\section{The Hardy Space}
\label{section:hardy-space}

\begin{definition}[Hardy Space]
\label{definition:hardy-space}
    Let $D = B_\complex(0, 1)$ and $p \in [1, \infty]$. For each $f \in H(D; \complex)$, let
    \[
    \norm{f}_{H^p(D)} = \sup_{r \in [0, 1)} \braks{\frac{1}{2\pi}\int_0^{2\pi}|f(re^{i\theta})|^p d\theta}^{1/p}
    \]

    if $p < \infty$, and $\norm{f}_{H^\infty(D)} = \norm{f}_u$ if $p = \infty$. Then the \textbf{Hardy space} is the space
    \[
    H^p(D) = \bracsn{f \in H(D; \complex)|\ \normn{f}_{H^p(D; \complex)} < \infty}
    \]

    In particular, $H^\infty(D)$ is a Banach algebra under pointwise multiplication. 
\end{definition}

\begin{proposition}
\label{proposition:hardy-space-h2}
    Let $D = B_\complex(0, 1)$ and $H^2(D)$ be the Hardy space, then:
    \begin{enumerate}
        \item $H^2(D)$ is a Hilbert space. For each $f, g \in H^2(D)$, 
        \[
        \dpn{f, g}{H^2(D)} = \lim_{r \upto 1}\frac{1}{2\pi}\int_0^{2\pi}f(re^{i\theta})\ol{g(re^{i\theta})} d\theta
        \]
        \item The mapping
        \[
        l^2(\natz; \complex) \to H^2(D) \quad x \mapsto \sum_{n = 0}^\infty x_nz^n
        \]

        is unitary. 
    \end{enumerate}
\end{proposition}
\begin{proof}
    (1): For each $r \in [0, 1)$ and $f \in H^2(D)$, let
    \[
    [f]_{r} = \braks{\frac{1}{2\pi}\int_0^{2\pi}|f(re^{i\theta})|^2 d\theta}^{1/2}
    \]

    Then, $[\cdot]_r$ is induced by the pseudo inner product
    \[
    \dpn{f, g}{r} = \frac{1}{2\pi}\int_0^{2\pi}f(re^{i\theta})\ol{g(re^{i\theta})} d\theta = \frac{1}{4}\sum_{k = 0}^3 i^k [f + i^kg]_{r}^2
    \]

    By the \hyperref[Maximum Modulus Theorem]{theorem:maximum-modulus-theorem}, $\norm{f}_{H^2(D)} = \lim_{r \upto 1}[f]_r$, so the inner product
    \begin{align*}
    \dpn{f, g}{H^2(D)} &= \lim_{r \upto 1}\dpn{f, g}{r} = \lim_{r \upto 1}\frac{1}{4}\sum_{k = 0}^3 i^k [f + i^kg]_r^2 \\
    &= \frac{1}{4}\sum_{k = 0}^3 i^k \normn{f + i^kg}_{H^2(D)}^2
    \end{align*}
    

    exists, and defines the norm on $H^2(D)$. 

    (2): For any $x \in l^2(\natz; \complex) \subset l^\infty(\natz; \complex)$, $\limsup_{n \to \infty}|x_n|^{1/n} \le 1$, so the given map is well-defined. 

    On the other hand, for any $f = \sum_{n = 0}^\infty x_nz^n, g = \sum_{n = 0}^\infty y_nz^n \in H^2(D)$ and $r \in (0, 1)$, 
    \begin{align*}
        \dpn{f, g}{r} &= \sum_{n = 0}^\infty \sum_{m = 0}^\infty x_n\ol{y_m}\dpn{z^n, z^m}{r} = \sum_{n = 0}^\infty x_n\ol{y_n}r^{2n} \\
        \dpn{f, g}{H^2(D)} &= \lim_{r \upto 1}\sum_{n = 0}^\infty x_n\ol{y_n}r^{2n} = \sum_{n = 0}^\infty x_n\ol{y_n}
    \end{align*}

    by the \hyperref[Dominated Convergence Theorem]{theorem:dct}. Thus $\bracsn{z^n}_0^\infty$ forms an orthonormal basis for $H^2(D)$ and the given map is unitary. 
\end{proof}

\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 $f \in A(D)$, $\phi_{\fU}(f) = f(x_0)$. 
        \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}


\begin{lemma}
\label{lemma:hardy-left-multiplication}
    Let $f \in H^\infty(D)$ and $M_f \in B(H^2(D))$ be the multiplication map $g \mapsto fg$, then $\norm{f}_{H^\infty(D)} = \norm{M_f}_{B(H^2(D))}$. 
\end{lemma}




\begin{proposition}
\label{proposition:toeplitz-triangular-norm}
    Let $\bracsn{a_j}_0 \subset \complex$ and
    \[
    T = \begin{bmatrix}
        a_0 &0 &\cdots &0 \\
        a_1 &a_0 &\ddots &\vdots \\
        \vdots &\ddots &\ddots &0 \\
        a_n &\cdots &a_1 &a_0
    \end{bmatrix} \in M_{n+1}(\complex)
    \]

    then
    \begin{enumerate}
        \item Let $p = \sum_{k = 0}^n a_k z^k$, $H^2_n(D) = \bracsn{q \in \complex[z]|\deg(q) \le n}$, $P \in B(H^2(D))$ be the orthogonal projection onto $H^2_n(D)$, and identify $H^2_n(D) = \complex^{n+1}$, then $Tf = Ppf$ for all $f \in H^2_n(D)$. 
        \item For any $q \in \complex[z]$, $P(p + z^{n+1}q)f = Ppf$ for all $f \in H^2(D)$. 
        \item For any $q \in \complex[z]$, $\norm{T}_{M_{n+1}(\complex)} \le \normn{p + z^{n+1}q}_{H^\infty(D)}$. 
    \end{enumerate}
\end{proposition}
\begin{proof}
    (1): Let $f = \sum_{k = 0}^n \xi_k z^k$, then
    \begin{align*}
        pf &= \braks{\sum_{k = 0}^n a_kz^k}\braks{\sum_{k = 0}^n \xi_k z^k} = \sum_{j, k = 0}^n a_j\xi_k z^{j+k} \\
        Ppf &= \sum_{k = 0}^n z^k\sum_{j = 0}^k a_{k-j}\xi_{j} = Tf
    \end{align*}
    
    (2): For any $f \in H^2(D)$, $z^{n+1}qf \perp H^2_n(D)$, so $P(p + z^{n+1}q)f = Ppf$.

    (3): For each $g \in H^\infty(D)$, let $M_g \in B(H^2(D))$ be the multiplication operator by $g$. By (1), $\norm{T}_{M_{n+1}(\complex)} = \normn{PM_p}_{B(H^2(D))}$. By (2), 
    \[
    \normn{PM_p}_{B(H^2(D))} = \normn{PM_{p + z^{n+1}q}}_{B(H^2(D))} \le \norm{M_{p+z^{n+1}q}}_{B(H^2(D))}
    \]

    By \autoref{lemma:hardy-left-multiplication}, 
    \[
        \norm{T}_{M_{n+1}(\complex)} = \normn{PM_p}_{B(H^2(D))} \le \normn{M_{p+z^{n+1}q}}_{B(H^2(D))} = \normn{p + z^{n+1}q}_{H^\infty(D)}
    \]
\end{proof}