Stone-Cech.

document.tex

@@ -16,6 +16,7 @@ Occasionally, I make up some definitions to play with. These definition blocks w
 \input{./src/measure/index}
 \input{./src/dg/index}
 \input{./src/op/index}
+\input{./src/dyn/index}
 %\input{./src/process/index}
 
 \bibliographystyle{alpha} % We choose the "plain" reference style

refs.bib

@@ -292,4 +292,20 @@
     note = {URL:https://math.stackexchange.com/q/150258 (version: 2012-05-27)},
     eprint = {https://math.stackexchange.com/q/150258},
     url = {https://math.stackexchange.com/q/150258}
+}
+
+@book{Paulsen,
+  title={Completely bounded maps and operator algebras},
+  author={Paulsen, Vern I and others},
+  volume={78},
+  year={2002},
+  publisher={Cambridge University Press}
+}
+
+@book{Lunardi,
+  title={Interpolation theory},
+  author={Lunardi, Alessandra and others},
+  volume={16},
+  year={2018},
+  publisher={Edizioni della normale Pisa}
 }

src/dyn/index.tex

@@ -0,0 +1,5 @@
+\part{Dynamical Systems}
+\label{part:dynamics}
+
+\input{./top/index.tex}
+\input{./notation.tex}

src/dyn/notation.tex

@@ -0,0 +1,12 @@
+\chapter{Notations}
+\label{chap:dyn-notations}
+
+\begin{tabular}{lll}
+    \textbf{Notation} & \textbf{Description} & \textbf{Source} \\
+    \hline
+
+    $G \curvearrowright X$ & Topological dynamical system. & \autoref{definition:topological-dynamical-system}\\
+    $N(x, U)$ & Return set of $x$ to $U$. & \autoref{definition:tds-return-set}
+
+\end{tabular}
+

src/dyn/top/def.tex

@@ -0,0 +1,162 @@
+\section{Topological Dynamical Systems}
+\label{section:topological-dynamical-systems}
+
+\begin{definition}[Topological Dynamical System]
+\label{definition:topological-dynamical-system}
+    Let $G$ be a group and $X$ be a (non-empty) compact Hausdorff space, then a \textbf{$G$-flow/topological dynamical system} is a homomorphism $\alpha: G \to \text{Aut}(X)$. In which case, the system is denoted by $G \curvearrowright X$. 
+\end{definition}
+
+\begin{definition}[Invariant]
+\label{definition:tds-invariant}
+    Let $G \curvearrowright X$ be a $G$-flow and $A \subset X$, then the following are equivalent:
+    \begin{enumerate}
+        \item $GA \subset A$.
+        \item $gA = A$ for all $g \in G$.
+    \end{enumerate}
+    
+    If the above holds, then $A$ is an \textbf{invariant} subset of $G$. 
+\end{definition}
+\begin{proof}
+    (1) $\Rightarrow$ (2): Let $g \in G$, then $gA \subset A$ and $g^{-1}A \subset A$ by (1). In which case, $A = gg^{-1}A \subset gA$, so $gA = A$. 
+\end{proof}
+
+
+\begin{definition}[Subflow]
+\label{definition:tds-sub}
+    Let $G \curvearrowright X$ be a $G$-flow and $A \subset X$ be closed\footnote{So $A$ is also compact Hausdorff. } and invariant, then $G \curvearrowright A$ under the restricted action is a \textbf{$G$-subflow}. 
+\end{definition}
+
+\begin{definition}[Orbit]
+\label{definition:tds-orbit}
+    Let $G \curvearrowright X$ be a $G$-flow and $x \in X$, then $Gx$ is the \textbf{orbit} of $x$ under $G$, and $\ol{Gx}$ is the \textbf{orbit closure} of $x$ under $G$. 
+\end{definition}
+
+\begin{definition}[Transitive]
+\label{definition:tds-transitive}
+    Let $G \curvearrowright X$ be a $G$-flow and $x \in X$, then $x$ is \textbf{topologically transitive} with respect to $G$ if $Gx$ is dense in $X$. 
+    
+    The flow $G \curvearrowright X$ is \textbf{point transitive} if it admits a topologically transitive point. 
+
+    Finally, $G \curvearrowright X$ is \textbf{topologically transitive} if for any $U, V \subset X$ non-empty and open, there exists $g \in G$ such that $gU \cap V \ne \emptyset$\footnote{Approximations of points may be translated into approximations of other points. }. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:tds-point-transitive}
+    Let $G \curvearrowright X$ be a point transitive $G$-flow, then $G \curvearrowright X$ is also topologically transitive. 
+\end{proposition}
+\begin{proof}
+    Let $x \in G$ be a topologically transitive point and $U, V \subset X$ be non-empty and open. Since $Gx$ is dense in $G$, $Gx \cap U \ne \emptyset$ and $Gx \cap V \ne \emptyset$. After translating $x$, assume without loss of generality that $x \in U$, then $GU \supset Gx \cap V \ne \emptyset$. Therefore there exists $g \in G$ such that $gU \cap V \ne \emptyset$. 
+\end{proof}
+
+\begin{definition}[Periodic]
+\label{definition:tds-periodic}
+    Let $G \curvearrowright X$ be a $G$-flow and $x \in X$, then $x$ is \textbf{periodic} with respect to $G$ if $Gx$ is finite. 
+\end{definition}
+
+\begin{definition}[Return Set]
+\label{definition:tds-return-set}
+    Let $G \curvearrowright X$ be a $G$-flow, $x \in X$, and $\emptyset \ne U \subset X$, then the \textbf{return set} of $x$ to $U$ is
+    \[
+    N(x, U) = \bracsn{g \in G|gx \in U}
+    \]
+\end{definition}
+
+\begin{definition}[Recurrence]
+\label{definition:tds-recurrence}
+    Let $G$ be an infinite group, $G \curvearrowright X$ be a $G$-flow, and $x \in X$, then $x$ is \textbf{recurrent} if for every $U \in \cn_X(x)$, $N(x, U)$ is infinite.
+\end{definition}
+
+\begin{definition}[Syndetic]
+\label{definition:group-syndetic}
+    Let $G$ be a group and $S \subset G$, then $S$ is \textbf{(left) syndetic} if there exists $F \subset G$ finite $FS = G$. 
+\end{definition}
+
+\begin{lemma}
+\label{lemma:subgroup-syndetic}
+    Let $G$ be a group and $H \subset G$ be a subgroup, then $H$ is syndetic if and only if the index of $H$ in $G$ is finite. 
+\end{lemma}
+% Proof omitted. 
+
+\begin{example}
+    \label{example:z-syndetic}
+    Let $S \subset \integer$, then $S$ has \textbf{bounded gaps} if
+    \[
+    \sup_{n \in \integer}d(n, S) < \infty
+    \]
+
+    The set $S$ has bounded gaps if and only if it is syndetic. 
+\end{example}
+% Proof omitted. 
+
+\begin{definition}[Minimality]
+\label{definition:tds-minimal}
+    Let $G \curvearrowright X$ be a $G$-flow, then the following are equivalent:
+    \begin{enumerate}
+        \item $G \curvearrowright X$ admits no proper subflows.
+        \item Every point in $X$ is topologically transitive. 
+        \item For every $x \in X$ and $\emptyset \ne U \subset X$ open, $N(x, U)$ is syndetic in $G$. 
+    \end{enumerate}
+    
+
+    If the above holds, then $G \curvearrowright X$ is \textbf{minimal}.
+\end{definition}
+\begin{proof}
+    (1) $\Rightarrow$ (2): Let $x \in X$, then $\ol{Gx}$ with the restricted action is a $G$-subflow of $X$. As $\emptyset \ne \ol{Gx}$ and $G \curvearrowright X$ admits no proper subflows, $\ol{Gx} = x$. 
+
+    (2) $\Rightarrow$ (1): Let $G \curvearrowright Y$ be a $G$-subflow of $G \curvearrowright X$. For any $x \in Y$, $Y \supset \ol{Gx} = X$, so $Y = X$. 
+
+    (2) $\Rightarrow$ (3): Let $\emptyset \ne U \subset X$ be open. Since $G \curvearrowright X$ is minimal, $Gx \cap U \ne \emptyset$ for all $x \in X$. Thus $X \subset G^{-1}U = GU$. By compactness of $X$, there exists $F \subset G$ finite such that $FU = X$. 
+
+    Let $x \in X$ be arbitrary, then for each $g \in G$, $gx \in X = FU$. Thus there exists $h \in F$ with $gx \in hU$ and $h^{-1}gx \in U$. Hence $h^{-1}g \in N(x, U)$, and $g \in hN(x, U) \subset FN(x, U)$.  
+\end{proof}
+
+\begin{lemma}
+\label{lemma:tds-minimal-recurrent}
+    Let $G$ be an infinite group and $G \curvearrowright X$ be minimal $G$-flow, then every point in $X$ is recurrent. 
+\end{lemma}
+\begin{proof}
+    Let $x \in X$ and $U \in \cn_X(x)$, then $N(x, U)$ is syndetic. As $G$ is infinite, so is $N(x, U)$. 
+\end{proof}
+
+
+
+\begin{theorem}[Existence of Minimal Subflows]
+\label{theorem:tds-minimal-existence}
+    Let $G \curvearrowright X$ be a $G$-flow, then there exists $Y \subset X$ closed such that $G \curvearrowright Y$ is minimal. 
+\end{theorem}
+% Proof omitted. 
+
+\begin{corollary}
+\label{corollary:tds-recurrent-existence}
+    Let $G$ be an infinite group and $G \curvearrowright X$ be a $G$-flow, then $X$ admits a recurrent point. 
+\end{corollary}
+\begin{proof}
+    By \autoref{theorem:tds-minimal-existence}, there exists $Y \subset X$ closed such that $G \curvearrowright X$ is minimal. In which case, \autoref{lemma:tds-minimal-recurrent} implies that every point in $Y$ is recurrent. 
+\end{proof}
+
+
+
+\begin{definition}[Bernoulli Shift]
+\label{definition:bernoulli-shift}
+    Let $G$ be a group, $X = 2^G$ equipped with the product topology, and
+    \[
+    \alpha: X \to X \quad \alpha_g(x)(h) = x(g^{-1}h)
+    \]
+
+    Viewing $2^G$ as the power set of $G$, $\alpha_g(A) = gA$ for any $A \subset G$. The $G$-flow $G \curvearrowright X$ is the \textbf{Bernoulli shift} of $G$, and $G$-subflows of $X$ are \textbf{subshifts} of $G$. 
+\end{definition}
+
+\begin{example}[The Shift]
+    \label{example:bernoulli-shift-z}
+
+    Let $\integer \curvearrowright 2^\integer$ be the Bernoulli shift of $\integer$, then:
+    \begin{enumerate}
+        \item For any $x, y \in 2^\integer$, $y \in \ol{\integer x}$ if and only if for each $A \subset \integer$, there exists $n \in \integer$ such that $y|_A = nx|_A$. In other words, $x$ contains a copy of any segment of $y$\footnote{Topological conditions in Bernoulli shifts translate into combinatorial information. }. 
+        \item $\integer \curvearrowright 2^\integer$ is point transitive, and hence topologically transitive\footnote{It is easier to show topological transitivity: move the trees until they are disjoint}. 
+        \item 
+    \end{enumerate}
+    
+\end{example}
+
+
+

src/dyn/top/index.tex

@@ -0,0 +1,4 @@
+\chapter{Topological Dynamical Systems}
+\label{chap:topological-dynamics}
+
+\input{./def.tex}

src/fa/interpolation/functors.tex

@@ -64,7 +64,7 @@
 
 \begin{definition}[Category of Compatible Couples]
 \label{definition:compatible-category}
-    Let $\catc$ be a subcategory of normed spaces over $K \in \RC$ and $(E_0, E_1)$ be a compatible couple, then $E_0 E_1$ are a \textbf{compatible couple in $\catc$} if $E_0, E_1, E_0 \cap E_1, E_0 + E_1 \in \catc$. 
+    Let $\catc$ be a subcategory of normed spaces over $K \in \RC$ and $(E_0, E_1)$ be a compatible couple, then $(E_0, E_1)$ is \textbf{compatible couple in $\catc$} if $E_0, E_1, E_0 \cap E_1, E_0 + E_1 \in \catc$. 
     
