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}

refs.bib

@@ -111,4 +111,14 @@
   number={10},
   pages={3211--3212},
   year={1996}
-}
+}
+@book{ConwayComplex,
+  title={Functions of One Complex Variable I},
+  author={Conway, J.B.},
+  isbn={9780387903286},
+  lccn={lc78018836},
+  series={Functions of one complex variable / John B. Conway},
+  url={https://books.google.ca/books?id=9LtfZr1snG0C},
+  year={1978},
+  publisher={Springer}
+}

src/cat/gluing/index.tex

@@ -22,6 +22,7 @@
     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:

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/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

@@ -1,25 +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 a covering ideal, $\mathcal{H} \subset B_\sigma(E; F)$ be a subspace, and $\mathcal{R}_\sigma = \mathcal{R}_\sigma(E; F)$.
 
-    Let $U \subset E$ be open, $f: U \to F$, $x_0 \in U$, and $n > 1$, then $f$ is \textbf{$n$-fold $\sigma$-differentiable at $x_0$} if
+\begin{definition}[Codomain of Derivatives]
-    \begin{enumerate}
+\label{definition:higher-derivatives-codomain}
-        \item There exists $V \in \cn_E(x_0)$ such that $f$ is $(n-1)$-fold differentiable on $V$.
+    Let $E, F$ be TVSs over $K \in \RC$, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, and $L^{(0)}_\sigma(E; F) = F$. For each $n \in \natp$, inductively define
-        \item The derivative $D_\sigma^{n-1}f: U \to B^{n-1}_\sigma(E; F)$ is derivative at $x_0$.
-    \end{enumerate}
-
-    In which case, $D_\sigma(D_\sigma^{n-1}f)(x_0) \in L(E; B^{n-1}_\sigma(E; F))$ is the \textbf{$n$-fold $\sigma$-derivative of $f$ at $x_0$}.
-
-    The mapping $f: U \to F$ is \textbf{$n$-fold $\sigma$-differentiable on $U$} if it is $n$-fold $\sigma$-differentiable at every point in $U$. Under the identification $B_\sigma(E; B^{n-1}_\sigma(E; F)) = B_\sigma^{n}(E; F)$ given by \autoref{proposition:multilinear-identify},
     \[
-    D_\sigma^{n}f: U \to B^{n-1}_\sigma(E; F)
+    L^{(n)}_\sigma(E; F) = L(E; L^{(n-1)}_\sigma(E; F)) \subset B_\sigma^n(E; F)
     \]
 
-    is the \textbf{$n$-fold $\sigma$-derivative of $f$ at $x_0$}.
+    and equip it with the $\sigma$-uniform topology, then under the identification
+    \[
+    I: L^{(n)}_\sigma(E; F) \to B_\sigma^n(E; F) \quad I\lambda(x_1, \cdots, x_n) = \lambda(x_1)\cdots(x_n)
+    \]
+
+    the space $L^{(n)}_\sigma(E; F)$ is a subspace of $B_\sigma^n(E; F)$. 
 \end{definition}
+\begin{proof}
+    By \autoref{proposition:multilinear-identify}. 
+\end{proof}
+
+
+
+\begin{definition}[$n$-Fold Differentiability]
+\label{definition:n-differentiable-sets}
+    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, and $\mathcal{R}_\sigma = \mathcal{R}_\sigma(E; F)$.
+
+    Let $U \subset E$ be open, $f: U \to F$, $x_0 \in U$, and $n > 1$, then $f$ is \textbf{$n$-fold $\tilde \sigma$-differentiable at $x_0$} if
+    \begin{enumerate}
+        \item There exists $V \in \cn_E(x_0)$ such that $f$ is $(n-1)$-fold $\tilde \sigma$-differentiable on $V$.
+        \item The derivative $D_\sigma^{n-1}f: U \to B^{(n-1)}_\sigma(E; F)$ is $\tilde \sigma$-differentiable at $x_0$.
+    \end{enumerate}
+
+    In which case, $D_\sigma(D_\sigma^{n-1}f)(x_0) = D_\sigma^{n}f(x_0)$ is the \textbf{$n$-fold $\tilde \sigma$-derivative of $f$ at $x_0$}.
+
+    If $f: U \to F$ is $n$-fold $\tilde \sigma$-differentiable at every point in $U$, then $f$ is \textbf{$n$-fold $\tilde \sigma$-differentiable on $U$}. Under the \hyperref[identification]{proposition:multilinear-identify} $B_\sigma(E; B_\sigma^{n}(E; F)) = B_\sigma^{(n)}(E; F)$, the mapping
+    \[
+    D_\sigma^{n}f: U \to B^{(n-1)}_\sigma(E; F)
+    \]
+
+    is the \textbf{$n$-fold $\tilde \sigma$-derivative of $f$}.
+
+    If for each $1 \le k \le n$, $D_\sigma^{k}f$ takes value in $L^{(k)}_\sigma(E; F)$, then $f$ is \textbf{$n$-fold $\sigma$-differentiable}, and $D_\sigma^{n}f$ is the \textbf{$n$-fold $\sigma$-derivative of $f$}.
+\end{definition}
+
+
+\begin{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}
@@ -31,7 +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)] \\
@@ -55,7 +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) \\
@@ -88,55 +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 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 $\sigma$-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]
+\begin{theorem}[Power Rule]
-\label{proposition:multilinear-derivative}
+\label{theorem:power-rule}
-    Let $E$ be a topological vector space, $\sigma \subset B(E)$ be an ideal that includes all bounded sets contained in finite-dimensional spaces, $F$ be a Hausdorff locally convex space, and
+    Let $E$ be a topological vector space, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, $F$ be a separated locally convex space, $T \in B^n_\sigma(E; F)$.
+
+    For each $1 \le m \le n$, let $\text{Inj}([m]; [n])$ be the set of injective mappings from $[m]$ to $[n]$. For any $\phi \in \text{Inj}([m]; [n])$, denote $\phi^c \in \text{Inj}([n-m]; [n] \setminus \phi([m]))$ as the unique increasing injective map. For each $h \in E^{n-m}$, $k \in E^m$, and $1 \le j \le n$, write
     \[
-    T \in \underbrace{L(E; L(E; \cdots L(E; F) \cdots ))}_{n \text{ times}} \subset B^n(E; F)
+    (h, k)_\phi = \begin{cases}
+        k_{\phi^{-1}(j)} &j \in \phi([m]) \\
+        h_{(\phi^c)^{-1}(j)} &j \not\in \phi([m])
+    \end{cases}
     \]
 
-    be symmetric. For any $x \in E$ and $1 \le k \le n$, let $x^{(k)}$ denote the tuple of $x$ repeated $k$ times, then:
+    For any $x \in E$, denote $x^{(m)}$ as the tuple of $x$ repeated $m$ times, then the mapping $f: E \to F$ defined by $x \mapsto T(x^{(n)})$ is infinitely $\tilde\sigma$-differentiable on $E$, where
     \begin{enumerate}
-        \item The mapping $f: E \to F \quad x \mapsto T(x^{(n)})$ is infinitely $\sigma$-differentiable on $E$.
+        \item For each $1 \le m \le n$, $x \in E$, and $h \in E^m$, 
-        \item For each $1 \le k \le n$ and $x, h \in E$,
         \[
-        Df(x)(h_1, \cdots, h_k) = \frac{n!}{(n-k)!} T(x^{(n-k)}, h_1, \cdots, h_k)
+        D^{m}_\sigma f(x)(h) = \sum_{\phi \in \text{Inj}([m]; [n])}T((x^{(n-m)}, h)_\phi]
         \]
 
