Added the Riesz decomposition.

.vscode/project.code-snippets

@@ -171,5 +171,10 @@
   "scope": "latex",
     "prefix": "rank",
     "body": ["\\text{rk}"]
+  },
+  "Cite Proof": {
+  "scope": "latex",
+    "prefix": "cproof",
+    "body": ["[Proof, {{\\cite[$1]{$2}}}. ]$0"]
   }
 }

.vscode/settings.json

@@ -3,9 +3,17 @@
         {
             "label": "Watch",
             "task": "Watch"
+        },
+        {
+            "label": "Conservative Watch",
+            "task": "Conservative"
+        },
+        {
+            "label": "Compile All",
+            "task": "Build"
         }
     ],
     "latex.linting.enabled": false,
     "latex-workshop.latex.autoBuild.run": "never",
-    "latex-workshop.latex.texDirs": ["${workspaceFolder}"]
+    "latex-workshop.latex.search.rootFiles.include": ["document.tex"]
 }

.vscode/tasks.json

@@ -4,7 +4,7 @@
       {
         "label": "Build",
         "type": "shell",
-          "command": "npx spec compile",
+        "command": "npx spec compile --all",
         "windows": {
           "options": {
             "shell": {
@@ -39,6 +39,19 @@
             }
           }
         }
+      },
+      {
+        "label": "Conservative",
+        "type": "shell",
+        "command": "npx spec watch --conservative",
+        "windows": {
+          "options": {
+            "shell": {
+              "executable": "cmd.exe",
+              "args": ["/d", "/c"]
+            }
+          }
+        }
       }
   ]
-  }
+}

document.tex

@@ -1,4 +1,4 @@
-%\documentclass{report}
+\documentclass{}
 \usepackage{amssymb, amsmath, hyperref}
 \usepackage{preamble}
 
@@ -11,10 +11,10 @@ Hello this is all my notes.
 \input{./src/fa/index}
 \input{./src/measure/index}
 \input{./src/dg/index}
+\input{./src/op/index}
 %\input{./src/process/index}
 
 \bibliographystyle{alpha} % We choose the "plain" reference style
 \bibliography{refs.bib} % Entries are in the refs.bib file
 
-
 \end{document}

refs.bib

@@ -112,3 +112,75 @@
   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}
+}
+@book{Zhu,
+  title={An Introduction to Operator Algebras},
+  author={Zhu, K.},
+  isbn={9780849378751},
+  lccn={93007172},
+  series={Studies in Advanced Mathematics},
+  url={https://books.google.ca/books?id=XHLj7bz8hOIC},
+  year={1993},
+  publisher={Taylor \& Francis}
+}
+@book{FollandHarmonic,
+  title={A Course in Abstract Harmonic Analysis},
+  author={Folland, G.B.},
+  isbn={9781498727150},
+  series={Textbooks in Mathematics},
+  url={https://books.google.ca/books?id=z-GYCgAAQBAJ},
+  year={2016},
+  publisher={CRC Press}
+}
+@book{Takesaki1,
+  title={Theory of Operator Algebras I},
+  author={Takesaki, M.},
+  isbn={9783540422488},
+  lccn={79013655},
+  series={Encyclopaedia of Mathematical Sciences},
+  url={https://books.google.ca/books?id=38QIwQEACAAJ},
+  year={2001},
+  publisher={Springer Berlin Heidelberg}
+}
+
+@unpublished{MarcouxNotes,
+  author       = {Laurent W. Marcoux},
+  title        = {An Introduction to Banach Algebras and Operator Algebras},
+  year         = {2021},
+  month        = apr,
+  url          = {https://www.math.uwaterloo.ca/~lwmarcou/notes/pmath810.pdf},
+  note         = {Unpublished lecture notes, Department of Pure Mathematics,
+                  University of Waterloo. Draft dated April 30, 2021.
+                  Potential victim of Google search scraping. },
+  institution  = {University of Waterloo},
+}
+
+
+
+@techreport{aronszajn1964interpolation,
+  title={Interpolation spaces and interpolation methods},
+  author={Aronszajn, Nachman and Gagliardo, Emilio},
+  year={1964}
+}
+@book{brudnyi1991interpolation,
+  title={Interpolation functors and interpolation spaces},
+  author={Brudnyi, Yu A and Krugljak, N Ya},
+  volume={47},
+  year={1991},
+  publisher={Elsevier}
+}
+@book{PietschHistory,
+  title={History of Banach spaces and linear operators},
+  author={Pietsch, Albrecht},
+  year={2007},
+  publisher={Springer}
+}

spec.toml

@@ -17,8 +17,17 @@ primaryColour = "violet"
 neutralColour = "grey"
 searchLimit = 16
 maxSearchPages = 48
-recentChanges = 0
-tableOfContentsDepth = 1
+
+recentChanges = 10
+tableOfContentsDepth = 3
+tableOfContentsUnfoldDepth = 2
+tableOfContentsShortDepth = 1
+
 hoverPreview = false
+
 copyLabelButton = false
+
 advertiseSpec = true
+
+displayComments = true
+allowComments = true

src/cat/cat/cat-func.tex

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

src/cat/cat/tensor.tex

@@ -34,6 +34,7 @@
         \]
     \end{enumerate}
 
+    
     The module $A = \bigoplus_{i \in I}A_i$ is the \textbf{direct sum} of $\seqi{A}$. 
 \end{definition}
 \begin{proof}
@@ -206,6 +207,7 @@
         \]
     \end{enumerate}
 
+    
     (1), (2): Let $\bigotimes_{j = 1}^n E_j = M/N$ and
     \[
     \iota: \prod_{j = 1}^n E_j \to \bigotimes_{j = 1}^n E_j \quad (x_1, \cdots, x_n) \mapsto (x_1, \cdots, x_n) + N

src/cat/cat/universal.tex

@@ -8,6 +8,7 @@
         \item \textbf{universally attracting} if for every $A \in \obj{\catc}$, there exists a unique $f \in \mor{A, P}$.
         \item \textbf{universally repelling} if for every $A \in \obj{\catc}$, there exists a unique $f \in \mor{P, A}$.
     \end{enumerate}
+
     If $P$ is universally attracting or repelling, then $P$ is a \textbf{universal object}.
 
     If $P, Q \in \obj{\catc}$ are both universally attracting/repelling, then they are isomorphic.
@@ -63,11 +64,13 @@
         \item For any $i \in I$, $i \lesssim i$.
         \item For any $i, j, k \in I$ such that $i \lesssim j$ and $j \lesssim k$, $i \lesssim k$.
     \end{enumerate}
+
     and one of the following holds:
     \begin{enumerate}
         \item[(3U)] For any $i, j \in I$, there exists $k \in I$ with $i, j \lesssim k$.
         \item[(3D)] For any $i, j \in I$, there exists $k \in I$ with $k \lesssim i, j$.
     \end{enumerate}
+
     The directed set is \textbf{upward-directed} if it satisfies (3U), and \textbf{downward-directed} if it satisfies (3D).
 \end{definition}
 
@@ -86,6 +89,7 @@
         \item For each $i, j \in I$ with $i \lesssim j$, $f^i_j \in \mor{A_i, A_j}$.
         \item For each $i, j, k \in I$ with $i \lesssim j \lesssim k$, $f^j_k \circ f^i_j = f^i_k$.
     \end{enumerate}
+
     If $I$ is upward/downward-directed, then $(\seqi{A}, \bracsn{f^i_j| i, j \in I, i \lesssim j})$ is upward/downward-directed.
 \end{definition}
 

src/cat/gluing/index.tex

@@ -9,6 +9,7 @@
         \item[(a)] $\bigcup_{i \in I}U_i = X$.
         \item[(b)] For each $i, j \in I$, either $U_i \cap U_j = \emptyset$, or $f_i|_{U_i \cap U_j} = f_j|_{U_i \cap U_j}$.
     \end{enumerate}
+
     then there exists a unique $f: X \to Y$ such that $f|_{U_i} = f_i$ for all $i \in I$.
 \end{lemma}
 \begin{proof}
@@ -17,9 +18,11 @@
         \item By assumption (a), $\bracs{x|(x, y) \in \Gamma} = \bigcup_{i \in I}U_i = X$.
         \item For any $x \in X$, there exists $y \in Y$ with $(x, y) \in \Gamma$, and $i \in I$ such that $(x, y) \in \Gamma_i$. If $(x, y') \in \Gamma_j \subset \Gamma$, then $x \in U_i \cap U_j \ne \emptyset$. By assumption (b), $y = y'$.
     \end{enumerate}
+
     Thus $\Gamma$ is the graph of a function $f: X \to Y$ with $f|_{U_i} = f_i$ for all $i \in I$.
 \end{proof}
 
+
 \begin{lemma}[Gluing for Linear Functions]
 \label{lemma:glue-linear}
     Let $E, F$ be vector spaces over a field $K$, $\fF$ be a family of subspaces of $E$, and $\bracs{T_V}_{V \in \fF}$ with $T_V \in \hom(V; F)$ for all $V \in \fF$. If:
@@ -28,6 +31,7 @@
         \item[(b)] For each $V, W \in \fF$, $T_V|_{V \cap W} = T_W|_{V \cap W}$.
         \item[(c)] $\fF$ is upward-directed with respect to includion.
     \end{enumerate}
+
     then there exists a unique $T \in \hom(E; F)$ such that $T|_{V} = T_V$ for all $V \in \fF$.
 \end{lemma}
 \begin{proof}

src/cat/notation/index.tex

@@ -13,6 +13,9 @@
     $\lim_{\longleftarrow} A_i$ & 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 \\
+    $R[x]$ & Ring of polynomials over $R$. & N/A \\
+    $F(x)$ & Field of fractions over $F$. & N/A \\ 
 \end{tabular}
 

src/cat/tricks/dyadic.tex

@@ -44,6 +44,7 @@
         \item[(b)] For each $x, y \in G$, $x + y \ge x, y$. 
     \end{enumerate}
 
+    
     For each $x \in \mathbb{D} \cap [0, 1)$, let $\phi(x) = \sum_{n \in M(x)}g_n$, then
     \begin{enumerate}
         \item For any $x, y \in \mathbb{D} \cap [0, 1)$ such that $x + y \in (0, 1)$, $\phi(x) + \phi(y) \le \phi(x + y)$.

src/dg/complex/derivative.tex

@@ -0,0 +1,289 @@
+\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 \hyperref[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}[Holomorphic]
+\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 $r > 0$ with $\ol{B(0, r)} \subset U$, there exists $\seq{a_n} \subset E$ such that for each $z \in B(z_0, r/2)$, 
+        \[
+        f(z) = \sum_{n = 0}^\infty a_n(z - z_0)^n
+        \]
+
+        where the radius of convergence of the series is 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{holomorphic/complex analytic} on $U$. 
+\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
+    \[
+    g(z) = \sum_{k = 0}^\infty \frac{1}{k!} D^kf(z_0)(z - z_0)^k
+    \]
+
+    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-integral} 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}
+
+\begin{remark}
+\label{remark:weak-holomorphy}
+    Since weak holomorphy is equivalent to strong holomorphy, most common results of scalar-valued holomorphic functions may be transferred to the vector valued case with little effort. Since this chapter represents my first attempt at learning complex analysis, and I happen to require vector-valued results, most results here are stated and proven in the vector-valued case. However, this is strictly unnecessary if one already knows the scalar-valued results. 
+\end{remark}
+
+
+

src/dg/complex/entire.tex

@@ -0,0 +1,92 @@
+\section{Entire Functions}
+\label{section:entire-holomorphic}
+
+\begin{definition}[Entire Function]
+\label{definition:entire-function}
+    Let $E$ be a complete separated locally convex space over $\complex$, then an \textbf{entire $E$-valued function} is an $E$-valued holomorphic function on $\complex$.
+\end{definition}
+
+\begin{proposition}
+\label{proposition:entire-power-series}
+    Let $E$ be a complete separated locally convex space over $\complex$, and $f \in H(\complex; E)$, then for any $z_0 \in \complex$ and $z \in \complex$,
+    \[
+    f(z) = \sum_{k = 0}^\infty \frac{1}{k!} D^kf(z_0)(z - z_0)^k
+    \]
+
+    where the radius of convergence of the series is infinite. 
+\end{proposition}
+\begin{proof}
+    See (4) of \autoref{definition:complex-analytic}. 
+\end{proof}
+
+\begin{theorem}[Liouville]
+\label{theorem:liouville}
+    Let $E$ be a complete separated locally convex space over $\complex$ and $f \in H(\complex; E)$. If $f$ is bounded, then $f$ is constant. 
+\end{theorem}
+\begin{proof}
+    Let $x_0 \in \complex$ and $[\cdot]_E: [E, \infty)$ be a continuous seminorm on $E$, then by \hyperref[Cauchy's Estimate]{corollary:cauchy-estimate},
+    \[
+    [Df(z_0)]_E \le \frac{1}{r} \sup_{z \in \ol{B(z_0, r)}}[f(z)]_E
+    \]
+
+    for all $r > 0$. Therefore $Df = 0$, and $f$ is constant. 
+\end{proof}
+
+\begin{theorem}[Fundamental Theorem of Algebra]
+\label{theorem:fundamental-theorem-of-algebra}
+    Let $p \in \complex[z]$ be a non-constant polynomial, then there exists $z \in \complex$ such that $f(z) = 0$. 
+\end{theorem}
+\begin{proof}
+    Let $p \in \complex[z]$ such that $p(z) \ne 0$ for all $z \in \complex$. Let $f = 1/p$, then $f \in H(\complex; \complex) \cap C_0(\complex; \complex)$, so $f$ is bounded. By \hyperref[Liouville's Theorem]{theorem:liouville}, $f$ and thus $p$ is constant. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:entire-logarithm}
+    Let $f \in H(\complex; \complex)$, then the following are equivalent:
+    \begin{enumerate}
+        \item $f(z) \ne 0$ for all $z \in \complex$. 
+        \item There exists $\ell \in H(\complex; \complex)$ such that $f = e^\ell$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1) $\Rightarrow$ (2): For each $z \in \complex$, let
+    \[
+    g(z) = \int_0^w \frac{f'(w)}{f(w)}dw
+    \]
+
+    then $g \in H(\complex; \complex)$ with $g' = f/f'$, and
+    \begin{align*}
+        \frac{d}{dz}\braks{f(z)e^{-g(z)}} &= f'(z)e^{-g(z)} - f(z)g'(z)e^{-g(z)} \\
+        &= f'(z)e^{-g(z)} - f'(z)e^{-g(z)} = 0
+    \end{align*}
+
+    Therefore $fe^{-g}$ is a non-zero constant. For any $\lambda \in \complex$ with $e^\lambda = fe^{-g}$, let $h = g + \lambda$, then $f = e^h$. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:entire-constant}
+    Let $f \in H(\complex; \complex)$ such that:
+    \begin{enumerate}[label=(\alph*)]
+        \item $f(0) = 1$. 
+        \item $f'(0) = 0$. 
+        \item $0 < |f| < |\exp|$. 
+    \end{enumerate}
+
+    then $f = 1$. 
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Lemma I.4.4]{Zhu}}}. ]
+    By (c) and \autoref{proposition:entire-logarithm}, there exists $g \in H(\complex; \complex)$ such that $f = e^g$. Since $f(0) = 1$, $g(0) = 0$. As $f'(0) = g'(0)f(0) = 0$, $g'(0) = 0$ as well. 
+
+    From (c), $|g(z)| \le e^{|z|}$ for each $z \in \complex$. Thus for every $r > 0$ and $z \in B_\complex(0, r)$, $|g(z)| \le |g(z) - 2r|$, and
+    \[
+    g_r: \ol{B_\complex(0, r)} \setminus \bracs{0} \to \complex \quad z \mapsto \frac{r^2}{z^2} \frac{g(z)}{2r - g(z)}
+    \]
+
+    extends to a holomorphic function on $B_\complex(0, r)$ by \autoref{proposition:zero-finite-multiplicity}. Since $|g_r(z)| \le 1$ for all $z \in \partial B_\complex(0, r)$, $g_r(z) \le 1$ for all $z \in B_\complex(0, r)$ by the \hyperref[Maximum Modulus Theorem]{theorem:maximum-modulus-theorem}. 
+
+    Finally, since $|g_r(z)| \le 1$ for all $z \in \complex$ and $r > |z|$, $g = 0$ and $f = 1$. 
+\end{proof}
+
+
+
+

src/dg/complex/index.tex

@@ -0,0 +1,12 @@
+\chapter{Complex Analysis}
+\label{chap:complex-analysis}
+
+
+\input{./derivative.tex}
+\input{./zero.tex}
+\input{./sphere.tex}
+\input{./space.tex}
+\input{./log.tex}
+\input{./entire.tex}
+\input{./runge.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/complex/runge.tex

@@ -0,0 +1,178 @@
+\section{Runge's Theorem}
+\label{section:runge}
+
+\begin{proposition}
+\label{proposition:existence-curves}
+    Let $K \subset \complex$ be compact and $U \in \cn_\complex(K)$, then there exists closed rectifiable curves $\seqf{\gamma_j}$ such that for any separated locally convex space $E$, 
+    \begin{enumerate}
+        \item For every $f \in H(U; E)$ and $z_0 \in K$, 
+        \[
+        f(z_0) = \frac{1}{2\pi i}\sum_{j = 1}^n \int_{\gamma_j} \frac{f(z)}{z - z_0}dz
+        \]
+        \item For every $W \in \cn_\complex(U)$, $f \in H(W; E)$, and $z_0 \in W \setminus U$, 
+        \[
+        \frac{1}{2\pi i}\sum_{j = 1}^n \int_{\gamma_j} \frac{f(z)}{z - z_0}dz = 0
+        \]
+    \end{enumerate}
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition VIII.1.1]{ConwayComplex}}}. ]
+    Via \hyperref[fattening]{proposition:distance-compact}, let $V \in \cn_\complex^o(K)$ precompact with $\ol V \subset U$. Identify $\complex = \real^2$, then since $V$ is precompact, there exists $\delta > 0$ and $\seqf{(x_j, y_j)} \subset U$ such that:
+    \begin{enumerate}
+        \item For each $1 \le j \le n$, $R_j = [x_j, x_j + \delta] \times [y_j, y_j + \delta] \subset U$. 
+        \item $\bigcup_{j = 1}^n R_j \supset V$. 
+        \item $x_j, y_j = 0 \mod \delta$. 
+    \end{enumerate}
+
+    In other words, $V$ is covered with a grid of squares with side-length $\delta$ and corners $\seqf{(x_j, y_j)}$. Now, for each $1 \le j \le n$ and $t \in [0, 1]$, let
+    \begin{align*}
+        \gamma_{j, \rightarrow}(t) &= (1 - t)(x_j, y_j) + t(x_j + \delta, y_j) \\
+        \gamma_{j, \uparrow}(t) &= (1 - t)(x_j + \delta, y_j) + t(x_j + \delta, y_j + \delta) \\
+        \gamma_{j, \leftarrow}(t) &= (1 - t)(x_j + \delta, y_j + \delta) + t(x_j, y_j + \delta) \\
+        \gamma_{j, \downarrow}(t) &= (1 - t)(x_j, y_j + \delta) + t(x_j, y_j) \\
+    \end{align*}
+
+    and $\gamma_j = \gamma_{j, \downarrow} \cdot \gamma_{j, \leftarrow} \cdot \gamma_{j, \uparrow} \cdot \gamma_{j, \rightarrow}$ be their concatenation, then for each $z \in U \setminus \partial R_j$, 
+    \[
+    \frac{1}{2\pi i}\int_{\gamma_j} \frac{f(z)}{z - z_0}dz = \begin{cases}
+        f(z_0) &z \in R_j^o \\
+        0 &z \in U \setminus R_j
+    \end{cases}
+    \]
+
+    by \hyperref[Cauchy's Integral Formula]{theorem:cauchy-formula}. Let $\seqf[N]{\mu_j}$ be an enumeration of
+    \[
+    \bigcup_{j = 1}^n \bracs{\gamma_{j, \downarrow}, \gamma_{j, \leftarrow}, \gamma_{j, \uparrow}, \gamma_{j, \rightarrow}}
+    \]
+
+    then for each $z_0 \in \bigcup_{j = 1}^n R_j^o$ and $f \in H(U; E)$, 
+    \[
+    f(z_0) = \frac{1}{2\pi i}\sum_{j = 1}^N \int_{\mu_j}\frac{f(z)}{z - z_0}dz
+    \]
+
+    From here, it is sufficient to eliminate line segments that intersect $K$ and ensure that the remaining segments form a collection of loops. Let $1 \le j \le k \le N$, then $(\mu_j, \mu_k)$ is \textit{redundant} if for each $t \in [0, 1]$, $\mu_j(t) = \mu_k(1 - t)$. If $(\mu_j, \mu_k)$ are redundant, then
+    \begin{enumerate}[start=3]
+        \item There exists no $1 \le l \le N$ with $l \ne j, k$ such that $(\mu_j, \mu_l)$ or $(\mu_k, \mu_l)$ is redundant. 
+        \item $\int_{\mu_j}\frac{f(z)}{z - z_0}dz + \int_{\mu_k}\frac{f(z)}{z - z_0}dz = 0$.
+    \end{enumerate}
+
+    so every line segment either cannot form a redundant pair, or forms a unique one. 
+
+    By relabeling, let $\bracs{\mu_j|1 \le j \le m}$ such that for each $1 \le j \le m$, there exists no $1 \le k \le n$ such that $(\mu_j, \mu_k)$ is redundant. By (5), for each $z_0 \in \bigcup_{j = 1}^n R_j^o$ and $f \in H(U; E)$, 
+    \[
+    f(z_0) = \frac{1}{2\pi i}\sum_{j = 1}^m \int_{\mu_j}\frac{f(z)}{z - z_0}dz
+    \]
+
+    Let $1 \le j \le N$ such that $\mu_j([0, 1]) \cap K \ne \emptyset$. Since $V \in \cn_\complex(K)$, there exists $1 \le k \le N$ such that $(\mu_j, \mu_k)$ are redundant by (2) and (3). Therefore $\bigcup_{j = 1}^m \mu_j([0, 1]) \cap K = \emptyset$. Since $\bigcup_{j = 1}^n R_j^o$ is dense in $K$, the above also holds for all $z \in K$. 
+
+    Finally, let $1 \le j \le m$. Since $\mu_j$ does not form a redundant pair, $\mu_j(1)$ intersects at most two distinct squares by (3). In which case, there must exist $1 \le k \le m$ such that $\mu_k(0) = \mu_j(1)$. Therefore $\seqf[m]{\mu_j}$ forms a collection of closed rectifiable curves. 
+\end{proof}
+
+
+\begin{lemma}
+\label{lemma:rational-curve-approximation}
+    Let $\gamma \in C([a, b]; \complex)$ be a rectifiable curve, $K \subset \complex$ such that $K \cap \gamma([a, b]) = \emptyset$, $f \in C(\gamma([a, b]); \complex)$, and $\eps > 0$, then there exists $R \in \complex(z)$ such that:
+    \begin{enumerate}
+        \item $R \in H(\complex \setminus \gamma([a, b]); \complex)$. 
+        \item For each $z_0 \in K$, 
+        \[
+        \abs{\int_{\gamma} \frac{f(z)}{z - z_0}dz - R(z)} < \eps
+        \]
+    \end{enumerate}
+\end{lemma}
+\begin{proof}[Proof, {{\cite[Lemma VIII.1.5]{ConwayComplex}}}. ]
+    Since the mapping
+    \[
+    \varphi: K \times [a, b] \to \complex \quad (z_0, t) \mapsto \frac{f \circ \gamma(t)}{\gamma(t) - z}
+    \]
+
+    is continuous, it is uniformly continuous by \autoref{proposition:uniform-continuous-compact}. Hence the mappings $\bracs{\varphi(z_0, \cdot)|t \in [a, b]}$ are uniformly equicontinuous. Thus there exists $\delta > 0$ such that for each $s, t \in [a, b]$ with $|s - t| \le \delta$, 
+    \[
+    \abs{\frac{f \circ \gamma(s)}{\gamma(s) - z_0} - \frac{f \circ \gamma(t)}{\gamma(t) - z_0}} < \eps
+    \]
+
+    for all $z_0 \in K$. 
+
+    Let $(P = \seqfz{t_j}) \in \scp([a, b])$ with $\sigma(t) < \delta$, and
+    \[
+    R(z) = \sum_{j = 1}^n f \circ \gamma(t_j)\frac{\gamma(t_j) - \gamma(t_{j-1})}{\gamma(t_j) - z}
+    \]
+
+    then $R \in \complex(z) \cap H(\complex \setminus \gamma([a, b]); \complex)$, and for each $z_0 \in K$, 
+    \begin{align*}
+        \abs{\int_\gamma \frac{f(z)}{z - z_0}dz - R(z)} &\le 
+        \sum_{j = 1}^n \int_{t_{j-1}}^{t_j}\abs{\frac{f \circ \gamma(t)}{\gamma(t) - z_0} - \frac{f \circ \gamma(t_j)}{\gamma(t_j) - z}}\gamma(dt) \\
+        &\le \eps \norm{\gamma}_{\text{var}}
+    \end{align*}
+\end{proof}
+
+\begin{lemma}[Pole Pushing]
+\label{lemma:pole-pushing}
+    Let $K \subset \complex$ be compact and $P \subset \complex_\infty \setminus K$ such that $P$ intersects every connected component of $\complex_\infty \setminus K$, then for every $a \in \complex \setminus K$ and $\eps > 0$, there exists $R \in \complex(z)$ such that:
+    \begin{enumerate}
+        \item $R \in H(\complex_\infty \setminus P; \complex)$. 
+        \item For every $z \in K$, $|R(z) - 1/(z - a)| < \eps$. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}[Proof, {{\cite[Lemma III.1.10]{ConwayComplex}}}. ]
+    First suppose that $\infty \not\in P$. Let $A \subset \complex_\infty \setminus K$ be the collection of elements such that the lemma holds\footnote{Under the identification that $1/(z - \infty) = 0$.}. 
+
+    Let $a_0 \in \complex \setminus K$, then there exists $V \in \cn_\complex(a)$ compact such that $V \cap K = \emptyset$. In which case, the function
+    \[
+    K \times V \to \complex \quad (z, a) \mapsto \frac{1}{z - a}
+    \]
+
+    is continuous, and hence uniformly continuous by \autoref{proposition:uniform-continuous-compact}. Thus $1/(z - a) \to 1/(z - a_0)$ uniformly on compact sets as $a \to a_0$, and if $a_0 \in \ol{A}$, then $a_0 \in A$ as well. Therefore $A$ is a closed subset of $\complex_\infty \setminus K$. 
+
+    Now, let $a_0 \in A \cap \complex$, then by \autoref{proposition:distance-compact}, $r = d(a_0, K) > 0$. Let $a \in B_\complex(a_0, r)$, then for each $z \in K$, 
+    \begin{align*}
+    \frac{1}{z - a} &= \frac{1}{z - a_0}\frac{z - a_0}{z - a} = \frac{1}{z - a_0}\braks{\frac{z - a}{z - a_0}}^{-1} \\
+    &= \frac{1}{z - a_0}\braks{1 - \frac{a - a_0}{z - a_0}}^{-1}
+    \end{align*}
+    
+    Since $\sup_{z \in K}|a - a_0|/|z - a_0| < 1$, 
+    \[
+    \frac{1}{z - a} = \frac{1}{z - a_0}\sum_{n = 0}^\infty \braks{\frac{a - a_0}{z - a_0}}^n
+    \]
+
+    where the convergence is uniform on $K$. Therefore $B_\complex(a_0, r) \subset A$ as well. 
+
+    Finally, for each $a \in \complex \setminus K$ with $|a| > \sup_{z \in K}|z|$, 
+    \[
+        \frac{1}{z - a} = -\frac{1}{a\paren{1 - \frac{z}{a}}} = -\frac{1}{a}\sum_{n = 0}^\infty \braks{\frac{z}{a}}^n
+    \]
+
+    where the convergence is uniform on $K$, so there exists a neighbourhood of $\infty$ that is contained in $A$. 
+    
+    Since $P \cup \bracs{\infty} \subset A$, and $A$ is an open and closed subset of $C_\infty \setminus K$, $A$ is a union of connected components of $\complex_\infty \setminus K$ by \autoref{lemma:union-connected-components}. Given that $P$ intersects every connected component of $\complex_\infty \setminus K$, the lemma holds for all $a \in \complex \setminus K$. 
+\end{proof}
+
+\begin{theorem}[Runge]
+\label{theorem:runge}
+    Let $K \subset \complex$ be compact and $P \subset \complex_\infty \setminus K$ such that $P$ intersects every connected component of $\complex_\infty \setminus K$, then $\complex(x) \cap H(\complex_\infty \setminus P; \complex)$ is dense in $H(K; \complex)$ with respect to the uniform topology. 
+\end{theorem}
+\begin{proof}
+    Let $U \in \cn_\complex(K)$, $f \in H(U; \complex)$, and $\eps > 0$. By \autoref{proposition:existence-curves}, there exists closed rectifiable curves $\seqf{\gamma_j}$ in $U \setminus V$ such that for each $z_0 \in K$, 
+    \[
+    f(z_0) = \frac{1}{2\pi i}\sum_{j = 1}^n \int_{\gamma_j}\frac{f(z)}{z - z_0}dz
+    \]
+
+    Let $T$ be the union of the images of $\seqf{\gamma_j}$, then by \autoref{lemma:rational-curve-approximation}, there exists $\seqf{R_j} \subset \complex(z) \cap H(\complex \setminus T; \complex)$ such that for each $z_0 \in K$ and $1 \le j \le n$, 
+    \[
+    \abs{\frac{1}{2\pi i}\int_{\gamma_j}\frac{f(z)}{z - z_0}dz - R_j(z)} < \frac{\eps}{n}
+    \]
+
+    Thus
+    \[
+    \abs{f(z_0) - \sum_{j = 1}^n R_j(z)} < \eps
+    \]
+    
+    for all $z_0 \in K$. By the \hyperref[pole pushing lemma]{lemma:pole-pushing}, there exists $S \in \complex(z) \cap H(\complex_\infty \setminus P; \complex)$ such that $|S(z_0) - \sum_{j = 1}^n R_j(z_0)| < \eps$ for all $z_0 \in K$. Therefore $|S(z_0) - f(z_0)| < 2\eps$ for all $z_0 \in K$.
+\end{proof}
+
+\begin{corollary}
+\label{corollary:runge-rational-approximation}
+    Let $K \subset \complex$ be compact, then $\complex(z) \cap H(K; \complex)$ is dense in $H(K; \complex)$. 
+\end{corollary}
+
+
+

