0
$\begingroup$

Assume that $g$ is a holomorphic function on a neighborhood of $\overline{B}(0,1)$ (the closed unit disc centered at $0$) such that $g(0)=0$. Let $s=sup_{|z|=1}Re(g(z))$. Consider the function $h(z)=\frac{g(z)}{2s-g(z)}$.
Show that the function $h$ maps ${B}(0,1)$ (the open unit disk centered at $0$) to ${B}(0,1)$ and that $h(0)=0$.

Any hint will be greatly appreciated. Thanks!

$\endgroup$

1 Answer 1

0
$\begingroup$

Assuming $g$ not identically zero (as otherwise $h$ doesn't make sense), we have by maximum modulus that $\Re g(z) < s, |z|<1$ and of course $s>0$ since $s > \Re g(0)=0$

But then $|2s-g(z)|^2-|g(z)|^2=4s^2-4s\Re g(z)=4s(s-\Re g(z)) >0$ for all $|z|<1$ so $|g(z)| <|2s-g(z)|$ hence $|h(z)|<1, |z|<1$, while $h(0)=0$ by defintion so we are done!

$\endgroup$
3
  • $\begingroup$ Great, thank you! If I may ask, does this imply that the function $h$ is holomorphic? $\endgroup$
    – user862167
    Commented Dec 14, 2020 at 23:03
  • $\begingroup$ yes as it is a ratio of holomorphic functions (denominator nonvanishing anywhere in $|z|<1$) $\endgroup$
    – Conrad
    Commented Dec 14, 2020 at 23:32
  • $\begingroup$ happy to be of help $\endgroup$
    – Conrad
    Commented Dec 14, 2020 at 23:43

You must log in to answer this question.