This is a problem from Complex analysis - Conway
Show that for a set $F \subset H(G)$, the followings are equivalent
(1) $F$ is normal family
(2) For every $\epsilon >0$, there is a number $c>0$ such that $\{\ cf : f \in F \}\ \subset B(0;\epsilon)$. ($B(0;\epsilon)$ is the open ball of radius $\epsilon$ with center at $0$ in $H(G)$)
I am struggling with both the sides. I have the following progress
Let $F$ be a normal family. Given $\epsilon > 0$, there is a $\delta >0$ and $K\subset G$, compact s.t
$$sup\{\ |f(z)||z\in K \}\ \implies f\in B(0;\epsilon) \hspace{2 cm} ..... (1)$$
Since $F$ is a normal family, for the above $\delta >0$ and $K \subset G$, compact we have $f_{1}, f_{2},... f_{n} \in F$ such that given any $f\in F$, there is a $f_{i} (1\leq i \leq n)$ such that $$sup \{\ |f(z)-f_{i}(z)| |z\in K \}\ < \delta \hspace{3 cm} ..... (2)$$
So from (1) and (2), what we have is for each $f\in F$ , there is a $f_{i}(1\leq i \leq n)$ such that $f-f_{i} \in B(0;\epsilon)$.
I am lost after this. I don't know how to find this $"c"$. Thanks in advance for any help!