Convergence at infinitely many points in a compact set and convergence in the whole region Let $\{f_n\}$ be a sequence of analytic functions on a region $\Omega$ with $|f_n| ≤ 1$ on $\Omega$. Let K be compact and contained in $\Omega$. Suppose $\{f_n\}$ converges at infinitely many points in K. Then is it true or false that $\{f_n\}$ necessarily converges at every point of $\Omega$?
 A: It is true. 


*

*$(f_n)$ is uniformly bounded, so by Montel's theorem it is normal: Every subsequence of $(f_{n_k})$  of $(f_n)$  has a (locally uniformly) convergent subsequence $(f_{n_{k_l}})$.

*On the other hand, every convergent subsequence has the same limit:
Assume that $f_{n_{1, k}} \to F_1$ and $f_{n_{2, k}} \to F_2$,
then $F_1(z) = F_2(z)$ for all points where the original sequence
is convergent, i.e. for infinitely many points in $K$. $K$ is compact,
so these points have an accumulation  point in $K$, and it follows from
the identity theorem that $F_1 = F_2$.
Now assume that $(f_n(a))$ is not convergent for some $a \in \Omega$.
Then there exist subsequences $(f_{n_{1, k}})$ and $(f_{n_{2, k}})$
such that both $(f_{n_{1, k}(a)})$ and $(f_{n_{2, k}}(a))$ are
both convergent but with a different limit.
According to (1), both sequences have subsequences which are convergent in $\Omega$.
According to (2),  those convergent subsequences have the same
limit. This is a contradiction.
The same argument can be used to show that $(f_n)$ is in fact locally uniformly
convergent, i.e. uniformly convergent on all compact subsets of
$\Omega$.