src/dg/complex/space.tex

@@ -0,0 +1,52 @@
+\section{Spaces of Holomorphic Functions}
+\label{section:holomorphic-function-space}
+
+\begin{definition}[Space of Holomorphic Functions]
+\label{definition:holomorphic-function-space}
+    Let $E$ be a complete separated locally convex space over $\complex$ and $U \subset \complex$ be open, then $H(U; E)$ is the \textbf{space of $E$-valued holomorphic functions on $U$}, equipped with the topology of uniform convergence on compact sets. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:holomorphic-complete}
+    Let $E$ be a complete separated locally convex space over $\complex$ and $U \subset \complex$ be open, then $H(U; E)$ is complete. 
+\end{proposition}
+\begin{proof}
+    By \hyperref[Cauchy's estimate]{corollary:cauchy-estimate}, uniform convergence on compact sets is equivalent to uniform convergence of derivatives of all orders on compact sets. Since $U$ is locally compact, uniform limits of holomorphic functions are holomorphic by \autoref{theorem:differentiable-uniform-limit}.
+\end{proof}
+
+\begin{theorem}[Montel]
+\label{theorem:montel}
+    Let $E$ be a complete separated locally convex space over $\complex$, $U \subset \complex$ be open, and $\cf \subset H(U; E)$, then the following are equivalent:
+    \begin{enumerate}[label=(B\arabic*)]
+        \item $\cf$ is equicontinuous, and for each $x \in U$, $\cf(x) = \bracs{f(x)|f \in \cf}$ is bounded.
+        \item $\cf$ is bounded in $H(U; E)$.
+    \end{enumerate}
+
+    and the following are equivalent:
+    \begin{enumerate}[label=(C\arabic*)]
+        \item $\cf$ is precompact in $H(U; E)$. 
+        \item $\cf$ is bounded in $H(U; E)$, and for each $x \in U$, $\cf(x) = \bracs{f(x)|f \in \cf}$ is precompact.  
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    (B1) $\Rightarrow$ (B2): Let $K \subset U$ be compact and $V \in \cn_E(0)$ be circled. Since $\cf$ is equicontinuous, for each $x \in K$, there exists $U_x \in \cn_U(x)$ such that $f(y) - f(x) \in V$ for all $y \in U_x$ and $f \in \cf$. By compactness of $K$, there exists $\seqf{x_j} \subset K$ such that $K \subset \bigcup_{j = 1}^n U_{x_j}$. Since $B = \bigcup_{j = 1}^n \cf(x_j)$ is bounded, there exists $\lambda > 0$ such that $\lambda V \supset B$. In which case, 
+    \[
+    (\lambda + 1)V \supset B + V \supset \bigcup_{x \in K}^n \cf(x)
+    \]
+
+    (B2) $\Rightarrow$ (B1): By \hyperref[Cauchy's Estimate]{corollary:cauchy-estimate}, $\bracsn{Df|f \in \cf}$ is also uniformly bounded on every compact set. Thus $\cf$ is equicontinuous. 
+
+    (C1) $\Leftrightarrow$ (C2): By the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}.
+\end{proof}
+
+\begin{definition}[Space of Holomorphic Functions Near a Set]
+\label{definition:holomorphic-function-space-near}
+    Let $E$ be a complete separated locally convex space over $\complex$ and $A \subset \complex$. Direct $\cn_{\complex}^o(A)$ under reverse inclusion, then the inductive limit
+    \[
+    H(A; E) = \varinjlim_{U \in \cn_{\complex}^o(A)} H(U; E)
+    \]
+
+    is the \textbf{space of holomorphic functions near} $A$, and is of type (LF). 
+\end{definition}
+
+

src/dg/complex/sphere.tex

@@ -0,0 +1,16 @@
+\section{The Riemann Sphere}
+\label{section:riemann-sphere}
+
+
+\begin{definition}[Extended Complex Plane]
+\label{definition:extended-complex-plane}
+    Let $\complex$ be the complex plane, then its one-point compactification $\complex_\infty = \complex \sqcup \bracs{\infty}$ is the \textbf{extended complex plane}. 
+\end{definition}
+
+\begin{definition}[Holomorphic on $\complex_\infty$]
+\label{definition:holomorphic-on-sphere}
+    Let $E$ be a separated locally convex space and $f \in C(\complex_\infty; E)$, then $f$ is \textbf{holomorphic at $\infty$} if $z \mapsto f(z^{-1})$ (under the identification that $1/0 = \infty$) is holomorphic at $0$. 
+\end{definition}
+
+
+

src/dg/complex/zero.tex

@@ -0,0 +1,96 @@
+\section{Zeroes of Holomorphic Functions}
+\label{section:complex-zeroes}
+
+\begin{definition}[Zero]
+\label{definition:complex-zero}
+    Let $E$ be a complete separated locally convex space over $\complex$, $U \subset \complex$ be open, $f \in H(U; E)$, and $z_0 \in U$, then $z_0$ is a \textbf{zero of $f$ of multiplicity $n \in \natp$} if there exists $g \in H(U; E)$ such that $f(z) = (z - a)^n g(z)$ for all $z \in \bracs{g \ne 0}$. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:complex-zero-gymnastics}
+    Let $E$ be a complete separated locally convex space over $\complex$, $U \subset \complex$ be open and connected, and $f \in H(U; E)$, then the following are equivalent:
+    \begin{enumerate}
+        \item $f = 0$. 
+        \item There exists $z_0 \in U$ such that $D^nf(z_0) = 0$ for all $n \in \natp$. 
+        \item There exists $z_0 \in U$ such that $z_0 \in \ol{\bracs{f = 0} \setminus \bracs{z_0}}$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (3) $\Rightarrow$ (2): Let $r > 0$ and $\bracs{a_n}_0^\infty \subset E$ such that $f(z) = \sum_{n = 0}^\infty a_n (z - z_0)^n$ for all $z \in B(z_0, r)$, where the series has a radius of convergence of at least $r$. By continuity of $f$, $a_0 = f(0) = 0$. 
+    
+    Suppose inductively that $a_k = 0$ for each $0 \le k \le n$. Let $[\cdot]_E: E \to [0, \infty)$ be a continuous seminorm, then for each $z \in B(z_0, r) \cap \bracs{f = 0} \setminus \bracs{z_0}$, 
+    \begin{align*}
+        f(z) &= \sum_{k = n + 1}^\infty a_k(z - z_0)^k \\
+        [a_{n+1}(z - z_0)^{n+1}]_E &\le \sum_{k = n+2}^\infty [a_k(z - z_0)^k]_E = \sum_{k = n+2}^\infty |z - z_0|^k[a_k]_E  \\
+        [a_{n+1}]_E &\le |z - z_0| \sum_{k = 0}^\infty |z - z_0|^{k}[a_{k+n+2}]_E
+    \end{align*}
+
+    Since the series $\sum_{n = 0}^\infty a_n (z - z_0)^n$ has a radius of convergence of at least $r$, $\limsup_{n \to \infty}[a_n]_E^{1/n} \le 1/r$, so there exists $N \ge n+2$ such that $[a_k]_E^{1/k} \le 2/r$ for all $k \ge N$. For each $s \in (0, r/4)$, $B(z_0, s) \cap \bracs{f = 0} \setminus \bracs{z_0} \ne \emptyset$, thus
+    \[
+        [a_{n+1}]_E \le s \braks{\sum_{k = 0}^N \paren{\frac{r}{4}}^k[a_{k+n+2}]_E + \sum_{k > N}\frac{1}{2^k}}
+    \]
+
+    As the above holds for all $s \in (0, r/4)$, $[a_{n+1}]_E = 0$. Therefore $a_n = 0$ for all $n \in \natp$. 
+
+    (2) $\Rightarrow$ (1): By continuity, $\bracs{f = 0}$ is closed. By local power series expansion and (3), $\bracs{f = 0}$ is open. As $U$ is connected, $U = \bracs{f = 0}$. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:zero-finite-multiplicity}
+    Let $E$ be a complete separated locally convex space over $\complex$, $U \subset \complex$ be connected, and $f \in H(U; E) \setminus \bracs{0}$, then for each $z_0 \in \bracs{f = 0}$, there exists $n \ge 1$ and $g \in H(U; \complex)$ such that:
+    \begin{enumerate}
+        \item $g(a) \ne 0$.
+        \item For each $z \in U$, $f(z) = (z - z_0)^ng(z)$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    Since $f \ne 0$, $n = \min\bracs{n \in \natp| D^nf(z_0) \ne 0} < \infty$ by \autoref{proposition:complex-zero-gymnastics}. Let
+    \[
+    g: U \to E \quad z \mapsto \begin{cases}
+        \frac{f(z)}{(z - z_0)^n} &z \ne z_0 \\
+        \frac{1}{n!}D^nf(z_0) &z = z_0
+    \end{cases}
+    \]
+
+    then $g$ is continuous on $U$ and holomorphic on $U \setminus \bracs{z_0}$. As $f \in H(U; E)$, there exists $r > 0$ and $\bracs{a_k}_0^\infty \subset E$ such that $f(z) = \sum_{k = 0}^\infty a_k (z - z_0)^k$ for all $z \in B(z_0, r)$, where the series has a radius of convergence of at least $r$. By assumption on $n$, for each $z \in B(z_0, r)$, 
+    \begin{align*}
+        f(z) &= \sum_{k = 0}^\infty a_k (z - z_0)^k = \sum_{k = n}^\infty a_k(z - z_0)^k \\
+        g(z) &= \sum_{k = 0}^\infty a_{k+n}(z - z_0)^{k}
+    \end{align*}
+
+    where the series $\sum_{k = 0}^\infty a_{k+n}(z - z_0)^{k}$ has the same radius of convergence. Therefore $g$ is holomorphic at $z_0$. 
+\end{proof}
+
+\begin{theorem}[Maximum Modulus Theorem]
+\label{theorem:maximum-modulus-theorem}
+    Let $E$ be a complete separated locally convex space over $\complex$, $U \subset \complex$ be open and connected, then
+    \begin{enumerate}
+        \item For each $f \in H(U; E)$ and continuous seminorm $[\cdot]_E: E \to [0, \infty)$, if there exists $z_0 \in U$ with
+        \[
+        [f(z_0)]_E = \sup_{z \in U}[f(z)]_E
+        \]
+
+        then $[f]_E$ is constant. 
+        \item For any $f \in H(U; \complex)$, if there exists $z_0 \in U$ with
+        \[
+        |f(z_0)| = \sup_{z \in U}|f(z)|
+        \]
+
+        then $f$ is constant. 
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    (1): Let $M = \sup_{z \in U}[f(z)]_E$. Since $[f]_E$ is continuous, $\bracsn{[f]_E = M}$ is closed. By \hyperref[Cauchy's Integral Formula]{theorem:cauchy-formula} and \autoref{proposition:rs-bound}, $\bracsn{[f]_E = M}$ is open. As $U$ is connected, $\bracsn{[f]_E = M} = U$. 
+
+    (2): By (1), $|f|$ is constant. After rescaling, assume without loss of generality that $\ol f = 1/f$. In this case, for each $z \in U$, 
+    \[
+    D\ol f(z) = \lim_{\substack{h \to 0 \\ h \in \real}}\frac{\ol{f(x + h) - f(x)}}{h} = \ol{Df}(z)
+    \]
+
+    and
+    \[
+    D\ol f(z) = \lim_{\substack{h \to 0 \\ h \in \real}}\frac{\ol{f(x + ih) - f(x)}}{ih} = -\ol{Df}(z)
+    \]
+
+    so $Df = 0$ and $f$ is constant. 
+\end{proof}

src/dg/derivative/euclid.tex

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

src/dg/derivative/higher.tex

@@ -1,24 +1,60 @@
 \section{Higher Derivatives}
 \label{section:higher-derivatives}
 
-\begin{definition}[$n$-Fold Differentiability]
-\label{definition:n-differentiable-sets}
-    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset B(E)$ be 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}
@@ -30,7 +66,7 @@
     A(h, k) = f(x + h + k) - f(x + h) - f(x + k) + f(x)
     \]
 
-    then there exists $r_1 \in \mathcal{R}_{B(E)}$ such that
+    then there exists $r_1 \in \mathcal{R}_{\mathfrak{B}(E)}$ such that
     \begin{align*}
         A(h, k) &= Df(x + h)(k) + Df(x)(k) \\
         &+ [f(x + h + k) - f(x + h) - Df(x + h)(k)] \\
@@ -54,7 +90,7 @@
     \end{align*}
     
 
-    Now, there exists $r_2, r_3 \in \mathcal{R}_{B(E)}$ such that for any $k \in B(0, r)$,
+    Now, there exists $r_2, r_3 \in \mathcal{R}_{\mathfrak{B}(E)}$ such that for any $k \in B(0, r)$,
     \begin{align*}
     DB_h(k) &= Df(x + h + k) - Df(x + k) - Df(x + h) + Df(x) \\
     &= D^2f(x)(h + k) + Df(x) - D^2f(x)(h) \\
@@ -87,55 +123,79 @@
 
 \begin{theorem}[Symmetry of Higher Derivatives]
 \label{theorem:derivative-symmetric}
-    Let $E$ be a topological vector space over $K \in \RC$, $\sigma \subset B(E)$ be an ideal that includes all bounded sets contained in finite-dimensional spaces, $F$ be a separated locally convex space over $K$, $U \subset E$ be open, and $f: E \to F$ be $n$-fold $\sigma$-differentiable at $x_0 \in U$, then $D_\sigma^nf(x_0) \in B_\sigma^n(E; F)$ is symmetric.
+    Let $E$ be a topological vector space over $K \in \RC$, $\sigma \subset \mathfrak{B}(E)$ be an ideal that includes all bounded sets contained in finite-dimensional spaces, $F$ be a separated locally convex space over $K$, $U \subset E$ be open, and $f: E \to F$ be $n$-fold $\tilde \sigma$-differentiable at $x_0 \in U$, then $D_\sigma^nf(x_0) \in L_\sigma^{(n)}(E; F)$ is symmetric.
 \end{theorem}
 \begin{proof}[Proof {{\cite[Proposition 4.5.14]{Bogachev}}}. ]
     Let $\seqf{h_j} \subset E$, $E_0$ be the subspace generated by $\seqf{h_j}$, and $g = f|_{E_0 \cap U}: E_0 \cap U \to F$. Since $\sigma$ includes all bounded sets contained in finite-dimensional spaces, for any $\phi \in F^*$, the mapping $\phi \circ g: E_0 \cap U \to K$ is $n$-times Fréchet-differentiable, with
     \[
-    D_{B(E_0)}^n(\phi \circ g)(x_0) = \phi \circ D_\sigma^n g(x_0)
+    D_{\mathfrak{B}(E_0)}^n(\phi \circ g)(x_0) = \phi \circ D_\sigma^n g(x_0)
     \]
 
-    by the \hyperref[chain rule]{proposition:chain-rule-sets-conditions}. By \autoref{theorem:derivative-symmetric-frechet}, $\phi \circ D_\sigma^n g(x_0) \in L^n(E_0; K)$ is symmetric. As this holds for any $\seqf{h_j} \subset E$ and $\phi \in F^*$, $D_{\sigma}^n g(x_0) \in B_\sigma^n(E; F)$ is symmetric by the \hyperref[Hahn-Banach theorem]{proposition:hahn-banach-utility}.
+    by the \hyperref[chain rule]{proposition:chain-rule-sets-conditions}. By \autoref{theorem:derivative-symmetric-frechet}, $\phi \circ D_\sigma^n g(x_0) \in L^n(E_0; K)$ is symmetric. As this holds for any $\seqf{h_j} \subset E$ and $\phi \in F^*$, $D_{\sigma}^n g(x_0) \in B_\sigma^{(n)}(E; F)$ is symmetric by the \hyperref[Hahn-Banach theorem]{proposition:hahn-banach-utility}.
 \end{proof}
 
-\begin{proposition}[Power Rule]
-\label{proposition:multilinear-derivative}
-    Let $E$ be a topological vector space, $\sigma \subset B(E)$ be an ideal that includes all bounded sets contained in finite-dimensional spaces, $F$ be a Hausdorff locally convex space, and
+\begin{theorem}[Power Rule]
+\label{theorem:power-rule}
+    Let $E$ be a topological vector space, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, $F$ be a separated locally convex space, $T \in B^n_\sigma(E; F)$.
+
+    For each $1 \le m \le n$, let $\text{Inj}([m]; [n])$ be the set of injective mappings from $[m]$ to $[n]$. For any $\phi \in \text{Inj}([m]; [n])$, denote $\phi^c \in \text{Inj}([n-m]; [n] \setminus \phi([m]))$ as the unique increasing injective map. For each $h \in E^{n-m}$, $k \in E^m$, and $1 \le j \le n$, write
     \[
-    T \in \underbrace{L(E; L(E; \cdots L(E; F) \cdots ))}_{n \text{ times}} \subset B^n(E; F)
+    (h, k)_\phi = \begin{cases}
+        k_{\phi^{-1}(j)} &j \in \phi([m]) \\
+        h_{(\phi^c)^{-1}(j)} &j \not\in \phi([m])
+    \end{cases}
     \]
 
-    be symmetric. For any $x \in E$ and $1 \le k \le n$, let $x^{(k)}$ denote the tuple of $x$ repeated $k$ times, then:
+    For any $x \in E$, denote $x^{(m)}$ as the tuple of $x$ repeated $m$ times, then the mapping $f: E \to F$ defined by $x \mapsto T(x^{(n)})$ is infinitely $\tilde\sigma$-differentiable on $E$, where
     \begin{enumerate}
-        \item The mapping $f: E \to F \quad x \mapsto T(x^{(n)})$ is infinitely $\sigma$-differentiable on $E$.
-        \item For each $1 \le k \le n$ and $x, h \in E$,
+        \item For each $1 \le m \le n$, $x \in E$, and $h \in E^m$, 
         \[
-        Df(x)(h_1, \cdots, h_k) = \frac{n!}{(n-k)!} T(x^{(n-k)}, h_1, \cdots, h_k)
+        D^{m}_\sigma f(x)(h) = \sum_{\phi \in \text{Inj}([m]; [n])}T((x^{(n-m)}, h)_\phi]
         \]
 
-        In particular, $D^kf = n! \cdot T$.
-        \item For each $k > n$ and $x \in E$, $Df(x) = 0$.
+        In particular, 
+        \[
+        D^{n}_\sigma f(x)(h) = \sum_{\phi \in S_n}T(h_{\phi(1)}, \cdots, h_{\phi(n)})
+        \]
+
+        \item For each $m > n$ and $x \in E$, $D^m_\sigma f(x) = 0$.
     \end{enumerate}
-\end{proposition}
+
+    Notably, if $T \in L^{(n)}(E; F)$ is symmetric or $T \in L^n(E; F)$, then $T$ is infinitely $\sigma$-differentiable on $E$. 
+\end{theorem}
 \begin{proof}
-    Suppose inductively that (2) holds for $0 \le k \le n$. Let $G =  B^{k}_\sigma(E; F)$, then $D^k_\sigma f \in B^{n-k}_\sigma(E; G)$ under the identification $B^n_\sigma(E; F) = B^{n-k}_\sigma(E; B^k_\sigma(E; F))$ in \autoref{proposition:multilinear-identify}. By \autoref{theorem:derivative-symmetric}, $D^k_\sigma f$ is also symmetric, so using the Binomial formula,
+    (1): Let $0 \le m \le n - 1$ and suppose inductively that (1) holds for $m$. For each $x, h \in E$, $S \subset [n-m]$, and $1 \le j \le n - m$,
+    \[
+    [(x, h)_S]_j = \begin{cases}
+        h &j \in S \\
+        x &j \not\in S
+    \end{cases}
+    \]
+    
+    By the Binomial formula, for each $h \in E$ and $k \in E^{m}$,
     \begin{align*}
-        D^k_\sigma f(x + h) &= \sum_{r = 0}^{n-k}{n - k \choose r}D^k_\sigma f(x^{(n-k-r)}, h^{(r)}) \\
-        &= f(x) + (n-k)D^k_\sigma f(x^{(n-k-1)}, h) \\
-        &+ \underbrace{\sum_{r = 2}^{n-k}{n - k \choose r}D^k_\sigma f(x^{(n-k-r)}, h^{(r)})}_{r(h)}
+        D^{m}_\sigma f(x + h)(k) &= \sum_{\phi \in \text{Inj}([m]; [n])}T[((x+h)^{(n-m)}, k)_\phi] \\
+        &= \sum_{\phi \in \text{Inj}([m]; [n])}\sum_{S \subset [n-m]}T[((x, h)_S, k)_\phi]
     \end{align*}
 
-    For each $k \ge 2$, let $A \in \sigma$ and $U \in \cn_F(0)$, then since $D^k_\sigma f \in B^{n-k}_\sigma(E; F)$, there exists $t > 0$ such that
+    For $\ell \ge 2$, maps in $B_\sigma^{\ell}(E; F)$ are $\sigma$-small, so
     \[
-    \frac{D^k_\sigma f(x^{(n-k)}, (sA)^{(k)})}{t} = s^{k-1}D^k_\sigma f(x^{(n-k)}, A^{(k)}) \subset U
+    r(h) = \sum_{\phi \in \text{Inj}([m]; [n])}\sum_{\substack{S \subset [n-m] \\ |S| \ge 2}}T[((x, h)_S, k)_\phi]
     \]
 
-    for all $s \in (0, t)$. Hence $r \in \mathcal{R}_\sigma(E; G)$, and
+    is $\sigma$-small. Hence
+    \begin{align*}
+        &D^{(m)}_\sigma f(x + h)(k) - D_\sigma^{(m)}f(x)(k) \\
+        &= \sum_{\phi \in \text{Inj}([m]; [n])}\sum_{j = 1}^{n-m}T[((x, h)_{\bracs{j}}, k)_\phi] + r(h) \\
+        &= \sum_{\phi \in \text{Inj}([m+1]; [n])}T[(x^{(n-m-1)}, (h, k))_\phi] + r(h)
+    \end{align*}
+
+    and for any $h \in E^{m+1}$,
     \[
-    D^{k+1}_\sigma f(x + h) = f(x) + \frac{n!}{(n-k-1)!}T(x^{(n-k-1)}, h_1, \cdots, h_{k+1}) + r(h)
+    D_\sigma^{(m+1)}f(x)(h) = \sum_{\phi \in \text{Inj}([m+1]; [n])}T[(x^{(n-m-1)}, h)_\phi]
     \]
 
-    by the inductive hypothesis.
-
-    (3): Since $D^n_\sigma f$ is constant, $D^k_\sigma f = 0$ for all $k > n$.
+    (2): By (1), $D^n_\sigma f$ is constant. 
 \end{proof}
+
+

src/dg/derivative/index.tex

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

src/dg/derivative/inverse.tex

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

src/dg/derivative/mvt.tex

@@ -18,6 +18,7 @@
         \item[(a)] $f, g$ are right-differentiable on $[a, b] \setminus N$.
         \item[(b)] For every $x \in [a, b] \setminus N$, $D^+f(x) \le D^+g(x)$.
     \end{enumerate}
+
     then for any $x \in [a, b]$, $f(x) - f(a) \le g(x) - g(a)$.
 \end{lemma}
 \begin{proof}
@@ -52,6 +53,7 @@
         \item For each $x \in [a, b] \setminus N$, $D^+f(x) \in D^+g(x)B$.
         \item $g$ is non-decreasing.
     \end{enumerate}
+
     then
     \[
     f(b) - f(a) \in [g(b) - g(a)]B
@@ -103,7 +105,7 @@
 
 \begin{theorem}[Mean Value Theorem]
 \label{theorem:mean-value-theorem}
-    Let $E$ be a topological vector space, $F$ be a separated locally convex space, $V \subset E$ be open and star shaped at $x \in V$, $f: V \to F$ be Gateau-differentiable on $V$, then for any $y \in V$,
+    Let $E$ be a topological vector space over $K \in \RC$, $F$ be a separated locally convex space over $K$, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, $V \subset E$ be open and star shaped at $x \in V$, and $f: V \to F$ be $\tilde \sigma$-differentiable on $V$, then for any $y \in V$,
     \[
     f(y) - f(x) \in \overline{\text{Conv}\bracs{Df(z)(y - x)|z \in (x, y)}}
     \]
@@ -111,7 +113,7 @@
     where $[x, y] = \bracs{(1 - t)x + ty|y \in [0, 1]}$.
 \end{theorem}
 \begin{proof}
-    Let $g: [0, 1] \to F$ be defined by $g(t) = f((1 - t)x + ty)$. Since $f$ is Gateaux-differentiable, $g$ is differentiable by the chain rule \autoref{proposition:chain-rule-sets-conditions} with $Dg(t) = Df((1 - t)x + ty)(y - x)$, and continuous by \autoref{proposition:derivative-sets-real}.
+    Let $g: [0, 1] \to F$ be defined by $g(t) = f((1 - t)x + ty)$, then $g$ is differentiable with $Dg(t) = Df((1 - t)x + ty)(y - x)$, and continuous by \autoref{proposition:derivative-sets-real}.
 
