Denote, by $ H^2= \left\{f\in L^2(\mathbb{T}): n\in\mathbb{N} \rightarrow \int \limits_{\mathbb{T}} f(z)z^nd\mu(z)=0 \right\} $ $ \Vert f\Vert_{2,\rho}=\left(\frac{1}{2\pi}\int\limits_{[-\pi,\pi]}|f(\rho e^{it})|^2d\mu(t)\right)^{1/2}$ $ A^2=\{f\in\mathcal{O}(\mathbb{D}):\lim\limits_{\rho\to 1-0}\Vert f\Vert_{2,\rho}<\infty\} $ where $\mathbb{T}=\{z\in\mathbb{C}: |z|=1\}$, $\mathbb{D}=\{z\in\mathbb{C}: |z|<1\}$. Please tell me how to construct bijection between $H^2$ and $A^2$.
Description of subspace in $L^2(\mathbb{T})$ of functions with vanishing positive Fourier coefficients
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1 Answers
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This is something often covered in harmonic analysis courses, such as chapter 3 of Katznelson's book. If $f(z) \in A^2(D)$, then $f^*(e^{it}) = \lim_{r \rightarrow 1} f(re^{it})$ is in $H^2(T)$, and you can recover $f(z)$ from $f^*(e^{it})$ via the Poisson integral ${\displaystyle f(z) = {1 \over 2\pi}\int_0^{2\pi}\operatorname{Re}\left({e^{it} + z \over e^{it} - z}\right)f^*(e^{it})\,dt}$.
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0Not a problem, and I deleted mine too... – 2011-12-07