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?


2 Answers 2


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 $\Omega = B_r(w)$.

  • 2
    $\begingroup$ 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 $\frac{\partial f}{\partial x} + i \frac{\partial f}{\partial y} = 0$. $\endgroup$ Dec 15, 2022 at 13:17
  • $\begingroup$ @StevenGubkin thanks. how is he saying $\int_{\Omega} |1/(z-w)dA(z)| \leq M$ when $w,z \in \Omega$ ? $\endgroup$
    – Balaji sb
    Dec 15, 2022 at 14:24
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    $\begingroup$ The way to justify this is using polar coordinates $z = w + re^{i \theta}$. $\int_{B(w,R)} \frac{1}{|z-w|} \textrm{d}A = \int_{r=0}^{r=R}\int_{\theta=0}^{\theta = 2\pi} \frac{1}{r} r \textrm{ d}\theta \textrm{d}r = 2\pi R$. $\endgroup$ Dec 15, 2022 at 14:32
  • $\begingroup$ @StevenGubkin edited my answer based on your inputs. Thanks. $\endgroup$
    – Balaji sb
    Dec 15, 2022 at 15:56
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    $\begingroup$ 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. $\endgroup$ 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

  • $\begingroup$ 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? $\endgroup$
    – Kiyoon Eum
    Dec 16, 2022 at 4:22
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    $\begingroup$ @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. $\endgroup$
    – Balaji sb
    Dec 16, 2022 at 7:11

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