate}
+    \begin{enumerate}[start=4]
-        \item[(4)] If $A$ is convex, then $[\cdot]_A$ is a sublinear functional.
+        \item If $A$ is convex, then $[\cdot]_A$ is a sublinear functional.
-        \item[(5)] If $A$ is convex and circled, then $[\cdot]_A$ is a seminorm.
+        \item If $A$ is convex and circled, then $[\cdot]_A$ is a seminorm.
     \end{enumerate}
 \end{definition}
 \begin{proof}
@@ -153,6 +175,13 @@
     \]
 
     then $(\lambda + \mu)^{-1} \in A$, and $\lambda + \mu \ge [x + y]_A$. Thus $[x + y]_A \le [x]_A + [y]_A$.
+
+    (4): Let $x \in \bracs{\rho \le 1}$, then $\lambda x \in A$ for all $\lambda \in (0, 1)$. Therefore
+    \[
+    x \in \overline{\bracs{\lambda x|\lambda \in (0, 1)}} \subset A
+    \]
+
+    so $x \in \overline{A}$. 
 \end{proof}
 
 \begin{definition}[Locally Convex Space]
@@ -163,12 +192,42 @@
         \item There exists a fundamental system of neighbourhoods at $0$ consisting of convex, circled, and radial sets.
         \item There exists a family of seminorms $\seqi{[\cdot]}$ that induces the topology on $E$.
     \end{enumerate}
+
     If the above holds, then $E$ is a \textbf{locally convex} space.
 \end{definition}
 \begin{proof}
-    $(1) \Rightarrow (2)$: Let $U \in \cn(0)$ be convex. By \autoref{proposition:tvs-good-neighbourhood-base}, there exists $V \in \cn(0)$ circled such that $V + V \subset U$. Let $W = \bracs{tx + (1 - t)y|x, y \in V}$ be the convex hull of $V$, then $W \subset U$ is convex and circled.
+    $(1) \Rightarrow (2)$: Let $U \in \cn(0)$ be convex. By \autoref{proposition:tvs-good-neighbourhood-base}, there exists $V \in \cn(0)$ circled such that $V + V \subset U$. Let $W = \text{Conv}(V)$ be the convex hull of $V$, then $W \subset U$ is convex and circled. 
 
     $(2) \Rightarrow (3)$: For each $V \in \cn(0)$ convex, circled, and radial, let $[\cdot]_V: E \to [0, \infty)$ be its gauge, then $[\cdot]_V$ is a seminorm. For each $x, y \in X$ and $r > 0$, $[x - y]_V < r$ if and only if $x - y \in rV$. Thus the uniformity induced by $[\cdot]_V$ corresponds to the uniformity generated by $\bracs{U_{rV}| r > 0}$, where $U_V = \bracs{(x, y) \in E|x - y \in V}$. Since this holds for all $U \in \cn(0)$, the topology of $E$ and the topology induced by $\bracs{[\cdot]_V| V \in \cn(0), \text{ convex, circled, radial}}$ coincide.
 
     $(3) \Rightarrow (1)$: For each $i \in I$ and $r > 0$, $\bracs{x \in E| [x]_i < r}$ is convex.
 \end{proof}
+
+
+
+\begin{definition}[Associated Normed Space]
+\label{definition:lc-associated-normed-space}
+    Let $E$ be a separated locally convex space and $A \subset E$ be convex and circled. Let $E_0 = \bigcup_{n \in \natp}nA$, $\rho_0: E_0 \to [0, \infty)$ be the gauge of $A$, and $(E_A, \rho_A)$ be the quotient of $E_0$ by $\bracs{\phi = 0}$, equipped with the quotient norm of $\rho_0$, then
+    \begin{enumerate}
+        \item $(E_A, \rho_A)$ is a normed space. 
+    \end{enumerate}
+
+    If $A$ is radial, then $E_0 = E$ and the map $\pi_A: E \to E_A$ is the \textbf{canonical projection}, and
+    \begin{enumerate}[start=1]
+        \item If $A \in \cn_E(0)$, then $\pi_A \in L(E; E_A)$.
+    \end{enumerate}
+
+    If $(E_0, \rho_0)$ is separated, then $(E_0, \rho_0) = (E_A, \rho_A)$, and the map $\iota_A: E_A \to E$ is the \textbf{canonical inclusion}. In particular, if $A$ is bounded, then
+    \begin{enumerate}[start=2]
+        \item $(E_0, \rho_0)$ is separated. 
+        \item $\iota_A \in L(E_A; E)$. 
+    \end{enumerate}
+
+    The space $(E_A, \rho_A)$ is the \textbf{normed space associated with} $A$. 
+\end{definition}
+\begin{proof}
+    (3): Let $x \in E_0 \setminus \bracs{0}$. Since $E$ is separated, there exists $U \in \cn_E(0)$ such that $x \not\in U$. As $A$ is bounded, there exists $\lambda > 0$ such that $\lambda U \supset A$. In which case, $x \not\in \lambda^{-1}A$, and $E_0$ is separated. 
+
+    (4): Let $U \in \cn_E(0)$, then there exists $\lambda > 0$ such that $\lambda U \supset A$, so $\iota_A^{-1}(U) \supset \lambda^{-1}A \in \cn_{E_A}(0)$, and $\iota_A$ is continuous by \autoref{proposition:tvs-convex-morphism}.
+\end{proof}
+

src/fa/lc/hahn-banach.tex

@@ -35,15 +35,15 @@
     \end{align*}
 \end{proof}
 
-\begin{theorem}[Hahn-Banach, Analytic Form {{\cite[Theorem 5.6, 5.7]{Folland}}}]
+\begin{theorem}[Hahn-Banach, Analytic Form]
 \label{theorem:hahn-banach}
-    Let $E$ be a vector space over $K \in \RC$, $\rho: E \to \real$ be a sublinear functional, $F \subsetneq E$ be a subspace,
+    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 $\Phi|_F = \phi$.
     \end{enumerate}
 \end{theorem}
-\begin{proof}
+\begin{proof}[Proof {{\cite[Theorem 5.6, 5.7]{Folland}}}. ]
     (1): Let $x_0 \in E \setminus F$, then by \autoref{lemma:hahn-banach}, there exists $\lambda \in \real$ such that if
     \[
     \phi_{x_0, \lambda}: F + \real x_0 \to \real \quad (x + tx_0) \mapsto x + \lambda t
@@ -77,7 +77,7 @@
 \begin{proof}
     By translation, assume without loss of generality that $0 \in A$. In which case, $A \in \cn^o(0)$ is convex.
 
-    Let $[\cdot]_A: E \to [0, \infty)$ be the \hyperref[gaugeg]{definition:gauge} of $A$, then $[\cdot]_A$ is a sublinear functional on $E$. For any $y, z \in E$ and $t > 0$ with $y, z \in tA$,
+    Let $[\cdot]_A: E \to [0, \infty)$ be the \hyperref[gauge]{definition:gauge} of $A$, then $[\cdot]_A$ is a sublinear functional on $E$. For any $y, z \in E$ and $t > 0$ with $y, z \in tA$,
     \[
     \abs{[y]_A - [z]_A} \le [y - z]_A \le t
     \]
@@ -96,21 +96,21 @@
     Finally, let $y \in A$, then $\dpb{y, \phi}{E} \le [y]_A < 1 = \dpb{x, \phi}{E}$.
 \end{proof}
 
-\begin{theorem}[Hahn-Banach, First Geometric Form {{\cite[Theorem 1.6]{Brezis}}}]
+\begin{theorem}[Hahn-Banach, First Geometric Form]
 \label{theorem:hahn-banach-geometric-1}
     Let $E$ be a TVS over $\real$, $A, B \subset E$ be non-empty convex sets, such that $A \cap B = \emptyset$ and $A$ is open, then there exists $\phi \in E^* \setminus \bracs{0}$ and $\alpha \in \real$ with $A \subset \bracs{\phi \le \alpha}$ and $B \subset \bracs{\phi \ge \alpha}$.
 \end{theorem}
-\begin{proof}
+\begin{proof}[Proof {{\cite[Theorem 1.6]{Brezis}}}. ]
     Let $C = A - B$, then $C \subset E$ is an open convex set by \autoref{lemma:convex-gymnastics}. Since $A \cap B = \emptyset$, $0 \not\in C$.
 
     By \autoref{lemma:hahn-banach-separation}, there exists $\phi \in E^*$ such that $C \subset \bracs{\phi < 0}$. In which case, for any $a \in A$ and $b \in B$, $\dpb{a - b, \phi}{E} < 0$ implies that $\dpb{a, \phi}{E} < \dpb{b, \phi}{E}$. Let $\alpha \in [\sup_{a \in A}\dpb{a, \phi}{E}, \inf_{b \in B}\dpb{b, \phi}{E}]$, then $A \subset \bracs{\phi \le \alpha}$ and $B \subset \bracs{\phi \ge \alpha}$.
 \end{proof}
 
-\begin{theorem}[Hahn-Banach, Second Geometric Form {{\cite[Theorem 1.7]{Brezis}}}]
+\begin{theorem}[Hahn-Banach, Second Geometric Form]
 \label{theorem:hahn-banach-geometric-2}
     Let $E$ be a locally convex space over $\real$, $A, B \subset E$ be non-empty convex sets such that $A \cap B = \emptyset$, $A$ is closed, and $B$ is compact, then there exists $\phi \in E^* \setminus \bracs{0}$, $\alpha \in \real$, and $r > 0$ with $A \subset \bracs{\phi \le \alpha - r}$ and $B \subset \bracs{\phi \ge \alpha + r}$.
 \end{theorem}
-\begin{proof}
+\begin{proof}[Proof {{\cite[Theorem 1.7]{Brezis}}}. ]
     Let $C = A - B$, then $C$ is closed by \autoref{proposition:tvs-set-operations} with $0 \not\in C$. Since $E$ is locally convex, there exists $U \in \cn^o(0)$ convex such that $U \cap C \ne \emptyset$.
 
     By \autoref{theorem:hahn-banach-geometric-1}, there exists $\phi \in E^* \setminus \bracs{0}$ and $\alpha \in \real$ such that $U \subset \bracs{f \le \alpha}$ and $C \subset \bracs{f \ge \alpha}$. In which case, for any $u \in U$, $a \in A$, and $b \in B$,
@@ -135,12 +135,13 @@
             \item $|\phi| \le \rho$.
             \item $\dpb{x, \phi}{E} = \inf_{y \in M}\rho(x + y)$.
         \end{enumerate}
+
         \item For any $x \in E$ and continuous seminorm $\rho: E \to [0, \infty)$, there exists $\phi \in E^*$ with $|\phi| \le \rho$ and $\dpb{x, \phi}{E} = \rho(x)$.
-        \item If $E$ is Hausdorff, then for any $x, y \in E$, there exists $\phi \in E^*$ with $\dpb{x, \phi}{E} \ne \dpb{y, \phi}{E}$.
+        \item If $E$ is separated, then for any $x, y \in E$, there exists $\phi \in E^*$ with $\dpb{x, \phi}{E} \ne \dpb{y, \phi}{E}$.
     \end{enumerate}
 \end{proposition}
 \begin{proof}
-    (1): Let $\rho_M: E \to [0, \infty)$ be the quotient of $\rho$ by $M$, then $\rho_M \le \rho$ is a continuous seminorm on $E$ by \autoref{definition:quotient-norm}. Let $\phi_0: Kx \to K$ be defined by $\lambda x \mapsto \lambda \rho_M(x)$. By the \hyperref[Hahn-Banach theorem]{theorem:hahn-banach}, there exists $\phi \in \hom{E; K}$ such that $\dpb{x, \phi}{E} = \rho_M(x)$ and $|\phi| \le \rho_M \le \rho$.
+    (1): Let $\rho_M: E \to [0, \infty)$ be the quotient of $\rho$ by $M$, then $\rho_M \le \rho$ is a continuous seminorm on $E$ by \autoref{definition:quotient-norm}. Let $\phi_0: Kx \to K$ be defined by $\lambda x \mapsto \lambda \rho_M(x)$. By the \hyperref[Hahn-Banach Theorem]{theorem:hahn-banach}, there exists $\phi \in \hom(E; K)$ such that $\dpb{x, \phi}{E} = \rho_M(x)$ and $|\phi| \le \rho_M \le \rho$.
 
     (2): By (1) applied to $M = \bracs{0}$.
 
@@ -149,7 +150,7 @@
 
 \begin{proposition}
 \label{proposition:seminorm-lsc}
-    Let $E$ be a locally convex space and $\rho: E \to [0, \infty)$ be a continuous seminorm, then $\rho: E_w \to [0, \infty)$ is lower semicontinuous and Borel measurable. 
+    Let $E$ be a locally convex space and $\rho: E \to [0, \infty)$ be a continuous seminorm, then $\rho: E_w \to [0, \infty)$ is lower semicontinuous and Borel measurable with respect to the weak topology.
 \end{proposition}
 \begin{proof}
     Let $x \in E$, then there exists $\phi_x \in E^*$ such that $\dpn{x, \phi_x}{E} = \rho(x)$ and $|\phi_x| \le \rho$. Thus

src/fa/lc/index.tex

@@ -4,9 +4,11 @@
 
 \input{./convex.tex}
 \input{./continuous.tex}
+\input{./barrel.tex}
 \input{./bornologic.tex}
 \input{./quotient.tex}
 \input{./projective.tex}
 \input{./inductive.tex}
 \input{./hahn-banach.tex}
 \input{./spaces-of-linear.tex}
+\input{./tensor.tex}

src/fa/lc/inductive.tex

@@ -3,7 +3,7 @@
 
 \begin{definition}[Inductive Locally Convex Topology]
 \label{definition:lc-inductive}
-    Let $\seqi{E}$ be locally convex spaces over $K \in \RC$, $\seqi{T}$ such that $T_i \in \hom(E_i; E)$ for all $i \in I$, and $E$ be a vector space over $K$, then there exists a topology $\topo$ on $E$ such that:
+    Let $\seqi{E}$ be locally convex spaces over $K \in \RC$, $E$ be a vector space over $K$, and $\seqi{T}$ such that $T_i \in \hom(E_i; E)$ for all $i \in I$, then there exists a topology $\topo$ on $E$ such that:
     \begin{enumerate}
         \item $(E, \topo)$ is a locally convex space over $K$.
         \item For each $i \in I$, $T_i \in L(E_i; E)$.
@@ -11,11 +11,18 @@
         \item For any locally convex space $F$ and $T \in \hom(E; F)$, $T \in L(E; F)$ if and only if $T \circ T_i \in L(E_i; F)$ for all $i \in I$.
         \item The family
         \[
-        \mathcal{B} = \bracs{U \subset E|U \text{ convex, radial, circled}, T_i^{-1}(U) \in \cn_{E_i}(0) \forall i \in I}
+        \mathcal{B} = \bracs{U \subset E|U \text{ convex, circled, radial}, T_i^{-1}(U) \in \cn_{E_i}(0) \forall i \in I}
+        \]
+
+        is a fundamental system of neighbourhoods for $E$ at $0$.
+        \item If $E$ is spanned by $\bigcup_{i \in I}T_i(E_i)$, then
+        \[
+        \fB = \bracs{\Gamma\paren{\bigcup_{i \in I}T_i(U_i)} \bigg | U_i \in \cn_{E_i}(0)}
         \]
 
         is a fundamental system of neighbourhoods for $E$ at $0$. 
     \end{enumerate}
+
     The topology $\topo$ is the \textbf{inductive locally convex topology} on $E$ induced by $\seqi{T}$.
 \end{definition}
 \begin{proof}
@@ -31,7 +38,44 @@
 
     (U): Let $U \in \cn_{(E, \mathcal{S})}(0)$ be convex, circled, and radial. By (2), $T_i^{-1}(U) \in \cn_{E_i}(0)$ for all $i \in I$. Thus the convex, circled, and radial neighbourhoods of $0$ in $(E, \mathcal{S})$ is a subset of $\mathcal{B}$.
 
-    (4): Let $i \in I$ and $U \in \cn_F(0)$ be convex, circled, and radial. Since $T \circ E_i \in L(E_i; F)$, $T_i^{-1}(T^{-1}(U)) \in \cn_{E_i}(0)$, so $T^{-1}(U) \in \mathcal{B} \subset \cn_E(0)$.
+    (5): Let $i \in I$ and $U \in \cn_F(0)$ be convex, circled, and radial. Since $T \circ E_i \in L(E_i; F)$, $T_i^{-1}(T^{-1}(U)) \in \cn_{E_i}(0)$, so $T^{-1}(U) \in \mathcal{B} \subset \cn_E(0)$.
+
+    (6): If $E$ is spanned by $\bigcup_{i \in I}T_i(E_i)$, then each set in $\fB$ is radial. Hence $\fB$ is a family of neighbourhoods of $E$ at $0$. 
+
+    Let $U \in \cn_E(0)$ be convex, circled, and radial, then for each $i \in I$, $T_i^{-1}(U) \in \cn_{E_i}(0)$, so $U \supset \bigcup_{i \in I}T_i[T_i^{-1}(U)]$. Since $U$ is convex and circled, $U \supset \Gamma\paren{\bigcup_{i \in I}T_i[T_i^{-1}(U)]} \in \fB$. Therefore $\fB$ forms a fundamental system of neighbourhoods for $E$ at $0$.
+\end{proof}
+
+\begin{definition}[Locally Convex Direct Sum]
+\label{definition:lc-direct-sum}
+    Let $\seqi{E}$ be locally convex spaces over $K \in \RC$, then there exists $(E, \seqi{\iota})$ such that:
+    \begin{enumerate}
+        \item $E$ is a locally convex space over $K$. 
+        \item For each $i \in I$, $\iota_i \in L(E_i; E)$. 
+        \item[(U)] For each $(F, \seqi{T})$ satisfying (1) and (2), there exists a unique $T \in L(E; F)$ such that the following diagram commutes:
+        \[
+        \xymatrix{
+        A \ar@{->}[r]^{T} & B \\
+        A_i \ar@{->}[u]^{\iota_i} \ar@{->}[ru]_{T_i} & 
+        }
+        \]
+        
+        \item The family
+        \[
+        \fB = \bracs{\Gamma\paren{\bigcup_{i \in I}\iota_i(U_i)} \bigg | U_i \in \cn_{E_i}(0)}
+        \]
+
+        is a fundamental system of neighbourhoods for $E$ at $0$. 
+    \end{enumerate}
+
+    
+    The space $E = \bigoplus_{i \in I}E_i$ is the \textbf{locally convex direct sum} of $\seqi{E}$. 
+\end{definition}
+\begin{proof}
+    Let $(E, \seqi{\iota})$ be the direct sum of $\seqi{E}$ as vector spaces, and equip it with the inductive topology induced by $\seqi{\iota}$, then $(E, \seqi{\iota})$ satisfies (1) and (2). 
+
+    (U): By (U) of the \hyperref[direct sum]{definition:direct-sum}, there exists a unique $T \in \hom(E; F)$ such that the diagram commutes. In which case, by (4) of \autoref{definition:lc-inductive}, $T \in L(E; F)$. 
+
+    (4): By (6) of \autoref{definition:lc-inductive}.
 \end{proof}
 
 \begin{definition}[Inductive Limit]
@@ -69,6 +113,7 @@
 
         is a fundamental system of neighbourhoods for $E$ at $0$.
     \end{enumerate}
+
     The pair $(E, \bracsn{T^i_E}_{i \in I})$ is the \textbf{inductive limit} of $(\seqi{E}, \bracsn{T^i_j| i, j \in I, i \lesssim j})$.
 \end{definition}
 \begin{proof}
@@ -88,6 +133,9 @@
     In addition to the neighbourhood construction given above, the inductive topology may also be constructed as the weak topology generated by all topologies satisfying certain properties. While this more non-constructive method is simpler, it does not directly provide an explicit fundamental system of neighbourhoods at $0$.
 \end{remark}
 
+\subsection{Strict Inductive Limits}
+\label{subsection:lc-induct-strict}
+
 \begin{lemma}[{{\cite[Lemma II.6.4]{SchaeferWolff}}}]
 \label{lemma:lc-induct-separate}
     Let $E$ be a locally convex space over $K$, $M \subset E$ be a subspace, $U \in \cn_M(0)$ be convex and circled, then
@@ -130,6 +178,7 @@
             \item[(a)] $B$ is bounded.
             \item[(b)] There exists $n \in \natp$ such that $B \subset E_n$ is bounded.
         \end{enumerate}
+
         \item If $E_n$ is complete for each $n \in \natp$, then $E$ is also complete.
     \end{enumerate}
 \end{proposition}
@@ -146,6 +195,7 @@
         \item For each $k \in \natp$, $U_k = U_{k+1} \cap E_{n_k}$.
         \item For each $k \in \natp$, $n^{-1}x_k \not\in U_k$.
     \end{enumerate}
+
     then $V = \bigcup_{k \in \natp}U_k \in \cn_E(0)$ with $V \cap E_{n_k} = U_k$ for all $k \in \natp$. For any $n \in \natp$, $x_k \not\in nU_k = nV \cap E_{n_k}$. Therefore $B$ is not bounded.
 
     (3), $(b) \Rightarrow (a)$: Let $U \in \cn_E(0)$, then $U \cap E_n \in \cn_{E_n}(0)$, so there exists $\lambda \in K$ with $\lambda (U \cap E_n) \supset B$.

src/fa/lc/projective.tex

@@ -25,7 +25,7 @@
 
 \begin{proposition}[{{\cite[II.5.4]{SchaeferWolff}}}]
 \label{proposition:complete-lc-projective-limit}
-    Let $E$ be a Hausdorff complete locally convex space over $K \in \RC$, $\mathcal{B} \subset \cn_E(0)$ be a fundamental system of neighbourhoods consisting of convex, circled, and radial sets, directed under inclusion.
+    Let $E$ be a separated complete locally convex space over $K \in \RC$, $\mathcal{B} \subset \cn_E(0)$ be a fundamental system of neighbourhoods consisting of convex, circled, and radial sets, directed under inclusion.
 
     For each $U \in \mathcal{B}$, let $[\cdot]_U$ be its gauge, $M_U = \bracs{x \in E|[\cdot]_U = 0}$, $E_U = E/M_U$, and $\norm{\cdot}_U: E_U \to [0, \infty)$ be the quotient of $[\cdot]_U$ by $M_U$, then
     \begin{enumerate}
@@ -51,7 +51,7 @@
 
     (3): Let $\lim E_U$ be the projective limit. For each $U \in \mathcal{B}$, let $p_U: \lim E_U \to E_U$ be the canonical map.
 
-    Let $x \in E$. Since $E$ is Hausdorff and $\mathcal{B}$ is a fundamental system of neighbourhoods at $0$, there exists $U \in \mathcal{B}$ such that $\pi_U(x) \ne 0$. In which case, $p_U \circ \pi(x) = \pi_U(x) \ne 0$, so $\pi$ is injective.
+    Let $x \in E$. Since $E$ is separated and $\mathcal{B}$ is a fundamental system of neighbourhoods at $0$, there exists $U \in \mathcal{B}$ such that $\pi_U(x) \ne 0$. In which case, $p_U \circ \pi(x) = \pi_U(x) \ne 0$, so $\pi$ is injective.
 
     Let $x \in \lim E_U$. For each $U \in \mathcal{B}$, let $x_U \in E$ such that $\pi_U(x_U) = p_U(x)$. For any $V \in \cn_E(0)$, there exists $W \in \mathcal{B}$ with $W \subset V$. In which case, for any $U \in \mathcal{B}$ with $U \subset W$,
     \[

src/fa/lc/quotient.tex

@@ -42,6 +42,7 @@
         If $F$ is a TVS over $K$ and $f \in L(E; F)$, then $\td f \in L(E; F)$.
         \item If $\seqi{\rho}$ is a family of seminorms that induces the topology on $E$, then their quotients by $M$ induces the topology on $\td E$.
     \end{enumerate}
+
     The space $\td E = E/M$ is the \textbf{quotient} of $E$ by $M$.
 \end{definition}
 \begin{proof}

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

@@ -3,18 +3,62 @@
 
 \begin{proposition}
 \label{proposition:lc-spaces-linear-map}
-    Let $T$ be a set, $E$ be a locally convex space defined by the seminorms $\seqi{[\cdot]}$, and $\mathfrak{S} \subset 2^T$ be an upward-directed family. For each $i \in I$ and $S \in \mathfrak{S}$, let
+    Let $T$ be a set, $\sigma \subset 2^T$ be an ideal, $E$ be a locally convex space over $K$, and $\cf \subset E^T$ be a subspace such that
+    \begin{enumerate}
+        \item[(B)] For each $f \in \cf$ and $S \in \sigma$, $f(S) \subset E$ is bounded. 
+    \end{enumerate}
+
+    
+    For each $S \in \sigma$ and continuous seminorm $\rho: E \to [0, \infty)$, let
     \[
-    [\cdot]_{S, i}: E^T \to [0, \infty) \quad f \mapsto \sup_{x \in S}[f(x)]_{S, i}
+    \rho_S: E^T \to [0, \infty] \quad f \mapsto \sup_{x \in S}\rho(f(x))
     \]
 
-    then the $\mathfrak{S}$-uniform topology on $E^T$ is defined by the seminorms
+    then the $\sigma$-uniform topology on $\cf$ is induced by seminorms of the form $\rho_S$, where $\rho$ is a continuous seminorm on $E$, and $S \in \sigma$. In which case, the $\sigma$-uniform topology on $\cf$ is locally convex. 
-    \[
-    \bracs{[\cdot]_{S, i}|S \in \mathfrak{S}, i \in I}
-    \]
-
-    and hence locally convex.
 \end{proposition}
 \begin{proof}
-    By \autoref{proposition:set-uniform-pseudometric}.
+    By \autoref{proposition:tvs-set-uniformity} and \autoref{proposition:set-uniform-pseudometric}.
+\end{proof}
+
+\begin{definition}[Saturated Ideal]
+\label{definition:saturated-ideal}
+    Let $E$ be a locally convex space over $K \in \RC$ and $\sigma \subset 2^E$ be an ideal, then $\sigma$ is \textbf{saturated} if:
+    \begin{enumerate}
+        \item For each $\lambda \in K$ and $S \in \sigma$, $\lambda S \in \sigma$. 
+        \item For each $S \in \sigma$, $\ol{\aconv}(S) \in \sigma$. 
+    \end{enumerate}
+
+    For any ideal $\sigma \subset 2^E$, the smallest saturated ideal $\ol \sigma$ containing it is the \textbf{saturated hull} of $\sigma$. 
+\end{definition}
+
+\begin{lemma}
+\label{lemma:locally-convex-saturated}
+    Let $E$, $F$ be locally convex spaces over $K \in \RC$, $\sigma \subset 2^E$ be an ideal, and $\ol \sigma$ be its saturated hull, then the $\sigma$-uniformity and $\ol \sigma$-uniformity on $L(E; F)$ coincide. 
+\end{lemma}
+\begin{proof}
+    Let $\tau \subset \ol \sigma$ be the collection of sets such that for each $S \in \tau$ and $U \in \cn_F(0)$,
+    \[
+    N(S, U) = \bracs{(S, T) \in L(E; F)| (S - T)(S) \subset U}
+    \]
+    
+    is an entourage in the $\sigma$-uniformity. 
+
+    For each $S \in \tau$, $U \in \cn_F(0)$, and $\lambda \in K$ with $\lambda \ne 0$, 
+    \begin{align*}
+    N(\lambda S, U) &= \bracs{(S, T) \in L(E; F)| (S - T)(\lambda S) \subset U} \\
+    &= \bracs{(S, T) \in L(E; F)| (S - T)(S) \subset \lambda^{-1}U}
+    \end{align*}
+
+    is another entourage in the $\sigma$-uniformity. If $\lambda = 0$, then $N(\lambda S, U) = L(E; F)$, which is also an entourage. 
+
+    Now, let $S \in \tau$ and $U \in \cn_F(0)$ be convex and circled, then by \autoref{proposition:closure-of-image},
+    \begin{align*}
+        N(\ol{\aconv}(S), U) &= \bracs{(S, T) \in L(E; F)| (S - T)(\ol{\aconv}(S)) \subset U} \\
+        &\supset \bracs{(S, T) \in L(E; F)| \overline{(S - T)(\aconv(S))} \subset U} \\
+        &= \bracs{(S, T) \in L(E; F)| \ol{\aconv}{(S - T)(S)} \subset U}
+    \end{align*}
+
+    so $N(\ol{\aconv}(S), U)$ contains an entourage in the $\sigma$-uniformity. 
+
+    Since $\tau$ is a saturated ideal that contains $\sigma$, $\tau = \ol \sigma$. Therefore the $\sigma$-uniformity and $\ol \sigma$-uniformity on $L(E; F)$ coincide. 
 \end{proof}

