TITLE: Uniform limit of holomorphic functions
QUESTION [31 upvotes]: Let $\{f_n\}$ be a sequence of holomorphic functions defined in a generic domain $D \subset \mathbb{C}$. Suppose that there exist $f$ such that $f_n \to f$ uniformly.
My question is: is it true that $f$ is holomorphic too?

REPLY [3 votes]: I just want to add a small precision : 

(If $f(z)$ is piecewise continuous) and $$\int_\gamma f(z) dz = 0$$ for every closed $C^1$ contour $\gamma \subset U$ some open, then $f(z)$ is holomorphic on $U$.
But if $U$ is closed, say $U = \{ \ |z| \le 1 \ \}$ then we only get that $f(z)$ is holomorphic on $U \setminus \partial U$.
Try with $$f(z) = \sum_{n=2}^\infty \frac{z^n}{n (n-1)}$$
then $\displaystyle\sum_{n=2}^\infty \left|\frac{1}{n (n-1)} \right|$ converges hence $\displaystyle\sum_{n=2}^\infty \frac{z^n}{n (n-1)}$ converges uniformly on $|z| \le 1$, but $$f'(z) = \sum_{n=1}^\infty \frac{z^n}{n} = -\log(1-z)$$
has a singularity at $z=1$ : $ \quad f(z)$ is holomorphic only on $|z| < 1$ and not on $|z| \le 1$.