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Show that the family of all analytic maps $f: B(0,1)\to\{z\in\Bbb C: Re(z)>0\}$ such that $|f(0)|\le1$ is normal.

I tried to use Azela-Ascoli Theorem. first show for each $z\in B(0,1)$,$f(z)$ has compact closure. Second is to show at each point $z$ the family of functions are equicontinuous.

But I do not know how to show $f(z)$ has compact closure for each $z$ and $f$ is equicontinuous. Is this the right direction to prove this question?

Could someone kindly help? Thanks!

Sherry
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    Are you aware of Cauchy's integral formula for the derivative? – Moishe Kohan Apr 04 '17 at 02:56
  • Not to be cheeky but could you not just pick a particular subfamily and show that it is uniformly convergent on any compact subset of $B(0, 1)$? I can think of at least one rational function whose iterates satisfy all of these properties. – M A Pelto Apr 04 '17 at 07:54
  • I just consulted L. V. Ahlfors and refreshed my memory on the definition of normal family (I seem to have developed particular thinking). This question is admittedly very general. I can actually think of more than one rational function and so more than one uniformly convergent (on compact subsets) sequence satisfying the aforementioned properties. Though like I mention this is an apparently useless endeavor here. Cohen seems to have the right idea. – M A Pelto Apr 04 '17 at 10:51
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    With Harnak or Schwartz we know $|f(z)|\leq\dfrac{1+|z|}{1-|z|}$ so this family is locally uniformly bounded. – Nosrati Apr 10 '17 at 13:36
  • Check out https://math.stackexchange.com/questions/495848/the-family-of-analytic-functions-with-positive-real-part-is-normal?rq=1 – Heisenberg Aug 15 '17 at 18:35
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    Possible duplicate of The family of analytic functions with positive real part is normal – Heisenberg Aug 15 '17 at 18:36

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