src/fa/lc/tensor.tex

@@ -0,0 +1,182 @@
+\section{The Projective Tensor Product}
+\label{section:projective-tensor-product}
+
+\begin{definition}[Projective Tensor Product]
+\label{definition:projective-tensor-product}
+    Let $E, F$ be locally convex spaces over $K \in \RC$, then there exists a pair $(E \otimes_\pi F, \iota)$ such that:
+    \begin{enumerate}
+        \item $E \otimes_\pi F$ is a locally convex space over $K$. 
+        \item $\iota \in L^2(E, F; E \otimes_\pi F)$ is a continuous bilinear map. 
+        \item[(U1)] For any $(G, \lambda)$ satisfying (1) and (2), there exists a unique $\Lambda \in L(E \otimes_\pi F; G)$ such that the following diagram commutes:
+        \[
+        \xymatrix{
+        E \times F \ar@{->}[rd]_{\lambda} \ar@{->}[r]^{\iota} & E \otimes F \ar@{->}[d]^{\Lambda} \\
+        & G
+        }
+        \]
+        \item[(U2)] For any topology $\topo$ on $E \otimes_\pi F$ satisfying (1) and (2), $\topo$ is coarser than the topology on $E \otimes_\pi F$. 
+
+        \item $E \otimes_\pi F$ is the linear span of $\iota(E \times F)$. 
+        \item For any $U \subset E$ and $V \subset F$, let $U \otimes V = \bracs{u \otimes v|u \in U, v \in V}$, then the convex circled hulls
+        \[
+        \fB = \bracsn{\aconv(U \otimes V)| U \in \cn_E(0), V \in \cn_F(0)}
+        \]
+
+        is a fundamental system of neighbourhoods at $0$ for $E \otimes_\pi F$.
+    \end{enumerate}
+
+    
+    The space $E \otimes_\pi F$ is the \textbf{projective tensor product} of $E$ and $F$, and the mapping $\iota \in L^2(E, F; E \otimes_\pi F)$ is the \textbf{canonical embedding}.
+
+    The space $E \widetilde{\otimes}_\pi F$ denotes the Hausdorff completion of $E \otimes_\pi F$.
+\end{definition}
+\begin{proof}
+    Let $E \otimes_\pi F = E \otimes F$ be the \hyperref[tensor product]{definition:tensor-product} of $E$ and $F$ as vector spaces. Let $\mathscr{T} \subset 2^{2^X}$ be the collection of all locally convex topologies satisfying (1) and (2), and let $\mathcal{S}$ be the projective topology on $E \otimes_\pi F$ generated by $\mathscr{T}$. 
+    
+    (1): By \autoref{proposition:lc-projective-topology}, $\mathcal{S}$ is a locally convex topology on $E \otimes_\tau F$. 
+
+    (2): Since $\iota: E \times F \to E \otimes_\pi F$ is continuous with respect to every topology in $\mathscr{T}$, it is also continuous with respect to $\mathcal{S}$. 
+
+    (U2): Since $\mathcal{T} \in \mathscr{T}$, $\mathcal{S} \supset \mathcal{T}$.
+
+    (U1): By (U) of the \hyperref[tensor product]{definition:tensor-product}, there exists a unique $\Lambda \in \hom(E \otimes_\pi F; G)$ such that the given diagram commutes. Since $\lambda$ is continuous, the projective topology generated by $\Lambda$ satisfies (1) and (2). By (U2), $\mathcal{S}$ contains the projective topology generated by $\Lambda$. Therefore $\Lambda \in L(E \otimes_\pi; F)$. 
+
+    (5): By (4) of the \hyperref[tensor product]{definition:tensor-product}. 
+
+    (6): Let $U \in \cn_E(0)$ and $V \in \cn_F(0)$ be balanced. For any $\sum_{j = 1}^n x_j \otimes y_j \in E \otimes_\pi F$, then there exists $\lambda > 0$ such that $\seqf{x_j} \subset \lambda U$ and $\seqf{y_j} \subset \lambda V$. In which case, $\sum_{j = 1}^n x_j \otimes y_j \subset \lambda \aconv (U \otimes V)$, so $\fB$ is a collection of convex, circled, and radial sets. Since $\fB$ defines a locally convex topology that satisfies (1) and (2), $\mathcal{S}$ contains the topology defined by $\fB$. 
+
+    On the other hand, for any convex and circled set $W \in \cn_{E \otimes_\pi F}(0)$, there exists $U \in \cn_E(0)$ and $V \in \cn_F(0)$ such that $U \otimes V \subset W$. In which case, $W \supset \aconv(U \otimes V)$, so $\fB$ is a fundamental system of neighbourhoods at $0$ for $E \otimes_\pi F$. 
+\end{proof}
+
+\begin{remark}
+\label{remark:projective-construction}
+    In constructing the \hyperref[projective tensor product]{definition:projective-tensor-product}, it may be more natural to obtain its topology as a projective topology using its universal property. However, doing so requires taking a least upper bound across \textit{all continuous linear maps defined on} $E \times F$, a collection too big to be a set. As such, constructing it as a projective topology is logically dubious, or at the very least beyond my abilities. 
+\end{remark}
+
+
+
+\begin{definition}[Cross Seminorm]
+\label{definition:cross-seminorm}
+    Let $E, F$ be locally convex spaces over $K \in \RC$. For any convex circled sets $U \in \cn_E(0)$ and $V \in \cn_F(0)$, let $p: E \to [0, \infty)$ and $q: F \to [0, \infty)$ be their \hyperref[gauges]{definition:gauge}. For any $z \in E \otimes_\pi F$, let
+    \[
+    \rho(z) = \inf\bracs{\sum_{j = 1}^n p(x_j)q(y_j) \bigg | \seqf{(x_j,y_j)} \subset E \times F, z = \sum_{j = 1}^n x_j \otimes y_j}
+    \]
+
+    then
+    \begin{enumerate}
+        \item $\rho$ is a continuous seminorm on $E \otimes_\pi F$.
+        \item $\rho$ is the gauge of $\aconv(U \otimes V)$. 
+        \item For any $x \in E$ and $y \in F$, $\rho(x \otimes y) = p(x)q(Y)$. 
+        \item $\rho$ is a norm if and only if $[\cdot]_U$ and $[\cdot]_V$ are norms. 
+    \end{enumerate}
+
+    and the seminorm $\rho = p \otimes q$ is the \textbf{cross seminorm} of $p$ and $q$. Moreover, 
+    \begin{enumerate}
+        \item[(5)] If the seminorms $\seqi{p}$ define the topology on $E$, and the seminorms $\seqj{q}$ define the topology on $F$, then the seminorms $\bracsn{p_i \otimes q_j| (i, j) \in I \times J}$ define the topology on $E \otimes_\pi F$. 
+    \end{enumerate}
+
+    
+\end{definition}
+\begin{proof}[Proof {{\cite[III.6.3]{SchaeferWolff}}}. ]
+    (1): Let $\lambda \in K$, then for any $\seqf{(x_j,y_j)} \subset E \times F$, 
+    \[
+    |\lambda| \sum_{j = 1}^n p(x_j)q(y_j) = \sum_{j = 1}^n p(\lambda x_j)q(y_j)
+    \]
+
+    and
+    \[
+    \lambda\sum_{j = 1}^n x_j \otimes y_j =  \sum_{j = 1}^n \lambda x_j \otimes y_j
+    \]
+
+    so for any $z \in E \otimes_\pi F$, $|\lambda|\rho(z) = \rho(\lambda z)$. 
+    
+    Let $z, z' \in E \otimes F$, $\seqf{(x_j,y_j)}, \bracsn{(x_j',y_j')}_1^m \subset E \times F$ such that $z = \sum_{j = 1}^n x_j \otimes y_j$ and $z' = \sum_{j = 1}^m x_j' \otimes y_j'$, then
+    \[
+    z + z' = \sum_{j = 1}^n x_j \otimes y_j + \sum_{j = 1}^m x_j' \otimes y_j'
+    \]
+
+    and
+    \[
+    \rho(z + z') \le \sum_{j = 1}^n p(x_j)q(y_j) + \sum_{j = 1}^m p(x_j')q(y_j')
+    \]
+
+    so $\rho$ satisfies the triangle inequality. 
+
+    (2): Let $z \in \aconv(U \otimes V)$, then there exists $\seqf{(x_j, y_j)} \subset U \times V$ and $\seqf{\lambda_j} \subset K$ such that $\sum_{j = 1}^n |\lambda_j| \le 1$ and $z = \sum_{j = 1}^n \lambda x_j \otimes y_j$. In which case, 
+    \begin{align*}
+        \rho(z) &\le \sum_{j = 1}^n p(\lambda x_j)q(y_j) = \sum_{j = 1}^n |\lambda_j|p(x_j)q(y_j) \\
+        &< \sum_{j = 1}^n |\lambda_j| \le 1
+    \end{align*}
+
+    so $\aconv(U \otimes V) \subset \bracs{\rho < 1}$.
+
+    Let $z \in \bracs{\rho < 1}$, then there exists $\seqf{(x_j, y_j)} \subset E \times F$ such that $z = \sum_{j = 1}^nx_j \otimes y_j$ and $\sum_{j = 1}^n p(x_j)q(x_j) < 1$. Let $\eps > 0$ such that $\sum_{j = 1}^n(p(x_j) + \eps)(q(x_j) + \eps) < 1$, then
+    \[
+    z = \sum_{j = 1}^n (p(x_j) + \eps)(q(x_j) + \eps) \cdot \underbrace{\frac{x_j}{p(x_j) + \eps}}_{\in \bracs{p < 1} = U} \otimes \underbrace{\frac{y_j}{q(x_j) + \eps}}_{\in \bracs{q < 1} = V} \in \aconv(U \otimes V)
+    \]
+
+    and $\aconv(U \otimes V) \supset \bracs{\rho < 1}$.
+    
+    (3): Let $x \in U$ and $y \in V$. By the \hyperref[Hahn-Banach Theorem]{proposition:hahn-banach-utility}, there exists $\phi \in E^*$ and $\psi \in F^*$ such that $\dpn{x, \phi}{E} = p(x)$, $\dpn{y, \psi}{F} = q(x)$, $|\phi| \le p$, and $|\psi| \le q$. By (U1) of the \hyperref[projective tensor product]{definition:projective-tensor-product}, there exists $\Phi \in (E \otimes_\pi F)^*$ such that the following diagram commutes
+    \[
+    \xymatrix{
+    E \times F \ar@{->}[rd]_{\phi \cdot \psi} \ar@{->}[r]^{\iota} & E \otimes_\pi F \ar@{->}[d]^{\Phi} \\
+    & K
+    }
+    \]
+    
+    For any $z \in E \otimes_\pi F$ and $\seqf{(x_j, y_j)} \subset E \times F$ such that $z = \sum_{j = 1}^n x_j \otimes y_j$, 
+    \[
+    \Phi(z) = \sum_{j = 1}^n \Phi(x_j \otimes y_j) = \sum_{j = 1}^n \phi(x_j)\psi(y_j) \le \sum_{j = 1}^n p(x_j)q(y_j)
+    \]
+
+    As the above holds for all such $\seqf{(x_j, y_j)} \subset E \times F$, $|\Phi| \le \rho$. Since $\Phi(x \otimes y) = p(x)q(y)$, $\rho(x \otimes y) = p(x)q(y)$ as well. 
+
+    (5): By (6) of \autoref{definition:projective-tensor-product}. 
+\end{proof}
+
+\begin{theorem}[{{\cite[III.6.4]{SchaeferWolff}}}]
+\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}
+        \item $\sum_{n \in \natp}|\lambda_n| < \infty$. 
+        \item $\limv{n}x_n = 0$ and $\limv{n}y_n = 0$.
+        \item $z = \sum_{n = 1}^\infty \lambda_n x_n \otimes y_n$. 
+    \end{enumerate}
+
+    
+\end{theorem}
+\begin{proof}
+    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
+    \begin{align*}
+        r_N(v_N) &= \td r_n(u_{N+1} - u_n) \le \td r_N(u - u_N) + \td r_{N}(u - u_{N+1}) \\
+        &\le \td r_N(u - u_N) + \td r_{N+1}(u - u_{N+1}) < 2^{-N+1}/n^2
+    \end{align*}
+
+    Since $r_N = p_N \otimes q_N$, there exists $\bracsn{(x_{N, k}, y_{N, k})}_{1}^{n_N} \subset X \times Y$ such that $v_N = \sum_{k = 1}^{n_N}x_{N, k} \otimes y_{N, k}$ and
+    \[
+        r_N(v_N) = \sum_{k = 1}^{n_N}p_N(x_{N, k})q_N(x_{N, k}) < 2^{-N+1}/n^2
+    \]
+    
+    By rescaling, assume without loss of generality that there exists $\bracsn{\lambda_{N, k}}_1^{n_N}$ such that\
+    \begin{enumerate}
+        \item $v_N = \sum_{k = 1}^{n_N}\lambda_{N, k}x_{N, k} \otimes y_{N, k}$.
+        \item For each $1 \le k \le n_N$, $p_N(x_{N, k}), q_N(x_{N, k}) \le 1/M$.
+        \item $\sum_{k = 1}^{n_N}|\lambda_k| \le 2^{-N+2}$.
+    \end{enumerate}
+
+    
+    From here, let $\seqf{(x_j, y_j)} \subset X \times Y$ such that $u_1 = \sum_{j = 1}^n x_j \otimes y_j$, then
+    \[
+    u = u_1 + \sum_{N = 1}^\infty v_N = \sum_{j = 1}^n x_j \otimes y_j + \sum_{N = 1}^\infty \sum_{k = 1}^{n_N}\lambda_{N, k}x_{N, k} \otimes y_{N, k}
+    \]
+
+    where $x_{N, k} \to 0$ and $y_{N, k} \to 0$ as $N \to \infty$, and $\sum_{N \in \natp}\sum_{k = 1}^{n_N}|\lambda_{N, k}| < \infty$.
+    
+\end{proof}
+
+
+
+

src/fa/lp/definition.tex

@@ -63,7 +63,7 @@
 
 \begin{theorem}[Minkowski's Inequality, {{\cite[6.5]{Folland}}}]
 \label{theorem:minkowski}
-    Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed space, $p \in [1, \infty]$, and $f, g \in \mathcal{L}^p(X; E)$, then
+    Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed vector space, $p \in [1, \infty]$, and $f, g \in \mathcal{L}^p(X; E)$, then
     \[
     \norm{f + g}_{L^p(X; E)} \le \norm{f}_{L^p(X; E)} + \norm{g}_{L^p(X; E)}
     \]
@@ -89,7 +89,7 @@
 
 \begin{definition}[$L^p$ Space]
 \label{definition:lp}
-    Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed space, and $p \in [1, \infty]$, then $\norm{\cdot}_{L^p(X; E)}$ is a seminorm on $\mathcal{L}^p(X; E)$. The quotient
+    Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed vector space, and $p \in [1, \infty]$, then $\norm{\cdot}_{L^p(X; E)}$ is a seminorm on $\mathcal{L}^p(X; E)$. The quotient
     \[
     L^p(X, \cm, \mu; E) = \mathcal{L}^p(X, \cm, \mu; E)/\bracs{f|f = 0\text{ a.e.}}
     \]
@@ -100,24 +100,77 @@
     By \hyperref[Minkowski's Inequality]{theorem:minkowski}.
 \end{proof}
 
+
 \begin{proposition}
-\label{proposition:lp-simple-dense}
+\label{proposition:dct-lp}
-    Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed space, and $p \in [1, \infty)$, then $\Sigma(X, \cm; E) \cap L^p(X; E)$ is dense in $L^p(X; E)$.
+    Let $(X, \cm, \mu)$ be a measure space, $E$ be a Banach space over $K \in \RC$, $p \in [1, \infty)$, $\seq{f_n} \subset L^p(X; E)$, and $f \in L^p(X; E)$. If
+    \begin{enumerate}
+        \item[(a)] $f_n \to f$ strongly pointwise.
+        \item[(b)] There exists $g \in L^p(X) \cap L^+(X)$ such that $\norm{f_n}_E \le g$ for all $n \in \natp$.
+    \end{enumerate}
+
+    
+    then $f_n \to f$ in $L^p(X; E)$.
 \end{proposition}
 \begin{proof}
-    Let $f \in L^p(X; E)$. By \autoref{definition:strongly-measurable}, there exists $\seq{f_n} \subset \Sigma(X, \cm; E)$ such that $\norm{f_n}_E \le \norm{f}_E$ and $\norm{f_n - f}_E \to 0$ pointwise as $n \to \infty$.
+    By assumptions $a$ and $b$, $\norm{f_n - f}_E \le 2g$ for all $n \in \natp$. Since $\seq{f_n} \subset L^p(X; E)$, $f \in L^p(X; E)$, and $g \in L^p(X)$, $\norm{f_n - f}_E^p, g^p \in L^1(X)$ for all $n \in \natp$. By the \hyperref[Dominated Convergence Theorem]{theorem:dct}, 
-
-    For each $n \in \nat$, $\norm{f_n}_E \le \norm{f}_E$, so $\norm{f_n}_{L^p(X; E)} \le \norm{f}_{L^p(X; E)} < \infty$, and $\norm{f_n - f}_E \le 2\norm{f}_E$. By the \hyperref[Dominated Convergence Theorem]{theorem:dct},
     \[
-    \limv{n}\int \norm{f_n - f}_E^p d\mu = \int \limv{n}\norm{f_n - f}_E^p d\mu = 0
+    \limv{n}\norm{f_n - f}_{L^p(X; E)}^p = \limv{n}\int \norm{f_n - f}_E d\mu  = 0
+    \]
+\end{proof}
+
+
+\begin{proposition}
+\label{proposition:lp-simple-dense}
+    Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed vector space, and $p \in [1, \infty)$, then $\Sigma(X, \cm; E) \cap L^p(X; E)$ is dense in $L^p(X; E)$.
+\end{proposition}
+\begin{proof}
+    Let $f \in L^p(X; E)$. By \autoref{definition:strongly-measurable}, there exists $\seq{f_n} \subset \Sigma(X, \cm; E)$ such that $\norm{f_n}_E \le \norm{f}_E$ and $\norm{f_n - f}_E \to 0$ strongly pointwise as $n \to \infty$. By the \hyperref[Dominated Convergence Theorem]{proposition:dct-lp}, $f_n \to f$ in $L^p(X; E)$.
+\end{proof}
+
+\begin{theorem}[{{\cite[III.6.5]{SchaeferWolff}}}]
+\label{theorem:l1-tensor}
+    Let $(X, \cm, \mu)$ be a measure space and $E$ be a Banach space over $K \in \RC$, then the map $L^1(X; K) \td{\otimes}_\mu E \to L^1(X; E)$ defined by extending
+    \[
+    L^1(X; K) \times E \to L^1(X; E) \quad f \otimes x \mapsto x \cdot f
     \]
     
-    Therefore $\norm{f_n - f}_{L^p(X; E)} \to 0$ as $n \to \infty$.
+    is an isometric isomorphism. 
+\end{theorem}
+\begin{proof}
+    By (U) of the \hyperref[tensor product]{definition:tensor-product}, the given map admits a unique extension
+    \[
+    M: L^1(X; K) \otimes E \to L^1(X; E) \quad \sum_{j = 1}^n f_j \otimes x_j \mapsto \sum_{j = 1}^n x_j \cdot f_j
+    \]
+
+    Restricting $M$ to the simple functions yields a linear isomorphism
+    \[
+    M: [L^1(X; K) \cap \Sigma(X; K)] \otimes E \to L^1(X; E) \cap \Sigma(X; E)
+    \]
+
+    For any $\phi \in L^1(X; E) \cap \Sigma(X; E)$, write
+    \[
+    \phi = \sum_{y \in \phi(X) \setminus \bracs{0}}y \cdot \one_{\bracs{\phi = y}} = M\braks{\sum_{y \in \phi(X) \setminus \bracs{0}}\one_{\bracs{\phi = y}} \otimes y}
+    \]
+
+    then
+    \[
+    \normn{M^{-1}\phi}_{L^1(X; K) \otimes E} \le \sum_{y \in \phi(X) \setminus \bracs{0}} \norm{y}_E \cdot \mu\bracs{\phi = y} = \int \norm{\phi}_E d\mu = \norm{\phi}_{L^1(X; E)}
+    \]
+
+    On the other hand, for any representation $M^{-1}\phi = \sum_{j = 1}^n a_j \one_{A_j}$, 
+    \[
+    \normn{\phi}_{L^1(X; E)} \le \sum_{j = 1}^n \norm{a_j}_E \mu(A_j) = \sum_{j = 1}^n \norm{a_j}_E \normn{\one_{A_j}}_{L^1(X; K)}
+    \]
+
+    As this holds for all such representations, $\normn{\phi}_{L^1(X; E)} = \normn{M^{-1}\phi}_{L^1(X; K) \otimes E}$. Therefore $M$ restricted to $[L^1(X; K) \cap \Sigma(X; K)] \otimes E$ is an isometry. By \autoref{proposition:lp-simple-dense}, $[L^1(X; K) \cap \Sigma(X; K)] \otimes E$ is dense in $L^1(X; K) \widehat{\otimes}_\pi E$, and $L^1(X; E) \cap \Sigma(X; E)$ is dense in $E$. By the \hyperref[Linear Extension Theorem]{theorem:linear-extension-theorem-normed}, $M$ extends uniquely into the given map on $L^1(X; K) \otimes E$, which then extends into an isometry $L^1(X; K) \otimes E \to L^1(X; E)$. 
 \end{proof}
 
+
+
 \begin{theorem}[Markov's Inequality]
 \label{theorem:markov-inequality}
-    Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed space, and $f: X \to E$ be a Borel measurable function, then
+    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 $\alpha > 0$, 
         \[
@@ -146,7 +199,7 @@
 
 \begin{proposition}
 \label{proposition:lp-in-measure}
-    Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed 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.
+    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.
 \end{proposition}
 \begin{proof}
     Let $\eps > 0$. If $p < \infty$, then by \hyperref[Markov's Inequality]{theorem:markov-inequality},

src/fa/norm/absolute.tex

@@ -3,17 +3,18 @@
 
 \begin{definition}[Absolute Convergence]
 \label{definition:absolute-convergence}
-    Let $E$ be a normed space, then a series $\sum_{n = 1}^\infty x_n$ with $\seq{x_n} \subset E$ \textbf{converges absolutely} if $\sum_{n \in \natp}\norm{x_n}_E < \infty$.
+    Let $E$ be a normed vector space, then a series $\sum_{n = 1}^\infty x_n$ with $\seq{x_n} \subset E$ \textbf{converges absolutely} if $\sum_{n \in \natp}\norm{x_n}_E < \infty$.
 \end{definition}
 
 \begin{lemma}
 \label{lemma:banach-absolute}
-    Let $E$ be a normed space, then the following are equivalent:
+    Let $E$ be a normed vector space, then the following are equivalent:
     \begin{enumerate}
         \item $E$ is a Banach space.
         \item For any absolutely convergent series $\sum_{n = 1}^\infty x_n$ with $\seq{x_n} \subset E$, there exists $x \in E$ such that $x = \sum_{n = 1}^\infty x_n$.
     \end{enumerate}
 
+    
 \end{lemma}
 \begin{proof}
     (2) $\Rightarrow$ (1): Let $\seq{x_n} \subset E$ be a Cauchy sequence, then there exists a subsequence ${n_k}$ such that $\norm{x_{n_{k+1}} - x_{n_k}} \le 2^{-k}$ for all $k \in \natp$. 
@@ -24,7 +25,7 @@
 
 \begin{definition}[Unconditional Convergence]
 \label{definition:unconditional-convergence}
-    Let $E$ be a normed space, then a series $\sum_{n = 1}^\infty x_n$ with $\seq{x_n} \subset E$ \textbf{converges unconditionally} if for any bijection $\sigma: \natp \to \natp$, 
+    Let $E$ be a normed vector space, then a series $\sum_{n = 1}^\infty x_n$ with $\seq{x_n} \subset E$ \textbf{converges unconditionally} if for any bijection $\sigma: \natp \to \natp$, 
     \[
     \sum_{n = 1}^\infty x_n = \sum_{n = 1}^\infty x_{\sigma(n)}
     \]
@@ -45,6 +46,7 @@
         \item If $\sum_{n \in N}x_n = -\infty$ but $\sum_{n \in P}x_n < \infty$, then $\sum_{n = 1}^\infty x_n$ converges to $-\infty$ unconditionally. 
         \item If $\sum_{n \in \natp}|x_n| < \infty$, then $\sum_{n = 1}^\infty x_n$ converges unconditionally.
     \end{enumerate}
+
     In other words, a series in $\real$ converges unconditionally if and only if its positive parts or its negative parts are finite. 
 \end{theorem}
 \begin{proof}

src/fa/norm/index.tex

@@ -1,7 +1,8 @@
-\chapter{Normed Spaces}
+\chapter{Normed Vector Spaces}
 \label{chap:normed-spaces}
 
 \input{./normed.tex}
 \input{./absolute.tex}
 \input{./linear.tex}
+\input{./separable.tex}
 \input{./multilinear.tex}

src/fa/norm/linear.tex

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

src/fa/norm/multilinear.tex

@@ -3,12 +3,13 @@
 
 \begin{proposition}
 \label{proposition:bilinear-separate}
-    Let $E, F, G$ be normed spaces and $T: E \times F \to G$ be a bilinear map. If:
+    Let $E, F, G$ be normed vector spaces and $T: E \times F \to G$ be a bilinear map. If:
     \begin{enumerate}
         \item For each $x \in E$, $y \mapsto T(x, y)$ is a continuous linear map from $F$ to $G$.
         \item For each $y \in F$, $x \mapsto T(x, y)$ is a continuous linear map from $E$ to $G$.
         \item $E$ is a Banach space.
     \end{enumerate}
+
     then $T \in L^2(E, F; G)$.
 \end{proposition}
 \begin{proof}

src/fa/norm/normed.tex

@@ -34,16 +34,18 @@
 
 \begin{theorem}[Successive Approximation]
 \label{theorem:successive-approximation}
-    Let $E, F$ be normed 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:
+    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}
         \item[(a)] $\norm{x}_E \le C\norm{y}_F$.
         \item[(b)] $\norm{y - Tx}_F \le \gamma \norm{y}_F$.
     \end{enumerate}
+
     then for any $y \in F$, there exists $\seq{x_n} \subset E$ such that:
     \begin{enumerate}
         \item $\sum_{n \in \natp}\norm{x_n}_E \le C\norm{y}_F/(1 - \gamma)$.
         \item $\sum_{n = 1}^\infty Tx_n = y$.
     \end{enumerate}
+
     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}
@@ -69,20 +71,38 @@
 
 \begin{theorem}[Uniform Boundedness Principle]
 \label{theorem:uniform-boundedness}
-    Let $E, F$ be normed spaces and $\mathcal{T} \subset L(E; F)$. If
+    Let $E, F$ be normed vector spaces and $\mathcal{T} \subset L(E; F)$. If
     \begin{enumerate}
-        \item For every $x \in E$, $\sup_{T \in \mathcal{T}}\norm{Tx}_F < \infty$.
+        \item[(B)] $E$ is a Banach space.
-        \item $E$ is a Banach space.
+        \item[(E2)] For every $x \in E$, $\sup_{T \in \mathcal{T}}\norm{Tx}_F < \infty$.
     \end{enumerate}
+
     then $\sup_{T \in \mathcal{T}}\norm{T}_{L(E; F)} < \infty$.
 \end{theorem}
 \begin{proof}
-    For each $n \in \natp$, let $A_n = \bracs{x \in X|\norm{Tx}_F \le n \forall T \in \mathcal{T}}$, then each $A_n$ is closed with $\bigcup_{n \in \natp}A_n = E$. By the \hyperref[Baire Category Theorem]{theorem:baire}, there exists $n \in \natp$ and $U \subset E$ open such that $\sup_{x \in U}\sup_{T \in \mathcal{T}}\norm{Tx}_{F} < \infty$.
+    By the \autoref{theorem:banach-steinhaus} theorem, $\mathcal{T}$ is equicontinuous. Therefore there exists $r > 0$ such that $\bigcup_{T \in \mathcal{T}}T[B_E(0, r)] \subset B_F(0, 1)$. In which case, $\sup_{T \in \mathcal{T}}\norm{T}_{L(E; F)} \le r^{-1}$. 
-
-    Let $x \in U$ and $r > 0$ such that $\overline{B(x, r)} \subset U$, then for any $y \in E$ with $\norm{y}_E \le r$ and $T \in \mathcal{T}$,
-    \[
-    \norm{Ty} = \norm{Ty + Tx - Tx}_E = \normn{T\underbrace{(x + y)}_{\in U}}_E + \norm{Tx}_E \le 2n
-    \]
-
-    so $\sup_{T \in \mathcal{T}}\norm{T}_{L(E; F)} \le 2n/r$.
 \end{proof}
+
+
+\begin{proposition}
+\label{proposition:dual-norm}
+    Let $E$ be a normed vector space, then for any $x \in E$, 
+    \[
+    \norm{x}_E = \sup_{\substack{\phi \in E^* \\ \norm{\phi}_{E^*} = 1}}\dpn{x, \phi}{E}
+    \]
+\end{proposition}
+\begin{proof}
+    For any $\phi \in E^*$ with $\norm{\phi}_{E^*} = 1$, $\dpn{x, \phi}{E} \le \norm{x}_E \cdot \norm{\phi}_{E^*} = \norm{x}_E$. On the other hand, by the \hyperref[Hahn-Banach Theorem]{proposition:hahn-banach-utility}, there exists $x \in E^*$ such that $\dpn{x, \phi}{E} = \norm{x}_E$ and $\norm{\phi}_{E^*} \le 1$. 
+\end{proof}
+
+
+% TODO: Replace this with a more general version involving polars in the future. 
+\begin{theorem}[Alaoglu's Theorem]
+\label{theorem:alaoglu}
+    Let $E$ be a normed vector space over $K \in \RC$, then $B^* = \bracsn{\phi \in E^*| \norm{x}_{E^*} \le 1}$ is compact in the weak*-topology. 
+\end{theorem}
+\begin{proof}
+    For each $x \in E$, $I_x = \bracsn{\dpn{x, \phi}{E}|\phi \in B^*}$ is compact. By \autoref{proposition:operator-space-completeness}, the closure of $B^*$ in $\prod_{x \in E}I_x$ is a subset of $\hom(E; K)$. Since $B^*$ is bounded, $I_x \subset \overline{B_K(0, 1)}$ for all $x \in E$, so the closure of $B^*$ is contained in $B^*$. By \autoref{theorem:tychonoff}, $\prod_{x \in E}I_x$ is compact. Therefore $B^*$ is compact with respect to the weak*-topology.
+\end{proof}
+
+

src/fa/norm/separable.tex

@@ -0,0 +1,49 @@
+\section{Separable Normed Vector Spaces}
+\label{section:separable-banach-space}
+
+\begin{proposition}
+\label{proposition:separable-dual}
+    Let $E$ be a separable normed vector space, then $E^*$ is separable with respect to the weak*-topology. 
+\end{proposition}
+\begin{proof}
+    Let $\seq{x_n} \subset E$ be a dense subset and $S = \bracsn{\phi \in E^*| \norm{\phi}_{E^*} \le 1}$. For each $N \in \natp$, let
+    \[
+    T_N: S \to \real^N \quad \phi \mapsto (\dpn{x_1, \phi}{E}, \cdots, \dpn{x_N, \phi}{E})
+    \]
+
+    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 $T_N(S)$.
+
+    Let $\phi \in S$, then for each $N \in \natp$, there exists $k_N \in \natp$ such that for each $1 \le n \le N$, 
+    \[
+    |\dpn{x_n, \phi_{N, k_N}}{E} - \dpn{x_n, \phi}{E}| \le \frac{1}{N}
+    \] 
+
+    Thus for each $N \in \natp$, $\dpn{x_n, \phi_{N, k_N}}{E} \to \dpn{x_n, \phi}{E}$ as $N \to \infty$. Since $\phi_{N, k_N} \to \phi$ pointwise on a dense subset of $E$, and $\bracsn{\phi_{N, k_N}|N \in \natp} \subset S$ is uniformly equicontinuous, $\phi_{N, k_N} \to \phi$ in the weak*-topology by \autoref{proposition:strong-operator-dense}. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:separable-banach-borel-sigma-algebra}
+    Let $E$ be a separable normed vector space, then the Borel $\sigma$-algebra on $E$ is generated by the following families of sets:
+    \begin{enumerate}
+        \item Open sets in $E$ with respect to the strong topology. 
+        \item $\bracs{B(x, r)|x \in E, r > 0}$.
+        \item $\bracsn{\ol{B(x, r)}|x \in E, r > 0}$. 
+        \item Open sets in $E$ with respect to the weak topology. 
+    \end{enumerate}
+
+    
+\end{proposition}
+\begin{proof}
+    (1) $\Leftrightarrow$ (2) $\Leftrightarrow$ (3): By \autoref{proposition:separable-metric-borel-sigma-algebra}.
+
+    (4) $\subset$ (1): Every weakly open set is strongly open. 
+
+    (2) $\subset$ (4): By \autoref{proposition:seminorm-lsc}, $\norm{\cdot}_E: E \to [0, \infty)$ is Borel measurable with respect to the weak topology. For any $x \in E$, let
+    \[
+    \phi_x: E \to [0, \infty) \quad y \mapsto \norm{x - y}_E
+    \]
+
+    then $\phi_x$ is Borel measurable with respect to the weak topology, so $B(x, r) = \bracs{\phi_x < r}$ is a Borel set with respect to the weak topology. 
+\end{proof}
+
+

src/fa/notation.tex

@@ -0,0 +1,53 @@
+\chapter{Notations}
+\label{chap:fa-notations}
+
+\begin{tabular}{lll}
+    \textbf{Notation} & \textbf{Description} & \textbf{Source} \\
+    \hline
+    % ---- Topological Vector Spaces ----
+    $E_A$ & Normed space associated with $A \subset E$. & \autoref{definition:lc-associated-normed-space} \\
+    $L(E; F)$ & Continuous linear maps $E \to F$. & \autoref{definition:continuous-linear} \\
+    $L^n(E_1,\ldots,E_n; F)$ & Continuous $n$-linear maps $\prod E_j \to F$. & \autoref{definition:continuous-multilinear} \\
+    $\mathfrak{B}(E)$ & Bounded subsets of TVS $E$. & \autoref{definition:bounded} \\
+    $B(T; E)$ & Bounded functions $T \to E$ with uniform topology. & \autoref{definition:bounded-function-space} \\
+    $B_\mathfrak{S}^k(E; F)$, $B(E; F)$ & $\mathfrak{S}$-bounded $k$-linear maps; bounded linear maps. & \autoref{definition:bounded-linear-map-space} \\
+    $E^*$ & Topological dual of TVS $E$. & \autoref{definition:topological-dual} \\
+    $E_w$ & $E$ equipped with the weak topology. & \autoref{definition:weak-topology} \\
+    $\langle x, \phi \rangle_E$ & Duality pairing between $x \in E$ and $\phi \in E^*$. & \autoref{proposition:polarisation-linear} \\
+    $L_s(E; F)$ & $L(E; F)$ with strong operator topology. & \autoref{definition:strong-operator-topology} \\
+    $L_w(E; F)$ & $L(E; F)$ with weak operator topology. & \autoref{definition:weak-operator-topology} \\
+    $L_b(E; F)$ & $L(E; F)$ with topology of bounded convergence. & \autoref{definition:bounded-convergence-topology} \\
+    $\widehat{E}$ & Hausdorff completion of TVS $E$. & \autoref{definition:tvs-completion} \\
+    % ---- Locally Convex ----
+    $\mathrm{Conv}(A)$ & Convex hull of $A$. & \autoref{definition:convex-hull} \\
+    $\aconv(A)$ & Convex circled hull of $A$. & \autoref{definition:convex-circled-hull} \\
+    $[\cdot]_A$ & Gauge of a radial set $A$. & \autoref{definition:gauge} \\
+    $\rho_M$ & Quotient of seminorm $\rho$ by subspace $M$. & \autoref{definition:quotient-norm} \\
+    $E \otimes_\pi F$ & 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} \\
+    % ---- 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} \\
+    $x \perp y$ & Disjointness $|x| \wedge |y| = 0$ in a vector lattice. & \autoref{definition:order-disjoint} \\
+    $[x, y]$ & Order interval $\{z \mid x \le z \le y\}$. & \autoref{definition:ordered-vector-space-interval} \\
+    $E^b$ & Order bounded dual of ordered vector space $E$. & \autoref{definition:order-bounded-dual} \\
+    $E^+$ & Order dual of $E$. & \autoref{definition:order-dual} \\
+    $f^+$, $f^-$ & Positive and negative parts $f \vee 0$ and $-(f \wedge 0)$. & \autoref{definition:positive-negative-parts} \\
+    % ---- Riemann--Stieltjes ----
+    $\mathscr{P}([a,b])$ & Set of all partitions of $[a,b]$. & \autoref{definition:partition-interval} \\
+    $\mathscr{P}_t([a,b])$ & Set of all tagged partitions of $[a,b]$. & \autoref{definition:tagged-partition} \\
+    $\sigma(P)$ & Mesh of a partition $P$. & \autoref{definition:mesh} \\
+    $V_{\rho,P}(f)$ & Variation of $f$ w.r.t.\ seminorm $\rho$ and partition $P$. & \autoref{definition:total-variation} \\
+    $[f]_{\mathrm{var},\rho}$ & Total variation of $f$ w.r.t.\ $\rho$. & \autoref{definition:total-variation} \\
+    $T_{f,\rho}(x)$ & Variation function of $f$ with respect to $\rho$. & \autoref{definition:variation-function} \\
+    $BV([a,b]; E)$ & Functions of bounded variation. & \autoref{definition:bounded-variation} \\
+    $S(P, c, f, G)$ & Riemann-Stieltjes sum $\sum_j f(c_j)[G(x_j)-G(x_{j-1})]$. & \autoref{definition:rs-sum} \\
+    $\int_a^b f dG$, $\int_a^b f(t) G(dt)$ & Riemann-Stieljes integral of $f$ with respect to $G$. & \autoref{definition:rs-integral}  \\
+    $RS([a,b], G)$ & Space of RS-integrable functions w.r.t.\ $G$. & \autoref{definition:rs-integral} \\
+    $\mathrm{Reg}([a,b], G; E)$ & 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}
+\end{tabular}
+

