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

Question

Let $\mathcal{F}$ be a family of analytic functions in a domain $G$. Denote by $\mathcal{F}':=\{ f' \mid f \in \mathcal{F}\}$. It does not suffice to know that $\mathcal{F}'$ is normal to ensure that $\mathcal{F}$ is normal (namely take $\mathcal{F} = \{ n \mid n \in \mathbb{N}\}$ as a counterexample) but someone assured me that if you assume at some fixed point $z_0 \in G$, if you know that $f(z_0) = 0$ for all $f \in \mathcal{F}$ and that $\mathcal{F}'$ is normal, then $\mathcal{F}$ is normal. But I can't see how to do it.

A friend tried something with line integrals to show that the $f$ are locally bounded on compact subsets, but he only considered compact sets that were the image of a path, which I don't think is sufficient. Which characterization of normality should I try (local boundedness? convergence in the space of functions?)? Does anyone have an idea?

Answer 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.

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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 $K$.

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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.

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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.)

Answer 2

I may have worked it out. If anyone finds an issue, let me know!

Let $\{f_n\}$ be a sequence in $\mathcal F$. It suffices to show that $\{ f_n\}$ has a subsequence which converges uniformly on compact subsets of the domain $G$. Let $\{f_n'\}$ be the corresponding sequence of derivatives. Since $\mathcal F'$ is a normal family, we have that $\{ f_n'\}$ has a convergent subsequence, $\{f_{n_k}'\}$ which converges uniformly on compact subsets to a function $g \in \mathcal F'$. By the definition of $\mathcal F'$, we must have that $g = f'$ for some $f \in \mathcal F$.

Now we'd like to show that $\{f_{n_k}\}$ converges uniformly on compact sets to $f$. But since the $f_{n_k}' \rightarrow f'$, applying the Fundamental Theorem of Calculus and continuity of the integral we have $\int f_{n_k}' \rightarrow \int f'$ so $f_{n_k}(z) + C_{n_k} \rightarrow f(z) + C$. But since each $f_{n_k}$ and $f$ is in $\mathcal F$, we have that $f(0) = f_{n_k}(0) = 0$ and so the $C_{n_k} \rightarrow C$. Thus $f_{n_k} \rightarrow f$ uniformly on compact subsets, and so since the sequence $\{f_n\}$ was arbitrary, $\mathcal F$ is a normal family.

I'm worried about playing with the constants of integration (basically the FTC step in general) but I think it should work.