-        In particular, $D^kf = n! \cdot T$.
+        In particular, 
-        \item For each $k > n$ and $x \in E$, $Df(x) = 0$.
+        \[
+        D^{n}_\sigma f(x)(h) = \sum_{\phi \in S_n}T(h_{\phi(1)}, \cdots, h_{\phi(n)})
+        \]
+
+        \item For each $m > n$ and $x \in E$, $D^m_\sigma f(x) = 0$.
     \end{enumerate}
-\end{proposition}
+
+    Notably, if $T \in L^{(n)}(E; F)$ is symmetric or $T \in L^n(E; F)$, then $T$ is infinitely $\sigma$-differentiable on $E$. 
+\end{theorem}
 \begin{proof}
-    Suppose inductively that (2) holds for $0 \le k \le n$. Let $G =  B^{k}_\sigma(E; F)$, then $D^k_\sigma f \in B^{n-k}_\sigma(E; G)$ under the identification $B^n_\sigma(E; F) = B^{n-k}_\sigma(E; B^k_\sigma(E; F))$ in \autoref{proposition:multilinear-identify}. By \autoref{theorem:derivative-symmetric}, $D^k_\sigma f$ is also symmetric, so using the Binomial formula,
+    (1): Let $0 \le m \le n - 1$ and suppose inductively that (1) holds for $m$. For each $x, h \in E$, $S \subset [n-m]$, and $1 \le j \le n - m$,
+    \[
+    [(x, h)_S]_j = \begin{cases}
+        h &j \in S \\
+        x &j \not\in S
+    \end{cases}
+    \]
+    
+    By the Binomial formula, for each $h \in E$ and $k \in E^{m}$,
     \begin{align*}
-        D^k_\sigma f(x + h) &= \sum_{r = 0}^{n-k}{n - k \choose r}D^k_\sigma f(x^{(n-k-r)}, h^{(r)}) \\
+        D^{m}_\sigma f(x + h)(k) &= \sum_{\phi \in \text{Inj}([m]; [n])}T[((x+h)^{(n-m)}, k)_\phi] \\
-        &= f(x) + (n-k)D^k_\sigma f(x^{(n-k-1)}, h) \\
+        &= \sum_{\phi \in \text{Inj}([m]; [n])}\sum_{S \subset [n-m]}T[((x, h)_S, k)_\phi]
-        &+ \underbrace{\sum_{r = 2}^{n-k}{n - k \choose r}D^k_\sigma f(x^{(n-k-r)}, h^{(r)})}_{r(h)}
     \end{align*}
 
-    For each $k \ge 2$, let $A \in \sigma$ and $U \in \cn_F(0)$, then since $D^k_\sigma f \in B^{n-k}_\sigma(E; F)$, there exists $t > 0$ such that
+    For $\ell \ge 2$, maps in $B_\sigma^{\ell}(E; F)$ are $\sigma$-small, so
     \[
-    \frac{D^k_\sigma f(x^{(n-k)}, (sA)^{(k)})}{t} = s^{k-1}D^k_\sigma f(x^{(n-k)}, A^{(k)}) \subset U
+    r(h) = \sum_{\phi \in \text{Inj}([m]; [n])}\sum_{\substack{S \subset [n-m] \\ |S| \ge 2}}T[((x, h)_S, k)_\phi]
     \]
 
-    for all $s \in (0, t)$. Hence $r \in \mathcal{R}_\sigma(E; G)$, and
+    is $\sigma$-small. Hence
+    \begin{align*}
+        &D^{(m)}_\sigma f(x + h)(k) - D_\sigma^{(m)}f(x)(k) \\
+        &= \sum_{\phi \in \text{Inj}([m]; [n])}\sum_{j = 1}^{n-m}T[((x, h)_{\bracs{j}}, k)_\phi] + r(h) \\
+        &= \sum_{\phi \in \text{Inj}([m+1]; [n])}T[(x^{(n-m-1)}, (h, k))_\phi] + r(h)
+    \end{align*}
+
+    and for any $h \in E^{m+1}$,
     \[
-    D^{k+1}_\sigma f(x + h) = f(x) + \frac{n!}{(n-k-1)!}T(x^{(n-k-1)}, h_1, \cdots, h_{k+1}) + r(h)
+    D_\sigma^{(m+1)}f(x)(h) = \sum_{\phi \in \text{Inj}([m+1]; [n])}T[(x^{(n-m-1)}, h)_\phi]
     \]
 
-    by the inductive hypothesis.
+    (2): By (1), $D^n_\sigma f$ is constant. 
-
-    (3): Since $D^n_\sigma f$ is constant, $D^k_\sigma f = 0$ for all $k > n$.
 \end{proof}
+
+

src/dg/derivative/index.tex

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

src/dg/derivative/inverse.tex

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

src/dg/derivative/mvt.tex

@@ -105,7 +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 $y \in V$,
     \[
     f(y) - f(x) \in \overline{\text{Conv}\bracs{Df(z)(y - x)|z \in (x, y)}}
     \]
@@ -113,7 +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
     \[
@@ -124,7 +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

@@ -3,7 +3,7 @@
 
 \begin{definition}[Small]
 \label{definition:differentiation-small}
-    Let $E, F$ be TVSs over $K \in \RC$, $\sigma \subset B(E)$ be a covering ideal, $r: E \to F$, 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 $F^E$.
@@ -17,39 +17,39 @@
 
 \begin{proposition}
 \label{proposition:differentiation-sets}
-    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset B(E)$ be a covering ideal, $B_\sigma(E; F)$ 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 a covering ideal, $U \subset E$ 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 a covering ideal, $U \subset E$ 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 covering ideals. 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 $s \in \mathcal{R}_\tau(F; G)$, $s \circ (T + r) \in \mathcal{R}_\sigma(E; G)$.
@@ -82,9 +82,9 @@
 
 \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}
 
@@ -128,35 +128,84 @@
     A method of extending this sense of differentiability is to require that \textit{every} extension of the function to some open set, or to the entire space is differentiable. Given that this paves way to confusion for related definitions of differentiability, this definition is not formally included here.
 \end{remark}
 
-
+\begin{theorem}[Interchange of Limits and Derivatives]
-
+\label{theorem:differentiable-uniform-limit}
-\begin{proposition}
+    Let $E$ be a TVS over $K \in \RC$, $F$ be a separated locally convex space over $K$, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, $U \subset E$ be open, and $n \in \natp$. Let $\fF \subset 2^{\tilde D_\sigma^n(U; F)}$ be a filter such that:
-\label{proposition:derivative-sets-real}
+    \begin{enumerate}[label=(\alph*)]
-    Let $E$ be a separated topological vector space and $\sigma \subset B(\real)$ be a covering ideal, then
+        \item There exists $f: U \to F$ such that $\fF \to f$ pointwise. 
-    \begin{enumerate}
+        \item For each $1 \le k \le n$, there exists $f^{(k)}: U \to B^{k}_\sigma(E; F)$ such that for all $x \in U$, and $A \in \sigma$ with $x + [0, 1]A \subset U$, $D_\sigma^k(\fF) \to f^{(k)}$ uniformly on $x + [0, 1]A$. 
-        \item $\mathcal{R}_{\sigma}(\real; E) = \mathcal{R}_{B(\real)}(\real; E)$. Hence, all forms of $\sigma$-differentiability on $\real$ are equivalent.
-        \item For any $U \subset \real$ open, $f: U \to E$, and $x_0 \in U$, $f$ is differentiable at $x_0$ if and only if
-        \[
-        \lim_{t \to 0}\frac{f(x + t) - f(x)}{t}
-        \]
-
-        exists. In which case, the above limit is identified with the derivative of $f$ at $0$.
-        \item For any $U \subset \real$ open, $f: U \to E$, and $x_0 \in U$, if $f$ is differentiable at $x_0$, then $f$ is continuous at $x_0$.
     \end{enumerate}
-\end{proposition}
-\begin{proof}
-    (1): Let $r \in \mathcal{R}_\sigma(\real; E)$. For any $R > 0$ and $U \in \cn_E(0)$, there exists $\delta > 0$ such that $t^{-1}r(tR), t^{-1}r(-tR) \in U$ for all $t \in (0, \delta)$. Thus $t^{-1}r(tB(0, R)) \subset U$, and $r \in \mathcal{R}_{B(\real)}(\real; E)$.
 
