Dtjytyk

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

Problem: Let $$mathcal G={f:mathbb Dtomathbb Cmid fin Hol(mathbb D ) ,f-is-one-to-one}$$
be the family of univalent analytic functions on the unit disk, and let $$mathcal F={finmathcal F mid 0notin f(mathbb D) } $$
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!

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.