uniform convergence of derivatives

if we have a series of functions $f_1, f_2, \cdots$ that uniformly converge to an $f$ on a domain $\Omega$, then all of their derivatives also do the same to $f'$ on any compact subset $K \subseteq \Omega$

this one is gonna use a good amount of stuff:

morera's theorem, i.e. integrating to 0 on any triangle implies that a function is holomorphic
and that $f$ is actually holomorphic at all, which follows from using the above on $f_i$

so we only have to prove uniform convergence on a compact subset, so i guess a natural thing to do would be to finitely cover $K$:

\[K=\bigcup_{i=1}^n B_{r_i}(z_i)\]

all $z_i$s here are in $K$, and all $r_i$s are chosen s.t. $B_{3r_i}(z_i) \subseteq \Omega$. you'll see why we make this the case later on

convergence on balls

now i'll show that the uniform convergence happens in each individual ball, so let's fix an $i$ and consider what happens to the $n$-th derivatives on $B_{r_i}(z_i)$:

\[\begin{align*} f^{(n)}(z)-f^{(n)}_k(z) &= \frac{n!}{2\pi i}\int_{|\zeta-z_i|=2r_i} \frac{f(\zeta)}{(\zeta-z)^{n+1}} \,d\zeta - \frac{n!}{2\pi i}\int_{|\zeta-z_i|=2r_i} \frac{f_k(\zeta)}{(\zeta-z)^{n+1}} \,d\zeta \\ &= \frac{n!}{2\pi i}\int_{|\zeta-z_i|=2r_i} \frac{f(\zeta)-f_k(\zeta)}{(\zeta-z)^{n+1}} \,d\zeta \\ &\to 0\text{ as }k \to \infty \end{align*}\]

notice that we're not integrating around the border of the ball itself, but a ball that's double the radius. this is necessary for the uniform convergence for all $z \in B_{r_i}(z_i)$.

if we just integrated around the border of $B_{r_i}(z_i)$ itself, then we could pick $z$ really really close to the border, and then have $\frac{1}{(\zeta-z)^{n+1}}$ get really big and cancel out whatever shrinking $f(\zeta)-f_k(\zeta)$ was doing

OTOH, with our funny bounds, the denominator becomes bounded:

\[\frac{1}{(\zeta-z)^{n+1}} \le \frac{1}{r_i^{n+1}}\ \forall z \in B_{r_i}(z_i), |\zeta-z_i|=2r_i\]

our funny choice of $r_i$ earlier on ensures $\overline{B_{2r_i}(z_i)} \subseteq \Omega$

and convergence overall

ok that was the hardest part

now that we know $f^{(n)}_k$ uniformly converges to $f^{(n)}$ on each $B_{r_i}(z_i)$, it's basically a given that the same thing happens on $K$

for any $\epsilon$, each $B_{r_i}(z_i)$ gives us a number $N$ that's needed for

\[\sup_{z \in B_{r_i}(z_i)} \left|f^{(n)}_k(z)-f^{(n)}(z)\right| < \epsilon\ \forall k \ge N\]

and since there's only finitely many of them, we can take the maximum of all of them and make that our actual $N$ for an error of less than $\epsilon$ on all of $K$! $\square$