src/fa/order/index.tex

@@ -0,0 +1,5 @@
+\chapter{Order Structures}
+\label{chap:order-structure}
+
+\input{./lattice.tex}
+\input{./norm.tex}

src/fa/order/lattice.tex

@@ -0,0 +1,305 @@
+\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]
+\label{definition:vector-lattice}
+    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 $x$. 
+\end{definition}
+
+\begin{lemma}
+\label{lemma:absolute-ge-0}
+    Let $(E, \le)$ be a vector lattice and $x \in E$, then $|x| \ge 0$.
+\end{lemma}
+\begin{proof}
+    For any $x \in E$,
+    \[
+    2|x| = 2(x \vee (-x)) \ge x + -x = 0
+    \]
+\end{proof}
+
+
+\begin{definition}[Disjoint]
+\label{definition:order-disjoint}
+    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{proposition}[{{\cite[V.1.1]{SchaeferWolff}}}]
+\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}
+        \item[(3)] $|\lambda x| = |\lambda| \cdot |x|$
+        \item[(4)] $|x + y| \le |x| + |y|$.
+    \end{enumerate}
+
+    
+    Finally, for any $x, y \in E$ with $x, y \ge 0$, 
+    \begin{enumerate}
+        \item[(5)] $[0, x] + [0, y] = [0, x + y]$.
+    \end{enumerate}
+
+    
+\end{proposition}
+\begin{proof}
+    (1): By \autoref{proposition:ordered-vector-space-properties},
+    \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),
+    \[
+    x  = x + 0 = x \vee 0 + x \wedge 0 = x^+ - x^-
+    \]
+
+    By \autoref{proposition:ordered-vector-space-properties} and \autoref{lemma:absolute-ge-0},
+    \[
+        x^+ + x^- = x \vee 0 + (-x \vee 0) = x \vee -x \vee 0 = |x|
+    \]
+    
+    and
+    \[
+    x^+ \vee x^- = x \vee 0 \vee (-x) \vee 0 = |x|
+    \]
+
+    Since $x^+, x^- \ge 0$, $|x^+| = x^+$ and $|x^-| = x^-$, so by (1),
+    \[
+    x^+ \wedge x^- = x^+ + x^- - x^+ \vee x^- = 0
+    \]
+
+    so $x^+ \perp x^-$. 
+
+    Now, let $y, z \in E$ with $y, z \ge 0$ such that $x = y - z$, then
+    \[
+    x^+ = x \vee 0 = (y - z) \vee 0 \le y \vee 0 = y
+    \]
+
+    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),
+    \[
+    |\lambda x| = (\lambda x) \vee (-\lambda x) = \lambda (x \vee -x) = \lambda |x|
+    \]
+
+    (4): For any $x, y \in E$, by \autoref{proposition:ordered-vector-space-properties},
+    \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)^+
+    \end{align*}
+    
+    Likewise, $x^- + y^- \ge (x + y)^-$. Thus
+    \[
+    |x+y| = (x+y)^+ + (x + y)^- \le x^+ + y^+ + x^- + y^- = |x| + |y|
+    \]
+
+    (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
+    \[
+        v = z - z \wedge x = z + (-z \vee -x) = (0 \vee z - x) \le z - x \le y
+    \]
+\end{proof}
+
+\begin{lemma}
+\label{lemma:positive-functional-extension}
+    Let $(E, \le)$ be a vector lattice, $C = \bracs{x \in E|x \ge 0}$ and $\phi: C \to [0, \infty)$ such that:
+    \begin{enumerate}
+        \item For any $x \in C$ and $\lambda \in \real$ with $\lambda \ge 0$, $\phi(\lambda x) = \lambda \phi(x)$.
+        \item For any $x, y \in C$, $\phi(x + y) = \phi(x) + \phi(y)$.
+    \end{enumerate}
+
+    
+    then the mapping
+    \[
+    \Phi: E \to \real \quad x \mapsto \phi(x^+) - \phi(x^-)
+    \]
+
+    is a positive linear functional on $E$. 
+\end{lemma}
+\begin{proof}
+    For any $\lambda \in \real$ with $\lambda \ge 0$, 
+    \begin{align*}
+    \Phi(\lambda x) &= \phi((\lambda x)^+) - \phi((\lambda x)^-) \\
+    &= \lambda\phi(x^+) - \lambda\phi(x^-) = \lambda\Phi(x)
+    \end{align*}
+    
+    Likewise, if $\lambda < 0$, then 
+    \begin{align*}
+    \Phi(\lambda x) &= \phi((\lambda x)^+) - \phi((\lambda x)^-) \\
+    &= -\lambda\phi(x^-) + \lambda\phi(x^+) = \lambda\Phi(x)
+    \end{align*}
+
+    For any $x, y \in E$, let $z = x + y$, then $z = z^+ - z^- = x^+ + y^+ - x^- - y^-$. Thus
+    \begin{align*}
+        z^+ + x^- + y^- &= z^- + x^+ + y^+ \\
+        \phi(z^+) + \phi(x^-) + \phi(y^-) &= \phi(z^-) + \phi(x^+) + \phi(y^+) \\
+        \phi(z^+) - \phi(z^-) &= \phi(x^+) - \phi(x^-) + \phi(y^+) - \phi(y^-) \\
+        \Phi(z) &= \Phi(x) + \Phi(y)
+    \end{align*}
+\end{proof}
+
+
+\begin{proposition}[{{\cite[V.1.4]{SchaeferWolff}}}]
+\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}
+        \item $E^b = E^+$.
+        \item The order bound dual $E^b$ equipped with its canonical ordering is a vector lattice.
+        \item $E^b$ is order complete. 
+        \item For any $\phi \in E^b$ and $x \in E$ with $x \ge 0$,
+        \[
+        |\phi|(x) = \sup\bracs{\phi(y)|y \in E, |y| \le x}
+        \]
+    \end{enumerate}
+
+    
+\end{proposition}
+\begin{proof}
+    (1): Let $C = \bracs{x \in E|x \ge 0}$, $\Phi^+ \in E^b$, and
+    \[
+    \Phi^+: C \to [0, \infty) \quad x \mapsto \sup(f([0, x]))
+    \]
+
+    then for any $x, y \in C$ and $\lambda \in \real$ with $\lambda \ge 0$,
+    \begin{align*}
+    \Phi^+(\lambda x + y) &= \sup(f([0, \lambda x + y])) = \sup(f(\lambda [0,x]) + f([0, y])) \\
+    &= \lambda \sup(f([0, x])) + \sup(f([0, y])) = \lambda \Phi^+(x) + \Phi^+(y)
+    \end{align*}
+
+    Let
+    \[
+    \Phi: E \to \real \quad x \mapsto \Phi^+(x^+) - \Phi^+(x^-)
+    \]
+
+    then $\Phi$ is a positive linear functional by \autoref{lemma:positive-functional-extension}.
+
+    For any $x \in C$, $\Phi(x) \ge \phi(x)$, so $\Phi \ge \phi$. On the other hand, let $\psi \in \hom(E; \real)$ such that $\psi \ge 0$ and $\psi \ge \phi$, then for each $x \in C$, $\psi(x) \ge \sup(\phi([0, x]))$. Therefore $\Phi = 0 \vee \phi$. 
+    
+    Since $0 \wedge \phi = -(0 \vee -\phi)$, $\phi = (0 \vee \phi) - (0 \vee -\phi)$ is a sum of two positive linear functionals, so $E^b = E^+$. 
+    
+    (2): For any $x, y \in E^b$, $x \wedge y = (x - y) \wedge 0 + y$ and $x \vee y = (x - y) \vee 0 + y$, so $E^b$ is a lattice. 
+
+    (3): Let $A \subset E^b$ be order bounded, then since $E^b$ is a lattice, assume without loss of generality that $A$ is upward-directed under the canonical ordering. Let
+    \[
+    \phi: C \to [0, \infty) \quad x \mapsto \sup_{f \in A}f(x)
+    \]
+
+    then for any $\lambda \ge 0$ and $x \in C$, $\phi(\lambda x) = \lambda \phi(x)$. Let $x, y \in C$, then for any $f \in A$, 
+    \[
+    f(x + y) = f(x) + f(y) \le \phi(x) + \phi(y)
+    \]
+
+    As this holds for all $f \in A$, $\phi(x + y) \le \phi(x) + \phi(y)$. On the other hand, for any $f, g \in A$, there exists $h \in A$ such that
+    \[
+    h(x + y) = h(x) + h(y) \ge f(x) + g(y)
+    \]
+
+    Since such an $h \in A$ exists for all $f, g \in A$, $\phi(x + y) \ge \phi(x) + \phi(y)$. 
+
+    By \autoref{lemma:positive-functional-extension}, the mapping 
+    \[
+    \Phi: E \to \real \quad x \mapsto \phi(x^+) - \phi(x^-)
+    \]
+
+    is a positive linear functional on $E$. By definition $\Phi \ge f$ for all $f \in A$. If $\psi \in \hom(E; \real)$ is a linear functional such that $\psi \ge f$ for all $f \in A$, then for any $x \in C$, $\psi(x) \ge \sup_{f \in A}f(x)$. Therefore $\Phi = \sup(A)$, and $E^b$ is order complete. 
+
+    (3): Let $\phi \in E^b$ and $x \in E$ with $x \ge 0$, then by (1), 
+    \begin{align*}
+    |\phi|(x) &= \sup(\phi([0, x])) + \sup(-\phi([0, x])) \\
+    &= \sup\bracs{\phi(u) - \phi(v)| u, v \in [0, x]}
+    \end{align*}
+
+    For any $u, v \in [0, x]$, $|u - v| \le x$, so
+    \[
+    |\phi|(x) \le \sup\bracs{\phi(y)|y \in E, |y| \le x}
+    \]
+
+    On the other hand, for any $y \in E$ with $|y| \le x$, $y^+, y^- \le x$, so
+    \[
+    \phi(y) = \phi(y^+) - \phi(y^-) \le \sup\bracs{\phi(u) - \phi(v)| u, v \in [0, x]} = |\phi|(x)
+    \]
+
+    Therefore
+    \[
+    |\phi|(x) = \sup\bracs{\phi(y)|y \in E, |y| \le x}
+    \]
+    
+\end{proof}
+
+
+
+
+

src/fa/order/norm.tex

@@ -0,0 +1,84 @@
+\section{Banach Lattices}
+\label{section:banach-lattice}
+
+\begin{definition}[Banach Lattice]
+\label{definition:banach-lattice}
+    Let $(E, \normn{\cdot}_E)$ be a Banach space over $\real$ and $\le$ be a partial order on $E$, then $(E, \normn{\cdot}_E, \le)$ is a \textbf{Banach lattice} if for any $x, y \in E$, $|x| \le |y|$ implies that $\normn{x}_E \le \normn{x}_E$.
+\end{definition}
+
+\begin{proposition}
+\label{proposition:banach-lattice-properties}
+    Let $(E, \normn{\cdot}_E, \le)$ be a Banach lattice, then:
+    \begin{enumerate}
+        \item For any $x \in E$, $\normn{\ |x|\ }_E = \normn{x}_E$.
+        \item For any $x \in E$, $\normn{x^+}_E, \normn{x^-}_E \le \normn{x}_E$.
+        \item For any $x \in E$ and positive linear functional $\phi \in E^*$, 
+        \[
+        \norm{\phi}_{E^*} = \sup\bracsn{\dpn{x, \phi}{E}|x \in E, x \ge 0, \norm{x}_E = 1}
+        \]
+    \end{enumerate}
+
+    % More to be added.
+
+\end{proposition}
+\begin{proof}
+    (1): Let $x \in E$, then $|\ |x|\ | = |x|$, so $\normn{\ |x|\ }_E = \normn{x}_E$.
+
+    (2): Let $x \in E$, then $|x| = x^+ + x^-$, so
+    \[
+    \normn{x}_E = \normn{x^+ + x^-}_E \ge \normn{x^+}_E, \normn{x^-}_E
+    \]
+
+    (3): Since $|x| = x \vee (-x)$,
+    \[
+    \norm{\phi}_{E^*} = \sup_{\substack{x \in E \\ \norm{x}_E = 1}}\dpn{x, \phi}{E} \le \sup_{\substack{x \in E \\ \norm{x}_E = 1}}\dpn{|x|, \phi}{E} \le \sup_{\substack{x \in E \\ x \ge 0 \\ \norm{x}_E = 1}}\dpn{x, \phi}{E} \le \norm{\phi}_{E^*}
+    \]
+\end{proof}
+
+\begin{proposition}
+\label{proposition:banach-lattice-dual}
+    Let $(E, \normn{\cdot}_E, \le)$ be a Banach lattice and $(E^*, \normn{\cdot}_{E^*}, \le)$ be its dual, equipped with the canonical ordering, then $E^*$ is a Banach lattice. 
+\end{proposition}
+\begin{proof}
+    Let $x, y \in E$ and $z \in [x, y]$, then
+    \[
+    |z| = z \vee (-z) \le y \vee (-x)
+    \]
+
+    so $[x, y]$ is bounded by $\norm{x}_E \vee \norm{y}_E$. Thus $E^* \subset E^b$. 
+
+    Let $\phi \in E^*$, then by \autoref{proposition:order-vector-dual}, $0 \vee \phi$ exists in $E^b$. For any $x \in E$ with $x \ge 0$, 
+    \[
+    (0 \vee \phi)(x) = \sup(\phi([0, x])) \le \norm{\phi}_{E^*} \cdot \norm{x}_E
+    \]
+
+    so for arbitrary $x \in E$,
+    \begin{align*}
+    (0 \vee \phi)(x) &= (0 \vee \phi)(x^+) - (0 \vee \phi)(x^-) \\
+    &\le \norm{\phi}_{E^*}(\normn{x^+}_E \vee \normn{x^-}_E) \le \norm{\phi}_{E^*} \cdot \norm{x}_E
+    \end{align*}
+    
+
+    and $\norm{0 \vee \phi}_{E^*} \le \norm{\phi}_{E^*}$. Therefore $E^*$ is a vector lattice. 
+
+    Now, let $x \in E$ with $\norm{x}_E = 1$, then
+    \begin{align*}
+        \dpn{x, \phi}{E^*} &= \dpn{x, \phi^+}{E} + \dpn{-x, \phi^-}{E} \\
+        &\le \dpn{|x|, \phi^+}{E} + \dpn{|x|, \phi^-}{E} \\
+        &= \dpn{|x|, \phi^+ + \phi^-}{E} \le \norm{\ |\phi|\ }_{E^*}
+    \end{align*}
+
+    On the other hand, by \autoref{proposition:order-vector-dual}, for any $x \in E$ with $x \ge 0$ and $\norm{x}_E = 1$,
+    \[
+        \dpn{x, |\phi|}{E^*} = \sup\bracsn{\dpn{y, \phi}{E}|y \in E, |y| \le x} \le \norm{\phi}_{E^*}
+    \]
+
+    so $\norm{\ |\phi|\ }_{E^*} \le \norm{\phi}_{E^*}$. Therefore for any $\phi, \psi \in E$ with $|\phi| \le |\psi|$, 
+    \begin{align*}
+        \norm{\phi}_{E^*} &= \norm{\ |\phi|\ }_{E^*} = \sup\bracsn{\dpn{x, |\phi|}{E}|x \in E, x \ge 0, \norm{x}_E = 1} \\
+        &\le \sup\bracsn{\dpn{x, |\psi|}{E}|x \in E, x \ge 0, \norm{x}_E = 1} \\
+        &= \norm{\ |\psi|\ }_{E^*} \norm{\psi}_E
+    \end{align*}
+\end{proof}
+
+

src/fa/rs/bv.tex

@@ -6,7 +6,7 @@
 \label{definition:total-variation}
     Let $E$ be a locally convex space, $\rho$ be a continuous seminorm on $E$, $f: [a, b] \to E$, and $P \in \scp([a, b])$ be a partition, then
     \[
-    V_{\rho, p}(f) = \sum_{j = 1}^n \rho(f(x_j) - f(x_{j - 1}))
+    V_{\rho, P}(f) = \sum_{j = 1}^n \rho(f(x_j) - f(x_{j - 1}))
     \]
 
     is the \textbf{variation} of $f$ with respect to $\rho$ and $P$. The supremum over all such partitions
@@ -16,10 +16,48 @@
 
     is the \textbf{total variation} of $f$ on $[a, b]$ with respect to $\rho$.
 
-    If $E$ is a normed space, then the variation and total variation of $f$ is taken with respect to its norm.
+    If $E$ is a normed vector space, then the variation and total variation of $f$ is taken with respect to its norm.
 \end{definition}
 
-\begin{definition}[Bounded Variation, {{\cite[Proposition X.1.1]{Lang}}}]
+\begin{definition}[Variation Function]
+\label{definition:variation-function}
+    Let $E$ be a locally convex space, $\rho$ be a continuous seminorm on $E$, $f: [a, b] \to E$, then the function
+    \[
+    T_{f, \rho}(x) = \sup_{P \in \scp([a, x])}V_{\rho, P}(f) = [f|_{[a, x]}]_{\text{var}, \rho}
+    \]
+
+    is the \textbf{variation function} of $f$ with respect to $\rho$, and:
+    \begin{enumerate}
+        \item $T_{f, \rho}: [a, b] \to [0, \infty]$ is a non-negative, non-decreasing function. 
+        \item If $f \in BV([a, b]; E)$, then for any $[c, d] \subset [a, b]$, $[f]_{\text{var}, \rho} = T_{f, \rho}(d) - T_{f, \rho}(c)$. 
+    \end{enumerate}
+\end{definition}
+\begin{proof}
+    (2): Let $P \in \scp([a, c])$ and $Q = \seqf{x_j} \in \scp([a, d])$ be partitions containing $P$, then
+    \[
+    V_{\rho, Q}(f) - V_{\rho, P}(f) = \sum_{x_j > c}\rho(f(x_j) - f(x_{j - 1})) \le [f]_{\text{var}, \rho}
+    \]
+
+    As this holds for all $Q \in \scp([a, d])$ containing $P$, 
+    \[
+    T_{f, \rho}(d) - T_{f, \rho}(c) \le T_{f, \rho}(d) - V_{\rho, P}(f) \le [f|_{[c, d]}]_{\text{var}, \rho}
+    \]
+
+    On the other hand, for any $R \in \scp([c, d])$, $P \cup R \in \scp([a, d])$ and contains $P$. Therefore
+    \[
+    T_{f, \rho}(d) - V_{\rho, P}(f) \ge V_{\rho, R \cup P}(f) - V_{\rho, P}(f) = V_{\rho, R}(f)
+    \]
+
+    Since this holds for all $P \in \scp([a, c])$, 
+    \[
+    T_{f, \rho}(d) - T_{f, \rho}(c) \ge V_{\rho, R}(f)
+    \]
+
+    and as the above holds for all $R \in \scp([c, d])$, $T_{f, \rho}(d) - T_{f, \rho}(c) \ge [f|_{[c, d]}]_{\text{var}, \rho}$. 
+\end{proof}
+
+
+\begin{definition}[Bounded Variation]
 \label{definition:bounded-variation}
     Let $E$ be a locally convex space, $\rho$ be a continuous seminorm on $E$, and $f: [a, b] \to E$. If $[f]_{\text{var}, \rho} < \infty$, then $f$ is of \textbf{bounded variation} with respect to $\rho$.
 
@@ -27,28 +65,17 @@
     \begin{enumerate}
         \item $BV([a, b]; E)$ is a vector space.
         \item For each continuous seminorm $\rho$ on $E$, $[\cdot]_{\text{var}, \rho}$ is a seminorm on $BV([a, b]; E)$.
-        \item Let $\fF$ be a filter on $BV([a, b]; E)$ and $f: [a, b] \to E$. If
+        \item For each continuous seminorm $\rho$ on $E$, $[\cdot]_{\text{var}, \rho}: E^{[a, b]} \to [0, \infty]$ is lower semicontinuous. In particular, for any $M > 0$, $\bracs{[\cdot]_{\text{var}, \rho} \le M} \subset E^{[a, b]}$ is closed. 
-        \begin{enumerate}
-            \item $\pi_x(\fF) \to f(x)$ for all $x \in [a, b]$.
-            \item For every continuous seminorm $\rho$ on $E$, there exists $U \in \fF$ such that $\sup_{g \in U}[g]_{\text{var}, \rho} = M_\rho < \infty$.
-        \end{enumerate}
-        then $f \in BV([a, b]; E)$ with $[f]_{\text{var}, \rho} \le M_\rho$.
         \item For any $f \in BV([a, b]; E)$ and continuous seminorm $\rho$ on $E$, $\sup_{x \in [a, b]}\rho(f(x)) \le \rho(f(a)) + [f]_{\text{var}, \rho}$.
     \end{enumerate}
-    If $(E, \norm{\cdot}_E)$ is a normed space, then
+
+    If $(E, \norm{\cdot}_E)$ is a normed vector space, then
     \begin{enumerate}
         \item[(5)] $f$ has at most countably many discontinuities.
     \end{enumerate}
 \end{definition}
-\begin{proof}
+\begin{proof}[Proof {{\cite[Proposition X.1.1]{Lang}}}. ]
-    (3): Let $\rho$ be a continuous seminorm on $E$ and $P \in \scp([a, b])$, then by assumption (a),
+    (3): For each $P \in \scp([a, b])$, the mapping $V_{P, \rho}: E^{[a, b]} \to [0, \infty]$ is continuous. Since $[\cdot]_{\text{var}, \rho} = \sup_{P \in \scp([a, b])}V_{P, \rho}$, $[\cdot]_{\text{var}, \rho}$ is lower semicontinuous by \autoref{proposition:semicontinuous-properties}. 
-    \[
-    V_{\rho, P}(f) = \sum_{j = 1}^n \rho(f(x_j) - f(x_{j - 1}))
-    = \lim_{g, \fF}\sum_{j = 1}^n \rho(g(x_j) - g(x_{j - 1}))
-    = \lim_{g \in \fF}V_{\rho, P}(g)
-    \]
-
-    By assumption (b), $[0, M_\rho]$ is in the filter generated by $V_{\rho, P}(\fF)$. Thus $V_{\rho, P}(f) \le M_\rho$. As this holds for all $P \in \scp([a, b])$, $V_{\rho, P}(f) \le M_\rho$, and $f \in BV([a, b]; E)$.
 
     (5): For each $n \in \nat^+$, let
     \[
@@ -62,9 +89,29 @@
         \item[(a)] $|E_k| \ge N - k$.
         \item[(b)] $E_k \subset I_k^o$.
     \end{enumerate}
+
     for $k = 1$.
 
     Let $k \le N$ and suppose inductively that $E_k, I_k$ have been constructed. Let $x_k \in E_k$, then by (b), there exists $\eps > 0$ such that $[x_k - \eps, x_k + \eps] \subset I_k$ and $|E_k \setminus [x_k - \eps, x_k + \eps]| \ge N - k$. Let $y_k \in [x_k - \eps, x_k + \eps]$ such that $\norm{f(x_k) - f(y_k)} \ge 1/n$, $I_{k + 1} = I_k \setminus [x_k - \eps, x_k + \eps]$, and $E_{k+1} = E_k \setminus [x_k - \eps, x_k + \eps]$, then $I_k$ and $E_k$ satisfies (a) and (b).
 
     Therefore there exists pairs $\bracs{(x_k, y_k)|1 \le k \le N}$ such that $\norm{f(x_k) - f(y_k)}_E \ge 1/n$ for all $n$, and the smallest interval containing each $(x_k, y_k)$ are pairwise disjoint. Thus $[f]_{\text{var}} \ge N/n$ for all $N \in \nat^+$, so $[f]_{\text{var}} = \infty$.
 \end{proof}
+
+\begin{proposition}
+\label{proposition:bounded-variation-one-side-limit}
+    Let $E$ be a complete locally convex space and $f \in BV([a, b]; E)$, then for each $x \in [a, b]$, the limits $\lim_{y \downto x}f(y)$ and $\lim_{y \upto x}f(y)$ exist. 
+\end{proposition}
+\begin{proof}
+    By flipping $f$, it is sufficient to consider the right-side limit $\lim_{y \downto x}f(y)$. 
+    
+    Let $\rho: E \to [0, \infty)$ be a continuous seminorm on $E$, and $T_{\rho, f}: [a, b] \to [0, \infty)$ be the variation function of $f$ with respect to $\rho$. For any $\eps > 0$, there exists $\delta > 0$ such that $T_{\rho, f}(z) - \lim_{y \downto x}T_{\rho, f}(y) < \eps$ for all $z \in (x, x + \delta)$. In which case, for any $x < y < z < x + \delta$, 
+    \[
+    \rho(f(z) - f(y)) \le [f|_{y, z}]_{\text{var}, \rho} \le T_{\rho, f}(z) - T_{\rho, f}(y) \le T_{\rho, f}(z) - \lim_{u \downto x}T_{\rho, f}(u) < \eps
+    \]
+
+    By completeness of $E$, the limit $\lim_{y \downto x}f(y)$ exists. 
+\end{proof}
+
+
+
+

src/fa/rs/index.tex

@@ -5,3 +5,6 @@
 \input{./bv.tex}
 \input{./rs.tex}
 \input{./rs-bv.tex}
+\input{./path.tex}
+\input{./regulated.tex}
+\input{./rs-measure.tex}

src/fa/rs/partition.tex

@@ -30,7 +30,7 @@
 
 \begin{definition}[Fine]
 \label{definition:partition-refinement}
-    Let $P = \seqfz[m]{x_j}, Q = \seqfz{y_j} \in \scp([a, b])$, then $Q$ is \textbf{finer} than $P$ if for every $0 \le j \le m$, there exists $0 \le k \le m$ such that $x_j = y_k$. For any $P, Q \in \scp([a, b])$, denote $P \le Q$ if $Q$ is finer than $P$, then
+    Let $P = \seqfz[m]{x_j}, Q = \seqfz{y_j} \in \scp([a, b])$, then $Q$ is \textbf{finer} than $P$ if for every $0 \le j \le m$, there exists $0 \le k \le n$ such that $x_j = y_k$. For any $P, Q \in \scp([a, b])$, denote $P \le Q$ if $Q$ is finer than $P$, then
     \begin{enumerate}
         \item $\scp([a, b])$/$\scp_t([a, b])$ equipped with $\le$ is a upward-directed set.
         \item If $P \le Q$, then $\sigma(P) \ge \sigma(Q)$.

src/fa/rs/path.tex

@@ -0,0 +1,152 @@
+\section{Path Integrals}
+\label{section:path-integrals}
+
+\begin{definition}[Rectifiable Path]
+\label{definition:rectifiable-path}
+    Let $[a, b] \subset \real$, $F$ be a locally convex space over $K \in \RC$, and $\gamma \in C([a, b]; F)$ be a path, then $\gamma$ is \textbf{rectifiable} if $\gamma \in BV([a, b]; F)$.
+\end{definition}
+
+\begin{definition}[Path Integral]
+\label{definition:path-integral}
+    Let $[a, b] \subset \real$, $E, F, H$ be locally convex spaces over $K \in \RC$, $E \times F \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map, and $\gamma \in C([a, b]; F)$ be a path. For any $f: \gamma([a, b]) \to E$, $f$ is \textbf{path-integrable with respect to $\gamma$} if $f \circ \gamma \in RS([a, b], \gamma; E)$. In which case, 
+    \[
+    \int_\gamma f = \int_a^b f(\gamma(t)) \gamma(dt)
+    \]
+
+    is the \textbf{path integral} of $f$ with respect to $\gamma$. The set $PI([a, b], \gamma; E)$ is the space of all functions path-integrable with respect to $\gamma$. 
+\end{definition}
+
+\begin{proposition}[Change of Variables]
+\label{proposition:path-integral-change-of-variables}
+    Let $[a, b], [c, d] \subset \real$, $E, F, H$ be locally convex spaces over $K \in \RC$, $E \times F \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map, $\gamma \in C([a, b]; F)$ be a path, and $\varphi: C([c, d]; [a, b])$ be non-decreasing with $\varphi(c) = a$ and $\varphi(d) = b$, then for any $f \in PI([a, b], \gamma; E)$, $f \in PI([c, d], \gamma \circ \varphi; E)$, and
+    \[
+    \int_\gamma f = \int_{\gamma \circ \varphi} f
+    \]
+\end{proposition}
+\begin{proof}
+    Since $\varphi(c) = a$, $\varphi(d) = b$, and $\varphi$ is continuous, it is surjective. As $\varphi$ is also non-decreasing, for any tagged partition $(P = \seqfz{x_j}, c = \seqf{c_j}) \in \scp_t([a, b])$, there exists a tagged partition $(Q = \seqfz{y_j}, d = \seqf{d_j}) \in \scp_t([c, d])$ such that $\varphi(y_j) = x_j$ for each $0 \le j \le n$ and $\varphi(d_j) = c_j$ for each $1 \le j \le n$. In addition,  
+    \begin{align*}
+        S(P, c, f \circ \gamma, \gamma) &= \sum_{j = 1}^n f \circ \gamma(c_j)[\gamma(x_j) - \gamma(x_{j - 1})] \\
+        &= \sum_{j = 1}^n f \circ \gamma \circ \varphi (d_j)[\gamma \circ \varphi(y_j) - \gamma \circ \varphi(y_{j-1})] \\
+        &= S(Q, d, f \circ \gamma \circ \varphi, \gamma \circ \varphi)
+    \end{align*}
+
+    Therefore if $f \in PI([a, b], \gamma; E)$, then $f \in PI([c, d], \gamma \circ \varphi; E)$, with $\int_\gamma f = \int_{\gamma \circ \varphi} f$.
+\end{proof}
+
+\begin{definition}[Curve]
+\label{definition:rs-curve}
+    Let $[a, b], [c, d] \subset \real$, $F$ be a locally convex space over $K \in \RC$, and $\gamma \in C([a, b]; F)$ and $\mu \in C([c, d]; F)$ be paths, then $\gamma$ and $\mu$ are \textbf{equivalent} if there exists a continuous, strictly increasing bijection $\varphi \in C([c, d]; [a, b])$ such that $\mu = \gamma \circ \varphi$. In which case, $\varphi$ is a \textbf{change of parameter} between $\gamma$ and $\mu$. 
+
+    A \textbf{curve} in $F$ is then an equivalence class of paths. 
+\end{definition}
+
+\begin{lemma}
+\label{lemma:rectifiable-piecewise-linear}
+    Let $[a, b] \subset \real$, $E, F, H$ be locally convex spaces over $K \in \RC$, $E \times F \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map, $\gamma \in C([a, b]; F)$ be a rectifiable path, and $U \in \cn_F(\gamma([a, b]))$.
+    
+    For each $P \in \scp([a, b])$, let $\Gamma_P \in C([a, b]; F)$ be the piecewise linear path obtained by interpolating values of $\gamma$ at points of $P$, then for any continuous seminorm $[\cdot]_G: G \to [0, \infty)$, $\eps > 0$, and $f \in C(U; E) \cap PI([a, b], \gamma; E)$, there exists $P \in \scp([a, b])$ such that for any $Q \in \scp([a, b])$ with $Q \ge P$,
+    \begin{enumerate}
+        \item $\Gamma_P(a) = \gamma(a)$ and $\Gamma_P(b) = \gamma(b)$. 
+        \item $\braks{\int_\gamma f - \int_{\Gamma_P} f}_F < \epsilon$. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    Let $[\cdot]_E: E \to [0, \infty)$ and $[\cdot]_F: F \to [0, \infty)$ such that for any $x \in E$ and $y \in F$, $[xy]_G \le [x]_E[y]_F$. Since $\gamma([a, b])$ is compact, by modifying $[\cdot]_F$, assume without loss of generality that there exists $V \in \cn_F(\gamma([a, b]))$ such that for any $x, y \in V$ with $[x - y]_F \le 1$, $[f(x) - f(y)]_E \le \eps$.
+
+    Since $f \in C(U; E)$, $f \in PI([a, b], \gamma; E)$ by \autoref{proposition:rs-bv-continuous}. Given that $\gamma$ is continuous, there exists $(P_0, c_0) \in \scp_t([a, b])$ such that for any $(P = \seqfz{x_j}, c) \in \scp_t([a, b])$ with
+    \begin{enumerate}[label=(\alph*)]
+        \item For each $1 \le j \le n$, 
+        \[
+        \gamma([x_{j-1}, x_j]) \subset \bracs{y \in F|[y - x_{j-1}]_F \le 1}
+        \]
+        \item $\braks{\int_\gamma f - S(P, c, f \circ \gamma, \gamma)}_G < \epsilon$.
+    \end{enumerate}
+
+    Let $\Gamma = \Gamma_P$, then for any $(Q, d) \in \scp_t([a, b])$ with $(Q, d) \ge (P, c)$, 
+    \[
+    \braks{S(P, c, f \circ \gamma, \gamma) - S(Q, d, f \circ \Gamma, \Gamma)}_G \le \eps [\gamma]_{\text{var}, [\cdot]_F}
+    \]
+
+    As $\Gamma$ is also of bounded variation, $f \in PI([a, b], \Gamma; E)$. Since the above holds for all refinements of $(Q, d)$, 
+    \[
+    \braks{\int_\gamma f - \int_\Gamma f}_G < \eps(1 + [\gamma]_{\text{var}, [\cdot]_F})
+    \]
+\end{proof}
+
+\begin{remark}
+\label{remark:piecewise-linear-remark}
+    Past me made the mistake of believing that in \autoref{lemma:rectifiable-piecewise-linear}, it is possible to approximate rectifiable curves with piecewise linear curves in \textit{total variation distance}. However, this is not possible: as every piecewise linear curve is absolutely continuous, and the limit of these curves in total variation distance must also be absolutely continuous. As such, this strong approximation exists if and only if the curve is absolutely continuous. 
+\end{remark}
+
+\begin{lemma}
+\label{lemma:rectifiable-smooth}
+    Let $[a, b] \subset \real$, $E$ be a separated locally convex space over $K \in \RC$, $F$ be a Banach space over $K$, $H$ be a complete locally convex space over $K$, all over $K \in \RC$, $E \times F \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map, $\gamma \in C([a, b]; F)$ be a piecewise $C^1$ curve that is constant on $[a, a + \eps)$ and $(b - \eps, b]$, and $U \in \cn_F(\gamma([a, b]))$. 
+
+    Extend $\gamma$ to $\real$ by 
+    \[
+    \ol \gamma : \real \to U \quad x \mapsto \begin{cases}
+        \gamma(a) &x \le a \\
+        \gamma(x) &x \in [a, b] \\
+        \gamma(b) &x \ge b
+    \end{cases}
+    \]
+
+    For each $\varphi \in C_c^\infty(\real; \real)$ with $\int_\real \varphi = 1$ and $t > 0$, let
+    \[
+    \gamma_t: [a, b] \to F \quad x \mapsto \frac{1}{t}\int_{\real} \ol \gamma(y) \varphi\braks{\frac{x - y}{t}} dy
+    \]
+
+    then
+
+    \begin{enumerate}
+        \item For each $t > 0$, $\gamma_t \in C^\infty([a, b]; F)$.
+        \item There exists $t > 0$ such that for any $s \in (0, t)$, $\gamma_s(a) = \gamma(a)$ and $\gamma_s(b) = \gamma(b)$. 
+        \item For any $f \in C(U; E)$, 
+        \[
+        \int_\gamma f = \lim_{t \downto 0} \int_{\gamma_t} f
+        \]
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    (1): By the \hyperref[Mean Value Theorem]{theorem:mean-value-theorem}, for each $x, y \in [a, b]$, 
+    \[
+    \norm{\frac{\varphi(x) - \varphi(y)}{x - y}}_F \le \sup_{z \in \real}\norm{D\varphi(z)}_F
+    \]
+
+    By the \hyperref[Dominated Convergence Theorem]{theorem:dct-bochner-vector}, $\gamma_t \in C^\infty([a, b]; F)$.
+
+    (2): For sufficiently small $t$, $\supp{\varphi} \subset (-\eps, \eps)$. In which case, by assumption, $\gamma_t(a) = \gamma(a)$ and $\gamma_t(b) = \gamma(b)$. 
+
+    (3): Since $\gamma$ is piecewise $C^1$ and $\gamma_t \in C^\infty([a, b]; F)$, 
+    \[
+    \int_\gamma f = \int_a^b f(t) D\gamma(t)dt = \lim_{t \downto 0}\int_a^b f(t) D\gamma_t(t) dt = \lim_{t \downto 0}\int_{\gamma_t}f
+    \]
+
+    by the \hyperref[Dominated Convergence Theorem]{theorem:dct-bochner-vector}.
+\end{proof}
+
+
+
+\begin{theorem}[Fundamental Theorem of Calculus for Path Integrals]
+\label{theorem:ftc-path-integrals}
+    Let $[a, b] \subset \real$, $E, F$ be separated locally convex spaces, $\sigma \subset \mathfrak{B}(F)$ be an ideal containing all compact sets, $\gamma \in C([a, b]; F)$ be a rectifiable path, and $U \in \cn_F(\gamma([a, b]))$, then for any $f \in C^1_\sigma(U; E)$, 
+    \[
+    \int_\gamma D_\sigma f = f(\gamma(b)) - f(\gamma(a))
+    \]
+
+    In particular, if $\gamma(a) = \gamma(b)$, then $\int_\gamma D_\sigma f = 0$. 
+\end{theorem}
+\begin{proof}
+    Using \autoref{lemma:rectifiable-piecewise-linear}, assume without loss of generality that $\gamma$ is piecewise smooth. By the \hyperref[Chain Rule]{proposition:chain-rule-sets-conditions}, $f \circ \gamma$ is piecewise $C^1$ with $D(f \circ \gamma)(t) = Df(\gamma(t)) \cdot D\gamma(t)$ on all but finitely many points. In which case, by \hyperref[change of variables formula]{theorem:rs-change-of-variables} and the \hyperref[Fundamental Theorem of Calculus]{theorem:ftc-riemann},
+    \begin{align*}
+    \int_\gamma D_\sigma f &= \int_a^b D_\sigma f (\gamma(t)) \cdot D\gamma(t)dt \\
+    &= \int_a^b D(f \circ \gamma)(t) dt =  f(\gamma(b)) - f(\gamma(a))
+    \end{align*}
+\end{proof}
+
+
+
+
+
+

