\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 algebra 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{definition:multiplicative-linear-functional-space}. 
\end{proof}