Equivalence of Local Uniform Convergence and Compact Convergence

Theorem
Let $U \subseteq \C$ be an open set.

Let $f_n \colon U \to \C$ be a sequence of functions which converges pointwise to $f \colon U \to \C$.

Then $f_n$ converges to $f$ uniformly on all compact subsets of $U$ if and only if $f_n$ converges locally uniformly on $U$.

Proof
Suppose that $f_n$ converges to $f$ uniformly on all compact subsets of $U$, and let $z \in U$.

Since $U$ is open, there is an open ball $B_\varepsilon \left( z \right)$, around $z$, of radius $\varepsilon$, with $\overline{B_\varepsilon \left( z \right)} \subset U$.

Since $\overline{B_\varepsilon \left( z \right)}$ is compact, we have by assumption that $f_n$ converges uniformly on $\overline{B_\varepsilon \left( z \right)}$.

Thus $f_n$ converges locally uniformly on $U$.

Now suppose that $f_n$ converges to $f$ locally uniformly.

That is, for every $z \in U$, there is an $r$ so that $B_r \left( z \right) \subset U$ and $f_n$ converges uniformly on $B_r \left( z \right)$.

Let $K \subset U$ be any compact subset, and $\left\{B_{r_z} \left( z \right)\right\}_{z \mathop \in K}$ an open covering of $K$ by open balls, one centered at each $z \in K$, such that $f_n$ converges uniformly on each ball.

Since $K$ is compact, there is a finite subset $\left\{B_{r_i} \left( z_i \right) \right\}_{i \mathop = 1}^n$ which covers $K$.

Let $\varepsilon > 0$, and $1 \leq i \leq n$.

Then there is an $N_i$ large enough so that $\displaystyle \sup_{z \mathop \in B_{r_i} \left( z_i \right)} \vert f_n \left( z \right) - f \left( z \right) \vert < \varepsilon$ for all $n > N_i$, since the convergence is uniform.

Let $N = \displaystyle \max_{1 \mathop \leq i \mathop \leq n} \left\{N_i\right\}$, and $B = \displaystyle \bigcup_{i \mathop = 1}^n B_{r_i} \left( z_i \right)$.

Then $\displaystyle \sup_{z \in B} \left\vert f_n \left( z \right) - f \left( z \right) \right\vert < \varepsilon$ for all $n > N$.

Thus $f_n$ converges uniformly on $B$, and so also on $K$.