src/fa/rs/regulated.tex

@@ -0,0 +1,108 @@
+\section{Regulated Functions}
+\label{section:regulated-function}
+
+
+\begin{proposition}
+\label{proposition:rs-interval}
+    Let $[a, b] \subset \real$, $E, F, H$ be TVSs over $K \in \RC$, and $E \times F \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map. 
+    
+    Let $G: [a, b] \to F$ and $[c, d] \subset [a, b]$ such that $G$ is continuous at $c$ and $d$, then for any $x \in E$, $x \cdot \one_{[c, d]} \in RS([a, b], G)$, and
+    \[
+    \int_a^b x \cdot \one_{[c, d]} dG = x \cdot [G(d) - G(c)]
+    \]
+\end{proposition}
+\begin{proof}
+    Assume without loss of generality that $a < c \le d < b$. Let $U \in \cn_H(0)$, then there exists $V \in \cn_F(0)$ such that $xV \subset U$. By continuity of $G$, there exists $\delta > 0$ such that $G((c - \delta, c]) - G(c) \subset V$ and $G([d, d + \delta)) - G(d) \subset V$. In which case, for any tagged partition $(P = \bracsn{x_j}_0^n, t = \seqf{t_j})$ that contains $\bracs{c - \delta, c, d, d + \delta}$, 
+    \begin{align*}
+        S(Q, t, x \cdot \one_{[c, d]}, G) &= x\sum_{a < x_j \le c - \delta}\one_{[c, d]}(t_j)[G(x_j) - G(x_{j - 1})] \\
+        &+ x \sum_{c - \delta < x_j \le c}\one_{[c, d]}(t_j)[G(x_j) - G(x_{j - 1})] \\
+        &+ x \sum_{c < x_j \le d}\one_{[c, d]}(t_j)[G(x_j) - G(x_{j-1})] \\
+        &+ x \sum_{d < x_j \le d + \delta}\one_{[c, d]}(t_j)[G(x_j) - G(x_{j - 1})] \\
+        &+ x \sum_{d + \delta < x_j \le b}\one_{[c, d]}(t_j)[G(x_j) - G(x_{j - 1})] \\
+        &= x \sum_{c - \delta \le x_j \le c}\one_{[c, d]}(t_j)[G(x_j) - G(x_{j - 1})] \\
+        &+ x \cdot [G(d) - G(c)] \\
+        &+ x \sum_{d < x_j \le d + \delta}\one_{[c, d]}[G(x_j) - G(x_{j - 1})] \\
+        &\in G(d) - G(c) + xG([c - \delta, c]) + xG([d, d + \delta]) \\
+        &\subset x \cdot [G(d) - G(c)] + xV + xV \subset [G(d) - G(c)] + U + U
+    \end{align*}
+\end{proof}
+
+
+\begin{definition}[Step Map]
+\label{definition:step-map}
+    Let $[a, b] \subset \real$, $E$ be a TVS, $f: [a, b] \to E$ be a function, and $P = \bracsn{x_j}_1^n \in \scp([a, b])$, then $f$ is a \textbf{step map} with respect to $P$ if for each $1 \le j \le n$, $f$ is constant on $(x_{j - 1}, x_j)$. 
+\end{definition}
+
+\begin{definition}[Regulated Function]
+\label{definition:regulated-function}
+    Let $[a, b] \subset \real$, $E, F, H$ be TVSs over $K \in \RC$, and $E \times F \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map. 
+
+    Let $G: [a, b] \to F$ and $f: [a, b] \to E$ be a step map, then $f$ is \textbf{regulated with respect to} $G$ if $G$ is continuous on all discontinuity points of $f$. Let $\text{Reg}([a, b], G; E)$ be closure of all regulated step maps with respect to the uniform topology, then:
+    \begin{enumerate}
+        \item Every regulated step map is in $RS([a, b], G)$. 
+        \item If $E$ is metrisable, then for any $f \in \text{Reg}([a, b], G; E)$, $f$ is continuous at all but at most countably many points, and $f$ does not share any discontinuity points with $E$. 
+        \item If $E, F, H$ are locally convex, $G \in BV([a, b]; F)$, and $H$ is complete, then $\text{Reg}([a, b], G; E) \subset RS([a, b], G)$. 
+        \item If $F$ is normed and $G \in BV([a, b]; F)$, then $\text{Reg}([a, b], G; E) \supset C([a, b]; E)$.
+    \end{enumerate}
+
+    The set $\text{Reg}([a, b], G; E)$ is the space of \textbf{regulated functions} with respect to $G$. If $G = \text{Id}$, then $\text{Reg}([a, b]; E) = \text{Reg}([a, b], G; E)$ denotes the space of regulated functions on $[a, b]$.
+\end{definition}
+\begin{proof}
+    (1): If $f: [a, b] \to E$ is a regulated step map, then there exists $\seqf{[a_j, b_j]} \subset 2^{[a, b]}$ and $\seqf{x_j}$ such that
+    \begin{enumerate}
+        \item $f = \sum_{j = 1}^n x_j \one_{[a_j, b_j]}$.
+        \item For each $1 \le j \le n$, $G$ is continuous at $a_j$ and $b_j$. 
+    \end{enumerate}
+
+    Thus $f \in RS([a, b], G)$ by \autoref{proposition:rs-interval}. 
+
+    (2): Let $f \in \text{Reg}([a, b], G; E)$, then there exists regulated step maps $\seq{f_n} \subset \text{Reg}([a, b], G; E)$ such that $f_n \to f$ uniformly. For each $n \in \natp$, let $D_n$ be the set of discontinuity points of $f_n$. Let $x \in [a, b] \setminus \bigcup_{n \in \natp}D_n$. For any symmetric neighbourhood $U \in \cn_E(0)$, there exists $n \in \natp$ such that $(f_n - f)([a, b]) \subset U$. Since $f_n$ is continuous at $x$, there exists $\eps > 0$ such that $f_n((x - \eps, x + \eps)) - f_n(x) \in U$. In which case,
+    \begin{align*}
+        f((x - \eps, x + \eps)) - f(x) &\subset (f_n - f)((x - \eps, x + \eps)) + (f_n - f)(x) \\
+        &+ f_n((x - \eps, x + \eps)) - f_n(x) \\
+        &\subset U + U + U 
+    \end{align*}
+
+    Therefore $f$ is continuous at $x$. Since each $D_n$ is finite, $\bigcup_{n \in \natp}D_n$ is countable. Given that $G$ is continuous on every point in each $D_n$, $G$ is also continuous on all the discontinuity points of $f$. 
+
+    (3): Suppose that $E, F, H$ are locally convex, $G \in BV([a, b]; F)$, and $H$ is complete. Let $f \in \text{Reg}([a, b], G; E)$, then there exists regulated step maps $\net{f} \subset \text{Reg}([a, b], G; E)$ such that $f_\alpha \to f$ uniformly. By \autoref{proposition:rs-complete}, $f \in RS([a, b], G)$. Therefore $\text{Reg}([a, b], G; E) \subset RS([a, b], G)$. 
+    
+    (4): If $F$ is normed and $G \in BV([a, b]; F)$, then $G$ has at most countably many discontinuities by \autoref{definition:bounded-variation}, so the continuity points of $G$ are dense in $[a, b]$. 
+
+    Let $f \in C([a, b]; E)$, then $f \in UC([a, b]; E)$ by \autoref{proposition:uniform-continuous-compact}. Let $U \in \cn_E(0)$, then there exists a partition $P = \bracsn{x_j}_1^n \in \scp([a, b])$ such that for each $1 \le j \le n$ and $x, y \in [x_{j-1}, x_j]$, $f(x) - f(y) \in U$. In which case, let
+    \[
+    \phi = f(a)\one_{\bracs{a}} + \sum_{j = 1}^n f(x_j)\one_{(x_{j - 1}, x_j]}
+    \]
+
+    then $\phi$ is a regulated step map with $(\phi - f)([a, b]) \in U$. 
+\end{proof}
+
+\begin{theorem}[Fundamental Theorem of Calculus for Riemann Integrals]
+\label{theorem:ftc-riemann}
+    Let $[a, b] \subset \real$ and $E$ be a separated locally convex space, then:
+    \begin{enumerate}
+        \item For any Riemann integrable function $f: [a, b] \to E$ and $x \in (a, b)$ such that $f$ is continuous at $x$, the function
+        \[
+        F: [a, b] \to E \quad y \mapsto \int_a^y f(t)dt
+        \]
+
+        is differentiable at $x$ with $DF(x) = f(x)$. 
+
+        \item For $F \in C^1([a, b]; E)$, 
+        \[
+        F(b) - F(a) = \int_a^b DF(t)dt
+        \]
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    (1): Let $x \in (a, b)$ such that $f$ is continuous at $x$, and $\delta > 0$ such that $(x_0 - \delta, x_0 + \delta) \subset (a, b)$, then for any $h > 0$, 
+    \begin{align*}
+        \frac{1}{h}\braks{\int_a^{x+h} f(t)dt - \int_a^{x} f(t)dt} &= \frac{1}{h}\int_x^{x+h}f(t)dt \in \overline{\text{Conv}(f([x, x +h)))} \\
+        -\frac{1}{h}\braks{\int_a^{x-h} f(t)dt - \int_a^{x} f(t)dt} &= \frac{1}{h}\int_{x-h}^{x}f(t)dt \in \overline{\text{Conv}(f((x - h, x]))}
+    \end{align*}
+
+    As $E$ is locally convex and separated, and $f$ is continuous at $x$, this implies that $[F(x + h) - F(x)]/h \to f(x)$.  
+
+    (2): Let $G(x) = \int_a^x DF(t)dt + F(a)$, then $G - F$ has derivative $0$. By the \hyperref[Mean Value Theorem]{proposition:zero-derivative-constant}, $G - F$ is constant. As $G(a) - F(a) = 0$, $G  = F$. 
+    
+\end{proof}

src/fa/rs/rs-bv.tex

@@ -1,40 +1,38 @@
-\section{Riemann-Stieltjes Integrals and Functions of Bounded Variation}
+\section{Integrators of Bounded Variation}
 \label{section:rs-bv}
 
 \begin{proposition}
 \label{proposition:rs-bound}
-    Let $[a, b] \subset \real$, $E_1, E_2, H$ be locally convex spaces, and $E_1 \times E_2 \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map, and $G: [a, b] \to E_2$.
+    Let $[a, b] \subset \real$, $E, F, H$ be locally convex spaces, $E \times F \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map, $G: [a, b] \to F$, and $f \in RS([a, b], G)$, then for any continuous seminorms $[\cdot]_E: E \to [0, \infty)$, $[\cdot]_F: F \to [0, \infty)$, and $[\cdot]_H: H \to [0, \infty)$ such that $[xy]_H \le [x]_E[y]_F$ for all $x \in E$ and $y \in F$, 
-
-    Let $[\cdot]_H$ be a continuous seminorm on $H$, then there exists continuous seminorms $[\cdot]_1$ on $E_1$ and $[\cdot]_2$ on $E_2$ such that for any $f \in RS([a, b], G)$,
     \[
-    \braks{\int_a^bf dG}_H \le \sup_{x \in [a, b]}[f]_1 \cdot [g]_{\text{var}, 2}
+    \braks{\int_a^bf dG}_H \le \sup_{x \in [a, b]}[f]_E \cdot [g]_{\text{var}, F}
     \]
 
 \end{proposition}
 \begin{proof}
-    By \autoref{proposition:tvs-convex-multilinear}, there exists continuous seminorms $[\cdot]_1$ on $E_1$ and $[\cdot]_2$ on $E_2$ such that $[xy]_H \le [x]_1[y]_2$ for all $(x, y) \in E_1 \times E_2$.
-
     Let $(P = \seqfz{x_j}, c = \seqf{c_j}) \in \scp_t([a, b])$, then
     \begin{align*}
-        [S(P, c, f, G)]_H &\le \sum_{j = 1}^n [f(c_j)[G(x_j) - G(x_{j - 1})]]_H \le \sum_{j = 1}^n [f(c_j)]_1[G(x_j) - G(x_{j - 1})]_2 \\
+        [S(P, c, f, G)]_H &\le \sum_{j = 1}^n [f(c_j)[G(x_j) - G(x_{j - 1})]]_H \\
-        &\le \sup_{x \in [a, b]}[f]_1 \cdot V_{2, P}(G) \le \sup_{x \in [a, b]}[f]_1 \cdot [g]_{\text{var}, 2}
+        &\le \sum_{j = 1}^n [f(c_j)]_E[G(x_j) - G(x_{j - 1})]_F \\
+        &\le \sup_{x \in [a, b]}[f]_E \cdot V_{2, P}(G) \le \sup_{x \in [a, b]}[f]_E \cdot [g]_{\text{var}, F}
     \end{align*}
 \end{proof}
 
 \begin{proposition}
 \label{proposition:rs-complete}
-    Let $[a, b] \subset \real$, $E_1, E_2, H$ be locally convex spaces, and $E_1 \times E_2 \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map, and $G \in BV([a, b]; E_2)$.
+    Let $[a, b] \subset \real$, $E, F, H$ be locally convex spaces, $E \times F \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map, and $G \in BV([a, b]; F)$.
 
-    For each continuous seminorm $\rho$ on $E_1$ and $f: [a, b] \to E_1$, define
+    For each continuous seminorm $\rho$ on $E$ and $f: [a, b] \to E$, define
     \[
     [f]_{u, \rho} = \sup_{x \in [a, b]}\rho(f(x))
     \]
 
     Let $\net{f} \subset RS([a, b], G)$ such that:
     \begin{enumerate}
-        \item[(a)] For each continuous seminorm $\rho$ on $E_1$, $[f_\alpha - f]_{u, \rho} \to 0$.
+        \item[(a)] For each continuous seminorm $\rho$ on $E$, $[f_\alpha - f]_{u, \rho} \to 0$.
         \item[(b)] $\lim_{\alpha \in A}\int_a^b f_\alpha dG$ exists.
     \end{enumerate}
+
     then $f \in RS([a, b], G)$ and $\int_a^b f dG = \lim_{\alpha \in A}\int_a^b f_\alpha dG$. In particular,
     \begin{enumerate}
         \item If $H$ is complete, then condition (b) may be omitted.
@@ -48,17 +46,19 @@
         &+ \rho\paren{\int_a^b f_\alpha dG - \lim_{\alpha \in A}\int_a^b f_\alpha dG} \\
         &+ \rho\paren{S(P, c, f_\alpha, G) - \int_a^b f_\alpha dG}
     \end{align*}
-    Let $\rho$ be a continuous seminorm on $H$, and $[\cdot]_1$ and $[\cdot]_2$ be continuous seminorms on $E_1$ and $E_2$ such that $\rho(xy) \le [x]_1[y]_2$ for all $(x, y) \in E_1 \times E_2$.
+    Let $\rho$ be a continuous seminorm on $H$, and $[\cdot]_E$ and $[\cdot]_F$ be continuous seminorms on $E$ and $F$ such that $\rho(xy) \le [x]_E[y]_F$ for all $(x, y) \in E \times F$.
 
     Let $\eps > 0$, then by assumption (a) and (b), there exists $\alpha \in A$ such that:
     \begin{enumerate}
-        \item $[f - f_\alpha]_1 < \eps/(3[G]_{\text{var}, 2})$.
+        \item $[f - f_\alpha]_E < \eps/(3[G]_{\text{var}, F})$.
         \item $\rho\paren{\int_a^b f_\alpha dG - \lim_{\alpha \in A}\int_a^b f_\alpha dG} < \eps/3$.
     \end{enumerate}
+
     Since $f_\alpha \in RS([a, b], G)$, there exists $P_0 \in \scp([a, b])$ such that if $P \ge P_0$,
     \begin{enumerate}
         \item[(3)] $\rho\paren{S(P, c, f_\alpha, G) - \int_a^b f_\alpha dG} < \eps/3$.
     \end{enumerate}
+
     Thus for any $(P, c) \in \scp_t([a, b])$ with $P \ge P_0$,
     \[
     \rho\paren{S(P, c, f, G) - \lim_{\alpha \in A}\int_a^b f_\alpha dG} < \eps
@@ -68,32 +68,79 @@
 
 \begin{proposition}
 \label{proposition:rs-bv-continuous}
-    Let $[a, b] \subset \real$, $E_1, E_2$ be locally convex spaces, $H$ be a sequentially complete locally convex space, and $E_1 \times E_2 \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map.
+    Let $[a, b] \subset \real$, $E, F$ be locally convex spaces, $H$ be a sequentially complete locally convex space, and $E \times F \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map.
 
-    Let $f \in C([a, b]; E_1)$, $G \in BV([a, b]; E_2)$, then
+    Let $f \in C([a, b]; E)$, $G \in BV([a, b]; F)$, then
     \begin{enumerate}
         \item $f \in RS([a, b], G)$.
-        \item For any $\seq{(P_n, t_n)} \subset \scp_t([a, b])$ with $\sigma(P_n) \to 0$,
+        \item For equicontinuous family $\cf \subset C([a, b]; E)$ and $\seq{(P_n, t_n)} \subset \scp_t([a, b])$ with $\sigma(P_n) \to 0$,
         \[
         \int_a^b fdG = \limv{n}S(P_n, t_n, f, G)
         \]
 
+        uniformly for all $f \in \cf$. 
+
     \end{enumerate}
 \end{proposition}
 \begin{proof}
-    Let $\rho$ be a continuous seminorm on $H$, and $[\cdot]_1$ and $[\cdot]_2$ be continuous seminorms on $E_1$ and $E_2$ such that $\rho(xy) \le [x]_1[y]_2$ for all $(x, y) \in E_1 \times E_2$.
+    Let $\rho$ be a continuous seminorm on $H$, and $[\cdot]_E$ and $[\cdot]_F$ be continuous seminorms on $E$ and $F$ such that $\rho(xy) \le [x]_E[y]_F$ for all $(x, y) \in E \times F$.
 
     Let $(P = \seqfz{x_j}, c = \seqf{c_j}), (Q = \seqfz[m]{y_j}, d = \seqf[m]{d_j}) \in \scp_t([a, b])$ with $Q \ge P$, then
     \begin{align*}
-        \rho(S(P, c, f, G) - S(Q, d, f, G)) &\le \sum_{j = 1}^n \sum_{y_k \in [x_{j - 1}, x_j]}[f(c_j) - f(d_k)]_1[G(y_k) - G(y_{k - 1})]_2 \\
+        &\rho(S(P, c, f, G) - S(Q, d, f, G)) \\\
-        &\le \sup_{\begin{array}{c} x, y \in [a, b] \\ |x - y| < \sigma(P) \end{array}}[f(x) - f(y)]_1 \cdot [G]_{\text{var}, 2}
+        &\le \sum_{j = 1}^n \sum_{y_k \in [x_{j - 1}, x_j]}[f(c_j) - f(d_k)]_E[G(y_k) - G(y_{k - 1})]_F \\
+        &\le \sup_{\begin{array}{c} x, y \in [a, b] \\ |x - y| < \sigma(P) \end{array}}[f(x) - f(y)]_E \cdot [G]_{\text{var}, F}
     \end{align*}
     Therefore for any two $(P, c), (Q, d) \in \scp_t([a, b])$,
     \[
-    \rho(S(P, c, f, G) - S(Q, d, f, G)) \le 2 \cdot \sup_{\begin{array}{c} x, y \in [a, b] \\ |x - y| < \max(\sigma(P), \sigma(Q)) \end{array}}[f(x) - f(y)]_1 \cdot [G]_{\text{var}, 2}
+    \rho(S(P, c, f, G) - S(Q, d, f, G)) \le 2 \cdot \sup_{\begin{array}{c} x, y \in [a, b] \\ |x - y| < \max(\sigma(P), \sigma(Q)) \end{array}}[f(x) - f(y)]_E \cdot [G]_{\text{var}, F}
     \]
 
-    by passing through a common refinement. Since $f \in C([a, b]; E_1)$, this bound tends to $0$ as $\max(\sigma(P), \sigma(Q))$ tends to $0$, so $\angles{S(P, c, f, G)}_{(P, c) \in \scp_t([a, b])}$ is a Cauchy net.
+    by passing through a common refinement. Since $f \in C([a, b]; E)$, this bound tends to $0$ as $\max(\sigma(P), \sigma(Q))$ tends to $0$, so $\angles{S(P, c, f, G)}_{(P, c) \in \scp_t([a, b])}$ is a Cauchy net.
 
     In addition, for any $\seq{(P_n, t_n)}$ as in (2), $\limv{n}S(P_n, t_n, f, G)$ exists by sequential completeness. Since $\angles{S(P, c, f, G)}_{(P, c) \in \scp_t([a, b])}$ is Cauchy, the limit $\lim_{(P, c) \in \scp_t([a, b])}S(P, c, f, G)$ exists as well and is equal to $\limv{n}S(P_n, t_n, f, G)$.
 \end{proof}
+
+
+\begin{theorem}[Fubini's Theorem for Riemann-Stieltjes Integrals]
+\label{theorem:rs-fubini}
+    Let $[a, b], [c, d] \subset \real$, $E, F, G, H$ be a locally convex space over $K \in \RC$ with $H$ being sequentially complete, $E \times F \times G \to H$ with $(x, y, z) \mapsto xyz$ be a $3$-linear map\footnote{$E, F, G$ are assumed to be disjoint, so the product is well-defined regardless of the order of the terms.}, $\alpha \in BV([a, b]; F)$, $\beta \in BV([c, d]; G)$, and $f \in C([a, b] \times [c, d]; E)$, then
+    \[
+    \int_a^b \int_c^d f(s, t) \beta(dt) \alpha(ds) = \int_c^d\int_a^b f(s, t) \alpha(ds) \beta(dt)
+    \]
+\end{theorem}
+\begin{proof}
+    Let
+    \[
+    g: [a, b] \to L(F; H) \quad s \mapsto \int_c^d f(s, t) \beta(dt)
+    \]
+
+    then for any $(P = \seqfz{x_j}, c = \seqf{c_j}) \in \scp_t([a, b])$,
+    \begin{align*}
+        S(P, c, g, \alpha) &= \sum_{j = 1}^n g(c_j) [\alpha(x_j) - \alpha(x_{j-1})] \\
+        &=  \sum_{j = 1}^n \int_c^d f(c_j, t) \beta(dt) [\alpha(x_j) - \alpha(x_{j-1})] \\
+        &= \int_c^d S(P, c, f(\cdot, t), \alpha) \beta(dt)
+    \end{align*}
+
+
+    Since $\alpha \in BV([a, b]; F)$, by \autoref{proposition:rs-bv-continuous}, for any $\seq{(P_n, c_n)} \subset \scp_t([a, b])$, 
+    \[
+    \int_a^b \int_c^d f(s, t) \beta(dt) \alpha(ds) = \limv{n}S(P_n, c, g, \alpha)
+    \]
+
+    and
+    \[
+    \limv{n}S(P_n, c_n, f(\cdot, t), \alpha) = \int_a^b f(s, t) \alpha(ds)
+    \]
+
+    uniformly for all $t \in [c, d]$. Since $f \in C([a, b] \times [c, d]; E)$, $f$ is uniformly continuous by \autoref{proposition:uniform-continuous-compact}, and $\bracs{f(\cdot, t)|t \in [c, d]} \subset C([a, b]; E)$ is uniformly equicontinuous. As $\beta \in BV([c, d]; G)$, 
+    \[
+    \int_c^d\int_a^b f(s, t) \alpha(ds) \beta(dt) = \limv{n}\int_c^d S(P_n, c_n, f(\cdot, t), \alpha) \beta(dt)
+    \]
+
+    by \autoref{proposition:rs-complete}. 
+\end{proof}
+
+
+
+

src/fa/rs/rs-measure.tex

@@ -0,0 +1,92 @@
+\section{The Lebesgue-Stieltjes Integral}
+\label{section:riemann-lebesgue-stieltjes}
+
+\begin{definition}[Lebesgue-Stieltjes Measure]
+\label{definition:riemann-lebesgue-stieltjes}
+    Let $[a, b] \subset \real$, $E$ be a Banach space, and $G \in BV([a, b]; E^*)$, then there exists a unique $\mu_G \in M_R([a, b]; E^*)$ such that:
+    \begin{enumerate}
+        \item For any $f \in \text{Reg}([a, b], G; E)$, 
+        \[
+        \int_a^b f(t) G(dt) = \int_{[a, b]} \dpn{f(t), \mu_G(dt)}{E}
+        \]
+
+        \item For any $a \le c < d < b$, 
+        \[
+        \mu_G((c, d]) = G(d+) - G(c+) = \lim_{z \downto d}G(z) - \lim_{z \downto c}G(z)
+        \]
+    \end{enumerate}
+
+    The measure $\mu_G$ is the \textbf{Lebesgue-Stieltjes measure} associated with $G$.
+\end{definition}
+\begin{proof}
+    By \autoref{proposition:rs-bv-continuous}, the mapping $f \mapsto \int f(t)G(dt)$ is a continuous linear functional on $C_0([a, b]; E) = C([a, b]; E)$. By \hyperref[Singer's Representation Theorem]{theorem:singer-representation}, there exists $\mu_G \in M_R([a, b]; E^*)$ such that (1) holds for $C([a, b], G; E)$.
+
+    (2): Fix $x \in [a, b)$. For each $u \in (x, b)$, there exists a non-increasing function $\phi_u \in C_c([a, b))$ such that $\phi_u|_{[a, x]} = 1$ and $\supp{\phi_u} \subset [a, u]$. By outer regularity of $\mu_G$, 
+    \[
+    \mu_G([a, x]) = \lim_{u \downto x}\mu_G([a, u]) = \lim_{u \downto x}\int_{[a, b]}\phi_u d\mu_G = \lim_{u \downto x}\int_a^b \phi_u dG
+    \]
+
+    For each $u > x$, using \hyperref[integration by parts]{theorem:rs-ibp}, 
+    \begin{align*}
+        \int_a^b \phi_u dG &= \phi(b)G(b) - \phi(a)G(a) - \int_a^b G d\phi_u \\
+        &= - G(a) - \int_a^b G d\phi_u
+    \end{align*}
+
+    Let $v \in (x, u)$ and $(P = \bracs{x_j}_0^n, c = \bracs{c_j}_1^n)$ be a tagged partition containing $\bracs{x, v, u}$, then since $\phi_u$ is non-increasing, 
+    \begin{align*}
+        \int_a^b Gd\phi_u &= \sum_{x < x_j \le v}G(c_j)[\phi(x_j) - \phi(x_{j - 1})] + \sum_{v < x_j \le u}G(c_j)[\phi(x_j) - \phi(x_{j - 1})] \\
+        &\in [\phi(v) - \phi(x)]G([x, u]) + [\phi(u) - \phi(v)]G((x, u]) \\
+        &\subset [\phi(v) - \phi(x)]G([x, u]) - G((x, u])
+    \end{align*}
+    
+
+    Since the above holds for all $v \in (x, u)$, $\int_a^b Gd\phi_u \in -\overline{G((a, u])}$, and
+    \[
+    \int_a^b \phi_u dG \in \overline{G(x, u]} - G(a)
+    \]
+    
+    
+    As $E^*$ is a Banach space, $G(x+)$ exists by \autoref{proposition:bounded-variation-one-side-limit}, so
+    \[
+    \mu_G([a, x]) = \lim_{u \downto x}\int_a^b \phi_u dG = \bigcap_{u > x}\overline{G((x, u])} - G(a) = G(x+) - G(a)
+    \]
+
+    
+    (1): Let $[c, d] \subset [a, b]$ such that $G$ is continuous at $c$ and $d$, then for any $x \in E$, $x \cdot \one_{[c, d]} \in RS([a, b], G)$ with
+    \[
+    \int_a^b x \cdot \one_{[c, d]} dG = x[G(d) - G(c)] = \int_{[a, b]}x \cdot \one_{[c, d]} d\mu_G
+    \]
+
+    By linearity, for any regulated step map $f: [a, b] \to E$, $\int_a^b f(t) G(dt) = \int_{[a, b]} \dpn{f(t), \mu_G(dt)}{E}$. Therefore (1) holds for $\text{Reg}([a, b], G; E)$ by the \hyperref[Linear Extension Theorem]{theorem:linear-extension-theorem-normed}. 
+\end{proof}
+
+
+% TODO: Strengthen this to the Lebesgue differentiation case in the future. 
+% It should work for ALL absolutely continuous functions. 
+
+\begin{proposition}
+\label{proposition:lebesgue-stieltjes-differentiable}
+    Let $[a, b] \subset \real$, $E$ be a normed space, $G \in C^1([a, b]; E^*)$, and $\mu_G \in M_R([a, b]; E^*)$ be the associated Lebesgue-Stieltjes measure, then
+    \[
+    \mu_G(dt) = DG(t)dt
+    \]
+\end{proposition}
+\begin{proof}
+    By the \hyperref[Fundamental Theorem of Calculus]{theorem:ftc-riemann}, for any $[c, d] \subset [a, b]$, 
+    \[
+    \int_a^b \one_{[c, d]}dG = \int_a^b \one_{[c, d]} DG(t)dt 
+    \]
+
+    By \autoref{definition:riemann-lebesgue-stieltjes}, applied to both $G$ and $\text{Id}$, 
+    \[
+    \int_{[a, b]}\one_{[c, d]}d\mu_G = \int_a^b \one_{[c, d]}dG = \int_a^b \one_{[c, d]} DG(t)dt = \int_{[a, b]} \one_{[c, d]}DG(t)dt
+    \]
+
+    Thus by linearity, for any regulated step map $f: [a, b] \to E$,
+    \[
+    \int_{[a, b]} f(t) \mu_G(dt) = \int_{[a, b]} \dpn{f(t), DG(t)}{E} dt
+    \]
+
+    By \autoref{definition:regulated-function}, the regulated step maps are dense in $\text{Reg}([a, b], G; E)$. Therefore by the \hyperref[Linear Extension Theorem]{theorem:linear-extension-theorem-normed}, the above holds for all $f \in \text{Reg}([a, b], G; E)$. By uniqueness of the \hyperref[Lebesgue-Stieltjes measure]{definition:riemann-lebesgue-stieltjes}, $\mu_G(dt) = DG(t)dt$. 
+\end{proof}
+

src/fa/rs/rs.tex

@@ -3,9 +3,9 @@
 
 \begin{definition}[Riemann-Stieltjes Sum, {{\cite[Section X.1]{Lang}}}]
 \label{definition:rs-sum}
-    Let $[a, b] \subset \real$, $E_1, E_2, H$ be TVSs over $F \in \RC$, $E_1 \times E_2 \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map, and $G: [a, b] \to E_2$.
+    Let $[a, b] \subset \real$, $E, F, H$ be TVSs over $K \in \RC$, $E \times F \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map, and $G: [a, b] \to F$.
 
