Fixed typo.

document.tex

@@ -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/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

@@ -50,6 +50,12 @@
     Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, $U \subset E$ be open, and $n \in \natp$, then $D_\sigma^k(U; F)$/$\tilde D_\sigma^k(U; F)$ is the \textbf{space of $n$-fold $\sigma$/$\tilde \sigma$-differentiable functions} from $U$ to $F$. 
 \end{definition}
 
+\begin{definition}[Space of Continuously Differentiable Functions]
+\label{definition:continuously-differentiable-space}
+    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, $U \subset E$ be open, and $n \in \natp$, then $C_\sigma^k(U; F)$/$\tilde C_\sigma^k(U; F)$ is the \textbf{space of $n$-fold continuously $\sigma$/$\tilde \sigma$-differentiable functions} from $U$ to $F$. 
+\end{definition}
+
+
 \begin{theorem}[Symmetry of Higher Derivatives]
 \label{theorem:derivative-symmetric-frechet}
     Let $E, F$ be Banach spaces, $U \subset E$ be open, $n \in \natp$, and $f: U \to F$ be a function $n$-times Fréchet-differentiable at $x \in U$, then $D^nf(x) \in L^n(E; F)$ is symmetric.
@@ -191,3 +197,5 @@
 
     (2): By (1), $D^n_\sigma f$ is constant. 
 \end{proof}
+
+

src/dg/derivative/index.tex

@@ -6,4 +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/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

@@ -3,7 +3,7 @@
 
 \begin{definition}[Power Series]
 \label{definition:power-series}
-    Let $E$ be a normed space over $K \in \RC$, $F$ be a Banach space over $K$, $\bracsn{T_n}_0^\infty$ 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
+    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
     \[
     f(x) = \sum_{n = 0}^\infty T_n(x - a)^{(n)}
     \]
@@ -13,25 +13,39 @@
 
 \begin{definition}[Radius of Convergence]
 \label{definition:radius-of-convergence}
-    Let $E$ be a normed space over $K \in \RC$, $F$ be a Banach space over $K$, and $\sum_{n = 0}^\infty T_n(x - a)^{(n)}$ be a power series about $a \in E$, then $R \in [0, \infty]$ be defined by\footnote{Under the abuse that $1/\infty = 0$ and $1/0 =\infty$.}
+    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
     \[
-    \frac{1}{R} = \limsup_{n \to \infty}\norm{T_n}_{L^n(E; F)}^{1/n}
+    [T]_{L^n(E; F), \rho} = \sup_{x \in B_E(0, 1)^n}\rho(Tx)
+    \]
+    
+    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 $\sum_{n = 0}^\infty T_n(x - a)^{(n)}$}. For each $0 < r < R$, the series converges uniformly and absolutely on $B_E(a, r)$. 
+    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 $x \in B_E(a, r)$, 
     \[
-        \sum_{n = 0}^\infty \normn{T_n(x - a)^{(n)}}_F \le \sum_{n \in \natz} \norm{T_n}_{L^n(E; F)} \norm{x - a}_E^n \le \sum_{n \in \natz} \norm{T_n}_{L^n(E; F)} r^n
+        \sum_{n = 0}^\infty \rho(T_n(x - a)^{(n)}) \le \sum_{n \in \natz} [T_n]_{L^n(E; F), \rho} \norm{x - a}_E^n \le \sum_{n \in \natz}  r^n[T_n]_{L^n(E; F), \rho}
     \]
 
     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, 
     \[
-    \sum_{n = 0}^\infty \norm{T_n}_{L^n(E; F)} r^n \le \sum_{n = 0}^N \norm{T_n}_{L^n(E; F)}r^n + \sum_{n \ge N}\frac{r^n}{s^n} < \infty
+    \sum_{n = 0}^\infty r^n[T_n]_{L^n(E; F), \rho} \le \sum_{n = 0}^N r^n[T_n]_{L^n(E; F), \rho} + \sum_{n \ge N}\frac{r^n}{s^n} < \infty
     \]
 
