# Locally bounded $\Leftrightarrow$ compactness

I am trying to prove the following

A set $\mathcal{F}$ in $H(G)$ is locally bounded if and only if for each compact set $K\subset G$ there is a constant $$|f(z)|\leq M$$ for all $f$ in $\mathcal{F}$ and $z$ in $K$. Where $G$ is an open subset of the complex plane and $H(G)$ is the collection of analytic functions on $G$.

My try goes as follows. Assuming that $\mathcal{F}$ is locally bounded then we have that $$|f(z)|\leq M$$ for $|z-a|<r$ for each point $a$ in $G$. Therefore we can take any compact closed disk that contains the open disk $|z-a|<r$. This would give $K\subset G$ compact and give the bounded condition.

However my reasoning doesn't seem to allow for ANY compact set. If our compact set is much bigger than the open disk then you couldn't be sure that all $z\in K$ would still satisfy the bounded condition. Could someone point me in the right direction?

$\textbf{EDIT:}$

The definition for locally bounded:

A set $\mathcal{F}\subset H(G)$ is locally bounded if for each point $a$ in $G$ there are constants $M$ and $r>0$ such that for all $f$ in $\mathcal{F}$ $$|f(z)|\leq M$$ for $|z-a|<r$ or equally $\mathcal{F}$ is locally bounded if there is an $r>0$ such that $$sup\{|f(z)|:|z-a|<r, f\in\mathcal{F}\}<\infty$$

• What do $G$ and $H(G)$ denote? – Crostul Feb 10 '16 at 13:24
• $G$ is an open subset of the complex plane and $H(G)$ is the collection of analytic functions on $G$. – George1811 Feb 10 '16 at 13:25
• I see. And what is your definition of "locally bounded"? I mean, what you are trying to prove is usually the definition. – Crostul Feb 10 '16 at 13:28
• Perhaps this is what you need: Every covering of a compact set by open sets has a finite subcover. – Martin R Feb 10 '16 at 13:33
• I'm trying to prove that $\mathcal{F}$ is uniformly bounded on compact sets if and only if it is locally bounded. – George1811 Feb 10 '16 at 13:35

In the following I will denote by $B_r(a)$ the open disk $\{ z \in \Bbb{C} : |z-a|<r\}$.
Let $K \subset G$ be any compact subset. In particular, the distance $d=d(K,\Bbb{C} \setminus G)$ is not zero. This means that you can consider the open cover of $K$ inside $G$ $$K \subset \bigcup_{a \in K} B_d(a) \subset G$$ Since $K$ is compact, you are allowed to pick a finite subcover of open disks $\{ B_d(a_j) \}_j$. For every such disk $B_d(a_j)$ you have a constant $M_j$, so the constant for $K$ is simply $M=\max_j M_j$.
The converse is quite simpler. For all $a \in G$, let $r > 0$ such that $B_{2r}(a) \subset G$. Now, since the closed disk $\overline{B_r(a)}$ is a compact set contained in $G$, you have a constant $M$. This constant works for the open disk $B_r(a)$, so you are done.
• In OP's definition of local boundedness – as I understand it – the radius $r$ depends on the the point $a$. So you cannot choose a fixed radius $d$ for the open covering. Of course this does not change the rest of your reasoning. – Martin R Feb 10 '16 at 17:17