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/Part 7: Operator Algebras/Chapter 40: Examples of Operator Algebras/Section 40.4: The Hardy Space

Proposition 40.4.6.label 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

  1. (1)

    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 $f \in H^{2}_{n}(D)$.

  2. (2)

    For any $q \in \complex[z]$, $P(p + z^{n+1}q)f = Ppf$ for all $f \in H^{2}(D)$.

  3. (3)

    For any $q \in \complex[z]$, $\norm{T}_{M_{n+1}(\complex)}\le \normn{p + z^{n+1}q}_{H^\infty(D)}$.

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 Lemma 40.4.5,

\[\norm{T}_{M_{n+1}(\complex)}= \normn{PM_p}_{B(H^2(D))}\le \normn{M_{p+z^{n+1}q}}_{B(H^2(D))}= \normn{p + z^{n+1}q}_{H^\infty(D)}\]

$\square$

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