A proof for a Hardy inequality for holomorphic functions in Hardy space $H^1$ Let $H^1$ be the Hardy space of holomorphic functions on the unit disk. 
Where can I find a proof (as simple and selfcontained as possible) of this result?
If $f\in H^1$, then
$$\sum_{n=0}^{\infty}\frac{\hat{f}(n)}{n+1}\leq C \|f\|_1$$
 A: Duren's book Theory of $H^p$ spaces has a stronger result as a Corollary on page 48. 

If $f(z) = \sum_{n=0}^\infty a_n z^n \in H^1$, then 
  $$
\sum_{n=0}^\infty \frac{|a_n|}{n+1} \le \pi \|f\|_1
$$
  where 
  $$
\|f\|_1 = \frac{1}{2\pi}\sup_{0<r<1}\int_0^{2\pi}|f(re^{it})|\,dt
$$

Proof: for $t\in [0, 2\pi)$ define 
$\psi(t) = ie^{-it} (\pi-t)$. Then $\|\psi\|_\infty = \pi$ and the Fourier coefficients are
$$
\widehat{\psi}(n) = \frac{1}{2\pi} \int_0^{2\pi} e^{-int} \psi(t)\,dt = \frac{1}{n+1}
$$
Given $f\in H^1$, factor it as $f=B\varphi $ where $B$ is a Blaschke product and $\varphi$ has no zeros in the unit disk. Let $g=B\sqrt{\varphi}$ and $h=\sqrt{\varphi}$, so that $f=gh$. Writing $g(z) = \sum_{n=0}^\infty b_n z^n$ and $h(z) = \sum_{n=0}^\infty c_n z^n  $ we get 
$$
a_n = \sum_{k=0}^n b_k c_{n-k}
$$
By the triangle inequality,
$$
\sum_{n=0}^N \frac{|a_n|}{n+1} = \sum_{n=0}^N \widehat{\psi}(n)|a_n|
\le \sum_{k, m = 0 }^N \widehat{\psi}(k+m)|b_k||c_m|
$$
Let $A(b, c)$ denote the right hand side. The key fact is that 
$$
|A(x, y)| \le \|\psi\|_\infty \|x\|_2 \|y\|_2 \tag1
$$
for any vectors $x, y$, where the vector norms are Euclidean. Once (1) is proved, the result follows because by Parseval, $\|b\|_2 \|c\|_2 = \|g\|_2\|h\|_2 = \|f\|_1$.

Proof of (1) begins by introducing, for any coefficients $x_n$, the function $P(t) = \sum_{n=0}^N x_n e^{-int}$. Note that $P(t)^2$ is a double sum of terms like $x_nx_me^{-i(n+m)t}$. Integrating this against $\psi$ shows that 
$$
|A(x,x)| =  \frac{1}{2\pi} \left|\int_0^{2\pi} P(t)^2 \psi(t)\,dt\right|
\le \|\psi\|_\infty \|x\|_2
$$
which gives a special case of (1). The general case follows by polarization:
$$
A(x, y) = \frac14(A(x+y,x+y) - A(x-y, x-y))
$$ 
hence
$$
|A(x, y)| \le \frac14\|\psi\|_\infty(\|x+y\|^2 + \|x-y\|^2)
= \frac12\|\psi\|_\infty(\|x\|^2 + \|y\|^2)
$$ 
The final step is to notice we can multiply $x$ by one scalar while dividing $y$ by the same scalar, thus reducing the problem to $\|x\|=\|y\|$, in which case 
$$
\frac12\|\psi\|_\infty(\|x\|^2 + \|y\|^2) = \|\psi\|_\infty \|x\|  \|y\|
$$
