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I'd love to get some help with the following problem:

Problem: Let $$\mathcal G=\{f:\mathbb D\to\mathbb C\mid f\in Hol(\mathbb D ) ,f-is-one-to-one\}$$ be the family of univalent analytic functions on the unit disk, and let $$\mathcal F=\{f\in\mathcal F \mid 0\notin f(\mathbb D) \} $$ be the family of functions in $\mathcal G$ that omit 0.

Prove that $\mathcal F$ is a normal family.

Hint: Consider family of square roots with appropriate branches.

My idea: I wanted to show that $\mathcal F$ should omit another value (except 0), and than apply Montel's theorem. However, I could not understand how I sould show that.

Any help will be appreciated!

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  • $\begingroup$ If $f$ is in $\mathcal F$ then $nf$ is in there for any $n$, but any subsequence of $nf$ obviously doesn't converge normally, so I think an extra condition is needed $\endgroup$
    – Conrad
    Mar 13 '19 at 22:45
  • $\begingroup$ Actually it does converge normally, maybe to the constant function $f=\infty$ (for example $f_n (z) = n(z+1)$). $\endgroup$ Mar 14 '19 at 10:23
  • $\begingroup$ if we accept normal convergence to infinity than the result follows as below $\endgroup$
    – Conrad
    Mar 14 '19 at 11:54
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Let $f_k(z)=a_k+b_kz+...$ be a sequence in $\mathcal F$; we will show that there are three alternatives:

1: $f_k$ contains a "proper" normally convergent subsequence - one converging normally to a non-constant, non-infinity function $f$; by Hurwitz $f$ is univalent and non-zero everywhere so it is in $\mathcal F$

2: $f_k$ has a subsequence which converges normally to a finite constant (and all its subsequences are like this or go to infinity)

3: $f_k$ converges normally to infinity (where we mean that in the extended plane sense, or equivalently that on any compact set, $f_k$ get uniformly big)

We use the following two facts:

1: $a_k, b_k \neq 0$ since $a_k=f_k(0), b_k=f_k'(0)$

2: $h_k(z)=\frac{f_k(z)-a_k}{b_k}$ is a normalized schlicht (univalent) function, the set of which is usually called $\mathcal S$, so it satisfies uniform boundness on compact sets, $|h_k(z)|\ \leq \frac{1}{1-r^2}, |z| \leq r<1$, and their image contains the disk of radius $\frac{1}{4}$ around the origin

Translating the second fact to $f_k$ it follows that the image of $f_k-a_k$ contains the (open) disk of radius $\frac{|b_k|}{4}$ and since zero is not in the image of $f$, it follows that $a_k$ is not in that disk, which is equivalent to $\frac{|a_k|}{|b_k|} \geq \frac{1}{4}$, so we have four mutually exclusive cases: $a_k$ converges to zero (hence $b_k$ does too), $a_k$ goes to infinity, a combination of the two ($a_k$ splits in subsequences converging to zero and infinity respectively - which we treat like either of the two previous cases, so it will follow, $f_k$ splits into subsequences converging normally to zero and infinity too) or $a_k$ has a subsequence which converges to some finite non-zero $a$, hence taking possibly a subsequence of it, we can assume $b_k$ also converges to some finite $b$ (could be zero now).

In the first case, the local uniform boundness of $h_k$ immediately implies $f_k$ converges normally to zero, while in the fourth case, applying Montel to the subsequence of the $h_k$ given in our assumptions (for which $a_k$ converges to $a$, $b_k$ converges to $b$) we get a $h$ in $\mathcal S$, s.t. the subsequence of $h_k$ converges normally to $h$, so the subsequence of $f_k$ converges normally to $a+bh$. If $b=0$, we get case 2 (constant) with non-zero $a$, if $b \neq 0$, Hurwitz shows as noted we are in the proper convergence case 1.

Assume now $a_k$ goes to infinity and take $g_k = \frac{1}{f_k}$ which is in $\mathcal F$; letting $g_k(z)=c_k+d_kz+...$, we get $a_kc_k=1$, so $c_k$ converges to zero (hence $d_k$ too as noted above), so $g_k$ converges normally to zero, hence $f_k$ converges "normally" to infinity, so we are done with this case too.

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  • $\begingroup$ Wow! Thank you very much for your detailed solution. I could not see in which part of the proof you used that fact that $a_k \neq 0$ (and the fact that $f_k$ are all omit 0). $\endgroup$ Mar 14 '19 at 19:55
  • $\begingroup$ Two places: first if $a_k$ could be zero, there is no restriction on $b_k$, restriction that follows from the zero omission everywhere btw, so they can go to infinity, but $f_k$ can oscillate, and second if $a_k$ goes to infinity, we may not be able to say anything about $f_k$ again without considering $\frac{1}{f_k}$ and that requires $f_k$ non zero everywhere not only at $0$ which follows from $a_k$ big; the whole space of univalent functions in the disc is not normal even accepting normal convergence at constant infinity $\endgroup$
    – Conrad
    Mar 14 '19 at 20:53
  • $\begingroup$ Think $kz, k(\frac{1}{2}-z)$, neither family is normal as it has a common zero while going to infinity everywhere else $\endgroup$
    – Conrad
    Mar 14 '19 at 21:15
  • $\begingroup$ Got it! Thank you very much, again :) $\endgroup$ Mar 14 '19 at 21:31

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