Fixed typo.

.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

@@ -1,4 +1,4 @@
-%\documentclass{report}
+\documentclass{}
 \usepackage{amssymb, amsmath, hyperref}
 \usepackage{preamble}
 
@@ -11,10 +11,10 @@ Hello this is all my notes.
 \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

@@ -224,3 +224,8 @@
 
 % Real or Complex Numbers
 \newcommand{\RC}{\bracs{\real, \complex}}
+
+% Convex Stuff
+\newcommand{\conv}{\text{Conv}}
+\newcommand{\aconv}{\text{AbsConv}}
+

refs.bib

@@ -112,3 +112,13 @@
   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}
+}

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/tensor.tex

@@ -34,6 +34,7 @@
         \]
     \end{enumerate}
 
+    
     The module $A = \bigoplus_{i \in I}A_i$ is the \textbf{direct sum} of $\seqi{A}$. 
 \end{definition}
 \begin{proof}
@@ -206,6 +207,7 @@
         \]
     \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

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}
 

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/notation/index.tex

@@ -13,6 +13,7 @@
     $\lim_{\longleftarrow} A_i$ & 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}
+    $M(x)$ & 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

@@ -44,6 +44,7 @@
         \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 $x + y \in (0, 1)$, $\phi(x) + \phi(y) \le \phi(x + y)$.

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

@@ -1,24 +1,60 @@
 \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{enumerate}
-        \item There exists $V \in \cn_E(x_0)$ such that $f$ is $(n-1)$-fold differentiable on $V$.
-        \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},
+\begin{definition}[Codomain of Derivatives]
+\label{definition:higher-derivatives-codomain}
+    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
     \[
-    D_\sigma^{n}f: U \to B^{n-1}_\sigma(E; F)
+    L^{(n)}_\sigma(E; F) = L(E; L^{(n-1)}_\sigma(E; F)) \subset B_\sigma^n(E; F)
     \]
 
-    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{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}
@@ -30,7 +66,7 @@
     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) \\
@@ -87,55 +123,79 @@
 
 \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}[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]
-\label{proposition:multilinear-derivative}
-    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
+\begin{theorem}[Power Rule]
+\label{theorem:power-rule}
+    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 $1 \le k \le n$ and $x, h \in E$,
+        \item For each $1 \le m \le n$, $x \in E$, and $h \in E^m$, 
         \[
-        Df(x)(h_1, \cdots, h_k) = \frac{n!}{(n-k)!} T(x^{(n-k)}, h_1, \cdots, h_k)
+        D^{m}_\sigma f(x)(h) = \sum_{\phi \in \text{Inj}([m]; [n])}T((x^{(n-m)}, h)_\phi]
         \]
 
-        In particular, $D^kf = n! \cdot T$.
-        \item For each $k > n$ and $x \in E$, $Df(x) = 0$.
+        In particular, 
+        \[
+        D^{n}_\sigma f(x)(h) = \sum_{\phi \in S_n}T(h_{\phi(1)}, \cdots, h_{\phi(n)})
+        \]
+
+        \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)}) \\
-        &= f(x) + (n-k)D^k_\sigma f(x^{(n-k-1)}, h) \\
-        &+ \underbrace{\sum_{r = 2}^{n-k}{n - k \choose r}D^k_\sigma f(x^{(n-k-r)}, h^{(r)})}_{r(h)}
+        D^{m}_\sigma f(x + h)(k) &= \sum_{\phi \in \text{Inj}([m]; [n])}T[((x+h)^{(n-m)}, k)_\phi] \\
+        &= \sum_{\phi \in \text{Inj}([m]; [n])}\sum_{S \subset [n-m]}T[((x, h)_S, k)_\phi]
     \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.
-
-    (3): Since $D^n_\sigma f$ is constant, $D^k_\sigma f = 0$ for all $k > n$.
+    (2): By (1), $D^n_\sigma f$ is constant. 
 \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 $y \in V$,
     \[
     f(y) - f(x) \in \overline{\text{Conv}\bracs{Df(z)(y - x)|z \in (x, y)}}
     \]
@@ -111,7 +113,7 @@
     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}, $f(y) - f(x) = g(1) - g(0)$ is contained in
     \[
@@ -122,7 +124,7 @@
 
 \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 $\mathcal{R}_\sigma^1(E; F)$.
@@ -16,43 +17,44 @@
 
 \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}[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)
@@ -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{proposition}
-\label{proposition:derivative-sets-real}
-    Let $E$ be a separated topological vector space and $\sigma \subset B(\real)$ be an upward-directed system that contains finite sets, then
-    \begin{enumerate}
-        \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 $x_0$.
+\begin{theorem}[Interchange of Limits and Derivatives]
+\label{theorem:differentiable-uniform-limit}
+    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:
+    \begin{enumerate}[label=(\alph*)]
+        \item There exists $f: U \to F$ such that $\fF \to f$ pointwise. 
+        \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 $x + [0, 1]A$. 
     \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) \\
-        \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)
+        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)} \\
+        &+ (f - \varphi)(x + h) - (f - \varphi)(x) \\
+        &+ (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

@@ -62,7 +62,7 @@
 
 \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
     \[
     g(x_0 + h) = g(x_0) + \sum_{k = 1}^n \frac{1}{k!}D^k_\sigma f(x_0)(h^{(k)}) + r(h)
     \]
@@ -74,7 +74,7 @@
     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

@@ -1,5 +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

@@ -11,6 +11,11 @@ Differential geometry is the study of things invariant under change of notation.
     $\mathcal{R}_\sigma^n(E; F)$, $\mathcal{R}_\sigma(E;F)$ & $\sigma$-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} \\
-    $x^{(k)}$ & Tuple of $x$ repeated $k$ times. & \autoref{proposition:multilinear-derivative} \\
-    $D^+f(x)$ & Right derivative of $f$ at $x$. & \autoref{definition:right-differentiable-mvt} \\
+    $D_\sigma^n(U; F)$ & $n$-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

@@ -8,4 +8,5 @@
 \input{./rs/index.tex}
 \input{./lp/index.tex}
 \input{./order/index.tex}
+\input{./duality/index.tex}
 \input{./notation.tex}

src/fa/lc/barrel.tex

@@ -25,14 +25,14 @@
 
 \begin{summary}
 \label{summary:barreled-space}
-    The following types of locally convex spaces are barreled:
+    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 barreled spaces. 
+        \item Inductive limits of barrelled spaces. 
         \item Spaces of type (LB) and (LF). 
-        \item The locally convex direct sum of barreled spaces. 
-        \item Products of barreled spaces.
+        \item The locally convex direct sum of barrelled spaces. 
+        \item Products of barrelled spaces.
     \end{enumerate}
 \end{summary}
 \begin{proof}
@@ -46,7 +46,7 @@
 
