Jerry's Digital Garden

BibliographyComments
/Part 7: Operator Algebras/Chapter 40: Examples of Operator Algebras

40.4 The Hardy Space

Definition 40.4.1 (Hardy Space).label 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 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.

Proposition 40.4.2.label Let $D = B_{\complex}(0, 1)$ and $H^{2}(D)$ be the Hardy space, then:

  1. (1)

    $H^{2}(D)$ is a Hilbert space. For each $f, g \in H^{2}(D)$,

    \[\dpn{f, g}{H^2(D)}= \lim_{r \upto 1}\frac{1}{2\pi}\int_{0}^{2\pi}f(re^{i\theta})\ol{g(re^{i\theta})}d\theta\]

  2. (2)

    The mapping

    \[l^{2}(\natz; \complex) \to H^{2}(D) \quad x \mapsto \sum_{n = 0}^{\infty} x_{n}z^{n}\]

    is unitary.

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

\[\dpn{f, g}{r}= \frac{1}{2\pi}\int_{0}^{2\pi}f(re^{i\theta})\ol{g(re^{i\theta})}d\theta = \frac{1}{4}\sum_{k = 0}^{3} i^{k} [f + i^{k}g]_{r}^{2}\]

By the 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^{k}g]_{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 $f = \sum_{n = 0}^{\infty} x_{n}z^{n}, g = \sum_{n = 0}^{\infty} y_{n}z^{n} \in H^{2}(D)$ 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 Dominated Convergence Theorem. Thus $\bracsn{z^n}_{0}^{\infty}$ forms an orthonormal basis for $H^{2}(D)$ and the given map is unitary.$\square$

Proposition 40.4.3.label Let $f \in H^{\infty}(D)$, then $\sigma(f) = \ol{f(D)}$.

Proposition 40.4.4.label 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:

  1. (1)

    If $\fU \to x_{0} \in \ol{D}$, then $f \in A(D)$, $\phi_{\fU}(f) = f(x_{0})$.

  2. (2)

    $\phi_{\fU}$ is a multiplicative linear functional on $H^{\infty}(D)$.

Proof. Let $f \in H^{\infty}(D)$, then by Proposition 5.2.4, $f(\fU)$ 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 Definition 37.7.4.$\square$

Lemma 40.4.5.label Let $f \in H^{\infty}(D)$ and $M_{f} \in B(H^{2}(D))$ be the multiplication map $g \mapsto fg$, then $\norm{f}_{H^\infty(D)}= \norm{M_f}_{B(H^2(D))}$.

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,

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

$\square$

Post a Comment

Name:Email:
Please enter the tag of the current page (WR) to post the comment.
Tag:

Direct References

Jerry's Digital Garden

BibliographyComments

Direct References