-    Let $f: [a, b] \to E_1$ and $(P = \seqfz{x_j}, c = \seqf{c_j}) \in \scp_t([a, b])$, then
+    Let $f: [a, b] \to E$ and $(P = \seqfz{x_j}, c = \seqf{c_j}) \in \scp_t([a, b])$, then
     \[
     S(P, c, f, G) = \sum_{j = 1}^n f(c_j)[G(x_j) - G(x_{j - 1})]
     \]
@@ -15,9 +15,9 @@
 
 \begin{definition}[Riemann-Stieltjes Integral, {{\cite[Section X.1]{Lang}}}]
 \label{definition:rs-integral}
-    Let $[a, b] \subset \real$, $E_1, E_2, H$ be TVSs over $F \in \RC$, $E_1 \times E_2 \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map, $G: [a, b] \to E_2$.
+    Let $[a, b] \subset \real$, $E, F, H$ be TVSs over $K \in \RC$, $E \times F \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map, $G: [a, b] \to F$.
 
-    Let $f: [a, b] \to E_2$, then $f$ is \textbf{Riemann-Stieltjes integrable} with respect to $G$ if the limit
+    Let $f: [a, b] \to F$, then $f$ is \textbf{Riemann-Stieltjes integrable} with respect to $G$ if the limit
     \[
     \int_a^b f dG = \int_a^b f(t)G(dt) = \lim_{(P, c) \in \scp_t([a, b])}S(P, c, f, G)
     \]
@@ -27,21 +27,23 @@
     The set $RS([a, b], G)$ is the vector space of all \textbf{Riemann-Stieltjes integrable functions} with respect to $G$.
 \end{definition}
 
-\begin{lemma}[Summation by Parts, {{\cite[Proposition 1.4]{Lang}}}]
+\begin{lemma}[Summation by Parts]
 \label{lemma:sum-by-parts}
-    Let $[a, b] \subset \real$, $E_1, E_2, H$ be TVSs over $F \in \RC$, $E_1 \times E_2 \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map, $f: [a, b] \to E_1$, $G: [a, b] \to E_2$, and $(P, c) \in \scp_t([a, b])$, then
+    Let $[a, b] \subset \real$, $E, F, H$ be TVSs over $K \in \RC$, $E \times F \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map, $f: [a, b] \to E$, $G: [a, b] \to F$, and $(P, c) \in \scp_t([a, b])$, then
     \[
     S(P, c, f, G) + S(P', c', G, f) = f(b)G(b) - f(a)G(a)
     \]
 
     where $P' = \seqfz[n+1]{y_j} = [a, c_1, \cdots, c_n, b]$ and $c' = \seqf[n+1]{d_j} = [x_0, \cdots, x_n]$.
 \end{lemma}
-\begin{proof}
+\begin{proof}[Proof {{\cite[Proposition 1.4]{Lang}}}. ]
     Denote $c_0 = a$ and $c_{n+1} = b$, then
     \begin{align*}
         S(P, c, f, G) &= \sum_{j = 1}^n f(c_j)[G(x_j) - G(x_{j - 1})]
-        = \sum_{j = 1}^n f(c_j)G(x_j) - \sum_{j = 1}^n f(c_j)G(x_{j - 1}) \\
+        \\
-        &= f(c_n)G(x_n)- f(c_0)G(x_0) + \sum_{j = 1}^n f(c_{j - 1})G(x_{j-1}) - \sum_{j = 1}^n f(c_j)G(x_{j - 1}) \\
+        &= \sum_{j = 1}^n f(c_j)G(x_j) - \sum_{j = 1}^n f(c_j)G(x_{j - 1}) \\
+        &= f(c_n)G(x_n)- f(c_0)G(x_0) + \sum_{j = 1}^n f(c_{j - 1})G(x_{j-1}) \\
+        &- \sum_{j = 1}^n f(c_j)G(x_{j - 1}) \\
         &= f(c_n)G(x_n)- f(c_0)G(x_0) - \sum_{j = 1}^n G(x_{j - 1})[f(c_j) - f(c_{j - 1})] \\
         &= f(c_{n+1})G(x_n) - f(c_0)G(x_0) - \sum_{j = 1}^{n+1}G(x_{j - 1})[f(c_j) - f(c_{j - 1})] \\
         &= f(b)G(b) - f(a)G(a) - S(P', c', G, f)
@@ -50,16 +52,16 @@
 
 \begin{theorem}[Integration by Parts]
 \label{theorem:rs-ibp}
-    Let $[a, b] \subset \real$, $E_1, E_2, H$ be TVSs over $F \in \RC$, and $E_1 \times E_2 \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map.
+    Let $[a, b] \subset \real$, $E, F, H$ be TVSs over $K \in \RC$, and $E \times F \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map.
 
-    Let $f: [a, b] \to E_1$ and $G: [a, b] \to E_2$, then $f \in RS([a, b], G)$ if and only if $G \in RS([a, b], f)$. In which case,
+    Let $f: [a, b] \to E$ and $G: [a, b] \to F$, then $f \in RS([a, b], G)$ if and only if $G \in RS([a, b], f)$. In which case,
     \[
     \int_a^b f dG + \int_a^b G df = f(b)G(b) - f(a)G(a)
     \]
 
 \end{theorem}
 \begin{proof}
-    Suppose that $f \in RS([a, b], G)$. Let $U \in \cn_F(0)$, then there exits $P_0 = \seqfz{x_j} \in \scp([a, b])$ such that $S(P, c, f, G) - \int_a^b fdG \in U$ for all $(P, c) \in \scp_t([a, b])$ with $P \ge P_0$. Let
+    Suppose that $f \in RS([a, b], G)$. Let $U \in \cn_H(0)$, then there exits $P_0 = \seqfz{x_j} \in \scp([a, b])$ such that $S(P, c, f, G) - \int_a^b fdG \in U$ for all $(P, c) \in \scp_t([a, b])$ with $P \ge P_0$. Let
     \[
     Q_0 = [x_0, x_1, x_1, \cdots, x_n, x_n]
     \]
@@ -67,8 +69,37 @@
     then for any $(Q = \seqfz[m]{y_j}, d = \seqf[m]{d_j}) \in \scp_t([a, b])$ with $Q \ge Q_0$,
     \[
     f(b)G(b) - f(a)G(a) - \int_a^b fdG - S(Q, d, G, f) =
-    \int_a^b fdG - S(Q', d', G, f)
+    S(Q', d', f, G) - \int_a^b fdG
     \]
 
-    by \autoref{lemma:sum-by-parts}, where $d$ and $Q'$ contain $\seqfz{x_j}$. Thus $(Q', d') \ge P_0$, and $\int_a^b fdG - S(Q', d', G, f) \in U$.
+    by \autoref{lemma:sum-by-parts}, where $d$ and $Q'$ contain $\seqfz{x_j}$. Thus $(Q', d') \ge P_0$, and $S(Q', d', f, G) - \int_a^b fdG \in U$.
+\end{proof}
+
+
+
+\begin{theorem}[Change of Variables]
+\label{theorem:rs-change-of-variables}
+    Let $[a, b] \subset \real$, $E, F, H$ be locally convex spaces over $K \in \RC$, $E \times F \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map, and $G \in C^1([a, b]; F)$, then for any bounded $f \in RS([a, b], G; E)$, 
+    \[
+    \int_a^b f(t) G(dt) = \int_a^b f(t) DG(t) dt
+    \]
+\end{theorem}
+\begin{proof}
+    Let $[\cdot]_H: H \to [0, \infty)$ be a continuous seminorm and $[\cdot]_E: E \to [0, \infty)$ and $[\cdot]_F: F \to [0, \infty)$ be continuous seminorms on $E$ and $F$, respectively, such that for any $x \in E$ and $y \in F$, $[xy]_H \le [x]_E[y]_F$. 
+
+    Since $G \in C^1([a, b]; F)$, $DG \in UC([a, b]; F)$ by \autoref{proposition:uniform-continuous-compact}. Thus there exists $\delta > 0$ such that $[DG(x) - DG(y)]_F < \eps$ for all $x, y \in [a, b]$ with $|x - y| \le \delta$. Let $(P = \seqfz{x_j}, c = \seqf{c_j}) \in \scp_t([a, b])$ with $\sigma(P) \le \delta$, then by the \hyperref[Mean Value Theorem]{theorem:mean-value-theorem-line}, for each $1 \le j \le n$, 
+    \begin{align*}
+        &G(x_j) - G(x_{j-1}) - (x_j - x_{j-1})DG(c_j) \\
+        &\in (x_j - x_{j-1})\ol{\text{Conv}}\bracs{DG(t) - DG(c_j)|t \in [x_{j-1}, x_j]}
+    \end{align*}
+
+    so
+    \[
+    [G(x_j) - G(x_{j-1}) - (x_j - x_{j-1})DG(c_j)]_F \le \eps(x_j - x_{j-1})
+    \]
+
+    and
+    \[
+        [S(P, c, f, G) - S(P, c, f \cdot DG, \text{Id})]_H \le \eps \cdot (b - a) \cdot \sup_{x \in [a, b]}[f(x)]_E
+    \]
 \end{proof}

src/fa/tvs/bounded.tex

@@ -3,12 +3,23 @@
 
 \begin{definition}[Bounded]
 \label{definition:bounded}
-    Let $E$ be a TVS over $K \in \RC$ and $B \subset E$, then $B$ is \textbf{bounded} if for every $U \in \cn(0)$, there exists $\lambda \in K$ such that $\lambda U \supset B$. The collection $B(E)$ is the set of all bounded sets of $E$.
+    Let $(E, \topo)$ be a TVS over $K \in \RC$ and $B \subset E$, then the following are equivalent:
+    \begin{enumerate}
+        \item For every $U \in \cn(0)$, there exists $\lambda \in K$ such that $\lambda U \supset B$. 
+        \item For every $\seq{x_n} \subset B$ and $\seq{\lambda_n} \subset K$ such that $\lambda_n \to 0$, $\lambda_n x_n \to 0$ as $n \to \infty$.
+    \end{enumerate}
+
+    If the above holds, then $B$ is \textbf{bounded}. The collection $\mathfrak{B}(E) = \mathfrak{B}(E, \topo)$ is the set of all bounded sets of $E$.
 \end{definition}
+\begin{proof}
+    (1) $\Rightarrow$ (2): Let $U \in \cn_E(0)$ be circled, then there exists $k \in \natp$ such that $kU \supset B$. Since $\lambda_n \to 0$ as $n \to \infty$, there exists $N \in \natp$ such that $|\lambda_n| \le 1/k$ for all $n \ge N$. In which case, $\lambda_n x_n \in \lambda_n B \subset U$ for all $n \ge N$. 
+
+    $\neg (1) \Rightarrow \neg (2)$: Let $U \in \cn_E(0)$ be circled such that $B \not\subset nU$ for all $n \in \natp$. For each $n \in \natp$, let $x_n \in B \setminus nU$, then $x_n/n \not\in U$ for all $n \in \natp$. 
+\end{proof}
 
 \begin{proposition}[{{\cite[I.5.1]{SchaeferWolff}}}]
 \label{proposition:bounded-operations}
-    Let $E$ be a TVS over $K \in \RC$ and $A, B \in B(E)$, then the following sets are bounded:
+    Let $E$ be a TVS over $K \in \RC$ and $A, B \in \mathfrak{B}(E)$, then the following sets are bounded:
     \begin{enumerate}
         \item Any $C \subset B$.
         \item The closure $\ol{B}$.

src/fa/tvs/complete-metric.tex

@@ -15,35 +15,39 @@
     Let $x = x_{n_1} + \limv{N}\sum_{k = 1}^N (x_{n_{k+1}} - x_{n_k})$, then $x = \limv{k}x_{n_k} \in E$. Since $\seq{x_n}$ is a Cauchy sequence that admits a convergent subsequence, it is convergent.
 \end{proof}
 
-\begin{theorem}[Successive Approximations {{\cite[Section III.2]{SchaeferWolff}}}]
+\begin{theorem}[Successive Approximations]
 \label{theorem:successive-approximations}
     Let $E, F$ be metric TVSs over $K \in \RC$ with pseudonorm $\rho$ and $\eta$, respectively. Let $T \in L(E; F)$, $r > 0$, $\gamma \in (0, 1)$, and $C \ge 0$. Suppose that for every $y \in B_F(0, r)$, there exists $x \in E$ such that:
     \begin{enumerate}
         \item[(a)] $\eta(y - Tx) \le \gamma \eta(y)$.
         \item[(b)] $\rho(x) \le C \eta(y)$.
     \end{enumerate}
+
     then for any $y \in F$, there exists $\seq{x_n} \subset E$ such that:
     \begin{enumerate}
         \item $\sum_{n \in \natp}\rho(x_n) \le C\eta(y)/(1 - \gamma)$.
         \item $y = \limv{N}\sum_{n = 1}^N Tx_n$.
     \end{enumerate}
+
     In particular,
     \[
     T\braks{B_E\paren{0, \frac{Cr}{(1 - \gamma)}}} \supset B_F(0, r)
     \]
 
 \end{theorem}
-\begin{proof}
+\begin{proof}[Proof {{\cite[Section III.2]{SchaeferWolff}}}. ]
     Let $y_0 = y$ and $x_0 = 0$. Let $N \in \natz$ and suppose inductively that $\seqf[N]{x_n} \subset E$ has been constructed such that:
     \begin{enumerate}
         \item[(I)] $\sum_{n = 1}^N\rho(x_n) \le C\eta(y)\sum_{n = 0}^{N-1}\gamma^{n}$.
         \item[(II)] $\eta\paren{y - \sum_{n = 1}^N Tx_n} \le \eta(y)\gamma^N$.
     \end{enumerate}
+
     By assumption, there exists $x_{N+1} \in E$ such that:
     \begin{enumerate}
         \item[(i)] $\eta\paren{y - \sum_{n = 1}^{N+1} Tx_n} \le \gamma \eta\paren{y - \sum_{n = 1}^N Tx_n} \le \gamma^{N+1}$.
         \item[(ii)] $\rho(x_{N+1}) \le C\eta\paren{y - \sum_{n = 1}^N Tx_n} \le C\eta(y)\gamma^N$.
     \end{enumerate}
+
     Combining (I) and (ii) shows that $\sum_{n = 1}^N \rho(x_n) \le C \eta(y) \sum_{n = 0}^N \gamma^n$. Therefore there exists $\seq{x_n} \subset E$ such that (I) and (II) holds for all $N \in \natp$.
 
     By (I), $\sum_{n \in \natp}\rho(x_n) \le C\eta(y)\sum_{n \in \natz}\gamma^n = C \eta(y)/(1 - \gamma)$. By (II), $\limv{N}\eta\paren{y - \limv{N}\sum_{n = 1}^N Tx_n} = \limv{N}\eta(y)\gamma^N = 0$.
@@ -56,6 +60,7 @@
         \item[(a)] For any $r > 0$, there exists $\delta(r) > 0$ such that $\overline{T(B_E(0, r))} \supset B_F(0, \delta(r))$.
         \item[(b)] $E$ is complete.
     \end{enumerate}
+
     then for every $s > r$, $T(B_E(0, s)) \supset B_F(0, \delta(r))$.
 \end{proposition}
 \begin{proof}
@@ -66,11 +71,13 @@
         \item[(iii)] For all $n \in \natp$, $\overline{T(B_E(0, s_n))} \supset B_F(0, \delta_n)$.
         \item[(iv)] $\rho_1 = \rho$.
     \end{enumerate}
+
     Let $y_0 \in B(0, r)$ and $x_0 = 0$. Let $N \in \natp$ and suppose inductively that $\bracs{x_n}_1^N \subset E$ has been constructed such that:
     \begin{enumerate}
         \item[(I)] For each $0 \le n \le N - 1$, $\rho(x_{n+1} - x_n) < s_n$.
         \item[(II)] For each $0 \le n \le N$, $\eta(Tx_n - y) \le \rho_{n+1}$.
     \end{enumerate}
+
     By density of $T(x_N + B_E(0, s_N))$ in $Tx_N + B_F(0, \rho_N)$, there exists $x_{N+1} \in T(x_N + B_E(0, s_N))$ such that $\eta(Tx_{N+1} - y) \le \rho_{N+2}$.
 
     By (I), $\seq{x_N}$ is a Cauchy sequence, so
@@ -89,6 +96,7 @@
         \item[(a)] For any $r > 0$, there exists $C \ge 0$ such that for any $y \in T(E)$, there exits $x \in T^{-1}(y)$ with $\rho(x) \le C\eta(y)$.
         \item[(b)] $E$ is complete.
     \end{enumerate}
+
     then $T(E)$ is closed.
 \end{proposition}
 \begin{proof}
@@ -111,7 +119,7 @@
     \end{enumerate}
 \end{theorem}
 \begin{proof}
-    Suppose that $T(E)$ is not meagre. Let $r_0 > 0$ and $r > 0$ such that $B_E(0, r) + B_E(0, r) \subset B_E(0, r_0)$, then since $B_E(0, r)$ is absorbing,
+    Suppose that $T(E)$ is not meagre. Let $r_0 > 0$ and $r > 0$ such that $B_E(0, r) + B_E(0, r) \subset B_E(0, r_0)$, then since $B_E(0, r)$ is radial,
     \[
     E = \bigcup_{n \in \natp}nB_E(0, r) \quad \overline{T(E)} = \bigcup_{n \in \natp}\overline{nT(B_E(0, r))}
     \]

src/fa/tvs/completion.tex

@@ -5,14 +5,16 @@
 \label{definition:tvs-completion}
     Let $E$ be a TVS over $K \in \RC$, then there exists $(\wh E, \iota)$ such that:
     \begin{enumerate}
-        \item $\wh E$ is a complete Hausdorff TVS.
+        \item $\wh E$ is a complete separated TVS.
         \item $\iota \in L(E; \wh E)$.
         \item[(U)] For any $(F, T)$ satisfying (1) and (2), there exists a unique $\ol{T} \in L(\wh E; F)$ such that the following diagram commutes:
     \end{enumerate}
+
     Moreover,
     \begin{enumerate}
         \item[(4)] $\iota(E)$ is dense in $\wh E$.
     \end{enumerate}
+
     The pair $(\wh E, \iota)$ is the \textbf{Hausdorff completion} of $E$.
 \end{definition}
 \begin{proof}

src/fa/tvs/continuous.tex

@@ -10,6 +10,7 @@
         \item $T \in C(E; F)$.
         \item $T$ is continuous at $0$.
     \end{enumerate}
+
     If the above holds, then $T$ is a \textbf{continuous linear map}. The set $L(E; F)$ denotes the vector space of all continuous linear maps from $E$ to $F$.
 \end{definition}
 \begin{proof}
@@ -49,5 +50,33 @@
         \]
 
     \end{enumerate}
+
     The uniformity $\fU$ and its induced topology are the \textbf{product uniformity/topology}, and $E$ equipped with $\fU$ is the \textbf{product TVS} of $\seqi{E}$.
 \end{definition}
+
+\begin{theorem}[Linear Extension Theorem (TVS)]
+\label{theorem:linear-extension-theorem-tvs}
+    Let $E$ be a TVS over $K \in \RC$, $F$ be a complete separated TVS over $K$, $D \subset E$ be a dense subspace, and $T \in L(D; F)$, then:
+    \begin{enumerate}
+        \item There exists an extension $\ol T \in L(E; F)$ such that $\ol T|_D = T$.
+        \item[(U)] For any $S \in C(E; F)$ satisfying (1), $S = \ol T$.
+    \end{enumerate}
+
+\end{theorem}
+\begin{proof}
+    By (3) of \autoref{definition:continuous-linear}, $T \in UC(D; F)$. By \autoref{theorem:uniform-continuous-extension}, there exists a unique $\ol T \in C(E; F)$ such that $\ol T|_D = T$. 
+
+    It remains to show that $\ol T$ is linear. Since $\ol T$ is continuous, the maps
+    \[
+    A: E \times E \to E \quad (x, y) \mapsto \ol Tx + \ol Ty - \ol T(x + y)
+    \]
+
+    and
+    \[
+    M: K \times E \to E \quad (\lambda x) \mapsto \lambda \ol Tx - Tx
+    \]
+
+    are continuous. Thus $\bracs{A = 0} \supset D \times D$ and $\bracs{M = 0} \supset K \times D$ are both closed. By density of $D$, $\bracs{A = 0} = E \times E$ and $\bracs{M = 0} = K \times E$. Therefore $T$ is linear. 
+\end{proof}
+
+

src/fa/tvs/definition.tex

@@ -9,6 +9,7 @@
         \item[(TVS1)] $E \times E \to E$ with $(x, y) \mapsto x + y$ is continuous.
         \item[(TVS2)] $K \times E \to E$ with $(\lambda, x) \mapsto \lambda x$ is continuous.
     \end{enumerate}
+
     then the pair $(E, \topo)$ is a \textbf{topological vector space}.
 \end{definition}
 
@@ -51,6 +52,7 @@
         \item There exists a unique translation-invariant uniformity $\fU$ on $E$ that induces the topology on $E$.
         \item For each neighbourhood $V \in \cn(0)$, let $U_V = \bracs{(x, y) \in E^2| x - y \in V}$, then for any fundamental system of neighbourhoods $\fB_0$ at $0$, $\fB = \bracs{U_V| V \in \fB_0}$ is a fundamental system of entourages for $\fU$.
     \end{enumerate}
+
     The space $E$ will always be assumed to be equipped with its translation-invariant uniformity.
 \end{proposition}
 \begin{proof}
@@ -137,7 +139,7 @@
 \label{proposition:tvs-good-neighbourhood-base}
     Let $E$ be a topological vector space over $K \in \RC$, then
     \begin{enumerate}
-        \item $E$ admits a fundamental system of neighbourhoods at $0$ consisting of circled and absorbing sets.
+        \item $E$ admits a fundamental system of neighbourhoods at $0$ consisting of circled and radial sets.
         \item The fundamental system of neighbourhoods in $(1)$ can be taken to be open or closed.
     \end{enumerate}
 \end{proposition}
@@ -163,11 +165,13 @@
         \item[(TVB1)] For each $U \in \fB$, there exists $V \in \fB$ such that $V + V \subset U$.
         \item[(TVB2)] For each $U \in \fB$, $U$ is circled and radial.
     \end{enumerate}
+
     Conversely, if $\fB \subset 2^E$ is a family of sets that contain $0$ and satisfies (TVB1) and (TVB2), then there exists a unique topology $\topo$ on $E$ such that:
     \begin{enumerate}
         \item $\topo$ is translation-invariant.
         \item $\fB$ is a fundamental system of neighbourhoods at $0$ for $\topo$.
     \end{enumerate}
+
     Moreover,
     \begin{enumerate}
         \item[(3)] $(E, \topo)$ is a TVS.
@@ -187,6 +191,7 @@
         \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 $U_W \circ U_W \subset U_V$.
     \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.
 
     (1): Since $\mathfrak{V}$ is translation-invariant, so is $\fU$.

src/fa/tvs/dual.tex

@@ -1,16 +1,17 @@
 \section{The Dual Space}
 \label{section:tvs-dual}
 
-\begin{proposition}[Polarisation, {{\cite[Proposition 5.5]{Folland}}}]
+\begin{proposition}[Polarisation]
 \label{proposition:polarisation-linear}
     Let $E$ be a vector space over $\complex$, $\phi \in \hom(E; \complex)$, and $u = \re{\phi}$, then
     \begin{enumerate}
         \item $u \in \hom(E; \real)$ when $E$ is viewed as a vector space over $\real$.
         \item For any $x \in E$, $\dpb{x, \phi}{E} = \dpb{x, u}{E} - i \dpb{ix, u}{E}$.
     \end{enumerate}
+
     Conversely, if $u \in \hom(E; \real)$ and $\phi \in \hom(E; \complex)$ is defined by $\dpb{x, \phi}{E} = \dpb{x, u}{E} - i \dpb{ix, u}{E}$ for all $x \in E$, then $f \in \hom(E; \complex)$.
 \end{proposition}
-\begin{proof}
+\begin{proof}[Proof {{\cite[Proposition 5.5]{Folland}}}. ]
     (1): Given that $\phi \in \hom(E; \real)$, $u \in \hom(E; \real)$. For any $x \in E$,
     \[
     \im{\dpb{x, \phi}{E}} = \re{-i \dpb{x, \phi}{E}} = -\dpb{ix, u}{E}

src/fa/tvs/equicontinuous.tex

@@ -0,0 +1,114 @@
+\section{Equicontinuous Families of Linear Maps}
+\label{section:equicontinuous-linear}
+
+\begin{proposition}[{{\cite[IV.4.2]{SchaeferWolff}}}]
+\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}
+        \item $\alg$ is uniformly equicontinuous. 
+        \item $\alg$ is equicontinuous. 
+        \item $\alg$ is equicontinuous at $0$. 
+        \item For each $V \in \cn_F(0)$, there exists $U \in \cn^o(E)$ such that $\bigcup_{T \in \alg}T(U) \subset V$. 
+        \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}
+    (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}
+
+\begin{proposition}
+\label{proposition:equicontinuous-bounded}
+    Let $E, F$ be TVSs over $K \in \RC$ and $\alg \subset L(E; F)$ be equicontinuous, then for any ideal $\sigma \subset \mathfrak{B}(E)$, $\alg$ is a bounded subset of $B_\sigma(E; F)$. 
+\end{proposition}
+\begin{proof}
+    Let $S \in \sigma$ and $U \in \cn_F(0)$, then there exists $V \in \cn_E(0)$ such that $\bigcup_{T \in \alg}T(V) \subset U$. Since $S$ is bounded, there exists $\lambda \in K$ such that $S \subset \lambda V$. Therefore $\bigcup_{T \in \alg}T(S) \subset \lambda U$, and $\alg$ is bounded in $B_\sigma(E; F)$. 
+\end{proof}
+
+
+\begin{proposition}[{{\cite[IV.4.3]{SchaeferWolff}}}]
+\label{proposition:equicontinuous-linear-closure}
+    Let $E, F$ be TVSs over $K \in \RC$ and $\alg \subset L(E; F)$ be equicontinuous, and $\alg'$ be the closure of $\alg$ in $F^E$ with respect to the product topology, then $\alg'$ is equicontinuous and hence $\alg' \subset L(E; F)$. 
+\end{proposition}
+\begin{proof}
+     By \autoref{proposition:operator-space-completeness}, $\alg' \subset \hom(E; F)$. By \autoref{theorem:arzela-ascoli}, $\alg'$ is equicontinuous.
+\end{proof}
+
+\begin{theorem}[Banach-Steinhaus]
+\label{theorem:banach-steinhaus}
+    Let $E, F$ be TVSs over $K \in \RC$ and $\alg \subset L(E; F)$. Suppose that one of the following holds:
+    \begin{enumerate}
+        \item[(B)] $E$ is a Baire space. 
+        \item[(B')] $E$ is barrelled and $F$ is locally convex.
+    \end{enumerate}
+
+    and that
+    \begin{enumerate}
+        \item[(E2)] For each $x \in E$, $\alg(x) = \bracs{Tx|T \in \alg}$ is bounded in $F$. 
+    \end{enumerate}
+
+    then
+    \begin{enumerate}
+        \item[(E1)] $\alg$ is equicontinuous.
+        \item[(C1)] The product topology and the compact-open topology on $\cf$ coincide. 
+        \item[(C2)] The closure of $\alg$ in $F^E$ is with respect to the product topology is an equicontinuous subset of $L(E; F)$. 
+    \end{enumerate}
+\end{theorem}
+\begin{proof}[Proof, {{\cite[IV.4.2]{SchaeferWolff}}}. ]
+    (B) + (E2) $\Rightarrow$ (E1): Let $V \in \cn_F(0)$ be closed and circled, then $U = \bigcap_{T \in \alg}T^{-1}(V)$ is circled and closed. By (E2), $U$ is absorbing, so $E = \bigcup_{n \in \natp}nU$. Since $E$ is Baire, there exists $n \in \natp$, $W \in \cn_E(0)$, and $x \in E$ such that $x + W \subset nU$. As $U$ is circled,
+    \[
+    W \subset nU - nU = nU + nU = 2nU
+    \]
+
+    so $U \in \cn_E(0)$, and $\alg$ is equicontinuous by \autoref{proposition:equicontinuous-linear}. 
+
+    (B') + (E2) $\Rightarrow$ (E1): Let $V \in \cn_F(0)$ be convex, circled, and closed, then $U = \bigcap_{T \in \alg}T^{-1}(V)$ is convex, circled, and closed. By (E2), $U$ is absorbing, and hence a barrel in $E$. By (B'), $U \in \cn_E(0)$, $\alg$ is equicontinuous by \autoref{proposition:equicontinuous-linear}. 
+
+    (E1) $\Rightarrow$ (C1) + (C2): By the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli} and \autoref{proposition:equicontinuous-linear-closure}.
+\end{proof}
+
+\begin{lemma}
+\label{lemma:equicontinuous-bilinear}
+    Let $E, F, G$ be TVSs over $K \in \RC$ and $\alg \subset L^2(E, F; G)$ be continuous bilinear maps, then the following are equivalent:
+    \begin{enumerate}
+        \item $\alg$ is equicontinuous.
+        \item $\alg$ is equicontinuous at $0$. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    (2) $\Rightarrow$ (1): For each $(x_0, y_0), (x, y) \in E \times F$ and $\lambda \in \alg$, 
+    \[
+    \lambda(x, y) - \lambda(x_0, y_0) = \lambda(x - x_0, y - y_0) + \lambda(x - x_0, y_0) + \lambda(x_0, y - y_0)
+    \]
+
+    For each $U \in \cn_G(0)$ circled, there exists circled neighbourhoods $V \in \cn_E(0)$ and $W \in \cn_F(0)$ such that $\lambda(V \times W) \subset U$ for all $\lambda \in \alg$. In which case, there exists $\mu > 0$ such that $y_0 \in \mu W$ and $x_0 \in \mu V$. Thus if $(x, y) - (x_0, y_0) \in \mu^{-1}(V \times W)$, then for every $\lambda \in \alg$,
+    \[
+    \lambda(x - x_0, y_0) \in \lambda(\mu^{-1} V \times \mu W) = \lambda(V \times W) \subset U
+    \]
+
+    and $\lambda(x_0, y - y_0) \in U$ as well. Therefore $\alg$ is equicontinuous at $(x_0, y_0)$. 
+\end{proof}
+
+
+
+\begin{theorem}
+\label{theorem:separate-joint-bilinear}
+    Let $E, F, G$ be TVSs over $K \in \RC$ and $\alg$ be separately continuous bilinear maps from $E \times F$ to $G$. If one of the following holds:
+    \begin{enumerate}
+        \item[(B)] $E$ is Baire.
+        \item[(B')] $E$ is barrelled and $G$ is locally convex. 
+    \end{enumerate}
+
+    and that
+    \begin{enumerate}
+        \item[(M)] $E$ and $F$ are both metrisable. 
+        \item[(E)] For each $x \in E$, $\bracsn{\lambda(x, \cdot)|\lambda \in \alg} \subset L(F; G)$ is equicontinuous. 
+    \end{enumerate}
+
+    then $\alg$ is equicontinuous. 
+\end{theorem}
+\begin{proof}[Proof, {{\cite[III.5.1]{SchaeferWolff}}}. ]
+    Let $\seq{(x_n, y_n)} \subset E \times F$ and $\seq{\lambda_n} \subset \alg$ such that $(x_n, y_n) \to 0$ as $n \to \infty$. Since $\seq{y_n}$ is convergent, for each $n \in \natp$ and $x \in E$, $\bracsn{\lambda_n(x, y_n)|n \in \natp}$ is bounded by (E) and \autoref{proposition:equicontinuous-net}. By (B) or (B') and the \hyperref[Banach-Steinhaus Theorem]{theorem:banach-steinhaus}, $\bracsn{\lambda_n(\cdot, y_n)|n \in \natp}$ is equicontinuous, and $\lambda_n(x_n, y_n) \to 0$ as $n \to \infty$ by \autoref{proposition:equicontinuous-net}. By (M) and \autoref{proposition:equicontinuous-net}, $\alg$ is equicontinuous at $0$, and hence equicontinuous by \autoref{lemma:equicontinuous-bilinear}. 
+\end{proof}
+
+
+

src/fa/tvs/index.tex

@@ -11,4 +11,6 @@
 \input{./complete-metric.tex}
 \input{./projective.tex}
 \input{./inductive.tex}
-\input{./spaces-of-linear.tex}
+\input{./vector-function.tex}
+\input{./space-of-linear.tex}
+\input{./equicontinuous.tex}

src/fa/tvs/inductive.tex

@@ -10,6 +10,7 @@
         \item[(U)] For any topology $\mathcal{S}$ on $E$ satisfying (1) and (2), $\mathcal{S} \subset T$.
         \item For any TVS $F$ and $T \in \hom(E; F)$, $T \in L(E; F)$ if and only if $T \circ T_i \in L(E_i; F)$ for all $i \in I$.
     \end{enumerate}
+
     The topology $\topo$ is the \textbf{inductive topology} on $E$ induced by $\seqi{T}$.
 \end{definition}
 \begin{proof}
@@ -62,6 +63,7 @@
         for all $i \in I$.
         \item For any TVS $F$ and $T \in \hom(E; F)$, $T \in L(E; F)$ if and only if $T \circ T^i_E \in L(E_i; F)$ for all $i \in I$.
     \end{enumerate}
+
     The pair $(E, \bracsn{T^i_E}_{i \in I})$ is the \textbf{inductive limit} of $(\seqi{E}, \bracsn{T^i_j| i, j \in I, i \lesssim j})$.
 \end{definition}
 \begin{proof}

src/fa/tvs/metric.tex

@@ -76,20 +76,21 @@
     $(4) \Rightarrow (1)$: By \autoref{definition:tvs-pseudonorm-topology}, for each $r > 0$, $\rho^{-1}([0, r)) \in \cn_E(0)$. Thus for any $x, y \in E$, if $x - y \in \rho^{-1}([0, r))$, then $\abs{\rho(x) - \rho(y)} \le r$. Therefore $\rho \in UC(E; [0, \infty))$.
 \end{proof}
 
