Added some example facts.

.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"]
   }
 }

src/op/example/bc.tex

@@ -0,0 +1,38 @@
+\section{$BC(X)$}
+\label{section:bounded-continuous-functions-algebra}
+
+\begin{definition}[Algebra of Bounded Continuous Functions]
+\label{definition:bc-algebra}
+    Let $X$ be a topological space, then $BC(X; \complex)$ is the \textbf{space of bounded continuous functions} on $X$. 
+    
+    Equipped with pointwise operations and the uniform norm, $BC(X; \complex)$ forms a commutative Banach algebra. 
+\end{definition}
+
+
+
+\begin{theorem}
+\label{theorem:multiplicative-functional-bc}
+    Let $X$ be a completely regular space and $\beta X$ be its \hyperref[Stone-Čech compactification]{definition:stone-cech}. For each $f \in BC(X; \complex)$, let $\beta f \in BC(\beta X; \complex)$ be its unique extension to $\beta X$, then the mapping
+    \[
+    E: \beta X \to \Omega(BC(X; \complex)) \quad E(x)(f) = (\beta f)(x)
+    \]
+
+    is a homeomorphism. 
+\end{theorem}
+\begin{proof}
+    Let $\phi \in BC(X; \complex)^* \setminus \ol{E(X)}$, then there exists $\seqf{f_k} \subset BC(X; \complex)$ and $\eps > 0$ such that for every $x \in X$, 
+    \[
+    f(x) = \sum_{k = 1}^n |f_k(x) - \dpn{f_k, \phi}{BC(X; \complex)}|^2 \ge \eps^2
+    \]
+
+    In which case, if $\phi \in \Omega(BC(X; \complex))$, then 
+    \[
+        \dpn{f, \phi}{BC(X; \complex)} = \sum_{k = 1}^n \phi(f_k - \phi(f_k))\phi(\ol{f_k - \phi(f_k)}) = 0
+    \]
+
+    Since $f(x) > 0$ for all $x \in X$, $f \in G(BC(X; \complex))$. As $\phi \in \Omega(BC(X; \complex))$, the above contradicts (3) of \autoref{proposition:multiplicative-unit}. Thus $\Omega(BC(X; \complex)) \setminus \ol{E(X)}$ is empty, and $E(X)$ is dense in $\Omega(BC(X; \complex))$. 
+
+    By (U) of the \hyperref[Stone-Čech compactification]{definition:stone-cech}, $E|_X$ extends uniquely to a surjective continuous map. Since $BC(\beta X; \complex)$ and $BC(X; \complex)$ are in bijection via the restriction map, $E$ is injective. Therefore $E$ is bijective, and $E$ is a homeomorphism by \autoref{proposition:compact-hausdorff-homeomorphism}. 
+\end{proof}
+
+

src/op/example/convolution.tex

@@ -17,7 +17,7 @@
     The space $\ell^1(\integer)$ is the \textbf{convolution algebra} on $\integer$. 
 \end{definition}
 \begin{proof}
-    For each $f, g \in \ell^1(\integer)$, 
+    For each $f, g \in \ell^1(\integer)$, by \hyperref[Fubini's Theorem]{theorem:fubini-tonelli},
     \begin{align*}
         \normn{f * g}_{\ell^1(\integer)} &= \sum_{n \in \integer} \abs{\sum_{k \in \integer}f(k)g(n - k)} \\
         &\le \sum_{n, k \in \integer}|f(k)| \cdot |g(n-k)| \le \sum_{k \in \integer}|f(k)| \cdot \sum_{n \in \integer}|g(n - k)| \\
@@ -45,4 +45,31 @@
     so $\normn{f^2}_{\ell^1(\integer)} = 7 < \normn{f}_{\ell^1(\integer)}^2$. By \autoref{proposition:gelfand-isometric}, the Gelfand transform is not isometric. 
 \end{proof}
 
+\begin{theorem}
+\label{theorem:convolution-integer-gelfand}
+    Let $\ell^1(\integer)$ be the convolution algebra on $\integer$, then the mapping
+    \[
+    F: \mathbf{S}^1 = \partial B_\complex(0, 1) \to \Omega(\ell^1(\integer)) \quad
+    (F(z))(f) = \sum_{n \in \integer}f(n)z^n
+    \]
+
+    is a homeomorphism. 
+\end{theorem}
+\begin{proof}{Proof, {{\cite[Theorem 6.3]{Zhu}}}. }
+    Let $z \in \mathbf{S}^1$ and $f, g \in \ell^1(\integer)$, then by \hyperref[Fubini's Theorem]{theorem:fubini-tonelli},
+    \begin{align*}
+        F(z)(f * g) &= \sum_{n \in \integer}z^n \sum_{k \in \integer}f(k)g(n-k) \\
+        &= \sum_{k \in \natp}f(k)z^k \sum_{n \in \integer}z^{n-k}g(n-k) \\
+        &= F(z)(f) \cdot F(z)(g)
+    \end{align*}
+
+    so $F(z)$ is a multiplicative functional for all $z \in \mathbf{S}^1$. 
+
+    For each $n \in \integer$, let $\delta_n = \one_{\bracs{k = n}}$, then for any $m, n \in \integer$, $\delta_m * \delta_n = \delta_{m + n}$. Let $\phi \in \Omega(\ell^1(\integer))$ and $z = \phi(\delta_1)$, then $\phi(\delta_n) = z^n$ for all $n \in \integer$. Since the span of $\bracsn{\delta_n|n \in \integer}$ is dense in $\ell^1(\integer)$, $\phi = F(z)$ by the \hyperref[Dominated Convergence Theorem]{theorem:dct}.
+
+    Finally, by the \hyperref[Dominated Convergence Theorem]{theorem:dct}, $F: \textbf{S}^1 \to \Omega(\ell^1(\integer))$ is continuous. Since $\textbf{S}^1$ is compact, $\Omega(\ell^1(\integer))$ is Hausdorff, and $F$ is a bijection, $F$ is a homeomorphism by \autoref{proposition:compact-hausdorff-homeomorphism}. 
+\end{proof}
+
+
+
 

