Can you please explain the equivalence of the following definitions of the Hardy $H^2$ space? The Hardy $H^2$ space is the class of holomorphic functions $f$ on the open unit disk satisfying:

First definition:

$$ {\displaystyle \sup _{0<r<1}\left({\frac {1}{2\pi }}\int _{0}^{2\pi }\left|f\left(re^{i\theta }\right)\right|^{2}\;\mathrm {d} \theta \right)^{\frac {1}{2}}<\infty } $$

Second definition:

$$ f(z) = \sum_{n=0}^\infty c_nz^n, \quad \sum_{n=0}^\infty |c_n|^2 < \infty $$

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    $\begingroup$ Use the orthogonality of the exponentials. $\endgroup$ – zhw. Feb 5 '17 at 19:41
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    $\begingroup$ This is classic maths. See the first chapter of Shapiro's Composition Operators and Classical Function Theory for a lovely and quick explanation. $\endgroup$ – user378947 Feb 5 '17 at 19:47

If $f(z)=\sum_{n=0}^{\infty}c_n z^n$ is holomorphic on the open unit disk, then the series converges absolutely and uniformly on any closed subspace of the open disk. For a fixed $0 \le r < 1$, the function $f(re^{i\theta})=\sum_{n=0}^{\infty}c_n r^n e^{in\theta}$ is an absolutely convergent Fourier series with coefficients $c_nr^n$ for $n \ge 0$ (the Fourier coefficients for negative $n$ are all $0$.) By Parseval's theorem for the Fourier series, the $L^2$ norm of the function is the sum of squares of the Fourier coefficients $$ \frac{1}{2\pi}\int_{0}^{2\pi}|f(re^{i\theta})|^2d\theta = \sum_{n=0}^{\infty}|c_n|^2r^{2n}. $$ The above is finite for any fixed $0 < r < 1$ because the series converges uniformly on any circle of radius $0 < r < 1$. By the Monotone Convergence Theorem applied to the sum, the following holds, whether the expressions are finite or infinite: $$ \sup_{0 < r < 1} \frac{1}{2\pi}\int_{0}^{2\pi}|f(re^{i\theta})|^2d\theta = \lim_{r\uparrow 1}\frac{1}{2\pi}\int_{0}^{2\pi}|f(re^{i\theta})|^2d\theta=\sum_{n=0}^{\infty}|c_n|^2. $$ So $f=\sum_{n=0}^{\infty}c_n z^n$ is in $H^2$ iff $\sum_{n}|c_n|^2 < \infty$. And, if $f\in H^2$, $\|f\|^2_{H^2}$ is equal to the above.

  • $\begingroup$ thank you very much for a perfectly clear proof! $\endgroup$ – Konstantin Feb 7 '17 at 18:13
  • $\begingroup$ @Konstantin : You're welcome. $\endgroup$ – DisintegratingByParts Feb 7 '17 at 18:17

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