# Proof verification) $f_n\rightarrow f$ uniformly and $\frac{\partial f_n}{\partial \bar{z}}\rightarrow 0$ uniformly implies that $f$ is holomorphic?

This is an exercise 26.1 from "Complex Made Simple" by David Ullrich.

Note that $$D$$ denotes an open subset of $$\mathbb{C}$$

$$\textbf{Proposition)}$$ Suppose that $$f_n\in C^1(D), f_n\rightarrow f$$ uniformly on compact subsets of $$D$$ (i.e. locally uniformly), and $$\partial f_n/\partial \bar{z} \rightarrow 0$$ uniformly on compact subsets of $$D$$. Then $$f$$ is holomorphic on $$D$$.

The text suggests that you can use the following integral formula for $$C^1$$ functions.

$$\textbf{Theorem)}$$ If $$f\in C^1(\bar{D})$$ and $$w\in D$$ then $$f(w) = \frac{1}{2\pi i}\int_{\partial D} \frac{f(z)}{z - w} dz\,-\frac{1}{\pi}\int_{D} \frac{1}{z-w}\frac{\partial f}{\partial \bar{z}} dA(z)$$ where $$dA(z)$$ denotes the Lebesgue measure for $$z$$-variable.

This is what I've tried:

$$\textbf{Proof)}$$ First fix $$\Omega \subset\subset D$$. Then on $$\,\bar{\Omega}$$,$$\,\,f_n\rightarrow f$$ and $$\partial f_n/\partial \bar{z} \rightarrow 0$$ unifomly.

Now let $$g_n(w)=\frac{1}{2\pi i}\int_{\partial \Omega} \frac{f(z)}{z - w} dz$$ for each n, $$w\in\Omega$$. Then each $$g_n$$ is holomorphic in $$\Omega$$, and $$\left| f_n(w)-g_n(w) \right| = \left| \frac{1}{\pi}\int_{\Omega} \frac{1}{z-w}\frac{\partial f_n}{\partial \bar{z}} dA(z)\right| \leq \frac{1}{\pi}\int_{\Omega} \left|\frac{1}{z-w}\frac{\partial f_n}{\partial \bar{z}} \right| dA(z) \leq M \max_{\bar{\Omega}}\left|\frac{\partial f_n}{\partial \bar{z}} \right|$$ for some constant $$M$$.

As $$\partial f_n/\partial \bar{z} \rightarrow 0$$ unifomly, $$\max_{\bar{\Omega}}\left|\frac{\partial f_n}{\partial \bar{z}} \right| \rightarrow 0$$ as $$n\rightarrow \infty$$. Since $$(f_n)$$ converges uniformly to a finite function, this implies that $$(g_n)$$ is uniformly bounded on $$\Omega$$. By Montel's theorem we can conclude that there exists a subsequence $$(g_{n_k})$$ which converge uniformly on compact subsets of $$\Omega$$ to holomorphic function $$g$$.

If we take a limit $$k \rightarrow \infty$$ of $$\left| f_{n_k}(w)-g_{n_k}(w) \right| \leq M \max_{\bar{\Omega}}\left|\frac{\partial f_{n_k}}{\partial \bar{z}} \right|,$$ we get $$\left|f(w)-g(w)\right|=0$$ and hence $$f\equiv g$$ on $$\Omega$$. Since $$g$$ is holomorphic on $$\Omega$$ and $$\Omega$$ was arbitrary, we conclude that $$f$$ is holomorphic on $$D$$. $$\qquad \square$$

Is my proof correct? Or is there any more concise proof of this fact?

If $$f_n$$ are holomorphic, and converges uniformly on compact subsets of an open set $$D$$ to a function $$f$$, then $$f$$ is continuous since $$|f(x)-f(y)| \leq |f(x)-f_n(x)|+|f_n(x)-f_n(y)|+|f_n(y)-f(y)| \leq \epsilon$$. Now Since $$f$$ is continuous, By morera's theorem, its enough to prove that $$\int_{\gamma} f(z) dz = 0$$. This is true since $$\int_{\gamma} f_n(z) dz = 0$$ and $$\int_{\gamma} (f(z)-f_n(z)) dz \rightarrow 0$$ as $$\gamma$$ is contained in a compact set.

Now see if above helps. Can u tell me what you mean by $$C^1(D)$$ ? Since $$f_n$$ is continuously differentiable in $$D$$, isnt it holomorphic in $$D$$ ?