 \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 barreled. 
+    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, 
@@ -59,9 +59,9 @@
 
 \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 barreled.  
+    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 barreled. 
+    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

@@ -9,6 +9,7 @@
         \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{bornological space}.
 \end{definition}
 \begin{proof}
@@ -66,10 +67,10 @@
 \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 $B(E, \mathcal{T}_B) \supset B(E, \mathcal{T})$. 
-        \item[(U)] For every $\mathcal{S}$ satisfying (1), $\mathcal{S} \subset \mathcal{T}_B$. In particular, $\mathcal{T} \subset \mathcal{T}_B$ and $B(E, \mathcal{T}_B) = B(E, \mathcal{T})$. 
+        \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 $\mathcal{B} \subset B(E, \topo)$ 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 $\bracsn{\iota_B}_{B \in \mathcal{B}}$. 
+        \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} $E$. 
@@ -77,14 +78,14 @@
 \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 $\mathcal{T}_B \supset \mathcal{T}$, $B(E, \mathcal{T}) \supset B(E, \mathcal{T}_B)$. 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 $B(E, \mathcal{T}) = B(E, \mathcal{S})$ for all $\mathcal{S} \in \mathscr{S}$, $\bigcap_{\mathcal{S} \in \mathscr{S}}U_{\mathcal{S}}$ absorbs $B$, so $U$ absorbs $B$ as well. 
+    (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 $B \in B(E, \mathcal{T})$, $U \in \cn_{\mathcal{S}}(0)$, 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), $\mathcal{T}_B \supset \mathcal{S}$. 
+    (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 $B(E, \mathcal{T}_B) = B(E, \mathcal{T})$, 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}, $\mathcal{S} \supset \mathcal{T}_B$. 
+    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/convex.tex

@@ -21,7 +21,7 @@
 \label{definition:convex-circled-hull}
     Let $E$ be a vector space over $K \in \RC$ and $A \subset E$, then the set
     \[
-    \Gamma(A) = \bracs{\sum_{j = 1}^n t_j x_j \bigg | \seqf{t_j} \subset K, \seqf{x_j} \subset E, \sum_{j = 1}^n |t_j| \le 1 }
+    \aconv(A) = \bracs{\sum_{j = 1}^n t_j x_j \bigg | \seqf{t_j} \subset K, \seqf{x_j} \subset E, \sum_{j = 1}^n |t_j| \le 1 }
     \]
 
     is the \textbf{convex circled hull} of $A$. 
@@ -139,6 +139,7 @@
         \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 $\bracsn{[\cdot]_j}_{j \in J}$ 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 $\seqi{[\cdot]}$.
@@ -160,6 +161,7 @@
         \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}[start=4]
         \item If $A$ is convex, then $[\cdot]_A$ is a sublinear functional.
@@ -190,6 +192,7 @@
         \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}

src/fa/lc/hahn-banach.tex

@@ -135,6 +135,7 @@
             \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 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}

src/fa/lc/inductive.tex

@@ -22,6 +22,7 @@
 
         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}
@@ -66,6 +67,7 @@
         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}
@@ -111,6 +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}
@@ -175,6 +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}
@@ -191,6 +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/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
-    \[
-    \bracs{[\cdot]_{S, i}|S \in \mathfrak{S}, i \in I}
-    \]
-
-    and hence locally convex.
+    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. 
 \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

@@ -19,12 +19,13 @@
         \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
         \[
-        \fB = \bracsn{\Gamma(U \otimes V)| U \in \cn_E(0), V \in \cn_F(0)}
+        \fB = \bracsn{\aconv(U \otimes V)| U \in \cn_E(0), V \in \cn_F(0)}
         \]
 
         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 $E \otimes_\pi F$.
@@ -42,9 +43,9 @@
 
     (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, $\sum_{j = 1}^n x_j \otimes y_j \subset \lambda \Gamma (U \otimes V)$, 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 $\fB$. 
+    (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, $W \supset \Gamma(U \otimes V)$, so $\fB$ is a fundamental system of neighbourhoods at $0$ for $E \otimes_\pi F$. 
+    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}
@@ -64,7 +65,7 @@
     then
     \begin{enumerate}
         \item $\rho$ is a continuous seminorm on $E \otimes_\pi F$.
-        \item $\rho$ is the gauge of $\Gamma(U \otimes V)$. 
+        \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}
@@ -74,6 +75,7 @@
         \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 $\seqf{(x_j,y_j)} \subset E \times F$, 
@@ -100,20 +102,20 @@
 
     so $\rho$ satisfies the triangle inequality. 
 
-    (2): Let $z \in \Gamma(U \otimes V)$, 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, 
+    (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 $\Gamma(U \otimes V) \subset \bracs{\rho < 1}$.
+    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
     \[
-    z = \sum_{j = 1}^n (p(x_j) + \eps)(q(x_j) + \eps) \cdot \underbrace{\frac{x_j}{p(x_j) + \eps}}_{\in \bracs{p < 1} = U} \otimes \underbrace{\frac{y_j}{q(x_j) + \eps}}_{\in \bracs{q < 1} = V} \in \Gamma(U \otimes V)
+    z = \sum_{j = 1}^n (p(x_j) + \eps)(q(x_j) + \eps) \cdot \underbrace{\frac{x_j}{p(x_j) + \eps}}_{\in \bracs{p < 1} = U} \otimes \underbrace{\frac{y_j}{q(x_j) + \eps}}_{\in \bracs{q < 1} = V} \in \aconv(U \otimes V)
     \]
 
-    and $\Gamma(U \otimes V) \supset \bracs{\rho < 1}$.
+    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
     \[
@@ -142,6 +144,7 @@
         \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 $E \td{\otimes}_\pi F$. 
@@ -164,6 +167,7 @@
         \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}

src/fa/lp/definition.tex

@@ -109,6 +109,7 @@
         \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}

src/fa/norm/absolute.tex

@@ -14,6 +14,7 @@
         \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 $k \in \natp$. 
@@ -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/multilinear.tex

@@ -9,6 +9,7 @@
         \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

@@ -39,11 +39,13 @@
         \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}
@@ -71,20 +73,14 @@
 \label{theorem:uniform-boundedness}
     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 $E$ is a Banach space.
+        \item[(B)] $E$ 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 $\sup_{x \in U}\sup_{T \in \mathcal{T}}\norm{Tx}_{F} < \infty$.
-
-    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 $\sup_{T \in \mathcal{T}}\norm{T}_{L(E; F)} \le 2n/r$.
+    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, $\sup_{T \in \mathcal{T}}\norm{T}_{L(E; F)} \le r^{-1}$. 
 \end{proof}
 
