Bohr’s theorem

(Bohr 1914).  If the power series${\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{a}_n{z}^n$ satisfies

(1)

in the unit disk  ,  then (1) and the inequality

(2)

is true in the disk  .  Here, the radius $\frac{1}{3}$ is the best possible.

Proof.  One needs Carathéodory’s inequality which says that if the real part of a holomorphic function

$$g(z):=\sum _{n=0}^{\mathrm{\infty }}{b}_n{z}^n$$

is positive in the unit disk, then

$$|b_n|\leqq \mathrm{\hspace{0.17em}2}\text{Re}{b}_0\mathit{\hspace{1em}}\text{for}n=1,\mathrm{\hspace{0.17em}2},\mathrm{\dots }$$

Choosing now  $g(z):=1-e^{i\phi }f(z)$  where $\phi$ is any real number and $f(z)$ the sum function of the series in the theorem, we get

$$|a_n|\leqq \mathrm{\hspace{0.33em}2}\text{Re}(1-e^{i\phi }{a}_0)=\mathrm{\hspace{0.33em}2}(1-a_0\cos \phi ),$$

and especially

$$|a_n|\leqq \mathrm{\hspace{0.33em}2}(1-|a_0|),\text{for}n=1,\mathrm{\hspace{0.17em}2},\mathrm{\dots }$$

If  $f(z)\not\equiv a_0$,  in the disk    we thus have

Take then in particular the function defined by

$$f(z):=\frac{z-c}{1-cz}$$

with  .  Its series expansion

$$f(z)=\sum _{n=0}^{\mathrm{\infty }}{a}_n{z}^n=-c+(1-c^2)z+(1-c^2)c{z}^2+(1-c^2)c^2{z}^3+\mathrm{\dots }$$

shows that

$$\sum _{n=0}^{\mathrm{\infty }}|a_n{z}^n|=f(|z|)+2c,$$

which last form can be seen to become greater than 1 for  $|z|>{\displaystyle \frac{1}{1+2c}}$.  Because $c$ may come from below arbitrarily to 1, one sees that the value $\frac{1}{3}$ in the theorem cannot be increased.