Exercise 1.7.

src/op/example/hardy.tex

@@ -3,9 +3,66 @@
 
 \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. 
+    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 $\norm{f}_{H^p(D)} = \norm{f}_u$ if $p = \infty$. Then the 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
+    \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 $\bracsn{z^n}_1^\infty$ 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 $\sigma(f) = \ol{f(D)}$. 
@@ -31,4 +88,49 @@
 \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 $p = \sum_{k = 0}^n a_k z^k$, $H^2_n(D) = \bracsn{q \in \complex[z]|\deg(p) \le n}$, $P \in B(H^2(D))$ 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}