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!
Best Answer
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!