TITLE: Show that $\sum_{n \geqslant 0} |a_n z^n|<1$ with $|z|<1/3$ .
QUESTION [7 upvotes]: Assume that $\sum_{n \geqslant 0} a_n z^n$ converges for $|z|<1$ and $$\left|\sum_{n \geqslant 0} a_n z^n \right|<1$$ 
  Show that for all $z \in \mathbb{C}$, with $|z|<1/3$ we have $$\sum_{n \geqslant 0} |a_n z^n|<1$$ 

I am stuck with this exercise, could someone give me some hint please ?

REPLY [6 votes]: This (beautiful) result is usually called Bohr's inequality.
An internet search will show you that there are lots of proofs for this inequality. Here is the simplest I know (unfortunately, I don't remember where I saw it; it is definitely not mine!).
The key point is the following 
Fact. We have $\vert a_m\vert\leq 2(1-\vert a_0\vert)\;$ for all $m\geq 1$.
To prove this, we may assume (multiplying $f(z)=\sum_0^\infty a_nz¨n$ by some constant $\omega$ with $\vert\omega\vert =1$) that $a_0\in\mathbb R^+$.
Define $g(z):=1-f(z)$, where $f(z)=\sum_0^\infty a_n z^n$. Note that ${\rm Re}(g)\geq 0$ because $\vert f\vert\leq 1$. Also,
$$g(z)=\sum_{m=0}^\infty b_m z^m\, ,$$
where $b_0=1-c_0$ and $b_m=-a_m$, $m\geq 1$. 
Since $g$ is holomorphic we have $$\int_0^{2\pi} \overline{g(e^{i\theta})}e^{-im\theta}\, d\theta=0$$ for all $m\geq 1$, and
$$ b_m=\frac1{2\pi}\int_0^{2\pi} g(e^{i\theta})e^{-im\theta}\, d\theta\, .$$
So we may write (for all $m\geq 1$)
$$b_m=\frac1{2\pi}\int_0^{2\pi}\left( g(e^{i\theta})+\overline{g(e^{i\theta})}\right)e^{-im\theta}\, d\theta $$
It follows (remembering that ${\rm Re}(g)\geq 0$ and using the mean value formula for the function ${\rm Re}(g)$) that for all $m\geq 1$ :
$$\vert a_m\vert=\vert b_m\vert\leq \frac1{2\pi}\int_0^{2\pi} 2{\rm Re}(g(e^{i\theta}))\, d\theta=2{\rm Re}(g(0))=2(1-a_0) \, . $$
This proves the Fact.
Now, by the Fact, we have for all $r>0$ :
$$\sum_{m=0}^\infty \vert a_m\vert r^m\leq \vert a_0\vert+2(1-\vert a_0\vert)\sum_{m=1}^\infty r^m\, . $$
The sum in the right hand side is equal to $\frac12$ if $r=\frac13$; so we get
$$\sum_{m=0}^\infty \vert a_m\vert \left(\frac13\right)^m\leq \vert a_0\vert +(1-\vert a_0\vert)=1.$$
Hence, $\sum_0^\infty\vert a_mz^m\vert\leq 1$ for all $z\in\mathbb C$ such that $\vert z\vert\leq 1/3$.