I was studying complex analysis when I encountered the following statement:
Let $f_n:D\rightarrow\mathbb{C}$, where $D$ is a open connected subset of $\mathbb{C},$ be holomorphic functions. Now, if there exists a functions $f,g:D\rightarrow\mathbb{C}$ such that for some $z_o\in D,$ $f_n\rightarrow f$ pointwise, and $f^\prime_n\rightarrow g$ uniformly, on $D,$ then $f_n\rightarrow f$ uniformly on $D$ and $g=f'.$
The corresponding theorem for real numbers for functions in $\mathcal{C}^1[a,b]$ is known (W. Rudin, Principles of Analysis Theorem 7.17). My instinct was to write a proof similar to that of the real case; the proof of the real case uses the Mean Value Theorem though. The only way to translate such proof is we additionally assume that $D$ is a star-shaped domain as well (proving for each line joining two points).
Keeping our initial assumption, we know that $D$ is path-connected. With a similar idea in mind, I thought of taking an arbitrary path $\gamma$ connecting two points in $D,$ and proving the claim for $f_n\circ\gamma,$ and other corresponding compositions; unfortunately, you cannot apply the real case method as your range in the entire complex plane.
I believe a key part in proving the claim is the fact that $f^\prime_n$ are continuous.
Is there a way to prove this simply using the result we arrived at in the real case? Or must something be done further? Is there a simpler way to prove this?
Any help and hints will be much appreciated! Thank you!
Best Answer
We need only to show that $f_n$ converges uniformly to some function $f$. The fact that $f' = g$ follows from the fact that $C^1(D, \mathbb{R})$ is complete (this is proven by restricting attention to a small ball and using partial derivatives and the fundamental theorem of calculus). If we add some hypotheses, a simple proof will work. Let's assume that for every $z \in D$, the line segment $\overline{z_0z} \subset D$ and that $D$ is bounded. By the fundamental theorem of calculus, we have $$f_n(z) = f_n(z_0) + \int_{0}^{1}f_n'(z_0 + t(z - z_0))(z - z_0)\,dt.$$ We have $$\left|\int_{0}^{1}f_n'(z_0 + t(z - z_0))(z - z_0)\,dt - \int_{0}^{1}g(z_0 + t(z - z_0))(z - z_0)\,dt\right| \leq |z - z_0|d(f_n', g) \leq \text{diam}(D)d(f_n', g).$$ Thus $\int_{0}^{1}f_n'(z_0 + t(z - z_0))(z - z_0)\,dt \to \int_{0}^{1}g(z_0 + t(z - z_0))(z - z_0)\,dt$ uniformly. Thus $f_n(z) \to \lim_{n \to \infty}f_n(z_0) + \int_{0}^{1}g(z_0 + t(z - z_0))(z - z_0)\,dt$ uniformly. This proof does not use the fact that $f_n$ are holomorphic though; it only uses the fact that they are $C^1$.