     By the \hyperref[Mean Value Theorem]{theorem:mean-value-theorem-line}, $f(y) - f(x) = g(1) - g(0)$ is contained in
     \[
@@ -122,7 +124,7 @@
 
 \begin{proposition}
 \label{proposition:zero-derivative-constant}
-    Let $E$ be a topological vector space, $F$ be a separated locally convex space, $V \subset E$ be open and connected, $f: V \to F$ be Gateaux-differentiable on $V$ such that $Df(x) = 0$ for all $x \in V$, then $f$ is constant.
+    Let $E$ be a topological vector space over $K \in \RC$, $F$ be a separated locally convex space over $K$, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, $V \subset E$ be open and connected, and $f: V \to F$ be $\tilde \sigma$-differentiable on $V$ with $Df = 0$, then $f$ is constant.
 \end{proposition}
 \begin{proof}
     Let $x \in V$, then for any $U \in \cn(0)$ circled with $U + x \subset V$ and $y \in U + x$,

src/dg/derivative/partial.tex

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

src/dg/derivative/power.tex

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

src/dg/derivative/sets.tex

@@ -3,12 +3,13 @@
 
 \begin{definition}[Small]
 \label{definition:differentiation-small}
-    Let $E, F$ be TVSs over $K \in \RC$, $\sigma \subset B(E)$ be a covering ideal, $r: E \to F$, and $n \in \natz$, then the following are equivalent:
+    Let $E, F$ be TVSs over $K \in \RC$, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, $r: E \to F$, and $n \in \natz$, then the following are equivalent:
     \begin{enumerate}
         \item For each $A \in \sigma$, $r(th)/t^n \to 0$ uniformly on $A$.
         \item If $r_t(x) = r(tx)/t^n$, then $r_t \to 0$ as $t \to 0$ with respect to the $\sigma$-uniform topology on $F^E$.
         \item For each $A \in \sigma$, $\seq{a_k} \subset A$, and $\seq{t_k} \subset K \setminus \bracs{0}$ with $t_k \to 0$ as $n \to \infty$, $r(t_ka_k)/t_k^n \to 0$ as $n \to \infty$.
     \end{enumerate}
+
     If the above holds, then $r$ is \textbf{$\sigma$-small of order $n$}.
 
     The set $\mathcal{R}_\sigma^n(E; F)$ is the $K$-vector space of all $\sigma$-small functions of order $n$ from $E$ to $F$. For simplicity, $\mathcal{R}_\sigma(E; F)$ denotes $\mathcal{R}_\sigma^1(E; F)$.
@@ -16,43 +17,44 @@
 
 \begin{proposition}
 \label{proposition:differentiation-sets}
-    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset B(E)$ be a covering ideal, $B_\sigma(E; F)$ be the space of linear operators bounded on sets in $\sigma$, and $\mathcal{R}_\sigma(E; F)$ be the space of $\sigma$-small functions, then $(B_\sigma(E; F), \mathcal{R}_\sigma(E; F))$ is a system of derivatives and remainders.
+    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, and $\mathcal{R}_\sigma(E; F)$ be the space of $\sigma$-small functions, and $\mathcal{H} \subset B_\sigma(E; F)$ be a subspace, $(\mathcal{H}, \mathcal{R}_\sigma(E; F))$ is a system of derivatives and remainders.
 \end{proposition}
 \begin{proof}
     Let $T \in B_\sigma(E; F)$ and suppose that there exists $V \in \cn_E(0)$ circled and $r \in \mathcal{R}_\sigma(E; F)$ such that $T|_V = r|_V$. For any $x \in V$, $\bracs{x} \in \sigma$, so $T(x) = \lim_{t \downto 0}T(tx)/t = 0$ as $F$ is separated.
 \end{proof}
 
 
-\begin{definition}[Derivative]
+\begin{definition}[$\sigma$-Derivative]
 \label{definition:derivative-sets}
-    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset B(E)$ be a covering ideal, $U \subset E$ be open, $f: U \to F$, and $x_0 \in U$, then $f$ is \textbf{$\sigma$-differentiable at $x_0$} if there exists $V \in \cn_E(0)$, $T \in L(E; F)$, and $r \in \mathcal{R}_\sigma(E; F)$ such that
+    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, $U \subset E$ be open, $f: U \to F$, and $x_0 \in U$, then $f$ is \textbf{$\tilde \sigma$-differentiable at $x_0$} if there exists $V \in \cn_E(0)$, $T \in B_\sigma(E; F)$, and $r \in \mathcal{R}_\sigma(E; F)$ such that
     \[
     f(x_0 + h) = f(x_0) + Th + r(h)
     \]
 
     for all $h \in V$.
 
-    The linear map $T \in L(E; F)$ is the \textbf{$\sigma$-derivative of $f$ at $x_0$}, denoted $D_{\sigma}f(x_0)$.
+    The linear map $T \in B_\sigma(E; F)$ is the \textbf{$\tilde \sigma$-derivative of $f$ at $x_0$}, denoted $D_{\tilde \sigma}f(x_0)$. If $T \in L(E; F)$, then $f$ is \textbf{$\sigma$-differentiable at $x_0$}, and $T$ is the \textbf{$\sigma$-derivative of $f$ at $x_0$}.
 \end{definition}
 
 \begin{definition}[Differentiable]
 \label{definition:differentiable-sets}
-    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset B(E)$ be a covering ideal, $U \subset E$ be open, and $f: U \to F$, then $f$ is \textbf{$\sigma$-differentiable on $U$} if it is $\sigma$-differentiable at every point in $U$. In which case, the map $D_\sigma f: U \to L(E; F)$ is the \textbf{$\sigma$-derivative} of $f$.
+    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, $U \subset E$ be open, and $f: U \to F$, then $f$ is \textbf{$\sigma$/$\tilde \sigma$-differentiable on $U$} if it is $\sigma$/$\tilde \sigma$-differentiable at every point in $U$. In which case, the map $D_\sigma f: U \to B_\sigma(E; F)$ is the \textbf{$\sigma$/$\tilde \sigma$-derivative} of $f$.
 \end{definition}
 
 \begin{definition}
 \label{definition:derivative-garden}
-    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma^E_{\text{Fin}}, \sigma^E_{c}, \sigma^E_b \subset 2^E$ be the collection of all finite, compact, and bounded subsets, respectively, then differentiability with respect to $\sigma^E_{\text{Fin}}, \sigma^E_{c}, \sigma^E_b$ correspond to \textbf{Gateaux}, \textbf{Hadamard}, and \textbf{Fréchet} differentiability.
+    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma^E_{\text{Fin}}, \sigma^E_{c}, \sigma^E_b \subset 2^E$ be the collection of all finite, precompact, and bounded subsets, respectively, then differentiability with respect to $\sigma^E_{\text{Fin}}, \sigma^E_{c}, \sigma^E_b$ correspond to \textbf{Gateaux}, \textbf{Hadamard}, and \textbf{Fréchet} differentiability.
 \end{definition}
 
 
 \begin{proposition}[Chain Rule]
 \label{proposition:chain-rule-sets}
-    Let $E$, $F$, $G$, be TVSs over $K \in \RC$ with $F, G$ being separated, $\sigma \subset B(E)$ and $\tau \subset B(F)$ be covering ideals. If:
+    Let $E$, $F$, $G$, be TVSs over $K \in \RC$ with $F, G$ being separated, $\sigma \subset \mathfrak{B}(E)$ and $\tau \subset \mathfrak{B}(F)$ be covering ideals. If:
     \begin{enumerate}
         \item[(a)] For any $r \in \mathcal{R}_\sigma(E; F)$ and $T \in L(F; G)$, $T \circ r \in \mathcal{R}_\sigma(E; G)$.
         \item[(b)] For any $r \in \mathcal{R}_\sigma(E; F)$, $T \in L(E; F)$, and $s \in \mathcal{R}_\tau(F; G)$, $s \circ (T + r) \in \mathcal{R}_\sigma(E; G)$.
     \end{enumerate}
+
     then for any $U \subset E$ and $V \subset F$ open, $f: U \to V$ $\sigma$-differentiable at $x_0 \in U$, $g: V \to F$ $\tau$-differentiable at $f(x_0) \in V$, $g \circ f: U \to F$ is $\sigma$-differentiable at $x_0$ with
     \[
     D_\sigma(g \circ f)(x_0) = D_\tau g(f(x_0)) \circ D_\sigma f(x_0)
@@ -80,16 +82,18 @@
 
 \begin{proposition}[{{\cite[Corollary 4.5.4]{Bogachev}}}]
 \label{proposition:chain-rule-sets-conditions}
-    Let $E$, $F$, $G$, be TVSs over $K \in \RC$ with $F, G$ being separated. If $\sigma \subset B(E)$ and $\tau \subset B(F)$ correspond to the following families of sets on $E$ and $F$:
+    Let $E$, $F$, $G$, be TVSs over $K \in \RC$ with $F, G$ being separated. If $\sigma \subset \mathfrak{B}(E)$ and $\tau \subset \mathfrak{B}(F)$ correspond to the following families of sets on $E$ and $F$:
     \begin{enumerate}
-        \item Compact sets.
+        \item Precompact sets.
         \item Bounded sets.
     \end{enumerate}
+
     then
     \begin{enumerate}
         \item For any $r \in \mathcal{R}_\sigma(E; F)$ and $T \in L(F; G)$, $T \circ r \in \mathcal{R}_\sigma(E; G)$.
         \item For any $r \in \mathcal{R}_\sigma(E; F)$, $T \in L(E; F)$, and $s \in \mathcal{R}_\tau(F; G)$, $s \circ (T + r) \in \mathcal{R}_\sigma(E; G)$.
     \end{enumerate}
+
     and by \autoref{proposition:chain-rule-sets}, $\sigma$-derivatives and $\tau$-derivatives satisfy the Chain rule.
 \end{proposition}
 \begin{proof}
@@ -124,35 +128,84 @@
     A method of extending this sense of differentiability is to require that \textit{every} extension of the function to some open set, or to the entire space is differentiable. Given that this paves way to confusion for related definitions of differentiability, this definition is not formally included here.
 \end{remark}
 
-
-
-\begin{proposition}
-\label{proposition:derivative-sets-real}
-    Let $E$ be a separated topological vector space and $\sigma \subset B(\real)$ be 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 $S \in \fF$, 
     \[
-    \lim_{t \to 0}\frac{f(x_0 + t) - f(x_0) - Tt}{t} = \lim_{t \to 0}\frac{f(x_0 + t) - f(x_0)}{t} - v = 0
+    (f - \varphi)(x + h) - (f - \varphi)(x) \in \overline{\bracs{(g - \varphi)(x + h) - (g - \varphi)(x)|g \in S}}
     \]
     
-    and $Df(x_0) = T$.
+    By the \hyperref[Mean Value Theorem]{theorem:mean-value-theorem}, for any $g \in \tilde D^1_\sigma(U; F)$, 
+    \[
+    (g - \varphi)(x + h) - (g - \varphi)(x) \in \ol{\conv}\bracs{D_\sigma(g - \varphi)(x + th)h|t \in [0, 1]}
+    \]
+
+    Hence
+    \[
+    (f - \varphi)(x + h) - (f - \varphi)(x) \in  \ol{\conv}\bracs{D_\sigma(g - \varphi)(x + th)h|g \in S, t \in [0, 1]}
+    \]
+
+    so for any $t \in (0, 1)$ and $A \in \sigma$, 
+    \begin{align*}
+    &(f - \varphi)(x + tA) - (f - \varphi)(x) \\
+    &\subset \ol{\conv}\bracs{D_\sigma(g - \varphi)(x + sth)th|g \in S, s \in [0, 1], h \in A} \\
+    &= t\ol{\conv}\bracs{D_\sigma(g - \varphi)(x + sth)h|g \in S, s \in [0, 1], h \in A}
+    \end{align*}
+
+    and
+    \begin{align*}
+    &t^{-1}[(f - \varphi)(x + tA) - (f - \varphi)(x)] \\
+    &\subset \ol{\conv}\bracs{D_\sigma(g - \varphi)(x + sth)h|g \in S, s \in [0, 1], h \in A} \\
+    &\subset \ol{\conv}\bracs{D_\sigma(g - \varphi)(x + sh)h|g \in S, s \in [0, 1], h \in A}
+    \end{align*}
+
+    In addition, since $D_\sigma(\fF) \to f^{(1)}$ pointwise, 
+    \[
+    t^{-1}(f^{(1)} - D_\sigma\varphi)(x)(tA) \subset \ol{\conv}\bracs{D_\sigma(g - \varphi)(x + sh)h|g \in S, s \in [0, 1], h \in A}
+    \]
+
+    as well. 
+
+    Now, let $V \in \cn_F(0)$ be convex and circled. Using assumption (b), let $S \in \fF$ such that for any $\varphi \in S$, 
+    \[
+    \ol{\conv}\bracs{D_\sigma(g - \varphi)(x + sh)h|g \in S, s \in [0, 1], h \in A} \subset V
+    \]
+    
+    Fix $\varphi \in S$, then as $\varphi$ is differentiable at $x$, there exists $\delta \in (0, 1)$ such that
+    \[
+    t^{-1}[\varphi(x + tA) - \varphi(x) - D_\sigma\varphi(x)(tA)] \subset V
+    \]
+
+    for all $t \in (0, \delta)$. 
+
+    So
+    \[
+    t^{-1}[f(x + tA) - f(x) - f^{(1)}(x)(tA)] \subset 3V
+    \]
+
+    for all $t \in (0, \delta)$. Therefore $f$ is $\tilde \sigma$-differentiable at $x$ with $D_\sigma f(x) = f^{(1)}(x)$. 
+    
 \end{proof}
+
+
+

src/dg/derivative/taylor.tex

@@ -62,19 +62,20 @@
 
 \begin{theorem}[Taylor's Formula, Peano Remainder]
 \label{theorem:taylor-peano}
-    Let $E$ be a topological vector space, $\sigma \subset B(E)$ be an upward-directed family that includes bounded sets contained in finite-dimensional subspaces, $F$ be a separated locally convex space, $U \subset E$ be open, and $f: U \to F$ be $n$-fold $\sigma$-differentiable at $x_0 \in U$, then there exists $r \in \mathcal{R}_\sigma^n(E; F)$ such that
+    Let $E$ be a topological vector space, $\sigma \subset \mathfrak{B}(E)$ be an ideal containing bounded subsets 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
     \[
-    g(x_0 + h) = g(x_0) + \sum_{k = 1}^n \frac{1}{k!}D^k_\sigma f(x_0)(h^{(k)}) + r(h)
+    f(x_0 + h) = f(x_0) + \sum_{k = 1}^n \frac{1}{k!}D^k_\sigma f(x_0)(h^{(k)}) + r(h)
     \]
 
+    for sufficiently small $h$. 
 \end{theorem}
 \begin{proof}[Proof {{\cite[Theorem 4.7.3]{Bogachev}}}. ]
     Let
     \[
-    r(h) = g(x_0 + h) - g(x) - \sum_{k = 1}^n \frac{1}{k!}D^k_\sigma f(x_0)(h^{(k)})
+    r(h) = f(x_0 + h) - f(x) - \sum_{k = 1}^n \frac{1}{k!}D^k_\sigma f(x_0)(h^{(k)})
     \]
 
-    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 \\
@@ -85,7 +86,7 @@
 
     so
     \[
-    D^k_\sigma r(0) = D^k_\sigma g(x_0) - D^k_\sigma(x_0) = 0
+    D^k_\sigma r(0) = D^k_\sigma g(x_0) - D^k_\sigma f(x_0) = 0
     \]
 
 
@@ -121,3 +122,41 @@
 
     Therefore $r \in \mathcal{R}_\sigma^{n+1}(E; F)$.
 \end{proof}
+
+
+\begin{theorem}[Taylor's Formula, Integral Remainder]
+\label{theorem:taylor-integral}
+    Let $E$ be a topological vector space, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, $F$ be a separated locally convex space, $U \subset E$ be open, and $f \in C^{n+1}_\sigma(E; F)$, then for any $x_0 \in U$ and $h \in E$ such that $[x_0, x_0 + h] \subset U$, then
+    \[
+    f(x_0 + h) = f(x_0) + \sum_{k = 1}^n \frac{1}{k!}D^k_\sigma f(x_0)(h^{(k)}) + r(h)
+    \]
+
+    where
+    \[
+    r(h) = \int_0^1 \frac{(1 - t)^{n}}{n!}D^{n+1}_\sigma f(x_0 + th)(h^{(n+1)}) dt
+    \]
+
+    In particular, for any continuous seminorm $[\cdot]_F: F \to [0, \infty)$, 
+    \[
+    [r(h)]_F \le \frac{1}{(n+1)!} \cdot \sup_{t \in [0, 1]}[D^{n+1}_\sigma f(x_0 + th)(h^{n+1})]
+    \]
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Section XIII.6]{Lang}}}. ]
+    Firstly, if $n = 0$, then by the \hyperref[Fundamental Theorem of Calculus]{theorem:ftc-riemann}, 
+    \[
+    f(x_0 + h) - f(x_0) = \int_0^1 D_\sigma f(x_0 + th)(h) dt
+    \]
+
+    Assume inductively that the theorem holds for $n \in \natz$. Let
+    \[
+    u(t) =  D^{n+1}_\sigma f(x + ty)(h^{(n+1)}) \quad v(t) = -\frac{(1 - t)^{n+1}}{(n+1)!}
+    \]
+
+    then $Dv(t) = (1 - t)^n/n!$, and using the \hyperref[change of variables formula]{theorem:rs-change-of-variables} and \hyperref[integration by parts]{theorem:rs-ibp},
+    \begin{align*}
+        r(h) &= \frac{1}{n!}\int_0^1 (1 - t)^{n}D^{n+1}_\sigma f(x_0 + th)(h^{(n+1)}) dt \\
+        &= \int_0^1 udv = u(1)v(1) - u(0)v(0) - \int_0^1 vdu \\
+        &= D_\sigma^{n+1}(x_0)(h^{(n+1)}) \\
+        &+ \int_0^1 \frac{(1 - t)^{n+1}}{(n+1)!}D^{n+2}_\sigma f(x_0 + th)(h^{(n+2)}) dt
+    \end{align*}
+\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

@@ -11,6 +11,14 @@ Differential geometry is the study of things invariant under change of notation.
     $\mathcal{R}_\sigma^n(E; F)$, $\mathcal{R}_\sigma(E;F)$ & $\sigma$-small functions of order $n$; order 1. & \autoref{definition:differentiation-small} \\
     $D_\sigma f(x_0)$ & $\sigma$-derivative of $f$ at $x_0$. & \autoref{definition:derivative-sets} \\
     $D_\sigma^n f$ & $n$-fold $\sigma$-derivative. & \autoref{definition:n-differentiable-sets} \\
-    $x^{(k)}$ & Tuple of $x$ repeated $k$ times. & \autoref{proposition:multilinear-derivative} \\
+    $D_\sigma^n(U; F)$ & $n$-fold $\sigma$-differentiable functions. & \autoref{definition:differentiable-space} \\
+    $\tilde D_\sigma^n(U; F)$ & $n$-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} \\
+    $\omega_{z, r}$ & Standard path of winding number 1. & \autoref{definition:winding-number-1} \\
+    $H(U; E)$ & Space of $E$-valued holomorphic functions on $U$. & \autoref{definition:holomorphic-function-space} \\
+    $H(A; E)$ & Space of $E$-valued holomorphic functions near $A$. & \autoref{definition:holomorphic-function-space-near} 
 \end{tabular}

src/fa/duality/polar.tex

@@ -22,14 +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/interpolation/functors.tex

@@ -0,0 +1,10 @@
+\section{Interpolation Functors}
+\label{section:interpolation-functors}
+
+\textit{"In the presence of so many different interpolation methods it seemed timely to study the general structure of all possible methods: to determine all of them and to analyze the properties which are common to all."} — \cite[Page 51]{aronszajn1964interpolation}. 
+
+\textit{"This is how things appeared in 1965. Fifteen years later, it was found that the number
+of interpolation methods at our disposal is not large."} — \cite[Page vi, Footnote 3]{brudnyi1991interpolation}. 
+
+The above quotes are taken from \cite[Page 427]{PietschHistory}.
+

src/fa/interpolation/index.tex

@@ -0,0 +1,6 @@
+\chapter{Interpolation Spaces}
+\label{chap:interpolation}
+
+\input{./functors.tex}
+
+

src/fa/lc/barrel.tex

@@ -25,14 +25,14 @@
 
 \begin{summary}
 \label{summary:barreled-space}
-    The following types of locally convex spaces are barreled:
+    The following types of locally convex spaces are barrelled:
     \begin{enumerate}
         \item Every locally convex space with the Baire property.
         \item Every Banach space and every Fréchet space. 
-        \item Inductive limits of barreled spaces. 
+        \item Inductive limits of barrelled spaces. 
         \item Spaces of type (LB) and (LF). 
-        \item The locally convex direct sum of barreled spaces. 
-        \item Products of barreled spaces.
+        \item The locally convex direct sum of barrelled spaces. 
+        \item Products of barrelled spaces.
     \end{enumerate}
 \end{summary}
 \begin{proof}
@@ -46,7 +46,7 @@
 
 \begin{proposition}
 \label{proposition:baire-barrel}
-    Let $E$ be a locally convex space over $K \in \RC$. If $E$ is a Baire space, then $E$ is barreled. 
+    Let $E$ be a locally convex space over $K \in \RC$. If $E$ is a Baire space, then $E$ is barrelled. 
 \end{proposition}
 \begin{proof}[Proof, {{\cite[II.7.1]{SchaeferWolff}}}. ]
     Let $D \subset E$ be a Barrel, then $E = \bigcup_{n \in \natp}nD$ is a countable union of closed sets. Since $E$ is Baire, there exists $n \in \natp$, $U \in \cn_E(0)$ circled, and $x \in E$ such that $x + U \in nB$. In which case, 
@@ -59,9 +59,9 @@
 
 \begin{proposition}
 \label{proposition:barrel-limit}
-    Let $\seqi{E}$ be locally convex spaces over $K \in \RC$, $E$ be a vector space over $K$, and $\seqi{T}$ such that $T_i \in \hom(E_i; E)$ for all $i \in I$, then the inductive locally convex topology on $E$ induced by $\seqi{T}$ is barreled.  
+    Let $\seqi{E}$ be locally convex spaces over $K \in \RC$, $E$ be a vector space over $K$, and $\seqi{T}$ such that $T_i \in \hom(E_i; E)$ for all $i \in I$, then the inductive locally convex topology on $E$ induced by $\seqi{T}$ is barrelled.  
 \end{proposition}
 \begin{proof}[Proof, {{\cite[II.7.2]{SchaeferWolff}}}. ]
-    Let $D \subset E$ be a barrel, then for each $i \in I$, $T_i^{-1}(D) \subset E_i$ is also a barrel, and thus a neighbourhood of $0$ in $E_i$. By (5) of \autoref{definition:lc-inductive}, $D$ is a neighbourhood of $0$ in $E$, so $E$ is barreled. 
+    Let $D \subset E$ be a barrel, then for each $i \in I$, $T_i^{-1}(D) \subset E_i$ is also a barrel, and thus a neighbourhood of $0$ in $E_i$. By (5) of \autoref{definition:lc-inductive}, $D$ is a neighbourhood of $0$ in $E$, so $E$ is barrelled. 
 \end{proof}
 

src/fa/lc/bornologic.tex

@@ -9,6 +9,7 @@
         \item For any seminorm $\rho: E \to [0, \infty)$ that is bounded on all bounded sets of $E$, $\rho$ is continuous. 
     \end{enumerate}
 
+    
     If the above holds, then $E$ is a \textbf{bornological space}.
 \end{definition}
 \begin{proof}
@@ -66,10 +67,10 @@
 \label{definition:associated-bornological}
     Let $(E, \mathcal{T})$ be a separated locally convex space over $K \in \RC$, then there exists a locally convex topology $\mathcal{T}_B \subset 2^E$ such that:
     \begin{enumerate}
-        \item $B(E, \mathcal{T}_B) \supset B(E, \mathcal{T})$. 
-        \item[(U)] For every $\mathcal{S}$ satisfying (1), $\mathcal{S} \subset \mathcal{T}_B$. In particular, $\mathcal{T} \subset \mathcal{T}_B$ and $B(E, \mathcal{T}_B) = B(E, \mathcal{T})$. 
+        \item $\mathfrak{B}(E, \mathcal{T}_B) \supset \mathfrak{B}(E, \mathcal{T})$. 
+        \item[(U)] For every $\mathcal{S}$ satisfying (1), $\mathcal{S} \subset \mathcal{T}_B$. In particular, $\mathcal{T} \subset \mathcal{T}_B$ and $\mathfrak{B}(E, \mathcal{T}_B) = \mathfrak{B}(E, \mathcal{T})$. 
         \item $(E, \mathcal{T}_B)$ is a bornological space. 
-        \item Let $\mathcal{B} \subset B(E, \topo)$ be the collection of all convex, circled, bounded, and closed subsets of $E$, ordered by inclusion. For each $B \in \mathcal{B}$, let $(E_B, \rho_B)$ be the normed space associated with $B$ and $\iota_B: E_B \to E$ be the canonical inclusion, then $\mathcal{T}_B$ is the inductive topology induced by $\bracsn{\iota_B}_{B \in \mathcal{B}}$. 
+        \item Let $\mathcal{B} \subset \mathfrak{B}(E, \topo)$ be the collection of all convex, circled, bounded, and closed subsets of $E$, ordered by inclusion. For each $B \in \mathcal{B}$, let $(E_B, \rho_B)$ be the normed space associated with $B$ and $\iota_B: E_B \to E$ be the canonical inclusion, then $\mathcal{T}_B$ is the inductive topology induced by $\bracsn{\iota_B}_{B \in \mathcal{B}}$. 
     \end{enumerate}
 
     The space $(E, \mathcal{T}_B)$ is the \textbf{bornological space associated with} $E$. 
@@ -77,14 +78,14 @@
 \begin{proof}
     Let $\mathscr{T}$ be the collection of all topologies satisfying (1), then $\mathcal{T} \in \mathscr{T}$, so $\mathscr{T} \ne \emptyset$. Let $\mathcal{T}_B$ be the projective topology induced by $\mathscr{T}$. 
 
