35 lines
1.2 KiB
TeX
35 lines
1.2 KiB
TeX
\section{The Hardy Space}
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\label{section:hardy-space}
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\begin{definition}[Hardy Space]
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\label{definition:hardy-space}
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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.
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\end{definition}
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\begin{proposition}
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\label{proposition:hardy-spectrum}
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Let $f \in H^\infty(D)$, then $\sigma(f) = \ol{f(D)}$.
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\end{proposition}
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\begin{proposition}
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\label{proposition:hardy-non-trivial-functional}
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Let $\fU \subset 2^{B_\complex(0, 1)}$ be an ultrafilter and
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\[
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\phi_{\fU}: H^\infty(D) \to \complex \quad f \mapsto \lim_{x, \fU} f(x)
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\]
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then:
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\begin{enumerate}
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\item If $\fU \to x_0 \in \ol{D}$, then $f \in A(D)$, $\phi_{\fU}(f) = f(z_0)$.
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\item $\phi_{\fU}$ is a multiplicative linear functional on $H^\infty(D)$.
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\end{enumerate}
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\end{proposition}
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\begin{proof}
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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.
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(2): By \autoref{definition:multiplicative-linear-functional-space}.
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\end{proof}
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