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The Hardy space $H^2$ can be viewed as a closed vector subspace of the complex $L^2$ space on the unit circle. Can you please explain why the $L^2$ space can be represented by bi-infinite sequences and $H^2$ can be represented by infinite sequences $f(z) = \sum_{n=0}^\infty a_n x^n, |z|<1$ ?

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The space $L^2(\mathbb{T})$ of functions on the unit circle $\mathbb{T}$ has inner product $$ \langle f,g\rangle = \int_{\mathbb{T}}f(z)\overline{g(z)}d\sigma(z) $$ where $\sigma$ is the Lebesgue measure on $\mathbb{T}$, normalized so that $\sigma(\mathbb{T})=1$. This space has an orthonormal basis $\{ z^n\}_{n=-\infty}^{\infty}$. The functions in $H^2(\mathbb{T})$ consist of those with holomorphic extensions to the unit disk $\mathbb{D}$. $H^2(\mathbb{D})$ has an orthonormal basis $\{ z^n \}_{n=0}^{\infty}$.

Every $f\in L^2(\mathbb{T})$ may be written as the $L^2$ convergent sum $$ f(z) = \sum_{n=-\infty}^{\infty}a_n z^n,\;\;\; z\in\mathbb{T}. $$ The functions in $H^2(\mathbb{T})$ are those for which $\langle f,z^n\rangle=0$ for $n < 0$. The functions in $H^2(\mathbb{T})$ have natural extensions to unit disk $\mathbb{D}$ as a power series $f(z) = \sum_{n=0}^{\infty}a_n z^n$, and it not difficult to show that $f_{r}(z)=f(rz)$ converges to $f$ in $L^2(\mathbb{D})$ and $r\uparrow 1$.

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    Thank you for the reply! Could you please explain why $\{ z^n\}_{n=-\infty}^{\infty}$ (Laurent series) is a basis for $L^2(\mathbb{T})$? Does $H^2(\mathbb{D})$ have an orthonormal basis $\{ z^n \}_{n=0}^{\infty}$ because it consists of holomorphic functions (no poles or essential singularities)? – 2017-02-06
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    @Konstantin : Because $\{e^{in\theta}\}_{n=-\infty}^{\infty}$ is an orthonormal basis, which is the same as $z^n$ on the unit circle. (I use the normalized inner product where $\|1\|=1$.) – 2017-02-06
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    @Konstantin : As for $H^2(\mathbb{D})$, there are multiple, equivalent ways to define the space. (1) You can define as holomorphic functions on the disk where the $L^2$ norms on concentric circles are uniformly bounded. Or (2) you can show this is the same as $L^2$ functions on the unit circle for which $\int f(e^{i\theta})e^{in\theta}d\theta=0$ for $n =1,2,3,\cdots$. These are equivalent. – 2017-02-06