For holomorphic function $f(z) = \sum_{n=0}^\infty a_n z^n$, $|a_n|r^n \le 2 \max_{|z|=r} \text{Re}f(z)-2 \text{Re} f(0)$. Let $f(z) = \sum_{n=0}^\infty a_n z^n \in H(B(0,R))$, for $r\in(0,R)$, let $A(r) = \max_{|z|=r} \text{Re}f(z)$.
Prove that $|a_n|r^n \le 2A(r)-2 \text{Re} f(0)$.
We have $a_n = \frac{f^{(n)}(0)}{n!} = \frac{1}{2\pi i}\int_{|\xi|=r} \frac{f(\xi)}{\xi^{n+1}}\, d\xi = \frac{1}{2\pi r^n}\int_{0}^{2\pi}f(re^{i\theta})e^{-in\theta}d\theta$, and
$\int_0^{2\pi} \overline{f(re^{i\theta})}e^{-in\theta} d\theta = 0$, so $a_nr^n = \frac{1}{\pi}\int_{0}^{2\pi}\text{Re} f(re^{i\theta}) \ e^{-in\theta} d\theta$.
How can we get that $\text{Re} f(0)$?
 A: Recall that for every $k\in\mathbb{C}, m\in\mathbb{Z}-\{0\}$ we have $$\int_0^{2\pi}ke^{im\vartheta}=0$$
So
$$ a_nr^n=\frac1\pi \int_0^{2\pi}\Re(f)(e^{i\vartheta})e^{-in\vartheta}d\vartheta=\frac1\pi\int_0^{2\pi}\left(\Re(f)(e^{i\vartheta})-\Re(f)(0)\right)e^{-in\vartheta}d\vartheta\\
|a_nr^n|\le 2|A(r)-\Re(f(0))|$$
As the right hand side must be positive by the maximum modulus principle for harmonic functions, the result follows:
$$|a_n|r^n\le 2A(r)-2\Re(f(0))$$
A: I am sure that this has been asked before but there is a truly beautiful proof (folklore, maybe Littlewood?) that goes like this - assume $f$ nonconstant
Using $g(z)=f(z/r)$ we can assume $r=1$ as nothing else changes and then we take $h(z)=f(z)-a_0$ so with $U=A(1), \alpha=\Re f(0), V=U-\alpha$ we have $\Re h \le V, h(0)=0$ and $h$ is analytic on the closed unit disc (hence in a smaller disc - continuity on the boundary would also have been enough by a limiting process).
Then $p(z)=\frac{h(z)}{h(z)-2V}$ self maps the unit disc into itself and $p(0)=0$ so $|p'(0)| \le 1$ by Schwarz. Since $p'(0)=h'(0)/(2V)=f'(0)/(2V)$ we get the inequality $|a_1| \le 2V=2(A(1)-\Re f(0)$ as required.
But now for $n \ge 2$ let $f_n(z)=\frac{f(z)+f(\omega z)+...f(\omega^{n-1}z)}{n}$ where $\omega^n=1$ is a nontrivial root of unity; it is easy to see that $f_n$ still satisfies the inequalities of $f$ and it is a function of $z^n$ so defining $g_n(z)=f_n(z^n)$ still nothing changes but now $g_n=a_0+a_nz+...$ so applying the previous case we get the result for $a_n$ too!