-    (2): Suppose that $f$ is differentiable at $x_0$, then there exists $r \in \mathcal{R}_\sigma$ such that for any $t \in \real$ with $x_0 + t \in U$,
+    then $f \in \tilde D_\sigma^n(U; F)$ and $D^k_\sigma f = f^{(k)}$ for all $1 \le k \le n$. In particular, if $\sigma$ is saturated, then $(b)$ may be replaced by
+    \begin{enumerate}
+        \item[(b)] For each $1 \le k \le n$, there exists $f^{(k)}: U \to B^{k}_\sigma(E; F)$ such that $D_\sigma^k(\fF) \to f^{(k)}$ uniformly on every $A \in \sigma$. 
+    \end{enumerate}
+    
+\end{theorem}
+\begin{proof}
+    Assume without loss of generality that $n = 1$. For any $\varphi \in \tilde D^1_\sigma(U; F)$, $x \in U$, and $h \in E$ such that $x + h \in U$, 
     \begin{align*}
-        f(x_0 + t) - f(x_0) &= Df(x_0)(t) + r(t) \\
+        f(x + h) - f(x) - f^{(1)}(x)h &= \underbrace{\varphi(x + h) - \varphi(x) - D_\sigma\varphi(x)h}_{\in \mathcal{R}_\sigma(E; F)} \\
-        \frac{f(x_0 + t) - f(x_0)}{t} &= Df(x_0)(1) + t^{-1}r(t) \\
+        &+ (f - \varphi)(x + h) - (f - \varphi)(x) \\
-        \lim_{t \to 0}\frac{f(x_0 + t) - f(x_0)}{t} &= Df(x_0)(1)
+        &+ (D_\sigma\varphi - f^{(1)})(x)h 
     \end{align*}
-    Now suppose that $v = \lim_{t \to 0}\frac{f(x + t) - f(x)}{t}$ exists. Let $T: \real \to E$ be defined by $t \mapsto tv$, then
+
+    Since $\fF \to f$ pointwise, for any $S \in \fF$, 
     \[
-    \lim_{t \to 0}\frac{f(x_0 + t) - f(x_0) - Tt}{t} = \lim_{t \to 0}\frac{f(x_0 + t) - f(x_0)}{t} - v = 0
+    (f - \varphi)(x + h) - (f - \varphi)(x) \in \overline{\bracs{(g - \varphi)(x + h) - (g - \varphi)(x)|g \in S}}
+    \]
+    
+    By the \hyperref[Mean Value Theorem]{theorem:mean-value-theorem}, for any $g \in \tilde D^1_\sigma(U; F)$, 
+    \[
+    (g - \varphi)(x + h) - (g - \varphi)(x) \in \ol{\conv}\bracs{D_\sigma(g - \varphi)(x + th)h|t \in [0, 1]}
     \]
 
-    and $Df(x_0) = T$.
+    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} \\
+    $D_\sigma^n(U; F)$ & $n$-fold $\sigma$-differentiable functions. & \autoref{definition:differentiable-space} \\
-    $D^+f(x)$ & Right derivative of $f$ at $x$. & \autoref{definition:right-differentiable-mvt} \\
+    $\tilde D_\sigma^n(U; F)$ & $n$-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/polar.tex

@@ -22,15 +22,108 @@
 \end{definition}
 
 \begin{proposition}
-\label{proposition:polar-properties}
+\label{proposition:polar-gymnastics}
     Let $\dpn{E, F}{\lambda}$ be a duality over $K \in \RC$, then:
     \begin{enumerate}
-        \item $\emptyset^\circ = \emptyset^\square = F$ and $F^\circ = F^\square = \bracs{0}$. 
+        \item $\emptyset^\circ = F$ and $E^\circ = \bracs{0}$. 
-        \item For any $A, B \subset E$ and $\lambda \ne 0$, if $\lambda A \subset B$, then $B^\circ \subset \lambda^{-1}A^\circ$. 
+        \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 $x \in E$,
-\end{proposition}
+    \[
-
+    \text{Re}\dpn{x, y}{\lambda} = \text{Re}\Phi(x) = \phi(x)
+    \]
 
+    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/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 The locally convex direct sum of barrelled spaces. 
-        \item Products of barreled 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

@@ -67,10 +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 $\mathfrak{B}(E, \mathcal{T}_B) \supset \mathfrak{B}(E, \mathcal{T})$. 
-        \item[(U)] For every $\mathcal{S}$ satisfying (1), $\mathcal{S} \subset \mathcal{T}_B$. In particular, $\mathcal{T} \subset \mathcal{T}_B$ and $B(E, \mathcal{T}_B) = B(E, \mathcal{T})$. 
+        \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} $E$. 
@@ -78,14 +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/spaces-of-linear.tex

@@ -24,7 +24,7 @@
 \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 $S \in \sigma$, $\lamdba S \in \sigma$. 
+        \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}
 

src/fa/norm/normed.tex

@@ -73,21 +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[(B)] $E$ is a Banach space.
-        \item $E$ is a Banach space.
+        \item[(E2)] For every $x \in E$, $\sup_{T \in \mathcal{T}}\norm{Tx}_F < \infty$.
     \end{enumerate}
 
     then $\sup_{T \in \mathcal{T}}\norm{T}_{L(E; F)} < \infty$.
 \end{theorem}
 \begin{proof}
-    For each $n \in \natp$, let $A_n = \bracs{x \in X|\norm{Tx}_F \le n \forall T \in \mathcal{T}}$, then each $A_n$ is closed with $\bigcup_{n \in \natp}A_n = E$. By the \hyperref[Baire Category Theorem]{theorem:baire}, there exists $n \in \natp$ and $U \subset E$ open such that $\sup_{x \in U}\sup_{T \in \mathcal{T}}\norm{Tx}_{F} < \infty$.
+    By the \autoref{theorem:banach-steinhaus} theorem, $\mathcal{T}$ is equicontinuous. Therefore there exists $r > 0$ such that $\bigcup_{T \in \mathcal{T}}T[B_E(0, r)] \subset B_F(0, 1)$. In which case, $\sup_{T \in \mathcal{T}}\norm{T}_{L(E; F)} \le r^{-1}$. 
-
-    Let $x \in U$ and $r > 0$ such that $\overline{B(x, r)} \subset U$, then for any $y \in E$ with $\norm{y}_E \le r$ and $T \in \mathcal{T}$,
-    \[
-    \norm{Ty} = \norm{Ty + Tx - Tx}_E = \normn{T\underbrace{(x + y)}_{\in U}}_E + \norm{Tx}_E \le 2n
-    \]
-
-    so $\sup_{T \in \mathcal{T}}\norm{T}_{L(E; F)} \le 2n/r$.
 \end{proof}
 
 

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} \\
@@ -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/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 $E$. 
@@ -105,5 +105,4 @@
 
     (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}
+\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 $[a, b] \subset \real$, $E, F, H$ be locally convex spaces, $E \times F \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map, $G: [a, b] \to F$, and $f \in RS([a, b], G)$, then for any continuous seminorms $[\cdot]_E: E \to [0, \infty)$, $[\cdot]_F: F \to [0, \infty)$, and $[\cdot]_H: H \to [0, \infty)$ such that $[xy]_H \le [x]_E[y]_F$ for all $x \in E$ and $y \in F$, 
-
-    Let $[\cdot]_H$ be a continuous seminorm on $H$, then there exists continuous seminorms $[\cdot]_E$ on $E$ and $[\cdot]_F$ on $F$ such that for any $f \in RS([a, b], G)$,
     \[
     \braks{\int_a^bf dG}_H \le \sup_{x \in [a, b]}[f]_E \cdot [g]_{\text{var}, F}
     \]
 