     Let $(E_0, E_1)$ and $(F_0, F_1)$ be compatible couples in $\catc$ and $T \in L(E_0 + E_1, F_0 + F_1)$, then $T$ is a \textbf{morphism of compatible couples} if $T|_{E_0} \in \text{Mor}_{\catc}(E_0; F_0)$ and $T|_{E_1} \in \text{Mor}_{\catc}(E_1; F_1)$. 
 

src/fa/interpolation/real.tex

@@ -0,0 +1,158 @@
+\section{The Real Interpolation Method}
+\label{section:real-interpolation}
+
+\begin{definition}[$L_*^p$]
+\label{definition:weighted-lp-interpolation}
+    Let $I \subset (0, \infty)$ be measurable and $p \in [1, \infty]$, then $L^p_*(I)$ denotes the Lebesgue space $L^p$ with respect to the measure $dt/t$ on $I$. 
+\end{definition}
+
+\begin{definition}[$K$-Functional]
+\label{definition:k-functional}
+    Let $(E_0, E_1)$ be a compatible couple of Banach spaces, then the \textbf{$K$-functional} of the couple $(E_0, E_1)$ is the mapping $(0, \infty) \times (E_0 + E_1) \to [0, \infty)$ defined by
+    \[
+    K(t, x) = K(t, x, E_0, E_1) = \inf_{\substack{x_0 \in E_0 \\ x_1 \in E_1 \\ x = x_0 + x_1}}\norm{x_0}_{E_0} + t\norm{x_1}_{E_1}
+    \]
+\end{definition}
+
+\begin{lemma}
+\label{lemma:k-functional-properties}
+    Let $(E_0, E_1)$ be a compatible couple of Banach spaces and $K: (0, \infty) \times (E_0 + E_1) \to [0, \infty)$ be its $K$-functional, then:
+    \begin{enumerate}
+        \item For each $t > 0$, the mapping $x \mapsto K(t, x)$ is a norm on $E_0 + E_1$. For any $s, t > 0$, 
+        \[
+        K(t, x) \le \paren{1 \vee \frac{s}{t}}K(s, x)
+        \]
+        
+        
+        In particular,
+        \[
+        K(t, x) \le (1 \vee t)K(1, x) = (1 \vee t) \norm{x}_{E_0 + E_1}
+        \]
+        \item For each $t > 0$ and $x \in E_0 \cap E_1$, 
+        \[
+        K(t, x) \le (1 \wedge t)\norm{x}_{E_0 \cap E_1}
+        \]
+        \item For each $x \in E_0 + E_1$, the mapping $t \mapsto K(t, x)$ is non-decreasing, concave, and hence continuous. 
+        \item $K$ is jointly continuous. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    (1): Let $x = x_0 + x_1 \in E_0 + E_1$, then
+    \[
+    K(t, x) \le \norm{x_0}_{E_0} + t\norm{x_1}_{E_1} \le \paren{1 \vee \frac{t}{s}}(\norm{x_0}_{E_0} + s\norm{x_1}_{E_1})
+    \]
+
+    As the above holds for all such decompositions of $x$, 
+    \[
+    K(t, x) \le \paren{1 \vee \frac{t}{s}}K(s, x)
+    \]
+
+
+    (3): For each $x \in E_0 + E_1$, the mapping $t \mapsto K(t, x)$ is an infimum of linear and hence concave functions, so it is concave. 
+
+    (4): Let $x, y \in E_0 + E_1$ and $s, t > 0$, then
+    \begin{align*}
+        K(s, x) - K(t, y) &= K(s, x) - K(t, x) + K(t, x) - K(t, y) \\
+        |K(s, x) - K(t, y)| &\le |K(s, x) - K(t, x)| + |K(t, x) - K(t, y)| \\
+        &\le |K(s, x) - K(t, x)| + (1 \vee s \vee t)\norm{x - y}_{E_0 + E_1}
+    \end{align*}
+
+    so $K(t, y) \to K(s, x)$ as $(t, y) \to (s, x)$. 
+    
+\end{proof}
+
+
+\begin{definition}[Method of Real Interpolation]
+\label{definition:real-interpolation}
+    Let $(E_0, E_1)$ be a compatible couple of Banach spaces, $K: (0, \infty) \times (E_0 + E_1) \to [0, \infty)$ be its $K$-functional, $\theta \in (0, 1)$, $p \in [1, \infty]$, and
+    \[
+    (E_0, E_1)_{\theta, p} = \bracsn{x \in E_0 + E_1| t \mapsto t^{-\theta}K(t, x) \in L_*^p((0, \infty))}
+    \]
+
+    with the norm
+    \[
+    \norm{x}_{(E_0, E_1)_{\theta, p}} = \normn{t^{-\theta}K(t, x)}_{L_*^p((0, \infty))}
+    \]
+
+    then:
+    \begin{enumerate}
+        \item $(E_0, E_1)_{\theta, p}$ is an intermediate Banach space between $E_0$ and $E_1$.
+        \item The mapping $K_{\theta, p}$ defined by $(E_0, E_1) \mapsto (E_0, E_1)_{\theta, p}$ is an interpolation functor of exact exponent $\theta$. 
+    \end{enumerate}
+
+    and the functor $K_{\theta, p}$ is the \textbf{method of real interpolation}. 
+\end{definition}
+\begin{proof}[Proof, {{\cite[Theorem 1.6]{Lunardi}}}. ]
+    (1): Let $x \in E_0 \cap E_1$, then by (2) of \autoref{lemma:k-functional-properties}, $K(t, x) \le (1 \wedge t)\norm{x}_{E_0 \cap E_1}$. If $p \in [1, \infty)$, then
+    \begin{align*}
+        \int_0^\infty |t^{-\theta} K(t, x)|^p \frac{dt}{t} &\le \norm{x}_{E_0 \cap E_1}^p\int_0^\infty |t^{-\theta}(1 \wedge t)|^p  \frac{dt}{t} \\
+        &\le \norm{x}_{E_0 \cap E_1}^p\braks{\int_0^1 t^{p(1 - \theta)} \frac{dt}{t} + \int_1^\infty t^{-p\theta} \frac{dt}{t}} \\
+        &\le \norm{x}_{E_0 \cap E_1}^p\braks{\frac{1}{1 - \theta} + \frac{1}{p\theta}}
+    \end{align*}
+
+    If $p = \infty$, then
+    \[
+    \sup_{t > 0}t^{-\theta}K(t, x) \le \norm{x}_{E_0 \cap E_1} \braks{\sup_{t \in (0, 1)}t^{1-\theta} \vee \sup_{t \ge 1}t^{-\theta}} \le  \norm{x}_{E_0 \cap E_1}
+    \]
+
+    so $E_0 \cap E_1$ is continuously embedded inside $(E_0, E_1)_{\theta, p}$. 
+    
+    On the other hand, let $x \in (E_0, E_1)_{\theta, p}$. If $p \in [1, \infty)$, then by (3) of \autoref{lemma:k-functional-properties},
+    \begin{align*}
+        \int_0^\infty |t^{-\theta} K(t, x)|^p \frac{dt}{t} &\ge \int_1^2 |t^{-\theta} K(t, x)|^p \frac{dt}{t} \ge \frac{1}{2}\int_1^2 |t^{-\theta} K(t, x)|^p dt \\
+        &\ge \frac{1}{2^{1+\theta p}} \int_1^2 |K(t, x)|^p dt \ge \frac{K(1, x)^p}{2^{1 + \theta p}} = \frac{\norm{x}_{E_0 + E_1}^p}{2^{1 + \theta p}}
+    \end{align*}
+
+    and if $p = \infty$, then $\norm{x}_{E_0 + E_1} = K(1, x) \le \norm{x}_{(E_0, E_1)_{\theta, p}}$. 
+
+    Finally, by (1) of \autoref{lemma:k-functional-properties}, $K(t, x + y) \le K(t, x) + K(t, y)$ for all $x, y \in (E_0, E_1)_{\theta, p}$, so $(E_0, E_1)_{\theta, p}$ is complete by \hyperref[Minkowski's inequality]{theorem:minkowski} and \autoref{lemma:banach-absolute}. 
+
+    Therefore $(E_0, E_1)_{\theta, p}$ is an intermediate Banach space between $E_0$ and $E_1$. 
+
+    (2): Let $(E_0, E_1)$ and $(F_0, F_1)$ be compatible couples of Banach spaces, and $T \in L(E_0 + E_1; F_0 + F_1)$ such that $T|_{E_0} \in L(E_0; F_0)$ and $T|_{E_1} \in L(E_1; F_1)$. 
+
+    Assume without loss of generality that $T \ne 0$. In particular, after possibly considering $(E_1, E_0)$, $(F_1, F_0)$, $(E_1, E_0)_{1 - \theta, p}$, and $(F_1, F_0)_{1 - \theta, p}$, assume without loss of generality that $\norm{T}_{L(E_0; F_0)} \ne 0$. 
+
+    Let $x = x_0 + x_1 \in E_0 + E_1$, $t > 0$, and $\lambda > \norm{T}_{L(E_1; F_1)}/\norm{T}_{L(E_0; F_0)}$, 
+    \begin{align*}
+        \norm{Tx_0}_{F_0} + t\norm{Tx_1}_{F_1} &\le \norm{T}_{L(E_0; F_0)}\norm{x_0}_{F_0} + t\norm{T}_{L(E_1; F_1)}\norm{x_1}_{F_1} \\
+        &= \norm{T}_{L(E_0; F_0)}\paren{\norm{x_0}_{F_0} + t\frac{\norm{T}_{L(E_1; F_1)}}{\norm{T}_{L(E_0; F_0)}}\norm{x_1}_{F_1}} \\
+        &\le \norm{T}_{L(E_0; F_0)}({\norm{x_0}_{F_0} + \lambda t\norm{x_1}_{F_1}})
+    \end{align*}
+
+    As the above holds for all decompositions of $x$, 
+    \[
+    K(t, Tx) \le \norm{T}_{L(E_0; F_0)}K(\lambda t, x)
+    \]
+
+    For any $p \in [1, \infty)$, 
+    \begin{align*}
+        \int_0^\infty t^{-\theta p}K(t, Tx)^p \frac{dt}{t} &\le \norm{T}_{L(E_0; F_0)}^p\int_0^\infty t^{-\theta p}K(\lambda t, Tx)^p \frac{dt}{t} \\
+        &= \norm{T}_{L(E_0; F_0)}^p\int_0^\infty \lambda^{\theta p} t^{-\theta p}K(t, Tx)^p \frac{dt}{t} \\
+        \norm{Tx}_{(F_0, F_1)_{\theta, p}}  &\le \norm{T}_{L(E_0; F_0)} \lambda^{\theta}\norm{x}_{(E_0, E_1)_{\theta, p}}
+    \end{align*}
+
+    Similarly, if $p = \infty$, 
+    \begin{align*}
+    t^{-\theta}K(t, Tx) &\le \norm{T}_{L(E_0; F_0)}t^{-\theta}K(\lambda t, x) \\
+    \sup_{t > 0}t^{-\theta}K(t, Tx) &\le \norm{T}_{L(E_0; F_0)}\sup_{t > 0}\lambda^{\theta}t^{-\theta}K(t, x) \\ 
+        \norm{Tx}_{(F_0, F_1)_{\theta, p}}  &\le \norm{T}_{L(E_0; F_0)} \lambda^{\theta}\norm{x}_{(E_0, E_1)_{\theta, p}}
+    \end{align*}
+
+    As the above holds for all $\lambda > \norm{T}_{L(E_1; F_1)}/\norm{T}_{L(E_0; F_0)}$, 
+    \begin{align*}
+        \norm{Tx}_{(F_0, F_1)_{\theta, p}}  &\le \norm{T}_{L(E_0; F_0)} \frac{\norm{T}_{L(E_1; F_1)}^\theta}{\norm{T}_{L(E_0; F_0)}^\theta}\norm{x}_{(E_0, E_1)_{\theta, p}} \\
+        \norm{T}_{L((E_0, E_1)_{\theta, p}; (F_0, F_1)_{\theta, p})}&\le \norm{T}_{L(E_0; F_0)}^{1 - \theta}\norm{T}_{L(E_1; F_1)}^\theta
+    \end{align*}
+\end{proof}
+
+\begin{proposition}
+\label{proposition:real-interpolation-inclusions}
+    Let $(E_0, E_1)$ be a compatible couple of Banach spaces, then:
+    \begin{enumerate}
+        \item For each $\theta \in (0, 1)$ and $1 \le p < q \le \infty$, $(E_0, E_1)_{\theta, p} \subset (E_0, E_1)_{\theta, q}$. 
+        \item For each $0 < \theta < \eta < 1$ and $p, q \in [1, \infty]$, $(E_0, E_1)_{\eta, p} \subset (E_0, E_1)_{\theta, q}$. 
+    \end{enumerate}
+\end{proposition}
+
+

src/fa/lc/hahn-banach.tex

@@ -163,3 +163,20 @@
     is lower semicontinuous and Borel measurable by \autoref{proposition:semicontinuous-properties}. 
 \end{proof}
 
