For holomorphic functions, if $\{f_n\}\to f$ uniformly on compact sets, then the same is true for the derivatives. Let $\Omega$ be an open subset in $\mathbb{C}$. Let $\{f_n\}$ be a sequence of holomorphic functions on $\Omega$ such that $f_n\to f$ pointwise and converges uniformly on any compact subset $K\subseteq \Omega$. Then by Cauchy Theorem and Morera Theorem, $f$ is holomorphic. Let $f_n$ and $f'$ be the derivatives of $f_n$ and $f$ respectively.
Prove that $f_n'\to f'$ uniformly on any compact subset $K\subseteq \Omega$. 
How to prove?
 A: Let $K \subset \Omega$ be compact. Since $\Omega$ is open, for some $\epsilon>0$, we have  $K_\epsilon = K+\overline{B(0,2\epsilon)} \subset \Omega$, and the set $K_\epsilon$ is compact.
Suppose $a \in K$. Then we have $f'(a) = {1 \over 2 \pi i} \int_{\gamma_a} { f(z) \over (z-a)^2 } dz$
and $f_n'(a) = {1 \over 2 \pi i} \int_{\gamma_a} { f_n(z) \over (z-a)^2 } dz$
where $\gamma_a$ is an anti-clockwise circle of radius $\epsilon$ centred at $a$.
Since $f_n \to f$ uniformly on $K_\epsilon$, and we have 
$|f'(a)-f_n'(a)| \le {1 \over 2 \pi } {l(\gamma_a) \over \epsilon^2} \sup_{z \in K_\epsilon} |f(z) -f_n(z)|$, we see that
$f_n' \to f'$ uniformly on $K$ as well.
It follows that $f_n^{(k)} \to f^{(k)}$ uniformly on $K$ for any $k$.
A: Not to be taken too seriously! The space $H(\Omega)$ of holomorphic functions on $\Omega$ endowed with the topology of uniform convergence on compact sets is complete and therefore a Frechet space. If you know that the derivative of a holomorphic function is again holomorphic you get from the closed graph theorem that the linear map $D:H(\Omega)\to H(\Omega)$, $f\mapsto f'$ is continuous. That $D$ has closed graph follows e.g. from its continuity as a map $H(\Omega)\to C(\Omega)$.
