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


*

*$A$ is open in $G$.

*$G \setminus A$ is open in $G$.

*$A$ 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.)
A: 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.