-\begin{lemma}[{{\cite[Theorem I.6.1]{SchaeferWolff}}}]
+\begin{lemma}[]
 \label{lemma:tvs-sequence-pseudonorm}
     Let $E$ be a vector space over $K \in \RC$, $\seq{U_n} \subset 2^E$ such that
     \begin{enumerate}
         \item[(a)] For each $n \in \natp$, $U_n$ is circled, radial, and contains $0$.
         \item[(b)] For each $n \in \natp$, $U_{n+1} + U_{n+1} \subset U_n$.
     \end{enumerate}
+
     then there exists a pseudonorm $\rho: E \to [0, \infty)$ such that for each $n \in \natp$,
     \[
     U_{n+1} \subset \rho^{-1}([0, 2^{-n})) \subset U_{n}
     \]
 
 \end{lemma}
-\begin{proof}
+\begin{proof}[Proof {{\cite[Theorem I.6.1]{SchaeferWolff}}}.]
     For each $H \subset \natp$ finite, let
     \[
     U_H = \sum_{n \in H}V_n \quad \rho_H = \sum_{n \in H}2^{-n}
@@ -111,6 +112,7 @@
         so $\rho(\lambda x) \le \rho(x)$.
         \item[(PN3)] Let $x, y \in X$ and $M, N \subset \natp$ finite such that $x \in U_M$ and $y \in U_N$. Assume without loss of generality that $\rho_M + \rho_N < 1$, then there exists a unique $P \subset \nat$ finite such that $\rho_P = \rho_M + \rho_N$. In which case, $U_P \supset U_M + U_N$ by assumption (b). Therefore $\rho(x + y) \le \rho(x) + \rho(y)$.
     \end{enumerate}
+
     For any $x \in U_{n+1}$, $\rho(x) \le 2^{-n+1} < 2^n$, so $U_{n+1} \subset \rho^{-1}([0, 2^{-n}))$ by \autoref{proposition:dyadic-semigroup-order}. On the other hand, for any $x \in E$ with $\rho(x) < 2^{-n}$, $x \in U_{2^{-n}} = U_n$. This allows showing the remaining seminorm axioms by considering neighbourhoods of the form $\bracs{U_n|n \in \natp}$.
     \begin{enumerate}
         \item[(PN4)] Let $x \in X$ and $n \in \natp$. By assumption (a), there exists $\alpha > 0$ such that for any $\lambda \in K$ with $\abs{\lambda} \ge \alpha$, $x \in \lambda U_n$. Therefore for any $\lambda \in K$ with $\abs{\lambda} \le \alpha^{-1}$, $\lambda x \in U_n$, and $\rho(x) \le 2^{-n}$.

src/fa/tvs/projective.tex

@@ -8,6 +8,7 @@
         \item For each $i \in I$, $T_i \in L(E; F_i)$.
         \item[(U)] If $\mathfrak{V}$ is a uniformity on $E$ satisfying $(1)$, then $\mathfrak{V} \supset \fU$.
     \end{enumerate}
+
     Moreover,
     \begin{enumerate}
         \item[(3)] $\fU$ is translation-invariant.
@@ -20,6 +21,8 @@
 
         is a fundamental system of neighbourhoods for $E$ at $0$.
     \end{enumerate}
+
+    
     The uniformity $\fU$ and its topology are the \textbf{projective uniformity/topology} induced by $\seqi{T}$.
 \end{definition}
 \begin{proof}
@@ -74,6 +77,7 @@
         for all $i \in I$.
         \item For any TVS $F$ over $K$ and $S \in \hom(F; E)$, $S \in L(F; E)$ if and only if $T^E_i \circ S \in L(F; E_i)$ for all $i \in I$.
     \end{enumerate}
+
     The pair $(E, \bracsn{T^E_i}_{i \in I})$ is the \textbf{projective limit} of $(\seqi{E}, \bracsn{T^i_j|i, j \in I, i \lesssim j})$.
 \end{definition}
 \begin{proof}

src/fa/tvs/quotient.tex

@@ -18,12 +18,13 @@
 
         If $F$ is a TVS over $K$ and $f \in L(E; F)$, then $\td f \in L(E; F)$.
     \end{enumerate}
+
     The space $\td E = E/M$ is the \textbf{quotient} of $E$ by $M$.
 \end{definition}
 \begin{proof}
     Let $\td E = E/M$ be the algebraic quotient of $E$ by $M$, and equip it with the quotient topology by $\pi$.
 
-    (1): By \autoref{definition:quotient-topology}, for each $\pi(U) \subset E/M$, $\pi(U)$ is open if and only if $U$ is open. Since the topology on $E$ is translation-invariant, so is the quotient topology on $E/M$. Let
+    (1): For each $U \subset E$ open, $\pi^{-1}\pi(U) = U + M$ is open, so $\pi(U)$ is open as well by \autoref{definition:quotient-topology}. Since the topology on $E$ is translation-invariant, so is the quotient topology on $E/M$. Let
     \[
     \fB = \bracs{\pi(U)| U \in \cn(0) \text{ circled and radial}}
     \]
@@ -33,6 +34,7 @@
         \item[(TVB1)] Let $U \in \cn(0)$ be circled and radial. For any $\lambda \in K$ with $\abs{\lambda} \le 1$, $\lambda \pi(U) = \pi(\lambda U) \subset \pi(U)$, so $\pi(U)$ is also circled. For any $x + M \in E/M$, there exists $\lambda \in K$ such that $x \in \lambda U$. In which case, $x \in \lambda U + M = \pi(U)$, so $\pi(U)$ is also radial.
         \item[(TVB2)] For any $U \in \cn(0)$ circled and radial, by \autoref{proposition:tvs-good-neighbourhood-base}, there exists $W \in \cn(0)$ such that $W + W \subset U$. In which case, $\pi(W) + \pi(W) \subset \pi(U)$.
     \end{enumerate}
+
     By \autoref{proposition:tvs-0-neighbourhood-base}, there exists a unique translation-invariant topology on $E/M$ such that $\fB$ is a fundamental system of neighbourhoods at $0$, which must be the quotient topology on $E/M$. In which case, the quotient topology is a vector space topology by (3) of \autoref{proposition:tvs-0-neighbourhood-base}.
 
     (2), (3), (U): By \autoref{definition:quotient-topology}.
@@ -40,7 +42,7 @@
 
 \begin{proposition}
 \label{proposition:tvs-quotient-hausdorff}
-    Let $E$ be a TVS over $K \in \RC$, $M \subset E$ be a subspace, then $E/M$ is Hausdorff if and only if $M$ is closed.
+    Let $E$ be a TVS over $K \in \RC$, $M \subset E$ be a subspace, then $E/M$ is separated if and only if $M$ is closed.
 \end{proposition}
 \begin{proof}
     The space $M$ is closed if and only if
@@ -48,5 +50,5 @@
     M = \bigcap_{V \in \cn(0)}M + V
     \]
 
-    which is equivalent to $E/M$ being Hausdorff.
+    which is equivalent to $E/M$ being separated.
 \end{proof}

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

@@ -0,0 +1,156 @@
+\section{Spaces of Linear Maps}
+\label{section:space-linear-map-new}
+
+\begin{definition}[Space of Bounded Linear Maps]
+\label{definition:bounded-linear-map-space}
+    Let $E, F$ be TVSs over $K \in \RC$, $\sigma \subset 2^E$ be an ideal, and $k \in \nat$. The space $B_{\sigma}^k(E; F)$ is the set of all $k$-linear maps $T: E^k \to F$ with $T(S^k) \in \mathfrak{B}(F)$ for all $S \in \sigma$, equipped with the $\bracsn{S^k| S \in \sigma}$-uniform topology.
+
+    Let $\fB \subset 2^E$ be the collection of all bounded subsets of $E$, then $B_{\sigma}(E; F) = B(E; F)$ is the \textbf{space of bounded linear maps} from $E$ to $F$. 
+\end{definition}
+
+\begin{proposition}[{{\cite[III.3.3]{SchaeferWolff}}}]
+\label{proposition:bounded-linear-map-space-bounded}
+    Let $E, F$ be TVSs over $K \in \RC$, $\sigma \subset 2^E$ be an ideal, and $A \subset B_\sigma(E; F)$, then the following are equivalent:
+    \begin{enumerate}
+        \item $A \subset B_\sigma(E; F)$ is bounded with respect to the $\sigma$-uniform topology. 
+        \item For each $V \in \cn_F(0)$, $\bigcap_{T \in A}T^{-1}(V)$ absorbs every $S \in \sigma$.
+        \item For every $S \in \sigma$, $\bigcup_{T \in A}T(A)$ is bounded in $F$. 
+    \end{enumerate}
+    
+\end{proposition}
+% Proof omitted because it is obvious. 
+
+
+\begin{proposition}
+\label{proposition:multilinear-identify}
+	Let $E, F$ be TVSs over $K \in \RC$, $\sigma \subset 2^E$ be a covering ideal, and $k \in \natp$, then
+	\begin{enumerate}
+        \item The map
+        \[
+        I: B_{\sigma}^k(E; B_{\sigma}(E; F)) \to B^{k+1}_{\sigma}(E; F)
+        \]
+
+        defined by
+        \[
+        (IT)(x_1, \cdots, x_{k+1}) = T(x_1, \cdots, x_k)(x_{k+1})
+        \]
+
+        is an isomorphism.
+        \item The map
+        \[
+        I: \underbrace{B_{\sigma}(E; B_{\sigma}(E; \cdots)))}_{k \text{ times}} \to B^k_{\sigma}(E; F)
+        \]
+
+        defined by
+        \[
+        IT(x_1, \cdots, x_k) = T(x_1)\cdots (x_k)
+        \]
+
+        is an isomorphism.
+    \end{enumerate}
+
+    which allows the identification
+    \[
+    \underbrace{B_{\sigma}(E; B_{\sigma}(E; \cdots)))}_{k \text{ times}} = B^k_{\sigma}(E; F)
+    \]
+
+    under the map $I$ in (2).
+\end{proposition}
+\begin{proof}
+    (1): To see that $I$ is surjective, let $T \in B_{\sigma}^{k+1}(E; F)$ and
+    \[
+    I^{-1}T: E \to B_{\sigma}(E; F) \quad x \mapsto T(x, \cdot)
+    \]
+
+    Let $(x_1, \cdots, x_k) \in E^k$ and $S \in \sigma$. Since $\sigma$ is a covering ideal, assume without loss of generality that $\bracsn{x_j}_1^k \subset S$. In which case,
+    \[
+    T(x_1, \cdots, x_k, S) \subset T(S^{k+1}) \in \mathfrak{B}(F)
+    \]
+
+    by assumption. Thus $I^{-1}T(x_1, \cdots, x_k) \in B_{\sigma}(E; F)$.
+
+    In addition, for any $S_1 \in \sigma$ and entourage $E(S_2, U)$ of $B_{\sigma}(E; F)$ where $S_2 \in \sigma$ and $U$ is an entourage of $F$, there exists $S \in \sigma$ with $S \supset S_1 \cup S_2$. Given that $T(S^{k+1}) \in \mathfrak{B}(F)$, there exists $\lambda > 0$ such that $T(S^{k+1}) \subset \lambda U(0)$. In which case, $I^{-1}T(S^k) \subset \lambda E(S, U)(0)$ and $I^{-1}T(S^k) \in B(B_{\sigma}(E; F))$. Thus $I^{-1}T \in B^k_{\sigma}(E; B_{\sigma}(E; F))$.
+
+    It remains to show that $I$ and $I^{-1}$ is continuous. To this end, let $S \in \sigma$ and $U$ be an entourage of $F$, then for any $T \in E(S^k, E(S, U))(0)$, $IT \in E(S^{k+1}, U)(0)$, so $I$ is continuous.
+
+    On the other hand, let $S_1 \in \sigma$ and $E(S_2, U)$ be an entourage of $B_{\sigma}(E; F)$ where $S_2 \in \sigma$ and $U$ is an entourage of $F$. Let $S \in \sigma$ with $S \supset S_1 \cup S_2$, then for any $T \in E(S^{k+1}, U)(0)$, $I^{-1}T \in E(S^{k}, E(S, U))(0)$. Thus $I^{-1}$ is continuous as well.
+
+    (2): The case for $k = 2$ is given by (1). If the proposition holds for $k \in \natp$, then
+    \[
+    \underbrace{B_{\sigma}(E; B_{\sigma}(E; \cdots)))}_{k+1 \text{ times}} = B^k_{\sigma}(E; B_{\sigma}(E; F)) = B^{k+1}_{\sigma}(E; F)
+    \]
+
+    Thus (2) holds for all $k \in \natp$.
+\end{proof}
+
+\begin{definition}[Strong Operator Topology]
+\label{definition:strong-operator-topology}
+    Let $E, F$ be TVSs over $K \in \RC$, $\fF \subset 2^E$ be the collection of finite subsets of $E$, then the $\fF$-uniform topology on $F^E$ is the \textbf{strong operator topology}.
+
+    The space $L_s(E; F)$ denotes $L(E; F)$ equipped with the strong operator topology.
+\end{definition}
+
+
+\begin{proposition}
+\label{proposition:strong-operator-dense}
+    Let $E, F$ be TVSs over $K \in \RC$ and $\net{T} \subset L(E; F)$ and $T \in L_s(E; F)$. If
+    \begin{enumerate}
+        \item[(a)] There exists a dense subset $S \subset E$ such that $T_\alpha x \to Tx$ strongly for all $x \in S$.
+        \item[(b)] $\bracs{T_\alpha|\alpha \in A}$ is uniformly equicontinuous. 
+    \end{enumerate}
+
+    
+    then $T_\alpha \to T$ in $L_s(E; F)$.
+\end{proposition}
+\begin{proof}
+    Let $x \in E$, $U \in \cn_F(Tx)$, and $V \in \cn_F(Tx)$ be balanced such that $V + V + V \subset U$. By (b), there exists a balanced neighbourhood $W \in \cn_E(0)$ such that $T(W) \cup \bigcup_{\alpha \in A}T_\alpha(W) \subset V$. By (a), there exists $y \in S \cap (x + W)$ and $\alpha_0 \in A$ such that for all $\alpha \ge \alpha_0$, $T_\alpha y - Ty \in V$. In which case, for any $\alpha \ge \alpha_0$,
+    \[
+    T_\alpha x - Tx = \underbrace{T_\alpha x - T_\alpha y}_{\in V} + \underbrace{T_\alpha y - Ty}_{\in V} + \underbrace{Ty - Tx}_{\in V} \in U
+    \]
+\end{proof}
+
+
+
+\begin{definition}[Weak Operator Topology]
+\label{definition:weak-operator-topology}
+    Let $E, F$ be TVSs over $K \in \RC$, $\fF \subset 2^E$ be the collection of finite subsets of $E$, then the $\fF$-uniform topology on $F_w^E$ is the \textbf{weak operator topology}.
+
+    The space $L_w(E; F) = L_s(E; F_w)$ denotes $L(E; F)$ equipped with the weak operator topology.
+\end{definition}
+
+\begin{definition}[Bounded Convergence Topology]
+\label{definition:bounded-convergence-topology}
+    Let $E, F$ be TVSs over $K \in \RC$, $\fB \subset 2^E$ be the collection of bounded subsets of $E$, then the $\fB$-uniform topology on $L(E; F)$ is the \textbf{topology of bounded convergence}.
+
+    The space $L_b(E; F)$ denotes $L(E; F)$ equipped with the topology of bounded convergence.
+\end{definition}
+
+\begin{proposition}
+\label{proposition:operator-space-completeness}
+    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, then:
+    \begin{enumerate}
+        \item $\hom(E; F)$ is a closed subspace of $F^E$ with respect to the product topology.
+        \item $B(E; F)$ is a closed subspace of $F^E$ with respect to the topology of bounded convergence. In particular, if $F$ is complete, then so is $B(E; F)$.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): For each $x, y \in E$ and $\lambda \in K$, the mappings
+    \[
+    \phi_{x, y}: F^E \to F \quad T \mapsto Tx + Ty - T(x + y)
+    \]
+
+    and
+    \[
+    \psi_{x, \lambda}: F^E \to F \quad T \mapsto T(\lambda x) - \lambda Tx
+    \]
+
+    are continuous with respect to the product topology. Since
+    \[
+    \hom(E; F) = \bigcap_{x, y \in E}\bracsn{\phi_{x, y} = 0} \cap \bigcap_{\substack{x \in E \\ \lambda \in K}}\bracsn{\psi_{x, \lambda} = 0}
+    \]
+
+    and $\bracs{0}$ is closed in $F$, $\hom(E; F)$ is a closed subspace of $F^E$. 
+
+    (2): By \autoref{definition:bounded-function-space} and (1), the space of bounded functions and the space of linear functions from $E$ to $F$ are closed subspaces of $F^E$ with respect to the topology of bounded convergence. Therefore $B(E; F)$ is also a closed subspace.
+
+\end{proof}

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

@@ -1,205 +0,0 @@
-\section{Vector-Valued Function Spaces}
-\label{section:spaces-linear-map}
-
-\begin{proposition}[{{\cite[III.3.1]{SchaeferWolff}}}]
-\label{proposition:tvs-set-uniformity}
-    Let $T$ be a set, $\mathfrak{S} \subset 2^T$ be an upward-directed system of sets, $F$ be a TVS over $K \in \RC$, then
-    \begin{enumerate}
-        \item The $\mathfrak{S}$-uniformity on $F^T$ (\autoref{definition:set-uniform}) is translation invariant.
-        \item The composition defined by
-        \[
-        T^F \times T^F \to T^F \quad (f + g)(x) = f(x) + g(x)
-        \]
-
-        is continuous.
-    \end{enumerate}
-    For any vector subspace $\cf \subset F^T$, the following are equivalent:
-    \begin{enumerate}
-        \item[(3)] The $\mathfrak{S}$-uniform topology on $\cf \subset F^E$ is a vector space topology.
-        \item[(4)] For each $S \in \mathfrak{S}$ and $f \in \cf$, $f(S) \subset E$ is bounded.
-    \end{enumerate}
-\end{proposition}
-\begin{proof}
-    (1): Let $U \subset F \times F$ be an entourage and $S \in \mathfrak{S}$. Using \autoref{proposition:tvs-uniform}, assume without loss of generality that $U$ is translation-invariant. For any $(f, g) \in E(S, U)$, $h \in F^T$, and $x \in S$, $(f(x) + h(x), g(x) + h(x)) \in U$. Thus $(f + h, g + h) \in E(S, U)$ and $E(S, U)$ is translation-invariant.
-
-    (2): Let $f, g, f', g' \in \cf$, $S \in \mathfrak{S}$, and $U \subset F \times F$ be an entourage. By (TVS1), there exists an entourage $V \subset F \times F$ such that for any $x, x', y, y' \in F$ with $(x, x'), (y, y') \in V$, $(x + y, x' + y') \in U$. If $(f, g), (f', g') \in E(S, V) \cap \cf$, then for any $x \in S$, $(f(x) + g(x), f'(x) + g'(x)) \in U$, so $(f + g, f' + g') \in E(S, U) \cap \cf$.
-
-    (3) $\Rightarrow$ (4): Let $f \in \cf$, $S \in \mathfrak{S}$ and $U \subset F \times F$ be a symmetric entourage, then $E(S, U)(0)$ is a neighbourhood of $0$ with respect to the $\mathfrak{S}$-uniform topology. By (TVS2), there exists $\lambda > 0$ such that $f \in \lambda E(S, U)(0)$. In which case, for any $x \in S$, $\lambda^{-1}f(x) \in U(0)$ and $f(x) \in \lambda U(0)$. Thus $f(S) \subset \lambda U(0)$, and $f(S)$ is bounded.
-
-    (4) $\Rightarrow$ (3): Let $f, g \in \cf$, $\lambda, \lambda' \in K$, and $S \in \mathfrak{S}$, then for any $x \in X$,
-    \begin{align*}
-    \lambda f(x) - \lambda' g(x) &= \lambda f(x) - \lambda' f(x) + \lambda' f(x) - \lambda' g(x) \\
-    &= (\lambda - \lambda')f(x) + \lambda' (f(x) - g(x))
-    \end{align*}
-    Let $U_0 \in \cn_F(0)$. By (TVS1) and \autoref{proposition:tvs-good-neighbourhood-base}, there exists $U \in \cn_F(0)$ circled such that $U + U \subset U_0$. By (TVS2), there exists $V \in \cn_F(0)$ such that $\lambda' V \subset U$. Since $f(S)$ is bounded, there exists $\eps > 0$ with $\eps f(S) \subset U$. In which case, if $\abs{\lambda - \lambda'} < \eps$ and $f(x) - g(x) \in V$ for all $x \in S$, then
-    \[
-    \lambda f(x) - \lambda' g(x) \in (\lambda - \lambda')f(S) + \lambda' V \subset U + U \subset U_0
-    \]
-
-    for all $x \in S$.
-\end{proof}
-
-
-\begin{definition}[Space of Bounded Functions]
-\label{definition:bounded-function-space}
-    Let $T$ be a set, $E$ be a TVS over $K \in \RC$, and $f: T \to E$ be a function, then $f$ is \textbf{bounded} if $f(T)$ is a bounded subset of $E$. The set $B(T; E) \subset E^T$ is the space of all bounded functions from $T$ to $E$, and:
-    \begin{enumerate}
-        \item $B(T; E)$ equipped with the uniform topology and pointwise operations is a TVS over $K \in \RC$. 
-        \item $B(T; E)$ is a closed subset of $E^T$ with respect to the uniform topology. In particular, if $E$ is complete, then so is $B(T; E)$.
-    \end{enumerate}
-\end{definition}
-\begin{proof}
-    (1): For any $f, g \in B(T; E)$ and $\lambda \in K$, $(\lambda f + g)(T) \subset (\lambda f)(T) + g(T)$, which is bounded by \autoref{proposition:bounded-operations}, so $B(T; E)$ is closed under addition and scalar multiplication. 
-    
-    Since $f(E)$ is bounded for all $f \in B(T; E)$, $B(T; E)$ forms a TVS over $K$ by \autoref{proposition:tvs-set-uniformity}.
-
-    (2): Let $f \in \ol{B(T; E)}$ and $U \in \cn_E(0)$ be circled, then there exists $g \in B(T; E)$ such that $(f - g)(E) \subset U$. Since $g \in B(T; E)$, there exists $\lambda \ge 0$ such that $g(E) \subset \lambda U$. In which case,
-    \[
-    f(E) \subset g(E) + (f - g)(E) \subset (\lambda + 1)U
-    \]
-    so $f \in B(T; E)$.
-
-    If $E$ is complete, then $E^T$ with the uniform topology is complete by \autoref{proposition:set-uniform-complete}. Thus $B(T; E)$ is also complete by \autoref{proposition:complete-closed}.
-\end{proof}
-
-
-\begin{definition}[Space of Bounded Continuous Functions]
-\label{definition:bounded-continuous-function-space}
-    Let $X$ be a topological space and $E$ be a TVS over $K \in \RC$, then $BC(X; E) = B(X; E) \cap C(X; E)$ is the space of bounded and continuous functions from $X$ to $E$, and
-    \begin{enumerate}
-        \item $BC(X; E)$ equipped with the uniform topology and pointwise operations is a TVS over $K \in \RC$. 
-        \item $BC(X; E)$ is a closed subspace of $C(X; E)$ and $B(X; E)$ with respect to the uniform topology. In particular, if $E$ is complete, then so is $BC(X; E)$.
-    \end{enumerate}
-\end{definition}
-\begin{proof}
-    (1): Since addition and scalar multiplication are continuous, $BC(X; E)$ is a subspace of $B(X; E)$, and hence a TVS over $K \in \RC$ by \autoref{definition:bounded-function-space}.
-
-    (2): Since $B(X; E)$ and $C(X; E)$ are both closed subspaces of $E^X$ by \autoref{definition:bounded-function-space}, $BC(X; E)$ is a closed subspace. 
-    
-    If $E$ is complete, then $B(X; E)$ and $C(X; E)$ are both complete under the uniform topology by \autoref{definition:bounded-function-space}. Therefore $BC(X; E)$ is also complete. 
-\end{proof}
-
-
-
-\begin{definition}[Space of Bounded Linear Maps]
-\label{definition:bounded-linear-map-space}
-    Let $E, F$ be TVSs over $K \in \RC$, $\mathfrak{S} \subset 2^E$ be an upward-directed system, and $k \in \nat$. The space $B_{\mathfrak{S}}^k(E; F)$ is the set of all $k$-linear maps $T: E^k \to F$ with $T(S^k) \in B(F)$ for all $S \in \mathfrak{S}$, equipped with the $\bracsn{S^k| S \in \mathfrak{S}}$-uniform topology.
-
-    Let $\fB \subset 2^E$ be the collection of all bounded subsets of $E$, then $B_{\mathfrak{S}}(E; F) = B(E; F)$ is the \textbf{space of bounded linear maps} from $E$ to $F$. 
-\end{definition}
-
-\begin{proposition}
-\label{proposition:multilinear-identify}
-	Let $E, F$ be TVSs over $K \in \RC$, $\mathfrak{S} \subset 2^E$ be an upward-directed system that contains all singletons, and $k \in \natp$, then
-	\begin{enumerate}
-        \item The map
-        \[
-        I: B_{\mathfrak{S}}^k(E; B_{\mathfrak{S}}(E; F)) \to B^{k+1}_{\mathfrak{S}}(E; F)
-        \]
-
-        defined by
-        \[
-        (IT)(x_1, \cdots, x_{k+1}) = T(x_1, \cdots, x_k)(x_{k+1})
-        \]
-
-        is an isomorphism.
-        \item The map
-        \[
-        I: \underbrace{B_{\mathfrak{S}}(E; B_{\mathfrak{S}}(E; \cdots)))}_{k \text{ times}} \to B^k_{\mathfrak{S}}(E; F)
-        \]
-
-        defined by
-        \[
-        IT(x_1, \cdots, x_k) = T(x_1)\cdots (x_k)
-        \]
-
-        is an isomorphism.
-    \end{enumerate}
-    which allows the identification
-    \[
-    \underbrace{B_{\mathfrak{S}}(E; B_{\mathfrak{S}}(E; \cdots)))}_{k \text{ times}} = B^k_{\mathfrak{S}}(E; F)
-    \]
-
-    under the map $I$ in (2).
-\end{proposition}
-\begin{proof}
-    (1): To see that $I$ is surjective, let $T \in B_{\mathfrak{S}}^{k+1}(E; F)$ and
-    \[
-    I^{-1}T: E \to B_{\mathfrak{S}}(E; F) \quad x \mapsto T(x, \cdot)
-    \]
-
-    Let $(x_1, \cdots, x_k) \in E^k$ and $S \in \mathfrak{S}$. Since $\mathfrak{S}$ is upward-directed and contains all singletons, assume without loss of generality that $\bracsn{x_j}_1^k \subset S$. In which case,
-    \[
-    T(x_1, \cdots, x_k, S) \subset T(S^{k+1}) \in B(F)
-    \]
-
-    by assumption. Thus $I^{-1}T(x_1, \cdots, x_k) \in B_{\mathfrak{S}}(E; F)$.
-
-    In addition, for any $S_1 \in \mathfrak{S}$ and entourage $E(S_2, U)$ of $B_{\mathfrak{S}}(E; F)$ where $S_2 \in \mathfrak{S}$ and $U$ is an entourage of $F$, there exists $S \in \mathfrak{S}$ with $S \supset S_1 \cup S_2$. Given that $T(S^{k+1}) \in B(F)$, there exists $\lambda > 0$ such that $T(S^{k+1}) \subset \lambda U(0)$. In which case, $I^{-1}T(S^k) \subset \lambda E(S, U)(0)$ and $I^{-1}T(S^k) \in B(B_{\mathfrak{S}}(E; F))$. Thus $I^{-1}T \in B^k_{\mathfrak{S}}(E; B_{\mathfrak{S}}(E; F))$.
-
-    It remains to show that $I$ and $I^{-1}$ is continuous. To this end, let $S \in \mathfrak{S}$ and $U$ be an entourage of $F$, then for any $T \in E(S^k, E(S, U))(0)$, $IT \in E(S^{k+1}, U)(0)$, so $I$ is continuous.
-
-    On the other hand, let $S_1 \in \mathfrak{S}$ and $E(S_2, U)$ be an entourage of $B_{\mathfrak{S}}(E; F)$ where $S_2 \in \mathfrak{S}$ and $U$ is an entourage of $F$. Let $S \in \mathfrak{S}$ with $S \supset S_1 \cup S_2$, then for any $T \in E(S^{k+1}, U)(0)$, $I^{-1}T \in E(S^{k}, E(S, U))(0)$. Thus $I^{-1}$ is continuous as well.
-
-    (2): The case for $k = 2$ is given by (1). If the proposition holds for $k \in \natp$, then
-    \[
-    \underbrace{B_{\mathfrak{S}}(E; B_{\mathfrak{S}}(E; \cdots)))}_{k+1 \text{ times}} = B^k_{\mathfrak{S}}(E; B_{\mathfrak{S}}(E; F)) = B^{k+1}_{\mathfrak{S}}(E; F)
-    \]
-
-    Thus (2) holds for all $k \in \natp$.
-\end{proof}
-
-
-
-\begin{definition}[Strong Operator Topology]
-\label{definition:strong-operator-topology}
-    Let $E, F$ be TVSs over $K \in \RC$, $\fF \subset 2^E$ be the collection of finite subsets of $E$, then the $\fF$-uniform topology on $L(E; F)$ is the \textbf{strong operator topology}.
-
-    The space $L_s(E; F)$ denotes $L(E; F)$ equipped with the strong operator topology.
-\end{definition}
-
-\begin{definition}[Weak Operator Topology]
-\label{definition:weak-operator-topology}
-    Let $E, F$ be TVSs over $K \in \RC$, $\fF \subset 2^E$ be the collection of finite subsets of $E$, then the $\fF$-uniform topology on $L(E; F_w)$ is the \textbf{weak operator topology}.
-
-    The space $L_w(E; F) = L_s(E; F_w)$ denotes $L(E; F)$ equipped with the weak operator topology.
-\end{definition}
-
-\begin{definition}[Bounded Convergence Topology]
-\label{definition:bounded-convergence-topology}
-    Let $E, F$ be TVSs over $K \in \RC$, $\fB \subset 2^E$ be the collection of bounded subsets of $E$, then the $\fB$-uniform topology on $L(E; F)$ is the \textbf{topology of bounded convergence}.
-
-    The space $L_b(E; F)$ denotes $L(E; F)$ equipped with the topology of bounded convergence.
-\end{definition}
-
-\begin{proposition}
-\label{proposition:operator-space-completeness}
-    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, then:
-    \begin{enumerate}
-        \item $\hom(E; F)$ is a closed subspace of $F^E$ with respect to the product topology.
-        \item $B(E; F)$ is a closed subspace of $F^E$ with respect to the topology of bounded convergence. In particular, if $F$ is complete, then so is $B(E; F)$.
-    \end{enumerate}
-\end{proposition}
-\begin{proof}
-    (1): For each $x, y \in E$ and $\lambda \in K$, the mappings
-    \[
-    \phi_{x, y}: F^E \to F \quad T \mapsto Tx + Ty - T(x + y)
-    \]
-
-    and
-    \[
-    \psi_{x, \lambda}: F^E \to F \quad T \mapsto T(\lambda x) - Tx
-    \]
-
-    are continuous with respect to the product topology. Since
-    \[
-    \hom(E; F) = \bigcap_{x, y \in E}\bracsn{\phi_{x, y} = 0} \cap \bigcap_{\substack{x \in E \\ \lambda \in K}}\bracsn{\psi_{x, \lambda} = 0}
-    \]
-
-    and $\bracs{0}$ is closed in $F$, $\hom(E; F)$ is a closed subspace of $F^E$. 
-
-    (2): By \autoref{definition:bounded-function-space} and (1), the space of bounded functions and the space of linear functions from $E$ to $F$ are closed subspaces of $F^E$ with respect to the topology of bounded convergence. Therefore $B(E; F)$ is also a closed subspace.
-
-\end{proof}
-
-