-    (1): Since $\mathcal{T}_B \supset \mathcal{T}$, $B(E, \mathcal{T}) \supset B(E, \mathcal{T}_B)$. Let $B \in (E, \mathcal{T})$ and $U \in \cn_{\mathcal{T}_B}(0)$, then there exists $\mathscr{S} \subset \mathscr{T}$ finite and $\bracsn{U_{\mathcal{S}}}_{\mathcal{S} \in \mathscr{S}}$ such that $\bigcap_{\mathcal{S} \in \mathscr{S}}U_{\mathcal{S}} \subset U$. In which case, since $B(E, \mathcal{T}) = B(E, \mathcal{S})$ for all $\mathcal{S} \in \mathscr{S}$, $\bigcap_{\mathcal{S} \in \mathscr{S}}U_{\mathcal{S}}$ absorbs $B$, so $U$ absorbs $B$ as well. 
+    (1): Since $\mathcal{T}_B \supset \mathcal{T}$, $\mathfrak{B}(E, \mathcal{T}) \supset \mathfrak{B}(E, \mathcal{T}_B)$. Let $B \in (E, \mathcal{T})$ and $U \in \cn_{\mathcal{T}_B}(0)$, then there exists $\mathscr{S} \subset \mathscr{T}$ finite and $\bracsn{U_{\mathcal{S}}}_{\mathcal{S} \in \mathscr{S}}$ such that $\bigcap_{\mathcal{S} \in \mathscr{S}}U_{\mathcal{S}} \subset U$. In which case, since $\mathfrak{B}(E, \mathcal{T}) = \mathfrak{B}(E, \mathcal{S})$ for all $\mathcal{S} \in \mathscr{S}$, $\bigcap_{\mathcal{S} \in \mathscr{S}}U_{\mathcal{S}}$ absorbs $B$, so $U$ absorbs $B$ as well. 
     
     (U): By (U) of the \hyperref[projective topology]{definition:tvs-initial}.
 
     (3): Let $\rho: E \to [0, \infty)$ be a seminorm that is bounded on bounded sets, then the topology induced by $\rho$ satisfies (1). By (U), $\rho$ is continuous with respect to $\mathcal{T}_B$, so $(E, \mathcal{T}_B)$ is a bornological space. 
 
-    (4): Let $\mathcal{S}$ be the inductive topology on $E$ induced by $\bracsn{\iota_B}_{B \in \mathcal{B}}$. Let $B \in B(E, \mathcal{T})$, $U \in \cn_{\mathcal{S}}(0)$, and $\rho: E \to [0, \infty)$ be the gauge of $U$. Since $\iota_{\ol B} \in L(E_{\ol B}; E, \mathcal{S})$, there exists $\lambda > 0$ such that $\rho \le \lambda \iota_{\ol B} \circ \rho_{\ol B}$. Thus $2\lambda U \supset B$ and $(E, \mathcal{S})$ satisfies (1). By (U), $\mathcal{T}_B \supset \mathcal{S}$. 
+    (4): Let $\mathcal{S}$ be the inductive topology on $E$ induced by $\bracsn{\iota_B}_{B \in \mathcal{B}}$. Let $B \in \mathfrak{B}(E, \mathcal{T})$, $U \in \cn_{\mathcal{S}}(0)$, and $\rho: E \to [0, \infty)$ be the gauge of $U$. Since $\iota_{\ol B} \in L(E_{\ol B}; E, \mathcal{S})$, there exists $\lambda > 0$ such that $\rho \le \lambda \iota_{\ol B} \circ \rho_{\ol B}$. Thus $2\lambda U \supset B$ and $(E, \mathcal{S})$ satisfies (1). By (U), $\mathcal{T}_B \supset \mathcal{S}$. 
 
-    On the other hand, since $B(E, \mathcal{T}_B) = B(E, \mathcal{T})$, for each $B \in \mathcal{B}$, $\iota_B: E_B \to (E, \mathcal{T}_B)$ is continuous. Therefore by (U) of the \hyperref[inductive topology]{definition:lc-inductive}, $\mathcal{S} \supset \mathcal{T}_B$. 
+    On the other hand, since $\mathfrak{B}(E, \mathcal{T}_B) = \mathfrak{B}(E, \mathcal{T})$, for each $B \in \mathcal{B}$, $\iota_B: E_B \to (E, \mathcal{T}_B)$ is continuous. Therefore by (U) of the \hyperref[inductive topology]{definition:lc-inductive}, $\mathcal{S} \supset \mathcal{T}_B$. 
 \end{proof}
 

src/fa/lc/compact.tex

@@ -0,0 +1,85 @@
+\section{Compact Convex Sets}
+\label{section:compact-convex}
+
+\begin{definition}[Extreme Point]
+\label{definition:extreme-point}
+    Let $E$ be a vector space over $\real$, $K \subset E$, and $x \in K$, then $x$ is \textbf{extremal} if there exists no $y, z \in K$ such that $x \in (y, z) \subset K$. 
+\end{definition}
+
+\begin{definition}[Extreme Subset]
+\label{definition:extreme-subset}
+    Let $E$ be a vector space over $\real$, $K \subset E$ be convex, and $A \subset K$, then $A$ is \textbf{extreme set} if for any $x \in A$ and $y, z \in K$ such that $x \in (y, z)$, $y, z \in A$ as well.  
+\end{definition}
+
+
+
+\begin{lemma}
+\label{lemma:extremal-face}
+    Let $E$ be a locally convex space over $\real$, $K \subset E$ be non-empty and compact, and $\phi \in E^*$. Let $\alpha = \sup\bracs{\dpn{x, \phi}{E}|x \in K}$, then $A = \bracs{\phi = \alpha} \cap K$ is a non-empty extreme subset of $K$. 
+\end{lemma}
+\begin{proof}
+    Since $K$ is compact, $\alpha < \infty$ and $A$ is non-empty by \autoref{proposition:compact-extensions}. Let $x \in A$ and $y, z \in K$ such that $x \in (y, z)$. By definition of $\alpha$, $\dpn{y, \phi}{E} = \dpn{z, \phi}{E} = \alpha$. Thus $y, z \in A$ as well. 
+\end{proof}
+
+\begin{theorem}[Krein-Milman]
+\label{theorem:krein-milman}
+    Let $E$ be a separated locally convex space over $\real$ and $K \subset E$ be a compact convex set, then $K$ is the closed convex hull of its extreme points. 
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem 1.12.5]{Bogachev}}}. ]
+    Assume without loss of generality that $K \ne \emptyset$. 
+    
+    Let $K_0 \subset K$ be a closed, extreme subset of $K$, and $\mathcal{E}(K_0) \subset 2^K$ be the collection of all non-empty closed extreme subsets of $K$ contained in $K_0$. Since $K_0 \in \mathcal{E}(K_0)$, $\mathcal{E}(K_0) \ne \emptyset$. Since $K$ is compact, for any chain $\mathcal{C} \subset \mathcal{E}$, $\bigcap_{A \in \mathcal{C}}A$ is also non-empty, closed, and extreme. 
+    
+    By Zorn's lemma, there exists a minimal element $A$ of $\mathcal{E}(K_0)$. Let $x, y \in A$, $\phi \in E^*$, and $\alpha = \sup_{z \in A}\dpn{z, \phi}{E}$, then $\bracs{\phi = \alpha} \cap K$ is a non-empty, closed, and extreme subset of $K$ by \autoref{lemma:extremal-face}, so $A \cap \bracs{\phi = \alpha}$ is also extreme. By minimality of $A$, $A \subset \bracs{\phi = \alpha}$. Thus by the \hyperref[Hahn-Banach Theorem]{proposition:hahn-banach-utility}, $A$ consists of exactly one point. In which case, for any $y, z \in E$ with $A \subset (y, z) \subset K$, $y = z \in A$. Therefore there exists an extreme point of $K$ in $K_0$. 
+    
+    Since $K$ itself is an extreme subset, a minimal element of $\mathcal{E}(K)$ represents an extreme point, so $K$ admits at least one extreme point. 
+    
+    Now, let $C$ be the collection of all extreme points in $K$. Assume for contradiction that $\ol{\conv}(C) \subsetneq K$, then by the \hyperref[Hahn-Banach Theorem]{theorem:hahn-banach-geometric-2}, there exists $x \in K$ and $\phi \in E^*$ such that $\sup_{y \in \ol{\conv}(C)}\dpn{y, \phi}{E} < \dpn{x, \phi}{E}$. Let $\alpha = \sup_{z \in K}\dpn{z, \phi}{E}$, then by \autoref{lemma:extremal-face}, $K \cap \bracs{\phi = \alpha}$ is a non-empty, closed, and extreme subset of $K$. By the preceding discussion, there exists an extreme point of $K$ in $K \cap \bracs{\phi = \alpha}$, which contradicts the fact that $\sup_{y \in \ol{\conv}(C)}\dpn{y, \phi}{E} < \alpha$. 
+\end{proof}
+
+\begin{theorem}[Markov-Kakutani Fixed Point Theorem]
+\label{theorem:markov-kakutani}
+    Let $E$ be a separated topological vector space over $\real$, $K \subset E$ be a compact convex set, and $\cf \subset C(K; K)$ such that:
+    \begin{enumerate}[label=(\alph*)]
+        \item For any $f, g \in \cf$, $f \circ g = g \circ f$. 
+        \item For each $f \in \cf$, $x, y \in K$, and $t \in [0, 1]$, 
+        \[
+        f(tx + (1 - t)y) = tf(x) + (1 - t)f(y)
+        \]
+    \end{enumerate}
+
+    then there exists $x_0 \in K$ such that $f(x_0) = x_0$ for all $f \in \cf$. 
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem 1.12.10]{Bogachev}}}. ]
+    For each $f \in \cf$ and $n \in \natp$, define $f^{(n)} = \frac{1}{n}\sum_{k = 0}^{n - 1}f^k$, then $f^{(n)} \in C(K; K)$ as well. Via a closure operation, assume without loss of generality that:
+    \begin{enumerate}
+        \item For any $f \in \cf$ and $n \in \natp$, $f^{(n)} \in \cf$. 
+        \item For any $f, g \in \cf$, $f \circ g \in \cf$. 
+    \end{enumerate}
+
+    Let $\bracsn{f_j}_1^N \subset \cf$ and $\bracsn{n_j}_1^N \subset \natp$, then
+    \[
+        \bigcap_{j = 1}^n f_j^{(n_j)}(K) \supset \braks{\prod_{j = 1}^n f_j^{(n_j)}}(K) \ne \emptyset 
+    \]
+
+    thus any finite intersections of elements in
+    \[
+    \mathcal{K} = \bracsn{f^{(n)}(K)| f \in \cf, n \in \natp}
+    \]
+
+    is non-empty. Since $E$ is separated, by \autoref{proposition:compact-extensions} and \autoref{proposition:compact-closed}, $\mathcal{K}$ is a family of closed sets satisfying the finite intersection property. Hence $\bigcap_{n \in \natp}\bigcap_{f \in \cf}f^{(n)}(K) \ne \emptyset$. 
+
+    Now, let $x \in \bigcap_{n \in \natp}\bigcap_{f \in \cf}f^{(n)}(K)$ and $U \in \cn_E(0)$, then there exists $N \in \natp$ such that $NU \supset K$. In which case, there exists $y \in K$ such that
+    \[
+        x  = \frac{1}{N}\sum_{j = 0}^{N-1}f^j(y)  \quad f(x) = \frac{1}{N}\sum_{j = 1}^{N}f^j(y)
+    \]
+
+    In which case, 
+    \[
+    f(x) - x = \frac{1}{N}(f^N(y) - y) \in \frac{1}{N}(K - K) \subset U - U
+    \]
+
+    As this holds for all $U \in \cn_E(0)$ and $E$ is separated, $f(x) = x$.   
+    
+\end{proof}
+

src/fa/lc/convex.tex

@@ -139,6 +139,7 @@
         \item The topology induced by $\seqi{d}$ makes $E$ a topological vector space.
         \item For each $i \in I$, $[\cdot]_i: E \to [0, \infty)$ is continuous.
     \end{enumerate}
+
     The topology induced by $\seqi{d}$ is the \textbf{vector space topology induced by} $\seqi{[\cdot]}$. In addition,
     \begin{enumerate}
         \item[(U)] For any family $\bracsn{[\cdot]_j}_{j \in J}$ of continuous seminorms on $E$, the vector space topology induced by $\bracsn{[\cdot]_j}_{j \in J}$ is contained in the vector space topology induced by $\seqi{[\cdot]}$.
@@ -160,6 +161,7 @@
         \item If $A$ is circled, then for any $x \in E$ and $\lambda \in K$, $[\lambda x]_A = \abs{\lambda}[x]_A$.
         \item If $A$ is circled, then $\bracs{\rho < 1} \subseteq A \subseteq \bracs{\rho \le 1} \subseteq \ol A$.
     \end{enumerate}
+
     In particular,
     \begin{enumerate}[start=4]
         \item If $A$ is convex, then $[\cdot]_A$ is a sublinear functional.
@@ -184,12 +186,13 @@
 
 \begin{definition}[Locally Convex Space]
 \label{definition:locally-convex}
-    Let $E$ be a TVS over $\RC$, then the following are equivalent:
+    Let $E$ be a TVS over $K \in \RC$, then the following are equivalent:
     \begin{enumerate}
         \item There exists a fundamental system of neighborhoods at $0$ consisting of convex sets.
         \item There exists a fundamental system of neighbourhoods at $0$ consisting of convex, circled, and radial sets.
         \item There exists a family of seminorms $\seqi{[\cdot]}$ that induces the topology on $E$.
     \end{enumerate}
+
     If the above holds, then $E$ is a \textbf{locally convex} space.
 \end{definition}
 \begin{proof}
@@ -200,6 +203,10 @@
     $(3) \Rightarrow (1)$: For each $i \in I$ and $r > 0$, $\bracs{x \in E| [x]_i < r}$ is convex.
 \end{proof}
 
+\begin{definition}[Fréchet Space]
+\label{definition:frechet-space}
+    Let $E$ be a locally convex space over $K \in \RC$, then $E$ is a \textbf{Fréchet space} if $E$ is first countable and complete. 
+\end{definition}
 
 
 \begin{definition}[Associated Normed Space]

src/fa/lc/hahn-banach.tex

@@ -135,6 +135,7 @@
             \item $|\phi| \le \rho$.
             \item $\dpb{x, \phi}{E} = \inf_{y \in M}\rho(x + y)$.
         \end{enumerate}
+
         \item For any $x \in E$ and continuous seminorm $\rho: E \to [0, \infty)$, there exists $\phi \in E^*$ with $|\phi| \le \rho$ and $\dpb{x, \phi}{E} = \rho(x)$.
         \item If $E$ is separated, then for any $x, y \in E$, there exists $\phi \in E^*$ with $\dpb{x, \phi}{E} \ne \dpb{y, \phi}{E}$.
     \end{enumerate}

src/fa/lc/index.tex

@@ -4,6 +4,7 @@
 
 \input{./convex.tex}
 \input{./continuous.tex}
+\input{./compact.tex}
 \input{./barrel.tex}
 \input{./bornologic.tex}
 \input{./quotient.tex}

src/fa/lc/inductive.tex

@@ -22,6 +22,7 @@
 
         is a fundamental system of neighbourhoods for $E$ at $0$. 
     \end{enumerate}
+
     The topology $\topo$ is the \textbf{inductive locally convex topology} on $E$ induced by $\seqi{T}$.
 \end{definition}
 \begin{proof}
@@ -53,8 +54,8 @@
         \item[(U)] For each $(F, \seqi{T})$ satisfying (1) and (2), there exists a unique $T \in L(E; F)$ such that the following diagram commutes:
         \[
         \xymatrix{
-        A \ar@{->}[r]^{T} & B \\
-        A_i \ar@{->}[u]^{\iota_i} \ar@{->}[ru]_{T_i} & 
+        E \ar@{->}[r]^{T} & F \\
+        E_i \ar@{->}[u]^{\iota_i} \ar@{->}[ru]_{T_i} & 
         }
         \]
         
@@ -66,16 +67,49 @@
         is a fundamental system of neighbourhoods for $E$ at $0$. 
     \end{enumerate}
 
+    
     The space $E = \bigoplus_{i \in I}E_i$ is the \textbf{locally convex direct sum} of $\seqi{E}$. 
 \end{definition}
 \begin{proof}
-    Let $(E, \seqi{\iota})$ be the direct sum of $\seqi{E}$ as vector spaces, and equip it with the inductive topology induced by $\seqi{\iota}$, then $(E, \seqi{\iota})$ satisfies (1) and (2). 
+    Let $(E, \seqi{\iota})$ be the direct sum of $\seqi{E}$ as vector spaces, and equip it with the locally convex inductive topology induced by $\seqi{\iota}$, then $(E, \seqi{\iota})$ satisfies (1) and (2). 
 
     (U): By (U) of the \hyperref[direct sum]{definition:direct-sum}, there exists a unique $T \in \hom(E; F)$ such that the diagram commutes. In which case, by (4) of \autoref{definition:lc-inductive}, $T \in L(E; F)$. 
 
     (4): By (6) of \autoref{definition:lc-inductive}.
 \end{proof}
 
+\begin{proposition}
+\label{proposition:finite-lc-product}
+    Let $\seqf{E_j}$ be TVSs over $K \in \RC$, then the following spaces coincide:
+    \begin{enumerate}
+        \item The product $\prod_{j = 1}^n E_j$.
+        \item The direct sum of $\seqf{E_j}$ as topological vector spaces.
+        \item The direct sum of $\seqf{E_j}$ as locally convex spaces. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    By \autoref{proposition:finite-tvs-product}, it is sufficient to show that (1) and (3) coincide. The proof is \textit{exactly} the same as \autoref{proposition:finite-tvs-product}, but included here for completeness.  
+
+    Let $1 \le k \le n$, then for each $1 \le k, l \le n$, $\pi_l \circ \iota_k \in L(E_k, E_l)$, so by (U) of the \hyperref[product]{definition:tvs-product}, $\iota_k \in L(E_k; \prod_{j = 1}^n E_j)$. Thus $\prod_{j = 1}^n E_j$ satisfies (1) and (2) of the \hyperref[direct sum]{definition:lc-direct-sum}. 
+
+    For any locally convex space $F$ over $K$ and $\seqf{T_j}$ with $T_j \in L(E_j; F)$ for each $1 \le j \le n$, let
+    \[
+    T: \prod_{j = 1}^n E_j \to F \quad (x_1, \cdots, x_n) \mapsto \sum_{j = 1}^n T_jx_j
+    \]
+
+    then $T \in L(\prod_{j = 1}^n E_j; F)$ is the unique continuous linear map such that the following diagram commutes:
+    \[
+    \xymatrix{
+    E \ar@{->}[r]^{T} & F \\
+    E_i \ar@{->}[u]^{\iota_i} \ar@{->}[ru]_{T_i} & 
+    }
+    \]
+
+    Hence $\prod_{j = 1}^n E_j$ satisfies (U) of the \hyperref[direct sum]{definition:lc-direct-sum}, so the spaces coincide. 
+\end{proof}
+
+
+
 \begin{definition}[Inductive Limit]
 \label{definition:lc-inductive-limit}
     Let $(\seqi{E}, \bracsn{T^i_j| i, j \in I, i \lesssim j})$ be an upward-directed system of locally convex spaces over $K \in \RC$, then there exists $(E, \bracsn{T^i_E}_{i \in I})$ such that:
@@ -111,6 +145,7 @@
 
         is a fundamental system of neighbourhoods for $E$ at $0$.
     \end{enumerate}
+
     The pair $(E, \bracsn{T^i_E}_{i \in I})$ is the \textbf{inductive limit} of $(\seqi{E}, \bracsn{T^i_j| i, j \in I, i \lesssim j})$.
 \end{definition}
 \begin{proof}
@@ -175,6 +210,7 @@
             \item[(a)] $B$ is bounded.
             \item[(b)] There exists $n \in \natp$ such that $B \subset E_n$ is bounded.
         \end{enumerate}
+
         \item If $E_n$ is complete for each $n \in \natp$, then $E$ is also complete.
     \end{enumerate}
 \end{proposition}
@@ -191,6 +227,7 @@
         \item For each $k \in \natp$, $U_k = U_{k+1} \cap E_{n_k}$.
         \item For each $k \in \natp$, $n^{-1}x_k \not\in U_k$.
     \end{enumerate}
+
     then $V = \bigcup_{k \in \natp}U_k \in \cn_E(0)$ with $V \cap E_{n_k} = U_k$ for all $k \in \natp$. For any $n \in \natp$, $x_k \not\in nU_k = nV \cap E_{n_k}$. Therefore $B$ is not bounded.
 
     (3), $(b) \Rightarrow (a)$: Let $U \in \cn_E(0)$, then $U \cap E_n \in \cn_{E_n}(0)$, so there exists $\lambda \in K$ with $\lambda (U \cap E_n) \supset B$.
@@ -231,3 +268,16 @@
 
     which contradicts the fact that $(F_n + U) \cap E_n = \emptyset$.
 \end{proof}
+
+\begin{definition}[Space of Type (LB)]
+\label{definition:lb-space}
+    Let $E$ be a locally convex space, then $E$ is of type \textbf{(LB)} if $E$ is the strict inductive limit of a countable system of Banach spaces. 
+\end{definition}
+
+\begin{definition}[Space of Type (LF)]
+\label{definition:lf-space}
+    Let $E$ be a locally convex space, then $E$ is of type \textbf{(LF)} if $E$ is the strict inductive limit of a countable system of Fréchet spaces. 
+\end{definition}
+
+
+

src/fa/lc/quotient.tex

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

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

@@ -8,6 +8,7 @@
         \item[(B)] For each $f \in \cf$ and $S \in \sigma$, $f(S) \subset E$ is bounded. 
     \end{enumerate}
 
+    
     For each $S \in \sigma$ and continuous seminorm $\rho: E \to [0, \infty)$, let
     \[
     \rho_S: E^T \to [0, \infty] \quad f \mapsto \sup_{x \in S}\rho(f(x))
@@ -23,7 +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/lc/tensor.tex

@@ -25,6 +25,7 @@
         is a fundamental system of neighbourhoods at $0$ for $E \otimes_\pi F$.
     \end{enumerate}
 
+    
     The space $E \otimes_\pi F$ is the \textbf{projective tensor product} of $E$ and $F$, and the mapping $\iota \in L^2(E, F; E \otimes_\pi F)$ is the \textbf{canonical embedding}.
 