EDIT : Based on @Steven Gubkin comments and your answer: It is clear that $$|f(w)-g_n(w)| \leq |f(w)-f_n(w)|+|f_n(w)-g_n(w)| \leq \epsilon$$. Hence $$g_n \rightarrow f$$. Now $$|g_n(w)-\int_{\partial \Omega} \frac{f(z)}{z-w} dz| \leq \int_{\partial \Omega} \frac{|f_n(z)-f(z)|}{|z-w|} dz \leq \epsilon$$. Hence $$g_n(w) \rightarrow \int_{\partial \Omega} \frac{f(z)}{z-w} dz$$. Hence we have $$f(w) = \int_{\partial \Omega} \frac{f(z)}{z-w} dz$$. Now use: Converse of Cauchy Integral Formula to conclude the proof.

You will need $$Length(\partial \Omega)$$ bounded. We can assume $$\Omega = B_r(w)$$.

• We do not know that $f_n$ are holomorphic. OP is asking about the generalization to the case that $\frac{\partial f_n}{\partial z} \to 0$ instead of $\frac{\partial f_n}{\partial z} = 0$ (the case you address). $f \in C^1(D)$ means that $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ are continuous functions. It does not assert that these partial derivatives satisfy the Cauchy-Riemann equations $\frac{\partial f}{\partial x} + i \frac{\partial f}{\partial y} = 0$. Dec 15, 2022 at 13:17
• @StevenGubkin thanks. how is he saying $\int_{\Omega} |1/(z-w)dA(z)| \leq M$ when $w,z \in \Omega$ ? Dec 15, 2022 at 14:24
• The way to justify this is using polar coordinates $z = w + re^{i \theta}$. $\int_{B(w,R)} \frac{1}{|z-w|} \textrm{d}A = \int_{r=0}^{r=R}\int_{\theta=0}^{\theta = 2\pi} \frac{1}{r} r \textrm{ d}\theta \textrm{d}r = 2\pi R$. Dec 15, 2022 at 14:32
• @StevenGubkin edited my answer based on your inputs. Thanks. Dec 15, 2022 at 15:56
• You don't need the length of the boundary bounded. Just make the argument relative to a disk $D(p,r)$ compactly contained in the domain. Since every point is in the interior of some such disk, you are done. Dec 15, 2022 at 16:41

Your proof looks correct but overly complicated.

First of all note that the Cauchy kernel "creates" holomorphic functions. That is to say that if $$\Omega$$ is a simply connected open set with boundary $$b\Omega$$, and $$g: b\Omega \to \mathbb{C}$$ is continuous* then the function $$f: \Omega \to \mathbb{C}$$ defined by

$$f(z) = \int_{b\Omega} \frac{g(\zeta)}{\zeta - z} \textrm{ d}\zeta$$

is holomorphic on $$\Omega$$.

Conceptually this is true just because $$\frac{g(\zeta)}{\zeta - z}$$ is a holomorphic function in $$z$$ for each fixed $$\zeta$$, and you are just summing up a bunch of those when you take the integral. This idea can be formalized into a proof by going back to the definition of the integral as a limit of a sequence of sums, and then extracting a uniform limit of holomorphic functions. Alternatively one could use Fubini's theorem and Morera's theorem together to reach the same conclusion.

It isn't hard to use your hypotheses together with Cauchy-Pompeiu to see that $$f$$ satisfies the conclusion of the Cauchy Integral formula for closed disks compactly contained in the domain. Thus $$f$$ is holomorphic.

* I am sure "continuity" could be relaxed substantially, but I don't want to work out the maximum level of generality here

• How can I use Cauchy-Pompeiu to $f$? We don't know a priori $f\in C^1$. Can you explain more detail of that process? Dec 16, 2022 at 4:22
• @Bergsom See edits section in my answer (based on comments from Steven Gubkin and your answer ). Since we have, $g_n(w) \rightarrow f(w)$ and $g_n(w) = \int_{\partial B_r(w)} \frac{f_n(z)}{z-w} dz \rightarrow g(w) = \int_{\partial B_r(w)} \frac{f(z)}{z-w} dz$, cauchy integral formula holds for $f$. Now if cauchy integral formula holds then the function is holomorphic. Dec 16, 2022 at 7:11