 

src/fa/norm/separable.tex

@@ -31,6 +31,7 @@
         \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}.

src/fa/notation.tex

@@ -8,7 +8,7 @@
     $E_A$ & 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} \\
-    $B(E)$ & Bounded subsets of TVS $E$. & \autoref{definition:bounded} \\
+    $\mathfrak{B}(E)$ & 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} \\
@@ -20,7 +20,7 @@
     $\widehat{E}$ & Hausdorff completion of TVS $E$. & \autoref{definition:tvs-completion} \\
     % ---- Locally Convex ----
     $\mathrm{Conv}(A)$ & Convex hull of $A$. & \autoref{definition:convex-hull} \\
-    $\Gamma(A)$ & Convex circled hull of $A$. & \autoref{definition:convex-circled-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} \\
@@ -43,8 +43,11 @@
     $T_{f,\rho}(x)$ & 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/lattice.tex

@@ -9,6 +9,7 @@
         \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}
@@ -19,6 +20,7 @@
         \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 $\sup(A) + \sup(B) = \sup(A + B)$.
@@ -105,11 +107,13 @@
         \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},
@@ -177,6 +181,7 @@
         \item For any $x, y \in C$, $\phi(x + y) = \phi(x) + \phi(y)$.
     \end{enumerate}
 
+    
     then the mapping
     \[
     \Phi: E \to \real \quad x \mapsto \phi(x^+) - \phi(x^-)
@@ -220,6 +225,7 @@
         \]
     \end{enumerate}
 
+    
 \end{proposition}
 \begin{proof}
     (1): Let $C = \bracs{x \in E|x \ge 0}$, $\Phi^+ \in E^b$, and

src/fa/rs/bv.tex

@@ -68,6 +68,7 @@
         \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. 
         \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 vector space, then
     \begin{enumerate}
         \item[(5)] $f$ has at most countably many discontinuities.
@@ -88,6 +89,7 @@
         \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).

src/fa/rs/index.tex

@@ -5,5 +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/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

@@ -4,7 +4,7 @@
 
 \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 linear 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 $G: [a, b] \to F$ 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
     \[
@@ -35,9 +35,9 @@
 
 \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 linear 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 $G: [a, b] \to F$ 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 norm, then:
+    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 $E$. 
@@ -106,4 +106,3 @@
     (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

@@ -1,19 +1,15 @@
-\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 $[a, b] \subset \real$, $E, F, H$ be locally convex spaces, and $E \times F \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map, and $G: [a, b] \to F$.
-
-    Let $[\cdot]_H$ be a continuous seminorm on $H$, then there exists continuous seminorms $[\cdot]_E$ on $E$ and $[\cdot]_F$ on $F$ such that for any $f \in RS([a, b], G)$,
+    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 $y \in F$, 
     \[
     \braks{\int_a^bf dG}_H \le \sup_{x \in [a, b]}[f]_E \cdot [g]_{\text{var}, F}
     \]
 
 \end{proposition}
 \begin{proof}
-    By \autoref{proposition:tvs-convex-multilinear}, there exists continuous seminorms $[\cdot]_E$ on $E$ and $[\cdot]_F$ on $F$ such that $[xy]_H \le [x]_E[y]_F$ for all $(x, y) \in E \times F$.
-
     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 \\
@@ -24,7 +20,7 @@
 
 \begin{proposition}
 \label{proposition:rs-complete}
-    Let $[a, b] \subset \real$, $E, F, H$ be locally convex spaces, and $E \times F \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map, and $G \in BV([a, b]; F)$.
+    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 $E$ and $f: [a, b] \to E$, define
     \[
@@ -36,6 +32,7 @@
         \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.
@@ -56,10 +53,12 @@
         \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
@@ -74,11 +73,13 @@
     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}
@@ -99,3 +100,47 @@
 
     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.tex

@@ -61,7 +61,7 @@
 
 \end{theorem}
 \begin{proof}
-    Suppose that $f \in RS([a, b], G)$. Let $U \in \cn_K(0)$, 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
     \[
     Q_0 = [x_0, x_1, x_1, \cdots, x_n, x_n]
     \]
@@ -69,10 +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

@@ -9,7 +9,7 @@
         \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 $B(E) = B(E, \topo)$ is the set of all bounded sets of $E$.
+    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 $n \ge N$. 
@@ -19,7 +19,7 @@
 
 \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

@@ -22,11 +22,13 @@
         \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)
@@ -39,11 +41,13 @@
         \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}

src/fa/tvs/completion.tex

@@ -9,10 +9,12 @@
         \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}
@@ -49,6 +50,7 @@
         \]
 
     \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}
 

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}
@@ -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

@@ -8,6 +8,7 @@
         \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}[Proof {{\cite[Proposition 5.5]{Folland}}}. ]

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

@@ -83,6 +83,7 @@
         \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}
@@ -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.
@@ -21,6 +22,7 @@
         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}
@@ -75,6 +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

@@ -18,6 +18,7 @@
 
         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}
@@ -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}.

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

@@ -1,225 +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 $F^E$ 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/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

@@ -54,6 +54,7 @@
         \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}
@@ -78,6 +79,7 @@
         \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}
@@ -98,6 +100,7 @@
         \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}

src/measure/bochner-integral/strongly.tex

@@ -12,6 +12,7 @@
             \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. 
@@ -41,6 +42,7 @@
         \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}. 

src/measure/lebesgue-integral/complex.tex

@@ -105,6 +105,7 @@
         \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}[Proof {{\cite[Theorem 2.24]{Folland}}}. ]

src/measure/measurable-maps/metric.tex

@@ -77,6 +77,7 @@
         \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}
@@ -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/kolmogorov.tex

@@ -17,12 +17,14 @@
         \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.
@@ -53,6 +55,7 @@
         \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}[Proof {{\cite[Theorem 1.14]{Baudoin}}}. ]
@@ -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

@@ -45,6 +45,7 @@
 
         \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}[Proof {{\cite[Theorem 1.16]{Folland}}}. ]

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}),
@@ -165,6 +169,7 @@
         \]
     \end{enumerate}
 
+    
 \end{definition}
 \begin{proof}
     (1): Preimage commutes with unions, intersections, and complements. 

src/measure/measure/outer.tex

@@ -116,6 +116,7 @@
             \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}[Proof {{\cite[Theorem 1.14]{Folland}}}. ]

src/measure/measure/product.tex

@@ -10,6 +10,7 @@
         \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}
@@ -59,6 +60,7 @@
         \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
@@ -101,8 +103,10 @@
         &= \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},

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

@@ -7,6 +7,7 @@
 
     $\sigma(\mathcal{E})$ & $\sigma$-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} \\

src/measure/radon/c0.tex

@@ -10,6 +10,7 @@
         \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 $I^+ \perp I^-$. 

src/measure/radon/radon.tex

@@ -116,6 +116,7 @@
         \item[(a)] For any $U \subset X$ open, $U$ is $\sigma$-compact.
         \item[(b)] For any $K \subset X$ compact, $\mu(K) < \infty$.
     \end{enumerate}
+
     then $\mu$ is a regular measure on $X$.
 \end{proposition}
 \begin{proof}

src/measure/radon/riesz.tex

@@ -72,6 +72,7 @@
 
         As this holds for all $\eps > 0$, $\mu^*(E) \le \sum_{n \in \natp}\mu^*(E_n)$.
     \end{enumerate}
+
     Therefore $\mu^*: 2^E \to [0, \infty]$ is an outer measure.
 