 \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
     \[
@@ -77,11 +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}
@@ -102,3 +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/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/space-of-linear.tex

@@ -1,228 +1,156 @@
-\section{Vector-Valued Function Spaces}
+\section{Spaces of Linear Maps}
-\label{section:spaces-linear-map}
+\label{section:space-linear-map-new}
-
+
-\begin{proposition}[{{\cite[III.3.1]{SchaeferWolff}}}]
+\begin{definition}[Space of Bounded Linear Maps]
-\label{proposition:tvs-set-uniformity}
+\label{definition:bounded-linear-map-space}
-    Let $T$ be a set, $\sigma \subset 2^T$ be an ideal, and $F$ be a TVS over $K \in \RC$, then
+    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.
-    \begin{enumerate}
+
-        \item The \hyperref[$\sigma$-uniformity]{definition:set-uniform} on $F^T$ is translation invariant.
+    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 $F$. 
-        \item The composition defined by
+\end{definition}
-        \[
+
-        T^F \times T^F \to T^F \quad (f + g)(x) = f(x) + g(x)
+\begin{proposition}[{{\cite[III.3.3]{SchaeferWolff}}}]
-        \]
+\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:
-        is continuous.
+    \begin{enumerate}
-    \end{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 $S \in \sigma$.
-    For any vector subspace $\cf \subset F^T$, the following are equivalent:
+        \item For every $S \in \sigma$, $\bigcup_{T \in A}T(A)$ is bounded in $F$. 
-    \begin{enumerate}
+    \end{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{proposition}
-    \end{enumerate}
+% Proof omitted because it is obvious. 
-\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.
+\begin{proposition}
-
+\label{proposition:multilinear-identify}
-    (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 $(f + g, f' + g') \in E(S, U) \cap \cf$.
+	Let $E, F$ be TVSs over $K \in \RC$, $\sigma \subset 2^E$ be a covering ideal, and $k \in \natp$, then
-
+	\begin{enumerate}
-    (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.
+        \item The map
-
+        \[
-    (T) $\Rightarrow$ (B): Let $f, g \in \cf$, $\lambda, \lambda' \in K$, and $S \in \sigma$, then for any $x \in X$,
+        I: B_{\sigma}^k(E; B_{\sigma}(E; F)) \to B^{k+1}_{\sigma}(E; F)
-    \begin{align*}
+        \]
-    \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))
+        defined by
-    \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
+        (IT)(x_1, \cdots, x_{k+1}) = T(x_1, \cdots, x_k)(x_{k+1})
-    \[
+        \]
-    \lambda f(x) - \lambda' g(x) \in (\lambda - \lambda')f(S) + \lambda' V \subset U + U \subset U_0
+
-    \]
+        is an isomorphism.
-
+        \item The map
-    for all $x \in S$.
+        \[
-\end{proof}
+        I: \underbrace{B_{\sigma}(E; B_{\sigma}(E; \cdots)))}_{k \text{ times}} \to B^k_{\sigma}(E; F)
-
+        \]
-
+
-\begin{definition}[Space of Bounded Functions]
+        defined by
-\label{definition:bounded-function-space}
+        \[
-    Let $T$ be a set, $E$ be a TVS over $K \in \RC$, and $f: T \to E$ be a function, then $f$ is \textbf{bounded} if $f(T)$ is a bounded subset of $E$. The set $B(T; E) \subset E^T$ is the space of all bounded functions from $T$ to $E$, and:
+        IT(x_1, \cdots, x_k) = T(x_1)\cdots (x_k)
-    \begin{enumerate}
+        \]
-        \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 $B(T; E)$.
+        is an isomorphism.
-    \end{enumerate}
+    \end{enumerate}
-\end{definition}
+
-\begin{proof}
+    which allows the identification
-    (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. 
+    \[
-    
+    \underbrace{B_{\sigma}(E; B_{\sigma}(E; \cdots)))}_{k \text{ times}} = B^k_{\sigma}(E; F)
-    Since $f(E)$ is bounded for all $f \in B(T; E)$, $B(T; E)$ forms a TVS over $K$ by \autoref{proposition:tvs-set-uniformity}.
+    \]
-
+
-    (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,
+    under the map $I$ in (2).
-    \[
+\end{proposition}
-    f(E) \subset g(E) + (f - g)(E) \subset (\lambda + 1)U
+\begin{proof}
-    \]
+    (1): To see that $I$ is surjective, let $T \in B_{\sigma}^{k+1}(E; F)$ and
-    so $f \in B(T; E)$.
+    \[
-
+    I^{-1}T: E \to B_{\sigma}(E; F) \quad x \mapsto T(x, \cdot)
-    If $E$ is complete, then $E^T$ with the uniform topology is complete by \autoref{proposition:set-uniform-complete}. Thus $B(T; E)$ is also complete by \autoref{proposition:complete-closed}.
+    \]
-\end{proof}
+
-
+    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,
-
+    \[
-\begin{definition}[Space of Bounded Continuous Functions]
+    T(x_1, \cdots, x_k, S) \subset T(S^{k+1}) \in \mathfrak{B}(F)
-\label{definition:bounded-continuous-function-space}
+    \]
-    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}
+    by assumption. Thus $I^{-1}T(x_1, \cdots, x_k) \in B_{\sigma}(E; F)$.
-        \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 $BC(X; E)$.
+    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 $I^{-1}T \in B^k_{\sigma}(E; B_{\sigma}(E; F))$.
-    \end{enumerate}
+
-\end{definition}
+    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.
-\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}.
+    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): 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. 
+    (2): The case for $k = 2$ is given by (1). If the proposition holds for $k \in \natp$, then
-    
+    \[
-    If $E$ is complete, then $B(X; E)$ and $C(X; E)$ are both complete under the uniform topology by \autoref{definition:bounded-function-space}. Therefore $BC(X; E)$ is also complete. 
+    \underbrace{B_{\sigma}(E; B_{\sigma}(E; \cdots)))}_{k+1 \text{ times}} = B^k_{\sigma}(E; B_{\sigma}(E; F)) = B^{k+1}_{\sigma}(E; F)
-\end{proof}
+    \]
-
+
-
+    Thus (2) holds for all $k \in \natp$.
-
+\end{proof}
-\begin{definition}[Space of Bounded Linear Maps]
+
-\label{definition:bounded-linear-map-space}
+\begin{definition}[Strong Operator Topology]
-    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 $T(S^k) \in B(F)$ for all $S \in \sigma$, equipped with the $\bracsn{S^k| S \in \sigma}$-uniform 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}.
-    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}
+    The space $L_s(E; F)$ denotes $L(E; F)$ equipped with the strong operator topology.
-
+\end{definition}
-\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{proposition}
-	\begin{enumerate}
+\label{proposition:strong-operator-dense}
-        \item The map
+    Let $E, F$ be TVSs over $K \in \RC$ and $\net{T} \subset L(E; F)$ and $T \in L_s(E; F)$. If
-        \[
+    \begin{enumerate}
-        I: B_{\sigma}^k(E; B_{\sigma}(E; F)) \to B^{k+1}_{\sigma}(E; F)
+        \item[(a)] There exists a dense subset $S \subset E$ such that $T_\alpha x \to Tx$ strongly for all $x \in S$.
-        \]
+        \item[(b)] $\bracs{T_\alpha|\alpha \in A}$ is uniformly equicontinuous. 
-
+    \end{enumerate}
-        defined by
+
-        \[
+    
-        (IT)(x_1, \cdots, x_{k+1}) = T(x_1, \cdots, x_k)(x_{k+1})
+    then $T_\alpha \to T$ in $L_s(E; F)$.
-        \]
+\end{proposition}
-
+\begin{proof}
-        is an isomorphism.
+    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 $\alpha \ge \alpha_0$,
-        \item The map
+    \[
-        \[
+    T_\alpha x - Tx = \underbrace{T_\alpha x - T_\alpha y}_{\in V} + \underbrace{T_\alpha y - Ty}_{\in V} + \underbrace{Ty - Tx}_{\in V} \in U
-        I: \underbrace{B_{\sigma}(E; B_{\sigma}(E; \cdots)))}_{k \text{ times}} \to B^k_{\sigma}(E; F)
+    \]
-        \]
+\end{proof}
-
+
-        defined by
+
-        \[
+
-        IT(x_1, \cdots, x_k) = T(x_1)\cdots (x_k)
+\begin{definition}[Weak Operator Topology]
-        \]
+\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}.
-        is an isomorphism.
+
-    \end{enumerate}
+    The space $L_w(E; F) = L_s(E; F_w)$ denotes $L(E; F)$ equipped with the weak operator topology.
-
+\end{definition}
-    which allows the identification
+
-    \[
+\begin{definition}[Bounded Convergence Topology]
-    \underbrace{B_{\sigma}(E; B_{\sigma}(E; \cdots)))}_{k \text{ times}} = B^k_{\sigma}(E; F)
+\label{definition:bounded-convergence-topology}
-    \]
+    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}.
-
+
-    under the map $I$ in (2).
+    The space $L_b(E; F)$ denotes $L(E; F)$ equipped with the topology of bounded convergence.
-\end{proposition}
+\end{definition}
-\begin{proof}
+
-    (1): To see that $I$ is surjective, let $T \in B_{\sigma}^{k+1}(E; F)$ and
+\begin{proposition}
-    \[
+\label{proposition:operator-space-completeness}
-    I^{-1}T: E \to B_{\sigma}(E; F) \quad x \mapsto T(x, \cdot)
+    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, then:
-    \]
+    \begin{enumerate}
-
+        \item $\hom(E; F)$ is a closed subspace of $F^E$ with respect to the product topology.
-    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,
+        \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 $B(E; F)$.
-    \[
+    \end{enumerate}
-    T(x_1, \cdots, x_k, S) \subset T(S^{k+1}) \in B(F)
+\end{proposition}
-    \]
+\begin{proof}
-
+    (1): For each $x, y \in E$ and $\lambda \in K$, the mappings
-    by assumption. Thus $I^{-1}T(x_1, \cdots, x_k) \in B_{\sigma}(E; F)$.
+    \[
-
+    \phi_{x, y}: F^E \to F \quad T \mapsto Tx + Ty - T(x + y)
-    In addition, for any $S_1 \in \sigma$ and entourage $E(S_2, U)$ of $B_{\sigma}(E; F)$ where $S_2 \in \sigma$ and $U$ is an entourage of $F$, there exists $S \in \sigma$ with $S \supset S_1 \cup S_2$. Given that $T(S^{k+1}) \in B(F)$, there exists $\lambda > 0$ such that $T(S^{k+1}) \subset \lambda U(0)$. In which case, $I^{-1}T(S^k) \subset \lambda E(S, U)(0)$ and $I^{-1}T(S^k) \in B(B_{\sigma}(E; F))$. Thus $I^{-1}T \in B^k_{\sigma}(E; B_{\sigma}(E; F))$.
+    \]
-
+
-    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.
+    and
-
+    \[
-    On the other hand, let $S_1 \in \sigma$ and $E(S_2, U)$ be an entourage of $B_{\sigma}(E; F)$ where $S_2 \in \sigma$ and $U$ is an entourage of $F$. Let $S \in \sigma$ with $S \supset S_1 \cup S_2$, then for any $T \in E(S^{k+1}, U)(0)$, $I^{-1}T \in E(S^{k}, E(S, U))(0)$. Thus $I^{-1}$ is continuous as well.
+    \psi_{x, \lambda}: F^E \to F \quad T \mapsto T(\lambda x) - \lambda Tx
-
+    \]
-    (2): The case for $k = 2$ is given by (1). If the proposition holds for $k \in \natp$, then
+
-    \[
+    are continuous with respect to the product topology. Since
-    \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)
+    \[
-    \]
+    \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}
-
+    \]
-    Thus (2) holds for all $k \in \natp$.
+
-\end{proof}
+    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.
-
+
-\begin{definition}[Strong Operator Topology]
+\end{proof}
-\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/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/vector/variation.tex

