Fixed typo.

document.tex

@@ -1,4 +1,4 @@
-%\documentclass{report}
+\documentclass{}
 \usepackage{amssymb, amsmath, hyperref}
 \usepackage{preamble}
 
@@ -11,6 +11,7 @@ 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

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/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, $\sigma \subset \mathfrak{B}(E)$ 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{enumerate}
-        \item There exists $V \in \cn_E(x_0)$ such that $f$ is $(n-1)$-fold differentiable on $V$.
-        \item The derivative $D_\sigma^{n-1}f: U \to B^{n-1}_\sigma(E; F)$ is derivative at $x_0$.
-    \end{enumerate}
-
-    In which case, $D_\sigma(D_\sigma^{n-1}f)(x_0) \in L(E; B^{n-1}_\sigma(E; F))$ is the \textbf{$n$-fold $\sigma$-derivative of $f$ at $x_0$}.
-
-    The mapping $f: U \to F$ is \textbf{$n$-fold $\sigma$-differentiable on $U$} if it is $n$-fold $\sigma$-differentiable at every point in $U$. Under the identification $B_\sigma(E; B^{n-1}_\sigma(E; F)) = B_\sigma^{n}(E; F)$ given by \autoref{proposition:multilinear-identify},
+\begin{definition}[Codomain of Derivatives]
+\label{definition:higher-derivatives-codomain}
+    Let $E, F$ be TVSs over $K \in \RC$, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, and $L^{(0)}_\sigma(E; F) = F$. For each $n \in \natp$, inductively define
     \[
-    D_\sigma^{n}f: U \to B^{n-1}_\sigma(E; F)
+    L^{(n)}_\sigma(E; F) = L(E; L^{(n-1)}_\sigma(E; F)) \subset B_\sigma^n(E; F)
     \]
 
-    is the \textbf{$n$-fold $\sigma$-derivative of $f$ at $x_0$}.
+    and equip it with the $\sigma$-uniform topology, then under the identification
+    \[
+    I: L^{(n)}_\sigma(E; F) \to B_\sigma^n(E; F) \quad I\lambda(x_1, \cdots, x_n) = \lambda(x_1)\cdots(x_n)
+    \]
+
+    the space $L^{(n)}_\sigma(E; F)$ is a subspace of $B_\sigma^n(E; F)$. 
 \end{definition}
+\begin{proof}
+    By \autoref{proposition:multilinear-identify}. 
+\end{proof}
+
+
+
+\begin{definition}[$n$-Fold Differentiability]
+\label{definition:n-differentiable-sets}
+    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, and $\mathcal{R}_\sigma = \mathcal{R}_\sigma(E; F)$.
+
+    Let $U \subset E$ be open, $f: U \to F$, $x_0 \in U$, and $n > 1$, then $f$ is \textbf{$n$-fold $\tilde \sigma$-differentiable at $x_0$} if
+    \begin{enumerate}
+        \item There exists $V \in \cn_E(x_0)$ such that $f$ is $(n-1)$-fold $\tilde \sigma$-differentiable on $V$.
+        \item The derivative $D_\sigma^{n-1}f: U \to B^{(n-1)}_\sigma(E; F)$ is $\tilde \sigma$-differentiable at $x_0$.
+    \end{enumerate}
+
+    In which case, $D_\sigma(D_\sigma^{n-1}f)(x_0) = D_\sigma^{n}f(x_0)$ is the \textbf{$n$-fold $\tilde \sigma$-derivative of $f$ at $x_0$}.
+
+    If $f: U \to F$ is $n$-fold $\tilde \sigma$-differentiable at every point in $U$, then $f$ is \textbf{$n$-fold $\tilde \sigma$-differentiable on $U$}. Under the \hyperref[identification]{proposition:multilinear-identify} $B_\sigma(E; B_\sigma^{n}(E; F)) = B_\sigma^{(n)}(E; F)$, the mapping
+    \[
+    D_\sigma^{n}f: U \to B^{(n-1)}_\sigma(E; F)
+    \]
+
+    is the \textbf{$n$-fold $\tilde \sigma$-derivative of $f$}.
+
+    If for each $1 \le k \le n$, $D_\sigma^{k}f$ takes value in $L^{(k)}_\sigma(E; F)$, then $f$ is \textbf{$n$-fold $\sigma$-differentiable}, and $D_\sigma^{n}f$ is the \textbf{$n$-fold $\sigma$-derivative of $f$}.
+\end{definition}
+
+
+\begin{definition}[Space of Differentiable Functions]
+\label{definition:differentiable-space}
+    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, $U \subset E$ be open, and $n \in \natp$, then $D_\sigma^k(U; F)$/$\tilde D_\sigma^k(U; F)$ is the \textbf{space of $n$-fold $\sigma$/$\tilde \sigma$-differentiable functions} from $U$ to $F$. 
+\end{definition}
+
+\begin{definition}[Space of Continuously Differentiable Functions]
+\label{definition:continuously-differentiable-space}
+    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, $U \subset E$ be open, and $n \in \natp$, then $C_\sigma^k(U; F)$/$\tilde C_\sigma^k(U; F)$ is the \textbf{space of $n$-fold continuously $\sigma$/$\tilde \sigma$-differentiable functions} from $U$ to $F$. 
+\end{definition}
+
 