     To see that all Borel sets are $\mu^*$-measurable, let $E \subset X$ and $U \in \topo$. First suppose that $E$ is open. Let $f \prec E \cap U$, then for any $g \prec E \setminus \supp{f}$, $\supp{f} \cap \supp{g} = \emptyset$ and $f + g \prec E$, so

src/measure/sets/algebra.tex

@@ -39,6 +39,7 @@
         \item For any $A, B \in \alg$, $A \cap B \in \alg$.
         \item For any $A, B \in \alg$, $A \setminus B \in \alg$.
     \end{enumerate}
+
     If $\alg$ is a $\sigma$-algebra, then:
     \begin{enumerate}
         \item[(1')] For any $\seq{A_n} \in \alg$, $\bigcap_{n \in \natp}A_n \in \alg$.

src/measure/sets/borel.tex

@@ -100,6 +100,7 @@
         \item $\bracsn{\ol{B(x, r)}|x \in X, r > 0}$. 
     \end{enumerate}
 
+    
 \end{proposition}
 \begin{proof}
     (1) $\subset$ (2): Let $U \subset X$ be open. By \autoref{definition:dense}, there exists a countable dense subset $S \subset U$. For each $x \in S$, let $r_x > 0$ such that $B(x, r) \subset U$, then $U = \bigcup_{x \in S}B(x, r_x)$ is a countable union of open balls.

src/measure/sets/elementary.tex

@@ -9,6 +9,7 @@
         \item[(P2)] For any $A, B \in \ce$, $A \cap B \in \ce$.
         \item[(E)] For any $E, F \in \ce$ with $E \subset F$, there exists $\seqf{E_j} \subset \ce$ such that $E \setminus F = \bigsqcup_{j = 1}^n E_j$.
     \end{enumerate}
+
     If $X \in \ce$, then (E) may be replaced with
     \begin{enumerate}
         \item[(E')] For any $E \in \ce$, there exists $\seqf{E_j} \subset \ce$ such that $E^c = \bigsqcup_{j = 1}^n E_j$.

src/measure/sets/lambda.tex

@@ -61,6 +61,7 @@
         \]
 
     \end{enumerate}
+
     so $\cm(\ce)$ is a $\lambda$-system.
 
     Since $\mathcal{P}$ is a $\pi$-system, $\cm(\mathcal{P}) \supset \mathcal{P}$, so $\cm(\mathcal{P}) = \lambda(\mathcal{P})$. Thus for any $E \in \lambda(\mathcal{P})$ and $F \in \lambda(\mathcal{P})$, $E \cap F \in \lambda(\mathcal{P})$. Therefore $\cm(\lambda(\mathcal{P})) \supset \mathcal{P}$, $\cm(\lambda(\mathcal{P})) = \lambda(\mathcal{P})$, and $\lambda(\mathcal{P})$ satisfies (P2). By \autoref{lemma:pi-lambda}, $\lambda(\mathcal{P})$ is a $\sigma$-algebra.

src/measure/vector/complex.tex

@@ -8,6 +8,7 @@
         \item[(M1)] $\mu(\emptyset) = 0$.
         \item[(M2)] For any $\seq{E_n} \subset \cm$ pairwise disjoint, $\mu\paren{\bigsqcup_{n \in \natp}E_n} = \sum_{n \in \natp} \mu(E_n)$ where the sum converges absolutely. 
     \end{enumerate}
+
     By \hyperref[Riemann's Rearrangement Theorem]{theorem:riemann-rearrangement}, (M2) implies that $\mu$ can only take at most one value in $\bracs{-\infty, \infty}$. 
 \end{definition}
 
@@ -103,6 +104,7 @@
         \item[(ii)] For each $1 \le k \le n - 1$, $A_{k+1} \subset A_k$ with $\mu(A_{k+1}) - \mu(A_k) > M_{k}/2$.
     \end{enumerate}
 
+    
     Let $A_n \in \cm$ with $A_{n+1} \subset A_n$ such that $\mu(A_{n+1}) - \mu(A_n) > M_n/2$. Since $A_{n+1} \cap P = \emptyset$ and $\mu(P) = M$, $A_{n+1}$ cannot be positive, so
     \[
     M_{n+1} = m(A_{n+1}) - \mu(A_{n+1}) > 0
@@ -159,6 +161,7 @@
         \item[(U)] For any other pair $(\nu^+, \nu^-)$ satisfying (1) and (2), $\mu^+ = \nu^+$ and $\mu^- = \nu^-$.
     \end{enumerate}
 
+    
 \end{theorem}
 \begin{proof}
     (1), (2): Let $X = P \sqcup N$ be a \hyperref[Hahn decomposition]{theorem:hahn-decomposition} of $\mu$. For each $E \in \cm$, let

src/measure/vector/variation.tex

@@ -98,6 +98,7 @@
         \item If $E$ is a Banach space, then so is $M(X, \cm; E)$.
     \end{enumerate}
 
+    
 \end{definition}
 \begin{proof}
     (1): For any $\mu, \nu \in M(X, \cm; E)$ and $\seqf{A_j} \subset \cm$ such that $X = \bigsqcup_{j = 1}^n A_j$, 
@@ -118,7 +119,5 @@
     \]
 
     by \hyperref[Fubini's theorem]{theorem:fubini-tonelli}.
-    
-    % TODO: Actually link Fubini once it's there.
 \end{proof}
 

src/measure/vector/vector.tex

@@ -9,6 +9,7 @@
         \item[(M2)] For any $\seq{A_n} \subset \cm$ pairwise disjoint, $\mu\paren{\bigsqcup_{n \in \natp}A_n} = \sum_{n \in \natp} \mu(A_n)$ where the sum converges absolutely. 
     \end{enumerate}
 
