Prove that the functions $g_k(z) = f_k \circ h_k(z)$ form a normal family.

Question

I am having a bit of trouble with the following complex analysis question which originates from a qual. Some help would be awesome.

Let $f_k :\mathbb{D} \rightarrow \mathbb{C}$ be a normal family of analytic functions and let $h_k :\mathbb{D}→\mathbb{D}$ be analytic functions satisfying $h_k(0) = 0.$ Prove that the functions $g_k(z) = f_k \circ h_k(z)$ form a normal family.

Answer 1

Hint: Use Schwartz lemma to show that $(h_k)$ is a normal family.

Answer 2

Since $\{ f_k : k \in \mathbb{N}\}$ and $\{ h_k : k \in \mathbb{N}\}$ are both normal families (the former by explicit assumption, the latter by Montel's little theorem), we can (by selecting subsequences) assume that $f_k \to f$ and $h_k \to h$ locally uniformly. The aim is then to show that the sequence $(f_k\circ h_k)_{k\in\mathbb{N}}$ is locally uniformly convergent.

Since $h_k(0) = 0$ for all $k$, it follows that $h\colon \mathbb{D}\to \mathbb{D}$ (without that constraint, it would have been possible that $h_k$ converges to a constant function with $\lvert h(z)\rvert = 1$ for all $z\in\mathbb{D}$).

We don't need the condition $h_k(0) = 0$ or the analyticity of the functions for anything else, although these facts can slightly simplify the proof.

Consider a compact $K \subset \mathbb{D}$. Then $h(K) \subset \mathbb{D}$ is also compact, and thus there is a $\delta > 0$ such that $B_{2\delta}(h(K)) = \{z\in\mathbb{C} : \operatorname{dist}(z, h(K)) < 2\delta\} \subset \mathbb{D}$. Then $M = \overline{B_\delta(h(K))}$ is a compact subset of $\mathbb{D}$.

Given $\varepsilon > 0$, there is, by the locally uniform convergence of the $f_k$, a $k_\varepsilon \in \mathbb{N}$ such that $\lvert f(z) - f_k(z)\rvert < \varepsilon/2$ for all $k \geqslant k_\varepsilon$ and $z \in M$. Further, by the uniform continuity of $f$ on the compact $M$, there is an $\eta > 0$ such that $\lvert f(z) - f(w)\rvert < \varepsilon/2$ for all $z,w\in M$ with $\lvert z-w\rvert \leqslant \eta$. If necessary, replace $\eta$ with $\min \{\delta,\eta\}$, so that $\eta \leqslant \delta$. By the locally uniform convergence of the $h_k$, there is a $k_\eta\in \mathbb{N}$ such that $\lvert h(z) - h_k(z)\rvert < \eta$ for all $z\in K$ and $k \geqslant k_\eta$, in particular it follows that $h_k(z)\in M$ for all $z\in K$ and $k\geqslant k_\eta$. Then, for $z \in K$ and $k \geqslant \max \{k_\varepsilon, k_\eta\}$, we have

$$\begin{align} \lvert f(h(z)) - f_k(h_k(z))\rvert &= \lvert f(h(z)) - f(h_k(z)) + f(h_k(z)) - f_k(h_k(z))\rvert\\ &\leqslant \lvert \underbrace{f(h(z)) - f(h_k(z))}_{\lvert h(z) - h_k(z)\rvert < \eta}\rvert + \lvert \underbrace{f(h_k(z)) - f_k(h_k(z))}_{\lVert f-f_k\rVert_{\infty,M} < \varepsilon/2}\rvert\\ &< \frac{\varepsilon}{2} + \frac{\varepsilon}{2}, \end{align}$$

which shows the uniform convergence of $f_k\circ h_k$ to $f\circ h$ on $K$. Since $K$ was an arbitrary compact subset of $\mathbb{D}$, the locally uniform convergence is established.