Added a handful of examples.

src/op/example/hardy.tex

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+\section{The Hardy Space}
+\label{section:hardy-space}
+
+\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. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:hardy-spectrum}
+    Let $f \in H^\infty(D)$, then $\sigma(f) = \ol{f(D)}$. 
+\end{proposition}
+
+\begin{proposition}
+\label{proposition:hardy-non-trivial-functional}
+    Let $\fU \subset 2^{B_\complex(0, 1)}$ be an ultrafilter and
+    \[
+    \phi_{\fU}: H^\infty(D) \to \complex \quad f \mapsto \lim_{x, \fU} f(x)
+    \]
+
+    then:
+    \begin{enumerate}
+        \item If $\fU \to x_0 \in \ol{D}$, then $f \in A(D)$, $\phi_{\fU}(f) = f(z_0)$. 
+        \item $\phi_{\fU}$ is a multiplicative linear functional on $H^\infty(D)$. 
+    \end{enumerate}
+    
+\end{proposition}
+\begin{proof}
+    Let $f \in H^\infty(D)$, then by \autoref{proposition:imagefilterbase}, $f(\fU)$ is an ultrafilter base. Since $f$ is bounded, $f(\fU)$ converges to exactly one element of $\complex$. Hence the limit is well-defined. 
+
+    (2): By \autoref{proposition:operator-space-completeness}. 
+\end{proof}
+
+
+