@@ -119,7 +119,5 @@
     \]
 
     by \hyperref[Fubini's theorem]{theorem:fubini-tonelli}.
-    
-    % TODO: Actually link Fubini once it's there.
 \end{proof}
 

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/topology/functions/ideal.tex

@@ -47,6 +47,25 @@
     (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/main/compactify.tex

@@ -0,0 +1,75 @@
+\section{Compactifications}
+\label{section:compactifications}
+
+\begin{definition}[Compactification]
+\label{definition:compactification}
+    Let $X$ be a topological space, then a \textbf{compactification} of $X$ is a pair $(Y, f)$ where
+    \begin{enumerate}
+        \item $Y$ is a compact Hausdorff space. 
+        \item $f \in C(X; Y)$ is an embedding.
+        \item $f(X)$ is dense in $Y$. 
+    \end{enumerate}
+\end{definition}
+
+\begin{definition}[Stone-Čech Compactification]
+\label{definition:stone-cech}
+    Let $X$ be a completely regular space, then there exists a pair $(\beta X, e)$ such that:
+    \begin{enumerate}
+        \item $(\beta X, e)$ is a compactification of $X$. 
+        \item[(U1)] For any $f \in C(X; [0, 1])$, there exists a unique $\beta f \in C(\beta X; [0, 1])$ such that the following diagram commutes:
+        \[
+        \xymatrix{
+        \beta X \ar@{->}[r]^{\beta f} & [0, 1] \\
+        X \ar@{->}[u]^{e} \ar@{->}[ru]_{f} & 
+        }
+        \]
+    \end{enumerate}
+
+    Moreover, if $(\beta X, e)$ is \textit{any} pair that satisfies (1) and (U1), then
+    \begin{enumerate}
+        \item[(U2)] For any pair $(Y, \varphi)$ satisfying (1), there exists a unique $\beta \varphi \in C(\beta X; Y)$ such that the following diagram commutes:
+        \[
+        \xymatrix{
+        \beta X \ar@{->}[r]^{\beta \varphi} & Y \\
+        X \ar@{->}[u]^{e} \ar@{->}[ru]_{\varphi} & 
+        }
+        \]
+    \end{enumerate}
+
+    The pair $(\beta X, e)$ is the \textbf{Stone-Čech compactification} of $X$. 
+\end{definition}
+\begin{proof}
+    Let $e: X \to [0, 1]^{C(X; [0, 1])}$ be the embedding of $X$ into $[0, 1]^{C(X; [0, 1])}$ associated with $C(X; [0, 1])$ in \autoref{definition:embedding-in-cube}, and $\beta X = \ol{e(X)}$. 
+
+    (1): By \autoref{theorem:tychonoff} and \autoref{proposition:product-hausdorff}, $[0, 1]^{C(X; [0, 1])}$ is a compact Hausdorff space. By definition, $e(X)$ is dense in $\beta X$. 
+
+    (U1): For each $f \in C(X; [0, 1])$, $\pi_f \in C([0, 1]^{C(X; [0, 1])}; [0, 1])$ is an extension of $f$ to $e(x)$. 
+
+    (U2): Let $(Y, \varphi)$ be a compactification of $X$. For each $f \in C(Y; [0, 1])$, by (U1), there exists a unique $\beta(f \circ \varphi) \in C(X; [0, 1])$ such that the following diagram commutes: 
+    \[
+    \xymatrix{
+    X \ar@{->}[d]_{\varphi} \ar@{->}[r]^{e} & \beta X \ar@{->}[d]^{\beta (f \circ \varphi)} \\
+    Y \ar@{->}[r]_{f} & [0, 1]
+    }
+    \]
+
+    Let $e': Y \to [0, 1]^{C(Y; [0, 1])}$ be the embedding of $Y$ into $[0, 1]^{C(Y; [0, 1])}$ associated with $C(Y; [0, 1])$, then by (U) of the \hyperref[product topology]{definition:product-topology}, there exists $\beta(e' \circ \varphi) \in C(\beta X; [0, 1])$ such that the following diagram commutes:
+    \[
+    \xymatrix{
+    X \ar@{->}[d]_{\varphi} \ar@{->}[r]^{e} & \beta X \ar@{->}[d]^{\beta (e' \circ \varphi)} \\
+    Y \ar@{->}[r]_{e'} & [0, 1]^{C(Y; [0, 1])}
+    }
+    \]
+
+    Since $Y$ is a compact Hausdorff space, $e'(Y)$ is closed by \autoref{proposition:compact-extensions} and \autoref{proposition:compact-closed}. As $e'$ is an embedding, identify $Y$ as a subspace of $[0, 1]^{C(Y; [0, 1])}$. Given that $e(X)$ is dense in $\beta X$, the the image of $\beta (e' \circ \varphi)$ lies in $Y$ by \autoref{proposition:closure-of-image}. Therefore under the identification, the following diagram commutes:
+    \[
+    \xymatrix{
+    X \ar@{->}[d]_{\varphi} \ar@{->}[r]^{e} & \beta X \ar@{->}[ld]^{\beta (e' \circ \varphi)} \\
+    Y & 
+    }
+    \] 
+    
+\end{proof}
+
+
+