src/fa/tvs/vector-function.tex

@@ -0,0 +1,85 @@
+\section{Vector-Valued Function Spaces}
+\label{section:spaces-linear-map}
+
+\begin{proposition}[{{\cite[III.3.1]{SchaeferWolff}}}]
+\label{proposition:tvs-set-uniformity}
+    Let $T$ be a set, $\sigma \subset 2^T$ be an ideal, and $F$ be a TVS over $K \in \RC$, then
+    \begin{enumerate}
+        \item The \hyperref[$\sigma$-uniformity]{definition:set-uniform} on $F^T$ is translation invariant.
+        \item The composition defined by
+        \[
+        T^F \times T^F \to T^F \quad (f + g)(x) = f(x) + g(x)
+        \]
+
+        is continuous.
+    \end{enumerate}
+
+    For any vector subspace $\cf \subset F^T$, the following are equivalent:
+    \begin{enumerate}
+        \item[(T)] The $\sigma$-uniform topology on $\cf \subset F^E$ is a vector space topology.
+        \item[(B)] For each $f \in \cf$ and $S \in \sigma$, $f(S) \subset E$ is bounded.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): Let $U \subset F \times F$ be an entourage and $S \in \sigma$. Using \autoref{proposition:tvs-uniform}, assume without loss of generality that $U$ is translation-invariant. For any $(f, g) \in E(S, U)$, $h \in F^T$, and $x \in S$, $(f(x) + h(x), g(x) + h(x)) \in U$. Thus $(f + h, g + h) \in E(S, U)$ and $E(S, U)$ is translation-invariant.
+
+    (2): Let $f, g, f', g' \in \cf$, $S \in \sigma$, and $U \subset F \times F$ be an entourage. By (TVS1), there exists an entourage $V \subset F \times F$ such that for any $x, x', y, y' \in F$ with $(x, x'), (y, y') \in V$, $(x + y, x' + y') \in U$. If $(f, g), (f', g') \in E(S, V) \cap \cf$, then for any $x \in S$, $(f(x) + g(x), f'(x) + g'(x)) \in U$, so $(f + g, f' + g') \in E(S, U) \cap \cf$.
+
+    (T) $\Rightarrow$ (B): Let $f \in \cf$, $S \in \sigma$ and $U \subset F \times F$ be a symmetric entourage, then $E(S, U)(0)$ is a neighbourhood of $0$ with respect to the $\sigma$-uniform topology. By (TVS2), there exists $\lambda > 0$ such that $f \in \lambda E(S, U)(0)$. In which case, for any $x \in S$, $\lambda^{-1}f(x) \in U(0)$ and $f(x) \in \lambda U(0)$. Thus $f(S) \subset \lambda U(0)$, and $f(S)$ is bounded.
+
+    (T) $\Rightarrow$ (B): Let $f, g \in \cf$, $\lambda, \lambda' \in K$, and $S \in \sigma$, then for any $x \in X$,
+    \begin{align*}
+    \lambda f(x) - \lambda' g(x) &= \lambda f(x) - \lambda' f(x) + \lambda' f(x) - \lambda' g(x) \\
+    &= (\lambda - \lambda')f(x) + \lambda' (f(x) - g(x))
+    \end{align*}
+    Let $U_0 \in \cn_F(0)$. By (TVS1) and \autoref{proposition:tvs-good-neighbourhood-base}, there exists $U \in \cn_F(0)$ circled such that $U + U \subset U_0$. By (TVS2), there exists $V \in \cn_F(0)$ such that $\lambda' V \subset U$. Since $f(S)$ is bounded, there exists $\eps > 0$ with $\eps f(S) \subset U$. In which case, if $\abs{\lambda - \lambda'} < \eps$ and $f(x) - g(x) \in V$ for all $x \in S$, then
+    \[
+    \lambda f(x) - \lambda' g(x) \in (\lambda - \lambda')f(S) + \lambda' V \subset U + U \subset U_0
+    \]
+
+    for all $x \in S$.
+\end{proof}
+
+
+\begin{definition}[Space of Bounded Functions]
+\label{definition:bounded-function-space}
+    Let $T$ be a set, $E$ be a TVS over $K \in \RC$, and $f: T \to E$ be a function, then $f$ is \textbf{bounded} if $f(T)$ is a bounded subset of $E$. The set $B(T; E) \subset E^T$ is the space of all bounded functions from $T$ to $E$, and:
+    \begin{enumerate}
+        \item $B(T; E)$ equipped with the uniform topology and pointwise operations is a TVS over $K \in \RC$. 
+        \item $B(T; E)$ is a closed subset of $E^T$ with respect to the uniform topology. In particular, if $E$ is complete, then so is $B(T; E)$.
+    \end{enumerate}
+\end{definition}
+\begin{proof}
+    (1): For any $f, g \in B(T; E)$ and $\lambda \in K$, $(\lambda f + g)(T) \subset (\lambda f)(T) + g(T)$, which is bounded by \autoref{proposition:bounded-operations}, so $B(T; E)$ is closed under addition and scalar multiplication. 
+    
+    Since $f(E)$ is bounded for all $f \in B(T; E)$, $B(T; E)$ forms a TVS over $K$ by \autoref{proposition:tvs-set-uniformity}.
+
+    (2): Let $f \in \ol{B(T; E)}$ and $U \in \cn_E(0)$ be circled, then there exists $g \in B(T; E)$ such that $(f - g)(E) \subset U$. Since $g \in B(T; E)$, there exists $\lambda \ge 0$ such that $g(E) \subset \lambda U$. In which case,
+    \[
+    f(E) \subset g(E) + (f - g)(E) \subset (\lambda + 1)U
+    \]
+    so $f \in B(T; E)$.
+
+    If $E$ is complete, then $E^T$ with the uniform topology is complete by \autoref{proposition:set-uniform-complete}. Thus $B(T; E)$ is also complete by \autoref{proposition:complete-closed}.
+\end{proof}
+
+
+\begin{definition}[Space of Bounded Continuous Functions]
+\label{definition:bounded-continuous-function-space}
+    Let $X$ be a topological space and $E$ be a TVS over $K \in \RC$, then $BC(X; E) = B(X; E) \cap C(X; E)$ is the space of bounded and continuous functions from $X$ to $E$, and
+    \begin{enumerate}
+        \item $BC(X; E)$ equipped with the uniform topology and pointwise operations is a TVS over $K \in \RC$. 
+        \item $BC(X; E)$ is a closed subspace of $C(X; E)$ and $B(X; E)$ with respect to the uniform topology. In particular, if $E$ is complete, then so is $BC(X; E)$.
+    \end{enumerate}
+\end{definition}
+\begin{proof}
+    (1): Since addition and scalar multiplication are continuous, $BC(X; E)$ is a subspace of $B(X; E)$, and hence a TVS over $K \in \RC$ by \autoref{definition:bounded-function-space}.
+
+    (2): Since $B(X; E)$ and $C(X; E)$ are both closed subspaces of $E^X$ by \autoref{definition:bounded-function-space}, $BC(X; E)$ is a closed subspace. 
+    
+    If $E$ is complete, then $B(X; E)$ and $C(X; E)$ are both complete under the uniform topology by \autoref{definition:bounded-function-space}. Therefore $BC(X; E)$ is also complete. 
+\end{proof}
+
+
+
+

src/measure/bochner-integral/bochner.tex

@@ -0,0 +1,110 @@
+\section{The Bochner Integral}
+\label{section:bochner-integral}
+
+\begin{definition}[Bochner Integral]
+\label{definition:bochner-integral}
+    Let $(X, \cm, \mu)$ be a measure space and $E$ be a Banach space over $K \in \RC$, then there exists a unique $I \in L(L^1(X; E); E)$ such that:
+    \begin{enumerate}
+        \item For any $x \in E$ and $A \in \cm$ with $\mu(A) < \infty$, $I(x \cdot \one_A) = x \cdot \mu(A)$. 
+        \item For all $f \in L^1(X; E)$, $\norm{If}_E \le \int \norm{f}_E d\mu$. 
+    \end{enumerate}
+
+    For any $f \in L^1(X; E)$, $If = \int f d\mu$ is the \textbf{Bochner integral} of $f$.
+\end{definition}
+\begin{proof}
+    (1): For any $\phi \in \Sigma(X; E) \cap L^1(X; E)$, let
+    \[
+    I\phi = \sum_{y \in \phi(X)}^n y \cdot \mu\bracs{\phi = y}
+    \]
+
+    For any $\lambda \in K$, if $\lambda \ne 0$, then the mapping $x \mapsto \lambda x$ is a bijection, so
+    \[
+    I(\lambda \phi) = \sum_{\lambda y \in \lambda \phi(X)}^n (\lambda y) \cdot \mu\bracs{\lambda \phi = \lambda y} 
+    = \lambda \sum_{y \in \phi(X)}y \cdot \mu\bracs{\phi = y} = \lambda I\phi
+    \]
+
+    If $\lambda = 0$, then $I(\lambda \phi) = I(0) = 0$. 
+
+    Let $\phi, \psi \in \Sigma(X; E) \cap L^1(X; E)$, then
+    \begin{align*}
+        I\phi + I\psi &= \sum_{y \in \phi(X)}y \cdot \mu\bracs{\phi = y} + \sum_{z \in \psi(X)}z \cdot \mu\bracs{\psi = z} \\
+        &= \sum_{y \in \phi(X)} \sum_{z \in \psi(X)} (y + z) \cdot \mu\bracs{\phi = y, \psi = z} \\
+        &= \sum_{y \in (\phi + \psi)(X)}\sum_{{z \in \phi(X) \atop {z' \in \psi(X) \atop z + z' = y}}}(z + z') \cdot \mu\bracsn{\phi = g, \psi = z'} \\
+        &= \sum_{y \in (\phi + \psi)(X)}y \cdot \mu\bracs{\phi + \psi = y} = I\phi + I\psi
+    \end{align*}
+
+    so $I$ is a linear operator on $\Sigma(X; E) \cap L^1(X; E)$ that satisfies (1).
+
+    (2): For any $\phi \in \Sigma(X; E) \cap L^1(X; E)$, 
+    \[
+    \norm{I\phi}_E \le \sum_{y \in \phi(X)}\norm{y}_E \cdot \mu\bracs{\phi = y} = \int \norm{\phi}_E d\mu = \norm{\phi}_{L^1(X; E)}
+    \]
+    
+    By \autoref{proposition:lp-simple-dense}, $\Sigma(X; E) \cap L^1(X; E)$ is dense in $L^1(X; E)$. Therefore by the \hyperref[Linear Extension Theorem]{theorem:linear-extension-theorem-normed}, $I$ admits a unique norm-preserving extension to $L^1(X; E)$. 
+    
+\end{proof}
+
+
+
+\begin{theorem}[Dominated Convergence Theorem]
+\label{theorem:dct-bochner}
+    Let $(X, \cm, \mu)$ be a measure spacs, $E$ be a Banach space over $K \in \RC$, $\seq{f_n} \subset L^1(X; E)$, and $f \in L^1(X; E)$. If
+    \begin{enumerate}
+        \item[(a)] $f_n \to f$ strongly pointwise.
+        \item[(b)] There exists $g \in L^1(X) \cap L^+(X)$ such that $\norm{f_n}_E \le g$ for all $n \in \natp$. 
+    \end{enumerate}
+
+    
+    then $\int f d\mu = \limv{n}\int f_n d\mu$. 
+\end{theorem}
+\begin{proof}
+    By the classical \hyperref[Dominated Convergence Theorem]{proposition:dct-lp}, $f_n \to f$ in $L^1(X; E)$. Since $h \mapsto \int h d\mu$ is a bounded linear operator, $\int f d\mu = \limv{n}\int f_n d\mu$. 
+\end{proof}
+
+\subsection{Vector Measure Version}
+\label{subsection:bochner-vector}
+
+
+
+\begin{definition}[Bochner Integral]
+\label{definition:bochner-integral-vector}
+    Let $(X, \cm)$ be a measurable space, $E, F$ be normed spaces over $K \in \RC$, $G$ be a Banach space over $K$,
+    \[
+    \lambda: E \times F \to G \quad (x, y) \mapsto xy
+    \]
+
+    be a bounded bilinear map, and $\mu: \cm \to F$ be a vector measure, then there exists a unique $I_\lambda \in L(L^1(X, |\mu|; E); G)$ such that:
+    \begin{enumerate}
+        \item For any $x \in E$ and $A \in \cm$, $I_\lambda(x \cdot \one_A) = x \mu(A)$.
+        \item For any $f \in L^1(X, |\mu|; E)$, $\normn{I_\lambda f}_{G} \le \norm{\lambda}_{L^2(E, F; G)} \cdot \norm{f}_{L^1(X, |\mu|; E)}$.
+    \end{enumerate}
+
+    
+    For any $f \in L^1(X; E)$, $I_\lambda f = \int \lambda(f, d\mu)$ is the \textbf{Bochner integral} of $f$ with respect to $\mu$ and $\lambda$. 
+\end{definition}
+\begin{proof}
+    Same as \autoref{definition:bochner-integral}.
+\end{proof}
+
+
+\begin{theorem}[Dominated Convergence Theorem]
+\label{theorem:dct-bochner-vector}
+    Let $(X, \cm, \mu)$ be a measure space, $E, F$ be normed spaces over $K \in \RC$, $G$ be a Banach space over $K$, 
+    \[
+    \lambda: E \times F \to G \quad (x, y) \mapsto xy
+    \]
+
+    be a bounded bilinear map, $\mu: \cm \to F$ be a vector measure, $\seq{f_n} \subset L^1(X, |\mu|; E)$, and $f \in L^1(X, |\mu|; E)$. If
+    \begin{enumerate}
+        \item[(a)] $f_n \to f$ strongly pointwise.
+        \item[(b)] There exists $g \in L^1(X) \cap L^+(X)$ such that $\norm{f_n}_E \le g$ for all $n \in \natp$. 
+    \end{enumerate}
+
+    
+    then $\int \lambda(f , d\mu) = \limv{n}\int \lambda(f_n, d\mu)$. 
+\end{theorem}
+\begin{proof}
+    By the classical \hyperref[Dominated Convergence Theorem]{proposition:dct-lp}, $f_n \to f$ in $L^1(X, |\mu|; E)$. Since $h \mapsto \int \lambda(f , d\mu)$ is a bounded linear operator, $\int \lambda(f , d\mu) = \limv{n}\int \lambda(f_n, d\mu)$. 
+\end{proof}
+
+

src/measure/bochner-integral/index.tex

@@ -2,3 +2,4 @@
 \label{chap:bochner-integral}
 
 \input{./strongly.tex}
+\input{./bochner.tex}

src/measure/bochner-integral/strongly.tex

@@ -6,16 +6,23 @@
     Let $(X, \cm)$ be a measurable space, $E$ be a normed vector space over $K \in \RC$, and $f: X \to E$, then the following are equivalent:
     \begin{enumerate}
         \item For each $\phi \in E^*$, $\phi \circ f$ is $(\cm, \cb_K)$-measurable and $f(X) \subset E$ is separable.
-        \item $f$ is $(\cm, \cb_E)$ measurable and $f(X) \subset E$ is separable.
+        \item $f$ is $(\cm, \cb_E)$-measurable and $f(X) \subset E$ is separable.
         \item There exists a sequence $\seq{f_n} \subset \Sigma(X, \cm; E)$ such that
         \begin{enumerate}
             \item[(a)] For each $n \in \natp$, $\norm{f_n}_E \le \norm{f}_E$.
             \item[(b)] $\norm{f_n(x) - f(x)}_E \to 0$ pointwise as $n \to \infty$.
         \end{enumerate}
+
     \end{enumerate}
+
+    If the above holds, then $f$ is a \textbf{strongly measurable} function. 
 \end{definition}
 \begin{proof}
-    (1) $\Rightarrow$ (2): TODO
+    (1) $\Rightarrow$ (2): First suppose that $E$ is separable. By \autoref{proposition:separable-banach-borel-sigma-algebra}, the Borel $\sigma$-algebra on $E$ coincides with the $\sigma$-algebra on $E$ generated by the weak topology. Thus if $\phi \circ f$ is $(\cm, \cb_K)$-measurable for all $\phi \in E^*$, then $f$ is $(\cm, \cb_E)$-measurable. 
+
+    Now suppose that $E$ is arbitrary. Let $F \subset E$ be the closure of the linear span of $f(X)$, then $F$ is a separable closed subspace of $E$. For any $\phi \in F^*$, by the \hyperref[Hahn-Banch Theorem]{theorem:hahn-banach}, there exists an extension $\Phi \in E^*$ of $\phi$. In which case, since $f(X) \subset F$, for any Borel set $B \in \cb_{K}$, $\bracs{\phi \circ f \in B} = \bracs{\Phi \circ f \in B} \in \cm$. Thus $\phi \circ f$ is $(\cm, \cb_K)$-measurable for all $\phi \in E^*$. 
+
+    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}.
 
@@ -26,3 +33,22 @@
 
     and each $f_n$ is finitely-valued, $f(X)$ is separable.
 \end{proof}
+
+\begin{proposition}
+\label{proposition:strongly-measurable-properties}
+    Let $(X, \cm)$ be a measurable space, $E$ be a normed vector space over $K \in \RC$, then:
+    \begin{enumerate}
+        \item For any strongly measurable functions $f, g: X \to E$ and $\lambda \in K$, $\lambda f + g$ is strongly measurable. 
+        \item For any strongly measurable functions $\bracs{f_n: X \to E|n \in \natp}$ and $f: X \to E$, if $f_n \to f$ strongly pointwise, then $f$ is strongly measurable. 
+    \end{enumerate}
+
+    
+\end{proposition}
+\begin{proof}
+    (1): Since $x \mapsto \lambda$ is continuous, $\lambda f$ is strongly measurable by (2) of \autoref{definition:strongly-measurable}. 
+
+    By (3) of \autoref{definition:strongly-measurable}, there exists $\seq{f_n}, \seq{g_n} \subset \Sigma(X, \cm; E)$ such that $f_n \to f$ and $g_n \to g$ strongly pointwise. In which case, $\seq{f_n + g_n} \subset \Sigma(X, \cm; E)$ and $f_n + g_n \to f + g$ strongly pointwise. Therefore $f + g$ is also strongly measurable. 
+
+    (2): By \autoref{proposition:metric-measurable-limit}, $f$ is $(\cm, \cb_E)$-measurable. Since $f(X) \subset \overline{\bigcup_{n \in \natp}}f_n(X)$, $f(X)$ is also separable, so $f$ is strongly measurable by (1) of \autoref{definition:strongly-measurable}.
+\end{proof}
+

src/measure/index.tex

@@ -8,3 +8,4 @@
 \input{./measurable-maps/index.tex}
 \input{./lebesgue-integral/index.tex}
 \input{./bochner-integral/index.tex}
+\input{./notation.tex}

src/measure/lebesgue-integral/complex.tex

@@ -28,7 +28,7 @@
 \label{definition:positive-negative-parts}
     Let $X$ be a set and $f: X \to \real$ be a function, then
     \[
-    f^+ = \max(f, 0) \quad f^- = -\min(f, 0)
+    f^+ = f \vee 0 \quad f^- = -(f \wedge 0)
     \]
 
     are the \textbf{positive} and \textbf{negative} parts of $f$, and $f = f^+ - f^-$.
@@ -56,27 +56,14 @@
     Let $(X, \cm, \mu)$ be a measure space, then the integral is a linear functional on $\mathcal{L}^1(X)$ such that for any $f \in \mathcal{L}^1(X)$, $\abs{\int f d\mu} \le \int \abs{f}d\mu$.
 \end{proposition}
 \begin{proof}
-    Let $f, g \in \mathcal{L}^1(X)$ and $\lambda \in \complex$. First suppose that $f, g$ are $\real$-valued and $\lambda \in \real$. In which case,
+    By \autoref{lemma:positive-functional-extension}, the mapping
     \[
-    \int \lambda f d\mu = \int (\lambda f)^+ d\mu - \int (\lambda f)^- d\mu
+    I: \mathcal{L}^1(X; \real) \to \real \quad f \mapsto \int f^+ d\mu - \int f^- d\mu
-    = \begin{cases}
-        \lambda\int f^+ d\mu - \lambda\int f^- d\mu &\lambda \ge 0 \\
-        -\lambda\int f^- d\mu + \lambda\int f^+ d\mu &\lambda < 0
-    \end{cases}
     \]
 
-    by \autoref{proposition:lebesgue-non-negative-properties}, so $\int \lambda f d\mu = \lambda \int f d\mu$.
+    is a $\real$-linear functional on $\mathcal{L}^1(X; \real)$. 
 
-    Let $h = f + g$, then $h = h^+ - h^- = f^+ + g^+ - f^- - g^-$, so
+    Let $f, g \in L^1(X; \complex)$ and $\lambda = \alpha + \beta i \in \complex$ with $\alpha, \beta \in \real$, then
-    \begin{align*}
-        h^+ + f^- + g^- &= h^- + f^+ + g^+ \\
-        \int h^+d\mu + \int f^- d\mu + \int g^-d\mu &= \int h^- d\mu + \int f^+ d\mu + \int g^+ d\mu \\
-        \int h^+ d\mu - \int h^- d\mu &= \int f^+ d\mu - \int f^- d\mu + \int g^+ d\mu - \int g^- d\mu \\
-        &= \int f d\mu + \int g d\mu
-    \end{align*}
-    by \autoref{proposition:lebesgue-non-negative-properties}, so $\int f + g d\mu = \int f d\mu + \int g d\mu$.
-
-    Now suppose that $f, g$ are $\complex$-valued and $\lambda = \alpha + \beta i \in \complex$ with $\alpha, \beta \in \real$, then
     \begin{align*}
         \int (\alpha f)d\mu &= \int \text{Re}(\lambda f)d\mu + i\int \text{Im}(\lambda f)d\mu \\
         &= \int \alpha \text{Re}(f) - \beta \text{Im}(f)d\mu + i\int \beta\text{Re}(f) + \alpha \text{Im}(f)d\mu \\
@@ -111,16 +98,17 @@
 	The construction of the Lebesgue integral using positive/negative and real/imaginary parts, and the deduction of properties from this definition both appear cumbersome. This comes from the lack of usage of its linearity on integrable simple functions. A more functional analytic approach will use this linearity and the density of simple functions to construct the integral. This is the route through which the Bochner integral will be presented.
 \end{remark}
 
-\begin{theorem}[Dominated Convergence Theorem {{\cite[Theorem 2.24]{Folland}}}]
+\begin{theorem}[Dominated Convergence Theorem]
 \label{theorem:dct}
     Let $(X, \cm, \mu)$ be a measure space, $\seq{f_n} \subset \mathcal{L}^1(X)$, and $f:X \to \complex$ such that
     \begin{enumerate}
         \item $f_n \to f$ pointwise.
         \item There exists $g \in \mathcal{L}^1(X)$ such that $\abs{f_n} \le \abs{g}$ for all $n \in \natp$.
     \end{enumerate}
+
     then $\int fd\mu = \limv{n}\int f_n d\mu$.
 \end{theorem}
-\begin{proof}
+\begin{proof}[Proof {{\cite[Theorem 2.24]{Folland}}}. ]
     By (1) and (2), $f$ is measurable with $\int \abs{f}d\mu \le \int \abs{g}d\mu < \infty$, so $f \in \mathcal{L}^1(X)$. Now, since $g + f, g - f \ge 0$, by \hyperref[Fatou's lemma]{lemma:fatou} and \autoref{proposition:lebesgue-integral-properties},
     \begin{align*}
         \int g d\mu + \int f d\mu &\le \liminf_{n \to \infty}\int g + f_n d\mu = \int g d\mu + \liminf_{n \to \infty}f_n d\mu \\

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

@@ -38,7 +38,7 @@
     the two sides are equal.
 \end{proof}
 
-\begin{theorem}[Monotone Convergence Theorem, {{\cite[Theorem 2.14]{Folland}}}]
+\begin{theorem}[Monotone Convergence Theorem]
 \label{theorem:mct}
     Let $(X, \cm, \mu)$ be a measure space, $\seq{f_n} \subset \mathcal{L}^+(X, \cm)$, and $f \in \mathcal{L}^+(X, \cm)$ such that $f_n \upto f$ pointwise, then
     \[
@@ -46,7 +46,7 @@
     \]
 
 \end{theorem}
-\begin{proof}
+\begin{proof}[Proof {{\cite[Theorem 2.14]{Folland}}}. ]
     By \autoref{definition:lebesgue-non-negative}, $\int f_n d\mu \le \int f d\mu$ for each $n \in \natp$.
 
     Let $\phi \in \Sigma^+(X, \cm)$ with $\phi|_{\bracs{f > 0}} < f|_{\bracs{f > 0}}$ and $\phi \le f$. Since $f_n \upto f$, $\bracs{f_n \ge \phi} \upto X$. In which case, since $A \mapsto \int \one_A \phi d\mu$ is a measure (\autoref{proposition:lebesgue-simple-properties}),
@@ -57,7 +57,7 @@
     by continuity from below (\autoref{proposition:measure-properties}). Therefore $\limv{n}\int f_n d\mu \ge \int f$ by \autoref{lemma:lebesgue-non-negative-strict}.
 \end{proof}
 
-\begin{lemma}[Fatou, {{\cite[Lemma 2.18]{Folland}}}]
+\begin{lemma}[Fatou]
 \label{lemma:fatou}
     Let $(X, \cm, \mu)$ be a measure space, $\seq{f_n} \subset \mathcal{L}^+(X, \cm)$, then
     \[
@@ -65,7 +65,7 @@
     \]
 
 \end{lemma}
-\begin{proof}
+\begin{proof}[Proof {{\cite[Lemma 2.18]{Folland}}}. ]
     For each $n \in \natp$, $\inf_{k \ge n}f_k \le f_n$. By the monotone convergence theorem,
     \[
     \int \liminf_{n \to \infty}f_n d\mu = \limv{n}\int \inf_{k \ge n}f_k d\mu

src/measure/lebesgue-integral/simple.tex

@@ -8,7 +8,7 @@
 
 \begin{definition}[Integral of Non-Negative Simple Functions]
 \label{definition:lebesgue-simple}
-    Let $(X, \cm, \mu)$ be a measure space and $f = \sum_{y \in f(X)}y \cdot \one_{\bracs{f = y}} \in \Sigma^+(X, \cm)$ be a non-negative simple function in standard form, then\footnote{With the convention that $0 \cdot \infty = 0$.}
+    Let $(X, \cm, \mu)$ be a measure space and $f = \sum_{y \in f(X)}y \cdot \one_{\bracs{f = y}} \in \Sigma^+(X, \cm)$ be a non-negative simple function in standard form, then (with the convention that $0 \cdot \infty = 0$)
     \[
     \int f d\mu = \int f(x) \mu(dx) = \sum_{y \in f(X)}y \cdot \mu(\bracs{f = y})
     \]

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

@@ -36,7 +36,7 @@
 
 \begin{proposition}[{{\cite[Theorem 2.30]{Folland}}}]
 \label{proposition:cauchy-in-measure-limit}
-    Let $(X, \cm, \mu)$ be a measure space, $(Y, d)$ be a complete metric space, and $\seq{f_n}$ be Borel measurable functions from $X \to Y$, 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 There exists a subsequence $\seq{n_k}$ such that $f_{n_k} \to f$ almost everywhere.

src/measure/measurable-maps/metric.tex

@@ -71,12 +71,13 @@
 
 \begin{proposition}
 \label{proposition:measurable-simple-separable}
-    Let $(X, \cm)$ be a measurable space, $Y$ be a separable metric space, and $N: Y \to 2^Y$\footnote{This mapping is typically obtained as slices of the level sets of a continuous function $Y \times Y \to \real$.} such that
+    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.
@@ -85,6 +86,7 @@
             \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}
@@ -99,9 +101,9 @@
     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 $k \in \natp$,
+    then for any $k \in \natp$, $\bracs{x \in X|k(n, x) \le k}$ is equal to
     \[
-    \bracs{x \in X|k(n, x) \le k} = \bigcup_{j = 1}^k\bracs{x \in X \bigg |y_j \in C(n, x), d(f(x), y_j) = \min_{m \in C(N, x)}d(f(x), y_m)}
+     \bigcup_{j = 1}^k\bracs{x \in X \bigg |y_j \in C(n, x), d(f(x), y_j) = \min_{m \in C(N, x)}d(f(x), y_m)}
     \]
 
     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},
@@ -125,7 +127,7 @@
 
 \begin{proposition}
 \label{proposition:measurable-simple-separable-norm}
-    Let $(X, \cm)$ be a measurable space, $(E, \norm{\cdot}_E)$ be a separable normed space, and $f: X \to E$, then the following are equivalent:
+    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.
@@ -147,5 +149,6 @@
         \]
 
     \end{enumerate}
+
     By \autoref{proposition:measurable-simple-separable}, (1) and (2) are equivalent.
 \end{proof}

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

@@ -25,6 +25,7 @@
         \item $g = \limsup_{n \to \infty}f_n$.
         \item $\limv{n}f_n$ (if it exists).
     \end{enumerate}
+
     In addition, if the above functions are $\real$-valued, then they are $(\cm, \cb_{\real})$-measurable.
 \end{proposition}
 \begin{proof}

src/measure/measure/complete.tex

@@ -21,6 +21,7 @@
         \item $(X, \ol{\cm}, \ol{\mu})$ is a complete measure space.
         \item[(U)] For any complete measure space $(X, \cf, \nu)$ where $\cf \supset \cm$ and $\nu|_\cm = \mu$, $\cf \supset \ol{\cm}$ and $\nu|_{\ol{\cm}} = \ol{\mu}$.
     \end{enumerate}
+
     and space $(X, \ol{\cm}, \mu)$ is the \textbf{completion} of $(X, \cm, \mu)$.
 \end{definition}
 \begin{proof}

src/measure/measure/index.tex

@@ -8,4 +8,5 @@
 \input{./regular.tex}
 \input{./outer.tex}
 \input{./lebesgue-stieltjes.tex}
+\input{./product.tex}
 \input{./kolmogorov.tex}

src/measure/measure/kolmogorov.tex

@@ -13,16 +13,18 @@
 \label{lemma:kolmogorov-compact-sequence}
     Let $\seq{X_n}$ be topological spaces where for each $n \in \natp$,
     \begin{enumerate}
-        \item[(a)] Every finite measure on $\prod_{j = 1}^n X_j$ is regular\footnote{A potential sufficient condition for this is that each $X_n$ is LCH where every open set is $\sigma$-compact. However, I have yet to verify if this condition persists over products.}.
+        \item[(a)] Every finite measure on $\prod_{j = 1}^n X_j$ is regular.
         \item[(b)] $X_n$ is Hausdorff.
         \item[(c)] $X_n$ is separable.
     \end{enumerate}
+
     Let $\bracs{\mu_{I}| I \subset \natp \text{ finite}}$ be consistent Borel probability measures, then for any $\seq{B_n}$ where:
     \begin{enumerate}
         \item[(d)] For each $n \in \nat$, $B_n \in \cb_{\prod_{j = 1}^n X_j}$.
         \item[(e)] For each $n \in \nat$, $B_{n+1} \subset B_n \times X_{n+1}$.
         \item[(f)] There exists $\eps > 0$ such that $\mu_{[n]}(B_n) > \eps$ for all $n \in \natp$.
     \end{enumerate}
+
     Then there exists $\seq{K_n}$ such that for every $n \in \natp$,
     \begin{enumerate}
         \item $K_n \subset \prod_{j = 1}^n X_j$ is compact.
@@ -45,7 +47,7 @@
     Thus $\mu_{[n]}(K_n) \ge \eps/2$.
 \end{proof}
 
-\begin{theorem}[Kolmogorov's Extension Theorem, {{\cite[Theorem 1.14]{Baudoin}}}]
+\begin{theorem}[Kolmogorov's Extension Theorem]
 \label{theorem:kolmogorov-extension}
     Let $\seqi{X}$ be topological spaces where for each $J \subset I$ finite,
     \begin{enumerate}
@@ -53,9 +55,10 @@
         \item[(b)] $X_j$ is Hausdorff.
         \item[(c)] $X_j$ is separable.
     \end{enumerate}
+
     Let $\bracs{\mu_{I}| I \subset \natp \text{ finite}}$ be consistent Borel probability measures, then there exists a unique probability measure $\mu: \bigotimes_{i \in I}\cb_{X_i} \to [0, 1]$ such that for any $J \subset I$ finite, $\mu = \mu_J \circ \pi_J^{-1}$.
 \end{theorem}
-\begin{proof}
+\begin{proof}[Proof {{\cite[Theorem 1.14]{Baudoin}}}. ]
     Let
     \[
     \alg = \bracs{\pi_J^{-1}(B)| B \in \bigotimes_{j \in J}\cb_{X_j}, J \subset I \text{ finite}}
@@ -72,6 +75,7 @@
         \item $K_{n+1} \subset K_n \times X_{n+1}$.
         \item $\mu(K_n) \ge \eps/2$.
     \end{enumerate}
+
     Let $N \in \natp$ and $x \in \prod_{j = 1}^N X_j$ such that $x \in \bigcap_{n \ge N}\pi_{[N]}(K_n)$. By compactness and (4), there exists $x_{N+1} \in X_{N+1}$ such that $(x_1, \cdots, x_N, x_{N+1}) \in \bigcap_{n > N}\pi_{[N+1]}(K_n)$. Thus there exists $x \in \prod_{i \in I}X_i$ such that $x \in \pi_{[n]}^{-1}(K_n)$ for all $n \in \natp$, and
     \[
     x \in \bigcap_{n \in \natp}\pi_{[n]}^{-1}(K_n) \subset \bigcap_{n \in \natp}\pi_{[n]}^{-1}(B_n) \ne \emptyset

src/measure/measure/lebesgue-stieltjes.tex

@@ -34,7 +34,7 @@
     (2): By \autoref{proposition:elementary-family-algebra}, $\alg$ is a ring.
 \end{proof}
 
-\begin{definition}[Lebesgue-Stieltjes Measure, {{\cite[Theorem 1.16]{Folland}}}]
+\begin{definition}[Lebesgue-Stieltjes Measure]
 \label{definition:lebesgue-stieltjes-measure}
     Let $\mu: \cb_\real \to [0, \infty]$ be a Borel measure on $\real$ such that for any $K \subset \real$ compact, $\mu(K) < \infty$, then
     \begin{enumerate}
@@ -45,9 +45,10 @@
 
         \item[(U)] For any function $G: \real \to \real$ satisfying (1), $F - G$ is constant.
     \end{enumerate}
+
     Conversely, if $F: \real \to \real$ is a Stieltjes function, then there exists a unique Borel measure $\mu_F: \cb_\real \to [0, \infty]$ satisfying (1), and $\mu_F$ is the \textbf{Lebesgue-Stieltjes measure} associated with $F$.
 \end{definition}
-\begin{proof}
+\begin{proof}[Proof {{\cite[Theorem 1.16]{Folland}}}. ]
     (1): Let
     \[
     F: \real \to \real \quad x \mapsto \begin{cases}

src/measure/measure/measure.tex

@@ -15,12 +15,14 @@
         \item[(M1)] $\mu(\emptyset) = 0$.
         \item[(M2)] For any $\seq{E_n} \subset \cm$ pairwise disjoint, $\mu\paren{\bigsqcup_{n \in \natp}E_n} = \sum_{n \in \natp} \mu(E_n)$.
     \end{enumerate}
+
     In which case, $(X, \cm, \mu)$ is a \textbf{measure space}.
 
