diff --git a/src/op/example/hardy.tex b/src/op/example/hardy.tex index 9b09fa5..5a282c2 100644 --- a/src/op/example/hardy.tex +++ b/src/op/example/hardy.tex @@ -8,7 +8,7 @@ \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 + if $p < \infty$, and $\norm{f}_{H^\infty(D)} = \norm{f}_u$ if $p = \infty$. Then the \textbf{Hardy space} is the space \[ H^p(D) = \bracsn{f \in H(D; \complex)|\ \normn{f}_{H^p(D; \complex)} < \infty} \] @@ -43,7 +43,7 @@ \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 + 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 @@ -60,7 +60,7 @@ \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. + 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} @@ -77,7 +77,7 @@ then: \begin{enumerate} - \item If $\fU \to x_0 \in \ol{D}$, then $f \in A(D)$, $\phi_{\fU}(f) = f(z_0)$. + \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} @@ -110,7 +110,7 @@ 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 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}