# Normal sequences and Montel's Theorem

I am currently stuck on an exam question involving normal sequences and Montel's theorem:

1. Give two examples of non-constant normal sequences one in the $(a)$ unit disk $\mathbb{D}$ and one in $(b)$ $\mathbb{C}$.

2. Let $(a_n)$ be a sequence of complex numbers satisfying $|a_n|\lt 1$ and an $a_n\to 0$ and consider the sequence of holomorphic functions $(f_n)$ in $\mathbb{D}$ deﬁned by:

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\large f_n(z)=\frac{z-a_n}{1-\overline{a_n}z}$

State and justify if $(f_n)$ is normal and whether it satisﬁes the condition of Montel’s theorem? Explain if $(f_n)$ converges uniformly in $\mathbb{D}$.

Firstly I will state Montel's theorem and then I will write what I have tried so far.

Montel's theorem: A family $\mathbb{F}\subset \mathbb{H}(U)$ is normal if and only if it is locally bounded. (Where $U\subset\mathbb{C}$, open, and $\mathbb{H}(U)$ is the space of holomorphic functions on $U$.

My examples for 1. are:

$(a)$ $g_n:\mathbb{D}\to\mathbb{C}, g_n(z)=\large e^{\frac{iz}{n}}$

$(b)$ $h_n:\mathbb{C}\to\mathbb{C}, h_n(z)=\large \frac{z}{n^2}$

Now for $2$. I have chosen to show that $f_n$ has a subsequence that converges uniformly on compact subsets of $\mathbb{D}$, so let $K\subset \mathbb{D}$ be compact, thus if $z\in K\Rightarrow |z|\le 1$.

I know that since $a_n\to 0$, $f_n(z)\to z$ pointwise.

Now: $\large|f_n(z)-z|=|\frac{z-a_n}{1-\overline{a_n}z}-z|=|\frac{z-a_n-z+\overline{a_n}z^2}{1-\overline{a_n}z}|$

$\large=|\frac{\overline{a_n}z^2-a_n}{1-\overline{a_n}z}|\le \frac{|\overline{a_n}z^2-a_n|}{1-|a_n|}\le \frac{|a_n|(|z|^2+1)}{1-|a_n|}\le \frac{2|a_n|}{1-|a_n|}$

Now we can construct a subsequence:

since $a_n\to 0$ it follows that $\exists N\in \mathbb{N}:\forall j\ge N\,:|a_j|\lt\frac{1}{N}$

So we consider the subsequence $\{f_j\,|j\ge N\}$

Here $|f_j(z)-z|\le\frac{2|a_j|}{1-|a_j|}\lt \frac{\frac{2}{N}}{1-\frac{1}{N}}=\frac{2}{N-1}\to 0$ as $N\to\infty$.

So the subsequence converges uniformly in $\mathbb{D}$, thus $(f_n)$ is normal, and thus by Montel's theorem it is locally bounded, so it satisfies the condition of montels theorem.

Now explain if $(f_n)$ is uniformly convergent on $\mathbb{D}$, this is the part that I am stuck on, and would really appreciate help with.

I would also like clarification if what I have done for the previous parts is at all correct?

Thanks!

There is a third part to the question:

1. Let $(f_n)$ be a sequence of holomorphic functions from $U$ to the upper half-plane. Prove that $(f_n)$ has a subsequence converging uniformly on all compact subsets of $U$ to either a holomorphic function $f$ or to $\infty$

My approach is to consider the sequence:

$g_n(z)=\large\frac{\frac{f_n(z)}{|f_n(z)|}-a_n}{1-\overline{a_n}\frac{f_n(z)}{|f_n(z)|}}$, here $g_n$ is of the form of part 2. and takes values in the unit disk.

• You have already proven that $f_n$ is uniformly convergent on $\mathbb{D}$. The bound $\frac{2\lvert a_n\rvert}{1-\lvert a_n\rvert}$ is valid on all of $\mathbb{D}$. Yup, all correct. – Daniel Fischer Apr 21 '14 at 21:40
• I thought that would only prove uniform convergence of subsequences? – Ellya Apr 21 '14 at 21:41
• No, you have $a_n \to 0$, hence $C_n := \frac{2\lvert a_n\rvert}{1-\lvert a_n\rvert} \to 0$, and you have $$\lvert f_n(z) - z\rvert \leqslant C_n$$ on all of $\mathbb{D}$ by your computation. – Daniel Fischer Apr 21 '14 at 21:43
• Ah yes, now you say it, it seems obvious, I've been going round in circles with this for a while. And you're sure that everything I've put is correct? Thanks! – Ellya Apr 21 '14 at 21:51
• Well, you may need to explain why a locally uniformly convergent sequence of holomorphic functions is normal. But that's pretty obvious, so you may not need to. – Daniel Fischer Apr 21 '14 at 21:55