src/topology/main/cube.tex

@@ -0,0 +1,63 @@
+\section{Embeddings in Cubes}
+\label{section:embeddings-in-cubes}
+
+\begin{definition}[Completely Regular]
+\label{definition:completely-regular}
+    Let $X$ be a topological space, then $X$ is \textbf{completely regular} if for any $E \subset X$ closed and $x \in X \setminus E$, there exists $f \in C(X; [0, 1])$ such that $f(x) = 1$ and $f|_E = 0$. 
+\end{definition}
+
+\begin{definition}[Separation of Points and Closed Sets]
+\label{definition:separate-points-closed-sets}
+    Let $X$ be a topological space and $\cf \subset C(X; [0, 1])$, then $\cf$ \textbf{separates points and closed sets} if for any $E \subset X$ closed and $x \in X \setminus E$, there exists $f \in \cf$ such that $f(x) \not\in \ol{f(E)}$. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:completely-regular-separate}
+    Let $X$ be a $T_1$ space, then the following are equivalent:
+    \begin{enumerate}
+        \item $X$ is completely regular. 
+        \item There exists $\cf \subset C(X; [0, 1])$ that separates points and closed sets.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (2) $\Rightarrow$ (1): Let $E \subset X$ closed and $x \in X \setminus E$, then there exists $f \in \cf$ such that $x \not\in \ol{f(E)}$. By \hyperref[Urysohn's lemma]{lemma:urysohn}, there exists $\phi \in C([0, 1]; [0, 1])$ such that $\phi(f(x)) = 1$ and $\phi(f(E)) = 0$. 
+\end{proof}
+
+\begin{definition}[Embedding in Cube]
+\label{definition:embedding-in-cube}
+    Let $X$ be a topological space, $\cf \subset C(X; [0, 1])$, and
+    \[
+    e: X \to [0, 1]^\cf \quad \pi_f(e(x)) = f(x)
+    \]
+
+    then:
+    \begin{enumerate}
+        \item $e \in C(X; [0, 1]^\cf)$. 
+        \item If $\cf$ separates points, then $e$ is injective.
+        \item If $X$ is $T_1$ and $\cf$ separates points and closed sets, then $e$ is an embedding. 
+    \end{enumerate}
+
+    The mapping $e$ is the \textbf{mapping of $X$ into the cube $[0, 1]^\cf$ associated with $\cf$}. 
+\end{definition}
+\begin{proof}[Proof, {{\cite[Proposition 4.53]{Folland}}}. ]
+    (1): By (U) of the \hyperref[product topology]{definition:product-topology}. 
+
+    (3): Since $X$ is $T_1$, $e$ is injective by (2). Let $x \in X$ and $U \in \cn_X^o(x)$, then there exists $f \in \cf$ such that $f(x) \not\in \ol{f(U^c)}$. In which case, there exists $V \in \cn_{[0, 1]}^o(f(x))$ such that $V \cap f(U^c) = \emptyset$. Thus for any $y \in X$ with $\pi_f(e(y)) \in V$, $f(y) \not\in f(U^c)$, so $y \in U$. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:completely-regular-uniformisable}
+    Let $X$ be a $T_1$ space, then the following are equivalent:
+    \begin{enumerate}
+        \item $X$ is completely regular.
+        \item There exists a uniformity $\fU$ on $X$ that induces the topology on $X$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1) $\Rightarrow$ (2): By \autoref{definition:uniform-separated}. 
+
+    (2) $\Rightarrow$ (1): By \autoref{definition:embedding-in-cube}, $X$ embeds into $[0, 1]^{C(X; [0, 1])}$, which is a uniform space. The subspace uniformity on $X$ then induces the topology on $X$. 
+\end{proof}
+
+
+

src/topology/main/index.tex

@@ -24,3 +24,5 @@
 \input{./c0.tex}
 \input{./semicontinuity.tex}
 \input{./baire.tex}
+\input{./cube.tex}
+\input{./compactify.tex}

src/topology/metric/metric.tex

@@ -36,10 +36,38 @@
 
     Let $J \subset I$ be countable such that for each $k \in K$, there exists $j \in J$ with $V_n \subset U_j$, then $X = \bigcup_{k \in K}V_k = \bigcup_{j \in J}V_j$.
 
-    % TODO: This stuff may be moved to more general spaces.
-
     (2) $\Rightarrow$ (3): Let $n \in \nat$, then $\bracs{B(x, 1/n)|x \in X}$ is an open cover of $X$, so there exists $\seq{x_{n, k}} \subset X$ such that $X = \bigcup_{k \in \natp}B(x_k, 1/n)$. Let $D = \bracs{x_{n, k}| n, k \in \natp}$, then for any $x \in X$ and $n \in \natp$, there exists $k \in \natp$ such that $x \in B(x_{n, k}, 1/n)$, so $x_{n, k} \in B(x, 1/n)$. Therefore $D$ is dense in $X$, and $X$ is separable. 
 
     (3) $\Rightarrow$ (1): Let $\seq{x_n} \subset X$ be a countable dense subset. Let $x \in X$ and $k \in \natp$, then there exists $x_n \in \natp$ such that $d(x, x_n) < 1/(2k)$. In which case, $x \in B(x_n, 1/(2k)) \subset B(x_n, 1/k)$. Therefore $\bracs{B(x_n, 1/k)|n, k \in \natp}$ forms a countable basis for $X$. 
 \end{proof}
 
+
+\begin{theorem}[Banach's Fixed Point Theorem]
+\label{theorem:banach-fixed-point}
+    Let $(X, d)$ be a metric space and $f: X \to X$. If there exists $C \in (0, 1)$ such that
+    \[
+    d(f(x), f(y)) \le Cd(x, y) \quad \forall x, y \in X
+    \]
+
+    then:
+    \begin{enumerate}
+        \item There exists a unique $x \in X$ such that $f(x) = x$. 
+        \item For any $y \in X$, $\limv{n}f^n(y) = x$. 
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    Let $x_0 \in X$ be arbitrary, and $x_n = f^n(x_0)$, then for ecah $n \in \natp$,
+    \[
+    d(x_n, x_{n+1}) \le C d(x_{n-1}, x_n) \le C^n d(x_0, x_1)
+    \]
+
+    Thus $\seq{x_n} \subset X$ is Cauchy, and converges to a point $x \in X$. 
+
+    (2): For any $y_0 \in X$, let $y_n = f^n(y_0)$, then $d(x_n, y_n) \to 0$ as $n \to \infty$, so $\limv{n}f^n(y_0) = x$. 
+
+    (1): Since $f$ is Lipschitz continuous, 
+    \[
+    f(x) = f\braks{\limv{n}f^n(x)} = \limv{n}f^{n+1}(x) = x
+    \]
+\end{proof}
+

src/topology/uniform/compact.tex

@@ -71,4 +71,12 @@
     Let $V$ be an entourage of $Y$. For each $x \in X$, let $U_x$ be an entourage of $X$ such that $(f(y), f(z)) \in V$ for all $y, z \in (U_x \circ U_x)(x)$. Since $X$ is compact, there exists $\seqf{x_j} \subset X$ such that $X = \bigcup_{j = 1}^nU_{x_j}(x_j)$. 
     
