Equivalence of Local Uniform Convergence and Compact Convergence

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Theorem

Let $U \subseteq \C$ be an open set of the complex plane.

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


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

Sufficient Condition

Suppose that $f_n$ converges to $f$ uniformly on all compact subsets of $U$.

Let $z \in U$.

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

Since $\overline {B_\varepsilon \left({z}\right)}$ is closed and bounded, it is compact, by the Heine-Borel Theorem.

We then have by assumption that $f_n$ converges uniformly on $\overline{B_\varepsilon \left({z}\right)}$.

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

$\Box$


Necessary Condition

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 an arbitrary compact subset.

Let $\left\{ {B_{r_z} \left({z}\right)}\right\}_{z \mathop \in K}$ be an open cover 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 $1 \le i \le n$.

Then there is an $N_i$ large enough so that:

$\displaystyle \sup_{z \mathop \in B_{r_i} \left({z_i}\right)} \left\vert{f_n \left({z}\right) - f \left({z}\right)}\right\vert < \varepsilon$

for all $n > N_i$, since the convergence is uniform.

Let $N = \displaystyle \max_{1 \mathop \le i \mathop \le n} \left\{N_i\right\}$.

Let $B = \displaystyle \bigcup_{i \mathop = 1}^n B_{r_i} \left( z_i \right)$.

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 $K$.

$\blacksquare$