# Prove that the family of functions that omit 0 is normal

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!

• 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 Mar 13 '19 at 22:45
• Actually it does converge normally, maybe to the constant function $f=\infty$ (for example $f_n (z) = n(z+1)$). Mar 14 '19 at 10:23
• if we accept normal convergence to infinity than the result follows as below Mar 14 '19 at 11:54

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.

• 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). Mar 14 '19 at 19:55
• 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 Mar 14 '19 at 20:53
• Think $kz, k(\frac{1}{2}-z)$, neither family is normal as it has a common zero while going to infinity everywhere else Mar 14 '19 at 21:15
• Got it! Thank you very much, again :) Mar 14 '19 at 21:31