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?

  • $\begingroup$ my thought is to contain $K$ within a finite family of closed paths, then use Cauchy on $f-f_n$. this should allow finding an $N$ which for $n \gt N$ sets an upper bound on $|f-f_n|$ for the whole of K. however i have not looked into the detail... $\endgroup$ – David Holden Aug 6 '14 at 4:28

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 $k$.

  • $\begingroup$ In the last two sentences, by "uniformly" you mean "uniformly on $K_\epsilon$", hence on $K$. (I suppose) $\endgroup$ – Thibaut Dumont Aug 6 '14 at 8:54
  • $\begingroup$ @ThibautDumont: Indeed you are correct. That is what I meant. $\endgroup$ – copper.hat Aug 6 '14 at 13:34
  • $\begingroup$ @copper.hat Could you explain the need for the compact set $k_{\epsilon}$? and the construction?. why can't we pick any arbitrary compact set say $A$ and let $\gamma_a=\partial A$ then apply Cauchy's Integral formula. And why do you have $\overline{B(0,2\epsilon)}$. I was thinking $\overline{B(z,2\epsilon)}$ for $z \in K$ $\endgroup$ – J. Kyei Dec 6 '17 at 1:43
  • $\begingroup$ @J.Kyei: For an arbitrary point in $K$ I want to take a small ball around the point in which $f_n \to f$ uniformly in the ball. $K_\epsilon$ just enlarges $K$ enough that a small ball exists around any point in $K$ that lies in $K_\epsilon$. $\endgroup$ – copper.hat Dec 6 '17 at 3:25
  • $\begingroup$ oh okay,I see. thank you $\endgroup$ – J. Kyei Dec 7 '17 at 4:38

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 $H(\Omega)\to C(\Omega)$.


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