+    
 \end{definition}
 
 \begin{proposition}

src/op/banach/definitions.tex

@@ -0,0 +1,18 @@
+\section{Banach Algebras}
+\label{section:banach-algebras}
+
+\begin{definition}[Banach Algebra]
+\label{definition:banach-algebra}
+    Let $A$ be an associative algebra over $\complex$ and $\norm{\cdot}_A: A \to [0, \infty)$ be a norm, then $A$ is a \textbf{Banach algebra} if:
+    \begin{enumerate}
+        \item $A$ is complete with respect to $\norm{\cdot}_A$. 
+        \item For any $x, y \in A$, $\norm{xy}_A \le \norm{x}_A\norm{y}_A$. 
+    \end{enumerate}
+\end{definition}
+
+\begin{definition}[Unital Banach Algebra]
+\label{definition:unital-banach-algebra}
+    Let $A$ be a Banach algebra, then $A$ is \textbf{unital} if there exists $1 \in A$ such that for any $x \in A$, $x1 = 1x = x$. In which case, $1$ is the unique \textbf{multiplicative identity} of $A$. 
+\end{definition}
+
+

src/op/banach/index.tex

@@ -0,0 +1,5 @@
+\chapter{$C*$-Algebras}
+\label{chap:banach-algebras}
+
+\input{./definitions.tex}
+\input{./invertible.tex}

src/op/banach/invertible.tex

@@ -0,0 +1,52 @@
+\section{Invertible Elements}
+\label{section:invertible-elements}
+
+
+\begin{definition}[Invertible]
+\label{definition:banach-algebra-invertible}
+    Let $A$ be a unital Banach algebra and $x \in A$, then $x$ is \textbf{invertible} if there exists $x^{-1} \in A$ such that $xx^{-1} = x^{-1}x = 1$. The set $G(A)$ denotes the collection of all invertible elements in $A$. 
+\end{definition}
+
+\begin{lemma}
+\label{lemma:neumann-series}
+    Let $A$ be a unital banach algebra and $x \in B_A(1, 1)$, then $x \in G(A)$ with
+    \[
+    x^{-1} = \sum_{n = 0}^\infty (1 - x)^n
+    \]
+\end{lemma}
+\begin{proof}
+    Since $\norm{1 - x}_A < 1$, the series converges absolutely. Let $y = \sum_{n = 0}^\infty (1 - x)^n$, then
+    \[
+    (1 - x) \sum_{n = 0}^\infty (1 - x)^n = \sum_{n = 0}^\infty (1 - x)^n - 1 = \sum_{n = 0}^\infty (1 - x)^n (1 - x)
+    \]
+
+    so $(1 - x)y = y - 1 = y(1 - x)$, and $xy = yx = 1$. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:banach-algebra-inverse}
+    Let $A$ be a unital Banach algebra, then:
+    \begin{enumerate}
+        \item $G(A)$ is open. 
+        \item For any $x \in G(A)$ and $y \in B_A(0, \normn{x^{-1}}_A^{-1})$, 
+        \[
+        (x - y)^{-1} = x^{-1}\sum_{n = 0}^\infty (yx^{-1})^n
+        \]
+
+        \item The map $G(A) \to G(A)$ defined by $x \mapsto x^{-1}$ is $C^\infty$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (2): For any $x \in G(A)$ and $y \in B(0, \normn{x^{-1}}_A^{-1})$, $(x - y) = (1 - yx^{-1})x$. By \autoref{lemma:neumann-series}, 
+    \[
+    (1 - yx^{-1})^{-1} = \sum_{n = 0}^\infty (yx^{-1})^n
+    \]
+
+    so
+    \[
+    (x - y)^{-1} = x^{-1}\sum_{n = 0}^\infty (yx^{-1})^n
+    \]
+
+    (3): Since the inversion map is locally a power series, it is $C^\infty$ by \autoref{theorem:termwise-differentiation}. 
+\end{proof}
+

src/op/index.tex

@@ -0,0 +1,5 @@
+\part{Operator Algebras}
+\label{part:operator-algebras}
+
+\input{./banach/index.tex}
+\input{./notation.tex}

src/op/notation.tex

@@ -0,0 +1,8 @@
+\chapter{Notations}
+\label{chap:op-notations}
+
+\begin{tabular}{lll}
+    \textbf{Notation} & \textbf{Description} & \textbf{Source} \\
+    \hline
+    $1$ & Identity element of a unital algebra. & \autoref{definition:unital-banach-algebra} \\
+\end{tabular}

src/process/markov/definition.tex

@@ -32,6 +32,7 @@
         \item For each $t \ge 0$, $\bp_t \one = \one$.
         \item For each $t \ge 0$, $\bp_t$ is a positive linear functional.
     \end{enumerate}
+
     known as the \textbf{semigroup} of $\bracs{P_t|t \ge 0}$.
 \end{definition}
 \begin{proof}

src/topology/functions/ideal.tex

@@ -0,0 +1,71 @@
+\section{Ideals}
+\label{section:set-ideal}
+
+
+\begin{definition}[Ideal]
+\label{definition:set-ideal}
+    Let $X$ be a set and $\sigma \subset 2^X$, then $\sigma$ is an \textbf{ideal} over $X$ if:
+    \begin{enumerate}[label=(I\arabic*)]
+        \item For any $E \in \sigma$ and $F \subset E$, $F \in \sigma$. 
+        \item For any $E, F \in \sigma$, $E \cup F \in \sigma$. 
+    \end{enumerate}
+\end{definition}
+
+\begin{definition}[Covering]
+\label{definition:set-ideal-covering}
+    Let $X$ be a set and $\sigma \subset 2^X$, then $\sigma$ \textbf{covers} $X$/is \textbf{covering} if $\bigcup_{E \in \sigma}E = X$. 
+\end{definition}
+
+\begin{definition}[Fundamental]
+\label{definition:set-ideal-fundamental}
+    Let $X$ be a set, $\sigma \subset 2^X$ be an ideal, and $\tau \subset \sigma$, then $\tau$ is \textbf{fundamental} with respect to $\sigma$ if for any $E \in \sigma$, there exists $F \in \tau$ such that $E \subset F$. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:set-ideal-fundamental-criterion}
+    Let $X$ be a set and $\tau \subset 2^X$, then the following are equivalent:
+    \begin{enumerate}
+        \item For any $E, F \in \tau$, there exists $G \in \tau$ such that $E \cup F \subset G$. 
+        \item There exists an ideal $\sigma \subset 2^X$ such that $\tau$ is fundamental with respect to $\sigma$. 
+    \end{enumerate}
+
+    If the above holds, then the ideal $\sigma$ in (2) is the \textbf{ideal generated by} $\tau$. 
+\end{proposition}
+\begin{proof}
+    (1) $\Rightarrow$ (2): Let
+    \[
+    \sigma = \bracs{F \subset X| \exists E \in \tau: F \subset E}
+    \]
+
+    then $\sigma$ satisfies (I1) by definition. For any $E, F \in \sigma$, there exists $E_0, F_0 \in \tau$ such that $E \subset E_0$ and $F \subset F_0$. By assumption, there exists $G \in \tau$ such that
+    \[
+    E \cup F \subset E_0 \cup F_0 \subset G
+    \]
+
+    so $\sigma$ satisfies (I2), and is an ideal. 
+
+    (2) $\Rightarrow$ (1): Let $E, F \in \tau$, then $E \cup F \in \sigma$. Since $\tau$ is fundamental, there exists $G \in \tau$ such that $E \cup F \subset G$. 
+\end{proof}
+
+\begin{definition}[Product Ideal]
+\label{definition:product-ideal}
+    Let $X, Y$ be sets, $\sigma \subset 2^X$ and $\tau \subset 2^Y$ be ideals, and
+    \[
+    \beta = \bracs{A \times B|A \in \sigma, B \in \tau}
+    \]
+
+    then there exists a unique ideal $\sigma \times \tau$ such that $\beta$ is fundamental with respect to $\sigma$. The ideal $\sigma \otimes \tau$ is the \textbf{product} of $\sigma$ and $\tau$. 
+\end{definition}
+\begin{proof}
+    For each $A_1, A_2 \in \sigma$ and $B_1, B_2 \in \tau$, 
+    \[
+    (A_1 \times B_1) \cup (A_2 \times B_2) \subset (A_1 \cup A_2) \times (B_1 \cup B_2)
+    \]
+
+    By \autoref{proposition:set-ideal-fundamental-criterion}, there exists an ideal $\sigma \otimes \tau$ such that $\beta$ is fundamental with respect to $\sigma \otimes \tau$. 
+\end{proof}
+
+
+
+
+

