Problem is, Let $\Omega$ be a simply connected Region Other than Whole plane with $0\in\Omega$.

Suppose $f(z)$ is a 1-1 holomorphic function from $\Omega$ into Unit Disk $\mathbb D$ such that $f(\Omega)$ is a proper subset of $\mathbb D$.

Show that there is a holomorphic function $g(z)$ from $\Omega$ into $\mathbb D$ such that $f'(0) < g'(0)$.

I think I should use Riemann mapping theorem, But I cannot proceed further. There are relation between value at 0 of derivative and 'size' of Codomain?

Here is my attempt.

There is conformal map $\psi:\Omega\rightarrow\mathbb D$ by Riemann Mapping Theorem.

Let $\phi :\mathbb D \rightarrow \Omega $ be a $\psi^{-1}$.

WLOG, We may assume $\phi(0)=0$.(Adjust by Moibius transformation if necessary.)

Then Clearly $(\psi\circ\phi)^{'}(0) =1$

Now $|(f\circ\psi)^{'}(0)|<1$ by Schwarz-pick lemma.

Note that $(\psi\circ\phi)^{'}(0) = \psi(\phi (0))\cdot \phi^{'}(0)$ and $(f\circ\psi)^{'}(0) = f(\phi (0))\cdot \phi^{'}(0)$

Since $\phi^{'}$ is never zero, We get the result.

Seems correct?

  • $\begingroup$ After using Riemann mapping theorem, My candidate is $g(z) = z$. From D to D.(Schwarz pick theorem.) But, '=' can hold. $\endgroup$ – nicksohn Jan 1 '16 at 12:25
  • $\begingroup$ Do you mean, $\left|f^\prime(0)\right|<\left|g^\prime(0)\right|$? $\endgroup$ – Jonathan Y. Jan 1 '16 at 12:57
  • $\begingroup$ I made a mistake.. $\endgroup$ – nicksohn Jan 1 '16 at 13:00
  • $\begingroup$ @JonathanY. I add my attempt. If you have time, Please check it. $\endgroup$ – nicksohn Jan 1 '16 at 13:43
  • $\begingroup$ Firstly, you haven't answered my question. Secondly, I don't follow how you get the result in the last sentence (which isn't to say there's a mistake; I simply didn't follow, if you care to elaborate). $\endgroup$ – Jonathan Y. Jan 1 '16 at 13:52

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