# Bohr’s theorem

(Bohr 1914).  If the power series  $\displaystyle\sum_{n=0}^{\infty}a_{n}z^{n}$ satisfies

 $\displaystyle\left|\sum_{n=0}^{\infty}a_{n}z^{n}\right|\;<\;1$ (1)

in the unit disk$|z|<1$,  then (1) and the inequality

 $\displaystyle\sum_{n=0}^{\infty}|a_{n}z^{n}|\;<\;1$ (2)

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

 $g(z)\;:=\;\sum_{n=0}^{\infty}b_{n}z^{n}$

is positive in the unit disk, then

 $|b_{n}|\;\leqq\,2\,\mbox{Re}\,b_{0}\quad\mbox{for}\;n=1,\,2,\,\ldots$

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

 $|a_{n}|\;\leqq\;2\,\mbox{Re}\,(1\!-\!e^{i\varphi}a_{0})\;=\;2(1\!-\!a_{0}\cos% \varphi),$

and especially

 $|a_{n}|\;\leqq\;2(1\!-\!|a_{0}|),\quad\mbox{for}\;n=1,\,2,\,\ldots$

If  $f(z)\not\equiv a_{0}$,  in the disk  $|z|<\frac{1}{3}$  we thus have

 $\sum_{n=0}^{\infty}|a_{n}z^{n}|\;<\;|a_{0}|+2(1\!-\!|a_{0}|)\sum_{n=1}^{\infty% }\left(\frac{1}{3}\right)^{n}\;=\;1.$

Take then in particular the function defined by

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

with  $0.  Its series expansion

 $f(z)\;=\;\sum_{n=0}^{\infty}a_{n}z^{n}\;=\;-c+(1\!-\!c^{2})z+(1\!-\!c^{2})cz^{% 2}+(1\!-\!c^{2})c^{2}z^{3}+\ldots$

shows that

 $\sum_{n=0}^{\infty}|a_{n}z^{n}|\;=\;f(|z|)+2c,$

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

## References

• 1 Harald Bohr: “A theorem concerning power series”. – Proc. London Math. Soc. 13 (1914).
• 2 Harold P. Boas: “Majorant series”. – J. Korean Math. Soc. 37 (2000).
Title Bohr’s theorem BohrsTheorem 2015-04-13 12:52:55 2015-04-13 12:52:55 pahio (2872) pahio (2872) 12 pahio (2872) Theorem msc 40A30 msc 30B10