+\begin{proposition}
+\label{proposition:closed-subspace-induction}
+    Let $E$ be a locally convex space over $K \in \RC$, $F \subset E$ be a closed subspace, and $y \in E \setminus F$, then $F + Ky$ is also a closed subspace. 
+
+    In particular, all finite-dimensional subspaces of $E$ are closed. 
+\end{proposition}
+\begin{proof}
+    Since $F$ is closed and $E$ is locally convex, there exists a continuous seminorm $\rho: E \to [0, \infty)$ such that $\rho(F) = \bracsn{0}$ and $\rho(y) \ne 0$. By the \hyperref[Hahn-Banach Theorem]{proposition:hahn-banach-utility}, there exists $\phi \in E^*$ such that $F \subset \ker(\phi)$ and $\dpn{y, \phi}{E} = 1$. In which case, $\phi$ induces an isomorphism
+    \[
+    E \to \ker(\phi) \times K \quad x \mapsto (x - \dpn{x, \phi}{E}y, \dpn{x, \phi}{E})
+    \]
+
+    whose restriction to $F + Ky$ is an isomorphism onto $F \times K$.
+
+    Let $\angles{x_\alpha + \lambda_\alpha y}_{\alpha \in A} \subset F + Ky$ be net convergent in $E$. For each $\alpha \in A$, $\dpn{x_\alpha + \lambda_\alpha y, \phi}{E} = \lambda_\alpha$. By continuity of $\phi$, $\angles{\lambda_\alpha}_{\alpha \in A}$ is convergent, and $\angles{x_{\alpha}}_{\alpha \in A}$ is also convergent. Since $\angles{x_\alpha}_{\alpha \in A} \subset F$ and $F$ is closed, $\lim_{\alpha \in A}x_\alpha \in F$. Therefore $\lim_{\alpha \in A}(x_\alpha + \lambda_\alpha y) \in F + Ky$, and $F + Ky$ is also a closed subspace. 
+\end{proof}
+

src/fa/lc/quotient.tex

@@ -24,6 +24,14 @@
     As this holds for all $y \in x + M$ and $y' \in x' + M$, $\rho_M(x + x') \le \rho_M(x) + \rho_M(x')$.
 \end{proof}
 
+\begin{lemma}
+\label{lemma:quotient-norm-optimiser}
+    Let $E$ be a vector space over $K \in \RC$, $\rho: E \to [0, \infty)$ be a seminorm, $M \subset E$ be a vector subspace, $\rho_M: E/M \to [0, \infty)$ be its quotient by $M$, and $x \in E$ with $\rho_M(x) < 1$, then there exists $x_0 \in B_E(0, 1)$ such that $x + M = x_0 + M$.
+\end{lemma}
+\begin{proof}
+    Since $\inf_{y \in M}\rho(x - y) = \rho_M(x) < 1$, there exists $y \in M$ with $\rho(x - y) < 1$. Let $x_0 = x - y$, then $\rho(x_0) < 1$ as well and $x_0 + M = (x - y) + M = x + M$. 
+\end{proof}
+
 \begin{definition}[Quotient Locally Convex Space]
 \label{definition:lc-quotient}
     Let $E$ be a locally convex space over $K \in \RC$ and $M \subset E$ be a vector subspace, then there exists $(\td E, \pi)$ such that:

src/fa/lp/definition.tex

@@ -52,23 +52,22 @@
 
 \begin{theorem}[Hölder's Inequality]
 \label{theorem:holder}
-    Let $(X, \cm, \mu)$ be a measure space, $E, F$ be a normed vector spaces, $p, q \in [1, \infty]$ be Hölder conjugates, then for any $f \in \mathcal{L}^p(X; E)$ and $g \in \mathcal{L}^q(X; F)$,
+    Let $(X, \cm, \mu)$ be a measure space and $p, q \in [1, \infty]$ be Hölder conjugates, then for any $f, g: \in L^+(X)$, 
     \[
-    \int \norm{f}_E \norm{g}_F d\mu \le \norm{f}_{L^p(X; E)}\norm{g}_{L^q(X; F)}
+    \int fg d\mu \le \norm{f}_{L^p(X; \real)}\norm{g}_{L^q(X; \real)}
     \]
 
 \end{theorem}
 \begin{proof}[Proof, {{\cite[Theorem 6.2]{Folland}}}. ]
     First suppose that $p = 1$ and $q = \infty$. In this case,
     \[
-    \int \norm{f}_E \norm{g}_F d\mu \le \norm{g}_{L^\infty(X; F)}\int \norm{f}_Ed\mu = \norm{f}_{L^1(X; E)}\norm{g}_{L^\infty(X; F)}
+    \int fg d\mu \le \norm{g}_{L^\infty(X; \real)}\int fd\mu = \norm{f}_{L^1(X; \real)}\norm{g}_{L^\infty(X; \real)}
     \]
 
-    Now suppose that $p, q \in (1, \infty)$ are Hölder conjugates. Assume without loss of generality that $\norm{f}_{L^p(X; E)} = \norm{g}_{L^q(X; F)} = 1$. By \hyperref[Young's inequality]{lemma:young-inequality},
+    Now suppose that $p, q \in (1, \infty)$ are Hölder conjugates, and assume without loss of generality that $\norm{f}_{L^p(X; \real)} = \norm{g}_{L^q(X; \real)} = 1$. In which case, by \hyperref[Young's inequality]{lemma:young-inequality},
     \[
-    \int \norm{f}_E \norm{g}_F d\mu \le \int \frac{\norm{f}_E^p}{p} + \frac{\norm{g}_F^q}{q} d\mu = \frac{1}{p}\int \norm{f}_E d\mu + \frac{1}{q}\int \norm{g}_F^q d\mu = 1
+    \int fg d\mu \le \int \frac{f^p}{p} + \frac{g^q}{q} d\mu = \frac{1}{p}\int f d\mu + \frac{1}{q}\int g d\mu = 1
     \]
-
 \end{proof}
 
 \begin{theorem}[Minkowski's Inequality, {{\cite[6.5]{Folland}}}]

src/fa/lp/ineq.tex

@@ -48,3 +48,50 @@
     \end{align*}
     
 \end{proof}
+
+
+\begin{theorem}[Minkowski's Inequality for Integrals]
+\label{theorem:minkowski-integral}
+    Let $(X, \cm, \mu)$ and $(Y, \cn, \nu)$ be $\sigma$-finite measure spaces, $E$ be a Banach space, and $f: X \times Y \to E$ be $(\cm \otimes \cn)$-strongly measurable. Then:
+    \begin{enumerate}
+        \item If $E = \real$ and $f \ge 0$, then for any $p \in [1, \infty)$, 
+        \[
+        \braks{\int_{X}\paren{\int_Y f(x, y)\nu(dy)}^{p}\mu(dx)}^{1/p} \le \int_{Y}\norm{f(\cdot, y)}_{L^p(X, \mu; E)} \nu(dy)
+        \]
+        \item For any $p \in [1, \infty]$, if
+        \begin{enumerate}[label=(\alph*)]
+            \item For $\nu$-almost every $y \in Y$, $f(\cdot, y) \in L^p(\mu; E)$.
+            \item The mapping $y \mapsto \norm{f(\cdot, y)}_{L^p(\nu; E)}$ is in $L^1(\mu; \real)$.
+        \end{enumerate}
+
+        then the mapping $x \mapsto \int_Y f(x, y)\nu(dy)$ is in $L^p(\mu; E)$, with
+        \[
+        \norm{\int_Y f(\cdot, y)\nu(dy)}_{L^p(X, \mu; E)} \le \int_{Y}\norm{f(\cdot, y)}_{L^p(X, \mu; E)} \nu(dy)
+        \]
+    \end{enumerate}
+    
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem 6.19]{Folland}}}. ]
+    (1): If $p = 1$, then by \hyperref[Tonelli's Theorem]{theorem:fubini-tonelli}, 
+    \[
+    \int_X\int_Y f(x, y) \nu(dy)\mu(dx) = \int_Y\int_X f(x, y)\mu(dx)\nu(dy)
+    \]
+
+    Now suppose that $p \in (1, \infty)$, assume without loss of generality that $\int_{Y}\norm{f(\cdot, y)}_{L^p(X, \mu; E)} \nu(dy) < \infty$. Let $q$ be the Hölder conjugate of $p$ and $\phi \in L^q(\mu; \real) \cap L^+(\mu)$ with $\phi \ge 0$, then by \hyperref[Tonelli's Theorem]{theorem:fubini-tonelli},
+    \begin{align*}
+        \int_X \int_Y f(x, y)\nu(dy) \phi(x) \mu(dx) &= \int_X\int_Y  f(x, y)\phi(x)\nu(dy) \mu(dx) \\
+        &= \int_Y\int_X f(x, y) \phi(x) \mu(dx) \nu(dy)
+    \end{align*}
+    
+
+    By \hyperref[Hölder's Inequality]{theorem:holder}, 
+    \[
+    \int_Y\int_X f(x, y) \phi(x) \mu(dx) \nu(dy) \le \norm{\phi}_{L^q(\mu; \real)}\int_Y \norm{f(\cdot, y)}_{L^p(\mu; \real)} \nu(dy)
+    \]
+    
+    As $f \ge 0$ and the above holds for all $\phi \in L^q(\mu; \real) \cap L^+(\mu)$, \autoref{theorem:lp-dual-function} implies that $x \mapsto \int_Y f(x, y)\nu(dy) \in L^p(\mu)$ with 
+    \[
+    \braks{\int_{X}\paren{\int_Y f(x, y)\nu(dy)}^{p}\mu(dx)}^{1/p} \le \int_{Y}\norm{f(\cdot, y)}_{L^p(X, \mu; E)} \nu(dy)
+    \]
+\end{proof}
+

src/fa/lp/weak.tex

@@ -0,0 +1,45 @@
+\section{Lorentz Spaces}
+\label{section:weak-lp}
+
+\begin{definition}[Distribution Function]
+\label{definition:weak-lp-distribution}
+    Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed vector space, and $f \in L^0(X; E)$ be strongly measurable, then the \textbf{distribution function} of $f$ is the mapping
+    \[
+    \lambda_f: (0, \infty) \to [0, \infty] \quad \alpha \mapsto \mu\bracsn{\norm{f}_E > \alpha}
+    \]
+\end{definition}
+
+\begin{proposition}
+\label{proposition:weak-lp-distribution-properties}
+    Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed space, and $f, g: X \to E$ be strongly measurable, then:
+    \begin{enumerate}
+        \item $\lambda_f$ is non-increasing and right-continuous.
+        \item If $|f| \le |g|$, then $\lambda_f \le \lambda_g$. 
+        \item For each $\alpha > 0$ and $t \in (0, 1)$, $\lambda_{f + g}(\alpha) \le \lambda_f(t\alpha) + \lambda_g((1 - t)\alpha)$. 
+        \item For any strongly measurable functions $\seq{f_n} \subset E^X$, if $\norm{f_n}_E \upto \norm{f}_{E}$ pointwise, then $\lambda_{f_n} \upto \lambda_f$ pointwise.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): Let $\alpha > 0$ and $\seq{\alpha_n} \subset (\alpha, \infty)$ such that $\alpha_n \downto \alpha$, then
+    \[
+        \mu\bracsn{\norm{f}_E > \alpha} = \mu\paren{\bigcup_{n \in \natp}\bracsn{\norm{f}_E > \alpha_n}} = \limv{n}\mu\bracsn{\norm{f}_E > \alpha}
+    \]
+
+    by \hyperref[continuity from below]{proposition:measure-properties}. 
+
+    (2): For each $\alpha > 0$ and $t \in (0, 1)$, 
+    \[
+    \bracsn{\norm{f + g}_E > \alpha} \subset \bracsn{\norm{f}_E > t\alpha} \cup \bracsn{\norm{g}_E > (1-t)\alpha}
+    \]
+
+    (3): For each $\alpha > 0$, $\bracsn{\norm{f}_E > \alpha} = 
+    
+    
+    Assume without loss of generality that $\fF \subset 2^{L^0(X; [0, \infty))}$ and $f \in L^0(X; [0, \infty))$. Fix $\alpha > 0$, then for each $t \in (0, 1)$, 
+    \[
+    \lambda_{f}(\alpha) \le \lambda_{f - g}(t\alpha) + \lambda_{g}((1 - t)\alpha)
+    \]
+\end{proof}
+
+
+

src/fa/norm/normed.tex

@@ -122,7 +122,7 @@
 
 \begin{proposition}
 \label{proposition:dual-norm}
-    Let $E$ be a normed vector space, then for any $x \in E$, 
+    Let $E$ be a normed vector space over $K \in \RC$, then for any $x \in E$, 
     \[
     \norm{x}_E = \sup_{\substack{\phi \in E^* \\ \norm{\phi}_{E^*} = 1}}\dpn{x, \phi}{E}
     \]
@@ -132,3 +132,32 @@
 \end{proof}
 
