Prove that if \(f_n\to f\) locally uniformly with \(f_n\) holomorphic, then \(f_n'\to f'\) locally uniformly and \(f'\) is holomorphic.
Solution
Simplifying step: for some reason, it suffices to assume \(f_n\to f\) uniformly on all of \(\Omega\)?
Take \(\Omega_R\) to be \(\Omega\) with a buffer of \(R\), so \(d(z, \bd \Omega) > R\) for every \(z \in \bar{\Omega_R}\).
It suffices to show the following bound for \(F\) any holomorphic function on \(\Omega\): \[ \sup_{z\in \Omega_R} \abs{F'(z)} \leq {1\over R} \sup_{\zeta \in \Omega} \abs{F(\zeta)} && \forall R ,\] where on the right we take the sup over all \(\Omega\).
- Then take \(F \da f_n-f\) and \(R\to 0\) to conclude, since the right-hand side is a constant not depending on \(\Omega_R\).
For any \(z\in \Omega_R\), we have \(\bar{D_R(z)} \subseteq \Omega_R\), so Cauchy’s integral formula can be applied: \[ \abs{F'(z)} &= \abs{ {1\over 2\pi i} \int_{\bd D_R(z)} {F(\xi) \over (\xi-z)^2 } \dxi } \\ &\leq {1\over 2\pi} \int_{\bd D_R(z)} { { \abs{F(\xi)} \over \abs{\xi-z}^2 }} \dxi \\ &\leq {1\over 2\pi} \int_{\bd D_R(z)} { { \sup_{\zeta\in \Omega} \abs{F(\zeta)} \over \abs{\xi-z}^2 }} \dxi \\ &= {1\over 2\pi} \sup_{\zeta\in \Omega} \abs{F(\zeta)} \int_{\bd D_R(z)} { { 1 \over R^2 }} \dxi \\ &= {1\over 2\pi} \sup_{\zeta\in \Omega} \abs{F(\zeta)} {1\over R^2} \int_{\bd D_R(z)} \dxi \\ &= {1\over 2\pi} \sup_{\zeta\in \Omega} \abs{F(\zeta)} {1\over R^2} 2\pi R \\ &\leq {1\over 2\pi} \sup_{\zeta\in \Omega} \abs{F(\zeta)} {1\over R^2}\qty{ 2\pi R} \\ &= {1\over R} \sup_{\zeta \in \Omega}\abs{F(\zeta)} .\]
Now \[ \norm{f_n' - f'}_{\infty, \Omega_R} \leq {1\over R} \norm{f_n - f}_{\infty, \Omega} ,\] where if \(R\) is fixed then by uniform convergence of \(f_n\to f\), for \(n\) large enough \(\norm{f_n - f} < \eps/R\).