-    As this estimate holds uniformly on $B_E(a, r)$, the series converges uniformly and absolutely on $B_E(a, r)$. 
+    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}
@@ -42,9 +56,9 @@
 
 \begin{theorem}[Termwise Differentiation]
 \label{theorem:termwise-differentiation}
-    Let $E$ be a normed space over $K \in \RC$, $F$ be a Banach 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 
+    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 $f \in C^\infty(B(a, R); F)$.
+        \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}})
@@ -54,14 +68,20 @@
     \end{enumerate}
 \end{theorem}
 \begin{proof}
-    (3): For each $n \in \natz$, let
+    (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 $\norm{S_n}_{L^n(E; L(E; F))} \le (n+1)\norm{T_{n+1}}_{L^{n+1}(E; F)}$. Since $(n+1)^{1/n}$ is convergent and $\{||T_n||_{L^n(E; F)}^{1/n}\}$ is bounded, 
+    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, 
     \[
-    \limsup_{n \to \infty} \norm{S_n}_{L^n(E; L(E; F))}^{1/n} \le \limsup_{n \to \infty}(n+1)^{1/n}\norm{T_{n+1}}_{L^{n+1}(E; F)}^{1/n} = \frac{1}{R}
+    \limsup_{n \to \infty} [S_n]_{L^n(E; L(E; F)), \rho}^{1/n} \le \limsup_{n \to \infty}(n+1)^{1/n}[T_{n+1}]_{L^{n+1}(E; F), \rho}^{1/n} \le \frac{1}{R}
     \]
 
     so the radius of convergence of the proposed series is at least $R$. 

src/dg/derivative/sets.tex

@@ -209,33 +209,3 @@
 
 
 
-\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}

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

@@ -12,6 +12,9 @@ Differential geometry is the study of things invariant under change of notation.
     $D_\sigma f(x_0)$ & $\sigma$-derivative of $f$ at $x_0$. & \autoref{definition:derivative-sets} \\
     $D_\sigma^n f$ & $n$-fold $\sigma$-derivative. & \autoref{definition:n-differentiable-sets} \\
     $D_\sigma^n(U; F)$ & $n$-fold $\sigma$-differentiable functions. & \autoref{definition:differentiable-space} \\
+    $\tilde D_\sigma^n(U; F)$ & $n$-fold $\tilde \sigma$-differentiable functions. & \autoref{definition:differentiable-space} \\
+    $C_\sigma^n(U; F)$ & $n$-fold continuously $\sigma$-differentiable functions. & \autoref{definition:continuously-differentiable-space} \\
+    $\tilde C_\sigma^n(U; F)$ & $n$-fold continuously $\tilde \sigma$-differentiable functions. & \autoref{definition:continuously-differentiable-space} \\
     $L^{(n)}_\sigma(E; F)$ & Codomain of derivatives. $L^{(0)}_\sigma(E; F) = F$, $L^{(n)}_\sigma(E; F) = L(E; L_\sigma^{(n-1)}(E; F))$, equipped with the $\sigma$-uniform topology. & \autoref{definition:higher-derivatives-codomain} \\
     $x^{(k)}$ & Tuple of $x$ repeated $k$ times. & \autoref{theorem:taylor-peano} \\
     $D^+f(x)$ & Right derivative of $f$ at $x$. & \autoref{definition:right-differentiable-mvt}

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 $[a, b] \subset \real$, $E, F, H$ be locally convex spaces, $E \times F \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map, $G: [a, b] \to F$, and $f \in RS([a, b], G)$, then for any continuous seminorms $[\cdot]_E: E \to [0, \infty)$, $[\cdot]_F: F \to [0, \infty)$, and $[\cdot]_H: H \to [0, \infty)$ such that $[xy]_H \le [x]_E[y]_F$ for all $x \in E$ and $y \in F$, 
-
-    Let $[\cdot]_H$ be a continuous seminorm on $H$, then there exists continuous seminorms $[\cdot]_E$ on $E$ and $[\cdot]_F$ on $F$ such that for any $f \in RS([a, b], G)$,
     \[
     \braks{\int_a^bf dG}_H \le \sup_{x \in [a, b]}[f]_E \cdot [g]_{\text{var}, F}
     \]
 
 \end{proposition}
 \begin{proof}
-    By \autoref{proposition:tvs-convex-multilinear}, there exists continuous seminorms $[\cdot]_E$ on $E$ and $[\cdot]_F$ on $F$ such that $[xy]_H \le [x]_E[y]_F$ for all $(x, y) \in E \times F$.
-
     Let $(P = \seqfz{x_j}, c = \seqf{c_j}) \in \scp_t([a, b])$, then
     \begin{align*}
         [S(P, c, f, G)]_H &\le \sum_{j = 1}^n [f(c_j)[G(x_j) - G(x_{j - 1})]]_H \\
@@ -24,7 +20,7 @@
 
 \begin{proposition}
 \label{proposition:rs-complete}
-    Let $[a, b] \subset \real$, $E, F, H$ be locally convex spaces, and $E \times F \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map, and $G \in BV([a, b]; F)$.
+    Let $[a, b] \subset \real$, $E, F, H$ be locally convex spaces, $E \times F \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map, and $G \in BV([a, b]; F)$.
 