 
+\begin{lemma}[Riesz]
+\label{lemma:riesz-almost-orthogonal}
+    Let $E$ be an infinite-dimensional normed vector space over $K \in \RC$ and $\seq{\alpha_n} \subset (0, 1)$, then there exists $\seq{x_n} \subset \partial B_E(0, 1)$ such that
+    \[
+    d(x_n, \text{span}\bracsn{x_j|1 \le j < n}) > \alpha_n \quad \forall n \in \natp
+    \]
+\end{lemma}
+\begin{proof}
+    Let $N \in \natp$ and suppose inductively that $\bracsn{x_n|1 \le n < N}$ have been constructed to satisfy the lemma for all $1 \le n < N$. Let $F_N = \text{span}\bracsn{x_n|1 \le n < N}$, then $F_N$ is finite-dimensional, and closed by \autoref{proposition:closed-subspace-induction}. As $E$ is infinite-dimensional, $E/F_N$ is non-trivial, and there exists $y_N \in E$ such that $\norm{y_N + F_N}_{E/F_N} > \alpha_N$. By \autoref{lemma:quotient-norm-optimiser}, there exists $z_N \in B_E(0, 1)$ such that $z_N + F_N = y_N + F_N$. As $z_N + F_N \ne 0$, $z_N \ne 0$ as well. Let $x_N = z_N/\norm{z_N}_{E} \in \partial B_E(0, 1)$, then
+    \[
+        \norm{x_N + F_N}_{E/F} > \norm{z_N + F_N}_{E/F} = \norm{y_N + F_N}_{E/F} > \alpha_N
+    \]
+
+    Thus $d(x_N, \text{span}\bracsn{x_n|1 \le n < N}) = \norm{x_N + F_N}_{E/F} > \alpha_N$.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:finite-dim-norm}
+    Let $K \in \RC$ and $d \in \natp$, then every norm on $K^d$ is equivalent to each other. 
+\end{proposition}
+\begin{proof}
+    Let $[\cdot]: K^d \to [0, \infty)$ be an arbitrary and
+    \[
+    \norm{\cdot}_{1}: K^d \to [0, \infty) \quad x \mapsto \sum_{j = 1}^n |x_j|
+    \]
+
+    then the identity $(K^d, \norm{\cdot}_1) \to (K^d, [\cdot])$ is continuous. Let $S = \bracsn{x \in K^d|\ \norm{x}_1 = 1}$, then $S$ is compact with respect to $\norm{\cdot}_1$ and hence compact with respect to $[\cdot]$. By continuity of $[\cdot]$, $\min_{x \in S}[x]$ exists and is non-zero. 
+\end{proof}
+

src/logic/index.tex

@@ -0,0 +1,7 @@
+\part{Mathematical Logic}
+\label{part:logic}
+
+This section is plagiarised from 
+
+\input{./tp/index.tex}
+\input{./notation.tex}

src/logic/notation.tex

@@ -0,0 +1,13 @@
+\chapter{Notations}
+\label{chap:logic-notations}
+
+\begin{tabular}{lll}
+    \textbf{Notation} & \textbf{Description} & \textbf{Source} \\
+    \hline
+    % ---- Topological Vector Spaces ----
+    $\mathcal{L} = (\mathcal{F}, \mathcal{R}, \mathcal{C})$ & Language $\mathcal{L}$ with function symbols $\mathcal{F}$, relation symbols $\mathcal{R}$, and constant symbols $\mathcal{C}$. & \autoref{definition:language} \\
+    $\mathcal{M} = (M, f^\cm, R^\cm, c^\cm)$ & $\mathcal{L}$-structure with functions $\bracsn{f^\cm|f \in \mathcal{F}}$, relations $\bracsn{R^\cm|R \in \mathcal{R}}$, and constants $\bracsn{c^\cm|c \in \mathcal{C}}$. & \autoref{definition:language-structure} \\
+    $\mathcal{T}(\mathcal{L})$ & Set of $\mathcal{L}$-terms. & \autoref{definition:logic-terms} \\
+    $\mathcal{T}_n(\mathcal{L})$ & Set of $\mathcal{L}$-terms of complexity at most $n$. & \autoref{lemma:logic-terms-induction}
+\end{tabular}
+

src/logic/tp/index.tex

@@ -0,0 +1,5 @@
+\chapter{Truth and Proof}
+\label{chap:truth-and-proof}
+
+\input{./lang.tex}
+

src/logic/tp/lang.tex

@@ -0,0 +1,56 @@
+\section{Languages}
+\label{section:languages}
+
+\begin{definition}[Language]
+\label{definition:language}
+    A \textbf{language} is a triplet of symbol sets $(\mathcal{F}, \mathcal{R}, \mathcal{C})$ and assignments $n_{\cdot}: \mathcal{F} \to \natp$ and $n_{\cdot}: \mathcal{R} \to \natp$, where:
+    \begin{enumerate}
+        \item For each $f \in \mathcal{F}$, $f$ is a function symbol in $n_{f}$ variables.
+        \item For each $R \in \mathcal{R}$, $R$ is a $n_R$-ary relation symbol. 
+        \item For each $C \in \mathcal{C}$, $C$ is a constant symbol.
+    \end{enumerate}
+\end{definition}
+
+\begin{definition}[$\mathcal{L}$-Structure]
+\label{definition:language-structure}
+    Let $L = (\mathcal{F}, \mathcal{R}, \mathcal{C})$ be a language, then a \textbf{\mathcal{L}-structure} $\mathcal{M}$ is a family
+    \[
+    \mathcal{M} = (M, f^\cm, R^\cm, c^\cm| f \in \mathcal{F}, R \in \mathcal{R}, c \in \mathcal{C})
+    \]
+
+    where
+
+    \begin{enumerate}
+        \item $M$ is a non-empty set, known as the \textbf{universe}, \textbf{domain}, or \textbf{underlying set} of $\mathcal{M}$. 
+        \item For each $f \in \mathcal{F}$, $f^\cm: M^{n_f} \to M$ is a function in $n$-variables in $M$. 
+        \item For each $R \in \mathcal{R}$, $R^\cm \subset M^{n_R}$ is a $n$-ary relation.
+        \item For each $c \in \mathcal{C}$, $c^\cm \in M$ is a constant.  
+    \end{enumerate}
+    
+    The assignments $f^\cm$, $R^\cm$, and $c^\cm$ are the \textbf{interpretations} of $(\mathcal{F}, \mathcal{R}, \mathcal{C})$. 
+\end{definition}
+
+\begin{definition}[Terms]
+\label{definition:logic-terms}
+    Let $\mathcal{L} = (\mathcal{F}, \mathcal{R}, \mathcal{C})$ be a language, then the set of \textbf{$\mathcal{L}$-terms} is the smallest set of expressions $\mathcal{T}(\mathcal{L})$ such that:
+    \begin{enumerate}
+        \item $\mathcal{C} \subset \mathcal{T}(\mathcal{L})$. 
+        \item For each $n \in \natp$, the variable symbol $v_n \in \mathcal{T}(\mathcal{L})$. 
+        \item For each $f \in \mathcal{F}$ and $\bracsn{t_j}_1^{n_f} \subset \mathcal{T}(\mathcal{L})$, the expression $f(t_1, \cdots, t_{n_f}) \in \mathcal{T}(\mathcal{L})$. 
+    \end{enumerate}
+\end{definition}
+
+\begin{lemma}
+\label{lemma:logic-terms-induction}
+    Let $\mathcal{L} = (\mathcal{F}, \mathcal{R}, \mathcal{C})$ be a language and $\mathcal{T}_0(\mathcal{L}) = \mathcal{C} \cup \seq{v_n}$. For each $n \in \natp$, inductively define
+    \[
+    \mathcal{T}_{n}(\mathcal{L}) = \mathcal{T}_{n-1}(\mathcal{L}) \cup \bracsn{f(t_1, \cdots, t_{n_f})|f \in \mathcal{F}, \bracsn{t_j}_1^{n_f} \in \mathcal{T}_{n-1}(\mathcal{L})}
+    \]
+
+    then $\mathcal{T}(\mathcal{L}) = \bigcup_{n \in \natz}\mathcal{T}_n(\mathcal{L})$. For each $n \in \natz$, $\mathcal{T}_n$ is the set of \textbf{$\mathcal{L}$-terms of complexity at most $n$}.
+\end{lemma}
+
+
+
+
+

