# Convergence is uniform on every compact subset of $\Omega$?

Suppose $\{f_n\}$ is a uniformly bounded sequence of holomorphic functions in $\Omega$ such that $\{f_n(z)\}$ converges for every $z \in \Omega$. Does it necessarily follow that the convergence is uniform on every compact subset of $\Omega$? Here, $\Omega$ denotes a plane open set.

Idea. Perhaps we should apply the dominated convergence theorem to the Cauchy formula for $f_n - f_m$?

It is true (assuming that $\Omega$ is an open subset of $\Bbb C$):
Assume that $f_n \to f$ pointwise, but not uniformly on a compact set $K \subset \Omega$. Then there exists an $\varepsilon > 0$, a subsequence $(f_{n_k})$ of $(f_n)$, and a sequence $(z_k)$ in $K$ such that $$\tag{*} |f_{n_k}(z_k) - f(z_k) | \ge \varepsilon \text{ for all } k \in \Bbb N \, .$$
$(f_n)$ is uniformly bounded, so by Montel's theorem it is normal: $(f_{n_k})$ has a locally uniformly convergent subsequence $(f_{n_{k_l}})$. In particular, $f_{n_{k_l}} \to f$ uniformly on $K$, which gives a contradiction to $(*)$.