 \begin{theorem}[Symmetry of Higher Derivatives]
 \label{theorem:derivative-symmetric-frechet}
@@ -88,7 +123,7 @@
 
 \begin{theorem}[Symmetry of Higher Derivatives]
 \label{theorem:derivative-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 $\sigma$-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
@@ -96,47 +131,71 @@
     D_{\mathfrak{B}(E_0)}^n(\phi \circ g)(x_0) = \phi \circ D_\sigma^n g(x_0)
     \]
 
-    by the \hyperref[chain rule]{proposition:chain-rule-sets-conditions}. By \autoref{theorem:derivative-symmetric-frechet}, $\phi \circ D_\sigma^n g(x_0) \in L^n(E_0; K)$ is symmetric. As this holds for any $\seqf{h_j} \subset E$ and $\phi \in F^*$, $D_{\sigma}^n g(x_0) \in B_\sigma^n(E; F)$ is symmetric by the \hyperref[Hahn-Banach theorem]{proposition:hahn-banach-utility}.
+    by the \hyperref[chain rule]{proposition:chain-rule-sets-conditions}. By \autoref{theorem:derivative-symmetric-frechet}, $\phi \circ D_\sigma^n g(x_0) \in L^n(E_0; K)$ is symmetric. As this holds for any $\seqf{h_j} \subset E$ and $\phi \in F^*$, $D_{\sigma}^n g(x_0) \in B_\sigma^{(n)}(E; F)$ is symmetric by the \hyperref[Hahn-Banach theorem]{proposition:hahn-banach-utility}.
 \end{proof}
 
-\begin{proposition}[Power Rule]
-\label{proposition:multilinear-derivative}
-    Let $E$ be a topological vector space, $\sigma \subset \mathfrak{B}(E)$ be an ideal that includes all bounded sets contained in finite-dimensional spaces, $F$ be a Hausdorff locally convex space, and
+\begin{theorem}[Power Rule]
+\label{theorem:power-rule}
+    Let $E$ be a topological vector space, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, $F$ be a separated locally convex space, $T \in B^n_\sigma(E; F)$.
+
+    For each $1 \le m \le n$, let $\text{Inj}([m]; [n])$ be the set of injective mappings from $[m]$ to $[n]$. For any $\phi \in \text{Inj}([m]; [n])$, denote $\phi^c \in \text{Inj}([n-m]; [n] \setminus \phi([m]))$ as the unique increasing injective map. For each $h \in E^{n-m}$, $k \in E^m$, and $1 \le j \le n$, write
     \[
-    T \in \underbrace{L(E; L(E; \cdots L(E; F) \cdots ))}_{n \text{ times}} \subset B^n(E; F)
+    (h, k)_\phi = \begin{cases}
+        k_{\phi^{-1}(j)} &j \in \phi([m]) \\
+        h_{(\phi^c)^{-1}(j)} &j \not\in \phi([m])
+    \end{cases}
     \]
 
