Vitali's Convergence Theorem

Theorem

Let $U$ be an open, connected subset of $\C$.

Let $S \subseteq U$ contain a limit point $\sigma$.

Let $\sequence {f_n}_{n \mathop \in \N}$ be a normal family of holomorphic mappings $f_n : U \to \C$.

Let $\sequence {f_n}_{n \mathop \in \N}$ converge to some holomorphic mapping $f : U \to \C$ at $\sigma$.

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

Proof

Aiming for a contradiction, suppose there exists some compact subset $K$ of $U$ such that $f_n$ does not converge uniformly to $f$ on $K$.

Consider $K^* := K \cup \set \sigma$.

From Subsets Inherit Uniform Convergence, $f_n$ does not converge uniformly to $f$ on $K^*$.

From Uniformly Convergent iff Difference Under Supremum Norm Vanishes, the above is equivalent to:

$\exists \epsilon > 0 : \forall N \in \N : \exists n \ge N : \norm {f_n - f}_{K^*} \ge \epsilon$

where $\norm {\cdot}_{K^*}$ denotes the supremum norm over $K^*$.

From Finite Union of Compact Sets is Compact, $K^{*}$ is compact.

Since $\sequence {f_n}$ is a normal family, there is some subsequence $\sequence {f_{n_r} }$ of $\sequence {f_n}$ and some mapping $g \in \map {\mathcal H} U$ such that:

$\sequence {f_{n_r} }$ converges uniformly to $g$ on $K^*$.

Further:

 $\displaystyle \map f \sigma$ $=$ $\displaystyle \lim_{n \mathop \to \infty} \map {f_n} \sigma$ Definition of $f$ at $\sigma$ $\displaystyle$ $=$ $\displaystyle \lim_{r \mathop \to \infty} \map {f_{n_r} } \sigma$ Limit of Subsequence equals Limit of Sequence $\displaystyle$ $=$ $\displaystyle \map g \sigma$ Definition of $g$ at $\sigma$

From the Identity Theorem, $f$ and $g$ agree on $U$.

$\exists N \in \N: r \ge N \implies \norm {f_{n_r} - f}_{K^*} < \epsilon$

$\forall N \in \N: \exists n \ge N: \norm {f_n - f}_{K^*} \ge \epsilon$

Hence the result, by Proof by Contradiction.

$\blacksquare$

Source of Name

This entry was named for Giuseppe Vitali.