     For each continuous seminorm $\rho$ on $E$ and $f: [a, b] \to E$, define
     \[
@@ -77,11 +73,13 @@
     Let $f \in C([a, b]; E)$, $G \in BV([a, b]; F)$, then
     \begin{enumerate}
         \item $f \in RS([a, b], G)$.
-        \item For any $\seq{(P_n, t_n)} \subset \scp_t([a, b])$ with $\sigma(P_n) \to 0$,
+        \item For equicontinuous family $\cf \subset C([a, b]; E)$ and $\seq{(P_n, t_n)} \subset \scp_t([a, b])$ with $\sigma(P_n) \to 0$,
         \[
         \int_a^b fdG = \limv{n}S(P_n, t_n, f, G)
         \]
 
+        uniformly for all $f \in \cf$. 
+
     \end{enumerate}
 \end{proposition}
 \begin{proof}
@@ -102,3 +100,47 @@
 
     In addition, for any $\seq{(P_n, t_n)}$ as in (2), $\limv{n}S(P_n, t_n, f, G)$ exists by sequential completeness. Since $\angles{S(P, c, f, G)}_{(P, c) \in \scp_t([a, b])}$ is Cauchy, the limit $\lim_{(P, c) \in \scp_t([a, b])}S(P, c, f, G)$ exists as well and is equal to $\limv{n}S(P_n, t_n, f, G)$.
 \end{proof}
+
+
+\begin{theorem}[Fubini's Theorem for Riemann-Stieltjes Integrals]
+\label{theorem:rs-fubini}
+    Let $[a, b], [c, d] \subset \real$, $E, F, G, H$ be a locally convex space over $K \in \RC$ with $H$ being sequentially complete, $E \times F \times G \to H$ with $(x, y, z) \mapsto xyz$ be a $3$-linear map\footnote{$E, F, G$ are assumed to be disjoint, so the product is well-defined regardless of the order of the terms.}, $\alpha \in BV([a, b]; F)$, $\beta \in BV([c, d]; G)$, and $f \in C([a, b] \times [c, d]; E)$, then
+    \[
+    \int_a^b \int_c^d f(s, t) \beta(dt) \alpha(ds) = \int_c^d\int_a^b f(s, t) \alpha(ds) \beta(dt)
+    \]
+\end{theorem}
+\begin{proof}
+    Let
+    \[
+    g: [a, b] \to L(F; H) \quad s \mapsto \int_c^d f(s, t) \beta(dt)
+    \]
+
+    then for any $(P = \seqfz{x_j}, c = \seqf{c_j}) \in \scp_t([a, b])$,
+    \begin{align*}
+        S(P, c, g, \alpha) &= \sum_{j = 1}^n g(c_j) [\alpha(x_j) - \alpha(x_{j-1})] \\
+        &=  \sum_{j = 1}^n \int_c^d f(c_j, t) \beta(dt) [\alpha(x_j) - \alpha(x_{j-1})] \\
+        &= \int_c^d S(P, c, f(\cdot, t), \alpha) \beta(dt)
+    \end{align*}
+
+
+    Since $\alpha \in BV([a, b]; F)$, by \autoref{proposition:rs-bv-continuous}, for any $\seq{(P_n, c_n)} \subset \scp_t([a, b])$, 
+    \[
+    \int_a^b \int_c^d f(s, t) \beta(dt) \alpha(ds) = \limv{n}S(P_n, c, g, \alpha)
+    \]
+
+    and
+    \[
+    \limv{n}S(P_n, c_n, f(\cdot, t), \alpha) = \int_a^b f(s, t) \alpha(ds)
+    \]
+
+    uniformly for all $t \in [c, d]$. Since $f \in C([a, b] \times [c, d]; E)$, $f$ is uniformly continuous by \autoref{proposition:uniform-continuous-compact}, and $\bracs{f(\cdot, t)|t \in [c, d]} \subset C([a, b]; E)$ is uniformly equicontinuous. As $\beta \in BV([c, d]; G)$, 
+    \[
+    \int_c^d\int_a^b f(s, t) \alpha(ds) \beta(dt) = \limv{n}\int_c^d S(P_n, c_n, f(\cdot, t), \alpha) \beta(dt)
+    \]
+
+    by \autoref{proposition:rs-complete}. 
+\end{proof}
+
+
+
+

