We have a course on complex analysis in the current semester.Our professor introduced a theorem in class which is as follows:
Theorem
Let $f_n:\Omega\to \mathbb C$ be a sequence of holomorphic functions such that $f_n\to f$ uniformly on each compact subset of $D$.Then $f$ is holomorphic and the sequence of derivatives $\{f_n'\}$ converges uniformly to $\{f'\}$ on each compact subset of $\Omega$.
I am trying to prove this theorem.I do not know how to start with.For the proof that $f$ is holomorphic,I think I need to use Morera's theorem but I do not know how to proceed and for the next part,I have no clue.I remember that I have proved a similar theorem for real differentiable functions which is as follows:
Theorem
Let $\{f_n\}$ be a sequence of real-valued differentiable functions on $[a,b]$ such that $\{f_n\}$ converges at least at one point $x_0\in [a,b]$ and $\{f_n'\}$ is uniformly convergent,then $f_n$ converges uniformly to some function $f$ which is differentiable and $f_n'\to f'$ on $[a,b]$.
But there I used mean value theorem which is not applicable for complex valued functions.So,can someone give me some idea?
Best Answer
As $f_n(z)$ tends to $f(z)$ almost uniformly, by the Morera theorem we get that $f(z)$ is holomorphic as well.
Assume $B_r=\{w\in \mathbb{C}\,:\, |w-a|\le r\}\subset \Omega.$ Then for any $|z-a|\le r-\delta$ we have $$f'_n(z)={1\over 2\pi i}\int\limits_{|w-a|=r}{f_n(w)\over (w-z)^2}\,dw$$ Thus $$|f_n'(z)-f_m'(z)|\le {1\over \delta^2}\max_{|w-a|= r}|f_n(w)-f_m(w)|\quad (*)$$ Therefore the sequence $f_n'(z)$ is convergent uniformly on $\{z\in \mathbb{C}\,:\, |z-a|\le r-\delta\}.$ Denote $g(z)=\lim_nf_n(z).$ Then $$g(z)={1\over 2\pi i}\int\limits_{|w-a|=r}{f(w)\over (w-z)^2}\,dw ,\quad |z-a|\le r-\delta$$ Therefore $f'(z)=g(z)$ for $|z-a|\le r-\delta.$
Taking the limit $m\to \infty$ in $(*)$ we get $$ |f_n'(z)-f'(z)|\le {1\over\delta^2} \max_{|w-a|= r}|f_n(w)-f(w)|,\quad |z-a|\le r-\delta$$ Therefore the sequence $f_n'$ is convergent to $f'$ uniformly on every ball $\{z\in \mathbb{C}\,:\,|z-a|\le r-\delta\}\subset \Omega.$ This implies the almost uniform convergence in $\Omega.$