-    be symmetric. For any $x \in E$ and $1 \le k \le n$, let $x^{(k)}$ denote the tuple of $x$ repeated $k$ times, then:
+    For any $x \in E$, denote $x^{(m)}$ as the tuple of $x$ repeated $m$ times, then the mapping $f: E \to F$ defined by $x \mapsto T(x^{(n)})$ is infinitely $\tilde\sigma$-differentiable on $E$, where
     \begin{enumerate}
-        \item The mapping $f: E \to F \quad x \mapsto T(x^{(n)})$ is infinitely $\sigma$-differentiable on $E$.
-        \item For each $1 \le k \le n$ and $x, h \in E$,
+        \item For each $1 \le m \le n$, $x \in E$, and $h \in E^m$, 
         \[
-        Df(x)(h_1, \cdots, h_k) = \frac{n!}{(n-k)!} T(x^{(n-k)}, h_1, \cdots, h_k)
+        D^{m}_\sigma f(x)(h) = \sum_{\phi \in \text{Inj}([m]; [n])}T((x^{(n-m)}, h)_\phi]
         \]
 
-        In particular, $D^kf = n! \cdot T$.
-        \item For each $k > n$ and $x \in E$, $Df(x) = 0$.
+        In particular, 
+        \[
+        D^{n}_\sigma f(x)(h) = \sum_{\phi \in S_n}T(h_{\phi(1)}, \cdots, h_{\phi(n)})
+        \]
+
+        \item For each $m > n$ and $x \in E$, $D^m_\sigma f(x) = 0$.
     \end{enumerate}
-\end{proposition}
+
+    Notably, if $T \in L^{(n)}(E; F)$ is symmetric or $T \in L^n(E; F)$, then $T$ is infinitely $\sigma$-differentiable on $E$. 
+\end{theorem}
 \begin{proof}
-    Suppose inductively that (2) holds for $0 \le k \le n$. Let $G =  B^{k}_\sigma(E; F)$, then $D^k_\sigma f \in B^{n-k}_\sigma(E; G)$ under the identification $B^n_\sigma(E; F) = B^{n-k}_\sigma(E; B^k_\sigma(E; F))$ in \autoref{proposition:multilinear-identify}. By \autoref{theorem:derivative-symmetric}, $D^k_\sigma f$ is also symmetric, so using the Binomial formula,
+    (1): Let $0 \le m \le n - 1$ and suppose inductively that (1) holds for $m$. For each $x, h \in E$, $S \subset [n-m]$, and $1 \le j \le n - m$,
+    \[
+    [(x, h)_S]_j = \begin{cases}
+        h &j \in S \\
+        x &j \not\in S
+    \end{cases}
+    \]
+    
+    By the Binomial formula, for each $h \in E$ and $k \in E^{m}$,
     \begin{align*}
-        D^k_\sigma f(x + h) &= \sum_{r = 0}^{n-k}{n - k \choose r}D^k_\sigma f(x^{(n-k-r)}, h^{(r)}) \\
-        &= f(x) + (n-k)D^k_\sigma f(x^{(n-k-1)}, h) \\
-        &+ \underbrace{\sum_{r = 2}^{n-k}{n - k \choose r}D^k_\sigma f(x^{(n-k-r)}, h^{(r)})}_{r(h)}
+        D^{m}_\sigma f(x + h)(k) &= \sum_{\phi \in \text{Inj}([m]; [n])}T[((x+h)^{(n-m)}, k)_\phi] \\
+        &= \sum_{\phi \in \text{Inj}([m]; [n])}\sum_{S \subset [n-m]}T[((x, h)_S, k)_\phi]
     \end{align*}
 
-    For each $k \ge 2$, let $A \in \sigma$ and $U \in \cn_F(0)$, then since $D^k_\sigma f \in B^{n-k}_\sigma(E; F)$, there exists $t > 0$ such that
+    For $\ell \ge 2$, maps in $B_\sigma^{\ell}(E; F)$ are $\sigma$-small, so
     \[
-    \frac{D^k_\sigma f(x^{(n-k)}, (sA)^{(k)})}{t} = s^{k-1}D^k_\sigma f(x^{(n-k)}, A^{(k)}) \subset U
+    r(h) = \sum_{\phi \in \text{Inj}([m]; [n])}\sum_{\substack{S \subset [n-m] \\ |S| \ge 2}}T[((x, h)_S, k)_\phi]
     \]
 