     The space $E \widetilde{\otimes}_\pi F$ denotes the Hausdorff completion of $E \otimes_\pi F$.
@@ -74,6 +75,7 @@
         \item[(5)] If the seminorms $\seqi{p}$ define the topology on $E$, and the seminorms $\seqj{q}$ define the topology on $F$, then the seminorms $\bracsn{p_i \otimes q_j| (i, j) \in I \times J}$ define the topology on $E \otimes_\pi F$. 
     \end{enumerate}
 
+    
 \end{definition}
 \begin{proof}[Proof {{\cite[III.6.3]{SchaeferWolff}}}. ]
     (1): Let $\lambda \in K$, then for any $\seqf{(x_j,y_j)} \subset E \times F$, 
@@ -142,6 +144,7 @@
         \item $z = \sum_{n = 1}^\infty \lambda_n x_n \otimes y_n$. 
     \end{enumerate}
 
+    
 \end{theorem}
 \begin{proof}
     Let $\seq{p_n}$ and $\seq{q_n}$ be increasing sequences of continuous seminorms that induce the topology on $E$ and $F$, respectively. For each $n \in \natp$, let $r_n = p_n \otimes q_n$, and $\td r_n$ be the continuous extension of $r_n$ to $E \td{\otimes}_\pi F$. 
@@ -164,6 +167,7 @@
         \item $\sum_{k = 1}^{n_N}|\lambda_k| \le 2^{-N+2}$.
     \end{enumerate}
 
+    
     From here, let $\seqf{(x_j, y_j)} \subset X \times Y$ such that $u_1 = \sum_{j = 1}^n x_j \otimes y_j$, then
     \[
     u = u_1 + \sum_{N = 1}^\infty v_N = \sum_{j = 1}^n x_j \otimes y_j + \sum_{N = 1}^\infty \sum_{k = 1}^{n_N}\lambda_{N, k}x_{N, k} \otimes y_{N, k}

src/fa/lp/definition.tex

@@ -109,6 +109,7 @@
         \item[(b)] There exists $g \in L^p(X) \cap L^+(X)$ such that $\norm{f_n}_E \le g$ for all $n \in \natp$.
     \end{enumerate}
 
+    
     then $f_n \to f$ in $L^p(X; E)$.
 \end{proposition}
 \begin{proof}

src/fa/norm/absolute.tex

@@ -14,6 +14,7 @@
         \item For any absolutely convergent series $\sum_{n = 1}^\infty x_n$ with $\seq{x_n} \subset E$, there exists $x \in E$ such that $x = \sum_{n = 1}^\infty x_n$.
     \end{enumerate}
 
+    
 \end{lemma}
 \begin{proof}
     (2) $\Rightarrow$ (1): Let $\seq{x_n} \subset E$ be a Cauchy sequence, then there exists a subsequence ${n_k}$ such that $\norm{x_{n_{k+1}} - x_{n_k}} \le 2^{-k}$ for all $k \in \natp$. 
@@ -45,6 +46,7 @@
         \item If $\sum_{n \in N}x_n = -\infty$ but $\sum_{n \in P}x_n < \infty$, then $\sum_{n = 1}^\infty x_n$ converges to $-\infty$ unconditionally. 
         \item If $\sum_{n \in \natp}|x_n| < \infty$, then $\sum_{n = 1}^\infty x_n$ converges unconditionally.
     \end{enumerate}
+
     In other words, a series in $\real$ converges unconditionally if and only if its positive parts or its negative parts are finite. 
 \end{theorem}
 \begin{proof}

src/fa/norm/multilinear.tex

@@ -9,6 +9,7 @@
         \item For each $y \in F$, $x \mapsto T(x, y)$ is a continuous linear map from $E$ to $G$.
         \item $E$ is a Banach space.
     \end{enumerate}
+
     then $T \in L^2(E, F; G)$.
 \end{proposition}
 \begin{proof}

src/fa/norm/normed.tex

@@ -39,11 +39,13 @@
         \item[(a)] $\norm{x}_E \le C\norm{y}_F$.
         \item[(b)] $\norm{y - Tx}_F \le \gamma \norm{y}_F$.
     \end{enumerate}
+
     then for any $y \in F$, there exists $\seq{x_n} \subset E$ such that:
     \begin{enumerate}
         \item $\sum_{n \in \natp}\norm{x_n}_E \le C\norm{y}_F/(1 - \gamma)$.
         \item $\sum_{n = 1}^\infty Tx_n = y$.
     \end{enumerate}
+
     In particular, if $E$ is a Banach space, then for every $y \in F$, there exists $x \in E$ such that $\norm{x}_E \le C\norm{y}_F/(1 - \gamma)$ and $Tx = y$.
 \end{theorem}
 \begin{proof}
@@ -71,20 +73,14 @@
 \label{theorem:uniform-boundedness}
     Let $E, F$ be normed vector spaces and $\mathcal{T} \subset L(E; F)$. If
     \begin{enumerate}
-        \item For every $x \in E$, $\sup_{T \in \mathcal{T}}\norm{Tx}_F < \infty$.
-        \item $E$ is a Banach space.
+        \item[(B1)] $E$ is a Banach space.
+        \item[(B2)] For every $x \in E$, $\sup_{T \in \mathcal{T}}\norm{Tx}_F < \infty$.
     \end{enumerate}
+
     then $\sup_{T \in \mathcal{T}}\norm{T}_{L(E; F)} < \infty$.
 \end{theorem}
 \begin{proof}
-    For each $n \in \natp$, let $A_n = \bracs{x \in X|\norm{Tx}_F \le n \forall T \in \mathcal{T}}$, then each $A_n$ is closed with $\bigcup_{n \in \natp}A_n = E$. By the \hyperref[Baire Category Theorem]{theorem:baire}, there exists $n \in \natp$ and $U \subset E$ open such that $\sup_{x \in U}\sup_{T \in \mathcal{T}}\norm{Tx}_{F} < \infty$.
-
-    Let $x \in U$ and $r > 0$ such that $\overline{B(x, r)} \subset U$, then for any $y \in E$ with $\norm{y}_E \le r$ and $T \in \mathcal{T}$,
-    \[
-    \norm{Ty} = \norm{Ty + Tx - Tx}_E = \normn{T\underbrace{(x + y)}_{\in U}}_E + \norm{Tx}_E \le 2n
-    \]
-
-    so $\sup_{T \in \mathcal{T}}\norm{T}_{L(E; F)} \le 2n/r$.
+    By the \hyperref[Banach-Steinhaus Theorem]{theorem:banach-steinhaus}, $\mathcal{T}$ is equicontinuous. Therefore there exists $r > 0$ such that $\bigcup_{T \in \mathcal{T}}T[B_E(0, r)] \subset B_F(0, 1)$. In which case, $\sup_{T \in \mathcal{T}}\norm{T}_{L(E; F)} \le r^{-1}$. 
 \end{proof}
 
 
@@ -100,13 +96,3 @@
 \end{proof}
 
 
-% TODO: Replace this with a more general version involving polars in the future. 
-\begin{theorem}[Alaoglu's Theorem]
-\label{theorem:alaoglu}
-    Let $E$ be a normed vector space over $K \in \RC$, then $B^* = \bracsn{\phi \in E^*| \norm{x}_{E^*} \le 1}$ is compact in the weak*-topology. 
-\end{theorem}
-\begin{proof}
-    For each $x \in E$, $I_x = \bracsn{\dpn{x, \phi}{E}|\phi \in B^*}$ is compact. By \autoref{proposition:operator-space-completeness}, the closure of $B^*$ in $\prod_{x \in E}I_x$ is a subset of $\hom(E; K)$. Since $B^*$ is bounded, $I_x \subset \overline{B_K(0, 1)}$ for all $x \in E$, so the closure of $B^*$ is contained in $B^*$. By \autoref{theorem:tychonoff}, $\prod_{x \in E}I_x$ is compact. Therefore $B^*$ is compact with respect to the weak*-topology.
-\end{proof}
-
-

src/fa/norm/separable.tex

@@ -31,6 +31,7 @@
         \item Open sets in $E$ with respect to the weak topology. 
     \end{enumerate}
 
+    
 \end{proposition}
 \begin{proof}
     (1) $\Leftrightarrow$ (2) $\Leftrightarrow$ (3): By \autoref{proposition:separable-metric-borel-sigma-algebra}.

src/fa/notation.tex

@@ -8,7 +8,7 @@
     $E_A$ & Normed space associated with $A \subset E$. & \autoref{definition:lc-associated-normed-space} \\
     $L(E; F)$ & Continuous linear maps $E \to F$. & \autoref{definition:continuous-linear} \\
     $L^n(E_1,\ldots,E_n; F)$ & Continuous $n$-linear maps $\prod E_j \to F$. & \autoref{definition:continuous-multilinear} \\
-    $B(E)$ & Bounded subsets of TVS $E$. & \autoref{definition:bounded} \\
+    $\mathfrak{B}(E)$ & Bounded subsets of TVS $E$. & \autoref{definition:bounded} \\
     $B(T; E)$ & Bounded functions $T \to E$ with uniform topology. & \autoref{definition:bounded-function-space} \\
     $B_\mathfrak{S}^k(E; F)$, $B(E; F)$ & $\mathfrak{S}$-bounded $k$-linear maps; bounded linear maps. & \autoref{definition:bounded-linear-map-space} \\
     $E^*$ & Topological dual of TVS $E$. & \autoref{definition:topological-dual} \\
@@ -43,8 +43,11 @@
     $T_{f,\rho}(x)$ & Variation function of $f$ with respect to $\rho$. & \autoref{definition:variation-function} \\
     $BV([a,b]; E)$ & Functions of bounded variation. & \autoref{definition:bounded-variation} \\
     $S(P, c, f, G)$ & Riemann-Stieltjes sum $\sum_j f(c_j)[G(x_j)-G(x_{j-1})]$. & \autoref{definition:rs-sum} \\
+    $\int_a^b f dG$, $\int_a^b f(t) G(dt)$ & Riemann-Stieljes integral of $f$ with respect to $G$. & \autoref{definition:rs-integral}  \\
     $RS([a,b], G)$ & Space of RS-integrable functions w.r.t.\ $G$. & \autoref{definition:rs-integral} \\
     $\mathrm{Reg}([a,b], G; E)$ & Regulated functions w.r.t.\ $G$ on $[a,b]$. & \autoref{definition:regulated-function} \\
     $\mu_G$ & Lebesgue-Stieltjes measure associated with $G$. & \autoref{definition:riemann-lebesgue-stieltjes} \\
+    $\int_\gamma f$, $\int_\gamma f(z)dz$ & Path integral of $f$ with respect to $\gamma$. & \autoref{definition:path-integral} \\
+    $PI([a, b], \gamma; E)$ & Space of path integrable functions with respect to $\gamma$. & \autoref{definition:path-integral}
 \end{tabular}
 

src/fa/order/lattice.tex

@@ -9,6 +9,7 @@
         \item[(LO2)] For any $x, y \in E$ and $\lambda > 0$, $x \le y$ implies that $\lambda x \le \lambda y$.
     \end{enumerate}
 
+    
 \end{definition}
 
 \begin{proposition}
@@ -19,6 +20,7 @@
         \item $\sup(A) = -\inf (-A)$
     \end{enumerate}
 
+    
 \end{proposition}
 \begin{proof}
     (1): For any $a \in A$ and $b \in B$, $a + b \le \sup(A) + \sup(B)$ by (LO1). Let $c \in E$ such that $c \ge a + b$ for all $a \in A$ and $b \in B$, then $c \ge a + \sup(B)$ for all $a \in A$, so $c \ge \sup(A) + \sup(B)$. Therefore $\sup(A) + \sup(B) = \sup(A + B)$.
@@ -105,11 +107,13 @@
         \item[(4)] $|x + y| \le |x| + |y|$.
     \end{enumerate}
 
+    
     Finally, for any $x, y \in E$ with $x, y \ge 0$, 
     \begin{enumerate}
         \item[(5)] $[0, x] + [0, y] = [0, x + y]$.
     \end{enumerate}
 
+    
 \end{proposition}
 \begin{proof}
     (1): By \autoref{proposition:ordered-vector-space-properties},
@@ -177,6 +181,7 @@
         \item For any $x, y \in C$, $\phi(x + y) = \phi(x) + \phi(y)$.
     \end{enumerate}
 
+    
     then the mapping
     \[
     \Phi: E \to \real \quad x \mapsto \phi(x^+) - \phi(x^-)
@@ -220,6 +225,7 @@
         \]
     \end{enumerate}
 
+    
 \end{proposition}
 \begin{proof}
     (1): Let $C = \bracs{x \in E|x \ge 0}$, $\Phi^+ \in E^b$, and

src/fa/rs/bv.tex

@@ -68,6 +68,7 @@
         \item For each continuous seminorm $\rho$ on $E$, $[\cdot]_{\text{var}, \rho}: E^{[a, b]} \to [0, \infty]$ is lower semicontinuous. In particular, for any $M > 0$, $\bracs{[\cdot]_{\text{var}, \rho} \le M} \subset E^{[a, b]}$ is closed. 
         \item For any $f \in BV([a, b]; E)$ and continuous seminorm $\rho$ on $E$, $\sup_{x \in [a, b]}\rho(f(x)) \le \rho(f(a)) + [f]_{\text{var}, \rho}$.
     \end{enumerate}
+
     If $(E, \norm{\cdot}_E)$ is a normed vector space, then
     \begin{enumerate}
         \item[(5)] $f$ has at most countably many discontinuities.
@@ -88,6 +89,7 @@
         \item[(a)] $|E_k| \ge N - k$.
         \item[(b)] $E_k \subset I_k^o$.
     \end{enumerate}
+
     for $k = 1$.
 
     Let $k \le N$ and suppose inductively that $E_k, I_k$ have been constructed. Let $x_k \in E_k$, then by (b), there exists $\eps > 0$ such that $[x_k - \eps, x_k + \eps] \subset I_k$ and $|E_k \setminus [x_k - \eps, x_k + \eps]| \ge N - k$. Let $y_k \in [x_k - \eps, x_k + \eps]$ such that $\norm{f(x_k) - f(y_k)} \ge 1/n$, $I_{k + 1} = I_k \setminus [x_k - \eps, x_k + \eps]$, and $E_{k+1} = E_k \setminus [x_k - \eps, x_k + \eps]$, then $I_k$ and $E_k$ satisfies (a) and (b).

src/fa/rs/index.tex

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

src/fa/rs/path.tex

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

src/fa/rs/regulated.tex

@@ -4,7 +4,7 @@
 
 \begin{proposition}
 \label{proposition:rs-interval}
-    Let $[a, b] \subset \real$, $E, F, H$ be TVSs over $K \in \RC$, and $E \times F \to H$ with $(x, y) \mapsto xy$ be a continuous linear map. 
+    Let $[a, b] \subset \real$, $E, F, H$ be TVSs over $K \in \RC$, and $E \times F \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map. 
     
     Let $G: [a, b] \to F$ and $[c, d] \subset [a, b]$ such that $G$ is continuous at $c$ and $d$, then for any $x \in E$, $x \cdot \one_{[c, d]} \in RS([a, b], G)$, and
     \[
@@ -35,9 +35,9 @@
 
 \begin{definition}[Regulated Function]
 \label{definition:regulated-function}
-    Let $[a, b] \subset \real$, $E, F, H$ be TVSs over $K \in \RC$, and $E \times F \to H$ with $(x, y) \mapsto xy$ be a continuous linear map. 
+    Let $[a, b] \subset \real$, $E, F, H$ be TVSs over $K \in \RC$, and $E \times F \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map. 
 
-    Let $G: [a, b] \to F$ and $f: [a, b] \to E$ be a step map, then $f$ is \textbf{regulated with respect to} $G$ if $G$ is continuous on all discontinuity points of $f$. Let $\text{Reg}([a, b], G; E)$ be closure of all regulated step maps with respect to the uniform norm, then:
+    Let $G: [a, b] \to F$ and $f: [a, b] \to E$ be a step map, then $f$ is \textbf{regulated with respect to} $G$ if $G$ is continuous on all discontinuity points of $f$. Let $\text{Reg}([a, b], G; E)$ be closure of all regulated step maps with respect to the uniform topology, then:
     \begin{enumerate}
         \item Every regulated step map is in $RS([a, b], G)$. 
         \item If $E$ is metrisable, then for any $f \in \text{Reg}([a, b], G; E)$, $f$ is continuous at all but at most countably many points, and $f$ does not share any discontinuity points with $E$. 
@@ -106,4 +106,3 @@
     (2): Let $G(x) = \int_a^x DF(t)dt + F(a)$, then $G - F$ has derivative $0$. By the \hyperref[Mean Value Theorem]{proposition:zero-derivative-constant}, $G - F$ is constant. As $G(a) - F(a) = 0$, $G  = F$. 
     
 \end{proof}
-

src/fa/rs/rs-bv.tex

@@ -1,19 +1,15 @@
-\section{Riemann-Stieltjes Integrals and Functions of Bounded Variation}
+\section{Integrators of Bounded Variation}
 \label{section:rs-bv}
 
 \begin{proposition}
 \label{proposition:rs-bound}
-    Let $[a, b] \subset \real$, $E, F, H$ be locally convex spaces, and $E \times F \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map, and $G: [a, b] \to F$.
-
-    Let $[\cdot]_H$ be a continuous seminorm on $H$, then there exists continuous seminorms $[\cdot]_E$ on $E$ and $[\cdot]_F$ on $F$ such that for any $f \in RS([a, b], G)$,
+    Let $[a, b] \subset \real$, $E, F, H$ be locally convex spaces, $E \times F \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map, $G: [a, b] \to F$, and $f \in RS([a, b], G)$, then for any continuous seminorms $[\cdot]_E: E \to [0, \infty)$, $[\cdot]_F: F \to [0, \infty)$, and $[\cdot]_H: H \to [0, \infty)$ such that $[xy]_H \le [x]_E[y]_F$ for all $x \in E$ and $y \in F$, 
     \[
     \braks{\int_a^bf dG}_H \le \sup_{x \in [a, b]}[f]_E \cdot [g]_{\text{var}, F}
     \]
 
 \end{proposition}
 \begin{proof}
-    By \autoref{proposition:tvs-convex-multilinear}, there exists continuous seminorms $[\cdot]_E$ on $E$ and $[\cdot]_F$ on $F$ such that $[xy]_H \le [x]_E[y]_F$ for all $(x, y) \in E \times F$.
-
     Let $(P = \seqfz{x_j}, c = \seqf{c_j}) \in \scp_t([a, b])$, then
     \begin{align*}
         [S(P, c, f, G)]_H &\le \sum_{j = 1}^n [f(c_j)[G(x_j) - G(x_{j - 1})]]_H \\
@@ -24,7 +20,7 @@
 
 \begin{proposition}
 \label{proposition:rs-complete}
-    Let $[a, b] \subset \real$, $E, F, H$ be locally convex spaces, and $E \times F \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map, and $G \in BV([a, b]; F)$.
+    Let $[a, b] \subset \real$, $E, F, H$ be locally convex spaces, $E \times F \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map, and $G \in BV([a, b]; F)$.
 
     For each continuous seminorm $\rho$ on $E$ and $f: [a, b] \to E$, define
     \[
@@ -36,6 +32,7 @@
         \item[(a)] For each continuous seminorm $\rho$ on $E$, $[f_\alpha - f]_{u, \rho} \to 0$.
         \item[(b)] $\lim_{\alpha \in A}\int_a^b f_\alpha dG$ exists.
     \end{enumerate}
+
     then $f \in RS([a, b], G)$ and $\int_a^b f dG = \lim_{\alpha \in A}\int_a^b f_\alpha dG$. In particular,
     \begin{enumerate}
         \item If $H$ is complete, then condition (b) may be omitted.
@@ -56,10 +53,12 @@
         \item $[f - f_\alpha]_E < \eps/(3[G]_{\text{var}, F})$.
         \item $\rho\paren{\int_a^b f_\alpha dG - \lim_{\alpha \in A}\int_a^b f_\alpha dG} < \eps/3$.
     \end{enumerate}
+
     Since $f_\alpha \in RS([a, b], G)$, there exists $P_0 \in \scp([a, b])$ such that if $P \ge P_0$,
     \begin{enumerate}
         \item[(3)] $\rho\paren{S(P, c, f_\alpha, G) - \int_a^b f_\alpha dG} < \eps/3$.
     \end{enumerate}
+
     Thus for any $(P, c) \in \scp_t([a, b])$ with $P \ge P_0$,
     \[
     \rho\paren{S(P, c, f, G) - \lim_{\alpha \in A}\int_a^b f_\alpha dG} < \eps
@@ -74,11 +73,13 @@
     Let $f \in C([a, b]; E)$, $G \in BV([a, b]; F)$, then
     \begin{enumerate}
         \item $f \in RS([a, b], G)$.
-        \item For any $\seq{(P_n, t_n)} \subset \scp_t([a, b])$ with $\sigma(P_n) \to 0$,
+        \item For equicontinuous family $\cf \subset C([a, b]; E)$ and $\seq{(P_n, t_n)} \subset \scp_t([a, b])$ with $\sigma(P_n) \to 0$,
         \[
         \int_a^b fdG = \limv{n}S(P_n, t_n, f, G)
         \]
 
+        uniformly for all $f \in \cf$. 
+
     \end{enumerate}
 \end{proposition}
 \begin{proof}
@@ -99,3 +100,47 @@
 
     In addition, for any $\seq{(P_n, t_n)}$ as in (2), $\limv{n}S(P_n, t_n, f, G)$ exists by sequential completeness. Since $\angles{S(P, c, f, G)}_{(P, c) \in \scp_t([a, b])}$ is Cauchy, the limit $\lim_{(P, c) \in \scp_t([a, b])}S(P, c, f, G)$ exists as well and is equal to $\limv{n}S(P_n, t_n, f, G)$.
 \end{proof}
+
+
+\begin{theorem}[Fubini's Theorem for Riemann-Stieltjes Integrals]
+\label{theorem:rs-fubini}
+    Let $[a, b], [c, d] \subset \real$, $E, F, G, H$ be a locally convex space over $K \in \RC$ with $H$ being sequentially complete, $E \times F \times G \to H$ with $(x, y, z) \mapsto xyz$ be a $3$-linear map\footnote{$E, F, G$ are assumed to be disjoint, so the product is well-defined regardless of the order of the terms.}, $\alpha \in BV([a, b]; F)$, $\beta \in BV([c, d]; G)$, and $f \in C([a, b] \times [c, d]; E)$, then
+    \[
+    \int_a^b \int_c^d f(s, t) \beta(dt) \alpha(ds) = \int_c^d\int_a^b f(s, t) \alpha(ds) \beta(dt)
+    \]
+\end{theorem}
+\begin{proof}
+    Let
+    \[
+    g: [a, b] \to L(F; H) \quad s \mapsto \int_c^d f(s, t) \beta(dt)
+    \]
+
+    then for any $(P = \seqfz{x_j}, c = \seqf{c_j}) \in \scp_t([a, b])$,
+    \begin{align*}
+        S(P, c, g, \alpha) &= \sum_{j = 1}^n g(c_j) [\alpha(x_j) - \alpha(x_{j-1})] \\
+        &=  \sum_{j = 1}^n \int_c^d f(c_j, t) \beta(dt) [\alpha(x_j) - \alpha(x_{j-1})] \\
+        &= \int_c^d S(P, c, f(\cdot, t), \alpha) \beta(dt)
+    \end{align*}
+
+
+    Since $f \in C([a, b] \times [c, d]; E)$, $f$ 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 $\alpha \in BV([a, b]; F)$, 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 $t \in [c, d]$. Finally, given that $\beta \in BV([c, d]; G)$, 
+    \[
+    \int_c^d\int_a^b f(s, t) \alpha(ds) \beta(dt) = \limv{n}\int_c^d S(P_n, c_n, f(\cdot, t), \alpha) \beta(dt)
+    \]
+
+    by \autoref{proposition:rs-complete}. 
+\end{proof}
+
+
+
+

src/fa/rs/rs.tex

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

src/fa/tvs/bounded.tex

@@ -9,7 +9,7 @@
         \item For every $\seq{x_n} \subset B$ and $\seq{\lambda_n} \subset K$ such that $\lambda_n \to 0$, $\lambda_n x_n \to 0$ as $n \to \infty$.
     \end{enumerate}
 
-    If the above holds, then $B$ is \textbf{bounded}. The collection $B(E) = B(E, \topo)$ is the set of all bounded sets of $E$.
+    If the above holds, then $B$ is \textbf{bounded}. The collection $\mathfrak{B}(E) = \mathfrak{B}(E, \topo)$ is the set of all bounded sets of $E$.
 \end{definition}
 \begin{proof}
     (1) $\Rightarrow$ (2): Let $U \in \cn_E(0)$ be circled, then there exists $k \in \natp$ such that $kU \supset B$. Since $\lambda_n \to 0$ as $n \to \infty$, there exists $N \in \natp$ such that $|\lambda_n| \le 1/k$ for all $n \ge N$. In which case, $\lambda_n x_n \in \lambda_n B \subset U$ for all $n \ge N$. 
@@ -19,7 +19,7 @@
 
 \begin{proposition}[{{\cite[I.5.1]{SchaeferWolff}}}]
 \label{proposition:bounded-operations}
-    Let $E$ be a TVS over $K \in \RC$ and $A, B \in B(E)$, then the following sets are bounded:
+    Let $E$ be a TVS over $K \in \RC$ and $A, B \in \mathfrak{B}(E)$, then the following sets are bounded:
     \begin{enumerate}
         \item Any $C \subset B$.
         \item The closure $\ol{B}$.

src/fa/tvs/complete-metric.tex

@@ -22,11 +22,13 @@
         \item[(a)] $\eta(y - Tx) \le \gamma \eta(y)$.
         \item[(b)] $\rho(x) \le C \eta(y)$.
     \end{enumerate}
+
     then for any $y \in F$, there exists $\seq{x_n} \subset E$ such that:
     \begin{enumerate}
         \item $\sum_{n \in \natp}\rho(x_n) \le C\eta(y)/(1 - \gamma)$.
         \item $y = \limv{N}\sum_{n = 1}^N Tx_n$.
     \end{enumerate}
+
     In particular,
     \[
     T\braks{B_E\paren{0, \frac{Cr}{(1 - \gamma)}}} \supset B_F(0, r)
@@ -39,11 +41,13 @@
         \item[(I)] $\sum_{n = 1}^N\rho(x_n) \le C\eta(y)\sum_{n = 0}^{N-1}\gamma^{n}$.
         \item[(II)] $\eta\paren{y - \sum_{n = 1}^N Tx_n} \le \eta(y)\gamma^N$.
     \end{enumerate}
+
     By assumption, there exists $x_{N+1} \in E$ such that:
     \begin{enumerate}
         \item[(i)] $\eta\paren{y - \sum_{n = 1}^{N+1} Tx_n} \le \gamma \eta\paren{y - \sum_{n = 1}^N Tx_n} \le \gamma^{N+1}$.
         \item[(ii)] $\rho(x_{N+1}) \le C\eta\paren{y - \sum_{n = 1}^N Tx_n} \le C\eta(y)\gamma^N$.
     \end{enumerate}
+
     Combining (I) and (ii) shows that $\sum_{n = 1}^N \rho(x_n) \le C \eta(y) \sum_{n = 0}^N \gamma^n$. Therefore there exists $\seq{x_n} \subset E$ such that (I) and (II) holds for all $N \in \natp$.
 
     By (I), $\sum_{n \in \natp}\rho(x_n) \le C\eta(y)\sum_{n \in \natz}\gamma^n = C \eta(y)/(1 - \gamma)$. By (II), $\limv{N}\eta\paren{y - \limv{N}\sum_{n = 1}^N Tx_n} = \limv{N}\eta(y)\gamma^N = 0$.
@@ -56,6 +60,7 @@
         \item[(a)] For any $r > 0$, there exists $\delta(r) > 0$ such that $\overline{T(B_E(0, r))} \supset B_F(0, \delta(r))$.
         \item[(b)] $E$ is complete.
     \end{enumerate}
+
     then for every $s > r$, $T(B_E(0, s)) \supset B_F(0, \delta(r))$.
 \end{proposition}
 \begin{proof}
@@ -66,11 +71,13 @@
         \item[(iii)] For all $n \in \natp$, $\overline{T(B_E(0, s_n))} \supset B_F(0, \delta_n)$.
         \item[(iv)] $\rho_1 = \rho$.
     \end{enumerate}
+
     Let $y_0 \in B(0, r)$ and $x_0 = 0$. Let $N \in \natp$ and suppose inductively that $\bracs{x_n}_1^N \subset E$ has been constructed such that:
     \begin{enumerate}
         \item[(I)] For each $0 \le n \le N - 1$, $\rho(x_{n+1} - x_n) < s_n$.
         \item[(II)] For each $0 \le n \le N$, $\eta(Tx_n - y) \le \rho_{n+1}$.
     \end{enumerate}
+
     By density of $T(x_N + B_E(0, s_N))$ in $Tx_N + B_F(0, \rho_N)$, there exists $x_{N+1} \in T(x_N + B_E(0, s_N))$ such that $\eta(Tx_{N+1} - y) \le \rho_{N+2}$.
 
