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?
Best Answer
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 $\Omega = B_r(w)$.