     If $\mu: \cm \to [0, \infty]$ instead satisfies (M1) and
     \begin{enumerate}
         \item[(M2')] For any $\seqf{E_j} \subset \cm$ pairwise disjoint, $\mu\paren{\bigsqcup_{j = 1}^n E_j} = \sum_{j = 1}^n \mu(E_j)$.
     \end{enumerate}
+
     then $\mu$ is a \textbf{finitely-additive measure}.
 \end{definition}
 
@@ -101,6 +103,7 @@
         \item[(b)] $\mu(E) = \nu(E)$ for all $E \in \mathcal{P}$.
         \item[(c)] There exists $\seq{E_n} \subset \mathcal{P}$ such that $E_n \upto X$ and $\mu(E_n) < \infty$ for all $n \in \natp$.
     \end{enumerate}
+
     then $\mu = \nu$.
 \end{theorem}
 \begin{proof}
@@ -123,6 +126,7 @@
 
         by continuity from below (\autoref{proposition:measure-properties}).
     \end{enumerate}
+
     so $\alg(F)$ is a $\lambda$-system. By (a) and Dynkin's $\pi$-$\lambda$ theorem (\autoref{theorem:pi-lambda}), $\alg(F) = \cm$.
 
     Let $\seq{E_n}$ as in assumption (c), then $\mu(E_n \cap F) = \mu(E_n \cap F)$ for all $n \in \natp$ and $F \in \cm$. Thus by continuity from below (\autoref{proposition:measure-properties}),
@@ -149,3 +153,27 @@
     \]
 \end{proof}
 
+\begin{definition}[Pushforward Measure]
+\label{definition:pushforward-measure}
+    Let $(X, \cm, \mu)$ be a measure space, $(Y, \cn)$ be a measurable space, and $f: X \to Y$ be a $(\cm, \cn)$-measurable map, then:
+    \begin{enumerate}
+        \item The mapping
+        \[
+        f_*\mu: \cn \to [0, \infty] \quad A \mapsto \mu\bracs{f \in A}
+        \]
+
+        is a measure on $(Y, \cn)$. 
+        \item For any Banach space $E$ and $g \in L^1(f_*\mu; E)$, 
+        \[
+        \int g df_*\mu = \int g \circ f d\mu
+        \]
+    \end{enumerate}
+
+    
+\end{definition}
+\begin{proof}
+    (1): Preimage commutes with unions, intersections, and complements. 
+
+    (2): By definition and linearity, (2) holds for $L^1(f_*\mu; E) \cap \Sigma(f^*\mu; E)$, which is dense in $L^1(f_*\mu; E)$ by \autoref{proposition:lp-simple-dense}. By continuity of the integral, (2) holds on $L^1(\mu_*\mu; E)$. 
+\end{proof}
+

src/measure/measure/outer.tex

@@ -44,7 +44,7 @@
 
 \end{definition}
 
-\begin{theorem}[Carathéodory, {{\cite[Theorem 1.11]{Folland}}}]
+\begin{theorem}[Carathéodory]
 \label{theorem:caratheodory}
     Let $X$ be a set, $\mu^*: 2^X \to [0, \infty]$ be an outer measure, and $\cm \subset 2^X$ be the collection of all $\mu^*$-measurable sets, then:
     \begin{enumerate}
@@ -57,7 +57,7 @@
         \item $(X, \cm, \mu^*|_\cm)$ is a complete measure space.
     \end{enumerate}
 \end{theorem}
-\begin{proof}
+\begin{proof}[Proof {{\cite[Theorem 1.11]{Folland}}}. ]
     (1): Let $N \in \nat$, then
     \[
     \mu^*\paren{F \cap \bigsqcup_{n \in \natp}E_n} \ge \mu^*\paren{F \cap \bigsqcup_{n =1}^N E_n} = \sum_{n = 1}^N \mu^*(F \cap E_n)
@@ -103,7 +103,7 @@
     \end{enumerate}
 \end{definition}
 
-\begin{theorem}[Carathéodory's Extension Theorem, {{\cite[Theorem 1.14]{Folland}}}]
+\begin{theorem}[Carathéodory's Extension Theorem]
 \label{theorem:caratheodory-extension}
     Let $X$ be a set, $\alg \subset 2^X$ be a ring, and $\mu_0: \alg \to [0, \infty]$ be a premeasure, then there exists a measure space $(X, \cm, \mu)$ such that
     \begin{enumerate}
@@ -116,9 +116,10 @@
             \item[(b)] For any $E \in \cm \cap \cn$ with $\mu(E) < \infty$, $\mu(E) = \nu(E)$.
             \item[(c)] If $\mu$ is $\sigma$-finite, then $\nu = \mu$.
         \end{enumerate}
+
     \end{enumerate}
 \end{theorem}
-\begin{proof}
+\begin{proof}[Proof {{\cite[Theorem 1.14]{Folland}}}. ]
     Let
     \[
     \mu^*: 2^X \to [0, \infty] \quad E \mapsto \inf\bracs{\sum_{n \in \natp}\mu_0(E_n) \bigg | \seq{E_n} \subset \alg, \bigcup_{n \in \natp}E_n \supset E}

src/measure/measure/product.tex

@@ -0,0 +1,133 @@
+\section{Product Measures}
+\label{section:product-measures}
+
+
+\begin{definition}[Product Measure]
+\label{definition:product-measure}
+    Let $(X, \cm, \mu)$ and $(Y, \cn, \nu)$ be measure spaces, then there exists a measure $\mu \otimes \nu: \cm \times \cn \to [0, \infty]$ such that:
+    \begin{enumerate}
+        \item For each $E \in \cm$ and $F \in \cn$, $\mu \otimes \nu(E \times F) = \mu(E)\nu(F)$. 
+        \item[(U)] For any measure $\lambda: \cm \otimes \cn \to [0, \infty]$, $\lambda \le \mu$. For any $A \in \cm \otimes \cn$ with $\mu(A) < \infty$, $\lambda(A) = \mu(A)$. In particular, if $\mu$ is $\sigma$-finite, then $\lambda = \mu$. 
+    \end{enumerate}
+
+    
+    The measure $\mu \otimes \nu$ is the \textbf{product} of $\mu$ and $\nu$. 
+\end{definition}
+\begin{proof}
+    Let
+    \[
+    \mathcal{R} = \bracs{E \times F|E \in \cm, F \in \cn}
+    \]
+
+    then $\mathcal{R}$ is an elementary family by \autoref{proposition:rectangle-elementary-family}. Let
+    \[
+    \alg = \bracs{\bigsqcup_{i = 1}^n E_j \times F_j \bigg | \seqf{E_j \times F_j} \subset \mathcal{R} \text{ pairwise disjoint}}
+    \]
+
+    then $\alg$ is a ring over $X \times Y$. For each pairwise disjoint collection $\seqf{E_j \times F_j} \subset \mathcal{R}$, let
+    \[
+    \mu \otimes \nu\paren{\bigsqcup_{j = 1}^n E_j \times F_j} = \sum_{j = 1}^n \mu(E_j) \mu(F_j)
+    \]
+
+    then $\mu \otimes \nu$ is well-defined and finitely additive on $\alg$. 
+    
+    Let $A \times B \in \mathcal{A}$ and $\seq{A_n \times B_n} \subset \mathcal{R}$ such that $A \times B = \bigsqcup_{n \in \natp}A_n \times B_n$, then for any $x \in X$ and $y \in Y$,
+    \[
+    \one_{A \times B}(x, y) = \sum_{n \in \natp}\one_{A_n \times B_n}(x, y) = \sum_{n \in \natp}\one_{A_n}(x)\one_{B_n}(y)
+    \]
+
+    By the \hyperref[Monotone Convergence Theorem]{theorem:mct}, for any $y \in Y$,
+    \begin{align*}
+        \mu(A)\one_{B}(y) &= \int_X \one_{A}(x)\one_B(y)\mu(dx) = \sum_{n \in \natp}\int_X \one_{A_n}(x)\one_{B_n}(y) \mu(dx) \\
+        &= \sum_{n = 1}^\infty \mu(A_n)\one_{B_n}(y)
+    \end{align*}
+
+    By the Monotone Convergence Theorem again, 
+    \begin{align*}
+        \mu(A)\nu(B) &= \int_Y \mu(A)\one_B(y) \nu(dy) = \sum_{n = 1}^\infty \int_Y \mu(A_n)\one_{B_n}\nu(dy) \\
+        &= \sum_{n = 1}^\infty \mu(A_n)\nu(B_n)
+    \end{align*}
+    
+    Therefore $\mu \otimes \nu$ is a premeasure on $\alg$. By \hyperref[Carathéodory's Extension Theorem]{theorem:caratheodory-extension}, there exists a measure $\mu \otimes \nu$ satisfying (1) and (U).
+    
+\end{proof}
+
+\begin{lemma}[{{\cite[Proposition 2.34]{Folland}}}]
+\label{lemma:section-measurable}
+    Let $(X, \cm)$ and $(Y, \cn)$ be measurable spaces, then:
+    \begin{enumerate}
+        \item For any $E \in \cm \otimes \cn$, $x \in X$, and $y \in Y$, $\bracs{z \in Y|(x, z) \in E} \in \cm$ and $\bracs{z \in X|(z, y) \in E} \in \cn$.
+        \item For any measure space $(Z, \cf)$, $(\cm \otimes \cn, \cf)$-measurable function $f: X \times Y \to Z$, $x \in X$, and $y \in Y$, $f(x, \cdot)$ is $(\cn, \cf)$-measurable and $f(\cdot, y)$ is $(\cm, \cf)$-measurable. 
+    \end{enumerate}
+
+    
+\end{lemma}
+\begin{proof}
+    (1): Let $\alg \subset \cm \otimes \cn$ be the collection of all sets that satisfy (1), then
+    \[
+    \alg \supset \bracs{E \times F| E \in \cm, F \in \cn}
+    \]
+
+    For any $E \in \alg$, $\bracs{z \in Y|(y, z) \in E}^c = \bracs{z \in Y|(y, z) \in E^c}$, so $E^c \in \alg$ as well. For any $\seq{E_n} \subset \alg$, 
+    \[
+    \bracs{z \in Y \bigg| (y, z) \in \bigcup_{n \in \natp}E_n} = \bigcup_{n \in \natp}\bracs{z \in Y | (y, z) \in E_n}
+    \]
+
+    so $\bigcup_{n \in \natp}E_n \in \alg$. Therefore $\alg$ is a $\sigma$-algebra, and $\alg = \cm \otimes \cn$. 
+
+    (2): For any $F \in \cf$, 
+    \[
+    \bracs{y \in Y|f(x, \cdot) \in F} = \bracs{y \in Y| (x, y) \in f^{-1}(F)} \in \cn
+    \]
+
+    by (1). 
+\end{proof}
+
+\begin{theorem}[Fubini-Tonelli Theorem]
+\label{theorem:fubini-tonelli}
+    Let $(X, \cm, \mu)$ and $(Y, \cn, \nu)$ be $\sigma$-finite measure spaces, then
+    \begin{enumerate}
+        \item For any $f \in L^+(X \times Y)$, $[x \mapsto \int f(x, y)\nu(dy)] \in L^+(X)$, $[y \mapsto \int f(x, y)\mu(dx)] \in L^+(Y)$, and
+        \begin{align*}
+        \int_{X \times Y}f(z)\mu \otimes \nu(dz) &= \int_X\int_Y f(x, y)\nu(dy)\mu(dx) \\
+        &= \int_{Y}\int_X f(x, y)\mu(dx)\nu(dy)
+        \end{align*}
+        
+        
+        \item For any normed space $E$ and $f \in L^1(X \times Y; E)$, 
+        \begin{enumerate}
+            \item[(a)] For almost every $x \in X$, $f(x, \cdot) \in L^1(Y; E)$. For almost every $y \in Y$, $f(\cdot, y) \in L^1(X; E)$.
+            \item[(b)] $[x \mapsto \int f(x, y)\nu(dy)] \in L^1(X; E)$ and $[y \mapsto \int f(x, y)\mu(dx)] \in L^1(Y; E)$.
+            \item[(c)] \begin{align*}
+        \int_{X \times Y}f(z)\mu \otimes \nu(dz) &= \int_X\int_Y f(x, y)\nu(dy)\mu(dx) \\
+        &= \int_{Y}\int_X f(x, y)\mu(dx)\nu(dy)
+        \end{align*}
+        \end{enumerate}
+
+    \end{enumerate}
+
+    
+\end{theorem}
+\begin{proof}
+    (1): First suppose that $\mu$ and $\nu$ are both finite. Let $\alg \subset \cm \otimes \cn$ be the collection of sets whose indicator functions satisfy (1), then $\alg$ contains all rectangles with finite measure. By linearity of the integral, for any $E, F\in \alg$, $E \setminus F \in \alg$. For any $\seq{E_n} \subset \alg$ and $E \in \alg$ such that $E_n \upto E$, by the \hyperref[Monotone Convergence Theorem]{theorem:mct},
+    \begin{align*}
+        \int_{X \times Y}\one_{E}\mu \otimes \nu(dz) &= \limv{n}\int_{X \times Y}\one_{E_n}\mu \otimes \nu(dz) \\
+        &= \limv{n}\int_X\int_Y \one_{E_n}(x, y)\nu(dy)\mu(dx) \\
+        &= \int_X\int_Y \one_{E_n}(x, y)\nu(dy)\mu(dx)
+    \end{align*}
+
+    so $\alg$ is a $\lambda$-system. By \hyperref[Dynkin's $\pi$-$\lambda$ Theorem]{theorem:pi-lambda}, $\alg = \cm \otimes \cn$. 
+
+    Now suppose that $\mu$ and $\nu$ are $\sigma$-finite, then $\alg$ contains all sets in $\cm \otimes \cn$ with finite measure. Since $\mu$ and $\nu$ are $\sigma$-finite, there exists rectangles $\seq{A_n \times B_n} \subset \alg$ such that $\mu \otimes \nu(A_n \times B_n)< \infty$ for all $n \in \natp$ and $A_n \times B_n \upto X \times Y$. Let $A \in \cm \otimes \cn$, then by the Monotone Convergence Theorem, 
+    \begin{align*}
+        \mu \otimes \nu(A) &= \limv{n}\mu \otimes \nu(A \cap (A_n \times B_n)) \\
+        &= \limv{n}\int_X\int_Y \one_{A \cap (A_n \times B_n)}(x, y)\nu(dy)\mu(dx) \\
+        &= \int_X\int_Y \one_{A}(x, y)\nu(dy)\mu(dx)
+    \end{align*}
+
+    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 $f$. 
+\end{proof}
+
+

src/measure/measure/regular.tex

@@ -32,5 +32,6 @@
         \item[(b)] Every open set of $X$ is $\sigma$-compact.
         \item[(c)] For any $K \subset X$ compact, $\mu(K) < \infty$.
     \end{enumerate}
+
     then $\mu$ is a regular measure.
 \end{theorem}

src/measure/measure/semifinite.tex

@@ -12,6 +12,7 @@
         \]
 
     \end{enumerate}
+
     If the above holds, then $\mu$ is a \textbf{semifinite measure}.
 \end{definition}
 \begin{proof}

src/measure/measure/sigma-finite.tex

@@ -8,5 +8,6 @@
         \item There exists $\seq{E_n} \subset \cm$ pairwise disjoint such that $\bigsqcup_{n \in \nat}E_n = X$ and $\mu(E_n) < \infty$ for all $n \in \nat$.
         \item There exists $\seq{E_n} \subset \cm$ such that $E_n \upto X$ and $\mu(E_n) < \infty$ for all $n \in \nat$.
     \end{enumerate}
+
     If the above holds, then $\mu$ is a \textbf{$\sigma$-finite measure}.
 \end{definition}

src/measure/notation.tex

@@ -0,0 +1,34 @@
+\chapter{Notations}
+\label{chap:measure-notations}
+
+\begin{tabular}{lll}
+    \textbf{Notation} & \textbf{Description} & \textbf{Source} \\
+    \hline
+
+    $\sigma(\mathcal{E})$ & $\sigma$-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} \\
+    % ---- 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} \\
+    $\chi_E = \mathbf{1}_E$ & 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} \\
+    $\mathcal{L}^+(X, \mathcal{M})$ & Space of non-negative measurable functions. & \autoref{definition:measurable-non-negative} \\
+    $f_*\mu$ & Pushforward of $\mu$ by $f$. & \autoref{definition:pushforward-measure} \\
+    $\mu \otimes \nu$ & Product measure. & \autoref{definition:product-measure} \\
+    $|\mu|$ & Total variation measure of a signed/vector measure. & \autoref{definition:total-variation-signed}, \autoref{definition:total-variation-vector} \\
+    $\mu = \mu^+ - \mu^-$ & Jordan decomposition of a signed measure. & \autoref{theorem:jordan-decomposition} \\
+    $\mu \perp \nu$ & Mutual singularity. & \autoref{definition:mutually-singular} \\
+    $\nu \ll \mu$ & $\nu$ is absolutely continuous w.r.t. $\mu$. & \autoref{definition:absolutely-continuous} \\
+    $M(X, \mathcal{M}; E)$,  & Space of finite $E$-valued measures. & \autoref{definition:vector-measure-finite-space} \\
+    $\|\mu\|_{\mathrm{var}}$ & Total variation of $\mu$. & \autoref{definition:vector-measure-finite-space} \\
+    $M_R(X; E)$ & Space of finite Radon $E$-valued measures on $X$. & \autoref{definition:space-radon-measures} \\
+    % ---- Lebesgue Integral ----
+    $\mathcal{L}^p(X, \mathcal{M}, \mu; E)$ & Space of $p$-integrable functions, without quotient. & \autoref{definition:lp-unequivalence} \\
+    $\|f\|_{L^p}$, & $L^p$ norm of $f$. & \autoref{definition:esssup} \\
+    $L^p(X, \mathcal{M}, \mu; E)$ & Space of $p$-integrable functions, modulo equality almost everywhere. & \autoref{definition:lp} \\
+
+\end{tabular}
+

src/measure/radon/c0.tex

@@ -0,0 +1,188 @@
+\section{Dual of $C_0$}
+\label{section:dual-of-c0}
+
+\begin{lemma}[Jordan Decomposition]
+\label{lemma:positive-linear-jordan}
+    Let $X$ be a topological space and $I \in C_0(X; \real)^*$, then there exists positive linear functionals $I^+, I^- \in C_0(X; \real)^*$ such that:
+    \begin{enumerate}
+        \item $I = I^+ - I^-$.
+        \item $I^+ \perp I^-$. 
+        \item $\norm{I^+}_{C_0(X; \real)^*}, \norm{I^-}_{C_0(X; \real)^*} \le \norm{I}_{C_0(X; \real)^*}$.
+    \end{enumerate}
+
+    
+\end{lemma}
+\begin{proof}
+    (1), (2): Since $C_0(X; \real)$ is a Banach lattice, $C_0(X; \real)^*$ is also a Banach lattice by \autoref{proposition:banach-lattice-dual}. Therefore there exists positive linear functionals $I^+, I^- \in C(X; \real)^*$ such that $I = I^+ - I^-$ and $I^+ \perp I^-$. 
+
+    (3): By \autoref{proposition:banach-lattice-properties}.
+\end{proof}
+
+\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. 
+\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 $K$. 
+\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. 
+\end{proof}
+
+
+\begin{theorem}[Riesz Representation Theorem]
+\label{theorem:riesz-radon-c0}
+    Let $X$ be an LCH space. For each $\mu \in M_R(X; \complex)$, let
+    \[
+    I_\mu: C_0(X; \complex) \to \complex \quad \dpn{f, I_\mu}{C_0(X; \complex)} = \int f d\mu
+    \]
+    
+    then the map
+    \[
+    M_R(X; \complex) \to C_0(X; \complex) \quad \mu \mapsto I_\mu
+    \]
+
+    is an isometric isomorphism. 
+\end{theorem}
+\begin{proof}[Proof {{\cite[Theorem 7.17]{Folland}}}. ]
+    Let $f \in C_0(X; \complex)$, then
+    \[
+    \abs{\int f d\mu} \le \int \norm{f}_u d\mu \le \norm{f}_u \cdot |\mu|(X)
+    \]
+
+    so $\norm{I_\mu}_{C_0(X; \complex)} \le \norm{\mu}_{\text{var}}$. 
+    
+    On the other hand, let $\seqf{A_j} \subset \cb_X$ such that $X = \bigsqcup_{j = 1}^nA_j$. Let $\eps > 0$, then by \autoref{proposition:radon-regular-sigma-finite} applied to $|\mu|$, there exists compact sets $\seqf{K_j}$ such that for each $1 \le j \le n$, $K_j \subset A_j$ and $|\mu(K_j) - \mu(A_j)| < \eps$. By outer regularity and \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, there exists $\seqf{\phi_j} \subset C_c(X; [0, 1])$ with disjoint support such that for each $1 \le j \le n$, $\norm{\phi_j - \one_{K_j}}_{L^1(|\mu|)} < \eps$. Let $\phi = \sum_{j = 1}^n \ol{\sgn(\mu(K_j))}\phi_j$, then $\norm{\phi}_u \le 1$ and
+    \[
+        \abs{\sum_{j = 1}^n |\mu(A_j)| - \int \phi d\mu} \le \sum_{j = 1}^n |\mu(A_j) - \mu(K_j)| + \sum_{j = 1}^n \norm{\phi_j - \one_{K_j}}_{L^1(|\mu|)} < 2n\eps
+    \]
+
+    so
+    \[
+    \abs{\int \phi d\mu} \ge \sum_{j = 1}^n |\mu(A_j)| - 2n\eps
+    \]
+
+    As such a $\phi$ exists for all $\eps > 0$, $\norm{I_\mu}_{C_0(X; \complex)} \ge \sum_{j = 1}^n |\mu(A_j)|$. Since this holds for all such partitions, $\norm{I_\mu}_{C_0(X; \complex)} \ge |\mu|(X)$. Therefore the map $\mu \mapsto I_\mu$ is isometric. 
+
+    Finally, let $I \in C_0(X; \complex)^*$, then there exists bounded linear functionals $I_r, I_i \in C_0(X; \real)^*$ such that for any $f \in C_0(X; \real)$, 
+    \[
+    \dpn{f, I}{C_0(X; \real)^*} = \dpn{f, I_r}{C_0(X; \real)} + i\dpn{f, I_i}{C_0(X; \real)}
+    \]
+
+    By \autoref{lemma:positive-linear-jordan}, there exists bounded positive linear functionals $I_r^+, I_r^-, I_i^+, I_i^-$ such that for any $f \in C_0(X; \real)$, 
+    \begin{align*}
+    \dpn{f, I}{C_0(X; \real)} &= \dpn{f, I_r^+}{C_0(X; \real)} - \dpn{f, I_r^-}{C_0(X; \real)} \\
+    &+ i\dpn{f, I_i^+}{C_0(X; \real)} - i\dpn{f, I_i^-}{C_0(X; \real)}
+    \end{align*}
+
+    Thus by the \hyperref[Riesz Representation Theorem]{theorem:riesz-radon}, there exists finite Radon measures $\mu_r^+, \mu_r^-, \mu_i^+, \mu_i^- \in M_R(X; \complex)$ such that for any $f \in C_0(X; \real)$, 
+    \[
+    \dpn{f, I}{C_0(X; \real)} = \int f d\mu_r^+ - \int f d\mu_r^- + i\int f d\mu_i^+ - i\int f d\mu_i^-
+    \]
+
+    Let $\mu = \mu_r^+ - \mu_r^- + i\mu_i^+ - i\mu_i^-$, then $I = I_\mu$, and the map $\mu \mapsto I_\mu$ is surjective. 
+    
+    
+\end{proof}
+
+
+\begin{theorem}[Singer's Representation Theorem]
+\label{theorem:singer-representation}
+    Let $X$ be an LCH space and $E$ be a normed space over $K \in \RC$. For each $\mu \in M_R(X; E^*)$, let
+    \[
+    I_\mu: C_0(X; E) \to K \quad \dpn{f, I_\mu}{C_0(X; E)} = \int \dpn{f, d\mu}{E}
+    \]
+    
+    then the map
+    \[
+    M_R(X; E^*) \to C_0(X; E)^* \quad \mu \mapsto I_\mu
+    \]
+
+    is an isometric isomorphism. 
+\end{theorem}
+\begin{proof}[Proof {{\cite{HensgenSinger}}}. ]
+    (Isometric): Let $\mu \in M_R(X; E^*)$, then for any $f \in C_0(X; E)$, 
+    \[
+    |\dpn{f, I_\mu}{C_0(X; E)}| \le \int \norm{f}_E d|\mu| \le \norm{f}_u \cdot \norm{\mu}_{\text{var}}
+    \]
+
+    so $\norm{I_\mu}_{C_0(X; E)^*} \le \norm{\mu}_{\text{var}}$. 
+    
+    On the other hand, let $\seqf{A_j} \subset \cb_X$ such that $\bigsqcup_{j = 1}^n A_j = X$ and $\eps > 0$.
+
+    By \autoref{proposition:radon-regular-sigma-finite} applied to $|\mu|$, there exists $\seqf{K_j}$ compact such that for each $1 \le j \le n$, $K_j \subset A_j$ and $\norm{\mu(K_j) - \mu(A_j)}_{E^*} < \eps$. By outer regularity and \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, there exists $\seqf{\phi_j} \subset C_c(X; [0, 1])$ with disjoint support such that for each $1 \le j \le n$, $\norm{\phi_j - \one_{K_j}}_{L^1(|\mu|)} < \eps$. 
+    
+    Let $\seqf{x_j} \subset \overline{B_E(0, 1)}$ such that for each $1 \le j \le n$, $\dpn{x_j, \mu(K_j)}{E} > \norm{\mu(K_j)}_{E^*} - \eps$. Define $\phi = \sum_{j = 1}^n x_j \phi_j$, then $\norm{\phi}_u \le 1$ and
+    \begin{align*}
+    \abs{\sum_{j = 1}^n \norm{\mu(A_j)}_{E^*} - \int \dpn{\phi, d\mu}{E}} &\le \sum_{j = 1}^n \norm{\mu(A_j) - \mu(K_j)}_{E^*} \\
+    &+ \sum_{j = 1}^n \norm{\phi_j - \one_{K_j}}_{L^1(|\mu|)} + n\eps < 3n\eps
+    \end{align*}
+    
+    
+    As such a $\phi \in C_0(X; E)$ exists for all $\eps > 0$ and $\seqf{A_j}$, $\norm{I_\mu}_{C_0(X; E)^*} \ge \norm{\mu}_{\text{var}}$. Therefore the map $\mu \mapsto I_\mu$ is isometric. 
+
+    (Surjective): Let $B = \bracsn{\phi \in E^*|\norm{\phi}_{E^*} \le 1}$ and equip it with the weak*-topology and
+    \[
+    T: C_0(X; E) \to C_0(X \times B; K) \quad (Tf)(x, \phi) = \dpn{f(x), \phi}{E}
+    \]
+
+    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 $f \in C_0(X \times B; K)$, 
+    \[
+    \dpn{f, \ol I}{C_0(X \times B; K)} = \int_{X \times B} f d\mu
+    \]
+    
+    Now, let
+    \[
+    \nu: \cb_X \to E^* \quad \dpn{y,\nu(A)}{E} = \int_{X \times B} \one_A(x) \cdot \dpn{y, \phi}{E} \mu(dx, d\phi)
+    \]
+
+
+    then for each $A \in \cb_X$ and $y \in E$, 
+    \[
+    |\dpn{y,\nu(A)}{E}| \le \int_{X \times B} \one_A(x) \cdot \norm{y}_E |\mu|(dx, d\phi) \le \norm{y}_E \cdot |\mu|(A \times B)
+    \]
+
+    As the above holds for all $y \in E$, $\norm{\nu(A)}_{E^*} \le |\mu|(A \times B)$. Moreover, for any pairwise disjoint sequence $\seq{A_n} \subset \cb_X$ and $A \in \cb_X$ such that $A = \bigsqcup_{n \in \nat}A_n$, 
+    \[
+    \limv{n}\norm{\nu(A) - \sum_{j = 1}^n \nu(A_n)}_{E^*} \le \limv{n}|\mu|\paren{\bigcup_{k > n}A_k \times B} = 0
+    \]
+
+    so $\nu$ is a vector measure on $\cb_X$. 
+    
+    Since $\norm{\nu(A)}_{E^*} \le |\mu|(A \times B)$ for all $A \in \cb_X$, $|\nu|(A) \le |\mu|(A \times B)$ for all $A \in \cb_X$, and $\nu$ is a Radon measure by \autoref{lemma:radon-compact-project}.
+
+    Finally, let $f \in C_0(X; K)$ and $y \in E$, then
+    \begin{align*}
+        \int_X \dpn{y \cdot f, d\nu}{E} &= \int_{X \times B} f(x)\dpn{y, \phi}{E} \mu(dx, d\phi) \\
+        &= \int_{X \times B}T(y \cdot f)(x, \phi)\mu(dx, d\phi) = \dpn{y \cdot f, I}{C_0(X; E)}
+    \end{align*}
+
+    Therefore for any $f \in C_0(X; K) \otimes E$, $\int_X \dpn{f, d\nu}{E} = \dpn{f, I}{C_0(X; E)}$. By \autoref{proposition:c0-tensor}, $C_0(X; K) \otimes E$ is a dense subspace of $C_0(X; E)$, so
+    \[
+    \int_X \dpn{f, d\nu}{E} = \dpn{f, I}{C_0(X; E)} \quad \forall f \in C_0(X; E)
+    \]
+    
+\end{proof}
+
+\begin{remark}
+\label{remark:singer-representation}
+    In the proof of \hyperref[Singer's Representation Theorem]{theorem:singer-representation}, the $E^*$-valued measure is constructed pointwise as
+    \[
+    \nu: \cb_X \to E^* \quad \dpn{y,\nu(A)}{E} = \int_{X \times B} \one_A(x) \cdot \dpn{y, \phi}{E} \mu(dx, d\phi)
+    \]
+
+    It may be tempting to use the strong formulation directly
+    \[
+    \nu: \cb_X \to E^* \quad \nu(A) = \int_{X \times B} \phi \cdot \one_A(x) \mu(dx, d\phi)
+    \]
+
+    However, without additional assumptions on $E^*$, $\phi \cdot \one_A(x)$ may not be strongly measurable, which prevents this direct use of the Bochner integral. Thus the weak formulation is a necessary complication in the proof. 
+\end{remark}
+
+

src/measure/radon/index.tex

@@ -3,4 +3,5 @@
 
 \input{./radon.tex}
 \input{./riesz.tex}
+\input{./c0.tex}