     Let $U = \bigcap_{j = 1}^n U_{x_j}$, then for any $(x, y) \in U$, there exists $1 \le j \le n$ such that $x \in U_{x_j}(x_j)$. In which case, $x, y \in (U_{x_j} \circ U_{x_j})(x_j)$, so $(f(x), f(y)) \in V$. 
-\end{proof}
+\end{proof}
+
+\begin{proposition}
+\label{proposition:compact-uniform-structure}
+    Let $X$ be a compact Hausdorff space, then there exists a unique uniformity on $X$ that induces its topology. 
+\end{proposition}
+\begin{proof}
+    By \autoref{proposition:completely-regular-uniformisable}, there exists a uniformity on $X$ that induces its topology. By \autoref{proposition:uniform-continuous-compact}, $X$ admits a unique uniformity. 
+\end{proof}

src/topology/uniform/definition.tex

@@ -270,19 +270,16 @@ V = (\bracs{x} \times V)(x) = (U \cup (\bracs{x} \times V))(x) = W(x) \in \cn(x)
         \item $X$ is T1.
         \item $X$ is Hausdorff.
         \item $X$ is regular.
+        \item $X$ is completely regular. 
         \item $\Delta = \bigcap_{U \in \fU}U$.
     \end{enumerate}
 
     If the above holds, then $X$ is \textbf{separated}.
 \end{definition}
 \begin{proof}
-    $(1) \Rightarrow (5)$: Let $x, y \in X$ with $x \ne y$. Assume without loss of generality that there exists $U(x) \in \cn(x)$ such that $y \not\in U$. In which case, $(x, y) \not\in U$ and $\Delta \supset \bigcap_{U \in \fU}U$.
+    (1) $\Rightarrow$ (6): Let $x, y \in X$ with $x \ne y$. Assume without loss of generality that there exists $U(x) \in \cn(x)$ such that $y \not\in U$. In which case, $(x, y) \not\in U$ and $\Delta \supset \bigcap_{U \in \fU}U$.
 
-    $(5) \Rightarrow (2)$: By \autoref{proposition:goodentourages}, $\ol \Delta \subset \bigcap_{U \in \fU}\ol U = \Delta$, so $\ol \Delta$ is closed. By (6) of \autoref{definition:hausdorff}, $X$ is Hausdorff.
+    (6) $\Rightarrow$ (5): Let $E \subset X$ be closed and $x \in X \setminus E$. Since $\Delta = \bigcap_{U \in \fU}U$, there exists $U \in \fU$ such that $U(x) \subset E^c$. By \autoref{theorem:uniform-pseudometric}, there exists a pseudometric $d: X \times X \to [0, \infty)$ such that $d(x, E) > 0$. Thus the function $y \mapsto d(x, y)$ is a continuous function that separates $x$ and $E$. 
-
-    $(1) \Rightarrow (4)$: $X$ is T1 and satisfies (2) of \autoref{definition:regular} by \autoref{proposition:uniform-neighbourhoods}, so $X$ is regular.
-
-    $(4) \Rightarrow (3) \Rightarrow (2) \Rightarrow (1)$:  (T3) $\Rightarrow $ (T2) $\Rightarrow$ (T1) $\Rightarrow$ (T0).
 \end{proof}
 
 

src/topology/uniform/equicontinuous.tex

@@ -8,12 +8,27 @@
     The set $\cf \subset C(X; Y)$ is \textbf{equicontinuous} if it is equicontinuous at every point in $x$. 
 \end{definition}
 
+\begin{proposition}
+\label{proposition:equicontinuous-net}
+    Let $X$ be a topological space, $(Y, \fU)$ be a uniform space, $\cf \subset Y^X$, and $x \in X$, then the following are equivalent:
+    \begin{enumerate}
+        \item $\cf$ is equicontinuous at $x$. 
+        \item For $\angles{x_\alpha}_{\alpha \in A} \subset X$ with $x_\alpha \to x$, $\angles{f_\alpha}_{\alpha \in A} \subset \cf$, and $U \in \fU$, there exists $\alpha_0 \in A$ such that $(f_\alpha(x_\alpha), f_\alpha(x)) \in U$ for all $\alpha \ge \alpha_0$.  
+        \item There exists a fundamental system of neighbourhoods $\fB \subset \cn_X(x)$ at $x$ such that for any $\angles{x_V}_{V \in \fB} \subset X$ with $x_\alpha \to x$, $\angles{f_V}_{V \in \fB} \subset \cf$, and $U \in \fU$, there exists $V_0 \in \fB$ such that $(f_V(x_V), f_V(x)) \in U$ for all $V \subset V_0$.  
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1) $\Rightarrow$ (2): Since $\cf$ is equicontinuous at $x$, there exists $V \in \cn_X(x)$ such that $(f_\alpha(y), f_\alpha(x)) \in U$ for all $y \in V$ and $\alpha \in A$. Given that $x_\alpha \to x$, there exists $\alpha_0 \in A$ such that $x_\alpha \in V$ for all $\alpha \ge \alpha_0$, so $(f_\alpha(x_\alpha), f_\alpha(x)) \in U$ for all $\alpha \ge \alpha_0$.  
+
+    $\neg (1) \Rightarrow \neg (3)$: If $\cf$ is not equicontinuous at $x$, then there exists $U \in \fU$ such that for every $V \in \fB$, there exists $f_V \in \cf$ and $x_V \in V$ with $(f_V(x_V), f_V(x)) \not\in U$. In which case, $x_V \to x$ but $(f_V(x_V), f_V(x)) \not\in U$ for all $V \in \cn_X(x)$. 
+\end{proof}
+
+
 \begin{definition}[Uniformly Equicontinuous]
 \label{definition:uniformly-equicontinuous}
     Let $(X, \fU)$ and $(Y, \fV)$ be uniform spaces, and $\cf \subset UC(X; Y)$, then $\cf$ is \textbf{uniformly equicontinuous} if for every $V \in \fV$, there exists $U \in \fU$ such that $(f \times f)(V) \subset \fU$ for all $f \in \cf$. 
 \end{definition}
 
-
 \begin{theorem}[Arzelà-Ascoli]
 \label{theorem:arzela-ascoli}
     Let $X$ be a topological space, $(Y, \fU)$ be a uniform space, and $\cf \subset C(X; Y)$. If
@@ -23,7 +38,8 @@
 
     then
     \begin{enumerate}[label=(C\arabic*)]
-        \item The product uniformity and the compact uniformity on $\cf$ coincide. 
+        \item The uniform structures of pointwise and compact convergence on $\cf$ coincide. 
+        \item The closure of $\cf$ in $Y^X$ with respect to the product topology is equicontinuous. 
     \end{enumerate}
 
     In addition, if $\cf$ satisfies (E1) and
@@ -32,11 +48,11 @@
     \end{enumerate}
 
     then
-    \begin{enumerate}[label=(C\arabic*), start=1]
+    \begin{enumerate}[label=(C\arabic*), start=2]
-        \item $\cf$ is a precompact subset of $Y^X$ with respect to the compact uniformity. 
+        \item $\cf$ is a precompact subset of $C(X; Y)$ with respect to the uniform structure of compact convergence. 
     \end{enumerate}
 
-    Conversely, if $X$ is a LCH space, then (C2) implies (E1) + (E2).
+    Conversely, if $X$ is a LCH space, then (C3) implies (E1) + (E2).
 \end{theorem}
 \begin{proof}
     (E1) $\Rightarrow$ (C1): By \autoref{proposition:compact-uniform-open}, the compact-open topology coincides with the compact-uniform topology on $C(X; Y)$ and thus $\cf$.
@@ -51,34 +67,20 @@
     \bigcap_{j = 1}^n E(\bracs{x_j}, U) \subset E(K, U \circ U \circ U)
     \]
 
-    so the product uniformity and the compact uniformity coincide. 
+    so the uniform structures of pointwise and compact convergence coincide. 
 
