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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 $sup_{\bar {B(z_0, r/2)}} |f_{n}(z)-f(z)| \to 0$?

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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:

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

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