H$^2$ and $l^2$ are isomorphic

Question

The Hardy space $H^2(\mathbb{D})$ is defined to be the space of all functions $f$ >holomorphic on the unit disk $\mathbb{D}$ with the norm $\lVert \cdot \rVert_H$

$\lVert f \rVert_H^2=\sup_{0<r<1}\int_0^{2\pi}|f(re^{i\theta})|^2 d\theta$ is finite.

Show that $H^2(\mathbb{D})$ and $l^2(\mathbb{Z})$ are isomorphic, where isomorphism is given by mapping $f \to c_n$ where $c_n$ is n-th coefficient in taylor expansion.

I've shown that if we take $f(z)=\sum_n c_nz^n$ then $\lVert f \rVert_H^2=2\pi \sum_n|c_n|^2$, but how it implies that they are isomorphic?

You should also prove the inverse equality, i.e. if ${c_n}_{n\in\mathbb{N}}\in\ell^2(\mathbb{Z})$ then $f(z)=\sum_n c_nz^n\in H^2(\mathbb{D})$. — Daniele Tampieri, Nov 07 '18 at 14:39

score 1 · Accepted Answer · answered Nov 07 '18 at 14:38

Since the Fourier coefficients of analytic functions vanish for negative integers, your computation actually shows that the spaces are isometrically isomorphic The map sending each $f$ to the sequence of its Fourier coefficients (multiplied by a suitable constant) is an isometric isomorphism.

H$^2$ and $l^2$ are isomorphic

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