-    for all $s \in (0, t)$. Hence $r \in \mathcal{R}_\sigma(E; G)$, and
+    is $\sigma$-small. Hence
+    \begin{align*}
+        &D^{(m)}_\sigma f(x + h)(k) - D_\sigma^{(m)}f(x)(k) \\
+        &= \sum_{\phi \in \text{Inj}([m]; [n])}\sum_{j = 1}^{n-m}T[((x, h)_{\bracs{j}}, k)_\phi] + r(h) \\
+        &= \sum_{\phi \in \text{Inj}([m+1]; [n])}T[(x^{(n-m-1)}, (h, k))_\phi] + r(h)
+    \end{align*}
+
+    and for any $h \in E^{m+1}$,
     \[
-    D^{k+1}_\sigma f(x + h) = f(x) + \frac{n!}{(n-k-1)!}T(x^{(n-k-1)}, h_1, \cdots, h_{k+1}) + r(h)
+    D_\sigma^{(m+1)}f(x)(h) = \sum_{\phi \in \text{Inj}([m+1]; [n])}T[(x^{(n-m-1)}, h)_\phi]
     \]
 
-    by the inductive hypothesis.
-
-    (3): Since $D^n_\sigma f$ is constant, $D^k_\sigma f = 0$ for all $k > n$.
+    (2): By (1), $D^n_\sigma f$ is constant. 
 \end{proof}
+
+

src/dg/derivative/index.tex

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

src/dg/derivative/inverse.tex

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

src/dg/derivative/mvt.tex

@@ -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

@@ -17,33 +17,33 @@
 
 \begin{proposition}
 \label{proposition:differentiation-sets}
-    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset \mathfrak{B}(E)$ 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, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, $U \subset E$ be open, $f: U \to F$, and $x_0 \in U$, then $f$ is \textbf{$\sigma$-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, $\sigma \subset \mathfrak{B}(E)$ 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}
 
 
@@ -84,7 +84,7 @@
 \label{proposition:chain-rule-sets-conditions}
     Let $E$, $F$, $G$, be TVSs over $K \in \RC$ with $F, G$ being separated. If $\sigma \subset \mathfrak{B}(E)$ and $\tau \subset \mathfrak{B}(F)$ correspond to the following families of sets on $E$ and $F$:
     \begin{enumerate}
-        \item 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{proposition}
-\label{proposition:derivative-sets-real}
-    Let $E$ be a separated topological vector space and $\sigma \subset B(\real)$ be a covering ideal, then
-    \begin{enumerate}
-        \item $\mathcal{R}_{\sigma}(\real; E) = \mathcal{R}_{B(\real)}(\real; E)$. Hence, all forms of $\sigma$-differentiability on $\real$ are equivalent.
-        \item For any $U \subset \real$ open, $f: U \to E$, and $x_0 \in U$, $f$ is differentiable at $x_0$ if and only if
-        \[
-        \lim_{t \to 0}\frac{f(x + t) - f(x)}{t}
-        \]
-
-        exists. In which case, the above limit is identified with the derivative of $f$ at $0$.
-        \item For any $U \subset \real$ open, $f: U \to E$, and $x_0 \in U$, if $f$ is differentiable at $x_0$, then $f$ is continuous at $x_0$.
+\begin{theorem}[Interchange of Limits and Derivatives]
+\label{theorem:differentiable-uniform-limit}
+    Let $E$ be a TVS over $K \in \RC$, $F$ be a separated locally convex space over $K$, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, $U \subset E$ be open, and $n \in \natp$. Let $\fF \subset 2^{\tilde D_\sigma^n(U; F)}$ be a filter such that:
+    \begin{enumerate}[label=(\alph*)]
+        \item There exists $f: U \to F$ such that $\fF \to f$ pointwise. 
+        \item For each $1 \le k \le n$, there exists $f^{(k)}: U \to B^{k}_\sigma(E; F)$ such that for all $x \in U$, and $A \in \sigma$ with $x + [0, 1]A \subset U$, $D_\sigma^k(\fF) \to f^{(k)}$ uniformly on $x + [0, 1]A$. 
     \end{enumerate}
-\end{proposition}
-\begin{proof}
-    (1): Let $r \in \mathcal{R}_\sigma(\real; E)$. For any $R > 0$ and $U \in \cn_E(0)$, there exists $\delta > 0$ such that $t^{-1}r(tR), t^{-1}r(-tR) \in U$ for all $t \in (0, \delta)$. Thus $t^{-1}r(tB(0, R)) \subset U$, and $r \in \mathcal{R}_{B(\real)}(\real; E)$.
 