src/measure/lcg/algebra.tex

@@ -0,0 +1,308 @@
+\section{The $L^1$ Group Algebra}
+\label{section:group-algebra-l1}
+
+\begin{definition}[Convolution of Measures]
+\label{definition:lcg-convolution-measures}
+    Let $G$ be a locally compact group, and $\mu, \nu \in M_R(G; \complex)$ be finite Radon measures, then the \textbf{convolution} of $\mu$ and $\nu$ is the Radon measure $\mu * \nu$ defined by
+    \[
+    \int_{G}\phi(x)(\mu * \nu)(dx) = \iint_{G \times G} \phi(xy)\mu(dx)\nu(dy)
+    \]
+
+    for all $\phi \in C_c(G; \complex)$\footnote{There seem to be some subtleties with having to take the Radon product here.}.
+\end{definition}
+
+
+\begin{proposition}
+\label{proposition:convolution-measures-property}
+    Let $G$ be a locally compact group, then:
+    \begin{enumerate}
+        \item For any $\mu, \nu, \sigma \in M_R(G; \complex)$, $(\mu * \nu) * \sigma = \mu * (\nu * \sigma)$.
+        \item For any $\mu, \nu \in M_R(G; \complex)$, $\norm{\mu * \nu}_{\text{var}} \le \norm{\mu}_{\text{var}}\norm{\nu}_{\text{var}}$.  
+        \item Convolution on $M_R(G; \complex)$ is commutative if and only if $G$ is commutative.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): Let $\phi \in C_c(G; \complex)$, then
+    \begin{align*}
+        \int_G \phi d(\mu * (\nu * \sigma)) &= \iint_{G \times G} \phi(xy)\mu(dx)(\nu * \sigma)(dy) \\
+        &= \iiint_{G \times G \times G}\phi(xyz) \mu(dx) \nu(dy)\sigma(dz) \\
+        &= \iint_{G \times G} \phi(yz) (\mu * \nu)(dy) \sigma(dz) = \int_G \phi d((\mu * \nu) * \sigma)
+    \end{align*}
+    
+    (2): Let $\phi \in C_c(G; \complex)$, then
+    \begin{align*}
+    \abs{\int_G \phi d(\mu * \nu)} &= \abs{\iint_{G \times G}\phi(xy)\mu(dx)\nu(dy)} \\
+    &\le \abs{\int_{G}\norm{\phi}_u\norm{\mu}_{\text{var}}d|\nu|} \le \norm{\phi}_u\norm{\mu}_{\text{var}}\norm{\nu}_{\text{var}}
+    \end{align*}
+
+    (3): For any $g, h \in G$ and $\phi \in C_c(G; \complex)$, 
+    \[
+    \int_G \phi d(\delta_g * \delta_h) = \iint_{G \times G} \phi(xy) \delta_g(dx)\delta_h(dy) = \phi(gh)
+    \]
+
+    so $\delta_g * \delta_h = \delta_{gh}$, and $\delta_g * \delta_h = \delta_h * \delta_g$ if and only if $gh = hg$. Therefore the convolution in $M_R(G; \complex)$ is commutative if and only if $G$ is commutative. 
+\end{proof}
+
+\begin{definition}[Measure Algebra]
+\label{definition:group-measure-algebra}
+    Let $G$ be a locally compact group. For each $\mu \in M_R(G; \complex)$, let
+    \[
+    \mu^*: \cb_G \to \complex \quad \mu^*(A) = \ol{\mu(A^{-1})}
+    \]
+
+    then $M_R(G; \complex)$ equipped with the convolution product and the above involution is an involutive unital Banach algebra with identity $\delta_1$. The algebra $M_R(G; \complex)$ is the \textbf{measure algebra} of $G$. 
+\end{definition}
+\begin{proof}
+    By (2) of \autoref{lemma:convolution-measures-property}, $M_R(G; \complex)$ with the convolution product is a Banach algebra. For any $\mu \in M_R(G; \complex)$ and $\phi \in C_c(G; \complex)$, 
+    \begin{align*}
+        \int_G \phi d(\mu * \delta_1) &= \iint_{G \times G} \phi(xy)\mu(dx)\delta_1(dy) \\
+        &= \iint_{G \times G} \phi(xy)\delta_1(dy)\mu(dx) = \int_G \phi(x)\mu(dx) \\
+        \int_G \phi d(\delta_1 * \mu) &= \iint_{G \times G}\phi(xy)\delta_1(dx)\mu(dy) = \int_G \phi(y)\mu(dy)
+    \end{align*}
+
+    so $\mu * \delta_1 = \mu = \delta_1 * \mu$, and $\delta_1$ is the identity. 
+
+    It remains to show that $\mu \mapsto \mu^*$ is an involution. Let $\mu, \nu \in M_R(G; \complex)$, then for any $\phi \in C_c(G; \complex)$, 
+    \begin{align*}
+        \int_G \phi d(\mu * \nu)^* &= \int_G \phi(x^{-1})d(\ol{\mu * \nu}) = \iint_{G \times G} \phi((xy)^{-1})\ol{\mu}(dx)\ol{\nu}(dy) \\
+        &= \iint_{G \times G} \phi(y^{-1}x^{-1}) \ol{\mu}(dx)\ol{\nu}(dy) = \iint_{G \times G}\phi(yx)\mu^*(dx)\nu^*(dy) \\
+        &= \iint_{G \times G}\phi(yx)\nu^*(dy)\mu^*(dx) = \int_{G}\phi d(\nu^* * \mu^*)
+    \end{align*}
+    
+    Therefore $(\mu * \nu)^* = \nu^* * \mu^*$. 
+\end{proof}
+
+\begin{definition}[Convolution]
+\label{definition:group-convolution-functions}
+    Let $G$ be a locally compact group and $f, g: G \to \complex$ be measurable functions, then the \textbf{convolution} of $f$ and $g$ is the function
+    \[
+    f * g(x) = \int_G f(y)g(y^{-1}x)dy
+    \]
+
+    defined wherever the integral exists.
+\end{definition}
+
+\begin{theorem}[Young's Inequality]
+\label{theorem:young-fourier}
+    Let $G$ be a locally compact group, $f \in L^1(G; \complex)$, $p \in [1, \infty]$, and $g \in L^p(G; \complex)$, then
+    \begin{enumerate}
+        \item If $p = \infty$, then $(f * g)(x) = \int_G f(y)g(y^{-1}x)dy$ exists for all $x \in G$, with
+        \[
+        \norm{f * g}_{L^\infty(G; \complex)}\le \norm{f}_{L^1(G; \complex)} \norm{g}_{L^\infty(G; \complex)}
+        \]
+
+        \item If $p \in [1, \infty)$, then $(f * g)(x)$ exists for almost every $x \in G$, and $f * g \in L^p(G; \complex)$ with 
+        \[
+        \norm{f * g}_{L^p(G; \complex)} \le \norm{f}_{L^1(G; \complex)} \norm{g}_{L^p(G; \complex)}
+        \] 
+    \end{enumerate}
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Proposition 2.40]{Folland}}}. ]
+    (1): By \autoref{definition:modular-function}, the measures $dx$ and $dx^{-1}$ are equivalent. Thus for each $x \in G$, $\normn{y \mapsto g(y^{-1}x)}_{L^\infty(G; \complex)} = \norm{g}_{L^\infty(G; \complex)}$. By \hyperref[Hölder's Inequality]{theorem:holder}
+    \[
+    |(f * g)(x)| = \abs{\int_G f(y)g(y^{-1}x)dy} \le \norm{f}_{L^1(G; \complex)}\norm{g}_{L^\infty(G; \complex)}
+    \]
+    
+    (2): Since $\bracsn{f \ne 0}$ and $\bracsn{g \ne 0}$ are $\sigma$-finite, assume without loss of generality that $dx$ is $\sigma$-finite. In which case, by \hyperref[Minkowski's Inequality for integrals]{theorem:minkowski-integral}, 
+    \begin{align*}
+        \norm{f * g}_{L^p(G; \complex)} &= \braks{\int_G \abs{\int_G f(y)g(y^{-1}x)dy}^pdx}^{1/p} \\
+        &\le \int_{G} \braks{\int_G|f(y)g(y^{-1}x)|^p dx}^{1/p} dy \\
+        &= \norm{g}_{L^p(G; \complex)}\int_G |f(y)|dy = \norm{f}_{L^1(G; \complex)} \norm{g}_{L^p(G; \complex)}
+    \end{align*}
+\end{proof}
+
+\begin{lemma}
+\label{lemma:convolution-measure-function-agree}
+    Let $G$ be a locally compact group, $f, g \in L^1(G; \complex)$, $\mu = fdx$, and $\nu = gdx$, then $\mu * \nu = (f * g)dx$. 
+\end{lemma}
+\begin{proof}
+    Let $\phi \in C_c(G; \complex)$, then
+    \begin{align*}
+        \int_G \phi(x) (f * g)(x)dx &= \iint_{G \times G} \phi(x) f(y)g(y^{-1}x)dydx \\
+        &= \iint_{G \times G} \phi(yx)g(x)dxf(y)dy \\
+        &= \iint_{G \times G} \phi(yx) \mu(dy)\nu(dx) = \int_G \phi d(\mu * \nu) 
+    \end{align*}
+    
+\end{proof}
+
+
+\begin{definition}[$L^1$ Group Algebra]
+\label{definition:l1-group-algebra}
+    Let $G$ be a locally compact group, then $L^1(G; \complex)$ equipped with the convolution product is an involutive Banach subalgebra of $M_R(G; \complex)$, where for any $f \in L^1(G; \complex)$, 
+    \[
+    f^*(x) = \Delta(x^{-1}) \ol{f(x^{-1})}
+    \]
+
+    The algebra $L^1(G; \complex)$ is the \textbf{$L^1$ group algebra} of $G$. 
+\end{definition}
+\begin{proof}
+    By \autoref{lemma:convolution-measure-function-agree}, the convolution operation on $L^1(G; \complex)$ agrees with the convolution on $M_R(G; \complex)$. By \hyperref[Young's Inequality]{theorem:young-fourier}, $L^1(G; \complex)$ is a Banach subalgebra of $M_R(G; \complex)$. 
+    
+    For any $f \in L^1(G; \complex)$, let $\mu = f dx$, then for each $A \in \cb_G$, 
+    \[
+        \mu^*(A) = \ol{\mu(A^{-1})} = \ol{\int_{A^{-1}}d\mu} = \int_{A^{-1}} \ol{f(x)}dx = \int_A \ol{f(x^{-1})}\Delta_G(x^{-1})dx
+    \]
+
+    so $f^*dx = d\mu^* = \Delta(x^{-1}) \ol{f(x^{-1})} dx$, and $f^* = \Delta(x^{-1}) \ol{f(x^{-1})}$. 
+\end{proof}
+
+
+\begin{proposition}
+\label{proposition:haar-translation}
+    Let $G$ be a locally compact group, $p \in [1, \infty)$, and $E$ be a normed vector space over $K \in \RC$, then
+    \begin{enumerate}
+        \item The mapping $G \times L^p(G; E) \to L^p(G; E)$ defined by $(x, f) \mapsto L_xf$ is jointly continuous. 
+        \item The mapping $G \times L^p(G; E) \to L^p(G; E)$ defined by $(x, f) \mapsto R_xf$ is jointly continuous. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition 2.42]{FollandHarmonic}}}. ]
+    (1): Let $\eps > 0$, $x, y \in G$, and $f, g \in L^p(\mu; E)$, then
+    \begin{align*}
+        \norm{L_xf - L_yg}_{L^p(\mu; E)} &\le \norm{L_xf - L_yf}_{L^p(\mu; E)} + \norm{L_yf - L_y g}_{L^p(\mu; E)} \\
+        &= \norm{L_xf - L_yf}_{L^p(\mu; E)} + \norm{f - g}_{L^p(\mu; E)}
+    \end{align*}
+
+    By \autoref{proposition:radon-cc-dense}, there exists $\phi \in C_c(G; E)$ such that $\norm{\phi - f}_{L^p(\mu; E)} < \eps$. In which case, 
+    \begin{align*}
+        \norm{L_xf - L_yf}_{L^p(\mu; E)} &\le \norm{L_xf - L_x \phi}_{L^p(\mu; E)} + \norm{L_x\phi - L_y\phi}_{L^p(\mu; E)} \\
+        &+ \norm{L_yf - L_y \phi}_{L^p(\mu; E)} \\
+        &= 2\norm{f - \phi}_{L^p(\mu; E)} + \norm{L_x\phi - L_y\phi}_{L^p(\mu; E)} \\
+        &\le 2\eps + \normn{L_{x^{-1}y}\phi - \phi}_u\mu(\bracs{\phi \ne 0} \cup x^{-1}y\bracs{\phi \ne 0})^{1/p}\\
+        &\le 2\eps + 2\normn{L_{x^{-1}y}\phi - \phi}_u\mu\bracs{\phi \ne 0}^{1/p}
+    \end{align*}
+
+    By \autoref{proposition:lcg-cc-uc}, there exists $V \in \cn_G(1)$ such that if $x^{-1}y \in V$, then $\norm{L_{x^{-1}y}\phi - \phi}_u < \eps/(2\mu\bracs{\phi \ne 0}^{1/p})$. Thus if $x^{-1}y \in V$, then
+    \[
+    \norm{L_xf - L_yg}_{L^p(\mu; E)} \le 3\eps + \norm{f - g}_{L^p(\mu; E)}
+    \]
+
+    (2): Let $V \in \cn_G(1)$ be compact, then since $\Delta_G: G \to (0, \infty)$ is a continuous homomorphism, $C := \sup_{y \in V}\Delta_G(y^{-1}) < \infty$ by \autoref{proposition:compact-extensions}. For any $f \in L^p(\mu; E)$ and $x \in V$,
+    \begin{align*}
+    \int_G \norm{R_xf(y)}_E^p dy &= \int_G \norm{f(yx)}_E^p dy = \int_G \Delta_G(x^{-1}) \norm{f(y)}_E^pdy \\
+    \norm{R_xf}_{L^p(G; E)} &= \Delta_G(x^{-1})^{1/p}\norm{f}_{L^p(G; E)} \le C^{1/p}\norm{f}_{L^p(G; E)} 
+    \end{align*}
+    
+    Let $f, g \in L^p(G; E)$ and $x \in V$, then
+    \begin{align*}
+        \norm{R_xg - f}_{L^p(G; E)} &\le \norm{R_xf - f}_{L^p(G; E)} + \norm{R_xg - R_xf}_{L^p(G; E)} \\
+        &\le \norm{R_xf - f}_{L^p(G; E)} + C^{1/p}\norm{f - g}_{L^p(G; E)} 
+    \end{align*}
+
+    By \autoref{proposition:radon-cc-dense}, there exists $\phi \in C_c(G; E)$ with $\norm{f - \phi}_{L^p(G; E)} < \eps$. Thus
+    \begin{align*}
+        \norm{R_xf - f}_{L^p(G; E)} &\le \norm{R_x(f - \phi)}_{L^p(G; E)} + \norm{f - \phi}_{L^p(G; E)} + \norm{R_x\phi - \phi}_{L^p(G; E)} \\
+        &\le (1 + C^{1/p})\eps + \norm{R_x\phi - \phi}_{L^p(G; E)}
+    \end{align*}
+
+    Since $V$ is compact, $\bracsn{\phi \ne 0}V^{-1}$ is relatively compact, and of finite measure. By \autoref{proposition:lcg-cc-uc}, $\phi$ is left and right uniformly continuous, so there exists $W \in \cn_G(1)$ such that $W \subset V$ and $\norm{R_x\phi - \phi}_u < \eps/\normn{\one_{\bracsn{\phi \ne 0}V^{-1}}}_{L^p(G; \real)}$. In which case, for any $x \in W$, 
+    \begin{align*}
+    \norm{R_x\phi - \phi}_{L^p(G; E)} &\le \norm{R_x\phi - \phi}_u \cdot \normn{\one_{\bracsn{\phi \ne 0} \cup \bracsn{\phi \ne 0}x^{-1}}}_{L^p(G; \real)} \\
+    &\le \norm{R_x\phi - \phi}_u \cdot \normn{\one_{\bracsn{\phi \ne 0}V^{-1}}}_{L^p(G; \real)} \le \eps
+    \end{align*}
+    
+    
+    and
+    \[
+    \norm{R_xg - f}_{L^p(G; E)} \le (2 + C^{1/p})\eps + C^{1/p}\norm{f - g}_{L^p(G; E)}
+    \]
+\end{proof}
+
+
+\begin{proposition}
+\label{proposition:convolution-function-gymnastics}
+    Let $G$ be a locally compact group and $f, g: G \to \complex$ be Borel measurable, then:
+    \begin{enumerate}
+        \item For any $x \in G$, $(f * g)(x) = \int_G f(y)L_yg(x) dy = \int_G R_yf(x) g(y^{-1}) dy$.
+        \item For any $z \in G$, $L_z(f * g) = (L_zf) * g$, and $R_z(f * g) = f * (R_z g)$. 
+        \item If $f \in L^1(G; \complex)$ and $g \in L^\infty(G; \complex)$, then $f * g$ is left uniformly continuous\footnote{Folland also claims that $g * f$ is right uniformly continuous \cite[Proposition 2.41, Proposition 2.43]{FollandHarmonic}. This is not true unless $G$ is unimodular. A counterexample is given by the $ax + b$ group, where $f = \one_{[1, \infty) \times [0, 1]}$ and $g = \one$.}.
+        \item If $G$ is unimodular, $p, q \in (1, \infty)$ are Hölder conjugates, $f \in L^p(G; \complex)$, and $g \in L^q(G; \complex)$, then $f * g \in C_0(G; \complex)$ with
+        \[
+        \norm{f * g}_u \le \norm{f}_{L^p(G; \complex)}\norm{g}_{L^q(G; \complex)}
+        \]
+    \end{enumerate}
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition 2.41, Proposition 2.43]{FollandHarmonic}}}. ]
+    (1): For any $x \in G$, by translation-invariance,
+    \begin{align*}
+    (f * g)(x) &= \int_G f(y)g(y^{-1}x) dy = \int_G f(y)L_yg(x)dy \\
+    &= \int_G f(xy)g(y^{-1})dy = \int_G R_yf(x)g(y^{-1})dy
+    \end{align*}
+
+    (2): Let $x, z \in G$, then by translation-invariance,
+    \begin{align*}
+        L_z(f * g)(x) &= (f * g)(z^{-1}x) = \int_G f(y)g(y^{-1}z^{-1}x) dy \\
+        &= \int_G f(z^{-1}y)g(y^{-1}x)dy = \int_G L_zf(y)g(y^{-1}x)dy \\
+        &= (L_zf * g)(x) \\
+        R_z(f * g)(x) &= (f * g)(xz) = \int_G f(y)g(y^{-1}xz)dy = (f * R_zg)(x)
+    \end{align*}
+
+    (3): By (2), $L_z(f * g) - f * g = (L_z f - f) * g$. By \hyperref[continuity of translation]{proposition:haar-translation} and \hyperref[Young's Inequality]{theorem:young-fourier}, 
+    \[
+        \lim_{z \to 1}\norm{L_z(f * g) - (f * g)}_{L^\infty(G; \complex)} \le \lim_{z \to 1}\norm{L_zf - f}_{L^1(G; \complex)} \cdot \norm{g}_{L^\infty(G; \complex)} = 0
+    \]
+
+    Therefore $(f * g)$ is left uniformly continuous. 
+
+    (4): Let $h: G \to \complex$ be defined by $h(y) = g(y^{-1})$, then for each $x \in G$, \hyperref[Hölder's Inequality]{theorem:holder} implies that
+    \begin{align*}
+        (f * g)(x) &= \int_G f(y)g(y^{-1}x) dy = \int_G f(xy)g(y^{-1}) dy \\
+        |(f * g)(x)| &\le \int_G |f(xy)h(y)| dy \le  \norm{f}_{L^p(G; \complex)}\norm{h}_{L^q(G; \complex)}
+    \end{align*}
+
+    Since $G$ is unimodular, $\norm{h}_{L^q(G; \complex)} = \norm{g}_{L^q(G; \complex)}$, so
+    \[
+    \norm{f * g}_u \le \norm{f}_{L^p(G; \complex)}\norm{h}_{L^q(G; \complex)}
+    \]
+
+    For any $f, g \in C_c(G; \complex)$, $\supp{f * g} \subset \supp{f}\supp{g}$, so $f * g \in C_c(G; \complex)$. By \autoref{proposition:radon-cc-dense}, $C_c(G; \complex)$ is dense in $L^p(G; \complex)$ and $L^q(G; \complex)$. By \autoref{proposition:closure-of-image}, 
+    \begin{align*}
+    L^p(G; \complex) * L^q(G; \complex) &= \ol{C_c(G; \complex)}^{L^p(G; \complex)} * \ol{C_c(G; \complex)}^{L^q(G; \complex)} \\
+    &\subset \ol{C_c(G; \complex) * C_c(G; \complex)}^{C_0(G; \complex)} = C_0(G; \complex)
+    \end{align*}
+\end{proof}
+
+Unfortunately, $L^1(G)$ is not a $C^*$-algebra, and its approximate identity requires a manual construction. Fortunately, the candidate is clear. 
+
+\begin{proposition}[Existence of Approximate Identity]
+\label{proposition:group-algebra-approximate-identity}
+    Let $G$ be a locally compact group and $\fB \subset \cn_G(1)$ be a fundamental system of neighbourhoods at $1$, directed under reverse inclusion. For each $U \in \fB$, let $\phi_U \in L^+(G)$ with $\ol{\bracsn{\phi_U \ne 0}} \subset U$ and $\int_G \phi_U = 1$, then:
+    \begin{enumerate}
+        \item For each $p \in [1, \infty)$ and $f \in L^p(G; \complex)$, $\phi_U * f \to f$ in $L^p(G; \complex)$.
+        \item For each left uniformly continuous $f \in C(G; \complex)$, $\phi_U * f \to f$ uniformly. 
+    \end{enumerate}
+
+    If in addition, $\phi_U(x^{-1}) = \phi_U(x)$ for all $x \in G$, then:
+    \begin{enumerate}[start=2]
+        \item For each $p \in [1, \infty)$ and $f \in L^p(G; \complex)$, $f * \phi_U \to f$ in $L^p(G; \complex)$.
+        \item For each right uniformly continuous $f \in C(G; \complex)$, $f * \phi_U \to f$ uniformly. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition 2.44]{FollandHarmonic}}}. ]
+    (1): Let $U \in \fB$, $f \in L^p(G; \complex)$, and $x \in G$, then since $\int_G \phi_U = 1$, 
+    \[
+    (\phi_U * f)(x) - f(x) = \int_G \phi_U(y)[L_yf(x) - f(x)]dy
+    \]
+
+    By \hyperref[Minkowski's Inequality for integrals]{theorem:minkowski-integral}, 
+    \begin{align*}
+        \norm{\phi_U * f - f}_{L^p(G; \complex)} &= \braks{\int_G \abs{\int_G \phi_U(y)[L_yf(x) - f(x)]dy}^p dx}^{1/p} \\
+        &\le \braks{\int_G \phi_U(y)\norm{L_yf - f}_{L^p(G; \complex)}^p dy }^{1/p} \\
+        &\le \sup_{y \in U}\norm{L_yf - f}_{L^p(G; \complex)}
+    \end{align*}
+    
+    Therefore by \hyperref[continuity of translation]{proposition:haar-translation}, $\phi_U * f \to f$ in $L^p(G; \complex)$. 
+
+    (3): Let $U \in \fB$, $f \in L^p(G; \complex)$, and $x \in G$, then since $\phi_U(y) = \phi_U(y^{-1})$ for all $y \in G$, 
+    \begin{align*}
+    (f * \phi_U)(x) - f(x) &= \int_G [R_yf(x) - f(x)]\phi_U(y^{-1})dy \\
+    &= \int_G \phi_U(y)[R_yf(x) - f(x)]dy
+    \end{align*}
+
+    By \hyperref[Minkowski's Inequality for integrals]{theorem:minkowski-integral}, 
+    \[
+    \norm{f * \phi_U - f}_{L^p(G; \complex)} \le \sup_{y \in U}\norm{R_yf - f}_{L^p(G; \complex)}
+    \]
+
+    and $f * \phi_U \to f$ in $L^p(G; \complex)$ by \hyperref[continuity of translation]{proposition:haar-translation}.
+\end{proof}
+

