# Show that the seqns $\{f_n\}$ converges uniformly on compacts in $D$ to $f$.

Suppose a sequence of holomorphic functions $F=\{f_n\}$ is a normal family on a domain $D$, $f$ is holomorphic on $D$ and $f_n(z) \to f(z)$ pointwise on a nonempty open set $U \subset D$. Show that the seqns $\{f_n\}$ converges uniformly on compacts in $D$ to $f$.

for $f_n$ there exists $f_{n_k}$ which converges uniformly to $f$ i.e $sup_{\overline{B(z_0, r/2)}} |f_{n_k}(z)-f(z)| \to 0$. Now how can I use the compactness & show $sup_{\bar {B(z_0, r/2)}} |f_{n}(z)-f(z)| \to 0$?

## 1 Answer

Take a compact set $K \in D$, and suppose it does not converge uniformly on $K$$f_n \not\rightrightarrows f$$ This means that there is some$\varepsilon >0$a sequence$(z_{n_k})_{n_k \geq 1} \subseteq K$and a subsequence of the$(f_n)$,$(f_{n_k})$such that for all$n_k \in \mathbb{N}$$$|f(z_{n_k})_{n_k} - f(z_{n_k})| \geq \varepsilon$$ Now, because$\mathcal{F}$is a normal family, and$(f_{n_k}) \subseteq \mathcal{F}$, there is some subsequence$f_{n_{k_j}} \rightrightarrows f$But this means that if$n_{k_j}\$ is big enough:

$$\varepsilon \leq |f(z_{n_{k_j}})_{n_{k_j}} - f(z_{n_{k_j}})| \leq \sup_{ z \in K} |f_{n_{k_j}}(z) - f(z)| < \varepsilon$$