-    (2): Suppose that $f$ is differentiable at $x_0$, then there exists $r \in \mathcal{R}_\sigma$ such that for any $t \in \real$ with $x_0 + t \in U$,
+    then $f \in \tilde D_\sigma^n(U; F)$ and $D^k_\sigma f = f^{(k)}$ for all $1 \le k \le n$. In particular, if $\sigma$ is saturated, then $(b)$ may be replaced by
+    \begin{enumerate}
+        \item[(b)] For each $1 \le k \le n$, there exists $f^{(k)}: U \to B^{k}_\sigma(E; F)$ such that $D_\sigma^k(\fF) \to f^{(k)}$ uniformly on every $A \in \sigma$. 
+    \end{enumerate}
+    
+\end{theorem}
+\begin{proof}
+    Assume without loss of generality that $n = 1$. For any $\varphi \in \tilde D^1_\sigma(U; F)$, $x \in U$, and $h \in E$ such that $x + h \in U$, 
     \begin{align*}
-        f(x_0 + t) - f(x_0) &= Df(x_0)(t) + r(t) \\
-        \frac{f(x_0 + t) - f(x_0)}{t} &= Df(x_0)(1) + t^{-1}r(t) \\
-        \lim_{t \to 0}\frac{f(x_0 + t) - f(x_0)}{t} &= Df(x_0)(1)
+        f(x + h) - f(x) - f^{(1)}(x)h &= \underbrace{\varphi(x + h) - \varphi(x) - D_\sigma\varphi(x)h}_{\in \mathcal{R}_\sigma(E; F)} \\
+        &+ (f - \varphi)(x + h) - (f - \varphi)(x) \\
+        &+ (D_\sigma\varphi - f^{(1)})(x)h 
     \end{align*}
-    Now suppose that $v = \lim_{t \to 0}\frac{f(x + t) - f(x)}{t}$ exists. Let $T: \real \to E$ be defined by $t \mapsto tv$, then
+
+    Since $\fF \to f$ pointwise, for any $S \in \fF$, 
     \[
-    \lim_{t \to 0}\frac{f(x_0 + t) - f(x_0) - Tt}{t} = \lim_{t \to 0}\frac{f(x_0 + t) - f(x_0)}{t} - v = 0
+    (f - \varphi)(x + h) - (f - \varphi)(x) \in \overline{\bracs{(g - \varphi)(x + h) - (g - \varphi)(x)|g \in S}}
+    \]
+    
+    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, $\sigma \subset \mathfrak{B}(E)$ be an upward-directed family that includes bounded sets contained in finite-dimensional subspaces, $F$ be a separated locally convex space, $U \subset E$ be open, and $f: U \to F$ be $n$-fold $\sigma$-differentiable at $x_0 \in U$, then there exists $r \in \mathcal{R}_\sigma^n(E; F)$ such that
+    Let $E$ be a topological vector space, $\sigma \subset \mathfrak{B}(E)$ be an ideal that includes bounded sets contained in finite-dimensional subspaces, $F$ be a separated locally convex space, $U \subset E$ be open, and $f: U \to F$ be $n$-fold $\tilde \sigma$-differentiable at $x_0 \in U$, then there exists $r \in \mathcal{R}_\sigma^n(E; F)$ such that
     \[
     g(x_0 + h) = g(x_0) + \sum_{k = 1}^n \frac{1}{k!}D^k_\sigma f(x_0)(h^{(k)}) + r(h)
     \]
@@ -74,7 +74,7 @@
     r(h) = g(x_0 + h) - g(x) - \sum_{k = 1}^n \frac{1}{k!}D^k_\sigma f(x_0)(h^{(k)})
     \]
 