     By (I), $\seq{x_N}$ is a Cauchy sequence, so
@@ -89,6 +96,7 @@
         \item[(a)] For any $r > 0$, there exists $C \ge 0$ such that for any $y \in T(E)$, there exits $x \in T^{-1}(y)$ with $\rho(x) \le C\eta(y)$.
         \item[(b)] $E$ is complete.
     \end{enumerate}
+
     then $T(E)$ is closed.
 \end{proposition}
 \begin{proof}

src/fa/tvs/completion.tex

@@ -9,10 +9,12 @@
         \item $\iota \in L(E; \wh E)$.
         \item[(U)] For any $(F, T)$ satisfying (1) and (2), there exists a unique $\ol{T} \in L(\wh E; F)$ such that the following diagram commutes:
     \end{enumerate}
+
     Moreover,
     \begin{enumerate}
         \item[(4)] $\iota(E)$ is dense in $\wh E$.
     \end{enumerate}
+
     The pair $(\wh E, \iota)$ is the \textbf{Hausdorff completion} of $E$.
 \end{definition}
 \begin{proof}

src/fa/tvs/complexify.tex

@@ -0,0 +1,153 @@
+\section{Complexification}
+\label{section:complexification}
+
+\begin{definition}[Complexification]
+\label{definition:complexification}
+    Let $E$ be a vector space over $\real$, then there exists a pair $(\complex(E), \iota)$ such that:
+    \begin{enumerate}
+        \item $\complex(E)$ is a vector space over $\complex$.
+        \item $\iota: E \to \complex(E)$ is a $\real$-linear map.
+        \item[(U)] For any pair $(F, T)$ satisfying (1) and (2), there exists a unique $\complex$-linear map $\complex(T): \complex(E) \to F$ such that the following diagram commutes:
+        \[
+        \xymatrix{
+        \mathbb{C}(E) \ar@{->}[r]^{\mathbb{C}(T)} & F \\
+        E \ar@{->}[u]^{\iota} \ar@{->}[ru]_{T} & 
+        }
+        \]
+        \item $\complex(E) = \iota(E) \oplus i\iota(E)$ as a vector space over $\real$. For each $z \in \complex(E)$ with $z = x + iy$, $x = \text{Re}(x)$ and $y = \text{Im}(y)$ are the \textbf{real} and \textbf{imaginary parts} of $z$. 
+    \end{enumerate}
+
+    The pair $(\complex(E), \iota)$ is the \textbf{complexification} of $E$, and
+    \begin{enumerate}
+        \item[(F)] For any vector space $F$ over $\real$ and $\real$-linear map $T: E \to F$, there exists a unique $\complex$-linear map $\complex(T): \complex(E) \to \complex(F)$ such that the following diagram commutes:
+        \[
+        \xymatrix{
+        \mathbb{C}(E) \ar@{->}[r]^{\mathbb{C}(T)} & \mathbb{C}(F) \\
+        E \ar@{->}[u]^{\iota} \ar@{->}[r]_{T} & F \ar@{->}[u]_{\iota}
+        }
+        \]
+
+        which is given by
+        \[
+        \complex(T)(x + iy) = Tx + iTy
+        \]
+    \end{enumerate}
+    
+\end{definition}
+\begin{proof}
+    (1): Let $\complex(E) = E \times E$ with coordinate-wise addition. For each $a, b \in \real$ and $x, y \in E$, let
+    \[
+    (a + bi)(x, y) = (ax - by, bx + ay)
+    \]
+
+    then $\complex(E)$ is a vector space over $\complex$. 
+
+    (2): Let $\iota: E \to \complex(E)$ be defined by $\iota(x) = (x, 0)$, then $\iota$ is $\real$-linear. 
+
+    (U): Let
+    \[
+    \complex(T): \complex(E) \to F \quad (x, y) \mapsto Tx + iTy
+    \]
+
+    then $\complex(T)$ is the unique $\complex$-linear map such that the given diagram commutes. 
+
+    (F): By (U) applied to $\iota \circ T$.
+\end{proof}
+
+\begin{definition}[Complex Conjugation]
+\label{definition:complex-conjugation}
+    Let $E$ be a vector space over $\complex$ and $*: E \to E$ be a $\real$-linear map, then $*$ is a \textbf{complex conjugation} if:
+    \begin{enumerate}
+        \item[(C1)] For each $\lambda \in \complex$, $(\lambda x)^* = \ol \lambda x^*$. 
+        \item[(C2)] For each $x \in E$, $x^{**} = x$. 
+    \end{enumerate}
+
+    In which case, $\text{Re}(E) = \bracs{x \in E| x^* = x}$ is the \textbf{real part} of $E$. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:complex-conjugation-properties}
+    Let $E$ be a vector space over $\complex$ and $*: E \to E$ be a complex conjugation, then:
+    \begin{enumerate}
+        \item $E = \complex(\text{Re}(E))$. 
+        \item For each $x \in E$, 
+        \[
+        \text{Re}(x) = \frac{x + x^*}{2} \quad \text{Im}(x) = \frac{x - x^*}{2i}
+        \]
+        \item For each $x \in E$, $x^* = \text{Re}(x) - i\text{Im}(x)$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (2): By properties of the complex conjugation, $\text{Re}(x), \text{Im}(x) \in \text{Re}(E)$. 
+
+    (1): For any $x, y \in \text{Re}(x)$ with $x = iy$, $x = -iy$ as well by (2) of the complex conjugation, so $x = y = 0$. Thus if $z = x + iy = x' + iy'$, then $x = x'$ and $y = y'$, and the decomposition is unique. 
+    
+\end{proof}
+
+
+
+
+
+\begin{definition}[Complexification of Topological Vector Space]
+\label{definition:complexification-tvs}
+    Let $E$ be a TVS over $\real$, then there exists a pair $(\complex(E), \iota)$ such that:
+    \begin{enumerate}
+        \item $\complex(E)$ is a TVS over $\complex$.
+        \item $\iota: E \to \complex(E)$ is a continuous $\real$-linear map. 
+        \item[(U)] For any pair $(F, T)$ satisfying (1) and (2), there exists a unique continuous $\complex$-linear map $\complex(T): \complex(E) \to F$ such that the following diagram commutes:
+        \[
+        \xymatrix{
+        \mathbb{C}(E) \ar@{->}[r]^{\mathbb{C}(T)} & F \\
+        E \ar@{->}[u]^{\iota} \ar@{->}[ru]_{T} & 
+        }
+        \]
+        \item $\complex(E) = \iota(E) \oplus i\iota(E)$ as a TVS over $\real$.
+    \end{enumerate}
+
+    The pair $(\complex(E), \iota)$ is the \textbf{complexification} of $E$ as a topological vector space, and
+    \begin{enumerate}[start=4]
+        \item If $E$ is locally convex, then so is $\complex(E)$. 
+        \item If $E$ is normed, then $\complex(E)$ is normable, and there exists a norm $\norm{\cdot}_{\complex(E)}: \complex(E) \to [0, \infty)$ such that $\iota: E \to \complex(E)$ is isometric. 
+        \item[(F)] For any vector space $F$ over $\real$ and continuous $\real$-linear map $T: E \to F$, there exists a unique continuous $\complex$-linear map $\complex(T): \complex(E) \to \complex(F)$ such that the following diagram commutes:
+        \[
+        \xymatrix{
+        \mathbb{C}(E) \ar@{->}[r]^{\mathbb{C}(T)} & \mathbb{C}(F) \\
+        E \ar@{->}[u]^{\iota} \ar@{->}[r]_{T} & F \ar@{->}[u]_{\iota}
+        }
+        \]
+
+        which is given by
+        \[
+        \complex(T)(x + iy) = Tx + iTy
+        \]
+    \end{enumerate}
+    
+\end{definition}
+\begin{proof}
+    (1), (2): Let $(\complex(E), \iota)$ be the complexification of $E$ as a vector space, and equip it with the \hyperref[direct sum]{definition:tvs-direct-sum} topology. 
+
+    (U): By (U) of the \hyperref[complexification]{definition:complexification}, there exists a $\complex$-linear map $\complex(T): \complex(E) \to F$ such that the given diagram commutes. Since $T \circ \iota$ and $iT \circ \iota$ are continuous, $T$ is continuous by (U) of the \hyperref[direct sum]{definition:tvs-direct-sum}.
+
+    (4): By \autoref{proposition:finite-lc-product}, the direct sum and product of finitely many locally convex spaces coincide. By \autoref{proposition:lc-projective-topology}, this topology is locally convex. 
+
+    (5): Let $\norm{\cdot}_E: E \to [0, \infty)$ be the norm of $E$, and define
+    \[
+    \norm{\cdot}_{\complex(E)}: \complex(E) \to [0, \infty) \quad (x, y) \mapsto \sup_{\theta \in [0, 2\pi]}\norm{\cos(\theta)x + \sin(\theta)y}_E
+    \]
+
+    then for any $\phi \in [0, 2\pi]$ and $x, y \in E$, 
+    \begin{align*}
+        \normn{e^{i \phi}(x, y)}_{\complex(E)} &= \normn{(\cos(\phi)x - \sin(\phi)y, \sin(\phi)x + \cos(\phi)y)}_{\complex(E)} \\
+        &= \sup_{\theta \in [0, 2\pi]}\norm{\cos(\theta - \phi)x + \sin(\theta - \phi)y}_E \\
+        &= \norm{(x, y)}_{\complex(E)}
+    \end{align*}
+
+    so $\norm{(x, y)}_{\complex(E)}$ is a norm. For any $x \in E$, 
+    \[
+    \norm{\iota x}_{\complex(E)} = \sup_{\theta \in [0, 2\pi]}\norm{\cos(\theta)x}_E = \norm{x}_E \\
+    \]
+
+    Therefore $\iota: E \to \complex(E)$ is isometric. 
+
+    (F): By (U) applied to $\iota \circ T$.
+\end{proof}

src/fa/tvs/continuous.tex

@@ -10,6 +10,7 @@
         \item $T \in C(E; F)$.
         \item $T$ is continuous at $0$.
     \end{enumerate}
+
     If the above holds, then $T$ is a \textbf{continuous linear map}. The set $L(E; F)$ denotes the vector space of all continuous linear maps from $E$ to $F$.
 \end{definition}
 \begin{proof}
@@ -49,6 +50,7 @@
         \]
 
     \end{enumerate}
+
     The uniformity $\fU$ and its induced topology are the \textbf{product uniformity/topology}, and $E$ equipped with $\fU$ is the \textbf{product TVS} of $\seqi{E}$.
 \end{definition}
 

src/fa/tvs/definition.tex

@@ -9,6 +9,7 @@
         \item[(TVS1)] $E \times E \to E$ with $(x, y) \mapsto x + y$ is continuous.
         \item[(TVS2)] $K \times E \to E$ with $(\lambda, x) \mapsto \lambda x$ is continuous.
     \end{enumerate}
+
     then the pair $(E, \topo)$ is a \textbf{topological vector space}.
 \end{definition}
 
@@ -51,6 +52,7 @@
         \item There exists a unique translation-invariant uniformity $\fU$ on $E$ that induces the topology on $E$.
         \item For each neighbourhood $V \in \cn(0)$, let $U_V = \bracs{(x, y) \in E^2| x - y \in V}$, then for any fundamental system of neighbourhoods $\fB_0$ at $0$, $\fB = \bracs{U_V| V \in \fB_0}$ is a fundamental system of entourages for $\fU$.
     \end{enumerate}
+
     The space $E$ will always be assumed to be equipped with its translation-invariant uniformity.
 \end{proposition}
 \begin{proof}
@@ -163,11 +165,13 @@
         \item[(TVB1)] For each $U \in \fB$, there exists $V \in \fB$ such that $V + V \subset U$.
         \item[(TVB2)] For each $U \in \fB$, $U$ is circled and radial.
     \end{enumerate}
+
     Conversely, if $\fB \subset 2^E$ is a family of sets that contain $0$ and satisfies (TVB1) and (TVB2), then there exists a unique topology $\topo$ on $E$ such that:
     \begin{enumerate}
         \item $\topo$ is translation-invariant.
         \item $\fB$ is a fundamental system of neighbourhoods at $0$ for $\topo$.
     \end{enumerate}
+
     Moreover,
     \begin{enumerate}
         \item[(3)] $(E, \topo)$ is a TVS.
@@ -187,6 +191,7 @@
         \item[(UB1)] For any $x \in E$ and $V \in \fB$, $x - x = 0 \in V$, so $\Delta \subset U_V$.
         \item[(UB2)] For any $V \in \fB$, by (TVB1), there exists $W \in \fB$ such that $W + W \subset V$. In which case, for any $x, y, z \in E$ with $x - y, y - z \in W$, $x - z \in V$. Therefore $U_W \circ U_W \subset U_V$.
     \end{enumerate}
+
     By \autoref{proposition:fundamental-entourage-criterion}, there exists a unique uniformity $\fU$ on $E$ for which $\mathfrak{V}$ is a fundamental system of entourages.
 
     (1): Since $\mathfrak{V}$ is translation-invariant, so is $\fU$.

src/fa/tvs/dual.tex

@@ -8,6 +8,7 @@
         \item $u \in \hom(E; \real)$ when $E$ is viewed as a vector space over $\real$.
         \item For any $x \in E$, $\dpb{x, \phi}{E} = \dpb{x, u}{E} - i \dpb{ix, u}{E}$.
     \end{enumerate}
+
     Conversely, if $u \in \hom(E; \real)$ and $\phi \in \hom(E; \complex)$ is defined by $\dpb{x, \phi}{E} = \dpb{x, u}{E} - i \dpb{ix, u}{E}$ for all $x \in E$, then $f \in \hom(E; \complex)$.
 \end{proposition}
 \begin{proof}[Proof {{\cite[Proposition 5.5]{Folland}}}. ]

src/fa/tvs/equicontinuous.tex

@@ -0,0 +1,129 @@
+\section{Equicontinuous Families of Linear Maps}
+\label{section:equicontinuous-linear}
+
+\begin{proposition}[{{\cite[IV.4.2]{SchaeferWolff}}}]
+\label{proposition:equicontinuous-linear}
+    Let $E, F$ be TVSs over $K \in RC$ and $\alg \subset \hom(E; F)$, then the following are equivalent:
+    \begin{enumerate}
+        \item $\alg$ is uniformly equicontinuous. 
+        \item $\alg$ is equicontinuous. 
+        \item $\alg$ is equicontinuous at $0$. 
+        \item For each $V \in \cn_F(0)$, there exists $U \in \cn^o(E)$ such that $\bigcup_{T \in \alg}T(U) \subset V$. 
+        \item For each $V \in \cn_F(0)$, $\bigcap_{T \in \alg}T^{-1}(V) \in \cn_E(0)$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (5) $\Rightarrow$ (1): Let $V \in \cn_F(0)$, then $U = \bigcap_{T \in \alg}T^{-1}(V) \in \cn_E(0)$. Thus for any $x, y \in E$ with $x - y \in U$, $Tx - Ty \in V$ for all $T \in \alg$. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:equicontinuous-bounded}
+    Let $E, F$ be TVSs over $K \in \RC$ and $\alg \subset L(E; F)$ be equicontinuous, then for any ideal $\sigma \subset \mathfrak{B}(E)$, $\alg$ is a bounded subset of $B_\sigma(E; F)$. 
+\end{proposition}
+\begin{proof}
+    Let $S \in \sigma$ and $U \in \cn_F(0)$, then there exists $V \in \cn_E(0)$ such that $\bigcup_{T \in \alg}T(V) \subset U$. Since $S$ is bounded, there exists $\lambda \in K$ such that $S \subset \lambda V$. Therefore $\bigcup_{T \in \alg}T(S) \subset \lambda U$, and $\alg$ is bounded in $B_\sigma(E; F)$. 
+\end{proof}
+
+
+\begin{proposition}[{{\cite[IV.4.3]{SchaeferWolff}}}]
+\label{proposition:equicontinuous-linear-closure}
+    Let $E, F$ be TVSs over $K \in \RC$ and $\alg \subset L(E; F)$ be equicontinuous, and $\alg'$ be the closure of $\alg$ in $F^E$ with respect to the product topology, then $\alg'$ is equicontinuous and hence $\alg' \subset L(E; F)$. 
+\end{proposition}
+\begin{proof}
+     By \autoref{proposition:operator-space-completeness}, $\alg' \subset \hom(E; F)$. By \autoref{theorem:arzela-ascoli}, $\alg'$ is equicontinuous.
+\end{proof}
+
+\begin{theorem}[Banach-Steinhaus]
+\label{theorem:banach-steinhaus}
+    Let $E, F$ be TVSs over $K \in \RC$ and $\alg \subset L(E; F)$. Suppose that one of the following holds:
+    \begin{enumerate}
+        \item[(B1)] $E$ is a Baire space. 
+        \item[(B1')] $E$ is barrelled and $F$ is locally convex.
+    \end{enumerate}
+
+    and that
+    \begin{enumerate}
+        \item[(B2')] For each $x \in E$, $\alg(x) = \bracs{Tx|T \in \alg}$ is bounded in $F$. 
+    \end{enumerate}
+
+    then
+    \begin{enumerate}
+        \item[(E1)] $\alg$ is equicontinuous.
+        \item[(C1)] The product topology and the compact-open topology on $\cf$ coincide. 
+        \item[(C2)] The closure of $\alg$ in $F^E$ is with respect to the product topology is an equicontinuous subset of $L(E; F)$. 
+    \end{enumerate}
+\end{theorem}
+\begin{proof}[Proof, {{\cite[IV.4.2]{SchaeferWolff}}}. ]
+    (B1) + (B2) $\Rightarrow$ (E1): Let $V \in \cn_F(0)$ be closed and circled, then $U = \bigcap_{T \in \alg}T^{-1}(V)$ is circled and closed. By (B2), $U$ is absorbing, so $E = \bigcup_{n \in \natp}nU$. Since $E$ is Baire, there exists $n \in \natp$, $W \in \cn_E(0)$, and $x \in E$ such that $x + W \subset nU$. As $U$ is circled,
+    \[
+    W \subset nU - nU = nU + nU = 2nU
+    \]
+
+    so $U \in \cn_E(0)$, and $\alg$ is equicontinuous by \autoref{proposition:equicontinuous-linear}. 
+
+    (B1') + (B2) $\Rightarrow$ (E1): Let $V \in \cn_F(0)$ be convex, circled, and closed, then $U = \bigcap_{T \in \alg}T^{-1}(V)$ is convex, circled, and closed. By (B2), $U$ is absorbing, and hence a barrel in $E$. By (B1'), $U \in \cn_E(0)$, $\alg$ is equicontinuous by \autoref{proposition:equicontinuous-linear}. 
+
+    (E1) $\Rightarrow$ (C1) + (C2): By the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli} and \autoref{proposition:equicontinuous-linear-closure}.
+\end{proof}
+
+\begin{lemma}
+\label{lemma:equicontinuous-bilinear}
+    Let $E, F, G$ be TVSs over $K \in \RC$ and $\alg \subset L^2(E, F; G)$ be continuous bilinear maps, then the following are equivalent:
+    \begin{enumerate}
+        \item $\alg$ is equicontinuous.
+        \item $\alg$ is equicontinuous at $0$. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    (2) $\Rightarrow$ (1): For each $(x_0, y_0), (x, y) \in E \times F$ and $\lambda \in \alg$, 
+    \[
+    \lambda(x, y) - \lambda(x_0, y_0) = \lambda(x - x_0, y - y_0) + \lambda(x - x_0, y_0) + \lambda(x_0, y - y_0)
+    \]
+
+    For each $U \in \cn_G(0)$ circled, there exists circled neighbourhoods $V \in \cn_E(0)$ and $W \in \cn_F(0)$ such that $\lambda(V \times W) \subset U$ for all $\lambda \in \alg$. In which case, there exists $\mu > 0$ such that $y_0 \in \mu W$ and $x_0 \in \mu V$. Thus if $(x, y) - (x_0, y_0) \in \mu^{-1}(V \times W)$, then for every $\lambda \in \alg$,
+    \[
+    \lambda(x - x_0, y_0) \in \lambda(\mu^{-1} V \times \mu W) = \lambda(V \times W) \subset U
+    \]
+
+    and $\lambda(x_0, y - y_0) \in U$ as well. Therefore $\alg$ is equicontinuous at $(x_0, y_0)$. 
+\end{proof}
+
+
+
+\begin{theorem}
+\label{theorem:separate-joint-bilinear}
+    Let $E, F, G$ be TVSs over $K \in \RC$ and $\alg$ be separately continuous bilinear maps from $E \times F$ to $G$. If one of the following holds:
+    \begin{enumerate}
+        \item[(B)] $E$ is Baire.
+        \item[(B')] $E$ is barrelled and $G$ is locally convex. 
+    \end{enumerate}
+
+    and that
+    \begin{enumerate}
+        \item[(M)] $E$ and $F$ are both metrisable. 
+        \item[(E)] For each $x \in E$, $\bracsn{\lambda(x, \cdot)|\lambda \in \alg} \subset L(F; G)$ is equicontinuous. 
+    \end{enumerate}
+
+    then $\alg$ is equicontinuous. 
+\end{theorem}
+\begin{proof}[Proof, {{\cite[III.5.1]{SchaeferWolff}}}. ]
+    Let $\seq{(x_n, y_n)} \subset E \times F$ and $\seq{\lambda_n} \subset \alg$ such that $(x_n, y_n) \to 0$ as $n \to \infty$. Since $\seq{y_n}$ is convergent, for each $n \in \natp$ and $x \in E$, $\bracsn{\lambda_n(x, y_n)|n \in \natp}$ is bounded by (E) and \autoref{proposition:equicontinuous-net}. By (B) or (B') and the \hyperref[Banach-Steinhaus Theorem]{theorem:banach-steinhaus}, $\bracsn{\lambda_n(\cdot, y_n)|n \in \natp}$ is equicontinuous, and $\lambda_n(x_n, y_n) \to 0$ as $n \to \infty$ by \autoref{proposition:equicontinuous-net}. By (M) and \autoref{proposition:equicontinuous-net}, $\alg$ is equicontinuous at $0$, and hence equicontinuous by \autoref{lemma:equicontinuous-bilinear}. 
+\end{proof}
+
+% TODO: Replace this with a more general version involving polars in the future. 
+\begin{theorem}[Banach-Alaoglu]
+\label{theorem:alaoglu}
+    Let $E$ be a locally convex space over $K \in \RC$ and $\alg \subset E^*$ be equicontinuous, then $\alg$ is precompact with respect to $\sigma(E^*, E)$.
+\end{theorem}
+\begin{proof}
+    For each $x \in E$, $\alg(x) = \bracsn{\dpn{x, \phi}{E}|\phi \in \alg}$ is precompact by \autoref{proposition:equicontinuous-bounded}. By the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}, 
+    \begin{enumerate}
+        \item[(C2)] The $\sigma(E^*, E)$-closure of $\alg$ in $\prod_{x \in E}K$ is equicontinuous. 
+        \item[(C3)] The $\sigma(E^*, E)$-closure of $\alg$ in $\prod_{x \in E}K$ is compact. 
+    \end{enumerate}
+
+    By \autoref{proposition:operator-space-completeness}, the $\sigma(E^*, E)$-closure of $\alg$ in $\prod_{x \in E}\ol{\alg(x)}$ is a subset of $\hom(E; K)$. Hence the $\sigma(E^*, E)$-closure of $\alg$ in $E^*$ is compact. 
+\end{proof}
+
+
+

src/fa/tvs/index.tex

@@ -2,6 +2,7 @@
 \label{chap:tvs}
 
 \input{./definition.tex}
+\input{./complexify.tex}
 \input{./metric.tex}
 \input{./bounded.tex}
 \input{./dual.tex}
@@ -11,4 +12,6 @@
 \input{./complete-metric.tex}
 \input{./projective.tex}
 \input{./inductive.tex}
-\input{./spaces-of-linear.tex}
+\input{./vector-function.tex}
+\input{./space-of-linear.tex}
+\input{./equicontinuous.tex}

src/fa/tvs/inductive.tex

@@ -10,6 +10,7 @@
         \item[(U)] For any topology $\mathcal{S}$ on $E$ satisfying (1) and (2), $\mathcal{S} \subset T$.
         \item For any TVS $F$ and $T \in \hom(E; F)$, $T \in L(E; F)$ if and only if $T \circ T_i \in L(E_i; F)$ for all $i \in I$.
     \end{enumerate}
+
     The topology $\topo$ is the \textbf{inductive topology} on $E$ induced by $\seqi{T}$.
 \end{definition}
 \begin{proof}
@@ -33,6 +34,58 @@
     (4): Let $U \in \cn_F(0)$ be circled and radial and $i \in I$. Since $T \circ E_i \in L(E_i; F)$, $T_i^{-1}(T^{-1}(U)) \in \cn_{E_i}(0)$, so $T^{-1}(U) \in \mathcal{B} \subset \cn_E(0)$.
 \end{proof}
 
+\begin{definition}[Direct Sum]
+\label{definition:tvs-direct-sum}
+    Let $\seqi{E}$ be TVSs over $K \in \RC$, then there exists $(E, \seqi{\iota})$ such that:
+    \begin{enumerate}
+        \item $E$ is a TVS over $K$. 
+        \item For each $i \in I$, $\iota_i \in L(E_i; E)$. 
+        \item[(U)] For each $(F, \seqi{T})$ satisfying (1) and (2), there exists a unique $T \in L(E; F)$ such that the following diagram commutes:
+        \[
+        \xymatrix{
+        E \ar@{->}[r]^{T} & F \\
+        E_i \ar@{->}[u]^{\iota_i} \ar@{->}[ru]_{T_i} & 
+        }
+        \]
+    \end{enumerate}
+
+    
+    The space $E = \bigoplus_{i \in I}E_i$ is the \textbf{direct sum} of $\seqi{E}$. 
+\end{definition}
+\begin{proof}
+    Let $(E, \seqi{\iota})$ be the direct sum of $\seqi{E}$ as vector spaces, and equip it with the inductive topology induced by $\seqi{\iota}$, then $(E, \seqi{\iota})$ satisfies (1) and (2). 
+
+    (U): By (U) of the \hyperref[direct sum]{definition:direct-sum}, there exists a unique $T \in \hom(E; F)$ such that the diagram commutes. In which case, by (4) of \autoref{definition:tvs-inductive}, $T \in L(E; F)$. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:finite-tvs-product}
+    Let $\seqf{E_j}$ be TVSs over $K \in \RC$, then
+    \[
+    \prod_{j = 1}^n E_j = \bigoplus_{j = 1}^n E_j
+    \]
+\end{proposition}
+\begin{proof}
+    Let $1 \le k \le n$, then for each $1 \le k, l \le n$, $\pi_l \circ \iota_k \in L(E_k, E_l)$, so by (U) of the \hyperref[product]{definition:tvs-product}, $\iota_k \in L(E_k; \prod_{j = 1}^n E_j)$. Thus $\prod_{j = 1}^n E_j$ satisfies (1) and (2) of the \hyperref[direct sum]{definition:tvs-direct-sum}. 
+
+    For any TVS $F$ over $K$ and $\seqf{T_j}$ with $T_j \in L(E_j; F)$ for each $1 \le j \le n$, let
+    \[
+    T: \prod_{j = 1}^n E_j \to F \quad (x_1, \cdots, x_n) \mapsto \sum_{j = 1}^n T_jx_j
+    \]
+
+    then $T \in L(\prod_{j = 1}^n E_j; F)$ is the unique continuous linear map such that the following diagram commutes:
+    \[
+    \xymatrix{
+    E \ar@{->}[r]^{T} & F \\
+    E_i \ar@{->}[u]^{\iota_i} \ar@{->}[ru]_{T_i} & 
+    }
+    \]
+
+    Hence $\prod_{j = 1}^n E_j$ satisfies (U) of the \hyperref[direct sum]{definition:tvs-direct-sum}, so the spaces coincide. 
+\end{proof}
+
+
+
 \begin{definition}[Inductive Limit]
 \label{definition:tvs-inductive-limit}
     Let $(\seqi{E}, \bracsn{T^i_j| i, j \in I, i \lesssim j})$ be an upward-directed system of TVSs over $K \in \RC$, then there exists $(E, \bracsn{T^i_E}_{i \in I})$ such that:
@@ -62,6 +115,7 @@
         for all $i \in I$.
         \item For any TVS $F$ and $T \in \hom(E; F)$, $T \in L(E; F)$ if and only if $T \circ T^i_E \in L(E_i; F)$ for all $i \in I$.
     \end{enumerate}
+
     The pair $(E, \bracsn{T^i_E}_{i \in I})$ is the \textbf{inductive limit} of $(\seqi{E}, \bracsn{T^i_j| i, j \in I, i \lesssim j})$.
 \end{definition}
 \begin{proof}

src/fa/tvs/metric.tex

@@ -83,6 +83,7 @@
         \item[(a)] For each $n \in \natp$, $U_n$ is circled, radial, and contains $0$.
         \item[(b)] For each $n \in \natp$, $U_{n+1} + U_{n+1} \subset U_n$.
     \end{enumerate}
+
     then there exists a pseudonorm $\rho: E \to [0, \infty)$ such that for each $n \in \natp$,
     \[
     U_{n+1} \subset \rho^{-1}([0, 2^{-n})) \subset U_{n}
@@ -111,6 +112,7 @@
         so $\rho(\lambda x) \le \rho(x)$.
         \item[(PN3)] Let $x, y \in X$ and $M, N \subset \natp$ finite such that $x \in U_M$ and $y \in U_N$. Assume without loss of generality that $\rho_M + \rho_N < 1$, then there exists a unique $P \subset \nat$ finite such that $\rho_P = \rho_M + \rho_N$. In which case, $U_P \supset U_M + U_N$ by assumption (b). Therefore $\rho(x + y) \le \rho(x) + \rho(y)$.
     \end{enumerate}
+
     For any $x \in U_{n+1}$, $\rho(x) \le 2^{-n+1} < 2^n$, so $U_{n+1} \subset \rho^{-1}([0, 2^{-n}))$ by \autoref{proposition:dyadic-semigroup-order}. On the other hand, for any $x \in E$ with $\rho(x) < 2^{-n}$, $x \in U_{2^{-n}} = U_n$. This allows showing the remaining seminorm axioms by considering neighbourhoods of the form $\bracs{U_n|n \in \natp}$.
     \begin{enumerate}
         \item[(PN4)] Let $x \in X$ and $n \in \natp$. By assumption (a), there exists $\alpha > 0$ such that for any $\lambda \in K$ with $\abs{\lambda} \ge \alpha$, $x \in \lambda U_n$. Therefore for any $\lambda \in K$ with $\abs{\lambda} \le \alpha^{-1}$, $\lambda x \in U_n$, and $\rho(x) \le 2^{-n}$.