src/topology/functions/index.tex

@@ -1,5 +1,6 @@
 \chapter{Function Spaces}
 \label{chap:function-spaces}
 
+\input{./ideal.tex}
 \input{./set-systems.tex}
 \input{./uniform.tex}

src/topology/functions/set-systems.tex

@@ -1,105 +1,107 @@
-\section{Topology With Respect to Families of Sets}
+\section{Topologies With Respect to Ideals}
 \label{section:pointwise}
 
 \begin{definition}[Set-Open Topology]
 \label{definition:set-open}
-    Let $T$ be a set, $\mathfrak{S} \subset 2^T$ be a non-empty family of sets, and $(X, \topo)$ be a topological space. For each $S \in \mathfrak{S}$ and $U \subset X$ open, let
+    Let $T$ be a set, $\sigma \subset 2^T$ be an ideal, and $(X, \topo)$ be a topological space. For each $S \in \sigma$ and $U \subset X$ open, let
     \[
     M(S, U) = \bracs{f \in X^T| f(S) \subset U}
     \]
 
     and
     \[
-    \ce(\mathfrak{S}, \topo) = \bracs{M(S, U)| S \in \mathfrak{S}, U \in \topo}
+    \ce(\sigma, \topo) = \bracs{M(S, U)| S \in \sigma, U \in \topo}
     \]
 
-    then the topology generated by $\ce$ is the \textbf{$\mathfrak{S}$-open topology} on $T^X$.
+    then the topology generated by $\ce$ is the \textbf{$\sigma$-open topology} on $T^X$.
 
-    If $\cb \subset \topo$ generates $\topo$, then $\ce(\mathfrak{S}, \cb)$ generates the $\mathfrak{S}$-open topology.
+    If $\cb \subset \topo$ generates $\topo$, then $\ce(\sigma, \cb)$ generates the $\sigma$-open topology.
 \end{definition}
 
 
 \begin{definition}[Set Uniformity]
 \label{definition:set-uniform}
-    Let $T$ be a set, $\mathfrak{S} \subset 2^T$ be a non-empty family of sets, and $(X, \fU)$ be a uniform space. For each $S \in \mathfrak{S}$ and $U \in \fU$, let
+    Let $T$ be a set, $\sigma \subset 2^T$ be an ideal, and $(X, \fU)$ be a uniform space. For each $S \in \sigma$ and $U \in \fU$, let
     \[
     E(S, U) = \bracs{(f, g) \in X^T|(f(x), g(x)) \in U \forall x \in S}
     \]
 
     and
     \[
-    \mathfrak{E}(\mathfrak{S}, \fU) = \bracs{E(S, U)| S \in \mathfrak{S}, U \in \fU}
+    \mathfrak{E}(\sigma, \fU) = \bracs{E(S, U)| S \in \sigma, U \in \fU}
     \]
 
     then
     \begin{enumerate}
-        \item $\mathfrak{E}(\mathfrak{S}, \fU)$ generates a uniformity $\fV$ on $X^T$.
-        \item The topology induced by $\fV$ is finer than the $\mathfrak{S}$-topology on $T^X$.
-        \item If $\mathfrak{S}$ is upward-directed with respect to inclusion, then $\mathfrak{E}(\mathfrak{S}, \fU)$ is forms a fundamental system of entourages for $\fV$.
+        \item $\mathfrak{E}(\sigma, \fU)$ generates a uniformity $\fV$ on $X^T$.
+        \item The topology induced by $\fV$ is finer than the $\sigma$-open topology on $T^X$.
+        \item If $\mathfrak{E}(\sigma, \fU)$ forms a fundamental system of entourages for $\fV$.
     \end{enumerate}
-    The uniformity $\fV$ is the \textbf{$\mathfrak{S}$-uniformity}, and the topology induced by $\fV$ is the \textbf{topology of uniform convergence on the sets $\mathfrak{S}$}/\textbf{$\mathfrak{S}$-uniform topology} on $X^T$.
+
+    The uniformity $\fV$ is the \textbf{$\sigma$-uniformity}, and the topology induced by $\fV$ is the \textbf{topology of uniform convergence on the sets $\sigma$}/\textbf{$\sigma$-uniform topology} on $X^T$.
 \end{definition}
 \begin{proof}
-    (1): Since $\Delta \subset E(S, U)$ for all $S \in \mathfrak{S}$ and $U \in \fU$, $\mathfrak{E}(\mathfrak{S}, \fU)$ generates a uniformity on $X^T$.
+    (1): Since $\Delta \subset E(S, U)$ for all $S \in \sigma$ and $U \in \fU$, $\mathfrak{E}(\sigma, \fU)$ generates a uniformity on $X^T$.
 
     (2): Let $U \subset X$ be open, then for each $x \in U$, there exists $V_x \in \fU$ such that $x \in V_x(x) \subset U$. In which case, $U = \bigcup_{x \in U}V_x(x)$ and $M(S, U) = \bigcup_{x \in U}M(S, V_x)(x)$.
 