-    For any $1 \le k \le n$, $D^k_\sigma(x_0) \in B^k_\sigma(E; F)$ is symmetric by \autoref{theorem:derivative-symmetric}. Let $T_k(h) = \frac{1}{k!}D^k_\sigma f(x_0)(h^{(k)})$, then by \autoref{proposition:multilinear-derivative}, for any $\bracs{t_j}_1^\ell \in E$,
+    For any $1 \le k \le n$, $D^k_\sigma(x_0) \in B^k_\sigma(E; F)$ is symmetric by \autoref{theorem:derivative-symmetric}. Let $T_k(h) = \frac{1}{k!}D^k_\sigma f(x_0)(h^{(k)})$, then by \autoref{theorem:power-rule}, for any $\bracs{t_j}_1^\ell \in E$,
     \[
     D^\ell_\sigma T_k(h)(t_1, \cdots, t_\ell) = \begin{cases}
         0 &\ell > k \\

src/dg/index.tex

@@ -1,5 +1,6 @@
-\part{Differential Geometry}
+\part{Calculus}
 \label{part:diffgeo}
 
 \input{./derivative/index.tex}
+\input{./complex/index.tex}
 \input{./notation.tex}

src/dg/notation.tex

@@ -11,6 +11,11 @@ Differential geometry is the study of things invariant under change of notation.
     $\mathcal{R}_\sigma^n(E; F)$, $\mathcal{R}_\sigma(E;F)$ & $\sigma$-small functions of order $n$; order 1. & \autoref{definition:differentiation-small} \\
     $D_\sigma f(x_0)$ & $\sigma$-derivative of $f$ at $x_0$. & \autoref{definition:derivative-sets} \\
     $D_\sigma^n f$ & $n$-fold $\sigma$-derivative. & \autoref{definition:n-differentiable-sets} \\
-    $x^{(k)}$ & Tuple of $x$ repeated $k$ times. & \autoref{proposition:multilinear-derivative} \\
-    $D^+f(x)$ & Right derivative of $f$ at $x$. & \autoref{definition:right-differentiable-mvt} \\
+    $D_\sigma^n(U; F)$ & $n$-fold $\sigma$-differentiable functions. & \autoref{definition:differentiable-space} \\
+    $\tilde D_\sigma^n(U; F)$ & $n$-fold $\tilde \sigma$-differentiable functions. & \autoref{definition:differentiable-space} \\
+    $C_\sigma^n(U; F)$ & $n$-fold continuously $\sigma$-differentiable functions. & \autoref{definition:continuously-differentiable-space} \\
+    $\tilde C_\sigma^n(U; F)$ & $n$-fold continuously $\tilde \sigma$-differentiable functions. & \autoref{definition:continuously-differentiable-space} \\
+    $L^{(n)}_\sigma(E; F)$ & Codomain of derivatives. $L^{(0)}_\sigma(E; F) = F$, $L^{(n)}_\sigma(E; F) = L(E; L_\sigma^{(n-1)}(E; F))$, equipped with the $\sigma$-uniform topology. & \autoref{definition:higher-derivatives-codomain} \\
+    $x^{(k)}$ & Tuple of $x$ repeated $k$ times. & \autoref{theorem:taylor-peano} \\
+    $D^+f(x)$ & Right derivative of $f$ at $x$. & \autoref{definition:right-differentiable-mvt}
 \end{tabular}