src/measure/lcg/haar.tex

@@ -216,39 +216,15 @@
     Thus $0 < \int g d\mu \le n\mu(U)$ and $0 < \int g d\mu \le n \int f d\mu$. 
 \end{proof}
 
-\begin{proposition}
+\begin{definition}
-\label{proposition:haar-translation}
+\label{definition:haar-measure-lp-notation}
-    Let $G$ be a locally compact group, $\mu: \cb_G \to [0, \infty]$ be a left Haar measure, $p \in [1, \infty)$, and $E$ be a normed vector space over $K \in \RC$, then the mapping
+    Let $G$ be a locally compact group and assume that $G$ is equipped with a fixed left Haar measure $\mu: \cb_G \to [0, \infty]$, then
-    \[
+    \begin{enumerate}
-    G \times L^p(\mu; E) \quad (x, f) \mapsto L_xf
+        \item For any normed vector space $E$ over $K \in \RC$, $L^p(G; E)$ denotes $L^p(G, \mu; E)$. 
-    \]
+        \item For any $f \in L^1(G; E) \cup L^+(G)$, $\int_G f(x)dx$ denotes integration with respect to $\mu$. 
+    \end{enumerate}
+\end{definition}
 
-    is jointly continuous. Similarly, if $\nu: \cb_G \to [0, \infty]$ is a right Haar measure, then
-    \[
-    G \times L^p(\mu; E) \quad (x, f) \mapsto R_xf
-    \]
 
-    is also jointly continuous.
-\end{proposition}
-\begin{proof}[Proof of the left case. ]
-    Let $\eps > 0$, $x, y \in G$, and $f, g \in L^p(\mu; E)$, then
-    \begin{align*}
-        \norm{L_xf - L_yg}_{L^p(\mu; E)} &\le \norm{L_xf - L_yf}_{L^p(\mu; E)} + \norm{L_yf - L_y g}_{L^p(\mu; E)} \\
-        &= \norm{L_xf - L_yf}_{L^p(\mu; E)} + \norm{f - g}_{L^p(\mu; E)}
-    \end{align*}
-
-    By \autoref{proposition:radon-cc-dense}, there exists $\phi \in C_c(G)$ such that $\norm{\phi - f}_{L^p(\mu; E)} < \eps$. In which case, 
-    \begin{align*}
-        \norm{L_xf - L_yf}_{L^p(\mu; E)} &\le \norm{L_xf - L_x \phi}_{L^p(\mu; E)} + \norm{L_x\phi - L_y\phi}_{L^p(\mu; E)} \\
-        &+ \norm{L_yf - L_y \phi}_{L^p(\mu; E)} \\
-        &= 2\norm{f - \phi}_{L^p(\mu; E)} + \norm{L_x\phi - L_y\phi}_{L^p(\mu; E)} \\
-        &\le 2\eps + \normn{L_{x^{-1}y}\phi - \phi}_u\mu\bracs{\phi \ne 0}
-    \end{align*}
-
-    By \autoref{proposition:lcg-cc-uc}, there exists $V \in \cn_G(1)$ such that if $x^{-1}y \in V$, then $\norm{L_{x^{-1}y}\phi - \phi}_u < \eps/\mu\bracs{\phi \ne 0}$. Thus if $x^{-1}y \in V$, then
-    \[
-    \norm{L_xf - L_yf}_{L^p(\mu; E)} \le 3\eps + \norm{f - g}_{L^p(\mu; E)}
-    \]
-\end{proof}
 
 

src/measure/lcg/index.tex

@@ -4,3 +4,4 @@
 \input{./lcg.tex}
 \input{./haar.tex}
 \input{./modular.tex}
+\input{./algebra.tex}

src/measure/lcg/modular.tex

@@ -42,8 +42,12 @@
     \end{align*}
 
     which contradicts the fact that $\lambda \ne 1$. Therefore $\lambda = 1$, and $\nu(dx) = \Delta_G(x^{-1})\mu(dx)$. 
-    
-    
 \end{proof}
 
+\begin{definition}[Unimodular]
+\label{definition:lcg-unimodular}
+    Let $G$ be a locally compact group, then $G$ is \textbf{unimodular} if $\Delta_G = 1$.
+\end{definition}
+
+
 

src/measure/notation.tex

@@ -32,6 +32,7 @@
     $\mathcal{L}^p(X, \mathcal{M}, \mu; E)$ & Space of $p$-integrable functions, without quotient. & \autoref{definition:lp-unequivalence} \\
     $\|f\|_{L^p}$, & $L^p$ norm of $f$. & \autoref{definition:esssup} \\
     $L^p(X, \mathcal{M}, \mu; E)$ & Space of $p$-integrable functions, modulo equality almost everywhere. & \autoref{definition:lp} \\
+    $A_r\mu(x)$ & Average of $\mu$ at $B_{\real^d}(x, r)$. & \autoref{definition:hl-average}
 