src/op/example/disk.tex

@@ -13,10 +13,17 @@
 
 \begin{proposition}
 \label{proposition:disk-algebra-multiplicative-functional}
-    Let $\phi \in A(D)^*$ be a multiplicative linear functional, then there exists $z_0 \in \ol D$ such that for every $f \in A(D)$, $\phi(f) = f(z_0)$. 
+    Let $D = B_\complex(0, 1)$ and $A(D)$ be the disk algebra, then the mapping
+    \[
+    E: \ol D \to \Omega(D) \quad E(z_0)(f) = f(z_0)
+    \]
+
+    is a homeomorphism. 
 \end{proposition}
 \begin{proof}
-    By \autoref{proposition:multiplicative-unit}, $\norm{\phi}_{A(D)^*} = 1$. Let $p$ be the identity polynomial, and $z_0 = \phi(p)$, then for every $q \in \complex[x]$, $\phi(q) = q(x_0)$. By density of polynomials in $A(D)$, $\phi(f) = f(z_0)$ for all $f \in A(D)$. 
+    Let $\phi \in \Omega(D)$. By \autoref{proposition:multiplicative-unit}, $\norm{\phi}_{A(D)^*} = 1$. Let $p$ be the identity polynomial, and $z_0 = \phi(p)$, then for every $q \in \complex[x]$, $\phi(q) = q(x_0)$. By density of polynomials in $A(D)$, $\phi(f) = f(z_0)$ for all $f \in A(D)$. Therefore $\ol D$ is in bijection with $\Omega(D)$. 
+
+    Since $A(D) \subset C(\ol D)$, $E$ is continuous. As $\ol D$ is compact and $\Omega(D)$ is Hausdorff, $E$ is a homeomorphism by \autoref{proposition:compact-hausdorff-homeomorphism}. 
 \end{proof}
 
 

src/op/example/hardy.tex

@@ -3,7 +3,7 @@
 
 \begin{definition}[Hardy Space]
 \label{definition:hardy-space}
-    Let $D = B_\complex(0, 1)$, then the \textbf{Hardy space} $H^\infty(D)$ is the space of all bounded holomorphic functions on $D$, equipped with the uniform norm. 
+    Let $D = B_\complex(0, 1)$, then the \textbf{Hardy space} $H^\infty(D)$ is the algebra of all bounded holomorphic functions on $D$, equipped with the uniform norm. 
 \end{definition}
 
 \begin{proposition}

src/topology/main/compact.tex

@@ -70,6 +70,15 @@
     Let $x \in E^c$. For each $y \in E$, there exists $U_y \in \cn(y)$ and $V_y \in \cn(x)$ such that $U_y \cap V_y = \emptyset$. By compactness, there exists $E_0 \subset E$ finite such that $\bigcup_{y \in E_0}U_y \supset E$. In which case, $E \cap \bigcap_{y \in E_0}V_y = \emptyset$ and $\bigcap_{y \in E_0}V_y \in \cn(x)$. Thus $E^c$ is open.
 \end{proof}
 
+\begin{proposition}
+\label{proposition:compact-hausdorff-homeomorphism}
+    Let $X$ be a compact topological space, $Y$ be a Hausdorff space, and $f \in C(X; Y)$ be a bijection, then $f$ is a homeomorphism. 
+\end{proposition}
+\begin{proof}
+    For each $K \subset X$ closed, $K$ and $f(K)$ are compact by \autoref{proposition:compact-extensions}. By \autoref{proposition:compact-closed}, $f(K)$ is closed. Thus $f^{-1}$ maps closed sets to closed sets, and hence open sets to open sets.
+\end{proof}
+
+
 \begin{proposition}
 \label{proposition:compact-hausdorff-normal}
     Let $X$ be a Hausdorff space, $A, B \subset X$ be compact with $A \cap B = \emptyset$, then there exists $U \in \cn(A)$ and $V \in \cn(B)$ such that $U \cap V = \emptyset$.

src/topology/main/compactify.tex

@@ -25,7 +25,7 @@
         \]
     \end{enumerate}
 
-    Moreover, if $(\beta X, e)$ is \textit{any} pair that satisfies (1) and (U1), then
+    Moreover, if $(\beta X, e)$ is \textit{any} pair that satisfies (U1), then
     \begin{enumerate}
         \item[(U2)] For any pair $(Y, \varphi)$ satisfying (1), there exists a unique $\beta \varphi \in C(\beta X; Y)$ such that the following diagram commutes:
         \[