src/fa/tvs/projective.tex

@@ -8,6 +8,7 @@
         \item For each $i \in I$, $T_i \in L(E; F_i)$.
         \item[(U)] If $\mathfrak{V}$ is a uniformity on $E$ satisfying $(1)$, then $\mathfrak{V} \supset \fU$.
     \end{enumerate}
+
     Moreover,
     \begin{enumerate}
         \item[(3)] $\fU$ is translation-invariant.
@@ -21,6 +22,7 @@
         is a fundamental system of neighbourhoods for $E$ at $0$.
     \end{enumerate}
 
+    
     The uniformity $\fU$ and its topology are the \textbf{projective uniformity/topology} induced by $\seqi{T}$.
 \end{definition}
 \begin{proof}
@@ -75,6 +77,7 @@
         for all $i \in I$.
         \item For any TVS $F$ over $K$ and $S \in \hom(F; E)$, $S \in L(F; E)$ if and only if $T^E_i \circ S \in L(F; E_i)$ for all $i \in I$.
     \end{enumerate}
+
     The pair $(E, \bracsn{T^E_i}_{i \in I})$ is the \textbf{projective limit} of $(\seqi{E}, \bracsn{T^i_j|i, j \in I, i \lesssim j})$.
 \end{definition}
 \begin{proof}

src/fa/tvs/quotient.tex

@@ -18,6 +18,7 @@
 
         If $F$ is a TVS over $K$ and $f \in L(E; F)$, then $\td f \in L(E; F)$.
     \end{enumerate}
+
     The space $\td E = E/M$ is the \textbf{quotient} of $E$ by $M$.
 \end{definition}
 \begin{proof}
@@ -33,6 +34,7 @@
         \item[(TVB1)] Let $U \in \cn(0)$ be circled and radial. For any $\lambda \in K$ with $\abs{\lambda} \le 1$, $\lambda \pi(U) = \pi(\lambda U) \subset \pi(U)$, so $\pi(U)$ is also circled. For any $x + M \in E/M$, there exists $\lambda \in K$ such that $x \in \lambda U$. In which case, $x \in \lambda U + M = \pi(U)$, so $\pi(U)$ is also radial.
         \item[(TVB2)] For any $U \in \cn(0)$ circled and radial, by \autoref{proposition:tvs-good-neighbourhood-base}, there exists $W \in \cn(0)$ such that $W + W \subset U$. In which case, $\pi(W) + \pi(W) \subset \pi(U)$.
     \end{enumerate}
+
     By \autoref{proposition:tvs-0-neighbourhood-base}, there exists a unique translation-invariant topology on $E/M$ such that $\fB$ is a fundamental system of neighbourhoods at $0$, which must be the quotient topology on $E/M$. In which case, the quotient topology is a vector space topology by (3) of \autoref{proposition:tvs-0-neighbourhood-base}.
 
     (2), (3), (U): By \autoref{definition:quotient-topology}.

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

@@ -1,94 +1,26 @@
-\section{Vector-Valued Function Spaces}
-\label{section:spaces-linear-map}
-
-\begin{proposition}[{{\cite[III.3.1]{SchaeferWolff}}}]
-\label{proposition:tvs-set-uniformity}
-    Let $T$ be a set, $\sigma \subset 2^T$ be an ideal, and $F$ be a TVS over $K \in \RC$, then
-    \begin{enumerate}
-        \item The \hyperref[$\sigma$-uniformity]{definition:set-uniform} on $F^T$ is translation invariant.
-        \item The composition defined by
-        \[
-        T^F \times T^F \to T^F \quad (f + g)(x) = f(x) + g(x)
-        \]
-
-        is continuous.
-    \end{enumerate}
-
-    For any vector subspace $\cf \subset F^T$, the following are equivalent:
-    \begin{enumerate}
-        \item[(T)] The $\sigma$-uniform topology on $\cf \subset F^E$ is a vector space topology.
-        \item[(B)] For each $f \in \cf$ and $S \in \sigma$, $f(S) \subset E$ is bounded.
-    \end{enumerate}
-\end{proposition}
-\begin{proof}
-    (1): Let $U \subset F \times F$ be an entourage and $S \in \sigma$. Using \autoref{proposition:tvs-uniform}, assume without loss of generality that $U$ is translation-invariant. For any $(f, g) \in E(S, U)$, $h \in F^T$, and $x \in S$, $(f(x) + h(x), g(x) + h(x)) \in U$. Thus $(f + h, g + h) \in E(S, U)$ and $E(S, U)$ is translation-invariant.
-
-    (2): Let $f, g, f', g' \in \cf$, $S \in \sigma$, and $U \subset F \times F$ be an entourage. By (TVS1), there exists an entourage $V \subset F \times F$ such that for any $x, x', y, y' \in F$ with $(x, x'), (y, y') \in V$, $(x + y, x' + y') \in U$. If $(f, g), (f', g') \in E(S, V) \cap \cf$, then for any $x \in S$, $(f(x) + g(x), f'(x) + g'(x)) \in U$, so $(f + g, f' + g') \in E(S, U) \cap \cf$.
-
-    (T) $\Rightarrow$ (B): Let $f \in \cf$, $S \in \sigma$ and $U \subset F \times F$ be a symmetric entourage, then $E(S, U)(0)$ is a neighbourhood of $0$ with respect to the $\sigma$-uniform topology. By (TVS2), there exists $\lambda > 0$ such that $f \in \lambda E(S, U)(0)$. In which case, for any $x \in S$, $\lambda^{-1}f(x) \in U(0)$ and $f(x) \in \lambda U(0)$. Thus $f(S) \subset \lambda U(0)$, and $f(S)$ is bounded.
-
-    (T) $\Rightarrow$ (B): Let $f, g \in \cf$, $\lambda, \lambda' \in K$, and $S \in \sigma$, then for any $x \in X$,
-    \begin{align*}
-    \lambda f(x) - \lambda' g(x) &= \lambda f(x) - \lambda' f(x) + \lambda' f(x) - \lambda' g(x) \\
-    &= (\lambda - \lambda')f(x) + \lambda' (f(x) - g(x))
-    \end{align*}
-    Let $U_0 \in \cn_F(0)$. By (TVS1) and \autoref{proposition:tvs-good-neighbourhood-base}, there exists $U \in \cn_F(0)$ circled such that $U + U \subset U_0$. By (TVS2), there exists $V \in \cn_F(0)$ such that $\lambda' V \subset U$. Since $f(S)$ is bounded, there exists $\eps > 0$ with $\eps f(S) \subset U$. In which case, if $\abs{\lambda - \lambda'} < \eps$ and $f(x) - g(x) \in V$ for all $x \in S$, then
-    \[
-    \lambda f(x) - \lambda' g(x) \in (\lambda - \lambda')f(S) + \lambda' V \subset U + U \subset U_0
-    \]
-
-    for all $x \in S$.
-\end{proof}
-
-
-\begin{definition}[Space of Bounded Functions]
-\label{definition:bounded-function-space}
-    Let $T$ be a set, $E$ be a TVS over $K \in \RC$, and $f: T \to E$ be a function, then $f$ is \textbf{bounded} if $f(T)$ is a bounded subset of $E$. The set $B(T; E) \subset E^T$ is the space of all bounded functions from $T$ to $E$, and:
-    \begin{enumerate}
-        \item $B(T; E)$ equipped with the uniform topology and pointwise operations is a TVS over $K \in \RC$. 
-        \item $B(T; E)$ is a closed subset of $E^T$ with respect to the uniform topology. In particular, if $E$ is complete, then so is $B(T; E)$.
-    \end{enumerate}
-\end{definition}
-\begin{proof}
-    (1): For any $f, g \in B(T; E)$ and $\lambda \in K$, $(\lambda f + g)(T) \subset (\lambda f)(T) + g(T)$, which is bounded by \autoref{proposition:bounded-operations}, so $B(T; E)$ is closed under addition and scalar multiplication. 
-    
-    Since $f(E)$ is bounded for all $f \in B(T; E)$, $B(T; E)$ forms a TVS over $K$ by \autoref{proposition:tvs-set-uniformity}.
-
-    (2): Let $f \in \ol{B(T; E)}$ and $U \in \cn_E(0)$ be circled, then there exists $g \in B(T; E)$ such that $(f - g)(E) \subset U$. Since $g \in B(T; E)$, there exists $\lambda \ge 0$ such that $g(E) \subset \lambda U$. In which case,
-    \[
-    f(E) \subset g(E) + (f - g)(E) \subset (\lambda + 1)U
-    \]
-    so $f \in B(T; E)$.
-
-    If $E$ is complete, then $E^T$ with the uniform topology is complete by \autoref{proposition:set-uniform-complete}. Thus $B(T; E)$ is also complete by \autoref{proposition:complete-closed}.
-\end{proof}
-
-
-\begin{definition}[Space of Bounded Continuous Functions]
-\label{definition:bounded-continuous-function-space}
-    Let $X$ be a topological space and $E$ be a TVS over $K \in \RC$, then $BC(X; E) = B(X; E) \cap C(X; E)$ is the space of bounded and continuous functions from $X$ to $E$, and
-    \begin{enumerate}
-        \item $BC(X; E)$ equipped with the uniform topology and pointwise operations is a TVS over $K \in \RC$. 
-        \item $BC(X; E)$ is a closed subspace of $C(X; E)$ and $B(X; E)$ with respect to the uniform topology. In particular, if $E$ is complete, then so is $BC(X; E)$.
-    \end{enumerate}
-\end{definition}
-\begin{proof}
-    (1): Since addition and scalar multiplication are continuous, $BC(X; E)$ is a subspace of $B(X; E)$, and hence a TVS over $K \in \RC$ by \autoref{definition:bounded-function-space}.
-
-    (2): Since $B(X; E)$ and $C(X; E)$ are both closed subspaces of $E^X$ by \autoref{definition:bounded-function-space}, $BC(X; E)$ is a closed subspace. 
-    
-    If $E$ is complete, then $B(X; E)$ and $C(X; E)$ are both complete under the uniform topology by \autoref{definition:bounded-function-space}. Therefore $BC(X; E)$ is also complete. 
-\end{proof}
-
-
+\section{Spaces of Linear Maps}
+\label{section:space-linear-map-new}
 
 \begin{definition}[Space of Bounded Linear Maps]
 \label{definition:bounded-linear-map-space}
-    Let $E, F$ be TVSs over $K \in \RC$, $\sigma \subset 2^E$ be an ideal, and $k \in \nat$. The space $B_{\sigma}^k(E; F)$ is the set of all $k$-linear maps $T: E^k \to F$ with $T(S^k) \in B(F)$ for all $S \in \sigma$, equipped with the $\bracsn{S^k| S \in \sigma}$-uniform topology.
+    Let $E, F$ be TVSs over $K \in \RC$, $\sigma \subset 2^E$ be an ideal, and $k \in \nat$. The space $B_{\sigma}^k(E; F)$ is the set of all $k$-linear maps $T: E^k \to F$ with $T(S^k) \in \mathfrak{B}(F)$ for all $S \in \sigma$, equipped with the $\bracsn{S^k| S \in \sigma}$-uniform topology.
 
     Let $\fB \subset 2^E$ be the collection of all bounded subsets of $E$, then $B_{\sigma}(E; F) = B(E; F)$ is the \textbf{space of bounded linear maps} from $E$ to $F$. 
 \end{definition}
 
+\begin{proposition}[{{\cite[III.3.3]{SchaeferWolff}}}]
+\label{proposition:bounded-linear-map-space-bounded}
+    Let $E, F$ be TVSs over $K \in \RC$, $\sigma \subset 2^E$ be an ideal, and $A \subset B_\sigma(E; F)$, then the following are equivalent:
+    \begin{enumerate}
+        \item $A \subset B_\sigma(E; F)$ is bounded with respect to the $\sigma$-uniform topology. 
+        \item For each $V \in \cn_F(0)$, $\bigcap_{T \in A}T^{-1}(V)$ absorbs every $S \in \sigma$.
+        \item For every $S \in \sigma$, $\bigcup_{T \in A}T(A)$ is bounded in $F$. 
+    \end{enumerate}
+    
+\end{proposition}
+% Proof omitted because it is obvious. 
+
+
 \begin{proposition}
 \label{proposition:multilinear-identify}
 	Let $E, F$ be TVSs over $K \in \RC$, $\sigma \subset 2^E$ be a covering ideal, and $k \in \natp$, then
@@ -116,6 +48,7 @@
 
         is an isomorphism.
     \end{enumerate}
+
     which allows the identification
     \[
     \underbrace{B_{\sigma}(E; B_{\sigma}(E; \cdots)))}_{k \text{ times}} = B^k_{\sigma}(E; F)
@@ -131,12 +64,12 @@
 
     Let $(x_1, \cdots, x_k) \in E^k$ and $S \in \sigma$. Since $\sigma$ is a covering ideal, assume without loss of generality that $\bracsn{x_j}_1^k \subset S$. In which case,
     \[
-    T(x_1, \cdots, x_k, S) \subset T(S^{k+1}) \in B(F)
+    T(x_1, \cdots, x_k, S) \subset T(S^{k+1}) \in \mathfrak{B}(F)
     \]
 
     by assumption. Thus $I^{-1}T(x_1, \cdots, x_k) \in B_{\sigma}(E; F)$.
 
-    In addition, for any $S_1 \in \sigma$ and entourage $E(S_2, U)$ of $B_{\sigma}(E; F)$ where $S_2 \in \sigma$ and $U$ is an entourage of $F$, there exists $S \in \sigma$ with $S \supset S_1 \cup S_2$. Given that $T(S^{k+1}) \in B(F)$, there exists $\lambda > 0$ such that $T(S^{k+1}) \subset \lambda U(0)$. In which case, $I^{-1}T(S^k) \subset \lambda E(S, U)(0)$ and $I^{-1}T(S^k) \in B(B_{\sigma}(E; F))$. Thus $I^{-1}T \in B^k_{\sigma}(E; B_{\sigma}(E; F))$.
+    In addition, for any $S_1 \in \sigma$ and entourage $E(S_2, U)$ of $B_{\sigma}(E; F)$ where $S_2 \in \sigma$ and $U$ is an entourage of $F$, there exists $S \in \sigma$ with $S \supset S_1 \cup S_2$. Given that $T(S^{k+1}) \in \mathfrak{B}(F)$, there exists $\lambda > 0$ such that $T(S^{k+1}) \subset \lambda U(0)$. In which case, $I^{-1}T(S^k) \subset \lambda E(S, U)(0)$ and $I^{-1}T(S^k) \in B(B_{\sigma}(E; F))$. Thus $I^{-1}T \in B^k_{\sigma}(E; B_{\sigma}(E; F))$.
 
     It remains to show that $I$ and $I^{-1}$ is continuous. To this end, let $S \in \sigma$ and $U$ be an entourage of $F$, then for any $T \in E(S^k, E(S, U))(0)$, $IT \in E(S^{k+1}, U)(0)$, so $I$ is continuous.
 
@@ -150,8 +83,6 @@
     Thus (2) holds for all $k \in \natp$.
 \end{proof}
 
-
-
 \begin{definition}[Strong Operator Topology]
 \label{definition:strong-operator-topology}
     Let $E, F$ be TVSs over $K \in \RC$, $\fF \subset 2^E$ be the collection of finite subsets of $E$, then the $\fF$-uniform topology on $F^E$ is the \textbf{strong operator topology}.
@@ -168,6 +99,7 @@
         \item[(b)] $\bracs{T_\alpha|\alpha \in A}$ is uniformly equicontinuous. 
     \end{enumerate}
 
+    
     then $T_\alpha \to T$ in $L_s(E; F)$.
 \end{proposition}
 \begin{proof}
@@ -222,5 +154,3 @@
     (2): By \autoref{definition:bounded-function-space} and (1), the space of bounded functions and the space of linear functions from $E$ to $F$ are closed subspaces of $F^E$ with respect to the topology of bounded convergence. Therefore $B(E; F)$ is also a closed subspace.
 
 \end{proof}
-
-

src/fa/tvs/vector-function.tex

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

src/measure/bochner-integral/bochner.tex

@@ -54,6 +54,7 @@
         \item[(b)] There exists $g \in L^1(X) \cap L^+(X)$ such that $\norm{f_n}_E \le g$ for all $n \in \natp$. 
     \end{enumerate}
 
+    
     then $\int f d\mu = \limv{n}\int f_n d\mu$. 
 \end{theorem}
 \begin{proof}
@@ -78,6 +79,7 @@
         \item For any $f \in L^1(X, |\mu|; E)$, $\normn{I_\lambda f}_{G} \le \norm{\lambda}_{L^2(E, F; G)} \cdot \norm{f}_{L^1(X, |\mu|; E)}$.
     \end{enumerate}
 
+    
     For any $f \in L^1(X; E)$, $I_\lambda f = \int \lambda(f, d\mu)$ is the \textbf{Bochner integral} of $f$ with respect to $\mu$ and $\lambda$. 
 \end{definition}
 \begin{proof}
@@ -98,6 +100,7 @@
         \item[(b)] There exists $g \in L^1(X) \cap L^+(X)$ such that $\norm{f_n}_E \le g$ for all $n \in \natp$. 
     \end{enumerate}
 
+    
     then $\int \lambda(f , d\mu) = \limv{n}\int \lambda(f_n, d\mu)$. 
 \end{theorem}
 \begin{proof}

src/measure/bochner-integral/strongly.tex

@@ -12,6 +12,7 @@
             \item[(a)] For each $n \in \natp$, $\norm{f_n}_E \le \norm{f}_E$.
             \item[(b)] $\norm{f_n(x) - f(x)}_E \to 0$ pointwise as $n \to \infty$.
         \end{enumerate}
+
     \end{enumerate}
 
     If the above holds, then $f$ is a \textbf{strongly measurable} function. 
@@ -41,6 +42,7 @@
         \item For any strongly measurable functions $\bracs{f_n: X \to E|n \in \natp}$ and $f: X \to E$, if $f_n \to f$ strongly pointwise, then $f$ is strongly measurable. 
     \end{enumerate}
 
+    
 \end{proposition}
 \begin{proof}
     (1): Since $x \mapsto \lambda$ is continuous, $\lambda f$ is strongly measurable by (2) of \autoref{definition:strongly-measurable}. 

src/measure/lebesgue-integral/complex.tex

@@ -105,6 +105,7 @@
         \item $f_n \to f$ pointwise.
         \item There exists $g \in \mathcal{L}^1(X)$ such that $\abs{f_n} \le \abs{g}$ for all $n \in \natp$.
     \end{enumerate}
+
     then $\int fd\mu = \limv{n}\int f_n d\mu$.
 \end{theorem}
 \begin{proof}[Proof {{\cite[Theorem 2.24]{Folland}}}. ]

src/measure/measurable-maps/metric.tex

@@ -77,6 +77,7 @@
         \item[(b)] $\bigcap_{y \in Y}N(y) \ne \emptyset$.
         \item[(c)] For any $y_0 \in Y$, $\bracs{y \in Y|y_0 \in N(y)} \in \cb_Y$.
     \end{enumerate}
+
     Then, for any $f: X \to Y$, the following are equivalent:
     \begin{enumerate}
         \item $f$ is $(\cm, \cb_Y)$-measurable.
@@ -85,6 +86,7 @@
             \item[(i)] For each $x \in X$ and $n \in \natp$, $f_n(x) \in N(f(x))$.
             \item[(ii)] $f_n \to f$ pointwise.
         \end{enumerate}
+
         \item There exists a sequence $\seq{f_n}$ of $(\cm, \cb_Y)$-measurable simple functions such that $f_n \to f$ pointwise.
     \end{enumerate}
 \end{proposition}
@@ -147,5 +149,6 @@
         \]
 
     \end{enumerate}
+
     By \autoref{proposition:measurable-simple-separable}, (1) and (2) are equivalent.
 \end{proof}

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

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

src/measure/measure/complete.tex

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

src/measure/measure/kolmogorov.tex

@@ -17,12 +17,14 @@
         \item[(b)] $X_n$ is Hausdorff.
         \item[(c)] $X_n$ is separable.
     \end{enumerate}
+
     Let $\bracs{\mu_{I}| I \subset \natp \text{ finite}}$ be consistent Borel probability measures, then for any $\seq{B_n}$ where:
     \begin{enumerate}
         \item[(d)] For each $n \in \nat$, $B_n \in \cb_{\prod_{j = 1}^n X_j}$.
         \item[(e)] For each $n \in \nat$, $B_{n+1} \subset B_n \times X_{n+1}$.
         \item[(f)] There exists $\eps > 0$ such that $\mu_{[n]}(B_n) > \eps$ for all $n \in \natp$.
     \end{enumerate}
+
     Then there exists $\seq{K_n}$ such that for every $n \in \natp$,
     \begin{enumerate}
         \item $K_n \subset \prod_{j = 1}^n X_j$ is compact.
@@ -53,6 +55,7 @@
         \item[(b)] $X_j$ is Hausdorff.
         \item[(c)] $X_j$ is separable.
     \end{enumerate}
+
     Let $\bracs{\mu_{I}| I \subset \natp \text{ finite}}$ be consistent Borel probability measures, then there exists a unique probability measure $\mu: \bigotimes_{i \in I}\cb_{X_i} \to [0, 1]$ such that for any $J \subset I$ finite, $\mu = \mu_J \circ \pi_J^{-1}$.
 \end{theorem}
 \begin{proof}[Proof {{\cite[Theorem 1.14]{Baudoin}}}. ]
@@ -72,6 +75,7 @@
         \item $K_{n+1} \subset K_n \times X_{n+1}$.
         \item $\mu(K_n) \ge \eps/2$.
     \end{enumerate}
+
     Let $N \in \natp$ and $x \in \prod_{j = 1}^N X_j$ such that $x \in \bigcap_{n \ge N}\pi_{[N]}(K_n)$. By compactness and (4), there exists $x_{N+1} \in X_{N+1}$ such that $(x_1, \cdots, x_N, x_{N+1}) \in \bigcap_{n > N}\pi_{[N+1]}(K_n)$. Thus there exists $x \in \prod_{i \in I}X_i$ such that $x \in \pi_{[n]}^{-1}(K_n)$ for all $n \in \natp$, and
     \[
     x \in \bigcap_{n \in \natp}\pi_{[n]}^{-1}(K_n) \subset \bigcap_{n \in \natp}\pi_{[n]}^{-1}(B_n) \ne \emptyset

src/measure/measure/lebesgue-stieltjes.tex

@@ -45,6 +45,7 @@
 
         \item[(U)] For any function $G: \real \to \real$ satisfying (1), $F - G$ is constant.
     \end{enumerate}
+
     Conversely, if $F: \real \to \real$ is a Stieltjes function, then there exists a unique Borel measure $\mu_F: \cb_\real \to [0, \infty]$ satisfying (1), and $\mu_F$ is the \textbf{Lebesgue-Stieltjes measure} associated with $F$.
 \end{definition}
 \begin{proof}[Proof {{\cite[Theorem 1.16]{Folland}}}. ]

src/measure/measure/measure.tex

@@ -15,12 +15,14 @@
         \item[(M1)] $\mu(\emptyset) = 0$.
         \item[(M2)] For any $\seq{E_n} \subset \cm$ pairwise disjoint, $\mu\paren{\bigsqcup_{n \in \natp}E_n} = \sum_{n \in \natp} \mu(E_n)$.
     \end{enumerate}
+
     In which case, $(X, \cm, \mu)$ is a \textbf{measure space}.
 
     If $\mu: \cm \to [0, \infty]$ instead satisfies (M1) and
     \begin{enumerate}
         \item[(M2')] For any $\seqf{E_j} \subset \cm$ pairwise disjoint, $\mu\paren{\bigsqcup_{j = 1}^n E_j} = \sum_{j = 1}^n \mu(E_j)$.
     \end{enumerate}
+
     then $\mu$ is a \textbf{finitely-additive measure}.
 \end{definition}
 
@@ -101,6 +103,7 @@
         \item[(b)] $\mu(E) = \nu(E)$ for all $E \in \mathcal{P}$.
         \item[(c)] There exists $\seq{E_n} \subset \mathcal{P}$ such that $E_n \upto X$ and $\mu(E_n) < \infty$ for all $n \in \natp$.
     \end{enumerate}
+
     then $\mu = \nu$.
 \end{theorem}
 \begin{proof}
@@ -123,6 +126,7 @@
 
         by continuity from below (\autoref{proposition:measure-properties}).
     \end{enumerate}
+
     so $\alg(F)$ is a $\lambda$-system. By (a) and Dynkin's $\pi$-$\lambda$ theorem (\autoref{theorem:pi-lambda}), $\alg(F) = \cm$.
 