-    (E1) + (E2) $\Rightarrow$ (C2): By \autoref{proposition:totally-bounded-product}, $\prod_{x \in X}\cf(x)$ is totally bounded. Since $\cf \subset \prod_{x \in X}\cf(x)$, $\cf$ is also totally bounded. As the pointwise and compact uniformities coincide, $\cf$ is totally bounded with respect to the compact uniformity. By \autoref{proposition:product-complete} and \autoref{proposition:compact-uniform}, $\prod_{x \in X}\ol{\cf(x)}$ is compact and hence complete. Therefore the closure of $\cf$ with respect to the compact uniformity is compact. 
+    (E1) $\Rightarrow$ (C2): Let $\cf'$ be the closure of $\cf$ in $Y^X$ with respect to the product topology. Let $x \in X$ and entourage $U$ of $Y$. Using \autoref{proposition:goodentourages}, assume without loss of generality that $U$ is closed. Since $\cf$ is equicontinuous, there exists $V \in \cn_X(x)$ such that $(f(x), f(y)) \in U$ for all $f \in \cf$ and $y \in V$. For any element $g \in \cf'$, $(g(x), g(y)) \in \ol U = U$ for all $y \in V$. Therefore $\cf'$ is also equicontinuous. 
 
-    (C2) $\Rightarrow$ (E1): Assume that $X$ is a LCH space. Let $x \in X$ and $U \in \fU$ be symmetric, then there exists a compact neighbourhood $V \in \cn_X(x)$. Since $\cf$ is totally bounded, there exists $\seqf{f_j} \subset \cf$ such that for each $g \in \cf$, there exists $1 \le j \le n$ such that $(f_j \times g)(V) \subset U$. For each $1 \le j \le n$, $f_j \in C(X; Y)$, so there exists $V_j \in \cn_X(x)$ with $V_j \subset V$ such that for any $y \in V_j$, $(f_j(x), f_j(y)) \in U$. Let $W = \bigcap_{j = 1}^n V_j$, then for any $g \in \cf$ with $(f_j \times g)(V) \subset U$ and $y \in W$,  
+    (E1) + (E2) $\Rightarrow$ (C3): Using (C2), assume without loss of generality that $\cf$ is closed in $Y^X$ with respect to the product topology. In which case, $\cf$ is a closed subset of $\prod_{x \in X}\ol{\cf(x)}$ with respect to the product topology. By \hyperref[Tychonoff's Theorem]{theorem:tychonoff} and \autoref{proposition:compact-extensions}, $\cf$ is compact in the product topology. By (C1), $\cf$ is also compact in the compact uniform topology.
+
+    (C3) $\Rightarrow$ (E1): Assume that $X$ is a LCH space. Let $x \in X$ and $U \in \fU$ be symmetric, then there exists a compact neighbourhood $V \in \cn_X(x)$. Since $\cf$ is totally bounded, there exists $\seqf{f_j} \subset \cf$ such that for each $g \in \cf$, there exists $1 \le j \le n$ such that $(f_j \times g)(V) \subset U$. For each $1 \le j \le n$, $f_j \in C(X; Y)$, so there exists $V_j \in \cn_X(x)$ with $V_j \subset V$ such that for any $y \in V_j$, $(f_j(x), f_j(y)) \in U$. Let $W = \bigcap_{j = 1}^n V_j$, then for any $g \in \cf$ with $(f_j \times g)(V) \subset U$ and $y \in W$,  
     \[
     (g(x), f_j(x)), (f_j(x), f_j(y)), (f_j(y), g(y)) \in U \circ U \circ U
     \]
 
     Therefore $g(W) \subset (U \circ U \circ U)(g(x))$, and $\cf$ is equicontinuous. 
 
-    (C2) $\Rightarrow$ (E2): Since the evaluation map is uniformly continuous with respect to the compact uniformity, $\cf(x)$ is totally bounded for all $x \in X$ by \autoref{proposition:totally-bounded-image}. 
+    (C3) $\Rightarrow$ (E2): Since the evaluation map is uniformly continuous with respect to the uniform structure of compact convergence, $\cf(x)$ is totally bounded for all $x \in X$ by \autoref{proposition:totally-bounded-image}. 
 \end{proof}
 
 
-\begin{corollary}
-\label{corollary:arzela-locally-compact}
-    Let $X$ be a LCH space, $Y$ be a uniform space, and $\cf \subset C(X; Y)$ such that:
-    \begin{enumerate}[label=(E\arabic*)]
-        \item $\cf$ is equicontinuous.
-        \item For each $x \in X$, $\cf(x) = \bracs{f(x)|f \in \cf}$ is precompact in $Y$.
-    \end{enumerate}
-
-    then $\cf$ is a precompact subset of $C(X; Y)$ with respect to the compact uniformity. 
-\end{corollary}
-\begin{proof}
-    By the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}, $\cf$ is a precompact subset of $Y^X$. By \autoref{proposition:lch-compactly-generated}, $C(X; Y)$ is a closed subset of $Y^X$. Therefore $\cf$ is a precompact subset of $C(X; Y)$. 
-\end{proof}
-
-
-

src/topology/uniform/uc.tex

@@ -51,7 +51,7 @@
     \end{enumerate}
 
 
-    known as the \textbf{initial uniformity} on $X$ generated by $\seqi{f}$.
+    The collection $\fU$ is the \textbf{initial uniformity} on $X$ generated by $\seqi{f}$.
 \end{definition}
 \begin{proof}
     (3): Since the diagonal is mapped to the diagonal and $\fB$ is closed under intersections, it is sufficient to verify (UB3) for $\fB$. Let $J \subset I$ be finite and $\bigcap_{j \in J}(f_j \times f_j)^{-1}(U_j) \in \fB$, then there exists $\bracs{V_j}_{j \in J}$ such that $V_j \circ V_j \subset U_j$ for each $j \in J$. In which case, for any $(x, y), (y, z) \in f_j^{-1}(V_j)$, $(f(x), f(y)), (f(y), f(z)) \in V_j$ and $(f(x), f(z)) \in U_j$. Thus $(f_j \times f_j)^{-1}(V_j) \circ (f_j \times f_j)^{-1}(V_j) \subset (f_j \times f_j)^{-1}(U_j)$, and
@@ -65,7 +65,7 @@
 
     (U): For any $i \in I$, $\mathfrak{V} \supset (f_i \times f_i)^{-1}(\fU_i)$. By (F2), $\mathfrak{V} \supset \fB$, so $\mathfrak{V} \supset \fU$.
 
-    (4): Let $J \subset I$ finite and $\seqj{U_j}$ such that $U_j \in \fU_j$ for each $j \in J$, then
+    (4): Let $J \subset I$ finite and $\seqj{U}$ such that $U_j \in \fU_j$ for each $j \in J$, then
     \[
     (f \times f)^{-1}\paren{\bigcap_{j \in J}(f_j \times f_j)^{-1}(U_j)} = \bigcap_{j \in J}[(f_j \circ f) \times (f_j \circ f)]^{-1}(U_j)
     \]