     (3): It is sufficient to verify
     \begin{enumerate}
-        \item[(FB1)] For any $S, S'  \in \mathfrak{S}$, there exists $T \in \mathfrak{S}$ with $T \supset S, S'$. In which case, for any $U, U' \in \fU$,
+        \item[(FB1)] For any $S, S'  \in \sigma$, there exists $T \in \sigma$ with $T \supset S, S'$. In which case, for any $U, U' \in \fU$,
         \[
         E(T, U \cap U') \subset E(S \cup S', U \cap U') \subset E(S, U) \cap E(S', U')
         \]
 
-        \item[(UB1)] For any $U \in \fU$, $\Delta \subset U$. Thus the diagonal in $X^T$ is in $E(S, U)$ for any $S \in \mathfrak{S}$.
-        \item[(UB2)] For any $U \in \fU$, there exists $V \in \fV$ with $V \circ V \subset U$. Thus for any $S \in \mathfrak{S}$,
+        \item[(UB1)] For any $U \in \fU$, $\Delta \subset U$. Thus the diagonal in $X^T$ is in $E(S, U)$ for any $S \in \sigma$.
+        \item[(UB2)] For any $U \in \fU$, there exists $V \in \fV$ with $V \circ V \subset U$. Thus for any $S \in \sigma$,
         \[
         E(S, V) \circ E(S, V) \subset E(S, V \circ V) \subset E(S, U)
         \]
 
     \end{enumerate}
+
     By \autoref{proposition:fundamental-entourage-criterion}, $\mathfrak{E}$ is a fundamental system of entourages for the uniformity that it generates.
 \end{proof}
 
 \begin{proposition}
 \label{proposition:set-uniform-pseudometric}
-    Let $T$ be a set, $\mathfrak{S} \subset 2^T$ be a non-empty family of sets, and $(X, \fU)$ be a uniform space whose uniformity is induced by the pseudometrics $\seqi{d}$. For each $i \in I$ and $S \in \mathfrak{S}$, let
+    Let $T$ be a set, $\sigma \subset 2^T$ be an ideal, and $(X, \fU)$ be a uniform space whose uniformity is induced by the pseudometrics $\seqi{d}$. For each $i \in I$ and $S \in \sigma$, let
     \[
     d_{i, S}: X^T \times X^T \quad (f, g) \mapsto \sup_{x \in S}d_i(f(x), g(x))
     \]
 
     then
     \begin{enumerate}
-        \item $\bracs{d_{i, S}| i \in I, S \in \mathfrak{S}}$ is a family of pseudometrics induces the $\mathfrak{S}$-uniformity on $X^T$.
-        \item If $\mathfrak{S}$ is upward-directed with respect to inclusion, then
+        \item $\bracs{d_{i, S}| i \in I, S \in \sigma}$ is a family of pseudometrics induces the $\sigma$-uniformity on $X^T$.
+        \item The family
         \[
-        \bracs{\bigcap_{j \in J}E(d_{j, S}, r)|J \subset I \text{ finite}, r > 0, S \in \mathfrak{S}}
+        \bracs{\bigcap_{j \in J}E(d_{j, S}, r)|J \subset I \text{ finite}, r > 0, S \in \sigma}
         \]
 
-        is a fundamental system of entourages for the $\mathfrak{S}$-uniformity on $X^T$.
+        is a fundamental system of entourages for the $\sigma$-uniformity on $X^T$.
     \end{enumerate}
 \end{proposition}
 \begin{proof}
-    (1): Let $S \in \mathfrak{S}$ and $U \in \fU$, then there exists $r > 0$ and $J \subset I$ finite such that $\bigcap_{j \in J}E(d_j, r) \subset U$, so
+    (1): Let $S \in \sigma$ and $U \in \fU$, then there exists $r > 0$ and $J \subset I$ finite such that $\bigcap_{j \in J}E(d_j, r) \subset U$, so
     \[
     \bigcap_{j \in J}E(d_{j, S}, r) \subset E\paren{S, \bigcap_{j \in J}E(d_j, r)} \subset E(S, U)
     \]
 
-    and the uniformity induced by $\bracs{d_{i, S}| i \in I, S \in \mathfrak{S}}$ contains the $\mathfrak{S}$-uniformity.
+    and the uniformity induced by $\bracs{d_{i, S}| i \in I, S \in \sigma}$ contains the $\sigma$-uniformity.
 
-    On the other hand, for any $i \in I$ and $r > 0$, $E(d_j, r/2) \in \fU$ by \autoref{definition:pseudometric-uniformity}. Therefore $E(S, E(d_j, r/2)) \subset E(d_{j, S}, r)$, so the $\mathfrak{S}$-uniformity contains the induced uniformity.
+    On the other hand, for any $i \in I$ and $r > 0$, $E(d_j, r/2) \in \fU$ by \autoref{definition:pseudometric-uniformity}. Therefore $E(S, E(d_j, r/2)) \subset E(d_{j, S}, r)$, so the $\sigma$-uniformity contains the induced uniformity.
 
-    (2): If $\mathfrak{S}$ is upward-directed with respect to inclusion, then by \autoref{definition:set-uniform},
+    (2): By \autoref{definition:set-uniform},
     \[
-    \bracs{E(S, U)| U \in \fU, S \in \mathfrak{S}}
+    \bracs{E(S, U)| U \in \fU, S \in \sigma}
     \]
 
-    Following the same steps in (1),
+    is a fundamental system of entourages for the $\sigma$-uniformity. Following the same steps in (1),
     \[
-    \bracs{\bigcap_{j \in J}E(d_{j, S}, r)|J \subset I \text{ finite}, r > 0, S \in \mathfrak{S}}
+    \bracs{\bigcap_{j \in J}E(d_{j, S}, r)|J \subset I \text{ finite}, r > 0, S \in \sigma}
     \]
 
-    is a fundamental system of entourages for the $\mathfrak{S}$-uniformity.
+    is a fundamental system of entourages for the $\sigma$-uniformity.
 \end{proof}
 
