Let $f_n :U \to \mathbb{C}$ be a sequence of analytic functions, where $U$ is open and connected. Suppose there exists a point $z_0 \in U$ such that for all $k \geq 0$ the sequence $f_n^{(k)}(z_0)$ converges.
How would you show that then the whole sequence $f_n$ converges locally uniformly (i.e. uniformly on compact subsets of U) to an analytic function $f$?
Best Answer
You cannot prove that, since it does not necessarily hold.
A simple counterexample is $U = \mathbb{C}$, and $f_n(z) = z^n$, with $z_0 = 0$.
We have $\lim\limits_{n\to\infty} f^{(k)}_n(z_0) = 0$ for all $k \geqslant 0$, but $(f_n)$ converges locally uniformly only in the open unit disk, not in all of $U$.
Choose $f_n(z) = n^n\cdot z^n$, and you don't even have locally uniform convergence in any neighbourhood of $z_0$.
If there is an additional hypothesis that $\{f_n : n \in \mathbb{N}\}$ is a normal family, say the sequence is locally bounded, things are of course different.
Then by normality, $(f_n)$ has a locally uniformly convergent subsequence $(f_{n_{k}})$. Let $f$ be the limit of this subsequence. It remains to see that the full sequence converges locally uniformly to $f$. Again by normality, every subsequence $(f_{n_m})$ has a locally uniformly convergent subsequence $\bigl(f_{n_{m_r}}\bigr)$. Let $g$ denote its limit. Since $f_n^{(k)}(z_0) \to f^{(k)}(z_0)$ for all $k$, and by Weierstraß' theorem $f_{n_{m_r}}^{(k)}(z_0) \to g^{(k)}(z_0)$, we have $f^{(k)}(z_0) = g^{(k)}(z_0)$ for all $k$, hence by the identity theorem, $g \equiv f$.
Thus every subsequence of $(f_n)$ has a subsequence converging locally uniformly to $f$, and that implies that the full sequence converges locally uniformly to $f$.