I'm having some difficulty with the following problem:
Show that for a set $\mathscr{F}\subset H(G)$, the following are equivalent conditions:
(1) $\mathscr{F}$ is normal;
(2) for every $\epsilon>0$, there is a number $c>0$ such that $\{cf\::\:f\in\mathscr{F}\}\subset B(0,\epsilon)$ (where $B(0,\epsilon)$ is the ball in $H(G)$ with center at zero and radius $\epsilon$).
I know that, by definition, a set $\mathscr{F}$ is normal if each sequence in $\mathscr{F}$ has a convergent subsequence, and I also know that Montel's theorem tells us that $\mathscr{F}$ is normal if and only if $\mathscr{F}$ is locally bounded.
Other than that, I'm not sure how to go about proving this result. Thanks in advance for any help!
Suppose $\mathcal F$ is normal and $\epsilon >0$. Choose $N$ such that $\sum_{N+1}^{\infty} (\frac 1 {2^{n}}) <\epsilon$. If 2) is false then $\{\frac 1 k f: f\in \mathcal F\}$ is not contained in $B(0,\epsilon)$ for any $k$.. These two facts together yield $\sum_{n=1}^{N} \frac {\rho_n (\frac 1 k f_n,0)} {1+\rho_n (\frac 1 k f_n,0)} >\epsilon /2$ for some sequence $\{f_n\}$ in $\mathcal F$. However ${\rho_n (\frac 1 k f_n,0)} \to 0$ for $1\leq n \leq N$ as $k \to \infty$ along some subsequence (because $\{f_n\}$ has a uniformly bounded subsequence on each $K_j$ which implies that $\frac 1 k f_n$ converges uniformly to 0 on these compact sets. This contradiction proves 1) implies 2). Converse is easier. Take $\epsilon = \frac 1 n$and choose $c_n$ corresponding to it. Just using the fact that each term in the definition of $\rho (f,g)$ is no more than the entire sum it follows easily that $\mathcal F$ is uniformly bounded on each $K_j$ and hence 1) holds.