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Let $\mathcal{F}$ be a family of holomorphic functions on a common domain $U \subset \mathbb C$. Suppose that $\mathcal{F}$ is pointwise bounded in the sense that for each $z \in U$, there is a constant $b > 0$ (potentially dependent on $z$) such that $|f(z)| \leq b$ for all $f \in \mathcal{F}$. Must it hold that $\mathcal{F}$ is uniformly bounded on each compact subset of $U$?

Using a covering argument, I can show that this holds provided $\mathcal{F}$ is equicontinuous. Moreover, the set $\mathcal{F} = \{f(z) =1/z\}$ is a counterexample if we don't require the sets on which the functions are bounded be compact. However, I am unsure about whether this is true as stated.

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up vote 3 down vote accepted

Let $U$ be the open unit disk. For each positive integer $n$, let $A_n = \{r e^{i\theta}: 0 \le r \le 1 - 1/n,\ 2/n \le \theta \le 2 \pi \}$. By Runge's theorem, there is a polynomial $p_n$ such that $|p_n(z)| \le 1$ on $A_n$ and $|p_n(e^{i/n}/2)| \ge n$. Let ${\cal F} = \{ p_n: n \in {\mathbb N}\}$. Each $z \in U$ is in all but finitely many $A_n$, so all but finitely many $|p_n| \le 1$, and thus $\cal F$ is pointwise bounded. On the other hand, $\cal F$ is not uniformly bounded on the compact set $\{z: |z| = 1/2 \}$.

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