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:
$$
{\displaystyle \sup _{0 Second definition: $$
f(z) = \sum_{n=0}^\infty c_nz^n, \quad \sum_{n=0}^\infty |c_n|^2 < \infty
$$
If $f(z)=\sum_{n=0}^{\infty}c_n z^n$ is holomorphic on the open unit disk, then the series converges absolutely and uniformly on any closed subspace of the open disk. For a fixed $0 \le r < 1$, the function $f(re^{i\theta})=\sum_{n=0}^{\infty}c_n r^n e^{in\theta}$ is an absolutely convergent Fourier series with coefficients $c_nr^n$ for $n \ge 0$ (the Fourier coefficients for negative $n$ are all $0$.) By Parseval's theorem for the Fourier series, the $L^2$ norm of the function is the sum of squares of the Fourier coefficients $$ \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.