The Hardy space is a Hilbert space 
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 $H^2(\mathbb{D})$ is a Hilbert space.

I have shown that if $f(z)=\sum_n c_nz^n$, then $\lVert f \rVert_H^2=2\pi \sum_n|c_n|^2$. How does this imply $H^2(\mathbb{D})$ is a Hilbert space? What is the inner product induced by the norm?
 A: Try using this fact (called the polarization identity): 

A Banach space $\mathcal{B}$ with norm $\parallel \ .\parallel$ is a Hilbert space iff  $$\forall f,g \in \mathcal{B}, \ \ \ \ \parallel f+g\ \parallel +\parallel f-g \ \parallel = 2\left(\parallel f \ \parallel +\parallel g \ \parallel \right)$$ 
  with $\langle f,g\rangle = \dfrac{1}{4} \left(\left(\parallel f+g \parallel - \parallel f-g \parallel \right) + i \left( \parallel f + ig \parallel - \parallel f - ig \parallel \right)  \right)$

In fact, letting $f(z) = \sum_{n}{a_n z^n}$, and $g(z) = \sum_{n}{b_n z^n}$, we have, $$|a_n + b_n|^2+|a_n - b_n|^2 = 2(|a_n|^2+|b_n|^2) \ \ \ \ \ \forall n\in \mathbb{N}$$, (check! [$a_n, b_n \in \mathbb{C}$]) and because all the terms in all the series are positive, we can write,  $$\parallel f+g\ \parallel +\parallel f-g \ \parallel = 2\pi \left(\sum_{n}{|a_n + b_n|^2}+ \sum_{n}{|a_n - b_n|^2}\right) = 2 \pi \left(2\left(\sum_{n}{|a_n|^2}+\sum_{n}{|b_n|^2}\right) \right) = 2\left(\parallel f \ \parallel +\parallel g \ \parallel \right)$$
Now you can try proving that the inner product you thus obtain is the same as the one @Fred has described!  In fact, check that $$\langle f,g \rangle = 2\pi \sum_{n}{a_n \overline{b_n}}$$
[Note that, even when you encounter a Banach space which you don't know the inner product formula for, you can thus check if it is a Hilbert space and describe an inner product.] 
To show completeness, do the following: 


*

*Let $\{f_n\}$ be Cauchy in $\mathbb{H}^2$.Say, $f_n(z) =
   \sum_{i}{a_{n,i} z^i}$. Using your formula for $\parallel .
   \parallel_{\mathbb{H}^2}$, $\{s_n\}:s_n = \{a_{n,i}\}_{i=1}^{\infty}$
is Cauchy in $l^2(\mathbb{N})$. 

*$l^2(\mathbb{N})$ is complete. $$\Rightarrow \ \ \ \exists \{a_i\} \in l^2(\mathbb{N}), \ \ \ \{a_{n,i}\} \xrightarrow[n\rightarrow \infty]{l^2(\mathbb{N})} \{a_i\}$$

*Define, $f(z) : = \sum_{i}{a_{i} z^i}$. Again, using your formula for $\parallel . \parallel_{\mathbb{H}^2}$, show that, $$f_{n} \xrightarrow[n\rightarrow \infty]{\mathbb{H}^2} f$$


This finishes the proof.
A: Inner product: $(f|g)=\sup_{0<r<1}\int_0^{2\pi}f(re^{i\theta}) \overline{g(re^{i\theta})}d\theta$