     Let $\seq{E_n}$ as in assumption (c), then $\mu(E_n \cap F) = \mu(E_n \cap F)$ for all $n \in \natp$ and $F \in \cm$. Thus by continuity from below (\autoref{proposition:measure-properties}),
@@ -165,6 +169,7 @@
         \]
     \end{enumerate}
 
+    
 \end{definition}
 \begin{proof}
     (1): Preimage commutes with unions, intersections, and complements. 

src/measure/measure/outer.tex

@@ -116,6 +116,7 @@
             \item[(b)] For any $E \in \cm \cap \cn$ with $\mu(E) < \infty$, $\mu(E) = \nu(E)$.
             \item[(c)] If $\mu$ is $\sigma$-finite, then $\nu = \mu$.
         \end{enumerate}
+
     \end{enumerate}
 \end{theorem}
 \begin{proof}[Proof {{\cite[Theorem 1.14]{Folland}}}. ]

src/measure/measure/product.tex

@@ -10,6 +10,7 @@
         \item[(U)] For any measure $\lambda: \cm \otimes \cn \to [0, \infty]$, $\lambda \le \mu$. For any $A \in \cm \otimes \cn$ with $\mu(A) < \infty$, $\lambda(A) = \mu(A)$. In particular, if $\mu$ is $\sigma$-finite, then $\lambda = \mu$. 
     \end{enumerate}
 
+    
     The measure $\mu \otimes \nu$ is the \textbf{product} of $\mu$ and $\nu$. 
 \end{definition}
 \begin{proof}
@@ -59,6 +60,7 @@
         \item For any measure space $(Z, \cf)$, $(\cm \otimes \cn, \cf)$-measurable function $f: X \times Y \to Z$, $x \in X$, and $y \in Y$, $f(x, \cdot)$ is $(\cn, \cf)$-measurable and $f(\cdot, y)$ is $(\cm, \cf)$-measurable. 
     \end{enumerate}
 
+    
 \end{lemma}
 \begin{proof}
     (1): Let $\alg \subset \cm \otimes \cn$ be the collection of all sets that satisfy (1), then
@@ -101,8 +103,10 @@
         &= \int_{Y}\int_X f(x, y)\mu(dx)\nu(dy)
         \end{align*}
         \end{enumerate}
+
     \end{enumerate}
 
+    
 \end{theorem}
 \begin{proof}
     (1): First suppose that $\mu$ and $\nu$ are both finite. Let $\alg \subset \cm \otimes \cn$ be the collection of sets whose indicator functions satisfy (1), then $\alg$ contains all rectangles with finite measure. By linearity of the integral, for any $E, F\in \alg$, $E \setminus F \in \alg$. For any $\seq{E_n} \subset \alg$ and $E \in \alg$ such that $E_n \upto E$, by the \hyperref[Monotone Convergence Theorem]{theorem:mct},

src/measure/measure/regular.tex

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

src/measure/measure/semifinite.tex

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

src/measure/measure/sigma-finite.tex

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

src/measure/notation.tex

@@ -7,6 +7,7 @@
 
     $\sigma(\mathcal{E})$ & $\sigma$-algebra generated by $\mathcal{E}$. & \autoref{definition:generated-sigma-algebra} \\
     $\lambda(\mathcal{E})$ & $\lambda$-system generated by $\mathcal{E}$. & \autoref{definition:generated-lambda-system} \\
+    $\sigma \otimes \tau$ & Product of ideals. & \autoref{definition:product-ideal} \\
     % ---- Measure Theory ----
     $\mathcal{B}_X$ & Borel $\sigma$-algebra on $X$. & \autoref{definition:borel-sigma-algebra} \\
     $\sigma(\{f_i \mid i \in I\})$ & $\sigma$-algebra generated by the maps $\{f_i\}$. & \autoref{definition:generated-sigma-algebra-function} \\

src/measure/radon/c0.tex

@@ -10,6 +10,7 @@
         \item $\norm{I^+}_{C_0(X; \real)^*}, \norm{I^-}_{C_0(X; \real)^*} \le \norm{I}_{C_0(X; \real)^*}$.
     \end{enumerate}
 
+    
 \end{lemma}
 \begin{proof}
     (1), (2): Since $C_0(X; \real)$ is a Banach lattice, $C_0(X; \real)^*$ is also a Banach lattice by \autoref{proposition:banach-lattice-dual}. Therefore there exists positive linear functionals $I^+, I^- \in C(X; \real)^*$ such that $I = I^+ - I^-$ and $I^+ \perp I^-$. 

src/measure/radon/radon.tex

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

src/measure/radon/riesz.tex

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

src/measure/sets/algebra.tex

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

src/measure/sets/borel.tex

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

src/measure/sets/elementary.tex

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

src/measure/sets/lambda.tex

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

src/measure/vector/complex.tex

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

src/measure/vector/variation.tex

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

src/measure/vector/vector.tex

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

src/op/banach/definitions.tex

@@ -0,0 +1,70 @@
+\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 $(A, \norm{\cdot}_A)$ 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, there exists an equivalent norm $\norm{\cdot}_{1}: A \to [0, \inftu)$ such that $\norm{1}_1 = 1$, and $A$ is always assumed to be equipped with this norm.  
+\end{definition}
+\begin{proof}[Proof, {{\cite[Proposition I.1.3]{Takesaki1}}}. ]
+    For each $x \in A$, let $L_x \in L(A; A)$ be defined by $y \mapsto xy$, and let $\norm{x}_1 = \norm{L_x}_{L(A; A)}$, then $\norm{x}_1 \le \norm{x}_A$ and $\norm{1}_1 = 1$. On the other hand, $\frac{\norm{x}_A}{\norm{1}_A} \le \norm{x}_1$, so $\norm{\cdot}_1$ is equivalent to $\norm{\cdot}_A$.
+\end{proof}
+
+\begin{definition}[Homomorphism]
+\label{definition:banach-algebra-homomorphism}
+    Let $A, B$ be Banach algebras and $\phi: A \to B$, then $\phi$ is a \textbf{homomorphism} if:
+    \begin{enumerate}
+        \item $\phi \in L(A; B)$. 
+        \item For each $x, y \in A$, $\phi(xy) = \phi(x)\phi(y)$. 
+    \end{enumerate}
+\end{definition}
+
+\begin{definition}[Unital Homomorphism]
+\label{definition:banach-algebra-unital-homomorphism}
+    Let $A, B$ be unital Banach algebras and $\phi: A \to B$ be a homomorphism, then $\phi$ is a \textbf{unital homomorphism} if $\phi(1) = 1$. 
+\end{definition}
+
+\begin{definition}[Unitisation]
+\label{definition:unitisation}
+    Let $A$ be a Banach algebra over $\complex$, and $\tilde A = \complex \oplus A$ with
+    \[
+    \iota: A \to \complex \oplus A \quad x \mapsto 0 + x
+    \]
+    
+    For each $\lambda + x, \mu + y \in \tilde A$, define
+    \[
+    (\lambda + x)(\mu + y) = \lambda \mu + (\lambda x + \mu x + xy)
+    \]
+    
+    and
+    \[
+    \norm{\lambda + x}_{\tilde A} = |\lambda| \norm{x}_A
+    \]
+
+    then
+    \begin{enumerate}
+        \item $\tilde A$ is a unital associative algebra over $\complex$.
+        \item $\iota: A \to \tilde A$ is a homomorphism. 
+        \item[(U)] For any pair $(B, \phi)$ satisfying (1) and (2), there exists a unique continuous unital homomorphism $\tilde \phi: \tilde A \to B$ such that $\phi(1) = 1$ and the following diagram commutes:
+        \xymatrix{
+        \tilde A \ar@{->}[r]^{\tilde \phi } & B \\
+        A \ar@{->}[u]^{\iota} \ar@{->}[ru]_{\phi} & 
+        }
+        \item $\iota(A)$ is a closed two-sided ideal of $\tilde A$. 
+    \end{enumerate}
+
+    The algebra $\tilde A$ is the \textbf{unitisation} of $A$. 
+\end{definition}
+\begin{proof}
+    (U): For each $\lambda + x \in \tilde A$, let $\tilde \phi(\lambda + x) = \lamdba + \phi(x)$. 
+\end{proof}
+
+

src/op/banach/fc.tex

@@ -0,0 +1,224 @@
+\section{The Holomorphic Functional Calculus}
+\label{section:holomorphic-functional-calculus}
+
+
+\begin{definition}[Holomorphic Functional Calculus]
+\label{definition:holomorphic-functional-calculus}
+    Let $A$ be a unital Banach algebra and $x \in A$, then there exists a unique continuous homomorphism
+    \[
+    H(\sigma_A(x); \complex) \to A \quad f \mapsto f(x)
+    \]
+
+    such that:
+    \begin{enumerate}
+        \item $1(x) = 1$. 
+        \item $\text{Id}(x) = x$. 
+    \end{enumerate}
+
+    Moreover, 
+    \begin{enumerate}[start=2]
+        \item For each $U \in \cn_\complex(\sigma_A(x))$ and closed rectifiable curves $\seqf{\gamma_j}$ on $U \setminus \sigma_A(x)$ such that $f(z_0) = \frac{1}{2\pi i}\sum_{j = 1}^n \int_{\gamma_j} f(z)/(z - z_0)dz$ for all $f \in H(U; \complex)$ and $z_0 \in \sigma_A(x)$, 
+        \[
+        f(x) = \frac{1}{2\pi i}\sum_{j = 1}^n \int_{\gamma_j} \frac{f(\lambda)}{\lambda - x}d\lambda
+        \]
+
+        for all $f \in H(U; \complex)$. 
+        \item For each $U \subset \complex$ open and $f \in H(U; \complex)$, the mapping
+        \[
+        \bracs{x \in A| \sigma_A(x) \subset U} \to A \quad x \mapsto f(x)
+        \]
+
+        is continuous. 
+    \end{enumerate}
+
+    The mapping $f \mapsto f(x)$ is the \textbf{holomorphic functional calculus} of $x$. 
+\end{definition}
+\begin{proof}[Proof, {{\cite[Proposition I.2.7]{Takesaki1}}}. ]
+    (Definition), (3): Let $U, V \in \cn_\complex(\sigma_A(x))$ such that $\ol V$ is a compact subset of $U$, then by \autoref{proposition:existence-curves}, there exists closed rectifiable curves $\seqf{\gamma_j}$ on $V \setminus \sigma_A(x)$ such that 
+    \begin{enumerate}[label=(\alph*)]
+        \item For all $f \in H(V; \complex)$ and $z_0 \in \sigma_A(x)$, 
+        \[
+        f(z_0) = \frac{1}{2\pi i}\sum_{j = 1}^n \int_{\gamma_j} \frac{f(z)}{z - z_0}dz
+        \]
+
+        \item For all $f \in H(U; \complex)$ and $z_0 \in U \setminus V$, 
+        \[
+        \frac{1}{2\pi i}\sum_{j = 1}^n \int_{\gamma_j} \frac{f(z)}{z - z_0}dz = 0
+        \]
+    \end{enumerate}
+    
+    For each $f \in H(U; \complex)$, define
+    \[
+    f(x) = \frac{1}{2\pi i}\sum_{j = 1}^n \int_{\gamma_j} \frac{f(\lambda)}{\lambda - x}d\lambda
+    \]
+
+    then by \hyperref[Cauchy's Theorem]{theorem:cauchy-homotopy}, the above definition is independent of the choice of curves satisfying (a). 
+
+    (Linearity): By \autoref{proposition:rs-bound}, the mapping $f \mapsto f(z)$ is a continuous linear map from $H(\sigma_A(x); \complex)$ to $A$. 
+
+    (Homomorphism): Let $T$ be the union of the image of $\seq{\gamma_j}$. By \autoref{proposition:existence-curves}, there exists closed rectifiable curves $\seqf[m]{\mu_j}$ on $U \setminus \ol V$ such that $g(z_0) = \frac{1}{2\pi i}\sum_{j = 1}^m \int_{\mu_j} g(z)/(z - z_0)dz$ for all $f \in H(U; \complex)$ and $z_0 \in \ol V$. Now, 
+    \begin{align*}
+        f(x)g(x) &= \frac{1}{(2\pi i)^2}\sum_{j = 1}^n \sum_{k = 1}^m \braks{\int_{\gamma_j} \frac{f(z)}{z - x}dz} \cdot \braks{\int_{\mu_k}\frac{g(w)}{w - x}dw} \\
+        &= \frac{1}{(2\pi i)^2}\sum_{j = 1}^n \sum_{k = 1}^m \int_{\gamma_j}\int_{\mu_k} \frac{f(z)g(w)}{(z - x)(w - x)}dwdz
+    \end{align*}
+    
+    For each $z, w \in U \setminus \sigma_A(x)$ with $z \ne w$, by the \hyperref[resolvent equation]{lemma:resolvent-equation}, 
+    \[
+        \frac{f(z)g(w)}{(z - x)(w - x)} = \frac{f(z)g(w)}{w - z}\braks{\frac{1}{z - x} - \frac{1}{w - x}}
+    \]
+
+    By assumptions on $\seqf[m]{\mu_j}$, for each $1 \le j \le n$, 
+    \[
+        \frac{1}{2\pi i }\int_{\gamma_j} \sum_{k = 1}^m \int_{\mu_k} \frac{f(z)g(w)}{(w - z)(z - x)}dwdz = \int_{\gamma_j} \frac{f(z)g(z)}{z - x}dz
+    \]
+
+    By assumption (b) and \hyperref[Fubini's Theorem]{theorem:rs-fubini}, for each $1 \le k \le m$, 
+    \[
+        \sum_{j = 1}^n\int_{\gamma_j}\int_{\mu_k}  \frac{f(z)g(w)}{(w - z)(z - x)}dwdz = 0
+    \]
+
+    Therefore
+    \[
+        f(x)g(x) = \frac{1}{2\pi i }\sum_{j = 1}^n \int_{\gamma_j} \frac{f(z)g(z)}{z - x}dz = (fg)(x)
+    \]
+
+    and the mapping $f \mapsto f(x)$ is a homomorphism. 
+
+    (1), (2): Let $R > 0$ such that $\sigma_A(x) \subset B_\complex(0, R)$, then by \autoref{proposition:rs-complete} and \hyperref[Cauchy's Integral Formula]{theorem:cauchy-formula}, 
+    \[
+        \text{Id}(x) = \frac{1}{2\pi i }\int_{\omega_{0, R}} \frac{1}{z - x}dz = \frac{1}{2\pi i }\sum_{n = 0}^\infty\int_{\omega_{0, R}}  z^{-n-1}x^n dz = 1
+    \]
+
+    and
+    \[
+        \text{Id}(x) = \frac{1}{2\pi i }\int_{\omega_{0, R}} \frac{z}{z - x}dz = \frac{1}{2\pi i }\sum_{n = 0}^\infty\int_{\omega_{0, R}}  z^{-n}x^n dz = x
+    \]
+
+    (Uniqueness): By (2), the homomorphism extends uniquely to $\complex(z) \cap H(\sigma_A(x); \complex)$. By \hyperref[Runge's Theorem]{corollary:runge-rational-approximation}, it extends uniquely to $H(\sigma_A(x); \complex)$ by continuity. 
+\end{proof}
+
+\begin{theorem}[Spectral Mapping Theorem (Holomorphic)]
+\label{theorem:spectral-mapping-holomorphic}
+    Let $A$ be a unital Banach algebra and $x \in A$, then:
+    \begin{enumerate}
+        \item For each $f \in H(\sigma_A(x); \complex)$, $\sigma_A(f(x)) = f(\sigma_A(x))$. 
+        \item For each $f \in H(\sigma_A(x); \complex)$ and $g \in H(\sigma_A(f(x)); \complex)$, $(f \circ g)(x) = f(g(x))$. 
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    (1): Let $\lambda \in \complex \setminus f(\sigma_A(x))$, then $1/(\lambda - f) \in H(\sigma_A(x); \complex)$. Since the holomorphic functional calculus is a homomorphism, $[1/(\lambda - f)](x) = (\lambda - f(x))^{-1}$. Thus $\sigma_A(f(x)) \subset f(\sigma_A(x))$. 
+
+    On the other hand, let $\lambda \in \sigma_A(x)$, then the function $h(\mu) = [f(\mu) - f(\lambda)]/(\mu - \lambda)$ extends to an element of $H(\sigma_A(x); \complex)$ by \autoref{proposition:zero-finite-multiplicity}. By the homomorphism property, 
+    \[
+    f(x) - f(\lambda) = (x - \lambda)h(x) = h(x)(x - \lambda)
+    \]
+
+    Since $(x - \lambda) \not\in G(A)$, $f(x) - f(\lambda) \not\in G(A)$ as well. Therefore $\sigma_A(f(x)) \supset f(\sigma_A(x))$. 
+
+    (2): If additionally $f, g \in \complex(z)$, then the theorem holds by (1) and (2) of \autoref{definition:holomorphic-functional-calculus}. 
+
+    Now suppose that $f \in H(\sigma_A(x); \complex)$ is arbitrary and $g \in \complex(z)$. For each $\eps > 0$, by \autoref{proposition:spectrum-continuous} and continuity of the holomorphic functional calculus, there exists $U \in \cn_{H(\sigma_A(x); \complex)}(f)$ such that for each $h \in U$, 
+    \begin{enumerate}[label=(\alph*)]
+        \item $g \in H(\sigma_A(h(x)); \complex)$.
+        \item $\norm{g(f(x)) - g(h(x))}_A < \eps$. 
+    \end{enumerate}
+    
+    By \hyperref[Runge's Theorem]{corollary:runge-rational-approximation}, there exists $h \in U \cap \complex(z)$. Thus $\norm{g(f(x)) - (g \circ f)(x)}_A < \eps$ as well. As this holds for all $\eps > 0$, $g(f(x)) = (g \circ f)(x)$ for any $f \in H(\sigma_A(x); \complex)$ and $g \in \complex(x) \cap H(f(\sigma_A(x)); \complex)$. 
+    
+    Through a second application of \hyperref[Runge's Theorem]{corollary:runge-rational-approximation} and continuity of the holomorphic functional calculus, the above also holds for all $f \in H(\sigma_A(x); \complex)$ and $g \in H(f(\sigma_A(x)); \complex)$. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:functional-calculus-exp}
+    Let $A$ be a unital Banach algebra and $x \in A$, then
+    \begin{enumerate}
+        \item $\exp(x) = \sum_{n = 0}^\infty \frac{x^n}{n!}$. 
+        \item $\exp(x) \in G_0(A)$ with $\exp(x)^{-1} = \exp(-x)$. 
+        \item For any $y \in A$ commuting with $x$, $\exp(x + y) = \exp(x)\exp(y)$. 
+        \item Let $\ell: \complex \setminus (-\infty, 0] \to \complex$ be the principal logarithm. If $\sigma_A(x) \subset \bracs{\lambda \in \complex| \text{Im}(\lambda) \in (-\pi, \pi)}$, then $\ell(\exp(x)) = x$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): By the power series representation of $\exp(x)$ and continuity of the holomorphic functional calculus. 
+
+    (2): By the homomorphism property of the functional calculus, $\exp(x)^{-1} = \exp(-x)$. Thus $\exp(x)$ is connected to $1$ by the path $t \mapsto \exp(tx)$. 
+
+    (3): Since the exponential is an entire function, the following series converges absolutely, and is eligible for arbitrary manipulations: 
+    \begin{align*}
+        \exp(x)\exp(y) &= \sum_{n = 0}^\infty \sum_{k = 0}^\infty \frac{x^ny^k}{n!k!} = \sum_{n = 0}^\infty \sum_{k = 0}^n \frac{x^ky^{n-k}}{k!(n-k)!} \\
+        &= \sum_{n = 0}^\infty \frac{1}{n!}\sum_{k = 0}^n {n \choose k}x^ky^{n-k} \\
+        &= \sum_{n = 0}^\infty \frac{(x + y)^n}{n!} = \exp(x + y)
+    \end{align*}
+    
+    
+    (4): By the \hyperref[Spectral Mapping Theorem]{theorem:spectral-mapping-holomorphic}. 
+\end{proof}
+
+\begin{theorem}[Riesz Decomposition]
+\label{theorem:riesz-decomposition}
+    Let $A$ be a unital Banach algebra, $x \in A$, and $\text{Clop}(\sigma_A(x))$ be the family of all relatively open and closed subsets of $\sigma_A(x)$, then there exists a mapping $P: \text{Clop}(\sigma_A(x)) \to A$ such that:
+    \begin{enumerate}
+        \item For each $K \in \text{Clop}(\sigma_A(x))$, $P(K)$ is idempotent and commutes with $x$. 
+        \item $P(\emptyset) = 0$ and $P(\sigma_A(x)) = 1$. 
+        \item For each $B, C \in \text{Clop}(\sigma_A(x))$, $P(B \cap C) = P(B)P(C)$. 
+        \item For pairwise disjoint sequence $\seqf{B_j} \in \text{Clop}(\sigma_A(x))$, $P(\bigsqcup_{j = 1}^n B_j) = \sum_{n = 1}^n P(B_j)$. 
+    \end{enumerate}
+
+    Let $E$ be a Banach space such that $A \subset B(E)$. For each $K \in \text{Clop}(\sigma_A(x))$, let $E_K = P(E)(K)$, then
+    \begin{enumerate}[start=4]
+        \item For any $B, C \in \text{Clop}(\sigma_A(x))$, $E_{B \cap C} = E_B \cap E_C$. 
+        \item For any $B, C \in \text{Clop}(\sigma_A(x))$, $E_{B \cup C} = E_B + E_C$. 
+        \item $E_K$ is a closed subspace of $E$.
+        \item $E = E_K \oplus E_{K^c}$ as a sum of complementary closed subspaces. 
+        \item $E_K$ is $x$-invariant. 
+    \end{enumerate}
+
+    Finally, let $x_k = x|_{E_K}$, then
+    \begin{enumerate}[start=9]
+        \item $\sigma_{B(E_K)}(x_K) = K$. 
+        \item For any $f \in H(\sigma_A(x); \complex)$, $f(a_K) = f(a)|_{E_K}$. 
+    \end{enumerate}
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Corollary 3.15]{MarcouxNotes}}}. ]
+    (1)-(4): Since $K$ is relatively open and closed, $\sigma_A(x) \setminus K$ is also closed. Therefore there exists $U \in \cn_{\complex}(K)$ and $V \in \cn_{\complex}(\sigma_A(x) \setminus K)$ with $U \cap V = \emptyset$. Let
+    \[
+    f_K: U \sqcup V \to \complex \quad \lambda \mapsto \begin{cases}
+        1 &\lambda \in U \\
+        0 &\lambda \in V
+    \end{cases}
+    \]
+
+    then $f_K \in H(U \sqcup V; \complex)$. Let $P(K) = f_K(x)$, then by properties of the holomorphic functional calculus, $P$ satisfies (1)-(4). 
+
+    (6): By (4), 
+    \[
+    P(B \cup C) = P(B \setminus C) + P(B \cap C) + P(C \setminus B)
+    \]
+
+    and by (3), 
+    \begin{align*}
+        E_{B \cup C} &\supset [P(B \setminus C) + P(B \cap C) + P(C \setminus B)](E_{B}) \\
+        &= [P(B \setminus C) + P(B \cap C) + P(C \setminus B)]P(B)(E_{B}) \\
+        &= [P(B \setminus C) + P(B \cap C)]P(B)(E_{B}) = E_B
+    \end{align*}
+
+    so $E_{B \cup C} \supset E_B + E_C$. On the other hand,
+    \begin{align*}
+        E_{B \cup C} &= [P(B \setminus C) + P(B \cap C) + P(C \setminus B)](E_{B \cup C}) \\
+        &\subset P(B \setminus C)(E_{B \cup C}) + P(B \cap C)(E_{B \cup C}) + P(C \setminus B)(E_{B \cup C}) \\
+        &\subset E_B + E_C - E_{B \cap C} \subset E_B + E_C
+    \end{align*}
+
+    (7), (8): By (2) and (3), $P(K^c)(E_K) = P(K)P(K^c)(E_K) = \bracs{0}$, so $E_K \subset \ker P(K^c)$. By (5) and (6), $E = E_K \oplus E_{K^c}$ as an algebraic direct sum, so $E_K = \ker P(K^c)$, and is closed. 
+
+    (9): By (1), $x(E_K) = xP(K)(E_K) = P(K)(xE_K) \subset E_K$. 
+
+    (10): By the \hyperref[Spectral Mapping Theorem]{theorem:spectral-mapping-holomorphic}, 
+    \[
+    \sigma_A(\text{Id} \cdot f_K(x)) = \text{Id} \cdot f_K(x)(\sigma_A(x)) = K
+    \]
+    
+    Since $E_K$ is $x$-invariant, $\sigma_A(\text{Id} \cdot f_K(x)) = \sigma_{B(E_K)}(x_K)$. 
+\end{proof}
+
+

src/op/banach/gelfand.tex

@@ -0,0 +1,59 @@
+\section{The Gelfand Transform}
+\label{section:gelfand-transform}
+
+\begin{definition}[Gelfand Transform]
+\label{definition:gelfand-transform}
+    Let $A$ be a unital Banach algebra, then the \textbf{Gelfand transform} is the contractive homomorphism
+    \[
+    \Gamma = \Gamma_A: A \to C(\Omega(A); \complex) \quad (\Gamma_Ax)(\varphi) = \varphi(x)
+    \]
+\end{definition}
+
+\begin{remark}
+\label{remark:gelfand-transform}
+    The Gelfand transform is limited in studying arbitrary Banach algebras, as they may admit no multiplicative functionals. However, these functionals come in abundance in the commutative case. 
+\end{remark}
+
+
+\begin{proposition}
+\label{proposition:gelfand-transform-gymnastics}
+    Let $A$ be a commutative unital Banach algebra and $x \in A$, then:
+    \begin{enumerate}
+        \item $\Gamma_A(1) = 1$. 
+        \item $x \in G(A)$ if and only if $\Gamma_A x \in G(C(\Omega(A); \complex))$. 
+        \item $(\Gamma_Ax)(\Omega(A)) = \sigma_A(x)$. 
+        \item $\norm{\Gamma_Ax}_u = [x]_{sp}$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Theorem 1.1.13]{FollandHarmonic}}}. ]
+    (1): For each $\phi \in \Omega(A)$, $\phi(1) = 1$, so $\Gamma_A(1) = 1$. 
+
+    (2): Since $A$ is commutative, $x \not\in G(A)$ if and only if the ideal generated by $x$ is proper, if and only if there exists a maximal ideal containing $x$, if and only if there exists $\phi \in \Omega(A)$ with $\phi(x) = 0$. 
+
+    (3): By (2), 
+    \[
+    (\Gamma_Ax)(\Omega(A)) = \sigma_{C(\Omega(A); \complex)}(\Gamma x) = \sigma_A(x)
+    \]
+\end{proof}
+
+\begin{proposition}
+\label{proposition:gelfand-isometric}
+    Let $A$ be a commutative unital Banach algebra, then the following are equivalent:
+    \begin{enumerate}
+        \item For each $x \in A$, $\normn{x^2}_A = \norm{x}_A^2$. 
+        \item $\Gamma_A$ is an isometry. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1) $\Rightarrow$ (2): For each $x \in A$, by the \hyperref[spectral radius formula]{proposition:spectral-radius-hadamard} and (4) of \autoref{proposition:gelfand-transform-gymnastics},
+    \[
+    \norm{\Gamma_A x}_u = [x]_{sp} = \norm{x}_A
+    \]
+
+    (2) $\Rightarrow$ (1): For each $x \in A$, by (4) of \autoref{proposition:gelfand-transform-gymnastics}, 
+    \[
+    \normn{x^2}_A \ge [x^2]_{sp} = \normn{\Gamma_A x^2}_u = \normn{\Gamma_A x}_u^2 = \normn{x}_A^2
+    \]
+\end{proof}
+
+