 \end{tabular}
 

src/measure/vector/index.tex

@@ -7,5 +7,6 @@
 \input{./ac.tex}
 \input{./ms.tex}
 \input{./rn.tex}
+\input{./ld.tex}
 \input{./fin.tex}
 

src/measure/vector/ld.tex

@@ -0,0 +1,84 @@
+\section{The Lebesgue Differentiation Theorem}
+\label{section:lebesgue-differentiation}
+
+\begin{lemma}[Vitali Covering Lemma]
+\label{lemma:vitali-covering}
+    Let $\bracsn{B_{\real^d}(x_i, \eps_i)}_{i \in I}$ be open balls in $\real^d$, $U = \bigcup_{i \in I}B_{\real^d}(x_i, \eps)$, and $m$ be the Lebesgue measure on $\real^d$, then for any $c \in [0, m(U))$, there exists $J \subset I$ finite such that:
+    \begin{enumerate}
+        \item $\bracsn{B_{\real^d}(x_j, \eps_j)}_{j \in J}$ is pairwise disjoint. 
+        \item $m\paren{\bigsqcup_{j \in J}B_{\real^d}(x_j, \eps_j)} > 3^{-d}c$. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}[Proof, {{\cite[Lemma 3.15]{Folland}}}. ]
+    By inner regularity of the Lebesgue measure, there exists $K \subset U$ compact such that $m(K) > c$. In which case, there exists $J_1 \subset I$ finite such that $K \subset \bigcup_{j \in J_1}B_{\real^d}(x_j, \eps_j)$. 
+
+    For each $n \in \natp$, if $J_n \ne \emptyset$, let $j_n \in J_n$ such that $\eps_{j_n} = \max_{j \in J_n}\eps_j$, and
+    \[
+    J_{n+1} = \bracs{j \in J \bigg | B_{\real^d}(x_j, \eps_j) \cap \bigcup_{k = 1}^n B_{\real^d}(x_{i_k}, \eps_{i_k}) = \emptyset}
+    \]
+
+    As $J$ is finite, there exists $n \in \natp$ such that $J_{n+1} = \emptyset$. The above construction then yields a pairwise disjoint family $\bracsn{B_{\real^d}(x_{i_k}, \eps_{i_k})}_1^n$. 
+
+    For each $j \in J$, let $1 \le k \le n$ be the smallest $k$ such that $B_{\real^d}(x_j, \eps_j) \cap B_{\real^d}(x_{i_k}, \eps_{i_k}) \ne \emptyset$, then $j \in J_k$, and $\eps_j \le \eps_{i_k}$. Thus $B_{\real^d}(x_j, \eps_j) \subset B_{\real^d}(x_{i_k}, 3\eps_{i_k})$. Therefore
+    \begin{align*}
+    c < m(K) &\le m\paren{\bigcup_{j \in J}B_{\real^d}(x_j, \eps_j)} \le m\paren{\bigcup_{k = 1}^n B_{\real^d}(x_{i_k}, 3\eps_{i_k})} \\
+    &\le \sum_{k = 1}^n m(B_{\real^d}(x_{i_k}, 3\eps_{i_k})) = 3^d\sum_{k = 1}^n m(B_{\real^d}(x_{i_k}, \eps_{i_k}))
+    \end{align*}
+\end{proof}
+
+\begin{definition}[Average]
+\label{definition:hl-average}
+    Let $\mu: \cb_{\real^d} \to [0, \infty]$ be a regular Borel measure on $\real^d$ and $m$ be the Lebesgue measure on $\real^d$, then
+    \[
+    (0, \infty) \times \real^d \to [0, \infty) \quad (r, x) \mapsto A_r\mu(x) = \frac{\mu(B_{\real^d}(x, r))}{m(B_{\real^d}(x, r))}
+    \]
+
+    is the \textbf{average} of $\mu$ on $B_{\real^d}(x, r)$. 
+\end{definition}
+
+\begin{lemma}
+\label{lemma:hl-average-continuous}
+    Let $\mu: \cb_{\real^d} \to [0, \infty]$ be a regular Borel measure on $\real^d$ such that $\mu(\partial B_{\real^d}(x, r)) = 0$ for all $x \in \real^d$ and $r > 0$, then the average $(r, x) \mapsto A_r\mu(x)$ is jointly continuous. 
+\end{lemma}
+\begin{proof}
+    Let $\seq{(r_n, x_n)} \subset (0, \infty) \times \real^d$ and $(r, x) \in (0, \infty) \times \real^d$ such that $(r_n, x_n) \to (r, x)$. Given that $\mu(\partial B_{\real^d}(y, s)) = 0$ for all $x \in \real^d$ and $s > 0$, $\one_{B_{\real^d}(x_n, r_n)} \to \one_{B_{\real^d}(x, r)}$ $\mu$-a.e. and Lebesgue-a.e. as $n \to \infty$. 
+    
+    Since $\seq{(r_n, x_n)}$ is convergent, there exists $K \subset \real^d$ compact such that $B_{\real^d}(x, r) \cup \bigcup_{n \in \natp}B_{\real^d}(x_n, r_n) \subset K$. By regularity of $\mu$, $\mu(K) < \infty$. Thus the \hyperref[Dominated Convergence Theorem]{theorem:dct} implies that $A_{r_n}\mu(x_n) \to A_{r}\mu(x)$ as $n \to \infty$. 
+\end{proof}
+
+
+\begin{definition}[Hardy-Littlewood Maximal Function]
+\label{definition:hl-maximal-function}
+    Let $\mu: \cb_{\real^d} \to [0, \infty]$ be a regular Borel measure on $\real^d$, then
+    \[
+    H\mu: \real^d \to [0, \infty) \quad x \mapsto \sup_{r > 0}A_r\mu(x)
+    \]
+
+    is the \textbf{Hardy-Littlewood maximal function} of $\mu$. 
+
+    If $\mu(\partial B_{\real^d}(x, r)) = 0$ for all $x \in \real^d$ and $r > 0$, then $H\mu$ is lower semicontinuous, and in particular Borel measurable. 
+\end{definition}
+\begin{proof}
+    If $\mu(\partial B_{\real^d}(x, r)) = 0$ for all $x \in \real^d$ and $r > 0$, then $(r, x) \mapsto A_{r}\mu(x)$ is jointly continuous by \autoref{lemma:hl-average-continuous}. Since $H\mu$ is a supremum of continuous functions, it is lower semicontinuous by \autoref{proposition:semicontinuous-properties}. 
+\end{proof}
+
+\begin{theorem}[The Maximal Theorem]
+\label{theorem:hl-maximal-theorem}
+    Let $\mu: \cb_{\real^d} \to [0, \infty)$ be a finite Borel measure, and $m^*$ be the Lebesgue outer\footnote{Measurability of $H\mu$ is not guaranteed.} measure on $\real^d$, then for each $\alpha > 0$, 
+    \[
+    m^*\bracsn{H\mu > \alpha} \le \frac{3^d \mu(\real^d)}{\alpha}
+    \]
+\end{theorem}
+\begin{proof}
+    For each $x \in \bracsn{H\mu > \alpha}$, there exists $r_x > 0$ such that $A_{r_x}\mu(x) > \alpha$. Let $U = \bigcup_{x \in B}B_{\real^d}(x, r_x)$, then $U \supset \bracsn{H\mu > \alpha}$. 
+    
+    Let $c \in [0, m(U))$, then by the \hyperref[Vitali Covering Lemma]{lemma:vitali-covering}, there exists $A \subset B$ finite such that $\bracsn{B_{\real^d}(x, r_x)}_{x \in A}$ is pairwise disjoint and $3^d\sum_{x \in A}m(B_{\real^d}(x, r_x)) \ge c$. In which case, 
+    \begin{align*}
+        c &\le 3^d \sum_{x \in A}m(B_{\real^d}(x, r_x)) \le \frac{3^d}{\alpha}\sum_{x \in A}\mu(B_{\real^d}(x, r_x)) \le \frac{3^d\mu(\real^d)}{\alpha}
+    \end{align*}
+
+    As the above holds for all $c \in [0, m(U))$, $m^*\bracsn{H\mu > \alpha} \le m(U) \le 3^d\mu(\real^d)/\alpha$. 
+\end{proof}
+
+
+

src/op/example/bounded.tex

@@ -81,3 +81,123 @@
 \end{proof}
 
 
+\begin{definition}[Unitary Dilation]
+\label{definition:unitary-dilation}
+    Let $H$ be a complex Hilbert space, $V \in B(H)$ be isometric, $P = I_H - VV^*$ be the projection onto $V(H)^\perp$, and
+    \[
+    U = \begin{bmatrix}
+        V & P \\
+        0 & V^*
+    \end{bmatrix}
+    \]
+
+    then:
+    \begin{enumerate}
+        \item $U^*U = UU^* = I_{H^2}$. In other words, $U$ is an isometry on $H^2$. 
+        \item Under the identification that $H = H \times \bracsn{0} \subset H^2$, let $Q \in B(H^2)$ be the projection onto $H$, then $V^n = QU^n|_H$ for all $n \ge 0$. 
+    \end{enumerate}
+    
+    The operator $U \in B(H^2)$ is a \textbf{unitary dilation} of $V$. 
+\end{definition}
+\begin{proof}
+    (1): By direct computation, 
+    \begin{align*}
+        UU^* &= \begin{bmatrix}
+            V & P \\
+            0 & V^*
+        \end{bmatrix} \cdot 
+        \begin{bmatrix}
+            V^* & 0 \\
+            P & V
+        \end{bmatrix} = \begin{bmatrix}
+            VV^* + P & PV \\
+            V^*P & V^*V
+        \end{bmatrix} = I_{H^2} \\
+        U^*U &= \begin{bmatrix}
+            V^* & 0 \\
+            P & V
+        \end{bmatrix} \cdot 
+        \begin{bmatrix}
+            V & P \\
+            0 & V^*
+        \end{bmatrix}
+         = \begin{bmatrix}
+            V^*V & V^*P \\
+            PV & VV^* + P
+        \end{bmatrix} = I_{H^2} \\
+    \end{align*}
+    
+
+    (2): For each $n \in \natp$, 
+    \[
+    U^n = \begin{bmatrix}
+            V & P \\
+            0 & V^*
+        \end{bmatrix}^n = \begin{bmatrix}
+            V^n & \cdots \\
+            0 & (V^*)^n
+        \end{bmatrix}
+    \]
+
+    so $QU^n|_{H} = V^n$.
+\end{proof}
+
+\begin{definition}[Isometric Dilation]
+\label{definition:isometric-dilation}
+    Let $H$ be a complex Hilbert space, $T \in B(H)$ with $\norm{T}_{B(H)} \le 1$, $D_T = (1 - T^*T)^{1/2}$, and
+    \[
+    V: l^2(\natp; H) \to l^2(\natp; H) \quad (Vx)_n = \begin{cases}
+        Tx_1 &n = 1 \\
+        D_Tx_1 &n = 2 \\
+        x_{n-1} &n \ge 3
+    \end{cases}
+    \]
+
+    then:
+    \begin{enumerate}
+        \item $V$ is isometric on $l^2(\natp; H)$. 
+        \item Under the identification that $H = H \oplus 0 \oplus \cdots$, let $Q$ be the orthogonal projection onto $H$, then $T^n = QV^n|_H$ for all $n \ge 0$. 
+    \end{enumerate}
+
+    The operator $V$ is an \textbf{isometric dilation} of $T$. 
+\end{definition}
+\begin{proof}
+    (1): Since $\norm{T}_H \le 1$, $1 - T^*T \ge 0$, and $(1 - T^*T)^{1/2}$ is well-defined. For each $x \in H$, 
+    \begin{align*}
+        \norm{Tx}_H^2 + \norm{D_Tx}_H^2 &= \dpn{T^*Tx, x}{H} + \dpn{(1 - T^*T)x, x}{H} \\
+        &= \dpn{x, x}{H} = \norm{x}_H^2
+    \end{align*}
+
+    Thus for each $x = \seq{x_n} \in l^2(H)$, 
+    \begin{align*}
+        \norm{Vx}_{l^2(\natp; H)}^2 &= \norm{Tx_1}_H^2 + \norm{D_Tx_1}_H^2 + \sum_{n = 3}^\infty \norm{x_{n-1}}_H^2 \\
+        &= \norm{x}_H^2 + \sum_{n = 2}^\infty \norm{x_{n}}_H^2 = \norm{x}_{l^2(\natp; H)}^2
+    \end{align*}
+\end{proof}
+
+\begin{theorem}[Sz.-Nagy's Dilation Theorem]
+\label{theorem:sznagy-dilation}
+    Let $H$ be a complex Hilbert space and $T \in B(H)$ with $\norm{T}_{B(H)} \le 1$, then there exists a Hilbert space $K \supset H$ and a unitary operator $U \in B(K)$ such that $T^n = P_HU^n|_H$ for all $n \in \natp$, where $P_H$ is the orthogonal projection onto $H$. 
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem 1.1]{Paulsen}}}. ]
+    Let $K = l^2(\natp; H)^2$ and identify $H = (H \oplus \bracsn{0} \oplus \cdot) \times \bracsn{0}$. Let $V \in B(l^2(\natp; H))$ be the isometric dilation of $T$, and $U \in B(l^2(\natp; H)^2)$ be the unitary dilation of $V$. Identify $l^2(\natp; H)$ as a subspace of $K$ as $l^2(\natp; H) \times \bracsn{0}$, and let $P_{l^2(\natp; H)}$ be the orthogonal projection onto $l^2(\natp; H)$, then for each $n \ge 0$, 
+    \begin{align*}
+        V^n &= P_{l^2(\natp; H)}U^n|_{l^2(\natp; H)} \\
+        T^n = P_{H}V^n|_{H} &= P_HU^n|_{H}
+    \end{align*}
+\end{proof}
+
+\begin{corollary}[Von Neumann's Inequality]
+\label{corollary:von-neumann-inequality-polynomial}
+    Let $H$ be a complex Hilbert space and $T \in B(H)$ with $\norm{T}_{B(H)} \le 1$, then for any $p \in \complex[x]$, 
+    \[
+    \norm{p(T)}_{B(H)} \le \sup_{z \in B_\complex(0, 1)}|p(z)|
+    \]
+\end{corollary}
+\begin{proof}[Proof, {{\cite[Corollary 1.2]{Paulsen}}}. ]
+    By \hyperref[Sz.-Nagy's Dilation Theorem]{theorem:sznagy-dilation}, there exists a Hilbert space $K \supset H$ and a unitary operator $U \in B(K)$ such that $T^n = P_HU^n|_H$ for all $n \in \natp$, where $P_H$ is the orthogonal projection onto $H$. 
+
+    Thus for any $p \in \complex[x]$, $p(T) = P_Hp(U)|_H$. As $U$ is unitary, the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus} shows that $\norm{p(U)}_{B(K)} \le \sup_{z \in B_\complex(0, 1)}|p(z)|$. Therefore $\norm{p(T)}_{B(H)} \le \sup_{z \in B_\complex(0, 1)}|p(z)|$. 
+\end{proof}
+
+