src/fa/duality/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 For any $A, B \subset E$ and $\lambda \ne 0$, if $\lambda A \subset B$, then $B^\circ \subset \lambda^{-1}A^\circ$. 
+        \item $\emptyset^\circ = F$ and $E^\circ = \bracs{0}$. 
+        \item For any $\alpha \in K \setminus \bracs{0}$ and $A \subset E$, $(\alpha A)^\circ = \alpha^{-1} \cdot A^\circ$.
+        \item For any $\seqi{A} \subset E$, $\paren{\bigcup_{i \in I}A_i}^\circ = \bigcap_{i \in I}A_i^\circ$. 
+        \item For any $A \subset B \subset E$, $A^\circ \supset B^\circ$. 
+        \item For any saturated ideal $\sigma \subset \mathfrak{B}(E, \sigma(E, F))$, $\bracs{S^\circ|S \in \sigma}$ is a fundamental system of neighbourhoods at $0$ for the $\sigma$-uniform topology on $F$.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (2): For any $\lambda \in K \setminus \bracs{0}$ and $A \subset E$, 
+    \begin{align*}
+        (\lambda A)^\circ &= \bracs{y \in F|\text{Re}\dpn{x, y}{\lambda} \le 1 \forall x \in \alpha A} \\
+        &= \bracs{y \in F|\text{Re}\dpn{x, y}{\lambda} \le 1 \forall x \in \alpha A} \\
+        &= \bracs{y \in F|\text{Re}\dpn{x, \alpha y}{\lambda} \le 1 \forall x \in A} \\
+        &= \bracs{y \in F|\text{Re}\dpn{\alpha x,  y}{\lambda} \le 1 \forall x \in A} \\
+        &= \bracs{\alpha^{-1} y \in F|\text{Re}\dpn{x,  y}{\lambda} \le 1 \forall x \in A} \\
+        &= \alpha^{-1} \cdot A^\circ
+    \end{align*}
+
+    (5): Let $S \in \sigma$, then
+    \[
+    (\aconv(S))^\circ = \bracs{y \in F|\ |\dpn{x, y}{\lambda}| \le 1 \forall x \in A} \subset S^\circ
+    \]
+
+    Since $\sigma$ is saturated, $\bracs{S^\circ|S \in \sigma}$ is a fundamental system of neighbourhoods at $0$ for the $\sigma$-uniform topology.     
+\end{proof}
+
+\begin{proposition}[{{\cite[IV.1.4]{SchaeferWolff}}}]
+\label{proposition:polar-properties}
+    Let $\dpn{E, F}{\lambda}$ be a duality over $K \in \RC$ and $A \subset E$, then
+    \begin{enumerate}
+        \item $A^\circ$ is a $\sigma(F, E)$-closed convex subset of $F$ containing $0$. 
+        \item If $A$ is circled, then so is $A^\circ$. 
+        \item If $A$ is a subspace of $E$, then
+        \[
+        A^\circ = A^\perp = \bracs{y \in F| \dpn{x, y}{E} = 0 \forall x \in A}
+        \]
+
+        and $A^\circ$ is a subspace of $F$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): For each $x \in E$, 
+    \[
+    \bracs{x}^\circ = \bracs{y \in F|\text{Re}\dpn{x, y}{\lambda} \le 1}
+    \]
+
+    is the sublevel set of a continuous $\real$-linear functional, so it is $\sigma(F, E)$-closed and convex. Since $A^\circ = \bigcap_{x \in A}\bracs{x}^\circ$, $A$ is also $\sigma(F, E)$-closed and convex. 
+
+    (2): If $A$ is circled, then by \autoref{proposition:polar-gymnastics},
+    \[
+    A^\circ = \bigcap_{\substack{\alpha \in K \\ |\alpha| \ge 1}}\alpha A^\circ
+    \]
+
+    For any $y \in A^\circ$ and $\alpha \in K \setminus \bracs{0}$ with $|\alpha| \le 1$. Since $y \in \alpha^{-1}A^\circ$, $\alpha y \in A^\circ$, so $A^\circ$ is circled. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:equicontinuous-polar}
+    Let $E$ be a TVS over $K \in \RC$, $\dpn{E, E^*}{E}$ be the canonical duality, and $A \subset E^*$, then the following are equivalent
+    \begin{enumerate}
+        \item $A$ is equicontinuous.
+        \item $A^\circ \in \cn_E(0)$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    By \autoref{proposition:equicontinuous-linear}, $A$ is equicontinuous if and only if
+    \[
+    \bigcap_{\phi \in A}\phi^{-1}(B_K(0, 1)) \in \cn_E(0)
+    \]
+
+    (1) $\Rightarrow$ (2): $A^\circ \supset \bigcap_{\phi \in A}\phi^{-1}(B_K(0, 1))$.
+
+    (2) $\Rightarrow$ (1): Since $A^\circ \in \cn_E(0)$, there exists $V \in \cn_E(0)$ circled with $V \subset A^\circ$, so $V \subset \bigcap_{\phi \in A}\phi^{-1}(B_K(0, 1))$, and $\bigcap_{\phi \in A}\phi^{-1}(B_K(0, 1)) \in \cn_E(0)$. 
+\end{proof}
+
+
+\begin{theorem}[Bipolar Theorem]
+\label{theorem:bipolar}
+    Let $\dpn{E, F}{\lambda}$ be a duality over $K \in \RC$. For each $A \subset F$,
+    \[
+    A^{\circ\circ} = \ol{\conv}(A \cup \bracs{0})
+    \]
+
+    with respect to the $\sigma(E, F)$-topology. 
+\end{theorem}
+\begin{proof}[Proof, {{\cite[IV.1.5]{SchaeferWolff}}}. ]
+    By \autoref{proposition:polar-properties}, $A^{\circ \circ}$ is a $\sigma(E, F)$-closed, convex set that contains $0$. Since $A^{\circ \circ} \supset A$, it is sufficient to show that $A^{\circ\circ} \subset \ol{\conv}(A \cup \bracs{0})$. 
+
+    Let $x_0 \in E \setminus \ol{\conv}(A \cup \bracs{0})$, then by the \hypreref[Hahn-Banach Theorem]{theorem:hahn-banach-geometric-2}, there exists $\phi: E \to \real$ such that:
+    \begin{enumerate}
+        \item $\phi$ is $\sigma(E, F)$-continuous.
+        \item $\phi(\ol{\conv}(A \cup \bracs{0})) \subset (-\infty, 1)$ and $\phi(x_0) > 1$. 
     \end{enumerate}
 
