When does the normality of the family of derivatives imply that the family itself is normal?

Solution 1:

Let $\mathcal{F}$ be a family of analytic functions in a (connected) domain $G$ and assume that

• $\mathcal{F}':=\{ f' \mid f \in \mathcal{F}\}$ is normal and
• $\{ f(a) \mid f \in \mathcal{F}\}$ is bounded for some $a \in G$.

Then $ \mathcal{F}$ is normal.

Part 1: First we prove this for the special case that $G = D$ is an open disk: Let $(f_n)$ be a sequence in $\mathcal{F}$. $\mathcal{F}'$ is normal, so there exists a subsequence $f_{n_k}$ such that $f_{n_k}'$ is locally uniformly convergent in $D$. $\{ f_{n_k}(a) \mid f \in \mathcal{F}\}$ is bounded and has a convergent subsequence, so by taking another subsequence we can assume that $$ f_{n_k}' \to g \text{ in } D \, , \quad f_{n_k}(a) \to w $$ Now define for $z \in D$ $$ G(z) = w + \int_{[a, z]} g(t) \, dt $$ where $[a, z]$ is the straight line from $a$ to $z$. $G$ is holomorphic in $D$ and $G' = g$, therefore $$ f_{n_k}(z) - G(z) = f_{n_k}(a) - w + \int_{[a, z]} (f_{n_k}'(t) - g(t)) \, dt $$ If $K \subset D$ is compact then choose a closed disc $B$ such that $a \in B$ and $K \subset B \subset D$. Then $[a, z] \subset B$ for all $z \in K$ and therefore $$ | f_{n_k}(z) - G(z)| \le | f_{n_k}(a) - w | + \text{diam}(D) \sup_{t \in B} | f_{n_k}'(t) - g(t)| $$ and that converges to zero uniformly on $K$.

Part 2: In order to generalize this to an arbitrary domain $G$ we define $$ A := \{ z \in G \mid \mathcal{F} \text{ is normal in a neighborhood of } z \} \, . $$ The aim is to show that $A = G$ by proving

1. $A$ is open in $G$.
2. $G \setminus A$ is open in $G$.
3. $A$ is not empty.

Now (1) follows directly from the definition of $A$. Part 1 shows that $a \in A$, this is (3).

To prove (2) assume that $b \in G \setminus A$. Let $D$ be a disk with center $b$ which is contained in $G$. We'll show that $D \subset G \setminus A$.

Assume that there exists a $c \in D \cap A$. Then $\mathcal{F}$ is normal in a neighborhood of $c$ and that implies that $\{ f(c) \mid f \in \mathcal{F}\}$ is bounded. (Otherwise $|f_n(c)| \to \infty$ for some sequence in $\mathcal{F}$, but $(f_n)$ has a subsequence which converges in particular at the point $c$.) From part 1 it follows that $\mathcal{F}$ is normal in $D$ in contradiction to the assumption that $b \in G \setminus A$.

This proves that $ G \setminus A$ is open.

Part 3: $G$ is connected, therefore from (1, 2, 3) above it follows that $A = G$, i.e. every $z \in G$ has a neighborhood $U(z)$ such that $\mathcal{F}$ is normal in $U(z)$.

Since $G$ is the countable union of compact subsets, and each compact subset is covered by finitely many $U(z)$, a classical diagonal argument shows that $ \mathcal{F}$ is normal in $G$. (Compare If a family of meromorphic functions is normal near each point in a region, then it's normal in the region.)