src/op/example/matrix.tex

@@ -82,5 +82,3 @@
     \]
 \end{example}
 
-
-

src/op/system/definitions.tex

@@ -0,0 +1,9 @@
+\section{Operator Systems}
+\label{section:operator-systems}
+
+\begin{definition}[Operator System]
+\label{definition:operator-system}
+    Let $A$ be a $C^*$-algebra and $S \subset A$, then $S$ is an \textbf{operator system} if $S$ is a linear subspace of $A$ and $S = S^*$. 
+\end{definition}
+
+

src/op/system/index.tex

@@ -0,0 +1,4 @@
+\chapter{Operator Systems}
+\label{chap:opertaor-systems}
+
+\input{./definitions.tex}

src/op/vn/commutative.tex

@@ -41,7 +41,7 @@
     Let $H$ be a separable Hilbert space and $A \subset B(H)$ be a commutative $C^*$-subalgebra with $I \in A$, then $A$ admits a separating vector. 
 \end{theorem}
 \begin{proof}[Proof, {{\cite[Theorem 22.3]{Zhu}}}. ]
-    Let $\seqj{x} \subset H$ be a maximal collection of non-zero vectors such that the spaces $\bracsn{Ax_j|j \in I}$ are mutually orthogonal. Such a collection exists by Zorn's lemma, and must be at most countable given that $H$ is separable. 
+    Let $\seqj{x} \subset H$ be a maximal collection of unit vectors such that the spaces $\bracsn{Ax_j|j \in J}$ are mutually orthogonal. Such a collection exists by Zorn's lemma, and must be at most countable given that $H$ is separable. 
 
     Let $\seq{x_n}$ be an enumeration of such a set, padding by zeroes if necessary, and $x = \sum_{n \in \natp}x_n/2^n$. For any $T \in A$, if $Tx = 0$, then as $\bracsn{Ax_n|n \in \natp}$ are mutually orthogonal, $Tx_n = 0$ for all $n \in \natp$. Since $A$ is commutative, $Ax_n \subset \ker(T)$ for all $n \in \natp$. By maximality of $\seq{x_n}$, $H = [l^2(\natp); \ol{Ax_n}]$, $H \subset \ker(T)$, and $T = 0$. Therefore $x$ is a separating vector. 
 \end{proof}

src/op/vn/topologies.tex

@@ -3,7 +3,7 @@
 
 Let $H$ be a complex Hilbert space. Thanks to its self-duality, there is a natural dual pairing
 \[
-B(H) \times (H \otimes H) \to \complex \quad \dpn{T, \phi \otimes x}{B(H)} =  \dpn{Tx, \phi}{H}
+B(H) \times (H \otimes \ol H) \to \complex \quad \dpn{T, \phi \otimes x}{B(H)} =  \dpn{Tx, \phi}{H}
 \]
 
 Depending on the topology placed on $H \otimes H$, and the corresponding completion, a handful of different topologies arise on $B(H)$. In fact, the above duality produces a predual for $B(H)$, being the trace class operators: 
@@ -111,7 +111,7 @@ Now, a few facts about the more familiar operator topologies:
 
 \begin{proposition}
 \label{proposition:spatial-isomorphism-sot-continuous}
-    Let $A$ be a $C^*$-algebra, $H_1, H_2$ be a complex Hilbert spaces, $\pi_1: A \to B(H_1)$ and $\pi_2: A \to B(H_2)$ be injective representations of $A$, and $U: H_1 \to H_2$ be an unitary equivalence, then the mapping
+    Let $A$ be a $C^*$-algebra, $H_1, H_2$ be complex Hilbert spaces, $\pi_1: A \to B(H_1)$ and $\pi_2: A \to B(H_2)$ be injective representations of $A$, and $U: H_1 \to H_2$ be a unitary equivalence, then the mapping
     \[
     \pi_1(A) \to \pi_2(A) \quad T \mapsto UTU^{-1}
     \]

src/op/vn/type-decomp.tex

@@ -25,7 +25,7 @@
 \begin{proof}
     (1) $\Rightarrow$ (2): Let $Q \in \text{Proj}(A)$ with $Q \le P$, then $Q = PQP$. As $PAP = \complex P$, either $PQP = 0$ or $PQP = P$. 
 
-    (2) $\Rightarrow$ (1): Given that there exists no projections strictly between $0$ and $P$, the only non-zero projection in $PAP$ is $P$ itself. By \autoref{theorem:vn-projection-norm-dense}, the linear span of projections in $PAP$ is norm-dense in $PAP$. Therefore $PAP = \complex P$. 
+    (2) $\Rightarrow$ (1): Given that there exist no projections strictly between $0$ and $P$, the only non-zero projection in $PAP$ is $P$ itself. By \autoref{theorem:vn-projection-norm-dense}, the linear span of projections in $PAP$ is norm-dense in $PAP$. Therefore $PAP = \complex P$. 
 \end{proof}
 
 \begin{lemma}
@@ -277,7 +277,7 @@
         &= \sum_{i, j \in I}\lambda_{i, j}\mu_{i, j}x_i \cdot e_j \cdot v \in l^2(I; \complex v)
     \end{align*}
 
-    Moreover, if $v \ne 0$, then $UTU^{-1}(xv) = 0$ for all $x \in l^2(I; \complex)$ implies that $\mu_{i, j} =0 $ for all $i, j \in I$, and $T = 0$. Thus for any $v \in H_0 \setminus \bracs{0}$, the mapping
+    Moreover, if $v \ne 0$, then $UTU^{-1}(xv) = 0$ for all $x \in l^2(I; \complex)$ implies that $\mu_{i, j} =0 $ for all $i, j \in I$, and $T = 0$. Thus for any $v \in H_0$ with $\norm{v}_{H_0} = 1$, the mapping
     \[
     \pi: A \to B(l^2(I; \complex v)) \quad T \mapsto UTU^{-1}|_{l^2(I; \complex v)}
     \]

src/topology/main/compactify.tex

@@ -71,6 +71,51 @@
     
 \end{proof}
 
+
+\begin{theorem}[Stone-Čech Compactification of Discrete Spaces]
+\label{theorem:stone-cech-ultrafilter}
+    Let $X$ be a discrete topological space and $\beta X$ be the set of all ultrafilters on $X$. For each $\emptyset \ne A \subset X$, let
+    \[
+    U_A = \bracsn{\fU \in \beta X| A \in \fU}
+    \]
+
+    Equip $\beta X$ with the topology generated by $\fU := \bracsn{U_A|\emptyset \ne A \subset X}$. For each $x \in X$, let $e(x)$ be the principal ultrafilter at $x$, then
+    \begin{enumerate}
+        \item $(\beta X, e)$ is a compactification of $X$. 
+        \item[(U1)] For any $f \in C(X; [0, 1])$, there exists a unique $\beta f \in C(\beta X; [0, 1])$ such that the following diagram commutes:
+        \[
+        \xymatrix{
+        \beta X \ar@{->}[r]^{\beta f} & [0, 1] \\
+        X \ar@{->}[u]^{e} \ar@{->}[ru]_{f} & 
+        }
+        \]
+    \end{enumerate}
+
+    In other words, $(\beta X, e)$ as constructed above is the Stone-Čech compactification of $X$. 
+\end{theorem}
+\begin{proof}
+    (1): Firstly, $\beta X = U_{X}$. For any $A, B \subset X$, $U_{A \cap B} = U_A \cap U_B$. Thus $\fU$ is a base for the topology that it generates. 
+
+    Let $\mathfrak{U}, \mathfrak{V} \in \beta X$ be distinct, then there exists $A \subset X$ with $A \in \mathfrak{U}$ and $A^c \in \mathfrak{V}$, so $\mathfrak{U} \in U_A$, $\mathfrak{V} \in U_{A^c}$, and $U_A \cap U_{A^c} = \emptyset$. Thus $\beta X$ is Hausdorff. 
+
+    Let $\mathcal{U} \subset 2^{\beta X}$ be an open cover of $X$. Since $\fU$ is a base for $\beta X$, assume without loss of generality that $\mathcal{U}$ is of the form $\bracsn{U_A|A \in \mathcal{S}}$, where $\mathcal{S} \subset 2^X \setminus \bracsn{\emptyset}$. Suppose for contradiction that $\mathcal{U}$ admits no finite subcover, then for any $F \subset \mathcal{S}$ finite,
+    \[
+    \bigcap_{A \in F}U_{A^c} = \bigcap_{A \in F}U_A^c = \braks{\bigcup_{A \in F}U_A}^c \ne \emptyset
+    \]
+    
+    In particular, $\bigcap_{A \in F}A^c \ne \emptyset$. As such, $\bracsn{A^c|A \in \mathcal{S}}$ generates an ultrafilter $\fU \in \beta X$. Since $\fU \not\in U_A$ for all $A \in \mathcal{S}$, $\mathcal{U}$ does not cover $X$, which is a contradiction. Thus $\beta X$ is compact. 
+
+    Finally, since $X$ is discrete, $e$ is automatically continuous. For any $\emptyset \ne A \subset X$, $e(x) \in U_A$ for all $x \in A$, so $U_A \cap e(X) \ne \emptyset$, and $e(X)$ is dense in $\beta X$. Therefore $e(X)$ is embedded as a dense subspace of $\beta X$. 
+
+    (U1): Let $f \in C(X; [0, 1])$, and let $\beta f: \beta X \to [0, 1]$ be defined by $\beta f(\fU) = \lim_{\fU}f$. Fix $\fU \in \beta X$ and $\eps > 0$, then there exists $A \in \fU$ such that $f(A) \subset B_{[0, 1]}(\lim_{\fU}f, \eps)$. Thus for any $\fV \in U_A$, 
+    \[
+    \beta f(\fV) = \lim_{\fV}f \in \ol{f(A)} \subset B_{[0, 1]}\paren{\lim_{\fU}f, \eps}
+    \]
+
+    and $|\beta f(\fU) - \beta f(\fV)| \le \eps$. Since $A \in \fU$, $U_A \in \cn_{\beta X}(\fU)$, so $\beta f$ is continuous at $\fU$. As this holds for all $\fU \in \beta X$, $\beta f$ is continuous. 
+\end{proof}
+
+
 \begin{lemma}
 \label{lemma:lch-compactification-open}
     Let $X$ be an LCH space and $(Y, \varphi)$ be a compactification of $X$, then $\varphi(X) \subset Y$ is open.