-    
-\end{proposition}
-
+    If $K = \real$, let $\Phi = \phi$. If $K = \complex$, then by \autoref{proposition:polarisation-linear}, the mapping $\Phi(x) = \phi(x) - i\phi(ix)$ is a $\sigma(E; F)$-continuous $K$-linear map on $E$ such that $\text{Re}(\Phi) = \phi$. By \autoref{lemma:duality-dual}, there exists $y \in F$ such that $\dpn{x, y}{\lambda} = \Phi(x)$ for all $x \in E$. In which case, for any $x \in E$,
+    \[
+    \text{Re}\dpn{x, y}{\lambda} = \text{Re}\Phi(x) = \phi(x)
+    \]
 
+    Therefore $y \in A^{\circ}$, but $\text{Re}\dpn{x_0, y}{\lambda} > 1$, so $x_0 \not\in A^{\circ\circ}$.
+\end{proof}
 

src/fa/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/notation.tex

@@ -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 $[\cdot]_H$ be a continuous seminorm on $H$, then there exists continuous seminorms $[\cdot]_E$ on $E$ and $[\cdot]_F$ on $F$ such that for any $f \in RS([a, b], G)$,
+    Let $[a, b] \subset \real$, $E, F, H$ be locally convex spaces, $E \times F \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map, $G: [a, b] \to F$, and $f \in RS([a, b], G)$, then for any continuous seminorms $[\cdot]_E: E \to [0, \infty)$, $[\cdot]_F: F \to [0, \infty)$, and $[\cdot]_H: H \to [0, \infty)$ such that $[xy]_H \le [x]_E[y]_F$ for all $x \in E$ and $y \in F$, 
     \[
     \braks{\int_a^bf dG}_H \le \sup_{x \in [a, b]}[f]_E \cdot [g]_{\text{var}, F}
     \]
 
 \end{proposition}
 \begin{proof}
-    By \autoref{proposition:tvs-convex-multilinear}, there exists continuous seminorms $[\cdot]_E$ on $E$ and $[\cdot]_F$ on $F$ such that $[xy]_H \le [x]_E[y]_F$ for all $(x, y) \in E \times F$.
-
     Let $(P = \seqfz{x_j}, c = \seqf{c_j}) \in \scp_t([a, b])$, then
     \begin{align*}
         [S(P, c, f, G)]_H &\le \sum_{j = 1}^n [f(c_j)[G(x_j) - G(x_{j - 1})]]_H \\
@@ -24,7 +20,7 @@
 
 \begin{proposition}
 \label{proposition:rs-complete}
-    Let $[a, b] \subset \real$, $E, F, H$ be locally convex spaces, and $E \times F \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map, and $G \in BV([a, b]; F)$.
+    Let $[a, b] \subset \real$, $E, F, H$ be locally convex spaces, $E \times F \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map, and $G \in BV([a, b]; F)$.
 
     For each continuous seminorm $\rho$ on $E$ and $f: [a, b] \to E$, define
     \[
@@ -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/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/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

@@ -41,3 +41,33 @@
     (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.
-
-    $(1) \Rightarrow (4)$: $X$ 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).
+    (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 $E$. 
 \end{proof}