src/fa/rs/rs.tex

@@ -61,7 +61,7 @@
 
 \end{theorem}
 \begin{proof}
-    Suppose that $f \in RS([a, b], G)$. Let $U \in \cn_K(0)$, then there exits $P_0 = \seqfz{x_j} \in \scp([a, b])$ such that $S(P, c, f, G) - \int_a^b fdG \in U$ for all $(P, c) \in \scp_t([a, b])$ with $P \ge P_0$. Let
+    Suppose that $f \in RS([a, b], G)$. Let $U \in \cn_H(0)$, then there exits $P_0 = \seqfz{x_j} \in \scp([a, b])$ such that $S(P, c, f, G) - \int_a^b fdG \in U$ for all $(P, c) \in \scp_t([a, b])$ with $P \ge P_0$. Let
     \[
     Q_0 = [x_0, x_1, x_1, \cdots, x_n, x_n]
     \]
@@ -69,10 +69,37 @@
     then for any $(Q = \seqfz[m]{y_j}, d = \seqf[m]{d_j}) \in \scp_t([a, b])$ with $Q \ge Q_0$,
     \[
     f(b)G(b) - f(a)G(a) - \int_a^b fdG - S(Q, d, G, f) =
-    \int_a^b fdG - S(Q', d', G, f)
+    S(Q', d', f, G) - \int_a^b fdG
     \]
 
-    by \autoref{lemma:sum-by-parts}, where $d$ and $Q'$ contain $\seqfz{x_j}$. Thus $(Q', d') \ge P_0$, and $\int_a^b fdG - S(Q', d', G, f) \in U$.
+    by \autoref{lemma:sum-by-parts}, where $d$ and $Q'$ contain $\seqfz{x_j}$. Thus $(Q', d') \ge P_0$, and $S(Q', d', f, G) - \int_a^b fdG \in U$.
 \end{proof}
 
 
+
+\begin{theorem}[Change of Variables]
+\label{theorem:rs-change-of-variables}
+    Let $[a, b] \subset \real$, $E, F, H$ be locally convex spaces over $K \in \RC$, $E \times F \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map, and $G \in C^1([a, b]; F)$, then for any bounded $f \in RS([a, b], G; E)$, 
+    \[
+    \int_a^b f(t) G(dt) = \int_a^b f(t) DG(t) dt
+    \]
+\end{theorem}
+\begin{proof}
+    Let $[\cdot]_H: H \to [0, \infty)$ be a continuous seminorm and $[\cdot]_E: E \to [0, \infty)$ and $[\cdot]_F: F \to [0, \infty)$ be continuous seminorms on $E$ and $F$, respectively, such that for any $x \in E$ and $y \in F$, $[xy]_H \le [x]_E[y]_F$. 
+
+    Since $G \in C^1([a, b]; F)$, $DG \in UC([a, b]; F)$ by \autoref{proposition:uniform-continuous-compact}. Thus there exists $\delta > 0$ such that $[DG(x) - DG(y)]_F < \eps$ for all $x, y \in [a, b]$ with $|x - y| \le \delta$. Let $(P = \seqfz{x_j}, c = \seqf{c_j}) \in \scp_t([a, b])$ with $\sigma(P) \le \delta$, then by the \hyperref[Mean Value Theorem]{theorem:mean-value-theorem-line}, for each $1 \le j \le n$, 
+    \begin{align*}
+        &G(x_j) - G(x_{j-1}) - (x_j - x_{j-1})DG(c_j) \\
+        &\in (x_j - x_{j-1})\ol{\text{Conv}}\bracs{DG(t) - DG(c_j)|t \in [x_{j-1}, x_j]}
+    \end{align*}
+
+    so
+    \[
+    [G(x_j) - G(x_{j-1}) - (x_j - x_{j-1})DG(c_j)]_F \le \eps(x_j - x_{j-1})
+    \]
+
+    and
+    \[
+        [S(P, c, f, G) - S(P, c, f \cdot DG, \text{Id})]_H \le \eps \cdot (b - a) \cdot \sup_{x \in [a, b]}[f(x)]_E
+    \]
+\end{proof}

src/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.
+    (6) $\Rightarrow$ (5): Let $E \subset X$ be closed and $x \in X \setminus E$. Since $\Delta = \bigcap_{U \in \fU}U$, there exists $U \in \fU$ such that $U(x) \subset E^c$. By \autoref{theorem:uniform-pseudometric}, there exists a pseudometric $d: X \times X \to [0, \infty)$ such that $d(x, E) > 0$. Thus the function $y \mapsto d(x, y)$ is a continuous function that separates $x$ and $E$. 
-
-    $(1) \Rightarrow (4)$: $X$ is T1 and satisfies (2) of \autoref{definition:regular} by \autoref{proposition:uniform-neighbourhoods}, so $X$ is regular.
-
-    $(4) \Rightarrow (3) \Rightarrow (2) \Rightarrow (1)$:  (T3) $\Rightarrow $ (T2) $\Rightarrow$ (T1) $\Rightarrow$ (T0).
 \end{proof}