 
@@ -109,9 +111,10 @@
     Let $T$ be a set and $X$ be a topological space, then the following topologies on $X^T$ coincide:
     \begin{enumerate}
         \item The product topology on $X^T$.
-        \item The $\mathfrak{S}$-open topology, where $\mathfrak{S} = \bracs{\bracs{x}| x \in X}$.
+        \item The $\sigma$-open topology, where $\sigma$ is the collection of all finite sets.
         \item (If $X$ is a uniform space) The $\mathfrak{F}$-uniform topology, where $\fF = \bracs{F| F \subset X \text{ finite}}$.
     \end{enumerate}
+
     This topology is the \textbf{topology of pointwise convergence} on $X^T$.
 \end{definition}
 \begin{proof}
@@ -125,10 +128,28 @@
 
 \begin{proposition}
 \label{proposition:set-uniform-complete}
-    Let $T$ be a set, $\mathfrak{S} \subset 2^T$ such that $\bigcup_{S \in \mathfrak{S}}S = T$, and $(X, \fU)$ be a complete uniform space, then $X^T$ equipped with the $\mathfrak{S}$-uniformity is complete.
+    Let $T$ be a set, $\sigma \subset 2^T$ be a covering ideal, and $(X, \fU)$ be a complete uniform space, then $X^T$ equipped with the $\sigma$-uniformity is complete.
 \end{proposition}
 \begin{proof}
-    Let $\fF \subset 2^{X^T}$ be a Cauchy filter. Let $x \in T$. If there exists $S \in \mathfrak{S}$ with $x \in S$, then $\pi_x(\fF) \subset 2^X$ is a Cauchy filter. By completeness of $X$, there exists $f: T \to X$ such that $\pi_x(\fF) \to f(x)$ for all $x \in \bigcup_{S \in \mathfrak{S}}S$.
+    Let $\fF \subset 2^{X^T}$ be a Cauchy filter. Let $x \in T$. If there exists $S \in \sigma$ with $x \in S$, then $\pi_x(\fF) \subset 2^X$ is a Cauchy filter. By completeness of $X$, there exists $f: T \to X$ such that $\pi_x(\fF) \to f(x)$ for all $x \in \bigcup_{S \in \sigma}S$.
 
-    Let $S \in \mathfrak{S}$ be non-empty and $V_0, V$ be symmetric entourages of $X$ with $V_0 \circ V_0 \subset V$. Since $\fF$ is Cauchy, there exists an $E(S, V_0)$-small set $F \in \fF$. Let $x \in S$, then $\pi_x(\fF) \to f(x)$ implies that $F_x = \pi_x^{-1}(V_0(f(x))) \in \fF$. Since $F \cap F_x \ne \emptyset$, $(f(x), g(x)) \in V_0 \circ V_0 \subset V$ for all $g \in F$. As this holds for all $x \in S$, $E \subset E(S, V)(f) \in \fF$. Thus $\fF \to f$.
+    Let $S \in \sigma$ be non-empty and $V_0, V$ be symmetric entourages of $X$ with $V_0 \circ V_0 \subset V$. Since $\fF$ is Cauchy, there exists an $E(S, V_0)$-small set $F \in \fF$. Let $x \in S$, then $\pi_x(\fF) \to f(x)$ implies that $F_x = \pi_x^{-1}(V_0(f(x))) \in \fF$. Since $F \cap F_x \ne \emptyset$, $(f(x), g(x)) \in V_0 \circ V_0 \subset V$ for all $g \in F$. As this holds for all $x \in S$, $E \subset E(S, V)(f) \in \fF$. Thus $\fF \to f$.
 \end{proof}
+
+\begin{proposition}
+\label{proposition:compact-uniform-open}
+    Let $X$ be a topological space, $\kappa \subset 2^X$ be the collection of all precompact sets in $X$, and $(Y, \fU)$ be a uniform space, then the $\kappa$-open topology and $\kappa$-uniform topology on $C(X; Y)$ coincide. 
+\end{proposition}
+\begin{proof}
+    By \autoref{definition:set-uniform}, the $\kappa$-uniform topology is finer than the $\kappa$-open topology. 
+
+    Let $K \in \kappa$ be compact, $U \in \fU$ be symmetric, and $f \in C(X; Y)$. For each $x \in K$, there exists $V_x \in \cn_X(x)$ such that $f(V_x) \subset U(f(x))$. Since $K$ is compact, there exists $\seqf{x_j} \subset X$ such that $K \subset \bigcup_{j = 1}^n V_{x_j}$. 
+
+    Let $g \in \bigcap_{j = 1}^n M(V_{x_j} \cap K, U(f(x_j)))$ and $y \in K$, then there exists $1 \le j \le n$ such that $y \in V_{x_j}$. In which case, since $f(y), g(y) \in U(f(x_j))$, $(f(y), g(y)) \in U \circ U$. Therefore
+    \[
+    f \in  \bigcap_{j = 1}^n M(V_{x_j} \cap K, U(f(x_j))) \subset [E(K, U \circ U)](f)
+    \]
+
+    and the $\kappa$-open topology is finer than the $\kappa$-uniform topology.
+\end{proof}
+

src/topology/functions/uniform.tex

@@ -24,6 +24,7 @@
         \item $C(X; Y) \subset Y^X$ is closed with respect to the uniform topology.
         \item If $X$ is a uniform space, then $UC(X; Y) \subset Y^X$ is closed with respect to the uniform topology. 
     \end{enumerate}
+
     In particular, if $Y$ is complete, then the above spaces are complete.
 \end{proposition}
 \begin{proof}
@@ -49,3 +50,12 @@
     If $Y$ is complete, then $Y^T$ with the uniform topology is complete by \autoref{proposition:set-uniform-complete}. Thus $C(T; X)$ and $UC(T; X)$ are both complete subspaces by \autoref{proposition:complete-closed}.
 \end{proof}
 
+
+\begin{corollary}
+\label{corollary:uniform-limit-continuous-generated}
+    Let $X$ be a topological space, $\sigma \subset 2^X$ be an ideal such that $X$ is $\sigma$-generated, and $Y$ be a uniform space, then $C(X; Y) \subset Y^X$ is closed with respect to the $\sigma$-uniformity.
+\end{corollary}
+\begin{proof}
+    Let $f \in \overline{C(X; Y)} \subset Y^X$ with respect to the $\sigma$-uniformity. By \autoref{proposition:uniform-limit-continuous}, $f \in C(S; Y)$ for all $S \in \sigma$, so $f \in C(X; Y)$ by (3) of \autoref{definition:final-topology}. 
+\end{proof}
+

src/topology/main/baire.tex

@@ -10,6 +10,7 @@
         \item For any $\seq{A_n} \subset 2^X$ closed with empty interior, $\bigcup_{n \in \nat^+}A_n$ has empty interior.
         \item For any $\seq{U_n} \subset 2^X$ open and dense, $\bigcap_{n \in \nat^+}U_n$ is dense.
     \end{enumerate}
+
     If the above holds, then $X$ is a \textbf{Baire space}.
 \end{definition}
 \begin{proof}
@@ -29,6 +30,7 @@
         \item[(a)] For all $n > 1$, $\ol V_n \subset U_n \cap V_{n - 1} \subset U$.
         \item[(b)] $\bigcap_{j \in \natp} \ol V_j$ is non-empty.
     \end{enumerate}
+
     then $X$ is a Baire space.
 \end{lemma}
 \begin{proof}